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 the triangular base of a right prism is 15 cm and radius of the in -circle of the triangular base is 3 cm. If the volume of the prism be 315 cm3, then the height of the prism is- एक समलम्ब प्रिज्म के त्रिभुजाकार आधार का परिमाप 15 सेमी. है और उस त्रिभुजाकार आधार के अन्तःवृत्त की त्रिज्या 3 सेमी. है। यदि उस प्रिज्म का आयतन 315 सेमी3 हो तो प्रिज्म की ऊंचाई होगी-

  1. 60 cm/सेमी.

  2. 15 cm/सेमी.

  3. 12 cm/सेमी.

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

For any triangle, area = (perimeter × inradius) / 2. Here base area = (15 × 3) / 2 = 22.5 cm². Volume of prism = base area × height, so 315 = 22.5 × h. Therefore height = 315 / 22.5 = 14 cm. The claimed answer D is correct.

Multiple choice

The circumference of two circles is 132 cm. and 308 cm. respectively. What is the difference between the area of the larger circle and that of the smaller circle? दो वृत्तों की परिधि क्रमशः 132 सेमी. और 308 सेमी. हैं। बड़े वृत्त के क्षेत्रफल और छोटे वृत्त के क्षेत्रफल में क्या अन्तर है ?

  1. 5936 cm.2/सेमी2

  2. 5879 cm.2/सेमी2

  3. 6160 cm.2/सेमी2

  4. 5286 cm.2/सेमी2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For circumferences C1 = 132, C2 = 308. Radii: r1 = 132/(2π) = 21, r2 = 308/(2π) = 49. Areas: A1 = π(21)² = 1386π, A2 = π(49)² = 2401π. Difference = π(2401 - 441) = 1960π = 6160 cm² (using π = 22/7). Direct formula: difference = π(r2² - r1²) = π(r2 - r1)(r2 + r1). Bigger circle has 49 cm radius, smaller has 21 cm radius.

Multiple choice

The total surface area of a solid hemisphere is (147\pi) cm2. The volume of the hemisphere is -
  1. $2156/3 cm3$
  2. $2512/7 cm3$
  3. $855 cm3$
  4. $521 cm3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total surface area of hemisphere = 3πr² = 147π, so r² = 49, r = 7 cm. Volume of hemisphere = (2/3)πr³ = (2/3)π × 343 = 686π/3 cm³. In terms of π: this is approximately 718.67 cm³. However, option A gives 2156/3 cm³. Let me verify: 2156/3 = 718.67, which matches. The claimed answer A is correct, as 2156/3 cm³ = (2/3)π × 343 when π ≈ 22/7 is used.

Multiple choice

A solid sphere of radius 9 cm is melted to form a sphere of radius 6 cm and a right circular cylinder of same radius. The height of the cylinder so formed is 9 सेमी. त्रिज्या का एक ठोस गोला पिघलाकर 6 सेमी. त्रिज्या का एक ठोस गोला और समान त्रिज्या का एक लम्बवृत्तीय बेलन बनाया गया है | इस तरह बने बेलन की ऊँचाई है

  1. 19 cm. /सेमी.

  2. 21 cm. /सेमी.

  3. 23 cm. /सेमी.

  4. 25 cm. /सेमी.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of original sphere = (4/3)π(9)³ = 972π cm³. Volume of new sphere = (4/3)π(6)³ = 288π cm³. Volume of cylinder = Volume remaining = 972π - 288π = 684π cm³. For cylinder: πr²h = 684π, where r = 6. So π(36)h = 684π, giving h = 684/36 = 19 cm.

Multiple choice

If the volume of a rectangular solid is to be increased by 50% without changing its base, its altitude must be- यदि बिना आधार में वृद्धि किये हुये एक आयताकार ठोस के आयतन में 50% की वृद्धि की जानी है, तो इसकी ऊंचाई होनी चाहिए-

  1. cubed/तिगुनी

  2. doubled/दोगुनी

  3. increased by 50%/ 50% की वृद्धि होती है।

  4. increased by 150%/ 150% की वृद्धि होती है।
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume = base × height. New volume = 1.5 × original volume. Base unchanged, so 1.5 = new height/old height. New height = 1.5 × old height, which is a 50% increase in height.

Multiple choice

From a solid right circular cylinder of height 24 cm and radius 7 cm, a cone of same height and same base is removed. The volume of the remaining solid is- 24 सेमी. ऊँचाई और 7 सेमी. त्रिज्या वाले एक ठोस लम्बवृत्तीय बेलन से, एक समान ऊँचाई और समान त्रिज्या का शंकु निकाला जाता है। शेष ठोस का आयतन है

  1. 6242 cm.3/सेमी3

  2. 1232 cm.3/सेमी3

  3. 2446 cm.3/सेमी3

  4. 2464 cm.3/सेमी3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Cylinder volume = πr²h = π × 49 × 24 = 1176π. Cone volume = (1/3)πr²h = (1/3) × 1176π = 392π. Remaining volume = 1176π - 392π = 784π ≈ 2464 (using π ≈ 22/7).

Multiple choice

Find the area of the maximum size of square which can be inscribed inside a semicircle of radius 12 cm. 12 सेमी त्रिज्या वाले अर्द्धवृत्त के भीतर बनने वाले सबसे बड़े वर्ग का क्षेत्रफल ज्ञात कीजिए।

  1. 122.5 cm2/सेमी2

  2. 171.4 cm2/सेमी2

  3. 115.2 cm2/सेमी2

  4. 99.6 cm2/सेमी2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a square inscribed in a semicircle of radius r = 12 cm, one side of the square lies along the diameter of the semicircle, and the opposite two vertices touch the semicircular arc. Let the square have side length s. If we draw the semicircle with the flat side (diameter) horizontal and the square sitting on it, the top vertices of the square lie on the semicircle. The center of the semicircle is at the midpoint of the diameter. For the square to be maximized, its top corners lie exactly on the semicircle. Using the property that for a square inscribed in a semicircle, the diagonal from the center to a top vertex forms a right triangle: (s/2)² + s² = r², giving s²/4 + s² = 144, or 5s²/4 = 144, so s² = 576/5 = 115.2. The area of the square is s² = 115.2 cm².

Multiple choice

The area enclosed between the circumference of two concentric circles is 16 π sq. cm and their radii are in the ratio of 5 : 3. What is the area of the outer circle? दो संकेंद्रीय वृत्तों की परिधि के बीच का क्षेत्रफल 16 π वर्ग सेमी है और उनकी त्रिज्याओं में अनुपात 5: 3 है । बाहरी वृत्त का क्षेत्रफल कितना है?

  1. 9 π sq. cm/वर्ग सेमी

  2. 16 π sq. cm/वर्ग सेमी

  3. 25 π sq. cm/वर्ग सेमी

  4. 36 π sq. cm/वर्ग सेमी
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Area between circles = πR² - πr² = 16π. Given R:r = 5:3, let R = 5k, r = 3k. Then π(25k² - 9k²) = 16πk² = 16π, so k² = 1, k = 1. Therefore R = 5 cm, and area of outer circle = π × 5² = 25π sq. cm. This tests understanding of concentric circles area relationship and ratio application.

Multiple choice

The curved surface of a cylinder is 1000 cm2 .A wire of diameters 5 mm. is wound around it ,so as to cover it completely .What is the length of the wire used?

  1. 22 m.

  2. 20 m.

  3. 18 m.

  4. 34 m.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

CSA of cylinder = 2πrh = 1000 cm². Wire diameter = 5 mm = 0.5 cm, so radius = 0.25 cm. When wire is wound, each turn uses 0.5 cm (diameter) of cylinder height. Number of turns = h/0.5. Length per turn = 2πr. Total length = 2πr × (h/0.5) = 2πrh/0.5 = 2×1000/0.5 = 4000 cm = 40 m. Wait - checking options again. Option B says 20 m. Let me reconsider: if CSA = πdh and wire length L × πd = πd × (h/0.5) = πdh/0.5 = CSA/0.5 = 1000/0.5 = 2000 cm = 20 m. Option B is correct.

Multiple choice

A solid cube of side 18.5 cm is heated such that its edges increase by 25%, then its surface area will increase by how much percent? एक 18.5 सेमी भुजा वाले ठोस घन को इस प्रकार गर्म किया गया कि इसकी भुजा में 25% की वृद्धि हुई, तो इसके पृष्ठीय क्षेत्रफल में कितने प्रतिशत की वृद्धि होगी?

  1. 25%

  2. 50%

  3. 62.5%

  4. 56.25%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Surface area of cube = 6a² where a is edge length. When edge increases by 25%, new edge = 1.25a. New surface area = 6(1.25a)² = 6 × 1.5625a² = 9.375a². Percentage increase = (9.375 - 6)/6 × 100 = 56.25%.

Multiple choice

The area of the region enclosed between two circles is 418 cm2. If the width of that region is 7 cm, find the radius of the bigger circle. दो वृत्त के मध्य स्थित भाग का क्षेत्रफल 418 सेमी2 है। यदि उस क्षेत्र की चौड़ाई 7 सेमी है, बड़े वृत्त की त्रिज्या ज्ञात कीजिए।

  1. 13 cm/सेमी

  2. 7 cm/सेमी

  3. 6 cm/सेमी

  4. 19 cm/सेमी
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let larger radius = R, smaller radius = r. Width of ring = R - r = 7 cm. Area of ring = π(R² - r²) = π(R - r)(R + r) = π × 7 × (R + r) = 418 cm². Therefore, R + r = 418/(7π) ≈ 19. Using R - r = 7 and R + r = 19: Adding equations: 2R = 26, so R = 13 cm. Option B (7 cm) is the ring width, not the radius. Option C (6 cm) would be the smaller radius. Option D (19 cm) is R + r, not R alone.

Multiple choice

Capacity of a cylindrical vessel is 25.872 litres. If the height of the cylinder is three times the radius of its base, what is the area of the base in square cm? एक बेलनाकार पात्र की क्षमता 25.872 लीटर है। यदि बेलन की ऊँचाई इसके आधार की त्रिज्या का तीन गुना है, आधार का क्षेत्रफल वर्ग सेमी. में क्या है?

  1. 336

  2. 1232

  3. 616

  4. 672
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume of cylinder = πr²h. Given h = 3r and volume = 25.872 L = 25872 cm³. Substituting: πr²(3r) = 25872, so r³ = 25872/(3π) ≈ 2746.6. This gives r ≈ 14 cm. Base area = πr² ≈ 616 cm².

Multiple choice

Radius and height of a cylinder is increased by 30%. Find the percentage change in its total surface area. एक बेलन की त्रिज्या और ऊंचाई में 30% की वृद्धि हुई है। इसके कुल पृष्ठ क्षेत्रफल में प्रतिशत परिवर्तन ज्ञात कीजिये।

  1. 69%

  2. 54%

  3. 81%

  4. 64%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When radius and height both increase by 30%, they become 1.3 times their original values. Total surface area = 2πr² + 2πrh. New TSA = 2π(1.3r)² + 2π(1.3r)(1.3h) = 1.69(2πr² + 2πrh) = 1.69 × original. The increase is (1.69 - 1) × 100% = 69%. Both terms of the surface area formula are multiplied by 1.3².