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Algebra questions are a common type of questions asked in competitive exams. These questions carry a weightage of 3-4 questions (6-8 marks) in SSC exams and 1-2 questions(1-2 marks) in bank exams. To get a good rank in competitive exams, you should have the concepts of Algebra on your tips, especially, its formulas and identities

Here are some tips for solving Algebra questions: Understand the basic concepts of Algebra, Practice solving a variety of algebra questions, Use identities and formulas. Systematically simplify equations and solve.

So, we have attached 10 questions of Algebra for you to practice with. You should aim to solve these questions in less than half a minute for each.

<h2 id="content-id-1700055491326-0">Practice Questions on Algebra</h2>

You can also download the Algebra questions and answers pdf. Just click on the Download PDF button. So let's start with the very first question.

Q:1 For three natural numbers x < y < z, (x2 + y2 + z2 + 2xy)/(2yz + 2xz) = 5/4, then what is the value of 2x + 2y – z?

1. 3x

2. 3y

3. 3z

4. (x + y - z)

(Difficulty: 3, Estimated Time: 20 Seconds) Not an easy one, be ready for more

Q:2 If the sum of the squares of the digits of a two-digit number is 13, then what would be the sum of all the possible combinations of the digits?

1. 44

2. 55

3. 24

4. 33

(Difficulty: 2, Estimated Time: 15 Seconds) This was also an easy one. Did you guess it right?

Q:3 3x2 + 4x + 2 is a quadratic equation with no real roots. b is the minimum positive value of the coefficient of x without changing other coefficients such that the roots become real. Then what are the roots of the quadratic equation 6x2 – b2x + b2 = 0?

1. 4, 4

2. 2, 2

3. 2, 4

4. 8, 4

(Difficulty: 3, Estimated Time: 20 Seconds) It is not an easy one but I think now you're prepared for it. Did you guess it right?

Q:4 Find the value of <img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Annotation 2020-07-05 210632.png" style="height:29px; width:100px">

1. 6

2. 7

3. - 6

4. - 7

(Difficulty: 3, Estimated Time: 20 Seconds) This was a test of your concepts!

Q:5 The points A, B, C, D, E, F, G, H and I are equidistant from each other on the number line. Find the fraction represented by AD on the number line.

1. 3/8

2. 1/2

3. 1/4

4. 7/8

(Difficulty: 3, Estimated Time: 20 Seconds) We're halfway through. Have you got all your questions correct so far?

Q:6 If a2 + b2 + c2 = 14 and ab + bc + ca = 11, find (a + b + c)3.

1. 216

2. -216

3. 36

4. both 1 & 2

(Difficulty: 4, Estimated Time: 30 Seconds) This was a hard nut to crack, be prepared for such questions in exam!

Q:7 Find the number of solutions of the pair of linear equations

x - 2y - 12 = 0 & -2x + 4y = -24

1. Only one

2. Zero

3. Infinitely many

4. Two 

(Difficulty: 4, Estimated Time: 25 Seconds) This was also a tricky one, but I think now you're prepared for it. 

Q:8 If the factors of the quadratic equation are (x + 3) and (x - 1/3). What will be the quadratic equation?

1. 3x2 - x - 8 = 0

2. 8x2 - 3x + 1 = 0

3. 8x2 + 3x - 1 = 0

4. 3x2 + 8x - 3 = 0

(Difficulty: 4, Estimated Time: 25 Seconds) This was a hard question!

Q:9 Find the number of solutions of the given pair of linear equations

2x + 6y = - 17 &

4x + 12y = - 34

1. Two

2. only one

3. No solution

4. infinite

(Difficulty: 2, Estimated Time: 15 Seconds) Another easy one! Let's score more...

Q:10 a, b and c are three real numbers such that a + b = √11, b + c = √22 and a + c = √8, then what is the value of (a + b + c)3 – a3 – b3 – c3?

1. 22

2. 44

3. 66

4. 132

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

 

<h2 id="content-id-1700055491347-1">Answer Key</h2>

Let's check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (3)</td>
			<td>2. (2)</td>
			<td>3. (2)</td>
			<td>4. (2)</td>
			<td>5. (1)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (3)</td>
			<td>8. (4)</td>
			<td>9. (4)</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't worry, we are here to help you master the subject.

Let's check the answers and solutions and try to find out what went wrong. 
 

<h2 id="content-id-1700055491347-2">Answers and Solutions</h2>

Q:1 The correct answer is Option 3 i.e. 3z.

⇒ (x2 + y2 + z2 + 2xy)/(2yz + 2xz) = 5/4

Using componendo and dividendo

⇒ (x2 + y2 + z2 + 2xy + 2yz + 2xz)/(x2 + y2 + z2 + 2xy - 2yz - 2xz) = 9/1

⇒ (x + y + z)2/(x + y – z)2 = 9/1

Taking square roots,

⇒ (x + y + z)/(x + y – z) = 3/1

⇒ x + y + z = 3x + 3y – 3z

⇒ 4z = 2x + 2y

⇒ 3z = 2x + 2y – z

 

Q:2 The correct answer is option 2 i.e. 55.

Let the number be 10x + y form

⇒ x2 + y2 = 13

⇒ x = 2 or 3 and y = 3 or 2

Sum of such numbers = 23 + 32 = 55

 

Q:3 The correct answer is option 2 i.e. 2, 2 

For a quadratic equation ax2 + bx + c = 0 to have real roots: b2 ≥ 4ac

The minimum value of the coefficient of x without changing other coefficients such that the roots are real:

b2 ≥ 4 × 3 × 2

b ≥ √24

⇒ 6x2 – b2x + b2 = 0

⇒ 6x2 - 24x + 24 = 0

⇒ 6x2 - 24x + 24 = 0 ÷6

⇒ x2 – 4x + 4 = 0

⇒ (x – 2)2 = 0

The roots are 2, 2

 

Q:4 The correct answer is option 2 i.e 7.

Assume the given series equal to x and solve the quadratic equation by splitting of middle term method.

let x = <img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Annotation 2020-07-05 210632.png" style="height:29px; width:100px">

⇒ x2 = 42 + <img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Annotation 2020-07-05 210632.png" style="height:29px; width:100px">

⇒ x2 = 42 + x

⇒ x2 - x - 42 = 0

⇒ x2 - 7x + 6x - 42 = 0 

⇒ x(x - 7) + 6(x - 7) = 0 

⇒ (x - 7)(x + 6) = 0 

⇒ x = 7 or - 6

we take only positive value 

So, x = 7

 

Q:5 The correct answer is option 1 i.e. 3/8.

Let us take each distance to be x

Total distance = 8x

Distance from A to D = 3x

Required distance = 3x/8x = 3/8

 

Q:6 The correct answer is option 4 i.e. both 1 & 2.

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

a2 + b2 + c2 = 14

ab + bc + ca = 11

Now, (a + b + c)2 = 14 + 2 × 11 = 36

a + b + c = +6 & -6

(a + b + c)3 = 216 & -216

 

 

Q:7 The correct answer is Option 3 i.e. Infinitely many.

If the pair of equations are given:

a1x + b1y + c1 = 0 & a2x + b2y + c2 = 0 and if

a1/a2 = b1/b2 = c1/c2 

Then, the number of solution are infinitely many.

Given pair of equations is:

x - 2y - 12 = 0 & -2x + 4y = -24

So,

a1 = 1 b1 = - 2 c1 = - 12

a2 = - 2, b2 = 4, c2 = 24

Now, 1/-2 = -2/4 = - 12/24 i.e 1/2 = 1/2 = 1/2

So, Number of possible solutions are infinitely many.

 

Q:8 The correct answer is option 4 i.e 3x2 + 8x - 3 = 0.

If a & b are the roots of the equation then the equation will be

x2 - x(a + b) + ab = 0

The factors of the quadratic equation are (x + 3) and (x - 1/3)

So,

Roots of the quadratic equation will be -3 & 1/3

The required equation will be:

x2 - (-3 + 1/3)x + (-3 × 1/3) = 0

⇒ x2 + 8/3x - 1 = 0 

⇒ 3x2 + 8x -3 = 0 

 

Q:9 The correct answer is option 4 i.e. infinite

The pair of linear equations are 

a1x + b1y = c1 & a2x + b2y = c2 

If a1/a2 = b1/b2 = c1/c2 then the number of solution is infinite

Given equations are

2x + 6y = - 17 & 4x + 12x = - 34

Here, 2/4 = 6/12 = -17/(-34) = 1/2 

So, the number of solutions is infinite.

 

Q:10 The correct answer is Option 4 i.e. 132.

We know that,

(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (a + c)

So,

(a + b + c)3 – a3 – b3 – c3 = 3 (a + b) (b + c) (a + c)

⇒ 3 × √11 × √22 × √8 = 132

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 Algebra.

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