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<html><body><h2 id="content-id-6250934781-2"><b>Mensuration 3D Questions Overview</b></h2><p>Mensuration 3D Questions have a good weightage in the Ssc Chsl Exam and the type of question asked in Ssc Chsl exam is similar to the question mentioned below. It has been solved and explained by KD Live Experts under the guidance of Neetu Mam and they have tried to elaborate the concept used in Mensuration 3D Questions.</p><h2 id="content-id-3208951467-3"><b>Question On Mensuration 3D </b></h2>An empty metallic right circular cylindrical vessel has an inner base radius of 14 cm and a total height of 40 cm. This vessel is initially filled with water up to a height of 21 cm.

Match the column


	
		
			Column I 
			Column II
		
	
	
		
			(A) A solid metallic sphere of radius 7 cm is completely immersed into the vessel, and a solid right circular cone is crafted such that its height equals the sphere's radius and its base radius equals the initial water height, giving an expression value of: (Volume of the crafted Cone) - 3,146.
			(P) The positive value of x2 - 40x - 4224 = 0
		
		
			(B) The total volume of the original cylindrical vessel is calculated, and this total value is then divided by the product of its own base radius and its total height.
			(Q) Three times the sum of two consecutive even numbers is 94 more than twice the sum of two consecutive odd numbers. If the larger odd number is 1 greater than the smaller even number, then the numerical value of the smaller even number.
		
		
			(C) The circular base of the cylinder is polished at the rate of Rs 2 per cm2, giving a value that represents the total cost incurred.
			(R) The numerical value which gives a quotient of 94 and a remainder of 10 when divided by 13
		
	



<div class="row show--dificulty--bar" style="margin-top:8px;"><div class="col-sm-6 show--dificulty--bar" style="display:inline-flex !important;border-right:0.1px solid #eee;"><h3 style="font-size:12px !important;">Difficulty : Easy </h3></div><div class="col-sm-6 show--dificulty--bar" style="display:inline-flex;"><h3 style="font-size:12px !important;"> Average Time : 83 Seconds </h3></div></div><h2 id="content-id-8153092746-4"><b> Mensuration 3D Questions Options</b></h2><ol><li><p>(A) → (R), (B) → (P), (C) → (R)</p>
</li><li><p>Only (A) → (P), (B) → (Q)</p>
</li><li><p>(A) → (P), (B) → (Q), (C) → (R)</p>
</li><li><p>Only (B) → (Q), (C) → (R)</p>
</li><li><p>(A) → (Q), (B) → (P), (C) → (R)</p>
</li></ol> <h2 id="content-id-4173850296-5"><b> Mensuration 3D Questions Solution</b></h2><p>The correct answer is<strong> option 3 </strong>i.e. <strong>(A) → (P), (B) → (Q), (C) → (R)</strong>.</p>

<p><strong>(A)</strong> Volume of the cone = (1/3) × (22/7) × 21<sup>2</sup> × 7 = 3234 cm<sup>3</sup></p>

<p>Required value = 3234 − 3146 = 88</p>

<p>For (P), x<sup>2</sup> − 40x − 4224 = 0</p>

<p>(x − 88)(x + 48) = 0</p>

<p>x = 88 or x = −48</p>

<p>Taking the positive value, x = 88</p>

<p>Hence, <strong>(A) → (P)</strong></p>

<p><strong>(B)</strong> Volume of the cylinder = (22/7) × 14<sup>2</sup> × 40 = 24640 cm<sup>3</sup></p>

<p>Required value = 24640 ÷ (14 × 40) = 24640 ÷ 560 = 44</p>

<p>For (Q), let the smaller even number = x and the larger even number = x + 2.</p>

<p>Let the smaller odd number = y and the larger odd number = y + 2.</p>

<p>Given, larger odd number is 1 greater than the smaller even number = y + 2 = x + 1</p>

<p>Therefore, y = x − 1</p>

<p>Sum of the even numbers = x + (x + 2) = 2x + 2</p>

<p>Three times their sum = 3(2x + 2) = 6x + 6</p>

<p>Sum of the odd numbers = y + (y + 2) = 2y + 2</p>

<p>Twice their sum = 2(2y + 2) = 4y + 4</p>

<p>According to the question, 6x + 6 = 4y + 4 + 94</p>

<p>6x + 6 = 4y + 98</p>

<p>6x − 4y = 92</p>

<p>3x − 2y = 46</p>

<p>Substituting y = x − 1, we get 3x − 2(x − 1) = 46</p>

<p>3x − 2x + 2 = 46</p>

<p>x + 2 = 46</p>

<p>x = 44</p>

<p>Therefore, the smaller even number = 44</p>

<p>Hence, <strong>(B) → (Q)</strong></p>

<p><strong>(C)</strong> Area of the circular base = (22/7) × 14<sup>2</sup> = 616 cm<sup>2</sup></p>

<p>Total polishing cost = 616 × 2 = Rs. 1232</p>

<p>For (R), required number = 13 × 94 + 10 = 1222 + 10 = 1232</p>

<p>Hence, <strong>(C) → (R)</strong></p>

<p>Final Matching:</p>

<p>(A) → (P), (B) → (Q), (C) → (R)</p>
<h2 id="content-id-5482796013-6"><b> Mensuration 3D Questions Tips and Tricks</b></h2><img loading="lazy" src="https://study.kdcampus.live/questions-blog-pdf/Mensuration_3D_Questions_Tips_and_Tricks.png" alt="Mensuration 3D Questions Tips and Tricks" height="260" width="480"><ol><li>Try to solve Mensuration 3D Questions without taking any help from the solutions.</li><li>Mensuration 3D Questions require proper usage of concept so firstly read the question thoroughly and then use the right concept. </li><li>In case you're not able to solve the question in less than 30 seconds in the exam then you should skip the question and move to the next question.</li></ol><h3><b>Daily Current Affairs </b></h3><img loading="lazy" src="https://study.kdcampus.live/questions-blog-pdf/Daily_Current_Affairs.png" alt="Daily Current Affairs" height="260" width="480"><p>KD Live is providing Current Affairs on a daily basis for SSC and a current affairs quiz too for the practice. 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The correct answer is option 3 i.e. (A) - Mensuration 3D Questions are asked in Ssc Chsl exams and your question has been explained by experts like Neetu Mam.

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