Quantitative Aptitude

Mensuration

2,507 Questions
All Direct Formula Application 1067 Shape Transformations 343 Mensuration Ratio Equations 277 Combined & Hollow Solids 247 Successive Percentage Changes 178 Inscribed & Circumscribed Figures 165 Cost Estimation Problems 100 Liquid Pour & Submergence 67 Unit Conversions 35 Path & Border Calculations 17 Time & Work Integration 7 Scale & Model Proportions 2

Mensuration Questions

Multiple choice

The perimeter of a prism of triangular base is 15 cm and incircle radius is of 3 cm made at its base. If volume of prism is 270 cm3 then find the height of the prism. एक त्रिभुजाकार आधार वाले प्रिज्म का परिमाप 15 सेमी है और उस त्रिभुजाकार आधार के अंतः वृत्त की त्रिज्या 3 सेमी है । यदि प्रिज्म का आयतन 270 सेमी.3 हो तो प्रिज्म की ऊँचाई ज्ञात कीजिये ।

  1. 14 cm/सेमी

  2. 12 cm/सेमी

  3. 13 cm/सेमी

  4. 7 cm/सेमी
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For triangle with perimeter 15 cm and incircle radius 3 cm: semi-perimeter s = 7.5 cm. Area = r × s = 3 × 7.5 = 22.5 cm². Volume of prism = Base Area × height, so 270 = 22.5 × h, giving h = 12 cm. Option B is correct.

Multiple choice

The diameter of a cylinder is 14 cm and its height is 16 cm the lateral surface area of cylinder is- बेलन का व्यास 14 सेमी है इसकी ऊँचाई 16 सेमी है तो उसका पार्श्व पृष्ठ क्षेत्रफल ज्ञात कीजिए।

  1. 704 cm2/सेमी2

  2. 700 cm2/सेमी2

  3. 710 cm2/सेमी2

  4. 696 cm2/सेमी2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Lateral surface area of a cylinder = 2πrh, where r is radius and h is height. Given diameter = 14 cm, so radius r = 7 cm. Height h = 16 cm. LSA = 2 × (22/7) × 7 × 16 = 2 × 22 × 16 = 704 cm². Option D (696) incorrectly uses π = 3.14 without proper calculation, while option B (700) is a rounding error.

Multiple choice

A solid cone of height 8 cm and base radius 6 cm is melted and recast into identical cones, each of height 2 cm and radius 1 cm. What is the number of cones formed? एक ठोस शंकु जिसकी ऊँचाई 8 सेमी और आधार की त्रिज्या 6 सेमी है जिसे पिघलाकर ,सर्वसम शंकुओं में पुनः ढाला जाता है प्रत्येक, शंकु की आधार त्रिज्या 1 सेमी और ऊँचाई 2 सेमी है तों निर्मित शंकुओं की संख्या क्या है?

  1. 36

  2. 72

  3. 144

  4. 180
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume of original cone = (1/3)π(6)²(8) = 96π. Volume of each small cone = (1/3)π(1)²(2) = 2π/3. Number of cones = 96π/(2π/3) = 96π × 3/(2π) = 144. Option C is correct.

Multiple choice

The radius of a circle is increased by 2% how much does the area of the circle increase? किसी वृत्त की त्रिज्या में 2% की वृद्धि की जाती है उस वृत्त के क्षेत्रफल में कितनी वृद्धि होगी?

  1. 2%

  2. 4.40%

  3. 4%

  4. 4.04%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Area of circle = πr². If radius increases by 2%, new radius = 1.02r. New area = π(1.02r)² = π × 1.0404 × r² = 1.0404 × πr². The increase = 1.0404 - 1 = 0.0404 = 4.04%. Option A (2%) would be true if area scaled linearly with radius. Option C (4%) neglects the compounding effect. Option B (4.40%) is incorrect.

Multiple choice

The diameter of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it melted and made of solid cylinder of length 8/3 cm, then the diameter ( in cm ) of the cylinder is : एक खोखले गोले का आन्तरिक और बाह्य का व्यास 6 सेमी और 10 सेमी है यदि इसको पिघलाकर एक ठोस बेलन बनाया जाता है जिसकी लम्बाई 8/3 सेमी है तो ठोस बेलन का व्यास क्या होगा ?

  1. 10

  2. 16

  3. 7

  4. 14
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of hollow sphere = (4/3)π(R³ - r³) = (4/3)π(5³ - 3³) = (4/3)π(98). This becomes volume of cylinder: πr²(8/3). Equating: (4/3)π × 98 = πr²(8/3). Solving: r² = 49, so r = 7 cm and diameter = 14 cm.

Multiple choice

A hemispherical bowl of internal radius 15 cm. contains a liquid. The liquid is to be filled in to a cylindrical-shaped bottles of diameter 5 cm. and height 6 cm. The number of bottles required to empty the bowl is- एक अर्धवृत्त कटोरा है जिसमें तरल पदार्थ भरा है और उसकी आन्तरिक त्रिज्या 15 सेमी. है। उस तरल पदार्थ को 5 सेमी. व्यास और 6 सेमी. ऊँचाई वाली बेलनाकार बोतलों में भरा जाता है। उस कटोरे को खाली करने के लिए कितनी बोतलों की आवश्यकता होगी?

  1. 20

  2. 30

  3. 50

  4. 60
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of hemisphere = (2/3)πr³ = (2/3)π(15)³ = 2250π cm³. Volume of each cylinder = πr²h = π(2.5)²(6) = 37.5π cm³. Number of bottles = 2250π ÷ 37.5π = 60. 20, 30, and 50 are incorrect as they don't match the volume ratio calculation.

Multiple choice

The radii of two cylinders are in the ratio 2 : 3 and their curved surface areas are in the ratio 5 : 3. What is the ratio of their volumes? दो बेलनों की त्रिज्यायें 2 : 3 के अनुपात में है, और उनके वक्र पृष्ठीय क्षेत्रफल का अनुपात 5 : 3 है। उनके आयतनों का अनुपात क्या है?

  1. 3 : 5

  2. 5 : 3

  3. 10 :11

  4. 10 : 9
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For cylinders: curved surface area = 2πrh and volume = πr²h. Given r₁:r₂ = 2:3, let r₁ = 2k, r₂ = 3k. CSA ratio: (2π·2k·h₁)/(2π·3k·h₂) = 5/3, so h₁/h₂ = 5/2. Volume ratio: (π·4k²·h₁)/(π·9k²·h₂) = (4/9)×(5/2) = 20/18 = 10/9.

Multiple choice

Find the area of the ring between two concentric circles, whose circumference are 132 cm and176 cm. दो संकेद्री वृत्तो जिनकी परिधि 132 सेमी और 176 सेमी है के बीच बने वलय का क्षेत्रफल ज्ञात कीजिये।

  1. 1080

  2. 1078

  3. 1060

  4. 1108
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the first circle: circumference = 2πr1 = 132, so r1 = 132/(2π) = 21 cm. For the second circle: circumference = 2πr2 = 176, so r2 = 176/(2π) = 28 cm. Area of ring = π(r2² - r1²) = π(28² - 21²) = π(784 - 441) = π × 343. Using π = 22/7: Area = (22/7) × 343 = 22 × 49 = 1078 sq cm.

Multiple choice

The total surface area of a solid right circular cylinder is twice that of a solid sphere. If they have the same radii,what will be the ratio of the volume of cylinder to that of the sphere. एक ठोस लम्बवृत्तीय बेलन के सम्पूर्ण पृष्ठ का क्षेत्रफल एक ठोस गोले के सम्पूर्ण पृष्ठ के क्षेत्रफल का दोगुना है। यदि उनकी त्रिज्यायें बराबर है तो बेलन के आयतन और गोले के आयतन का अनुपात क्या होगा?

  1. 3 : 2

  2. 9 : 4

  3. 1 : 2

  4. 1 : 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Cylinder surface area = 2πr(r+h), sphere surface area = 4πr². Given cylinder SA = 2 × sphere SA, so 2πr(r+h) = 8πr², giving r+h = 4r and h = 3r. Volume of cylinder = πr²h = 3πr³. Volume of sphere = (4/3)πr³. Ratio = 3πr³ : (4/3)πr³ = 9 : 4. 1:2 would be if heights were equal; 3:2 incorrectly uses diameter.

Multiple choice

From a right circular cylinder of radius 10 cm. and height 21 cm., a right circular cone of same base-radius is removed. If the volume of the remaining portion is 4400 cm3 then find the height of the removed cone. एक लम्ब वृत्तीय बेलन की त्रिज्या 10 सेमी. और ऊँचाई 21 सेमी. है। इनमें से एक लम्बवृत्तीय शंकु उसी त्रिज्या का काटा जाता है। यदि शेष भाग का आयतन 4400 सेमी3. हो तो काटे गये शंकु की ऊँचाई ज्ञात कीजिए।

  1. 14

  2. 21

  3. 18

  4. 16
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Cylinder volume = πr²h = π(100)(21) = 2100π ≈ 6600 cm³. Cone volume = (1/3)πr²h_cone = (100π/3)h. Remaining volume = 6600 - (100π/3)h = 4400. Solving: (100π/3)h = 2200, h = 2200 × 3/(100π) = 66/π ≈ 21. Height 14 would give remaining volume 5133, 18 gives 4800, 16 gives 5026.

Multiple choice

If the radius of the base of a cone be 7 cm and its curved surface area be 550 sq. cm, then the volume of the cone is? यदि एक शंकु के आधार की त्रिज्या 7 सेमी. हो और उसका वक्र पृष्ठीय क्षेत्रफल 550 वर्ग सेमी. हो तो, शंकु का आयतन कितना होगा?

  1. 1132 sq.cm/वर्ग सेमी.

  2. 1324 sq.cm/वर्ग सेमी.

  3. 1232 sq.cm/वर्ग सेमी.

  4. 1024 sq.cm/वर्ग सेमी.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Curved surface area of cone = πrl = 550 cm². With r = 7 cm: π × 7 × l = 550, so l = 550/(7π) = 25 cm. Using Pythagorean theorem: h = √(l² - r²) = √(625 - 49) = √576 = 24 cm. Volume = (1/3)πr²h = (1/3) × π × 49 × 24 = 392π = 1232 cm³. Note: options incorrectly show sq.cm instead of cu.cm.

Multiple choice

A rectangular box P is inscribed in a sphere of radius r cm. The surface area of P is 384 cm2. and the sum of the length of its 12 edges is 112 cm. Then r (in cm.) is equal to- एक आयताकार बाक्स P, एक r सेमी. त्रिज्या वाले गोले में खींचा जाता है। P का पृष्ठीय क्षेत्रफल 384 सेमी2. है और इसके 12 कोरो की लम्बाई का योग 112 सेमी. है, तो r (सेमी. में) बराबर है-

  1. 8

  2. 10

  3. 12

  4. 14
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a rectangular box with dimensions l, b, h: 4(l+b+h) = 112, so l+b+h = 28. Surface area: 2(lb+bh+lh) = 384. If edges are 8, 12, 16: sum = 28 ✓, surface area = 2(96+192+128) = 2(416) = 832 ✗. If edges are 8, 12, 8: surface area = 2(96+96+64) = 512 ✗. For a box inscribed in a sphere, the sphere's diameter equals the box's diagonal. With edges summing to 28 and surface area 384, the dimensions satisfy these conditions, giving a diagonal of 20, so radius r = 10.

Multiple choice

A Solid metallic cone of height 10 cm, radius of base 20 cm is melted to make spherical balls each of 4 cm. diameter. How many such balls can be made? एक धातुई शंकु जिसकी ऊंचाई 10 सेमी है और आधार की त्रिज्या 20 सेमी है इसको पिघलाकर 4 सेमी व्यास के कितनी गेंदे बनाये जा सकते है?

  1. 120

  2. 75

  3. 50

  4. 125
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The volume of the cone equals the total volume of all spherical balls combined. Cone volume = (1/3)πr²h = (1/3)π(20)²(10) = (4000/3)π cm³. Each ball has radius 2 cm (half of 4 cm diameter), so volume = (4/3)π(2)³ = (32/3)π cm³. Number of balls = (4000/3)π ÷ (32/3)π = 4000/32 = 125. The claimed answer is correct.

Multiple choice

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 :5, the ratio of the radius and their height. एक बेलन और शंकु की आधार की त्रिज्या तथा इनकी ऊंचाई समान हैं | यदि इनके वक्रपृष्ठ के क्षेत्रफल का अनुपात 8 : 5 है तो इनकी त्रिज्या और ऊंचाई का अनुपात ज्ञात कीजिये ?

  1. 1:2

  2. 1:3

  3. 2:3

  4. 3:4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let radius = r and height = h. Curved surface area of cylinder = 2πrh. Curved surface area of cone = πrl where l = √(r²+h²). Given 2πrh / πr√(r²+h²) = 8/5, we get 2h / √(r²+h²) = 8/5. Squaring: 4h² / (r²+h²) = 64/25, which simplifies to 100h² = 64r² + 64h², giving 36h² = 64r². Taking square root: 6h = 8r, so r/h = 6/8 = 3/4. The ratio of radius to height is 3:4.

Multiple choice

Find the surface area of largest possible sphere inscribed in a cone of radius 21 cm and slant height equal to the diameter of the base. 21 सेमी.त्रिज्या के एक शंकु के अंर्तगत, जिसकी तिर्यक ऊँचाई उसके आधार के व्यास के बराबर है, गोले का पृष्ठीय क्षेत्रफल ज्ञात कीजिये।

  1. 1650 cm2/सेमी.2

  2. 1830 cm2/सेमी.2

  3. 1848 cm2/सेमी.2

  4. 1770 cm2/सेमी.2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Cone: radius r=21, slant height l=2r=42, so height h=√(l²-r²)=21√3. Inscribed sphere touches base and sides. Largest inscribed sphere has radius R = r×h/(r+√(r²+h²)) = 21×21√3/(21+42) = 7√3. Surface area = 4πR² = 4π×147 = 1848. Answer C is correct.