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Time, Speed and Distance questions are the very popular type of questions in competitive exams. These questions carry a weightage of 2-3 questions (4-6 marks) in SSC exams and 1-2 questions in bank exams. To get a good rank in competitive exams, it is important to be know the distance speed time formula quickly and accurately.

Here are some tips for solving Time, Speed and Distance questions: Know the formulas and relationships, be careful with units, work with the given information to find the unknown, use a diagram or table to help visualize the problem.

So, we have attached 10 questions of Time, Speed and Distance for you to practice with. You should aim to solve these questions in less than half a minute for each.

<h2>Practice Questions on Time, Speed and Distance</h2>

You can also download the Time, Speed and Distance questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 In a racing competition two men, A and B are participating. The speed of A is 300 m/min and the speed of B is 400 m/min. If the racing competition is for 700 m, find who will win the race and in how much time. 

1. B, 1 min 75 sec

2. A, 2 min 20 sec

3. A, 2 min 33 sec

4. B, 1.75 min

5. B, 1 min 45 sec

(Difficulty: 2, Estimated Time: 15 Seconds) It was very easy, right?

Q:2 The speed of a boat in still water is 5 km/hr. The boat was going downstream when the speed of the stream was 2 km/hr. The boat had covered 3/8 of the total distance when the speed of the stream increased by 2 km/hr. What was the average speed of the boat for the whole journey?

1. 7.56 km/hr

2. 8.13 km/hr

3. 9 km/hr

4. 6.90 km/hr

5. None of these

(Difficulty: 4, Estimated Time: 30 Seconds) Try decreasing your time used in calculations!

Q:3 A person covers 5/11th of the total distance with a speed of 5 km/hr and the rest with a speed of 3 km/hr. What is the average speed of the person?

1. 3 km/hr

2. 4.5 km/hr

3. 3.67 km/hr

4. 4 km/hr

5. None of these

(Difficulty: 3, Estimated Time: 20 Seconds) This was quite easier than the previous one but not that much easy.

Q:4 A bike rider covers a 175 km distance and returns up to 1/5th of that distance in a total of 7.5 hours. Find the average speed of the journey in m/s.

1. 50/3 m/s

2. 60/9 m/s

3. 70/9 m/s

4. 40/3 m/s

5. None of these

(Difficulty: 3, Estimated Time: 20 Seconds) Now you have a good practice of such questions.

Q:5 2 dogs run towards each other after seeing a cat between their path. The speeds of the dogs are 20 km/hr and 25 km/hr. They reach to cat at the same time but the faster dog covers 80 meters more than the slower dog. Find the distance between the cat and the slower dog, initially.

1. 400 m

2. 320 m

3. 480 m

4. 560 m

5. 240 m

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. We will increase the difficulty level from now.

Q:6 Mahesh started from point A towards B with a speed of 25 km/h and after 3 hours, Manish started from point B towards point A with a speed of 40 km/h. Mahesh stopped midway and started waiting for Manish. If Mahesh waited for 45 minutes before Manish arrived then find the distance AB.

1. 400 km

2. 240 km

3. 320 km

4. 300 km

5. None of these

(Difficulty: 3, Estimated Time: 20 Seconds) Have you got all your questions correct so far?

Q:7 Rajdhani express normally reaches its destination at 80 km/h in 40 hours. Find the speed at which it travels to reduce the time by 15 hours.

1. 108 km/h

2. 115 km/h

3. 128 km/h

4. 100 km/h

5. 150 km/h

(Difficulty: 2, Estimated Time: 15 Seconds) This was a piece of cake!

Q:8 A man can row upstream at 19 km/h and downstream at 27 km/h. The speed of water current of the river is:

1. 13 km/h

2. 15 km/h

3. 11 km/h

4. 7 km/h

5. 4 km/h

(Difficulty: 2, Estimated Time: 15 Seconds) You might have wrapped it up in 10 seconds!

Q:9 A person got his leg injured so his speed became 1/3 of his usual speed and he became late 10 mins late. Find the usual time taken taken by him to cover that distance.

1. 10 mins

2. 14 mins

3. 20 mins

4. 5 mins

5. 12 mins

(Difficulty: 4, Estimated Time: 25 Seconds) It is different type of question. But you&#39;ll get these type of questions in the exam too. So, prepare yourself!

Q:10 A man covered a certain distance by his bicycle at a rate of 8 km/hr and walked back at the rate of 6 km/hr. The whole journey took 7 hours. Find the distance he travelled. 

1. 44 km

2. 32 km

3. 24 km

4. 54 km

5. 60 km

(Difficulty: 3, Estimated Time: 20 Seconds) Did you guess them all correctly? 
 

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (5)</td>
			<td>2. (2)</td>
			<td>3. (3)</td>
			<td>4. (3)</td>
			<td>5. (2)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (3)</td>
			<td>8. (5)</td>
			<td>9. (4)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 5 i.e. B, 1 min 45 sec

Since the speed of A = 300 m/min

So, Time is taken by A = 1/300 × 700 = 2 min 20 sec

Now, Since the speed of B = 400 m/min

So,Time is taken by B = 1/400 × 700 = 1 min 45 sec

Hence, B completes the race first in 1 min 45 sec

 

Q:2 The correct answer is option 2 i.e. 8.13 km/hr

Average speed = Total distance covered/Total time taken

Let the total distance be x

Time taken to cover 3/8 of x when speed of stream was 2 km/hr = (3/8)x/(5 + 2) = 3x/56

Time taken to cover 5/8 of x when speed of stream was 4 km/hr = (5/8)x/(5 + 4) = 5x/72

Average speed = x/[(3x/56) + (5x/72)] = 1/(3/56 + 5/72) = 8.13 km/hr

 

Q:3 The correct answer is option 3 i.e. 3.67 km/hr 

Average speed = total distance covered/total time taken

Let the total distance covered be x km

time = distance/speed

time taken to cover 5/11 of total distance = t1 = (5/11)x/5 = 5x/55 = x /11

time taken to cover 6/11 of total distance t2 = (6/11)x/3 = 2x/11

Average speed = x/(t1 + t2) = x/(x/11 + 2x/11) = 11/3 = 3.67 km/hr

 

Q:4 The correct answer is option 3 i.e. 70/9 m/s. 

From the question:

Total distance covered by the bike rider = 175 + 175/5 = 175 + 35 = 210 km

Total time = 7.5 hours

So,Average speed of the journey = 210/7.5 = 28 km/h = 28 × 5/18 = 70/9 m/s

 

Q:5 The correct answer is option 2 i.e. 320 m

Suppose they reach to cat after 1 minute:

So, difference of distance covered by 2 dogs in 1 minute = (25 - 20) × 1/60 = 5/60 km = 500/6 m

Since the fast dog covers 80 meters more than the slower dog

So, Time in which they reach the cat = 80/(500/6) = 0.96 minute

Hence, Distance between the cat and the slower dog = 0.96/60 × 20 = 0.32 km or 320 m

 

Q:6 The correct answer is option 4 i.e. 300 km

Suppose total distance AB = x km

So, Time taken by Mahesh to reach midway = (x/2)/25 = x/50 hours

Time taken by Manish to reach midway = (x/2)/40 = x/80 hours

Since Manish started 3 hours after Mahesh and Mahesh waited for 45 minutes (= 0.75 hours) also.

So, x/50 + 0.75 = x/80 + 3

(x/50 – x/80) = 2.25

3x/400 = 2.25

x = 300

Hence, Distance AB = 300 km

 

Q:7 The correct answer is option 3 i.e. 128 km/h

Given 

Speed = 80 km/h

Time = 40 hours

Distance = Speed × Time = 80 × 40 = 3200 km

New time = 40 - 15 = 25 h

∴ Required speed = 200/25 = 128 km/h 

 

Q:8 The correct answer is option 5 i.e. 4 km/h

Let the speed of man and water current be x and y respectively

According to question

x - y = 19    ---(1)

x + y = 27    ---(2)

From equation (1) and (2)

x = 23 and y = 4

∴ The speed of water current of the river

4 km/h

 

Q:9 The correct answer is Option 4 i.e. 5 mins.

If a person changes his speed to m/n of its usual speed and becomes late by T minutes, then the usual time taken by him is T/(n/m - 1)

Therefore the usual time = T/(n/m - 1) = 10/ (3 - 1) = 5 mins.

 

Q:10 The correct answer is Option 3 i.e. 24 km.

If a man travels certain distance at x km/hr and comes back at the rate of y km/hr and the whole journey takes t hours, then the distance travelled is {x × y/ (x + y)} × t 

Therfore, Distance travelled = (8 × 6/ 8 + 6 ) × 7 = 24 km.

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Time, Speed and Distance.

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Practice Questions on Time, Speed and Distance!

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Time, Speed and Distance Questions - Download PDF now!

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Comparison of Quantities questions are a major type of questions regularly asked in competitive exams. These questions carry a weightage of 1-5 questions (1-5 marks) in bank exams. To achieve good marks in banking exams, especially in mains you should have a good practice of Comparison of Quantities questions.

Here are some tips for solving Comparison of Quantities questions: Clear all the arithmetic topics extensively, because you may need to solve 2 statements from any of the topic like time and work, interest, profit and loss etc. and then compare them wisely and a slightest of error can cost you precious marks.

So, we have attached 10 questions of Comparison of Quantities for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Comparison of Quantities</h2>

You can also download the Comparison of Quantities questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Directions(Q1-Q10): In each of these questions, two quantities (A) and (B) are given. You have to solve both quantities and mark the appropriate answer.

Q:1 Quantity A: Average of remaining two numbers. The average of 5 numbers is 554 and the average of three numbers of them is 605.

Quantity B: Length of the platform (in m). A train of length 120 m crosses a pole in 10 seconds while it crosses the platform in 50 seconds.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A < Quantity B

4. Quantity A > Quantity B

5. Quantity A = Quantity B or no relationship can be established

(Difficulty: 3, Estimated Time: 20 Seconds) These questions take time, isn&#39;t it?

Q:2 Quantity A: Find the area swept by minute hand of circular 12-hour clock in 40 minutes. Length of minute hand is 10.5 cm.

Quantity B: Find the difference between Simple Interest and Compound Interest on Rs. 5775 for 2 years at 20% per annum.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A < Quantity B

4. Quantity A > Quantity B

5. Quantity A = Quantity B or no relationship can be established

(Difficulty: 3, Estimated Time: 20 Seconds) Do you got stuck in the first statement?

Q:3 Quantity A: Volume of the prism (in cu. cm). The base is an equilateral triangle with sides of 8√3 cm and a height is 20. Take √3 = 1.7.

Quantity B: 4/9 of the new salary of the man. The expenditure of the man increases in ratio 3 : 5. Previous salary is Rs 3000 and the expenditure percentage is 75%. His new expenditure is 83.33% of his new salary.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A < Quantity B

4. Quantity A > Quantity B

5. Quantity A = Quantity B or no relationship can be established

(Difficulty: 2, Estimated Time: 20 Seconds) Another time taking one but easier..

Q:4 Quantity A: Profit percent on another book. The cost price of a book is Rs 180 and is sold at 35% profit. The selling price of another book is Rs 55 more than 33.33% less than the selling price of this book and the cost price is 83.33% of this book.

Quantity B: Equivalent discount. Three successive discounts of 15, 18 and 12 percent are given.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A < Quantity B

4. Quantity A > Quantity B

5. Quantity A = Quantity B or no relationship can be established

(Difficulty: 2, Estimated Time: 20 Seconds) This was an easy one!

Q:5 Quantity A: Probability of selecting a diamond or a jack, if 1 card is picked at random.

Quantity B: Fraction. Denominator is 5 more than numerator. When the number is doubled, and the denominator is increased by 1, the resultant is 1/2.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A < Quantity B

4. Quantity A > Quantity B

5. Quantity A = Quantity B or no relationship can be established

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 Quantity A: Find discount percent, if marked price of article is Rs 2400 and the selling price is Rs 1980.

Quantity B: Find rate percent, if a sum of Rs 1610 is invested at simple interest for 2 years and it amounts to Rs 2173.5

1. Quantity A < Quantity B

2. Quantity A ≥ Quantity B

3. Quantity A ≤ Quantity B

4. Quantity A = Quantity B or no relationship can be established

5. Quantity A > Quantity B

(Difficulty: 2, Estimated Time: 20 Seconds) Don&#39;t let the calculations be the obstacle!

Q:7 Quantity A: Number of 4-digit numbers divisible by 4 that can be formed by 2, 4, 6 and 8 if repetition of digits is not allowed.

Quantity B: Positive root of x2 + 5x - 84 = 0

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A = Quantity B or no relationship can be established

4. Quantity A > Quantity B

5. Quantity A < Quantity B

(Difficulty: 3, Estimated Time: 20 Seconds) This one could be solved without lifting the pen!

Q:8 Quantity A: Ratio of speed of stream to speed of boat. Upstream speed of boat is 40% of downstream speed of boat.

Quantity B: Probability if getting at least 2 heads when 3 coins are simultaneously thrown.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A = Quantity B or no relationship can be established

4. Quantity A > Quantity B

5. Quantity A < Quantity B

(Difficulty: 3, Estimated Time: 20 Seconds) This was a piece of cake....

Q:9 Quantity A: Find the original number. The number when increased and decreased by 12% becomes 1540000.

Quantity B: 10 bikes are bought at Rs 70000 each and sold at profit of 8% on each. Two cars are bought at Rs 450000 each and sold at loss of 4% on both. Find the total selling price.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A = Quantity B or no relationship can be established

4. Quantity A > Quantity B

5. Quantity A < Quantity B

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks to save time!

Q:10 Quantity A: A Three digit number which when divided by 4, 5, 6 and 7 gives remainder 1.

Quantity B: 1287 bananas are bought at Rs. 2 per banana and sold at Rs. 30 per dozen. The remaining bananas are sold at Rs. 4 per banana. Find the profit.

1. Quantity A ≥ Quantity B

2. Quantity A ≤ Quantity B

3. Quantity A = Quantity B or no relationship can be established

4. Quantity A > Quantity B

5. Quantity A < Quantity B

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (3)</td>
			<td>2. (5)</td>
			<td>3. (3)</td>
			<td>4. (4)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (5)</td>
			<td>7. (4)</td>
			<td>8. (5)</td>
			<td>9. (5)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 3 i.e. Quantity A < Quantity B.

Quantity A:

Sum of 5 numbers = Average × 5 = 554 × 5 = 2770

Sum of 3 numbers = 605 × 3 = 1815

Sum of remaining two numbers = 2770 - 1815 = 955

Average = 955/2 = 477.5

Quantity B:

Distance covered while crossing pole = length of train = 120 m

Speed = Distance/time = 120/10 = 12 m/s

Distance travelled while crossing platform = length of train + length of platform = 12 × 50 = 600 m

Length of platform = 600 - 120 = 480

Quantity A < Quantity B

 

Q:2 The correct answer is option 5 i.e. Quantity A = Quantity B or no relationship can be established.

Quantity A:

Angle swept in 40 minutes = (40/60) × 360° = 240°

Radius of circle = length of hand = 10.5 cm

Area swept = (240/360) × (22/7) × 10.5 × 10.5 = 231 sq cm

Quantity B:

Difference between CI and SI for 2 years = P(r/100)2 = 5775 × (20/100)2 = 5775 × 1/5 × 1/5 = Rs. 231

Quantity A = Quantity B

 

Q:3 The correct answer is option 3 i.e. Quantity A < Quantity B.

Quantity A:

Volume of prism = Base area × height

Base area = (√3/4) × 8√3 × 8√3 = 81.6

Volume = 81.6 × 20 = 1632 cu.cm

Quantity B:

Previous salary = Rs 3000

Previous expenditure = 75% of 3000 = Rs 2250

New expenditure = (5/3) × 2250 = Rs 3750

Let new salary = b

83.33% of b = 3750

(5/6) × b = 3750

b = 4500

Quantity B = (4/9) × 4500 = 2000

Quantity A < Quantity B

 

Q:4 The correct answer is option 4 i.e. Quantity A > Quantity B.

Quantity A:

Cost price of book = Rs 180

Selling price = (100 + 35)% of 180 = Rs 243

Cost price of another book = 83.33% of 180 = Rs 150

Selling price of another book = (100 - 33.33)% of 243 + 55 = 217

Profit percent = [(217 - 150)/150] × 100 = 44.67%

Quantity B:

Let Marked price be 100

After 1st discount = (100 - 15) = 85

After second discount = 85 × (100 - 18)% = 69.7

After third discount = (100 - 12)% of 69.7 = 61.336

Discount percent = discount (MP 100) = 100 - 61.336 = 38.664

Quantity A > Quantity B

 

Q:5 The correct answer is option 4 i.e. Quantity A > Quantity B.

Quantity A:

Probability = Probability of picking a diamond + probability of picking a jack - probability of picking a jack of a diamond.

Quantity A: - P = (13/52) + (4/52) - (1/52) = 16/52 = 4/13

Quantity B

Let the numerator be x

Denominator = (x + 5)

According to the conditions,

2x/(x + 5 + 1) = 1/2

4x = x + 6

x = 2

Fraction = x/(x + 5) = 2/7 = 4/14

Quantity A > Quantity B

 

Q:6 The correct answer is Option 5 i.e. Quantity A > Quantity B.

Quantity A

Let discount percent be x.

(100 - x) % of 2400 = 1980

(100 - x)/100 = 0.825

100 - x = 82.5

Quantity A: x = 17.5

Quantity B:

Principal = Rs 1610

Amount = Rs 2173.5

Time = 2 years

Interest = 2173.5 - 1610 = Rs 563.5

Interest = (P × r × t)/100

563.5 × 100 = 1610 × r × 2

Quantity B: r = 17.48

A > B

 

Q:7 The correct answer is Option 4 i.e. Quantity A > Quantity B

Quantity A.

If the number formed by two last digits of a number is divisible by 4, whole number is divisible by 4.

Of 2, 4, 6, 8 only numbers ending in 24, 28, 48, 64, 68, 84 will be divisible by 4.

With each number only two 4 digit numbers will be formed like, 6824 or 8624

Likewise two numbers each will be formed from other 5 sets of numbers.

A: total numbers = 6 × 2 = 12

Quantity B:

x2 + 5x - 84 = 0

x2 + 12x - 7x - 84 = 0

x(x + 12) - 7(x + 12) = 0

(x - 7) (x + 12) = 0

B: positive root = 7

A > B

 

Q:8 The correct answer is Option 5 i.e. Quantity A < Quantity B

Quantity A:

Speed of boat = x

Speed of stream = y

Upstream speed = x - y

Downstream speed = x + y

40% of (x + y) = x - y

4x + 4y = 10x - 10y

6x = 14y

Ratio: y/x = 6/14 = 3/7

Quantity B:

Total cases when three coins are thrown = 8 (HHH, TTT, HHT, THH, HTH, TTH, THT, HTT)

Favourable cases (atleast 2 heads) = 4

B: Probability = 4/8 = 1/2

B > A

 

Q:9 The correct answer is Option 5 i.e. Quantity A < Quantity B

Quantity A:

Let the number be X

(100 - 12) % of (100 + 12) % of X = 1540000

(88/100) × (112/100) × X = 1540000

A = X = 1562500

Quantity B:

Cost price of 10 bikes = 70000 × 10 = 700000

Selling price of 10 bikes = (100 + 8) % of 700000 = Rs 756000

Cost price of 2 cars = 2 × 450000 = 900000

Selling price of 2 cars = (100 - 4) % of  900000 = 864000

B: Total selling price = 864000 + 756000 = 1620000

B > A

 

Q:10 The correct answer is Option 3 i.e. Quantity A = Quantity B or no relationship can be established. 
Quantity A:

Three-digit number divisible by 4, 5, 6 and 7 will be LCM of 4, 5, 6 and 7.

LCM of 4, 5, 6 and 7 = 420

So the number which will give remainder 1 is 420 + 1 , 840 + 1, 1260 + 1 etc.

Since, we need only three-digit numbers.

A = 421 or 841

Quantity B:

Cost of 1287 bananas = Rs. 2 × 1287 = Rs. 2574

1287 = 12 × 107 + 3

Selling price of 107 dozens = Rs. 30 × 107 = Rs. 3210

Selling price of remaining 3 bananas = 3 × 4 = 12

Total selling price = 3210 + 12 = Rs. 3222

Profit = 3222 - 2574 = Rs. 648

A has two values, one of which is smaller than B and the other is greater.

Hence, no relationship can be established.

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Mensuration.

Practice Questions on Comparison of Quantities!

मात्राओं की तुलना पर अभ्यास प्रश्न!

Comparison of Quantities - Download PDF Now!

मात्राओं की तुलना - अभी पीडीएफ डाउनलोड करें!

Percentage questions are a common type of questions regularly asked in competitive exams. These questions carry a weightage of 1-2 questions(2-4 marks) in SSC exams and 1-2 questions(1-2 marks) in bank exams. To perform well in competitive exams, concepts of percentage should be on your tips.

Here are some tips for solving Percentage questions: Try converting percentages into fractions, memorise common percentage values, break down complex problems into simpler steps, and practice consistently to sharpen your skills with real-world scenarios.

So, we have attached 10 questions of Percentage for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Percentage</h2>

You can also download the Percentage questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 Initially a shopkeeper had n chocolates. A customer bought 14.28% chocolate from n then another customer bought 16.67% of the remaining chocolates. Finally, shopkeeper is left with 240 chocolates in his shop. How many chocolates were there initially in his shop?

1. 432

2. 336

3. 442

4. 342

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks, fraction values of percentages

Q:2 Rakesh gives 15% of his salary to his daughter. He then gives 20% of the remaining salary to his son. Of the remaining salary, he gives 37.5% to his wife and keeps rest of the salary. If the difference between the amounts that his son and daughter got is Rs 15270, find the money that Rakesh is left with himself.

1. Rs 415244.5

2. Rs 321467.5

3. Rs 318765.5

4. Rs 324487.5

(Difficulty: 3, Estimated Time: 20 Seconds) This was a simple one, don&#39;t get stuck in unnecessary calculations! 

Q:3 In an election two candidates participated, 10% voters did not vote, 6800 votes were declared invalid, the winner get 55% of the valid votes and wins by 4000 votes. Find the number of voters in the voting list.

1. 60000

2. 48000

3. 36000

4. 52000

(Difficulty: 3, Estimated Time: 20 Seconds) This is a common type of question asked..

Q:4 The cost of a Table is Rs. 900 and of a chair is Rs. 400. The price of the table and chair increases by 5% and 10% respectively. Find the percentage increase in the price of 2 Tables and 5 chairs.

1. 8.43

2. 8.72

3. 7.63

4. 7.38

(Difficulty: 3, Estimated Time: 20 Seconds) It is a bit calculative one...

Q:5 The salary of Dinesh in 2018 was Rs, 15,00,000 lakhs. Dinesh got an increment of 10% and 20% for the years 2019 and 2020. The present salary of Dinesh is equal to Sarita’s salary. What is Sarita’s present salary?

1. Rs. 18,00,000

2. Rs. 18,60,000

3. Rs. 19,20,000

4. Rs. 19,80,000

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 The cost price of an article is Rs. 1200 and the marked price is Rs. 1920. On selling the article at a discount of 25%, the shopkeeper still makes a profit of x%. What is the value of x?

1. 10%

2. 12%

3. 15%

4. 20%

(Difficulty: 3, Estimated Time: 20 Seconds) You should not spend more than 20 seconds in attempting this question...

Q:7 Amisha being a cunning shopkeeper laid out a trick to sell her products. She is marketing her product by saying that she is selling the product at a 20% loss per kg. The reality is she is weighing 30% less in each packet to cover this loss. What is the total profit percent she is getting through this trick? (Approx)

1. 11%

2. 13%

3. 14%

4. 9%

(Difficulty: 4, Estimated Time: 25 Seconds) This was a bit hard. Did you get it right?

Q:8 India played total of 24 matches in 2011, 36 matches in 2012 and 15 matches in 2013. If out of the total matches in 2011, 2012 and 2013, 50%, 75% and 80% respectively are won by India, then what percent of total matches played in all three years are won by India?

1. 60%

2. 68%

3. 46%

4. 55%

(Difficulty: 2, Estimated Time: 15 Seconds) This was a piece of cake....

Q:9 A man deposits 10% of his salary in PF. He saves 30% of the remaining. Out of the remaining salary after saving, he spends on medicine and groceries in 3 : 4. If his expenses on medicine were Rs. 8100. Find his monthly salary.

1. Rs. 30,000

2. Rs. 25,000

3. Rs. 20,000

4. Rs. 35,000

(Difficulty: 3, Estimated Time: 20 Seconds) You might have wrapped it up in 15 seconds...

Q:10 A number, when multiplied by 2 instead of dividing by 4, gives a result which is 525 more. If the same number is divided by 6 instead of multiplying by 3, then by approximately what percent the resultant value will be decreased?

1. 81.42%

2. 88.37%

3. 94.44%

4. 79.67%

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (2)</td>
			<td>2. (4)</td>
			<td>3. (4)</td>
			<td>4. (3)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (3)</td>
			<td>8. (2)</td>
			<td>9. (1)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is Option 2 i.e. 336.

14.28% = 1/7, which means 6/7 chocolates were left

16.67% = 1/6, which means 5/6 chocolates were left

According to question,

n × 6/7 × 5/6 = 240

⇒ n = 336

∴ Initially the number of chocolates = 336

 

Q:2 The correct answer is option 4 i.e. Rs 324487.5

Let his salary be 100x

Money given to daughter = 15% = 15x

Money given to son = 20% of (100x - 15x) = 17x

Remaining amount = (100 - 15 - 17) x = 68x

Money given to wife = 37.5% of 68x = 3/8 × 68x = 25.5x

Money left with Rakesh = 68x - 25.5x = 42.5x

Difference between amounts given to son and daughter = 17x - 15x = 2x

2x = 15270

x = 7635

Money left with Rakesh = 42.5x = 42.5 × 7635 = Rs 324487.5

 

Q:3 The correct answer is Option 4 i.e. 52000

Let the number of total voters in the voting list be x

Winner gets 55% of valid votes

So, loser gets 45% of valid votes

Difference = 55 - 45 = 10%

According to question, (90x/100 - 6800)10/100 = 4000

⇒ 9x/10 - 6800 = 40000

⇒ 9x/10 = 46800

⇒ x = 52000

∴ Total number of voters = 52000

 

Q:4 The correct answer is option 3 i.e. 7.63

Cost of one Table = Rs. 900

Cost of One Chair = Rs. 400

New price of one table = 900 + 5% of 900 = 900 + 45 = 945

New price of one chair = 400 + 10% of 400 = 400 + 40 = 440

Now, Total cost of 2 tables and 5 chairs,

Old price = (2 × 900) + (5 × 400) = 1800 + 2000 = 3800

New price = (2 × 945) + (5 × 440) = 1890 + 2200 = 4090

% increase = {(4090 - 3800)/3800} × 100 = (290/3800) × 100 = 7.63% 

 

Q:5 The correct answer is Option 4 i.e. Rs. 19,80,000.

Successive increase of a% and b% on a certain value &#39;x&#39; will result the original value to x (1 + a%)(1 + b%)

Given x = Rs. 15,00,000

a = 10%

b = 20%

Sarita&#39;s salary = 15,00,000 × (1 + 10%) × (1 + 20%)

= 1500000 × 1.1 × 1.2

= 1500000 × 1.32

= Rs. 1980000 

 

Q:6 The correct answer is option 4 i.e. 20%

Selling Price = Marked Price (1 - Discount %) 

Also, Selling Price = Cost Price (1 + Profit %)

Given Cost Price = Rs. 1200 
Marked Price = Rs. 1920 
Discount = 25% 
Selling Price = 1920 (1 - 25%)

= 1920 [1 - (25/100)]

= 1920 [1 - (1/4)]

= (3/4) × 1920

= Rs. 1440 
Also Selling Price = Cost Price (1 + Profit %)

1440 = 1200 (1 + Profit %)

(1 + Profit %) = (1440/1200)

(1 + Profit %) = 1.2

Profit % = 1.2 - 1 = 0.2

Profit = 20%

 

Q:7 The correct answer is Option 3 i.e. 14%

Let the cost price per kg be x.

selling price per kg = x - 0.2x = 0.8x 

Now, in each kg, she actually weights 30% less i.e. 0.7 kg.

Hence, she is selling 0.7 kg at 0.8x

So, she will be selling 1 kg at: 0.8x/0.7 = 1.142x 

Hence, Profit percentage = [(1.142x - x)/x] × 100 = 14% (Approx)

 

Q:8 The correct answer is option 2 i.e. 68%.

Total matches played by India in all three years = 24 + 36 + 15 = 75

Matches won in 2011 = 50% of 24 = 12

Matches won in 2012 = 75% of 36 = 27

Matches won in 2013 = 80% of 15 = 12

Total matches won = 12 + 27 + 12 = 51

Required percent = (51/75) × 100 = 68%

 

Q:9 The correct answer is Option 1 i.e. Rs. 30,000.

Suppose his monthly salary = Rs. 100x

According to the question:

Remaining salary after savings = 100x × 0.9 × 0.7= 63x 
Suppose he spends 3y and 4y on medicine and groceries respectively.

So, 7y = 63x

y = 9x

Expenditure on medicine was Rs. 8100

So, 3 × 9x = 8100

x = 300

Hence, Monthly salary = Rs. 30000

 

Q:10 The correct answer is Option 3 i.e. 94.44%

Let the number be ‘x’.

From the question:

x × 2 – x/4 = 525

8x – x = 2100

x = 300

The number = 300

Now, When the number is multiplied by 3, the resultant value = 300 × 3 = 900

When the number is divided by 6, the resultant value = 300/6 = 50

Difference = 900 – 50 = 850

Required percentage = (850/900) × 100 = 94.44%

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Percentage.

Practice Questions on Percentage!

Percentage Questions - Download PDF now!

प्रतिशत प्रश्न - अभी पीडीएफ डाउनलोड करें!

Partnership questions are a common type of questions regularly asked in competitive exams. These questions carry a weightage of 1-2 questions(2-4 marks) in SSC exams and 1-2 questions(1-2 marks) in bank exams. To perform well in competitive exams, concepts of partnership should be on your tips.

Here are some tips for solving Partnership questions: Start by understanding the basics and defining variables like investments and profit-sharing ratios. Utilize the partnership formula to calculate each partner&#39;s share based on their investment and time contributions. Calculate the total investment and ensure that the individual shares add up to the total profit or loss. Be prepared to handle special cases and convert ratios to fractions for easier computation.

So, we have attached 10 questions of Partnership for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Partnership</h2>

You can also download the Partnership questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 In a partnership, A invested Rs. 4000 and B invested Rs.1000 more than A. After eight months, C invested Rs 3000. If at the end of the year, C gets profit of Rs. 700, then find the total profit.

1. Rs. 8400

2. Rs. 7000

3. Rs. 8800

4. Rs. 6400

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks, fraction values of percentages

Q:2 A and B are partners in a business. They invest initially in the ratio 5 : 6. At the end of 8 months, A withdraws his investment. If in the end, they receive profits in the ratio of 5 : 9, find how long B’s investment was used.

1. 12 months

2. 10 months

3. 15 months

4. 14 months

(Difficulty: 3, Estimated Time: 20 Seconds) This was a simple one, don&#39;t get stuck in unnecessary calculations! 

Q:3 A and B started a business with initial investments in the ratio 5 : 7. After one year their profits were in the ratio 1 : 2 . If period for A’s investment was 7 months, B invested the money for :

1. 6 months

2. 2 \(1\over 2\) months

3. 10 months

4. 4 months

(Difficulty: 3, Estimated Time: 20 Seconds) This is a common type of question asked..

Q:4 A invests Rs. 12,000 for X months while B invests Rs. 16,000 for 9 months in a scheme. The profit share of B is Rs. 12,000 out of total profit Rs. 21,000. Find the value of X.

1. 6 months

2. 9 months

3. 8 months

4. 7 months

(Difficulty: 3, Estimated Time: 20 Seconds) It is a bit calculative one...

Q:5 A, B, C and D enter into a partnership. They invest Rs. 5,000, Rs. 8,000, Rs. 9,000 and Rs. 13,000 respectively. At the end of the first year, B withdraws Rs. 4,000, while at the end of the second year, C withdraws Rs. 8,000. In what ratio will the profit be shared at the end of 5 years?

1. 25 : 8 : 7 : 65

2. 25 : 24 : 21 : 65

3. 15 : 14 : 21 : 15

4. 25 : 14 : 11 : 16

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 Two friends Angel and Venger start a business with an investment of Rs.23000 and Rs.46230. If the profit of Venger is Rs. 8040 at the end of the year then, find the profit of Angel.

1. 3000

2. 3350

3. 4000

4. 3520

(Difficulty: 3, Estimated Time: 20 Seconds) You should not spend more than 20 seconds in attempting this question...

Q:7 There are two friends Satyam and Rahul, they started a business with an investment of Rs 90000 and Rs 85000 respectively. After 6 months two friends Pinku and Raja join them with an investment of Rs 75000 and Rs 80000. If the profit of Pinku is Rs 5400 at the end of the year then find the profit of the Satyam and Rahul together.

1. Rs. 325614

2. Rs. 102500

3. Rs. 25200

4. Rs. 963256

(Difficulty: 4, Estimated Time: 25 Seconds) This was a bit hard. Did you get it right?

Q:8 If P started a business with Rs 8000 and after 2 months Q joined the business with Rs 6000. Find the ratio of their profit.

1. 4 : 7

2. 3 : 8

3. 7 : 8

4. 8 : 5

(Difficulty: 2, Estimated Time: 15 Seconds) This was a piece of cake....

Q:9 A, B, and C are partners. They share profit in the ratio of 5 : 8 : 7. If C earns Rs. 8750 as his share of profit, then the share of A is:

1. 6250

2. 6500

3. 7250

4. 6520

(Difficulty: 3, Estimated Time: 20 Seconds) You might have wrapped it up in 15 seconds...

Q:10 A and B invested Rs.4800 and Rs.7200, respectively. B invested for two more months than A. If the total annual profit is Rs.5750 and the share of A in the profit is Rs.2000, then for how many months did B invest?

1. 16 months

2. 8 months

3. 10 months

4. 14 months

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (2)</td>
			<td>2. (1)</td>
			<td>3. (3)</td>
			<td>4. (2)</td>
			<td>5. (2)</td>
		</tr>
		<tr>
			<td>6. (3)</td>
			<td>7. (3)</td>
			<td>8. (4)</td>
			<td>9. (1)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is Option 2 i.e. Rs. 7000

B invested, 4000 + 1000 = Rs. 5000

Ratio of investment of A, B, and C in order = 4000 × 12 : 5000 × 12 : 3000 × 4

⇒ 4 : 5 : 1

⇒ Total = 10 units

and C, 1 unit = Rs. 700 then

⇒ 10 × 700 = Rs. 7000 

 

Q:2 The correct answer is option 1 i.e. 12 months

5x : 6x, Let B investment was used for y months

8 × 5x : 6x × y = 5 : 9

\(\frac{40x}{ 6xy}\) = \(\frac{5 }{9}\) ⇒ y = 12

 

Q:3 The correct answer is option 3 i.e. 10 months

Let B invested money for x months.

∴ 5 × 7 ∶ 7 × x = 1 ∶ 2

∴ 35 ∶ 7x = 1 ∶ 2

7x = 35 × 2

x = 10 months

 

Q:4 The correct answer is option 2 i.e. 9 months.

Profit of B = 12000

Profit of A = 21,000 - 12,000 = 9,000

The profit share ratio = A : B = 12000 × x : 16000 × 9 = x : 12

ATQ,

A : B = x/12 = 9000/12000 
⇒ x = 9 months.

 

Q:5 The correct answer is option 2 i.e. 25 : 24 : 21 : 65

Profit Share of A : B : C : D =  Respective Amount invested × Respective time periods

A : B : C : D = (5000 × 5) : (8000 × 1 + 4000 × 4) : (9000 × 2 + 1000 × 3) : (13000 × 5)

= 25000 : 24000 : 21000 : 65000

= 25 : 24 : 21 : 65

 

Q:6 The correct answer is option 3 i.e. 4000.

Profit/Loss = Investment × time

<table border="1" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>Angel</td>
			<td>Venger</td>
		</tr>
		<tr>
			<td>23000 × 12</td>
			<td>46230 × 12</td>
		</tr>
		<tr>
			<td>100</td>
			<td>201</td>
		</tr>
	</tbody>
</table>

Ratio: 100 : 201

Now, If 201 units = 8040

So, 100 units = 8040/201 × 100 = 4000

 

Q:7 The correct answer is Option 3 i.e. Rs. 25200

Profit = investment × time

<table border="1" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>Satyam</td>
			<td>Rahul</td>
			<td>Pinku</td>
			<td>Raja</td>
		</tr>
		<tr>
			<td>90000 × 12</td>
			<td>85000 × 12</td>
			<td>75000 × 6</td>
			<td>80000 × 6</td>
		</tr>
		<tr>
			<td>90 × 2</td>
			<td>85 × 2</td>
			<td>75</td>
			<td>80</td>
		</tr>
		<tr>
			<td>18 × 2</td>
			<td>17 × 2</td>
			<td>15</td>
			<td>16</td>
		</tr>
	</tbody>
</table>

Now, 15 units = 5400

So, (36 + 34) units = 5400/15 × 70 = 25200

 

Q:8 The correct answer is option 4 i.e. 8 : 5.

The ratio of their profit will also same as the money invested by them.

So, according to the question

P : Q = 8000 × 12 : 6000 × 10 = 96 : 60 = 8 : 5

 

Q:9 The correct answer is option 1 i.e. Rs 6250.

The ratio of A : B : C = 5 : 8 : 7

Profit of C = 7 units = 8750

= 1 unit = 1250

So, the share of A = 1250 × 5 = Rs 6250

 

Q:10 The correct answer is Option 3 i.e. 10 months

Let time for which A and B invested their capital is ‘x’ and ‘x + 2’ months, respectively.

Ratio of profit share of A to B = [4800 × x] : [7200 (x + 2)] = 2000 : (5750 – 2000)

2x : (3x + 6) = 2000 : 3750 = 8 : 15

30x = 24x + 48

6x = 48

x = 8

Hence, time for which B invested = x + 2 = 10 months

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Partnership.

Practice Questions on Partnership!

साझेदारी पर प्रश्नों का अभ्यास करें!

Partnership Questions - Download PDF now!

साझेदारी प्रश्न - अभी पीडीएफ डाउनलोड करें!

Profit and Loss questions are a common type of questions regularly asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams and 1-2 questions in bank exams. To get a good rank in competitive exams, you have a good grasp on the concepts of Profit and Loss as they are asked in almost every exam.

Here are some tips for solving Profit and Loss questions: Clear the basic concepts, Use the appropriate formula as there are many different formulas, Practice with diverse examples to enhance calculation speed. Practice Percentages, if you have a good hold on percentages, you will save your precious seconds.

So, we have attached 10 questions of Profit and Loss for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Profit and Loss</h2>

You can also download the Profit and Loss questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 A shopkeeper marked the price of an article 75% above the cost price and sold it after a discount of 30%. Instead of that if he had sold it after two successive discounts of 20% and 10%, he would have made a profit of Rs 168 more. Find the cost price of the article.

1. Rs. 4800

2. Rs. 4200

3. Rs. 5400

4. Rs. 4500

(Difficulty: 2, Estimated Time: 15 Seconds) This was an easy one. Did get it right?

Q:2 Pritesh purchased a tablet at Rs. 7500 and marked it ‘x’% over its cost price and sold it after giving three successive discounts of 15%, 10% and 20% to Anita. If amount paid by Anita to Pritesh is Rs. 7344, then what is the value of ‘x’?

1. 50

2. 55

3. 60

4. 70

(Difficulty: 3, Estimated Time: 20 Seconds) This was a simple one, don&#39;t get stuck in unnecessary calculations! 

Q:3 A shopkeeper gives 3 successive discounts of 10%, 12% and 15% on a travel bag. The customer pays Rs. 4039.2 for the bag then find the cost price of the bag.

1. Rs. 5000

2. Rs. 6500

3. Rs. 6000

4. Rs. 5500

(Difficulty: 2, Estimated Time: 15 Seconds) A question of seconds.....

Q:4 On a iPad, 2 successive discounts of 10% and 15% are offered by the shopkeeper. If the shopkeeper gives a single discount of 25% on it then he receives Rs. 525 less. Find the cost price of the iPad.

1. Rs. 30000

2. Rs. 3500

3. Rs. 45000

4. Rs. 35000

(Difficulty: 3, Estimated Time: 20 Seconds) It is not an easy one but I think now you&#39;re prepared for it.

Q:5 Suresh sold his car to Mukesh at a profit of 15% and Mukesh sold the same car to Deepika at a loss of 10%. If the purchase price for Deepika is Rs. 414000, then at what profit percent Suresh must sell the car, so that Mukesh faced no profit and no loss?

1. 6.5%

2. 5.5%

3. 1.5%

4. 3.5%

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 The selling price of an article becomes Rs 2160 after giving two successive discounts of x% and 25% and the marked price of an article are Rs 3600. If there is a profit of x% on selling the article after giving two successive discounts. Find the cost price of the article.

1. Rs 1500

2. Rs 1600

3. Rs 1700

4. Rs 1800

(Difficulty: 2, Estimated Time: 15 Seconds) Should we raise the level of questions?

Q:7 A shopkeeper earned 11.6% profit on cost price even after selling a laptop after two consecutive discounts of 20% and 10% on the marked price. What would have been profit percentage if the laptop is sold without any discount?

1. 45%

2. 55%

3. 60%

4. 65%

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks to save time..

Q:8 A shopkeeper cheats by marking the cost price of a duplicate bag equal to the cost price of original bag. He offered 20% discount on that price but still earned 37.5% profit by giving the duplicate bag. What is the cost price of duplicate bag if the difference of cost prices of original and duplicate bags is Rs. 690?

1. Rs. 1010

2. Rs. 1060

3. Rs. 950

4. Rs. 960

(Difficulty: 3, Estimated Time: 20 Seconds) This was a bit calculative one...

Q:9 A shopkeeper provides 2 successive discounts of 10% and 10% on a football and 12% and 8% on a cricket bat. If the marked prices of both are the same and the difference of selling prices is 30 paisa then, find the marked price.

1. Rs. 720

2. Rs. 700

3. Rs. 800

4. Rs. 750

(Difficulty: 3, Estimated Time: 20 Seconds) Try decreasing time in your calculations

Q:10 A person purchased a wooden sofa at Rs. 60000 and marked it ‘x%&#39; above its cost price and sold after allowing discount of ‘x – 25%&#39;. If the profit per cent is 16% and it is sold at &#39;x – 20%&#39; profit then what will be the selling price of the wooden sofa?

1. Rs. 70000

2. Rs. 75000

3. Rs. 72000

4. Rs. 64000

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (1)</td>
			<td>2. (3)</td>
			<td>3. (3)</td>
			<td>4. (4)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (2)</td>
			<td>8. (4)</td>
			<td>9. (4)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 1 i.e. Rs. 4800

Suppose the cost price of the article = Rs. x

So,

Marked price of the article = Rs. 1.75x

If discount of 30% is offered:

Selling price = 1.75x × 0.70 = 1.225x

If two consecutive discounts of 20% and 10% are offered:

Selling price = 1.75x × 0.80 × 0.90 = Rs. 1.26x

According to question:

1.26x – 1.225x = 168

0.035x = 168

x = 4800

So,

Cost price of the article = Rs. 4800

 

Q:2 The correct answer is option 3 i.e. 60

Given:

Cost price of tablet for Pritesh = Rs. 7500

Selling price of tablet for Pritesh = Cost price for Anita = Rs. 7344

Since, three successive discounts of 15%, 10% and 20% are provided

Marked price of tablet for Pritesh

= 7344 × (100/85) × (100/90) × (100/80)

= 8640 × (100/90) × (100/80)

= 9600 × (100/80)

= Rs. 12000

Marked up amount by Pritesh = 12000 – 7500 = Rs. 4500

Hence,

Marked up percent = x% = (4500/7500) × 100 = 60%

So, x = 60

 

Q:3 The correct answer is option 5 i.e. Rs. 6000

Suppose the cost price of the bag = Rs. x

So, Price after successive discounts = 0.9 × 0.88 × 0.85 × x = 0.6732x

Given:

The customer pays Rs. 4039.2 for the bag

So, 0.6732x = 4039.2

x = 6000

Hence, the CP of the bag = Rs. 6000

 

Q:4 The correct answer is option 4 i.e. Rs. 35000.

2 successive discounts of 10% and 15%

So,

Cumulative discount

⇒ 10 + 15 – (10 × 15)/100 = 23.5%

Suppose CP = Rs. A

Then,

⇒ A × (0.25 – 0.235) = 525

⇒ A × 0.015 = 525

⇒ A = 35000

Hence, CP = Rs. 35000

 

Q:5 The correct answer is option 4 i.e. 3.5%.

Let the purchase price for Suresh be x

So, the purchase price for Mukesh will be 1.15x

Deepika purchased the car from Mukesh at 90% of 1.15x which is equal to 414000.

So, 0.9 × 1.15x = 414000

On Solving, we get x = Rs. 400000

The cost of the car for Suresh is Rs. 400000

If Suresh sold the car to Mukesh at Rs. 414000, then Mukesh would have faced no profit and no loss.

Therefore, required profit = (414000 - 400000)/400000 × 100 = 3.5%

 

Q:6 The correct answer is Option 4 i.e. Rs. 1800.

We know that the discount is given on the marked price.

After applying 25% discount,

⇒ 3600 × 0.75 = Rs. 2700

After applying x% discount,

⇒ 2700 × (1 - x/100) = 2160

⇒ 1 - x/100 = 4/5

⇒ x/100 = 1/5 

⇒ x = 20%

By calculating the profit,

⇒ Cost Price × 1.2 = 2160

⇒ Cost Price = Rs. 1800

 

Q:7 The correct answer is option 2 i.e. 55%

Suppose the cost price of the laptop = Rs. X

So,

Selling price = 1.116 × X

Since the laptop is sold after two consecutive discounts of 20% and 10% on the marked price:

Marked price = (1.116 × X)/(0.8 × 0.9)

= 1.55X

Hence,

The profit percentage if the laptop is sold without any discount

= [(1.55X – X)/X] × 100

= 55%

 

Q:8  The correct answer is option 4 i.e. Rs. 960.

Suppose the cost prices of original and duplicate bags are ‘a’ and ‘b’ respectively.

He offered 20% discount on that price but still earned 37.5% profit by giving the duplicate bag.

So,

0.8a = 1.375b

a/b = 1.375/0.8

a/b = 55/32

Difference of cost prices of original and duplicate bags is Rs. 690

So,

Cost price of duplicate bag

= [690/(55 – 32)] × 32

= 30 × 32

= Rs. 960

 

Q:9 The correct answer is option 4 i.e. Rs. 750.

2 successive discounts of 10% and 10% on a football are to be converted into equivalent discount

So, Equivalent discount = 10 + 10 – 100/100 = 19%

2 successive discounts of 12% and 8% on a cricket bat are to be converted into equivalent discount

So, Equivalent discount = 12 + 8 – 96/100 = 19.04%

Suppose the marked price = Rs. X

The difference of selling prices is 30 paisa:

So, ⇒ X × (0.04)/100 = 0.3

⇒ X = Rs. 750

Hence, Marked price = Rs. 750

 

Q:10 The correct answer is option 2 i.e. Rs.75000

Given: Cost price = Rs. 60000

So, Marked price = (100 + x)% of 60000 = 600(100 + x)

Since the sofa is sold after allowing discount of ‘x – 25’%

Selling price = (100 – x + 25)% of 600(100 + x) = 6(100 + x)(125 – x)

The profit percent is 16%,

So, 6(100 + x)(125 – x) = 116% of 60000

(100 + x)(125 – x) = 11600

x = 45

Now the value of ‘x – 20’% profit = 25% profit

So, Selling price of the sofa when sold at ‘x – 20’% profit = 125% of 60000 = Rs.75000

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Profit and Loss.

Practice Questions on Profit and Loss!

लाभ और हानि&nbsp;पर&nbsp;अभ्यास&nbsp;प्रश्न!

Profit and Loss Questions - Download PDF now!

लाभ और हानि के प्रश्न - अभी पीडीएफ डाउनलोड करें!

Probability questions are a type of questions that can provide you valuable marks in competitive exams. These questions carry a weightage of 1-2 questions(2-4 marks) in SSC exams and 1-2 questions(1-2 marks) in bank exams. To perform well in competitive exams, you should clear your concepts of probability and have a good practice as there are a variety of questions.

Here are some tips for solving Probability questions: Understand basics of various concepts like dice, coin, cards etc. You should have a good hold on conepts of permutation and combinations, read carefully the question and evaluate all the possibilites. 

So, we have attached 10 questions of Probability for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Probability</h2>

You can also download the Probability questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 Find the probability of getting a sum of at least 8 while throwing 2 dice.

1. 11/36

2. 7/36

3. 1/3

4. 2/9

(Difficulty: 2, Estimated Time: 15 Seconds) This was an easy one, isn&#39;t it?

Q:2 What is the probability of getting a co-prime when two dices are thrown?

1. 13/36

2. 17/36

3. 31/36

4. 23/36

(Difficulty: 3, Estimated Time: 20 Seconds) Sharpen your concepts to solve such questions...

Q:3 In a general survey of 832 people, it was found that 624 owned a car. If a person is selected randomly, what is the probability that the person will not be an owner of a car?

1. 1.33

2. 0.25

3. 0.75

4. 0.40

(Difficulty: 2, Estimated Time: 15 Seconds) This was a calk walk....

Q:4 A bag contains 3 red, 5 white and 4 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

1. 23/29

2. 31/39

3. 17/27

4. 14/33

(Difficulty: 3, Estimated Time: 20 Seconds) This is a common type asked in exams..

Q:5 A bag contains 3 white balls and 5 red balls. If two balls are taken out without replacement and the first ball was found to be white, then what is the probability that second ball is red?

1. 5/8

2. 15/56

3. 5/7

4. 3/8

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 A survey of 250 students of a school was conducted and it was found that 145 students like coffee and 105 students dislike it. Out of these students, one student is selected at random, what is the probability that the selected student like coffee?

1. 0.58

2. 0.42

3. 1

4. None of these

(Difficulty: 2, Estimated Time: 15 Seconds) You should not spend more than 10 seconds in attempting this question...

Q:7 The probability that A is telling the truth is 2/5 and of B telling the truth is 4/5. Find the probability that both will contradict each other in a situation.

1. 11/25

2. 12/25

3. 13/25

4. 14/25

(Difficulty: 4, Estimated Time: 25 Seconds) This was a bit hard. Did you get it right?

Q:8 A box contains 39 cards numbered from 1 to 39. Only 1 number is numbered on each card. While choosing a card, what is the probability of selecting a card whose face number is divisible by 3 but not by 6 or 12?

1. 5/39

2. 7/39

3. 11/39

4. 17/39

(Difficulty: 3, Estimated Time: 20 Seconds) This was not an easy one...be prepared for such questions

Q:9 In a bag, there are 3 green balls, 5 white balls, and 8 orange balls. What is the minimum number of balls that must be drawn such that there are at least four balls of the same color?

1. 8

2. 9

3. 10

4. 11

(Difficulty: 3, Estimated Time: 20 Seconds) This is a similar one like you have solved....

Q:10 There are 4 red pens, and 7 green pens. If one randomly draws a pen what is the probability of getting a red pen?

1. 3/2

2. 4/11

3. 5/9

4. 2/7

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (2)</td>
			<td>2. (4)</td>
			<td>3. (2)</td>
			<td>4. (4)</td>
			<td>5. (3)</td>
		</tr>
		<tr>
			<td>6. (1)</td>
			<td>7. (4)</td>
			<td>8. (2)</td>
			<td>9. (3)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 2 i.e. 7/36.

Sample set on throwing two dices = (1, 1) (1, 2)...... (6, 6) = 36

Favourable cases (sum of 8 or more) = (4, 4) (4, 5) (5, 4) (5, 5) (5, 6) (6, 5) (6, 6) = 7

Probability = Number of favourable cases/sample set

Probability = 7/36

 

Q:2 The correct answer is option 4 i.e. 23/26.

Two numbers are coprime if their highest common factor (or greatest common divisor ) is 1

Here we have two dice, so the total number of outcomes = 6 × 6 = 36

Here are the favorable outcomes of getting coprime

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,5), (3,1), (3,2), (3,4), (3,5), (4,1), (4,3), (4,5), (5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,5)

Number of favorable outcomes = 23

∴ Probability of getting a coprime = 23/36

 

Q:3 The correct answer is option 2 i.e. 0.25.

Total number of people  = 832

Number of people who own a car = 624

Number of people who don&#39;t have a car = 832 - 624 = 208

The probability that the person will not have a car = 208/832 = 0.25

 

Q:4 The correct answer is Option 4 i.e. 14/33.

Total number of balls = (3 + 5 + 4) = 12

Probability = n(E)/n(S)

Let S be the sample spaces

E = number of ways of drawing 2 balls other than blue

n(S) = Number of ways of drawing 2 balls out of 12

n(S) = 12C2 = (12 × 11)/2 = 66

n(E) = Number of ways of drawing two balls out of 8 balls

n(E) = 8C2 = (8 × 7)/2 = 28

P(E) = n(E)/n(S) = 28/66 = 14/33

 

Q:5 The correct answer is Option 3 i.e. 5/7

If first ball is white, the remaining balls = 2 white and 5 red balls

P (second ball is red) = Number of red balls/Total number of balls

⇒ 5/(5 + 2)

⇒ 5/7

 

Q:6  The correct answer is Option 1 i.e. 0.58

Total number of students = 250

Number of students who like coffee = 145

Number of students who dislike coffee = 105

So, P (random that selected like coffee) = 145/250 = 0.58

 

Q:7 The correct answer is option 4 i.e. 14/25

Probability of A telling truth = P(A) = 2/5

Probability of B telling truth = P(B) = 4/5

Probability of A telling a lie = P(A&#39;) = 1 - (2/5) = 3/5

Probability of B telling a lie = P(B&#39;) = 1 - (4/5) = 1/5

Required Probability = P(A) × P(B&#39;) + P(B) × P(A&#39;) = (2/5) × (1/5) + (4/5) × (3/5) = (2/25) + (12/25) = 14/25

 

Q:8 The correct answer is Option 2 i.e. 7/39.

Card numbers divisible by 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39.

Card numbers that are divisible by 6 or 12 = 6, 12, 18, 24, 30, 36.

We will now remove the cards with numbers divisible by 6 or 12.

Card numbers left = 3, 9, 15, 21, 27, 33, 39

Probability = Number of favourable outcomes/Total number of outcomes

Number of favourable outcomes = 7

Total number of outcomes = 39

Probability of choosing a card on which the number marked is divisible by 3 but not by 6 or 12 = 7/39

 

Q:9 The correct answer is option 3 i.e. 10

Think the worst-case scenario :

We would pick green balls first, then pick 3 white and 3 orange balls. Now whatever ball we will pick that will be either white or orange that makes four balls of the same color, so our condition satisfies.

⇒ 3 + 3 + 3 + 1 = 10 balls

 

Q:10 The correct answer is Option 2 i.e. 4/11.

Probability = (No. of Successful Attempts / No. of total Attempts)

4 red pen & 7 green pen hence total of 11 pens

If we draw 1 pen out of the bag randomly the no. of attempts = 11C1 = No. of total Attempts

No of Successful Attempts = Attempts where a red pen is drawn out = 4C1

Probability = (4C1/11C1) = 4/11

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Probability.

Practice Questions on Probability!

Probability Questions - Download PDF now!

प्रायिकता प्रश्न - अभी पीडीएफ डाउनलोड करें!

Practice Questions on Data Interpretation (Table Chart)!

Data Interpretation (Table Chart) Questions - Download PDF now!

डेटा व्याख्या (तालिका आलेख) प्रश्न - अभी पीडीएफ डाउनलोड करें!

Missing Number series questions are the popular type of questions in competitive exams. Number series questions carry a weightage of 5-6 questions(5 - 6 marks) in bank exams. To get a good rank in competitive exams, it is important to be able to solve number series questions quickly and accurately.

Here are some tips for solving missing number series questions: identify the type of question, look for patterns in the numbers, use elimination to rule out incorrect answers, and practice regularly.

So, we have attached 10 missing number series questions for you to practice with. You should aim to solve these questions in less than half a minute for each. 
 

<h2>Practice Questions on Missing Number Series</h2>

You can also download the missing number series questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question. 
 

Directions(Q1-Q10): What will come in the place of the question mark (?) in the following number series?

Q:1 8, 6, 8, 14, 30, ?

1. 75

2. 76

3. 77

4. 78

5. 79

(Difficulty: 2, Estimated Time: 20 Seconds) It was very easy, right?

Q:2 8, 10, 20, 50, ? , 248

1. 115

2. 103

3. 113

4. 108

5. 118

(Difficulty: 2, Estimated Time: 15 Seconds) This is also too easy. Did you guess it right?

Q:3 2880, 480, 96, ?, 8, 4

1. 16

2. 24

3. 20

4. 28

5. 32

(Difficulty: 2, Estimated Time: 15 Seconds) Should we raise the level of questions?

Q:4 18, 8, 30, 20, 42, ?

1. 38

2. 36

3. 28

4. 32

5. 30

(Difficulty: 3, Estimated Time: 25 Seconds) Now, this was some good. Have you got all your questions correct so far?

Q:5 1, 11, 59, 239, 719, ?

1. 1438

2. 1439

3. 1428

4. 1429

5. 1419

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. We will increase the difficulty level from now.

Q:6 4, 36, 45, 48, 58, ?, 73, 78

1. 61

2. 60

3. 70

4. 62

5. 68

(Difficulty: 3, Estimated Time: 20 Seconds) The time is ticking. Hurry up!

Q:7 82, 86, 102, ?, 422, 1446

1. 144

2. 136

3. 166

4. 212

5. 236

(Difficulty: 2, Estimated Time: 15 Seconds) This was a tricky one. Right?

Q:8 56, 30, 49, 128.5, ?

1. 457.75

2. 375.75

3. 432.25

4. 437.75

5. 397.25

(Difficulty: 3, Estimated Time: 15 Seconds) It is a hard question but I think now you&#39;re prepared for it. Did you guess it right?

Q:9 13, 36, 105, ?, 933, 2796

1. 310

2. 315

3. 313

4. 305

5. 312

(Difficulty: 4, Estimated Time: 25 Seconds) It is different type of question. But you&#39;ll get these type of questions in the exam too. So, prepare yourself!

Q:10 ?, 1890, 630, 126, 18, 2

1. 3780

2. 2835

3. 2380

4. 1890

5. 2670

(Difficulty: 3, Estimated Time: 20 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (3)</td>
			<td>2. (5)</td>
			<td>3. (2)</td>
			<td>4. (4)</td>
			<td>5. (2)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (3)</td>
			<td>8. (1)</td>
			<td>9. (5)</td>
			<td>10. (4)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong.

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 3 i.e. 77

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent_26.01.2022_D40.png" />

 

Q:2 The correct answer is option 5 i.e. 118

8 + (13 + 1) = 10

10 + (23 + 2) = 20

20 + (33 + 3) = 50

50 + (43 + 4) = 118 = ?

118 + (53 + 5) = 248

 

Q:3 The correct answer is option 2 i.e. 24

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent_26.01.2022_D42.png" />

 

Q:4 The correct answer is option 4 i.e. 32

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent_26.01.2022_D43.png" />

 

Q:5 The correct answer is option 2 i.e. 1439

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent_26.01.2022_D44.png" />

 

Q:6 The correct answer is option 4 i.e. 62

Here we have two alternate series:

Series 1 :

34 + 11 = 45

45 + 13 = 58

58 + 15 = 73

Series 2 :

36 + 12 = 48

48 + 14 = 62

62 + 16 = 78

Hence, missing term = 62

 

Q:7 The correct answer is option 3 i.e. 166.

⇒ 82 + 41 = 86

⇒ 86 + 42 = 102

⇒ 102 + 43 = 166

⇒ 166 + 44 = 422

⇒ 422 + 45 = 1446

Hence, the missing term = 166

 

Q:8 The correct answer is option 1 i.e. 457.75.

⇒ 56 × 1/2 + 2 = 30

⇒ 30 × 3/2 + 4 = 49

⇒ 49 × 5/2 + 6 = 128.5

⇒ 128.5 × 7/2 + 8 = 457.75

Hence, the missing term = 457.75

 

Q:9 The correct answer is option 5 i.e. 312.

⇒ 13 × 3 – 3 = 36

⇒ 36 × 3 – 3 = 105

⇒ 105 × 3 – 3 = 312

⇒ 312 × 3 – 3 = 933

⇒ 933 × 3 – 3 = 2796

Hence, missing term = 312

 

Q:10 The correct answer is option 4 i.e. 1890.

⇒ 1890 ÷ 1 = 1890

⇒ 1890 ÷ 3 = 630

⇒ 630 ÷ 5 = 126

⇒ 126 ÷ 7 = 18

⇒ 18 ÷ 9 = 2

Hence, the missing term = 1890

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of missing number series.

Practice Questions on Missing Number Series!

Missing Number Series Questions - Download PDF now!

लुप्त संख्या श्रृंखला प्रश्न - अभी पीडीएफ डाउनलोड करें!

Practice Questions on Data Interpretation (Caselet)!

Data Interpretation (Caselet) Questions - Download PDF now!

डेटा व्याख्या (केसलेट) प्रश्न - अभी पीडीएफ डाउनलोड करें!

Approximation questions are a major type of questions asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams and 5-10 questions in bank exams. To get a good rank in competitive exams, it is important to know how to solve approximation questions in a speedy way.

Here are some tips for solving Approximation questions: Identify the important digits, focus on the most significant digits, use rounding off rules, after converting all the approximate values into integers, solve them using basic simplification rules. Remember in bank exams they can go up to 10 questions in a set,  hence maintaining speed and accuracy while solving these questions will award you some major marks.

So, we have attached 10 questions of Approximation for you to practice with. You should aim to solve these questions in less than half a minute for each. 
 

<h2>Practice Questions on Approximation</h2>

You can also download the Approximation questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Direction (1 - 10): What approximate value will come in place of question mark (?) in the following question?

Q:1 74.79% of 1344.11 + 12.48% of 128.20 = ?

1. 1048

2. 1024

3. 1072

4. 1096

5. 1120

(Difficulty: 3, Estimated Time: 20 Seconds) Try converting percentages to fractions..

Q:2 33.30003% of 509.99 = ?

1. 140

2. 185

3. 155

4. 170

5. 100

(Difficulty: 2, Estimated Time: 15 Seconds) This was a cake walk....

Q:3 24.987% × 639.97 + 45.21% of 359 = ?

1. 358

2. 378

3. 322

4. 302

5. 288

(Difficulty: 3, Estimated Time: 20 Seconds) A similar one...

Q:4 √2499.99 + 14.97% of 14 =?

1. 40

2. 45

3. 52

4. 58

5. 64

(Difficulty: 3, Estimated Time: 20 Seconds) Do you remember squares up to 50? Yes they save your precious seconds

Q:5 8399.99 × 14.996 ÷ 374.982 + √16.011 = ?

1. 564

2. 340

3. 320

4. 324

5. 384

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 3699.98 ÷ 99.97 + 2640.02 ÷ 29.98 − \(\frac{9.9 }{11.9}\) of 95.89 = ?

1. 76

2. 66

3. 32

4. 36

5. 45

(Difficulty: 3, Estimated Time: 20 Seconds) Remember the rule of BODMAS...

Q:7 √1294 + √674 + 3√729.12 3 = 71.01% of ?

1. 168

2. 192

3. 78

4. 718

5. 100

(Difficulty: 2, Estimated Time: 15 Seconds) The time is ticking. Hurry up!

Q:8 11.89 ÷ 2.87 + 124.9 ÷ 4.98 = 9.9% of ?

1. 490

2. 390

3. 190

4. 290

5. 590

(Difficulty: 3, Estimated Time: 20 Seconds) Don&#39;t make mistakes while rounding off...

Q:9 34.98 % of 400.01 + 49.97 % of 249.98 = 499% of ?

1. 226

2. 53

3. 26

4. 216

5. 136

(Difficulty: 2, Estimated Time: 15 Seconds) Another easy one! Let&#39;s score more...

Q:10 21.01² + 31.9² – 40.01² = ? – 26.9²

1. 485

2. 459

3. 594

4. 694

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly? 
 

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (2)</td>
			<td>2. (4)</td>
			<td>3. (3)</td>
			<td>4. (3)</td>
			<td>5. (2)</td>
		</tr>
		<tr>
			<td>6. (5)</td>
			<td>7. (5)</td>
			<td>8. (4)</td>
			<td>9. (2)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 2 i.e. 1024

74.79% of 1344.11 + 12.48% of 128.20 = ?

Taking approximate values,

75% of 1344 + 12.5% of 128 = ?

3/4 × 1344 + 1/8 × 128 = ?

1008 + 16 = ?

1024 = ?

 

Q:2 The correct answer is option 4 i.e. 170

33.30003% of 509.99 = ?

Taking approximate values,

33.33% of 510 = ?

510/3 = ?

? = 170

 

Q:3 The correct answer is option 3 i.e. 322

24.987% × 639.97 + 45.21% of 359 = ?

Taking approximate values,

? = 25% × 640 + 45% of 360 

? = 160 + 162 = 322

 

Q:4 The correct answer is option 3 i.e. 52

√2499.99 + 14.97% of 14 = ?

Taking approximate values,

√2500 + 15/100 × 14 = ?

50 + 2.1 = ?

52 ≈ ?

 

Q:5 The correct answer is option 2 i.e. 340

8399.99 × 14.996 ÷ 374.982 + √16.011 = ?

Taking approximate values,

(8400 × 15)/375 + √16 = ?

(84 × 100)/25 + 4 = ?

336 + 4 = ?

340 = ?

 

Q:6 The correct answer is option 5 i.e. 45.

3699.98 ÷ 99.97 + 2640.02 ÷ 29.98 − \(\frac{9.9 }{11.9}\) of 95.89 = ?

Taking approximate values,

? = 3700/100 +  \(\frac{3700}{100}+\frac{2640}{30}-\frac{10}{12}× 96\)

? = 37 + 88 – 80 = 45

 

Q:7 The correct answer is option 5 i.e. 100.

√1294 + √674 + 3√729.12 = 71.01% of ?

Taking approximate values,

\(\frac{71}{100}\) × ? = 36 + 26 + 9

⇒ ? = 100

 

Q:8 The correct answer is option 4 i.e. 290.

11.89 ÷ 2.87 + 124.9 ÷ 4.98 = 9.9% of ?

Taking approximate values,

 \(\frac{?}{10}\) = 12 ÷ 3 + 125 ÷ 5

⇒ ? = 290

 

Q:9 The correct answer is option 2 i.e. 53.

34.98 % of 400.01 + 49.97 % of 249.98 = 499% of ?

Taking approximate values,

\(\frac{500}{100}\) × ? =  \(\frac{35}{100}\) × 400 + \(\frac{50}{100}\) × 250

⇒ ? = 265/5 ⇒ ? = 53

 

Q:10 The correct answer is option 3 i.e. 594.

21.01² + 31.9² – 40.01² = ? – 26.9²

Taking approximate values,

? = 21² + 32² – 40² + 27²

? = 441 + 1024 – 1600 + 729 = 594

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Approximation.

Approximation Questions - Download the PDF now!

सन्निकटन के प्रश्न - पीडीएफ डाउनलोड करें!

Simplification questions are regularly asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams and 5-15 questions(10-15 marks) in banking exams. To get a good rank in competitive exams, you should have a great practice of a variety of simplification as they are scoring questions.

Here are some tips for solving Simplification questions: Clear the basic concepts of BODMAS, percentage, power and exponents, Memorise squares, cubes and tables, Learn short tricks, get a good hold on percentages. Practice as much as you can on improving your calculation speed and accuracy.

So, we have attached 10 questions of Simplification for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Simplification</h2>

You can also download the Simplification questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Directions(Q1-Q10): What value should come in place of question mark (?) in the following questions?

Q:1 4\(2\over3\)− 3\(1\over6 \) + 5\(5\over9\) − 2\(7\over12\) = ?

1. 4\(17\over 36\)

2. 4\(19\over36\)

3. 4\(7\over12\)

4. 4\(1\over2\)

5. 4\(4\over9\)

(Difficulty: 3, Estimated Time: 20 Seconds) Try adding the whole numbers and fractions separately...

Q:2 (√11664 + 3√74088) × ? = 125 × 12

1. 12

2. 13

3. 12.5

4. 15

5. 10

(Difficulty: 4, Estimated Time: 25 Seconds) Square and cube roots questions are very common...

Q:3 [(238 ÷ 4) + 36.5] ÷ 12 = ?% of 64

1. 12 ½%

2. 15%

3. 16%

4. 8%

5. 10%

(Difficulty: 3, Estimated Time: 20 Seconds) Don&#39;t forget using BODMAS....

Q:4 <img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/057.png" />

1. 30

2. 24

3. 36

4. 48

5. 18

(Difficulty: 3, Estimated Time: 20 Seconds) Should we increase the level?

Q:5 35% of 48% of 2/5 of 15000 = ?

1. 1004

2. 996

3. 1020

4. 1008

5. 1012

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 1528 + 525 ÷ 25 – 840 = 510 + ?

1. 199

2. 299

3. 159

4. 189

5. 165

(Difficulty: 2, Estimated Time: 15 Seconds) This was a an easy one....

Q:7 √1225 ÷ 7 + 18.5 × 16 – 18% of 10800 = ? – 1800

1. 259

2. 169

3. 157

4. 129

5. 141

(Difficulty: 3, Estimated Time: 20 Seconds) Be prepared for such questions in exam!

Q:8 65% of 180 + ?% of 210 = 80% of 225

1. 45

2. 30

3. 40

4. 50

5. 25

(Difficulty: 2, Estimated Time: 15 Seconds) Try converting percentages to fractions to fasten the calculation

Q:9 1500 + ? + 17.5 × 8 – 5% × 20 = 1764

1. 145

2. 115

3. 120

4. 135

5. 125

(Difficulty: 3, Estimated Time: 20 Seconds) Yes, You know how to solve such questions!

Q:10 13/17 of 8/156 of 153 = ?

1. 8

2. 12

3. 7

4. 6

5. 4

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (1)</td>
			<td>2. (5)</td>
			<td>3. (1)</td>
			<td>4. (3)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (1)</td>
			<td>7. (3)</td>
			<td>8. (2)</td>
			<td>9. (5)</td>
			<td>10. (4)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 1 i.e. 4\(17\over 36\)

? = (4 − 3 + 5 − 2) + ( 2/3 − 1/6 + 5/9 − 7/12)

? = 4 + ((24 − 6 + 20 − 21)/36) = 4\(17\over 36\)

 

Q:2 The correct answer is option 5 i.e. 10

(√11664 + 3√74088) × ? = 125 × 12

(108 + 42) × ? = 1500

⇒ ? = 1500/150 = 10

 

Q:3 The correct answer is option 1 i.e. 12 ½%

[(238 ÷ 4) + 36.5] ÷ 12 = (?% × 64)/100

⇒ (59.5 + 36.5) ÷ 12 = (?% × 64)/100

⇒ 8 = (?% × 64)/100

⇒ ?% = 12 ½%

 

Q:4 The correct answer is option 3 i.e. 36

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/057.png" />

√24 × 435 ÷ ? + 16 + 18 = (18)2

⇒ 24 × 435 ÷ ? + 34 = (18)2 = 324

⇒ (24 × 435)/? = 290

⇒ ? = (24 × 435)/290 = 36

 

Q:5 The correct answer is option 4 i.e. 1008

35% of 48% of 2/5 of 15000 = ?

? = 35/100 × 48/100 × 2/5 × 15000 = 1008

 

Q:6 The correct answer is option 1 i.e. 199

1528 + 21 – 840 – 510 = ?

? = 1549 – 1350

? = 199

 

Q:7 The correct answer is option 3 i.e. 157

35/7 + 296 – 1944 = ? – 1800

301 + 1800 – 1944 = ?

? = 157

 

Q:8 The correct answer is option 2 i.e. 30

65/100 × 180 + ?/100 × 210 = 80/100 × 225

?/100 × 210 = 180 – 117

? = (63 × 100)/210 = 30

 

Q:9 The correct answer is option 5 i.e. 125

1500 + 140 – 1 + ? = 1764

? = 1764 – 1639

? = 125

 

Q:10 The correct answer is option 4 i.e. 6

13/17 × 8/156 × 153 = ?

⇒ ? = 6

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Simplification.

Practice Questions on Simplification!

Simplification Questions - Download the PDF now!

सरलीकरण प्रश्न - पीडीएफ डाउनलोड करें!

Data Sufficiency questions are a typical type of questions asked in competitive exams. These questions carry a weightage of 3-5 questions (3-5 marks) in bank exams. To get a good rank in competitive exams, it is important to know how to solve Data Sufficiency questions in a speedy and easier way, as these questions test your knowledge of basics.

Here are some tips for solving Data Sufficiency questions: Carefully read and understand the problem statement, Evaluate the information provided in both statements individually for sufficiency, determine relationships between unknowns and known quantities. Be wary of distractors; the goal is to find if the data is sufficient, not to solve the problem. As familiarity with common patterns is crucial; practice a variety of examples.

So, we have attached 10 questions of Data Sufficiency for you to practice with. You should aim to solve these questions in less than half a minute for each.

<h2>Practice Questions on Data Sufficiency</h2>

You can also download the Data Sufficiency questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 Directions: Given below is a question along with two statements, you have to tell which statement, A or B is sufficient to answer the question or if both A and B together are sufficient to answer the question. Use your knowledge of mathematics to answer the questions.

Average weight of group before addition of new boy?

Statement A: On addition of new boy weighing 40 kg, average of group increases by 1 kg.

Statement B: Number of boys before addition of new boy was 24.

1. Statement A alone is sufficient while statement B is not.

2. Statement B alone is sufficient while statement A is not.

3. Statement A and B together are needed.

4. Both statements A and B together are not sufficient even with the combined information.

5. Both statements A and B alone are sufficient to answer the question.

(Difficulty: 3, Estimated Time: 20 Seconds) It was a basic question...

Q:2 Directions: Given below is a question along with two statements, you have to tell which statement, A or B is sufficient to answer the question or if both A and B together are sufficient to answer the question. Use your knowledge of mathematics to answer the questions.

Will the sum of two natural numbers, say a and b be odd?

A: Difference of the numbers is even.

B: Product of numbers is even.

1. Statement A alone is sufficient while statement B is not.

2. Statement B alone is sufficient while statement A is not.

3. Statement A and B together are needed.

4. Both statements A and B together are not sufficient even with the combined information.

5. Both statements A and B alone are sufficient to answer the question.

(Difficulty: 3, Estimated Time: 20 Seconds) A test of your concepts.....

Q:3 Directions: Given below is a question along with two statements, you have to tell which statement, A or B is sufficient to answer the question or if both A and B together are sufficient to answer the question. Use your knowledge of mathematics to answer the questions.

Find the Interest if the sum of  Rs 24000 is given at compound interest.

A: It is given at 12% per annum for 2 years

B: Interest is compounded semi-annually.

1. Statement A alone is sufficient while statement B is not.

2. Statement B alone is sufficient while statement A is not.

3. Statement A and B together are needed.

4. Both statements A and B together are not sufficient even with the combined information.

5. Both statements A and B alone are sufficient to answer the question.

(Difficulty: 3, Estimated Time: 20 Seconds) First analyse both statements, then answer...

Q:4 Directions: Given below is a question along with two statements, you have to tell which statement, A or B is sufficient to answer the question or if both A and B together are sufficient to answer the question. Use your knowledge of mathematics to answer the questions.

Time taken by train to cross a platform of length 300 m

Statement A: It crosses a train twice its length travelling in opposite direction in 25 seconds.

Statement B: Length of train is 150 m.

1. Statement A alone is sufficient while statement B is not.

2. Statement B alone is sufficient while statement A is not.

3. Statement A and B together are needed.

4. Both statements A and B together are not sufficient even with the combined information.

5. Both statements A and B alone are sufficient to answer the question.

(Difficulty: 3, Estimated Time: 20 Seconds) An easy one....

Q:5 Directions: Given below is a question along with two statements, you have to tell which statement, A or B is sufficient to answer the question or if both A and B together are sufficient to answer the question. Use your knowledge of mathematics to answer the questions.

Percentage of profit X received out of the total profit in a business with only two partners, X and Y

Statement A: Ratio of investments of X and Y is 3 : 4

Statement B: Ratio of time of investment of X and Y is 2 : 3

1. Statement A alone is sufficient while statement B is not.

2. Statement B alone is sufficient while statement A is not.

3. Statement A and B together are needed.

4. Both statements A and B together are not sufficient even with the combined information.

5. Both statements A and B alone are sufficient to answer the question.

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 Directions: The following questions are accompanied by three statements (I), (II) and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the questions.

What is the marked price of the suitcase?

I. When a discount of 15% is offered, the profit earned is 10.5%.

II. The cost price of the suitcase is Rs. 1500.

III. The marked price is 30% above the cost price.

1. Only Statement I and Statement III are sufficient

2. Either Statement I and Statement II or Statement II and Statement III are sufficient

3. Only Statement II and Statement III are sufficient

4. All statements together are necessary to answer the question 

5. All three statements are not sufficient 

(Difficulty: 3, Estimated Time: 20 Seconds) The more the statements, the more combinations possible.....

Q:7 Directions: The following questions are accompanied by three statements (I), (II) and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the questions.

A, B, and C together start a business with a total investment of Rs. 15000. At the end of the year, the total profit is Rs. 3000. What is A&#39;s share in the profit?

I. A&#39;s investment is 3/2 times B&#39;s.

II. B&#39;s investment is twice that of C.

III. A&#39;s investment is thrice that of C.

1. I and II only

2. II and III only

3. All I, II, and III

4. Any two of the three

5. None of these

(Difficulty: 3, Estimated Time: 20 Seconds) The time is ticking. Hurry up!

Q:8 Directions: The following questions are accompanied by three statements (I), (II) and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the questions.

What is the area of the rectangle?

I: The difference between the sides is 5 cm.

II: The measure of the diagonal is 10 cm.

III: There are two diagonals in the rectangle.

1. Only Statement I and Statement II are sufficient

2. Either Statement I and Statement II or Statement II and Statement III are sufficient

3. Only Statement II and Statement III are sufficient

4. All statements together are necessary to answer the question 

5. All three statements are not sufficient 

(Difficulty: 3, Estimated Time: 20 Seconds) Just a matter of formulas......

Q:9 Directions: The following questions are accompanied by three statements (I), (II) and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the questions.

What is R&#39;s share of profit in a joint venture?

I. Q started a business investing Rs. 80000.

II. R joined him after 3 months.

III. P joined after 4 months with a capital of Rs. 120000 and got Rs. 6000 as his share of profit.

1. Only Statement I and Statement III are sufficient

2. Either Statement I and Statement II or Statement II and Statement III are sufficient

3. Only Statement II and Statement III are sufficient

4. All statements together are necessary to answer the question 

5. All three statements are not sufficient 

(Difficulty: 4, Estimated Time: 25 Seconds) Try maintaining high accuracy in such questions...

Q:10 Directions: The following questions are accompanied by three statements (I), (II) and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the questions.

What was the discount percentage?

I. 23.5% profit was earned by selling an almirah for Rs. 12350.

II. If there was no discount, the earned profit would have been 30%.

III. The cost price of the almirah was Rs. 10000.

1. I and II only

2. II and III only

3. I and III only

4. All I, II, and III

5. None of these

(Difficulty: 3, Estimated Time: 20 Seconds) Did you guess them all correctly? 
 

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (3)</td>
			<td>2. (1)</td>
			<td>3. (3)</td>
			<td>4. (4)</td>
			<td>5. (3)</td>
		</tr>
		<tr>
			<td>6. (2)</td>
			<td>7. (4)</td>
			<td>8. (1)</td>
			<td>9. (5)</td>
			<td>10. (1)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is Option 3 i.e. Statement A and B together are needed.

Statement A:

Let average before the addition of new boy be x

Number of persons = y

(x + 1) × (y + 1) = xy + 40

Two variables and 1 equation, information insufficient.

Statement B:

Number of boys before = 24

New number = 25

25 × (x + 1) = 24x + y

Still insufficient.

Combining both,

(x + 1) × (y + 1) = xy + 40

(x + 1) × 25 = 24x + 40

25x + 25 = 24x + 40

x = 15 kg

Statement A and B together are needed.

 

Q:2 The correct answer is Option 1 i.e. Statement A alone is sufficient while Statement B is not.

Statement A:

Difference between the numbers is even.

Numbers can either be both odd or both even to get a difference even.

Sum = o + o or e + e = odd

Hence statement A alone is sufficient.

Statement B:

Product of numbers is even.

This can happen if at least one number is even.

e + o = o

e + e = e

We can get either odd or even sum, hence Statement B is insufficient.

 

Q:3 The correct answer is Option 3 i.e. Statement A and B together are needed.

Principal = Rs 24000

Rate = 12%

Time = 2 years

But we don&#39;t have compounding period.

So statement A alone is insufficient.

Statement B: talks of only the compounding period.

Though statement B alone cannot alone answer the question, it gives the only information that was missing from Statement A.

Now if we combine both statements we can get the interest as,

Rate = 12%/2 = 6%(by B)

Hence time = 4

Compound interest = 24000(1 + 6/100)4 = Rs. 30300(approx.)

 

Q:4 The correct answer is option 4 i.e. Both statements A and B together are not sufficient even with the combined information.

Length of platform = 300 m

Statement A:

Let length of train = x m

Length of other train = 2x m

Let speed of train = a m/s

Speed of other train = b m/s

Relative speed = (a + b) m/s

x + 2x = 25 × (a + b)

Three Variables and 1 equation, statement A is insufficient.

Statement B:

Length of train = 150 m

Length of platform = 300 m

Total distance = 150 + 300 = 450 m

But we don&#39;t know the speed to find the time.

Combining both,

450 = (a + b) × 25

(a + b) = 18 m/s

But we still can&#39;t find the speed of the train to find time.

Hence, even A and B together are insufficient.

 

Q:5 The correct answer is Option 3 i.e. Statement A and B together are needed.

Statement A:

Let their investments be 3m and 4m respectively

But the time is unknown

Statement B:

Let time be 2t and 3t respectively.

But here investment is not given, 

Hence combining both,

Ratio of profit = Ratio of investments = (3m × 2t)/(4m × 3t) = 1/2

X will get 1/(2 + 1) of total profit or 33.33%

Hence both A and B are needed

 

Q:6 The correct answer is option 2 i.e. Either Statement I and Statement II or Statement II and Statement III are sufficient

From Statements I and II:

Cost price = 1500

Gain = 10.5%

Selling price = 110.5% of cost price = (110.5 × 1500/100) = 1657.50

Discount = 15%

Marked price = (100 × 1657.5/85) = 1950

From Statements II and III:

Marked price = 130% of cost price = (130 × 1500/100) = 1950

 

Q:7 The correct answer is option 4 i.e. any two of the three

Let C&#39;s contribution = x

From Statement I and II, we get C = x, B = 2x, A = 3x

From Statement  II and III, we get C = x, B = 2x, A = 3x

From Statement  I and III, we get C = x, A = 3x, B = 2x

Thus, A : B : C = 3 : 2 : 1

So, A&#39;s share = 3000 × 3/6 = 1500

 

Q:8 The correct answer is option 1 i.e. Only Statement I and Statement II are sufficient

We know that area of the rectangle = Length × breadth = LB

Statement I:

Let the length and breadth of the rectangle be L and B

L - B = 5

Statement II:

⇒ (Diagonal)2 = L2 + B2 = 100

Using the formula of (a - b)2 we get

⇒ (L - B)2 = L2 + B2 - 2LB

⇒ 25 = 100 - 2LB

⇒ 2LB = 75

⇒ LB = 37.5 cm2

Statement III : It is a basic property of rectangle and does not provide any help in finding the area.

Hence, Only Statement I and Statement II are sufficient

 

Q:9 The correct answer is option 5 i.e. All three statements are not sufficient.

From the statments I, II, and III we get:

⇒ P : Q : R = (120000 × 8) : (80000 × 12) : (x × 9)

Since R&#39;s investment is not given, the above ratio cannot be found.

R&#39;s share of profit cannot be found.

 

Q:10 The correct answer is option 1 i.e. I and II only.

Statement I: Selling price = Rs. 12350

Gain = 23.5%

Cost price = (100/123.5 × 12350) = Rs. 10000

Statement II: Marked price = 130% of cost price = 10000 × 1.3 = 13000

From I and II, discount = 13000 - 12350 = 650

Discount % = (650/13000 × 100) = 5%

So, I and II are sufficient to find the answer.

Statement III: Cost price = Rs. 10000

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Data Sufficiency.

Practice Questions on Data Sufficiency!

Data Sufficiency Questions - Download the PDF now!

डेटा पर्याप्तता प्रश्न - अभी पीडीएफ डाउनलोड करें!

Circles questions are a major part of geometry asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams. To get a good rank in competitive exams, you should have a conceptual clarity of questions of circles as they also are part of mixed geometry questions asked.

Here are some tips for solving Circles questions: Grasp circle fundamentals. Learn geometry rules for tangents, secants, and chords. Understand inscribed angles and central angles, and practice accurate diagram drawing for visualization. Utilize Pythagoras&#39; theorem in right triangles associated with circles.

So, we have attached 10 questions of Circles for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Circles</h2>

You can also download the Circles questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 In the given figure, O is the centre of the circle. If ∠AOB = 88° then what is the value of ∠ACB?

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406488,.png" />

1. 130°

2. 132°

3. 134°

4. 136°

(Difficulty: 3, Estimated Time: 20 Seconds) A good question to start with....

Q:2 In the given figure, QS and RS are two tangents to the circle with center O. If ∠QSR = 80° then what is the value of ∠QPR?

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406477,.png" />

1. 40°

2. 45°

3. 50°

4. 55°

(Difficulty: 3, Estimated Time: 20 Seconds) You won&#39;t get easy questions in exam...

Q:3 In the given figure, QT is tangent to the circle with radius 5 cm. If T is at a distance of 11 cm from the centre O and TR is 6 cm, then what is the length of chord PR?

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406474,.png" />

1. 5 cm 

2. 10 cm

3. 15 cm 

4. 20 cm

(Difficulty: 3, Estimated Time: 20 Seconds) Do you remember the relation.....

Q:4 In the given figure, if a = 70° then what is the value of b?

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406358,.png" />

1. 20°

2. 30°

3. 40°

4. 50°

(Difficulty: 2, Estimated Time: 15 Seconds) This was quite easy.....

Q:5 in the given figure O is the center of the circle and ∠XPY = 30°. Find the average of the ∠XOY and ∠ XZY.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407619,.png" style="height:129px; width:260px" />

1. 150°

2. 255°

3. 127.5°

4. 105°

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 MN and QR are two chords in a circle with center O and MR is the circle&#39;s diameter. When MN and QR are further extended, it meets at a point P outside the Circle. if the ∠RMN = 38°, ∠MPR = 32° then find the angle ∠QNR.

1. 35°

2. 25°

3. 30°

4. 20°

(Difficulty: 4, Estimated Time: 25 Seconds) You should have a good practice to solve such.....

Q:7 PQ and MN are the two chords that intersect on point O in the circle. If the length of MN  is 11cm, PO is 4 cm, and QO is 7.5 cm then, find the difference between the ON and OM.

1. 1

2. 2

3. 3

4. 4

(Difficulty: 3, Estimated Time: 20 Seconds) Be prepared for such questions in exam!

Q:8 If the radii of the two circles, not touching each other, are 8 cm and 7 cm respectively and the distance between the centres of the circles is 25 cm then find the length of the common transversal tangent. 

1. 15 cm

2. 20 cm

3. 25 cm

4. 30 cm

(Difficulty: 2, Estimated Time: 15 Seconds) This was not a hard question!

Q:9 If two circles with radius 25 cm and 16 cm respectively touch each other externally, then find the length of the common tangent to both the circles.

1. 20 cm

2. 30 cm

3. 40 cm

4. 50 cm

(Difficulty: 3, Estimated Time: 20 Seconds) A similar one....

Q:10 In the given figure, what will be the value of x (in cm)?

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407298,.png" style="height:131px; width:226px" />

1. 1

2. 2

3. 3

4. 4

(Difficulty: 3, Estimated Time: 20 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (4)</td>
			<td>2. (3)</td>
			<td>3. (2)</td>
			<td>4. (1)</td>
			<td>5. (3)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (1)</td>
			<td>8. (2)</td>
			<td>9. (3)</td>
			<td>10. (4)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 4 i.e. 136°.

Given:

∠AOB = 88°

Concept Used:

Central angle is double the angle of the major arc.

The sum of the opposite angles of a cyclic quadrilateral is 180°.

Explanation:

In the given figure,

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_05.06.2023_D2.png" style="height:150px; width:147px" />

The lines from A and B touches the circumference of the circle at point D, such that ∠ADB = ½ × ∠AOB 

∠AOB = 88° 

Then, ∠ADB = ½ × 88° 

⇒ ∠ADB = 44°

Now, In quadrilateral ABCD, ∠C + ∠D = 180°

⇒ ∠C + 44° = 180°

⇒ ∠C = 180° - 44°

⇒ ∠C = 136°

Hence the value of ∠ACB is 136°.

 

Q:2 The correct answer is option 3 i.e. 50°

Given:

∠QSR = 80°

Concept Used:

The angle between the radius of the circle and the tangent to the circle is 90°.

Sum of all angle of a quadrilateral is 360°.

The angle on the circumference of the circle is half the angle at the center of the circle, when subtended from the same arc.

Explanation:

In the figue given,

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/(Ans) HRQ-406477,.png" />

∠Q = ∠R = 90°

Now, in quadrilateral QORS, ∠Q + ∠O + ∠R + ∠S = 360°

⇒ 90° + ∠O + 90° + 80° = 360°

⇒ ∠O + 260° = 360°

⇒ ∠O = 100°

Now, ∠QPR = ½ × ∠O

⇒ ∠QPR = ½ × 100°

⇒ ∠QPR = 50°

Hence the value of ∠QPR is 50°.

 

Q:3 The correct answer is option 2 i.e. 10 cm.

Given:

OT = 11 cm

Radius, OC = 5 cm

TR = 6 cm

Concept Used:

If any two chords, say AB and CD, of a circle, intersect each other at a point E outside a circle, then AE × EB = EC × ED

Explanation:

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/(Ans) HRQ-406474,.png" />

In the above figure, TA = OT + OA = 11 + 5 = 16 cm

TC = TA - AC = 16 - 10 = 6 cm

Now, TR × TP = TC × TA

⇒ 6 × TP = 6 × 16

⇒ TP = 16 cm

So, PR = TP - TR

⇒ PR = 16 - 6 

⇒ PR = 10 cm

Hence the length of PR is 10 cm

 

Q:4 The correct answer is option 1 i.e. 20°.

Given:

a = 70°

Concept Used:

The tangent to the circle makes a right angle with tha radius of the circle.

Sum of all angles of a triangle is 180°.

Explanation:

∠P = ∠Q = 90°  {PR and QR are tangents to the circle}

Now, in triangle POR,

∠P + ∠O + ∠R = 180°

⇒ 90° + 70° + ∠R = 180°

⇒ ∠R = 180° - 160°

⇒ ∠R = 20°

Hence the value of b is 20°.

 

Q:5 The correct answer is option 3 i.e. 127.5°.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407619,.png" style="height:109px; width:220px" />

⇒ ∠XPY = 30°

⇒ ∠OXP = ∠OYZ = 90° (Tangent on a circle)

⇒ ∠YOX + ∠OXP + ∠OYZ + ∠XPY = 360°

⇒ ∠YOX = 360° - (90° + 90° + 30) = 150°

⇒ ∠XZY = 90° + ∠XPY/2 = 90 + 15 = 105°

Required average = (150° + 105°)/2 = 127.5°

 

Q:6 The correct answer is option 4 i.e. 20°.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-407613,.png" style="height:119px; width:220px" />

∠RMN = 38°, ∠MPR = 32° Given

∠MNR = 90° (The angle subtended in a semicircle is a right angle)

∠MRQ = 32° + 38° = 70° (The exterior angle of the given triangle equals to the sum of the two opposite interior angles present in that triangle)

∠MRQ = ∠QNM = 70°

∠QNR = 90° - 70° = 20°

 

Q:7 The correct answer is option 2 i.e.1.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_05.06.2023_D5.png" />

We know that if the two chords in a circle intersect each other on point O then,

⇒ OP × OQ = OM × ON

⇒ 4 × 7.5 = x(11 - x)

⇒ 30 = 11x - x2

⇒ x2 - 11x + 30 = 0

⇒ x = 6, 5

If x = 6,

Then MO = 6, NO = (11 - 6) = 5

So, the difference = (6 - 5) = 1

 

Q:8 The correct answer is option 2 i.e.  20 cm.

Formula Used:

AB2 = (OO&#39;)2 - (OA + O&#39;B)2

Explanation:

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_05.06.2023_D6.png" />

From the formula, we get

AB2 = (25)2 - (8 + 7)2

⇒ AB2 = 625 - 225

⇒ AB2 = 400

⇒ AB = \(\sqrt{400}\)

⇒ AB = 20

Hence, the value of a common transverse tangent is 20 cm

 

Q:9 The correct answer is option 3 i.e. 40 cm.

Given:

R1 = 25 cm

R2 = 16 cm

Formula Used:

(Length of common tangent)2 = (Distance between the centers of the circle)2 - (R1 - R2)2 

Explanation:

In the given question, the two circles touch each other externally, then the distance between their centres is equal to the sum of their radii.

Now, according to the formula

(Common Tangent)2 = (25 + 16)2 - (25 - 16)2

⇒ (Common Tangent)2 = (41)2 - (9)2

⇒ (Common Tangent)2 = (41 + 9)(41 - 9)

⇒ (Common Tangent)2 = 50 × 32

⇒ (Common Tangent)2 = 1600

⇒ Common tangent = \(\sqrt{1600}\)

⇒ Common Tangent = 40

Hence the value of common tangent is 40 cm.

 

Q:10 The correct answer is option 4 i.e. 4 cm.

Given :

AD = 7 cm

BD = 5 cm

BE = 6 cm

Concept Used:

If the two chords, say AB and CD, of a circle, intersect each other at point E outside the circle, then 

EA × EB = EC × ED

Explanation:

In the given question,

BD × BA = BE × BC

⇒ 5 × (7 + 5) = 6 × (6 + x)

⇒ 5 × 12 = 6 × (6 + x)

⇒ 6 + x = 10

⇒ x = 4

Hence, the value of x is 4 cm

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Circles.

Circles Questions PDF with detailed solutions

वृत्त के प्रश्नों की पीडीएफ विस्तृत उत्तरों के साथ

Hello everyone, simplification is a topic that is asked in the quantitative aptitude section of almost every competitive exam. To score good marks you need to excel in this topic, so here are some key concepts which will let you learn about it more clearly.

Here, we are going to learn all about the BODMAS rule, what it is, where to use it along with some mind-blowing formulae which will transform you into human calculator.

E.g. 221 ÷ 13 × √576 + (10)2 = ?

We are completely clueless about where to start right now, so let’s start from left

⇒ 221 ÷ 13 × 24 + 100 = ? ⇒ 221 ÷13 × 124 = ?

⇒ 221 ÷ 1612 = ? ⇒ 7.290 (Approx.)

Which is wrong, so how to approach these types of questions? Is there a specific method to approach them?

Yes, VBODMAS rule.

<h2>VBODMAS Rule</h2>

Full form of VBODMAS is:

<table cellspacing="0" style="width:300px">
	<tbody>
		<tr>
			<td>
			Sr. No.
			</td>
			<td>
			Alphabet
			</td>
			<td>
			Significance
			</td>
		</tr>
		<tr>
			<td>
			1.
			</td>
			<td>
			V
			</td>
			<td>
			Vinculum or Bar
			</td>
		</tr>
		<tr>
			<td>
			2.
			</td>
			<td>
			B
			</td>
			<td>
			Brackets
			</td>
		</tr>
		<tr>
			<td>
			3.
			</td>
			<td>
			O
			</td>
			<td>
			Of
			</td>
		</tr>
		<tr>
			<td>
			4.
			</td>
			<td>
			D
			</td>
			<td>
			Division
			</td>
		</tr>
		<tr>
			<td>
			5.
			</td>
			<td>
			M
			</td>
			<td>
			Multiplication
			</td>
		</tr>
		<tr>
			<td>
			6.
			</td>
			<td>
			A
			</td>
			<td>
			Addition
			</td>
		</tr>
		<tr>
			<td>
			7.
			</td>
			<td>
			S
			</td>
			<td>
			Subtraction
			</td>
		</tr>
	</tbody>
</table>

For every question we have to do each step from up to bottom. Let’s take an example and try to solve it.

E.g. 221 ÷ 13 × √576 + (10)2 = ?

Step-1: First try to Solve √576 and (10)2

⇒ 221 ÷ 13 × 24 + 100 = ?

Step-2: Division

⇒ 221 ÷ 13 × 24 + 100 = ?

⇒ 17 × 24 + 100 = ?

Step-3: Multiplication

⇒ 17 × 24 + 100 = ? ⇒ 408 + 100 = ?

Step-4: Addition

 408 + 100 = ? ⇒ 508 (Ans.)

Tip: “While solving the brackets in any question, first we solve the small brackets (), then we solve the curly brackets

{} and at last we solve the big brackets [].”

E.g. [(2211 ÷ 67)2 – 21 × √(256 )] ÷ (549 – 213) = ? ÷ 1344

Let’s apply VBODMAS rule here

Step-1: Let’s Solve √256

⇒ [(2211 ÷ 67)2 - 21 × 16] ÷ (549 - 213) = ? ÷ 1344

Step-2: Start opening the Brackets

⇒ [(2211 ÷ 67)2 - 21 × 16] ÷ (549 – 213) = ? ÷ 1344

⇒ [(33)2 – 21 × 16] ÷ 336 = ? ÷ 1344

Step-3: Now we have to first simplify equation in bracket then open it by VBODMAS

⇒ [1089 – 21 × 16] ÷ 336 = ? ÷ 1344

⇒ [1089 – 336] ÷ 336 = ? ÷ 1344

Step-4: Now open the bracket

⇒ [1089 –336] ÷ 336 = ? ÷ 1344

⇒ 753 ÷ 336 = ? ÷ 1344

<h3>Step-5: Division</h3>

⇒ 753 ÷ 336 = ? ÷ 1344

⇒ 753 × 1344 ÷ 336 = ?

⇒ ? = 753 × 4 = 3012 (Ans.)

E.g. 1/6 of 355 of 1/5 of 2160 + √3969 - 448.98 = ?

Step-1: As there are no brackets, so we first solve of

1/5 of 2160 ⇒ 1/5 × 2160 ⇒ 432

And 1/6 of 355 ⇒ 355/6 So, we got,

⇒ 355/6 of 432 + 37 – 448.98

⇒ 355/6 of 432 + 37 – 448.98

⇒ 355/6 × 432 + 37 – 448.98

Step-2: Division

⇒ 355 × 72 + 37 – 448.98

Step-3: Multiplication

⇒ 355 × 72 + 37 – 448.98 ⇒ 25560 + 37 – 448.98

Step-4: Addition

⇒ 25560 + 37 – 448.98

⇒ 25597 – 448.98

Step-5: Subtraction

⇒ 25597 – 448.98 = 25,148.02 (Ans.)

E.g. 28/9 ×144/12 ÷ 17/5 + 13/17

Step-1: By BODMAS rule we would firstly (D) Divide

⇒ 28/9 × 12 ÷ 17/5 + 13/17

⇒ 28/9 × 60/17 + 13/17

Step-2: Multiplication

⇒ 28/9 × 60/17 + 13/17

⇒ 560/51 + 13/17

Step-3 Addition

⇒ (560 + 39)/51 = 599/51 (Ans.)

So here it concludes our blog. Hope you have a learnt a lot and the next thing to excel in it, is to pratice more and more. So practice and all the best!

Learn the key concepts of Simplification

Simplification - Learn and Practice Key concepts

सरलीकरण - मुख्य अवधारणाओं को सीखें और अभ्यास करें

Data Interpretation (Bar Graph) questions are a type of questions which are frequently asked in competitive exams. These questions carry a weightage of 2-3 questions(3-6 marks) in SSC exams and 5-10 questions(5-10 marks) in bank exams. To perform well in competitive exams, you should have a good practice of data interpretation questions.

Here are some tips for solving Data Interpretation (Bar Graph) questions: Read the Graph Carefully, Start by understanding the directions, labels, and units. Decide which questions to tackle first. Usually, start with straightforward ones. Clearly write down your calculations, so you can retrace your steps if needed. Practice Regularly to maintain speed and accuracy

So, we have attached 10 questions of Data Interpretation (Bar Graph) for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Data Interpretation (Bar Graph)</h2>

You can also download the Data Interpretation (Bar Graph) questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Directions (1 - 5):- The following chart shows the overall transaction (in thousands) deposit and withdrawal of cash in the month of April, May, June, and July at a small town bank. Examine the following chart carefully and answer the given questions.

 <img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-388592,.png" style="height:132px; width:280px" />

Q:1 The month which has the highest deposit and the month which has the lowest deposit is what percentage of the total transaction in the month of April? 

1. 74.44

2. 95.02

3. 93.33

4. 86.45

(Difficulty: 2, Estimated Time: 15 Seconds) This question was a matter of seconds....

Q:2 Total transaction in May and June is what percentage of the total transaction in July and April?

1. 75.54

2. 87.77

3. 60.47

4. 83.87

(Difficulty: 2, Estimated Time: 15 Seconds) Another easy one....

Q:3 The total withdrawal in all months is what percentage of the total deposit in all months?

1. 60.45

2. 72.72

3. 75.57

4. 88.21

(Difficulty: 2, Estimated Time: 15 Seconds) This was a cake walk....

Q:4 The sum of the average of deposits in months (April, May, and June) and withdrawals in months (May, June, and July)? 

1. Rs.45500

2. Rs.85000

3. Rs.65000

4. Rs.50500

(Difficulty: 3, Estimated Time: 20 Seconds) This is a common type asked in exams..

Q:5 The difference between total deposits and total withdrawals is what percentage of total deposits?

1. 36.63

2. 25.52

3. 45.54

4. 27.72

(Difficulty: 2, Estimated Time: 15 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Directions (6 - 10): The following graph shows the data of living and extinct percentages of three birds(swan, robin, and hawk). The total number of birds of each type is 4500. Examine the graph carefully and answer the given questions.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Nishchay_05_06_23_03.png" style="height:159px; width:280px" />

Q:6 The number of Swans extinct is what percentage of the number of Swans living?

1. 25

2. 35

3. 40

4. 46.66

(Difficulty: 3, Estimated Time: 25 Seconds) This was a bit calculative....

Q:7 Which bird has the most living number and how much more than bird Robin?

1. Hawk,900

2. Hawk,450

3. Hawk,1125

4. Robin,900

(Difficulty: 3, Estimated Time: 20 Seconds) This was a similar one...

Q:8 What will be the average of all living birds?

1. 1050

2. 1005

3. 1500

4. 1750

(Difficulty: 3, Estimated Time: 20 Seconds) Just basic concepts.....

Q:9 What will be the percentage of the average of all extinct birds to all living birds?

1. 25

2. 35

3. 45

4. 40

(Difficulty: 3, Estimated Time: 20 Seconds) You might have had a good score today....

Q:10 The number of swans extinct is what percentage of the total number of extinct birds Robin and Hawk together?

1. 16.52

2. 14.28

3. 12.48

4. 14.44

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (3)</td>
			<td>2. (4)</td>
			<td>3. (2)</td>
			<td>4. (3)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (1)</td>
			<td>7. (2)</td>
			<td>8. (3)</td>
			<td>9. (4)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is Option 3 i.e. 93.33.

The highest deposit in the month of July = Rs.60000

The lowest deposit in the month of June = Rs.10000

Total = (60000 + 10000) = Rs.70000

Total transactions in the month of April = (45000 + 30000) = Rs.75000

Percentage = (70000/75000) × 100 = 93.33%

 

Q:2 The correct answer is option 4 i,e. 83.87.

Total transaction in may and june = (50000 + 30000 + 10000 + 40000) = Rs.130000

Total transaction in the month of July and april = (60000 + 20000 + 45000 + 30000) = Rs.155000

Percentage = (130000/155000) × 100 = 83.87%  

 

Q:3 The correct answer is option 2 i.e. 72.72.

Total withdrawal = (30000 + 30000 + 40000 + 20000) = Rs.120000

Total deposit = (45000 + 50000 + 10000 + 60000) = Rs.165000

Percentage = (120000/165000) × 100

⇒ (120/165) × 100 = 72.72%

 

Q:4 The correct answer is Option 3 i,e. Rs.65000.

Total deposit in April, May, June = (45000 + 50000 + 10000) = Rs.105000

Average = 105000/3 = 35000

Total withdrawal in May,June and July = (30000 + 40000 + 20000) = Rs.90000 

Average = 90000/3 = Rs.30000

Sum = (35000 + 30000) = Rs.65000

 

Q:5 The correct answer is Option 4 i.e. 27.72.

Total deposits = (45000 + 50000 + 10000 + 60000) = Rs.165000

Total Withdrawal = (30000 + 30000 + 40000 + 20000) = Rs.120000

Difference = (165000 - 120000) = Rs.45000

Percentage = (45000/165000) × 100 = 27.27%

 

Q:6 The correct answer is option 1 i.e. 25.

The number of Swans Extinct = (5/100) × 4500 = 225

The number of Swans living = (20/100) × 4500 = 900

Percentage of Swans extinct to Swans living 

⇒ (225/900) × 100 

⇒ 25%

 

Q:7 The correct answer is option 2 i.e Hawk,450.

Living number of Swan = (20/100) × 4500 = 900

Living number of Robin = (35/100) × 4500 = 1575

Living number of Hawk = (45/100) × 4500 = 2025

Hence, Hawk has the most living number i.e. 2025

⇒ (2025 - 1575) = 450 more than bird Robin

 

Q:8 The correct answer is option 3 i.e. 1500.

Number of Swan living = (20/100) × 4500 = 900

Number of Robin living = (35/100) × 4500 = 1575

Number of Hawk living = (45/100)  × 4500 = 2025

Average = Sum of observations/number of observations  

Average = (900 + 1575 + 2025)/3 = 4500/3 = 1500

 

Q:9 The correct answer is option 4 i.e. 40.

Number of extinct of Swan = (5/100) × 4500 = 225

Number of extinct of Robin = (10/100) × 4500 = 450

Number of extinct of Hawk = (25/100) × 4500 = 1125

Average = (225 + 450 + 1125)/3 = 1800/3 = 600

Now,

Number of living of Swan = (20/100) × 4500 = 900

Number of living of Robin = (35/100) × 4500 = 1575

Number of living of Hawk = (45/100) × 4500 = 2025

Average = (900 + 1575 + 2025)/3 = 4500/3 = 1500

The percentage of the average of all extinct birds to all living birds

⇒ (600/1500) × 100 = 40%

 

Q:10 The correct answer is option 2 i.e. 14.28.

No. of swan extinct = (5/100) × 4500 = 225

No. of Robin extinct = (10/100) × 4500 = 450

No. of Hawk extinct = (25/100)  × 4500 = 1125

Total number of extinct birds Robin and Swan together = (450 + 1125) = 1575

Percentage = (225/1575) × 100 = 14.28%

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Data Interpretation (Bar Graph).

Practice Questions on Data Interpretation (Bar Graph)!

Data Interpretation (Bar Graph) Questions - Download PDF now!

डेटा व्याख्या (दंड आलेख) प्रश्न - अभी पीडीएफ डाउनलोड करें!

Triangles questions are a major part of geometry asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams. To get a good rank in competitive exams, you should have a conceptual clarity of questions of triangles as they also are part of mixed geometry questions asked.

Here are some tips for solving Triangles questions: Understand triangle types (e.g., equilateral, isosceles, scalene), the Pythagorean theorem for right triangles, Use the Law of Sines and Cosines for non-right triangles, and the area formula (1/2 × base × height).Practice visualization and diagram drawing for clarity.

So, we have attached 10 questions of Triangles for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Triangles</h2>

You can also download the Triangles questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

 Q:1 A scalene triangle ABC is given with sides 15cm, 13cm, and 12cm, and a circle is drawn in the triangle having radius r. Find r.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Nishchay_03_05_23_03.png" style="height:139px; width:150px" />

1. \({\sqrt47}\over3\)

2. \({\sqrt42}\over2\)

3. \({\sqrt15}\over3\)

4. \({\sqrt29}\over2\)

(Difficulty: 3, Estimated Time: 20 Seconds) A good question to start with....

 Q:2 The perimeter of an isosceles right angle triangle is \((16 +18\sqrt2)\) cm. Find the area of the triangle.

1. 34

2. 54

3. 43

4. 32

(Difficulty: 3, Estimated Time: 20 Seconds) You won&#39;t get easy questions in exam...

Q:3 In triangle ABC, if AB = 16 cm. AC = 10 cm and ∠BAC = 60o, then find the length of the side BC.

1. 12 cm

2. 15 cm

3. 14 cm

4. 18 cm

(Difficulty: 3, Estimated Time: 20 Seconds) Do you remember the relation.....

Q:4 A triangle is drawn inside a circle such that the sides of a triangle are given as 9 cm, 40 cm, and 41 cm then, find the radius (r) inside the triangle

1. 5

2. 5.85

3. 4

4. 7.6

(Difficulty: 2, Estimated Time: 15 Seconds) This was quite easy.....

Q:5 A circle is drawn outside a triangle and the sides of the triangle are 12 cm, 35 cm, and 37 cm then, find the radius of the circle.

1. 17.5

2. 18.5

3. 18.7

4. 16.5

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

 Q:6 In an equilateral triangle, ABC having side a, three perpendiculars P1, P2, and P3 are drawn to BC, AB, and AC from the center. Also the sum of lengths of perpendiculars equal to 63 cm. Find the value of a. (in cm.)

1. 54

2. 63

3. \({45\sqrt7}\)

4. \({42\sqrt3}\)

(Difficulty: 4, Estimated Time: 25 Seconds) You should have a good practice to solve such.....

Q:7 In a right-angle triangle, two circles are drawn inside and outside the triangle having radius r and R. The relation between Perpendicular and Hypotenuse is given as (P - H) = 17 and (P + H) = 27 and Base = 3. Find the value of (r + 10R).

1. 35

2. 45

3. 25

4. 39

(Difficulty: 3, Estimated Time: 20 Seconds) Be prepared for such questions in exam!

Q:8 In triangle ABC, AB = 24 cm and AC = 30 cm. In ∠BAC is 60o, then find the area of triangle ABC.

1. 180√3 cm2

2. 160√3 cm2

3. 150√3 cm2

4. None of these

(Difficulty: 2, Estimated Time: 15 Seconds) This was not a hard question!

 Q:9 &#39;K&#39; is a point in the interior of an equilateral triangle. The perpendicular distance from &#39;K&#39; to the sides are \(\sqrt6\)cm, 4\(\sqrt6\)cm. and 6\(\sqrt6\)cm. Then find the ratio between the Perimeter and area of the triangle.

1. \(\sqrt6\) : 12

2. \(\sqrt6\) : 10

3. \(\sqrt6\) : 11

4.  \(\sqrt6\) : 19

(Difficulty: 3, Estimated Time: 20 Seconds) A similar one....

Q:10 If the lengths of two sides of a triangle are x and y such that their product is 30 where x and y are integers, then find the number of such possible triangles.

1. 10

2. 9

3. 8

4. 7

(Difficulty: 3, Estimated Time: 20 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (2)</td>
			<td>2. (4)</td>
			<td>3. (3)</td>
			<td>4. (3)</td>
			<td>5. (2)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (1)</td>
			<td>8. (1)</td>
			<td>9. (3)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is Option 2 i.e. \({\sqrt42}\over2\)

We know that, 

r = area/s  (s = semi-perimeter)

s = (a + b + c)/2

Here, a, b, and c are sides of the triangle

s = (15 + 13 + 12)/2 = 40/2 = 20

Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Area = \(\sqrt{20(20-15)(20-13)(20-12)}\)

Area = \(\sqrt{20\times 5 \times 7 \times 6 }\)

Area = \(10\sqrt42\)

Hence,

r = \({10\sqrt42}\over20\) = \({\sqrt42}\over2\)

 

Q:2 The correct answer is option 4 i.e. 32.

We know that the perimeter of an isosceles right angle triangle is = \((2a +\sqrt2a)\) or \({\sqrt2}a(\sqrt2 + 1)\)

Comparing with the given value of the perimeter 

⇒ \((16 +18\sqrt2)\) = \((2a +\sqrt2a)\)

⇒ (8 × 2 + \(18\sqrt2\)) = \((2a +\sqrt2a)\)

So, a = 8 cm

Area = 1/2 a2

Area = 1/2 × 8 × 8 = 32 cm2

 

Q:3 The correct answer is option 3 i.e. 14 cm

Using cosine rule,

We have, BC2 = AB2 + AC2 – 2 × AB × AC × cos60o

Or, BC2 = 162 + 102 – 2 ×16 × 10 × (1/2)

BC2 = 256 + 100 – 160

BC = √196

BC = 14 cm {Since, length cannot be negative}

 

Q:4 The correct answer is option 3 i.e. 4.

The sides of a triangle are given as 9, 40, and 41

So, it is a right-angle triangle

H = 41, B = 9, and P = 40

Formulae for the radius of the circle inside the triangle 

⇒ r = (P + B - H)/2

⇒ r = (40 + 9 - 41)/2 

⇒ r = (49 - 41)/2 = 8/2 = 4

 

Q:5 The correct answer is option 2 i.e. 18.5.

The sides of a triangle are 12,35 and 37 and its forms a triplet 

Hence, it is right angle triangle

⇒ H = 37, B = 35 and P = 12

The radius of the circle outside the triangle 

⇒ R = H/2

⇒ R = 37/2 = 18.5

 

Q:6 The correct answer is option 4 i.e. \({42\sqrt3}\)

We know that, in an equilateral triangle

⇒ a = \({2\over \sqrt3} \)(P1 + P2 + P3)

⇒ a = \({2\over \sqrt3}\times 63\)

⇒ a = \({2\over \sqrt3}\times 63 \times {\sqrt3\over \sqrt3}\)

⇒ a = \({{2\times 63\sqrt3}\over3}\)

⇒ a = \({2\times21\sqrt3}\)

⇒ a = \({42\sqrt3}\) cm

 

Q:7 The correct answer is option 1 i.e. 35.

Here, P = perpendicular, H = Hypotenuse and B = base

We know that

⇒ r = (P + B - H)/2

⇒ r = (17 + 3)/2 = 10

Given,

⇒ (P - H) = 17.....(1) and,

⇒ (P + H) = 27.....(2)

Solving equations (1) and (2)

⇒ 2P = 44

⇒ P = 22 so H = 5

⇒ R = H/2 = 5/2

Now, ⇒ (r + 10R) = (10 + 10 × 5/2)

⇒ (10 + 25) = 35

 

Q:8 The correct answer is option 1 i.e. 180√3 cm2.

Area of triangle = (1/2) × sin θ × product of adjacent sides

Area of triangle = (1/2) × sin 60o × 24 × 30

⇒ (1/2) × (√3/2) × 24 × 30 = 180√3 cm2

 

Q:9 The correct answer is option 3 i.e. \(\sqrt6\) : 11.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/HRQ-406363,.png" />

We know that 

Height of the equilateral triangle = h1 + h2 + h3

⇒ H =   \(\sqrt6\) cm + 4\(\sqrt6\) cm + 6\(\sqrt6\) cm = 11\(\sqrt6\) cm

⇒ H = \(\sqrt3\over2\)a

⇒  11\(\sqrt6\)  =   \(\sqrt3\over2\)a

⇒ a = 22\(\sqrt2\) cm

So, the required ratio = \(3\times a\over\frac{\sqrt3a^2}{4}\) = \(\frac{12}{\sqrt3\times22\sqrt2}\)= \(\sqrt6\)/11

 

Q:10 The correct answer is option 2 i.e. 9.

The possible sides of the triangle:

<table border="1" cellpadding="1" cellspacing="1" style="width:300px">
	<tbody>
		<tr>
			<td>x</td>
			<td>y</td>
		</tr>
		<tr>
			<td>1</td>
			<td>30</td>
		</tr>
		<tr>
			<td>2</td>
			<td>15</td>
		</tr>
		<tr>
			<td>3</td>
			<td>10</td>
		</tr>
	</tbody>
</table>

 
The possible number of triangle:

Let&#39;s assume the third side is z.

|x - y| < z< |x + y|

29 < z < 31 = 1 triangle

Here, the possible value of z is 30.

13 < z < 17 = 3 triangles

Here, the possible value of z is 14, 15, 16.

7 < z < 13 = 5 triangles

Here, the possible value of z is 8, 9, 10, 11, 12.

So, the total number of triangles = 1 + 3 + 5 = 9

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Triangles.

Triangles Questions PDF with detailed solutions

त्रिभुज के प्रश्नों की पीडीएफ विस्तृत उत्तरों के साथ

Hello friends, we usually encounter problems on surds and indices in competitive exams. So lets study and clear our doubts. Lets learn some basic properties of indices:

<ul>
	<li>am = a × a × a × a ×…..(m times)</li>
	<li>a0 = 1 </li>
	<li>am × an = am+n</li>
	<li>a m/ an = am-n</li>
	<li>(am) n = am × n</li>
	<li>a1/m = m√a</li>
	<li>an/m = m√a n</li>
</ul>

Note: If the power of any number is negative, the reciprocal of the same number makes the power positive.

Let us see some cases:

(a)-m = 1/(a)m

1/(a)-m = (a)m

(a)-m/(b)-n = (b)n/(a)m

(a/b)-m = (b/a)m

Let’s solve some problems on these properties.

E.g. Find the value of y, if 37 × 82 = 92 × 42 × 12y

Sol: We know that 9 = 32 , 4 = 22 and 8 = 23 then,

3 7 × (23 ) 2 = (32 ) 2 × (22 ) 2 × 12y

3 7 × 26 = 34 × 24 × 12y

3 3 × 22 = 12y

27 × 4/12 = y

y = 9 (Ans.)

E.g. Find the value of y, if

(27/64)2 × (3/4)2 × (9/16)2 × y + 1 = (81/256)3 × y

Sol: Now, (27/64)2 × (3/4)2 × (9/16)2 × y + 1 = (81/256)3 × y

(3/4)3 × 2 × (3/4)2 × (3/4)2 × (2 × y + 1) = (3/4)4 × 3 × y

(3/4)6 + 2 + 2 + 4 × y = (3/4)12 × y

(3/4)10 + 4 × y = (3/4)12 × y

10 + 4 × y = 12 × y

8 × y = 10 

y = 10/8 = 1.25 (Ans.)

Now, let’s learn some vividly used powers of numbers.

<h2>1. Squares</h2>

Squares refer to the numbers which are obtained when a number is multiplied by itself or raised to the power 2 i.e. Square of a = a × a = a2. The table below shows the squares of numbers from 1 to 30.

<table cellspacing="0" class="Table" style="border-collapse:collapse; border:none; width:300px">
	<tbody>
		<tr>
			<td>
			1
			</td>
			<td>
			121
			</td>
			<td>
			441
			</td>
		</tr>
		<tr>
			<td>
			4
			</td>
			<td>
			144
			</td>
			<td>
			484
			</td>
		</tr>
		<tr>
			<td>
			9
			</td>
			<td>
			169
			</td>
			<td>
			529
			</td>
		</tr>
		<tr>
			<td>
			16
			</td>
			<td>
			196
			</td>
			<td>
			576
			</td>
		</tr>
		<tr>
			<td>
			25
			</td>
			<td>
			225
			</td>
			<td>
			625
			</td>
		</tr>
		<tr>
			<td>
			36
			</td>
			<td>
			256
			</td>
			<td>
			676
			</td>
		</tr>
		<tr>
			<td>
			49
			</td>
			<td>
			289
			</td>
			<td>
			729
			</td>
		</tr>
		<tr>
			<td>
			64
			</td>
			<td>
			324
			</td>
			<td>
			784
			</td>
		</tr>
		<tr>
			<td>
			81
			</td>
			<td>
			361
			</td>
			<td>
			841
			</td>
		</tr>
		<tr>
			<td>
			100
			</td>
			<td>
			400
			</td>
			<td>
			900
			</td>
		</tr>
	</tbody>
</table>

<h2>2. Cubes</h2>

Cubes refers to the numbers which are obtained when a number is multiplied by itself thrice or raised to the power 3 i.e. Cube of a = a × a × a = a3

The table given below shows the cubes of numbers from 1 to 20.

<table cellspacing="0" style="width:300px">
	<tbody>
		<tr>
			<td>
			1
			</td>
			<td>
			1331
			</td>
		</tr>
		<tr>
			<td>
			8
			</td>
			<td>
			1728
			</td>
		</tr>
		<tr>
			<td>
			27
			</td>
			<td>
			2197
			</td>
		</tr>
		<tr>
			<td>
			64
			</td>
			<td>
			2744
			</td>
		</tr>
		<tr>
			<td>
			125
			</td>
			<td>
			3375
			</td>
		</tr>
		<tr>
			<td>
			216
			</td>
			<td>
			4096
			</td>
		</tr>
		<tr>
			<td>
			343
			</td>
			<td>
			4913
			</td>
		</tr>
		<tr>
			<td>
			512
			</td>
			<td>
			5832
			</td>
		</tr>
		<tr>
			<td>
			729
			</td>
			<td>
			6859
			</td>
		</tr>
		<tr>
			<td>
			1000
			</td>
			<td>
			8000
			</td>
		</tr>
	</tbody>
</table>

 

Hope this blog clear your concept. For more such blogs, stay connected with us. 

Thank you

&nbsp;Learn the&nbsp;key concepts of surds and indies

Surds and indices- Quant Study Notes for Competitive Exams

सूचकांक और सूचकांक- प्रमुख अवधारणाओं को सीखें और अभ्यास करें

Hello our dear learners, today we are going to learn about approximation. Many of you might have doubts, so let&#39;s clear all your problems.

We use approximation in our daily life to ease down our day-to-day work. In mathematical expressions which include division and multiplication of decimal values of large numbers. It becomes quite complex to solve these problems, so to solve these we use approximation.

When we approximate the final result obtained is not equal to the exact result, but it is very close to the exact result.

Let’s try one.

E.g. ? = 32.01 + 128.01 × 1023.99 + 7.99

Sol: Solving by BODMAS rule, as learned earlier

? = 32.01 + 131,080.9599 + 7.99

? = 32.01 + 131,080.9599 + 7.99 = 131,120.9599

E.g. 10% of 1350 + ? = 365 

10/100 × 1350 + ? = 365

135 + ? = 365

? = 365 – 135

? = 230

E.g. 78 × 98 – 25% of 1376

78 × 98 – 25/100 × 1376

7644 – 34400/100 = 7644 – 344

7300 (it requires tedious calculation)

After this tedious calculation (128.01 × 1023.99) without a calculator, we are not left with time, in the exam hall we will prefer to leave this question.

In Approximation, to solve the complex mathematical expression, take the nearest value of numbers given in the expression. Try to make unit digit 0’ in most cases.

E.g. ? = 32.01 + 128.01 × 1023.99 + 7.99

Let’s try to round off to nearest integer

? = 32 + 128 × 1024 + 8

Or we can say,

? = 25 + 27 × 210 + 23

? = 25 + 217 + 23

? = 23 (22 + 214 + 1)

? = 8 (16389) = 131112

By calculator we would get 131112.09

Here it is, we got the solution.

E.g. 393 × 197 + 5600 × 5/4 + 8211.80 = ?

393 × 197 + 5600 × 5/4 + 8211.80 = ?

390 × 200 + 5600 × 5/4 + 8200 = ?

390 × 200 + 5600 × 5/4 + 8200 = ?

(By BODMAS rule)

390 × 200 + 7000 + 8200 = ?

78000 + 7000 + 8200 = 93200 (Ans.)

We got 92,632.8 by calculator. It is quite near its original value.

Tip: For finding 10% of a number simply move the decimal to one digit left.

For finding 25% simply divides the number by 4.

<h2>Short- cut method for percentage</h2>

<table cellspacing="0" class="Table" style="border-collapse:collapse; border:none; margin-left:27px; width:300px">
	<tbody>
		<tr>
			<td>
			Sr.

			No.
			</td>
			<td>
			Required Percentage

			(%)
			</td>
			<td>
			Number

			divided by
			</td>
		</tr>
		<tr>
			<td>
			1.
			</td>
			<td>
			200
			</td>
			<td>
			½
			</td>
		</tr>
		<tr>
			<td>
			2.
			</td>
			<td>
			50
			</td>
			<td>
			2
			</td>
		</tr>
		<tr>
			<td>
			3.
			</td>
			<td>
			33.33
			</td>
			<td>
			3
			</td>
		</tr>
		<tr>
			<td>
			4.
			</td>
			<td>
			25
			</td>
			<td>
			4
			</td>
		</tr>
		<tr>
			<td>
			5.
			</td>
			<td>
			20
			</td>
			<td>
			5
			</td>
		</tr>
		<tr>
			<td>
			6.
			</td>
			<td>
			10
			</td>
			<td>
			10
			</td>
		</tr>
		<tr>
			<td>
			7.
			</td>
			<td>
			5
			</td>
			<td>
			20
			</td>
		</tr>
		<tr>
			<td>
			8.
			</td>
			<td>
			1
			</td>
			<td>
			100
			</td>
		</tr>
	</tbody>
</table>

E.g. 10% of 1350 + ? = 365

? = 365 – 10% of 1350

? = 365 – 135.0 = 230.0 (Ans).

E.g. 78 × 98 – 25% of 1376

78 × 98 – 344 (Using approximation and table)

8000 – 344 = 7656 (Ans.)

This can be done in mind without the use of pen and paper.

E.g. 34.02% of 550.09 + ? = 297.07 + √728.95

34.02% of 550.09 + ? = 297.07 + √728.95

34% of 550 + ? = 300 + √729

(25 + 10 – 1)% of 550 + ? = 300 + 27

25% of 550 + 10% of 550 – 1% of 550 + ? = 327

137.5 + 55 – 5.5 + ? = 327

132 + ? = 327

? = 327 – 132 = 195 (Ans.)

E.g. (? + 9.97) × 12.8 = 20.12% of 1319.97

(? + 10.00) × 13.0 = 20.00% of 1320.00

(Using approximation)

(? + 10.00) × 13.0 = 1/5 × 1320.00

(? + 10.00) × 13.0 = 264 

(? + 10.00) = 260/13

(Using approximation) 

? = 20 – 10 = 10 (Ans.)

Hope your doubts are clear. Practice some more questions for better understanding of this topic and stay connected.

Learn the key concepts of Approximation

Approximation- Quant Study Notes for Competitive Exams

सन्निकटन - प्रमुख अवधारणाओं को सीखें और अभ्यास करें

We all have studied number system since our childhood. So, today we are going to discuss about basic classification of numbers. Let&#39;s get started.

Numbers: The basic pillar of Mathematics, the language of the universe. The thing which we learnt in our first mathematics class. The thing we deal with, in our day-to-day life. The thing which is almost everywhere.

Real Numbers: These are the numbers whose squares are positive numbers.

Imaginary Numbers: These are the numbers whose squares are negative numbers. They are represented by iota ‘i’. (i = √(-1), i2 = -1, i3 = -i, i4 = 1) E.g. If the number 6i is given, it means it’s value is 6 × √(-1) = 6√(-1) = √(-36)

Rational Numbers: The numbers which can be written in the form p/q where p and q are co-prime integers and q not equal to 0. i.e. In decimal form, they are either terminating or recurring numbers. Let’s find out how we can convert a rational number from decimal form to fraction form.

<h2>Classification</h2>

Numbers can be classified into two groups, i.e. Real and imaginary numbers. Real numbers can be represented on number line. Real numbers can further be classified into two groups, i.e. rational and irrational numbers. Rational numbers can also be classified as terminating and non-terminating. And so on...

E.g. Express 0.82828282…. in the form of a fraction.

Sol: Let x = 0.82828282… _____(1)

As the period contains 2 digits, we multiply by 102 = 100,

100x = 82.82828282…. ______(2)

Now, (2) – (1), we get, 99x = 82 ⇒ x = 82/99.

E.g. Express 0.024024024024024024….. in the form of a fraction.

Sol: Let x = 0.024024024024…. ____(1)

As the period is contains 3 digits, we multiply with 103 = 1000

1000x = 24.02404024…. ____(2)

Now, (2) – (1), we get, 999x = 24 ⇒ x = 24/999 = 8/333.

Irrational Numbers: The numbers which cannot be written as p/q. i.e. In decimal form, these numbers are non-terminating and non-recurring numbers. e.g. √2.

Integers: The numbers whose value after decimal is 0.

Decimals: The numbers whose value after decimal is non zero. e.g. 3.456

Negatives: The numbers whose value is less than 0.

Positives: The numbers whose value is more than 0.

Natural Numbers:- The integers more than 0.

Whole Numbers: Whole numbers are basically natural numbers, including 0. These are also called non-negative integers.

Prime Numbers: The number which has exactly two factors (1 and the number itself).

Composite Numbers: The number which has more than two factors. [Note: 1 is neither Prime nor Composite number.]

<h2>Prime numbers</h2>

As we have discussed earlier, these are the numbers which have exactly two factors. These two factors are 1 and the number itself. From 1 to 50, we have 15 prime factors. From 1 to 100, we have 25 prime factors. And these are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Prime numbers have some properties like:

<ol>
	<li>No integer can divide them completely other than 1 and the number itself (that’s why they have only two factors).</li>
	<li>Every prime number should be in the form of [4n + 1 or 4n – 1] for less than 5 and [6n + 1 or 6n – 1] for more than or equal to 5. But the reverse of this statement is not true. e.g. 25 can be written as 6(4) + 1 but it is not a prime number.</li>
	<li>Every integer, greater than 1, can be written as the product of prime numbers.</li>
</ol>

Since, there is no formula to find out whether a number is prime or not. But for small numbers, we can use this trick. Take the square root of the number (approximately) and check its divisibility by the prime numbers smaller than its square root.

1) E.g. Check whether 97 is a prime number or not. Sol: Square root of 97 will be between 9 and 10. So the prime numbers smaller than 9 are 2, 3, 5 and 7. Since 97 is not divisible by any of these, hence it is a prime number. 2) E.g. Check whether 91 is a prime number or not. Sol: Square root of 91 will be between 9 and 10. So the prime numbers smaller than 9 are 2, 3, 5 and 7. Since 91 is divided by 7 (13 × 7 = 91). 91 is not a prime number.

Let me make this thing very clear that this is the most important topic of the quant section. Not only this one, but the entire Number System. Hope this blog helped you to learn well. Stay tuned for more.

Learn the key concepts of&nbsp;classification of numbers

Classification of numbers - Quant Study Notes for Competitive Exams

संख्याओं का बुनियादी वर्गीकरण - प्रमुख अवधारणाओं को सीखें और अभ्यास करें

Hello our dear readers, presently we will be talking about divisibility rules. Many of you have doubts related to this topic. So, let&#39;s discuss.

<h2>Divisible rule</h2>

Divisibility of 2: A number is always divisible by 2 if it is an even number i.e. its last digit is 0, 2, 4, 6 or 8.

Divisibility of 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

Divisibility of 4: A number is divisible by 4 if the last two digits of the number are divisible by 4.

Divisibility of 5: A number is divisible by 5 if its last digit is 0 or 5.

Divisibility of 6: A number is divisible by 6 if it is divisible by both 2 and 3.

Divisibility of 8: A number is divisible by 8 if its last 3 digits are divisible by 8.

Divisibility of 10: A number is divisible by 10 if its last digit is 0.

Divisibility of 12: A number is divisible by 12 if it is divisible by both 3 and 4. Now, we will discuss the divisibility rule of 7, 9, 11 and 13.

Divisibility Rule of 9 and 11:

Divisibility Rule of 9: A number is divisible by 9 if the sum of the digits of the number is multiple of 9. E.g. 356211 has sum of the digits, 3 + 5 + 6 + 2 + 1 + 1 = 18 and 18 had sum of the digits, 8 + 1 = 9, so 356211 is divisible by 9.

Divisibility Rule of 11: If the difference between the sum of the digits at odd places (from right) and the sum of the digits at even place (from right) is 0 or a multiple of 11, then it is divisible by 11. E.g. 356213 has the sum of the digits at odd places 3 + 2 + 5 = 10 and the sum of the digits at even places, 1 + 6 + 3 = 10. So 10 – 10 = 0. Hence it is divisible by 11.

The most interesting fact is, the divisibility rule of 9 and 11 is also applicable for 99, 999, 101, 1001, etc. but in a slightly different way. Let’s understand this thing by taking an example of a number ‘abcdefg’ . This number is divisible by 9 if: (a + b + c + d + e + f + g) is 9 at the end (or a multiple of 9). This number is divisible by 99 if: (a + bc + de + fg) is divisible by 99. [Pairing from right side] This number is divisible by 999 if: (a + bcd + efg) is divisible by 999. [Tripling from right side] And so on....

Also, This number is divisible by 11 if: (g + e + c + a) – (d + f + b) is 0 or 11. This number is divisible by 101 if: (fg + bc) – (de + a) is 0 or a multiple of 101 [Pairing from right side] . This number is divisible by 1001 if: (efg + a) – (bcd) is 0 or a multiple of 1001 [Tripling from right side] And so on…

Let us take some examples to understand it clearly. E.g. Check whether 4636665 is divisible by 9, 99 and 999 or not. Sol: Divisibility by 9: 4 + 6 + 3 + 6 + 6 + 6 + 5 = 36 and 3 + 6 = 9 So, yes it is divisible by 9.

Divisibility of 99: (Pairing from right side) 4 + 63 + 66 + 65 = 198 So, yes it is divisible by 99 [Note: Since the number is divisible by 99. Hence, it is also divisible by all the factors of 99. And so, this number is divisible by 9]

Divisibility of 999: (Tripling from right side) 4 + 636 + 665 = 1305 So, no it is not divisible by 999. E.g. Check whether 8064143 is divisible by 11, 101, 1001. Sol: Divisibility by 11: (3 + 1 + 6 + 8) – (4 + 4 + 0) = 18 – 8 = 10 So, it is not divisible by 11.

Divisibility of 101: (Pairing from right side) (43 + 06) – (41 + 8) = 49 – 49 = 0 So, it is divisible by 101.

Divisibility of 1001: (Tripling from right side) (143 + 8) – (064) = 151 – 64 = 87 So, it is not divisible by 1001.

Divisibility of 7 and 13: We have a rule for 7 that if the difference between the number of tens in the number and twice the unit digit is divisible by 7. But it has some limitations. It is good for some small numbers like 651, 994, 1078 etc. but it is a very time consuming process for numbers like 456745. So, we can do some additional stuff here. We know how to find whether a number is divisible by 1001 or not. And, 1001 = 7 × 11 × 13 So, if a number is divisible by 1001 it will also be divisible by 7 and 13.

Divisibility rules help you in solving Number system and simplification questions in a quick way. Excel them to get good marks in exams.

Learn the key concepts of Divisibility&nbsp;rules!

Divisiblilty Rules -- Quant Study Notes for Competitive Exams

विभाज्यता नियम - प्रमुख अवधारणाओं को सीखें और अभ्यास करें

Practice Questions on Quadratic Equations!

द्विघात समीकरण पर अभ्यास प्रश्न!

Quadratic Equations PDF with detailed solutions

द्विघात समीकरण की पीडीएफ विस्तृत उत्तरों के साथ

Average questions are a major type of questions asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams and 1-2 questions in bank exams. To get a good rank in competitive exams, you should have a good hold on the concepts of Average.

Here are some tips for solving Average questions: Identify the key information in the question, Use the appropriate formula as there are many different formulas that can be used to calculate the average. Practice solving a variety of average questions, Use shortcuts and formulas. 

So, we have attached 10 questions of Average for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Average</h2>

You can also download the Average questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 In a company, the average salary of male employees is ₹x, and that of female employees is ₹(x + 8000). The ratio of male to female employees in the company is 9 : 7. If the average salary of all the employees in the company is ₹27500, then find the average of total male employees and total female employees.

1. 40

2. 32

3. 44

4. Can’t be determined

(Difficulty: 4, Estimated Time: 25 Seconds) This was a tricky question!

Q:2 MP’s from 5 different parties A, B, C, D and E are present in the Parliament. The average number of MP’s from parties A and C is 100 and that from parties B and E is 90. The number of MP’s from party B is 50 more than that of party C. The total number of MP’s from parties A and E is 230. The number of MP’s of party D is 20 more than that of party B. Find the total number of MP’s from parties A, E and D.

1. 280

2. 600

3. 350

4. 440

(Difficulty: 4, Estimated Time: 25 Seconds) It is not an easy one but I think now you&#39;re prepared for it. Did you guess it right?

Q:3 The average number of people living in each house in a society having 47 houses is 4. If this average decreased by 0.2 when 2 families left the society then, what is the total number of people in 2 families which left?

1. 19

2. 14

3. 17

4. 15

(Difficulty: 3, Estimated Time: 20 Seconds) Another calculative one, but you will have to solve such questions within time!

Q:4 In a class, the number of boys is 10 more than that of girls. The average weight of a boy and a girl is 48 kg and 32 kg, respectively and the average weight of each student in the class is 41 kg. Find the total number of students (boys + girls) in the class.

1. 65

2. 80

3. 55

4. 34

(Difficulty: 2, Estimated Time: 15 Seconds) This was quite easy!

Q:5 The average number of boys in classes &#39;A&#39; and &#39;B&#39; together is 46 and class &#39;A&#39; has 8 less boys than class &#39;B&#39;. The average weight of a boy in class &#39;A&#39; is 45 kg and the total weight of all boys in class &#39;B&#39; is 410 kg more that the total weight of all boys in class &#39;A&#39;. Find the average weight of a boy in class &#39;B&#39;.

1. 40 kg

2. 52 kg

3. 33 kg

4. 46 kg

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 The present average age of Tanish and Vikesh is 18 years. The present average age of Tanish, Vikesh, and Ashish is 20 years. The difference between the present ages of Tanish and Dinesh is 12 years. Find the present age of Tanish if the present average age of Tanish, Vikesh, Dinesh, and Ashish is 24 years.

1. 24 years

2. 20 years

3. 32 years

4. 40 years

(Difficulty: 4, Estimated Time: 30 Seconds) This was also a tricky one!

Q:7 The average age of a group of 60 students is 15 years. If 3 students whose ages are 17 years, 24 years and _____ years is replaced by 3 new students whose ages are ____ years, 22 years and 31 years, then the new average age of the group increases by 0.5 years.

The values given in which of the following options will fill the blanks in the same order in which is it given to make the above statement true:

A. 27, 45

B. 44, 64

C. 19, 37

1. Only A

2. Only B

3. Only A and C

4. All A, B and C

(Difficulty: 4, Estimated Time: 25 Seconds) This was a new type of question, but I think now you&#39;re prepared for it. 

Q:8 The average age of a family having five members is 42 years and the age of the oldest member of the family is 76 years. In the same year, a member whose age is 32 years has married a girl whose age is 30 years and after marriage, the average age of the family (including the bride) is reduced by 2 years. If the ratio of the age of the other three members (other than the oldest member, the member who got married and his bride) is 5 : 14 : 15, respectively then find the difference between the age of the oldest and the age of the youngest member.

1. 56 years

2. 61 years

3. 38 years

4. 65 years

(Difficulty: 4, Estimated Time: 25 Seconds) Tough questions alert!

Q:9 In a class, there are ‘x’ students and the average weight of the students is ‘y’ kg. If one new student whose weight is 50 kg is added, then the average weight of the class is increased by 1 kg. If one more student whose weight is 50 kg is added, then the average weight of the class increases by 1.5 kg over the original average. Which of the following can be found according to the given information?

(i) If Anant joins the class, the average weight of the class increases by 0.5 kg, find the weight of Anant.

(ii) The number of students in the class.

(iii) Find the value of (y – 3x)

1. Only (i)

2. Only (ii)

3. Only (iii)

4. All (i), (ii) and (iii)

(Difficulty: 3, Estimated Time: 20 Seconds) Practice and get fast with your calculations!

Q:10 A cricketer has played &#39;a&#39; matches and scored average of &#39;b&#39; runs. If the cricketer plays &#39;c&#39; more matches and scores average of &#39;d&#39; runs in those matches, what will be the cricketer&#39;s average runs per match?

1. \(\frac{b+a d}{d+a}\)

2. \(\frac{a b+a d}{a+c}\)

3. \(\frac{a +b d}{a+c}\)

4. None of these

(Difficulty: 2, Estimated Time: 15 Seconds) This was an easy one. Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (4)</td>
			<td>2. (3)</td>
			<td>3. (3)</td>
			<td>4. (2)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (1)</td>
			<td>7. (3)</td>
			<td>8. (2)</td>
			<td>9. (4)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 4 i.e. Can’t be determined.

Let the number of male employees and female employees in the company be ‘9y’ and ‘7y’ respectively.

So,

⇒ (9yx + 7y(x + 8000))/(9y + 7y) = 27500

⇒ 9x + 7x + 56000 = 440000

⇒ x = 24000

Since we can’t find the value of ‘y’.

So, we can’t determine the required average

 

Q:2 The correct answer is option 3 i.e. 350

Let there are a, b, c, d, and e number of MP’s are there from A, B, C, D and E respectively.

According to the question,

(a + c)/2 = 100

Or, a + c = 200.... (1)

(b + e)/2 = 90

Or, b + e = 180.... (2)

Adding equation (1) and (2)

a + c + b + e = 380..... (3)

b – c = 50..... (4)

Adding equation (3) and (4), we get

a + 2b + e = 430

Or, 2b = 430 – 230 (Since, a + e = 230)

Or, 2b = 200

Or, b = 100

c = 100 – 50 = 50

a = 200 – 50 = 150

e = 180 – 100 = 80

d = b + 20 = 120

Total number of MP’s from A, E and D = 150 + 80 + 120 = 350

 

Q:3 The correct answer is option 3 i.e. 17.

The average number of people living in each house in a society having 47 houses is 4

So, Total number of people living in 47 houses = 47 × 4 = 188

The average decreased by 0.2 when 2 families left the society

So, Total number of people in 45 houses now = 45 × (4 – 0.2) = 171

Hence, Total number of people in 2 families which left = 188 – 171 = 17

 

Q:4 The correct answer is option 2 i.e. 80

Let the number of girls in the class = ‘x’

Then, number of boys in the class = (x + 10)

Total weight of all boys in the class = (x + 10) × 48 = (48x + 480) kg

Total weight of all girls in the class = x × 32 = 32x kg

So, total weight of all the students in the class = (48x + 32x + 480) = (80x + 480) kg

Also, total weight of all the students in the class = (x + x + 10) × 41 = (82x + 410) kg

So, 80x + 480 = 82x + 410

Or, x = (70/2) = 35

So, total number of students in the class = x + x + 10 = 35 + 35 + 10 = 80

 

Q:5 The correct answer is option 4 i.e. 46 kg.

The sum of the number of boys in classes &#39;A&#39; and B&#39; = (46 × 2) = 92

Let the number of boys in class &#39;A&#39; = &#39;x&#39;

Then, the number of boys in class &#39;B&#39; = (x + 8)

So, x + x + 8 = 92

⇒ x = (92 - 8) ÷ 2 = 42

Therefore, the number of boys in classes &#39;A&#39; and &#39;B&#39; is 42 and 50  respectively

Total weight of all boys in class &#39;A&#39; = (42 × 45) = 1890 kg

Total weight of all boys in class &#39;B&#39; = (1890 + 410) = 2300 kg

So, the average weight of a boy in class &#39;B&#39; = 2300 ÷ 50 = 46 kg

 

Q:6 The correct answer is option 1 i.e. 24 years.

The sum of the present ages of Tanish, Vikesh, Dinesh, and Ashish

⇒ 24 × 4 = 96 years

The sum of the present ages of Tanish and Vikesh = (18 × 2) = 36 years

The sum of the present age of Tanish, Vikesh, and Ashish

⇒ (3 × 20) = 60 years

So, the present age of Dinesh = (96 – 60) = 36 years

The present age of Tanish = is either (36 - 12) or (36 + 12) = either 24 years or 48 years

As the sum of the ages of Tanish and Vikesh is 36 years, so the present age of Tanish is less than 36 years.

So, the present age of Tanish is 24 years

 

Q:7 The correct answer is option 3 i.e. Only A and C

Since, the new average increases by 0.5 years.

So, sum of the ages of the added students – sum of the ages of the students who left = 60 × 0.5 = 30 years

Option A:

(45 + 22 + 31) – (17 + 24 + 27) = 30 years

So, option A can be the answer.

Option B:

(64 + 22 + 31) – (17 + 24 + 44) = 32 years

So, option B cannot be the answer.

Option C:

(37 + 22 + 31) – (17 + 24 + 19) = 30 years

So, option C can be the answer.

 

Q:8 The correct answer is option 2 i.e. 61 years

Sum of the age of the members before marriage = 42 × 5 = 210 years

Sum of the age of the members after marriage = 40 × 6 = 240 years

Let, the age of the rest three members be 5x, 14x, and 15x, respectively.

Sum of the age of the rest three members = 240 – (76 + 32 + 30) = 102 years

So, 34x = 102

x = 3

Age of the youngest member = 5 × 3 = 15 years

Required difference = 76 – 15 = 61 years

 

Q:9 The correct answer is option 4 i.e. All (i), (ii), and (iii).

Total weight of the class = x × y = xy kg

According to the question,

⇒ xy + 50 = (x + 1)(y +1)

⇒ xy + 50 = xy + x + y + 1

⇒ x + y = 49

⇒ x = 49 – y ….(1)

Again, one student joins the class,

⇒ xy + 50 + 50 = (x + 2)(y + 1.5)

⇒ xy + 100 = xy + 2y + 1.5x + 3

⇒ 97 = 2y + 1.5x

Putting the value of x:

⇒ 1.5(49 – y) + 2y = 97

⇒ 73.5 – 1.5y + 2y = 97

⇒ 0.5y = 23.5

⇒ y = 47

Putting the value of y in Equation (1)

⇒ x = 49 – 47 = 2

(i) If Anant joins the class, the average weight of the class increases by 0.5 kg, find the weight of Anant.

Let the weight of Anant = a kg

According to the question,

⇒ (47 × 2 + x)/3 = 47.5

⇒ 94 + x = 142.5

⇒ x = 48.5 kg

(ii) The number of students in the class = 2

(iii) Find the value of (y – 3x)

⇒ (y – 3x) = (47 – 3 × 2) = 47 – 6 = 41

Hence, option All (i), (ii), and (iii)

 

Q:10 The correct answer is option 2 i.e.\(\frac{a b+a d}{a+c}\).

Total runs scored by a cricketer in ‘a’ matches = ab

Total runs scored by a cricketer in ‘c’ matches = cd

Total matches played = (a + c)

Total runs scored in (a + c) matches = ab + cd

Required average = \(\frac{a b+a d}{a+c}\)

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Average.

औसत&nbsp;पर&nbsp;अभ्यास&nbsp;प्रश्न!

Average Questions PDF with detailed solutions

औसत के प्रश्न विस्तृत उत्तरों के साथ पीडीएफ

Data Interpretation (Pie Chart) questions are a type of questions which are frequently asked in competitive exams. These questions carry a weightage of 2-3 questions(3-6 marks) in SSC exams and 5-10 questions(5-10 marks) in bank exams. To perform well in competitive exams, you should have a good practice of data interpretation questions.

Here are some tips for solving Data Interpretation (Pie Chart) questions: Read the Graph Carefully, Start by understanding the directions, labels, and units. Decide which questions to tackle first. Usually, start with straightforward ones. Clearly write down your calculations, so you can retrace your steps if needed. Practice Regularly to maintain speed and accuracy

So, we have attached 10 questions of Data Interpretation (Pie Chart) for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Data Interpretation (Pie Chart)</h2>

You can also download the Data Interpretation (Pie Chart) questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Directions (1 - 5): 4 workers A, B, C, and D together work to finish a project, and the below pie chart shows the percentage of total wages received by 4 workers for that work. Study the chart carefully and answer the questions that follow. Total wages are Rs. 7500.

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_04.09.2020_D3.png" />

Q:1 A alone can complete the work in 30 days, then find the time in which B will complete the work alone.

1. 24 days

2. 36 days

3. 18 days

4. 20 days

5. 26 days

(Difficulty: 3, Estimated Time: 20 Seconds) If your concepts are clear, you might have done it easily!

Q:2 Total wages received by A and D is what percentage of total wages received by B and C?

1. 66.67%

2. 33.33%

3. 50%

4. 62.5%

5. 67.5%

(Difficulty: 3, Estimated Time: 20 Seconds) Time, Work, Wages....

Q:3 D alone can complete the work in 20 days then, find the time in which C will complete the work alone.

1. 15 days

2. 10 days

3. 12 days

4. 18 days

5. 8 days

(Difficulty: 3, Estimated Time: 20 Seconds) A similar one

Q:4 Find the ratio of time in which A, B, C and D will finish the work individually.

1. 6 : 10 : 5 : 4

2. 10 : 12 : 6 : 5

3. 24 : 15 : 30 : 10

4. 5 : 4 : 2 : 3

5. 15 : 12 : 6 : 10

(Difficulty: 3, Estimated Time: 20 Seconds) Solve carefully....

Q:5 Find the difference between the total wages received by A and C and the total wages received by B and D.

1. Rs. 800

2. Rs. 850

3. Rs. 1000

4. Rs. 900

5. Rs. 1100

(Difficulty: 2, Estimated Time: 15 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Direction (6 - 10): The pie chart given below shows the total number of passed candidates in different subjects in a class. Study the following pie chart carefully & answer the following questions.

Total number of passed students = 12000

<img alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_05.04.2020_D34.png" style="height:198px; width:200px" />

Q:6 If the ratio of total number of passed to total number of students in class is 2 : 5 and the percentage distribution of failed students is same as that of the passed students in the class, then find average number of failed student in chemistry & English.

1. 2200

2. 3200

3. 1500

4. 1700

5. 2700

(Difficulty: 3, Estimated Time: 25 Seconds) This was a bit calculative....

 Q:7 Total number of passed students in English & physics together is what percent more/less than total number of passed students in chemistry & Hindi together?

1. \(57\frac{1}{11}%\)%

2.  \(59\frac{1}{11}%\)%

3.  \(51\frac{2}{3}%\)%

4.  \(43\frac{2}{5}%\)%

5.  \(47\frac{1}{9}%\)%

(Difficulty: 3, Estimated Time: 20 Seconds) This was a similar one...

Q:8 If the total number of passed students are 25% of the total number of student in the class, then total number of failed student is how much more/less than total number of passed students in computer, Math & Hindi together?

1. 28,800

2. 27,400

3. 26,200

4. 29,400

5. 29,600

(Difficulty: 3, Estimated Time: 20 Seconds) Just basic concepts.....

Q:9 If total number of student who had taken exam of physics are 4000, then find ratio of number of failed students in physics to number of passed students in chemistry. 

1. 11 : 6

2. 13 : 6

3. 11 : 5

4. 3 : 2

5. 2 : 1

(Difficulty: 3, Estimated Time: 20 Seconds) You might have had a good score today....

Q:10 If the number of failed students in Hindi is 25% more than number of passed students in Hindi, then number of failed students in Hindi is what percent of number of passed students in English?

1. 80%

2. 75%

3. 55%

4. 60%

5. 70%

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (1)</td>
			<td>2. (1)</td>
			<td>3. (3)</td>
			<td>4. (5)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (5)</td>
			<td>7. (2)</td>
			<td>8. (4)</td>
			<td>9. (1)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 1 i.e. 24 days.

From the pie chart:

The ratio of wages received by A, B, C, and D = 16 : 20 : 40 : 24 = 4 : 5 : 10 : 6

We know that the ratio of wages shared will be equal to the ratio of their efficiencies

So,

The ratio of efficiencies of A and B = 4 : 5

Since efficiency and time are inversely proportional:

The ratio of time in which A and B will finish the work working alone = 5 : 4

Now, A alone can complete the work in 30 days.

So,

Time taken by B to complete the work alone = 30 × 4/5 = 24 days

 

Q:2 The correct answer is option 1 i.e. 66.67%.

From the pie chart:

Total wages received by A and D = 16 + 24 = 40%

And,

Total wages received by B and C = 20 + 40 = 60%

Hence,

Required percentage = 40/60 × 100 = 66.67%

 

Q:3 The correct answer is option 3 i.e. 12 days.

From the pie chart:

The ratio of wages received by A, B, C, and D = 16 : 20 : 40 : 24 = 4 : 5 : 10 : 6

We know that the ratio of wages shared will be equal to the ratio of their efficiencies.

So,

The ratio of efficiencies of C and D = 10 : 6

Since efficiency and time are indirectly proportional:

The ratio of time in which C and D will finish the work working alone = 6 : 10

Now, D alone can complete the work in 20 days

So,

Time taken by C to complete the work alone = 20 × 6/10 = 12 days

 

Q:4 The correct answer is option 5 i.e. 15 : 12 : 6 : 10.

From the pie chart:

The ratio of wages received by A, B, C, and D = 16 : 20 : 40 : 24 = 4 : 5 : 10 : 6

We know that the ratio of wages shared will be equal to the ratio of their efficiencies.

So,

The ratio of efficiencies of A, B, C, and D = 4 : 5 : 10 : 6

Since efficiency and time are indirectly proportional:

The ratio of time in which A, B, C, and D will finish the work individually

⇒ 1/4 : 1/5 : 1/10 : 1/6

⇒ 15 : 12 : 6 : 10

 

Q:5 The correct answer is option 4 i.e. Rs. 900.

From the pie chart:

Required difference = [(16 + 40) – (20 + 24)]/100 × 7500 = Rs. 900

 

Q:6 The correct answer is option 5 i.e. 2700.

Total failed student = 12000/2 × 3 = 18000

Required average = 1/2 [(10 + 20)/100] × 18000 = 2700

 

Q:7 The correct answer is option 2 i.e. \(59\frac{1}{11}%\)%.

Required percentage = \(\frac{(20+15)-(10+12)}{10+12} \) x 100

⇒ \(\frac{35-22}{22}\) x 100 = 59\(\frac{1}{11}% \)%

 

Q:8 The correct answer is option 4 i.e. 29,400.

25% of total student = 12,000

Total student = 48,000

Total failed student = 48,000 – 12,000 = 36,000

Total passed student in Computer, Math, and Hindi together

⇒ 55/100 × 12,000 = 6,600

Required difference = 36,000 – 6,600 = 29,400

 

Q:9 The correct answer is option 1 i.e. 11 : 6.

Total passed student in Physics = 15/100 × 12,000 = 1800

Passed student in Chemistry = 10/100 × 12,000 = 1200

Required Ratio = (4000 - 1800)/1200 = 2200/1200 = 11 ∶ 6

 

Q:10 The correct answer is option 2 i.e. 75%.

Passed student in Hindi = 12/100 × 12,000 = 1440

Failed student in Hindi = 1440 × 125/100 = 1800

Passed student in English = 12,000 × 20/100 = 2400

Required percentage = 1800/2400 × 100 = 75%

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Data Interpretation (Pie Chart).

Practice Questions on Data Interpretation (Pie Chart)!

Data Interpretation (Pie Chart) Questions PDF with detailed solutions

डेटा व्याख्या (वृत आलेख) के प्रश्न की पीडीएफ विस्तृत उत्तरों के साथ

Hi guys, presently we are here to talk about Factors and multiples. Let&#39;s get started. Firstly, let’s just know the meaning of factors and multiples.

Factors: Factors of a number are the numbers by which when the number is divided, the remainder will be 0.

Multiples: Multiples of a number are the numbers which are if divided by the number, the remainder will be 0. E.g. 8 has the factors 1, 2, 4 and 8 and has the multiples 8, 16, 24, etc.

Now, let’s talk about factors.

Any number greater than 1 can be written as the product of prime factors. Let’s generalize this term then we will explore this topic.

Let N is a composite number. So, by above statement, N can be written as, N = ap × bq × cr × … (Where a, b, c are distinct prime factors of N and p, q, r are integers) So, the general formula to calculate the number of factors of N is: Total number of factors of N = (p + 1) (q + 1) (r + 1) …

E.g. Find the total number of factors of 756.

Sol: 756 can be written as, 756 = 22 × 33 × 7 So, the total number of factors of 756 = (2 + 1) (3 + 1) (1 + 1) = 3 × 4 × 2 = 24.

E.g. How many factors of 756 are multiple of 12?

Sol: 756 = 22 × 33 × 7 and 12 = 22 × 3 ⇒756 = 22 × 31 [32 × 71 ] Since we have to find out the multiples of 12 in it. So, this [32 × 71 ] is important for us. No. of factors = (2 + 1) (1 + 1) = 3 + 2 = 6.

Co-Prime Numbers: Co-Prime numbers are the two numbers which have no common factor other than 1. e.g. 24 and 25 are co-prime numbers because 24 has factors 1, 2, 3, 4, 6, 8, 12, 24 and 25 has factors 1, 5, 25. No factor is common other than 1.

Now let us take a look at some interesting points:

<ol>
	<li>The product of two consecutive integers is always even. Because one of them is even and one of them is odd, so the product will always be even.</li>
	<li>The product of three consecutive integers is always divisible by 6. Because one of them is always even and one of them is always divisible by 3.</li>
	<li>Like that, the product of four consecutive integers is always divisible by 24.</li>
	<li>The product of n consecutive integers will always be divisible by n.</li>
	<li>Total one digit numbers are 9. Total two digit numbers are 90 (from 10 to 99). So, they have 90 × 2 = 180 digits. Total three digit numbers are 900 (from 100 to 999). So, they have 900 × 3 = 2700 digits. 35 Total four digit numbers are 9000 (from 1000 to 9999). So, they have 9000 × 4 = 36000 digits. And so on.</li>
</ol>

E.g. Divya typed the first n natural numbers on a keyboard without any spaces. If she had to press keys 1692 times, find n.

Sol: 1 digit numbers have 9 digits. 2 digit numbers are 90 and have 180 digits. So, 1692 – (180 + 9) = 1503. Since 3 digit numbers are 900 and have 2700 digits but we left only 1503 digits, it means the number is a 3 digit number. Now, 1503/3 = 501, means it is the 501th 3 digit number. So, the number is, 501 + 90 + 9 = 600.

Let us look at two examples and if you are now confident about this section, just try these questions by yourself before looking at my solution. It would be fun.

E.g. How many natural numbers divide 35999 but do not divide 35998?

Sol: It wants us to find the factors of 35999 which are not the factors of 35998. In simple words, (No. of factors of 35999) – (No. of factors of 35998), Because these are the only factors which are greater than 35998 and, so that, they do not divide 35998 . Now, 35999 = (5 × 7)999 = 5999 × 7999 ⇒No. of factors = (999 + 1) (999 + 1) = 1000 × 1000 = 10002 Also, 35998 = (5 × 7)998 = 5998 × 7998 ⇒No. of factors = (998 + 1) (998 + 1) = 999 × 999 = 9992 ⇒required answer = 10002 – 9992 = (1000 + 999) (1000 – 999) = 1999 × 1 = 1999.

E.g. In an NGO, there are 70 workers. All the workers visited an old age home having 70 senior citizens. The first worker donated Rs. 1000 to each senior citizen. The second worker donated Rs. 1000 to every second senior citizen starting from the second senior citizen. The third worker donated Rs. 1000 to every third senior citizen starting from the third senior citizen and so on. How many senior citizens received donations from an odd number of workers?

Sol: Read the question twice and thrice and try to understand what this question wants to say. The first worker gives rupees to all 70 citizens. But the 2nd worker gives only to 2nd, 4th, 6th , 8th, etc. citizens. Same thing is done by 3rd, he gives rupees to only 3rd, 6th, 9th, 12th, etc. citizens. Same for 4th, 5th, 6th workers till 70th. It means, a worker whose number is x is giving money to all the citizens whose number is a multiple of x. And hence, the citizen, whose number is y, is receiving the donations from all the workers whose number is a factor of y. i.e. is receiving as many donations as the number of factors of y. So, we just have to find out, from 1 to 70, how many numbers have odd numbers of factors. And from the previous discussion, we know that only perfect squares have an odd number of factors. And from 1 to 70, there are 8 perfect squares, which are 1, 4, 9, 16, 25, 36, 49 and 64. So, the answer is, only 8 senior citizens received donations from an odd number of worker.

Hope all your doubts have been clear after going through the blog. Stay connected for more.

Learn the key concepts of Factors and multiples

Factors and multiples- Quant Study Notes for Competitive Exams

गुणनखंड और गुणज- प्रमुख अवधारणाओं को सीखें और अभ्यास करें

Today we are going to learn about Percentages. We saw 2-3 questions are asked every year from percentages. Let’s first learn what we mean by percentage.

The percentage is derived from the Latin word ”percentum” which means ”per hundred” and it is denoted by %. A fraction with denominator 100 is called percentage. Similarly, a fraction with denominator 10 is called a decimal.

<h2>Fraction into percentage</h2>

To convert a fraction ‘a/b’ in percentage, multiply it with 100 and put a ‘%’ sign.

E.g. Convert 4 / 5 into a percentage.

Sol: We have to multiply this fraction to 100 first ⇒4 / 5 × 100 = 400 / 5 = 80 Now we have to put a ‘%’ sign at the end  80% (Ans.)

There are some frequently occurring fractions so we should remember them. Let us learn those important fractions.

<table cellspacing="0" class="Table" style="border-collapse:collapse; border:none; margin-left:16px; width:300px">
	<tbody>
		<tr>
			<td>
			1/1 = 100%
			</td>
			<td>
			1/2 = 50%
			</td>
			<td>
			1/3 = 33.33%
			</td>
			<td>
			1/4 = 25%
			</td>
			<td>
			1/5 = 20%
			</td>
		</tr>
		<tr>
			<td>
			1/6 = 16.66%
			</td>
			<td>
			1/7 = 14.28%
			</td>
			<td>
			1/8 = 12.5%
			</td>
			<td>
			1/9 = 11.11%
			</td>
			<td>
			1/10 = 10%
			</td>
		</tr>
		<tr>
			<td>
			1/11 = 9.09%
			</td>
			<td>
			1/12 = 8.33%
			</td>
			<td>
			1/13 = 7.69%
			</td>
			<td>
			1/14 = 7.14%
			</td>
			<td>
			1/15 = 6.66%
			</td>
		</tr>
		<tr>
			<td>
			1/16 = 6.25%
			</td>
			<td>
			1/17 = 5.88%
			</td>
			<td>
			1/18 = 5.55%
			</td>
			<td>
			1/19 = 5.26%
			</td>
			<td>
			1/20 = 5%
			</td>
		</tr>
	</tbody>
</table>

<h2>Percentage into fraction</h2>

To convert the percentage into fraction, divide the number by 100 and remove the % sign.

E.g. Convert 9.09% into fraction form.

Sol: First, we have to divide this by 100 ⇒ 9.09 / 100 = (100 / 11%) / 100 (As we know 9.09 can be written as 100 / 11) ⇒1 / 11% Now remove the % sign ⇒1 / 11 (Ans.)

E.g. A student scored 85% marks. The total marks were 400. How much did he score?

Sol: As his marks are 85% of total marks and total marks are 400 ⇒His score = 85% of 400 ⇒His score = (85 / 100) × 400 ⇒His score = (85) × 4 = 340 (Ans.)

Effect of Percentage Change on Any Quantity:

Let a number ‘p’ be increased by ‘x%’ then new quantity will be ⇒{(100 + x) / 100} × p Similarly, when ‘p’ decreases by ‘x%’ then new quantity will be ⇒{(100 - x) / 100} × p

E.g. The present salary of Amit is 3000. His salary will be increased by 15% next year, find his increased salary.

Sol: His salary is increased by 15% so here x is 15 According to formula, {(100 + x) / 100} × p By substituting the values ⇒{(100 + 15) / 100} × 3000 ⇒{115 / 100} × 3000 = 115 × 30 = 3450 (Ans.)

Here, we saw there could be an increase or decrease in quantity by some percentage.

<h2>Succesive increase or decrease of percentage</h2>

What if the quantity (p) is changed in two steps like in the first step it increases or decreases by ‘x%’ and in second step it increases or decreases by ‘y%’ then total percentage change would be ⇒ [{(100 ± x)} / 100] × [{(100 ± y)} / 100] × p

The sign will be positive when value increases and negative when value decreases. Let’s take an example to understand this.

E.g. If the length of the rectangle increases by 30% and the breadth of the rectangle decreases by 12%. Find the overall change in the area of a rectangle.

Sol: Let the area of the rectangle be ‘a’ square unit, and we know that values of x and y are 30 and 12 respectively. 

According to our formula ⇒[{(100 ± x)} / 100] × [{(100 ± y)} / 100] × p. By substituting all values, [{(100 + 30)} / 100] × [{(100 -12)} / 100] × a ⇒[130 / 100] × [88 / 100] × a [13 / 10] × [22 / 25] × a (286 / 250) × a = 1.144x Our difference would be 1.144x – x = 0.144 OR 14.4% (Ans.)

For this type of question, we don’t have to opt these long conventional methods, we have to just spell a magic formula, Total percent change = x + y + xy/100. Where ‘x’ and ‘y’ are percentage changes on quantity. Signs of ‘x’ and ‘y’ would depend on fluctuation in quantity.

E.g. If the length of the rectangle increases by 30% and the breadth of the rectangle decreases by 12%. Find the overall change in the area of the rectangle.

Sol: Here x = +30 and y = -12 By putting values, x + y + xy/100 = 30 - 12 - 30 × 12/100 = 18 – 36/10 = 18 – 3.6 = 14.4% (Ans.)

E.g. If the length of the rectangle increases by 30% and the area doesn’t change. Find the change in the breadth of the rectangle.

Sol: So here the question is twisted little bit, we have given total change as 0% and x = 30 and we have to find ‘y’ Total change = x + y + xy/100 ⇒ 0 = 30 + y + (30 × y)/100 ⇒ -30 = y + (3y) / 10 ⇒ -30 = (1 + 0.3) × y ⇒y = -30 / 1.3 = -23.07% (Ans.)

We saw a decrease of 23.07% When one factor of a product increases by ‘p%’ then the other factor will decrease by {p / (100 + p)} × 100%, similarly when one factor of a product decreases by ‘p%’ then the other factor will increases by {p / (100 - p)} × 100% so the total change would be zero in both cases.

E.g. If the price of coffee is increased by 10% then by how much percentage a housewife decreases the consumption, to have no extra expenditure.

Sol: We know that Expenditure = Consumption × Price By substituting the values, {p / (100 + p)} × 100% ⇒{10 / (100 + 10)} × 100% ⇒ {10 / 110} × 100% = (1 / 11) × 100% = 9.09% (Ans.)

E.g. If the numerator of a fraction is increased by 200% and the denominator of the fraction is increased by 150%, the resultant fraction is 9/25, find the original fraction.

Sol: Let the numerator be ‘x’ and denominator be ‘y’ resultant would be: {x + 2x} / {y + (3 / 2 × y)} which is equivalent to 9/25 ⇒3x / 2.5y = 9 / 25 ⇒x / y = 3/10 (Ans.)

E.g. If the income of Shyam is 50% more than Ram, then how much percent of Ram’s income is less than Shyam?

Sol: Ram’s income would be less by, ⇒{p / (100 + p)} × 100% = {50 / (100 + 50)} × 100% = {50 / 150} × 100% = {1 / 3} × 100% = 33.33% (Ans.)

To solve all the mathematics problems, it is crucial to master the concept of percentages, which is a fundamental component of the quantitative section. To excel in this area, one must be able to apply the relevant concepts with accuracy and speed. So, keep practicing and stay tuned for more.

Learn the key concepts of percentages