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

What would be the cost (in Rs.) of building a fence around a square plot with area 506.25 sq.ft. at the rate of Rs. 44 per foot ?

  1. 3460

  2. 3640

  3. 3690

  4. 3960

  5. 4000
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First find the side length of the square by taking the square root of the area: sqrt(506.25) = 22.5 ft. The perimeter of a square is 4 times the side length: 4 × 22.5 = 90 ft. The total cost is perimeter × rate: 90 × 44 = Rs. 3960. This matches option D.

Multiple choice

Three solid metallic spheres of radius 12 cm, 16 cm, and 20 cm are melted and recast into a new solid sphere. Find the diameter of the new sphere. 12सेमी , 16 सेमी एवं 20 सेमी त्रिज्या वाले गोलो को गलाया गया एवं नया गोला बनाया गया। नये गोले की त्रिज्या ज्ञात कीजिए।

  1. 24 cm /सेमी

  2. 36 cm /सेमी

  3. 48 cm /सेमी

  4. 12 cm /सेमी
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume is conserved when melting and recasting. Total volume = (4/3)π(12³ + 16³ + 20³) = (4/3)π(1728 + 4096 + 8000) = (4/3)π(13824). For the new sphere with radius R: (4/3)πR³ = (4/3)π(13824), so R³ = 13824 = 24³. Therefore, R = 24 cm. (Note: The English text asks for 'diameter' but Hindi asks for 'त्रिज्या' (radius); the answer 24 cm refers to the radius).

Multiple choice

What is the length of the uniform wire of diameter 0.2cm that can be drawn from a solid sphere of radius 3 cm.? 3 सेमी त्रिज्या वाले धातु के एक ठोस गोले से 0.2 सेमी व्यास वाला कितने मीटर लम्बा तार बनाया जा सकता है।

  1. 72 m./मी.

  2. 343 m./मी.

  3. 36 m./मी.

  4. 18 m./मी.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume of sphere = (4/3)πr³ = (4/3)π(3)³ = 36π cm³. Volume of wire = πr²h, where r = 0.1 cm (diameter 0.2 cm). Equating volumes: π(0.1)² × h = 36π, so 0.01πh = 36π, giving h = 3600 cm = 36 m. The wire length is 36 meters.

Multiple choice

A circle circumscribe a rectangle with side 15 cm and 8 cm. What is the area of the circle एक वृत्त में एक आयत खींचा गया है, जिसकी भुजाएं 15 सेमी और 8 सेमी है। वृत्त का क्षेत्रफल है।

  1. 48 π sq. cm./वर्ग सेमी.

  2. 100 π sq. cm./वर्ग सेमी.

  3. 72.25 π sq. cm./वर्ग सेमी.

  4. 200 π sq. cm./वर्ग सेमी.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a rectangle is circumscribed by a circle, the diagonal of the rectangle equals the diameter of the circle. Diagonal = √(15² + 8²) = √(225 + 64) = √289 = 17 cm. Radius = diagonal/2 = 8.5 cm. Area of circle = πr² = π(8.5)² = π × 72.25 = 72.25π sq. cm.

Multiple choice

If the slant height of a right pyramid with square base is 4 m and the slant surface area is 12 m2, then the ratio of slant surface area and total surface area is- यदि वर्गाकार आधार वाले एक पिरामिड की तिर्यक ऊँचाई 4 मी. है और इसका तिर्यक पृष्ठ क्षेत्रफल 12 मी2 है तो इसके तिर्यक पृष्ठ क्षेत्रफल और सम्पूर्ण पृष्ठ क्षेत्रफल का अनुपात ज्ञात कीजिए।

  1. 16 : 19

  2. 16 : 17

  3. 17 : 19

  4. 17 : 20
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a square pyramid with slant height 4 m and slant surface area 12 m², we find: slant surface area = (1/2) × perimeter × slant height, so 12 = (1/2) × 4p, giving perimeter p = 6 m. Each side of base = 6/4 = 1.5 m, so base area = 1.5 × 1.5 = 2.25 m². Total surface area = 12 + 2.25 = 14.25 m². Ratio = 12/14.25 = 48/57 = 16/19.

Multiple choice

The area of the biggest circle which can be drawn in side a square with a side of 42 cm. is - 42 सेमी भुजा वाले एक वर्ग में से वह बड़ी से बड़ी वृत्त का क्षेत्रफल क्या है जो उस वर्ग से निकाला जा सके। -

  1. 344.5 cm2

  2. 1386 cm2

  3. 512 cm2

  4. 1024 cm2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Largest circle in a square has diameter equal to side of square. Diameter = 42 cm, so radius = 21 cm. Area = πr² = (22/7) × 21² = 1386 cm².

Multiple choice

A sector containing an angle of 120o is cut off from a circle of radius 21 cm. and folded in to a cone. Find the curved surface area of the cone. 21 सेमी. त्रिज्या के 120o कोण का एक त्रिज्यखण्ड काटा गया और एक शंकु के रूप में मोड़ा गया। तो शंकु का तिर्यक पृष्ठ क्षेत्रफल ज्ञात कीजिए।

  1. 462 cm2./सेमी2

  2. 426 cm2./सेमी2

  3. 246 cm2./सेमी2

  4. 642 cm2./सेमी2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a sector of angle 120° is folded into a cone, the radius of the sector becomes the slant height (l = 21 cm), and the arc length becomes the circumference of the base. Arc length = (120°/360°) × 2π × 21 = 14π cm. This equals 2πr (base circumference), so r = 7 cm. Curved surface area = πrl = π × 7 × 21 = 462 cm².

Multiple choice

The total area of a circle and a square is 2611 cm2. The diameter of the circle is 42 cm. What is the sum of the circumference of the circle and the perimeter of the square (cm)?

  1. 272

  2. 276

  3. 250

  4. 266

  5. 259
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Circle diameter = 42 cm, so radius = 21 cm. Circle area = πr² = π(21)² ≈ 1385.44 cm². Square area = 2611 - 1385.44 ≈ 1225.56 cm². Square side = √1225.56 ≈ 35 cm. Circle circumference = 2πr = 2π(21) ≈ 132 cm. Square perimeter = 4×35 = 140 cm. Sum = 132 + 140 = 272 cm.

Multiple choice

The ratio of radii of two right circular cylinders is 2:3.and their heights are in the ratio 5:4 .find the ratio of their curved surface areas. दो लंब वृत्तीय बेलन की त्रिज्या का अनुपात 2:3 है और उनकी ऊंचाई का अनुपात 5:4 है | उनके वक्र पृष्ठ का अनुपात ज्ञात कीजिए |

  1. 5:6

  2. 3:4

  3. 4:5

  4. 2:3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Curved surface area of cylinder = 2πrh. For two cylinders, CSA1:CSA2 = (2πr1h1):(2πr2h2) = r1h1:r2h2. Given r1:r2 = 2:3 and h1:h2 = 5:4, so CSA1:CSA2 = (2×5):(3×4) = 10:12 = 5:6.

Multiple choice

Find the area of the iron sheet required to prepare a cone 20 cm. high with base radius 21 cm. 20 सेमी. ऊंचे शंकु और 21 सेमी. आधार त्रिज्या वाले शंकु को बनाने के लिए आवश्यक लोहे की चादर का क्षेत्रफल ज्ञात कीजिए।

  1. 2200 cm2/ सेमी2

  2. 3300 cm2/ सेमी2

  3. 2277 cm2/ सेमी2

  4. 3366 cm2/ सेमी2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the area of iron sheet needed, we calculate the total surface area of the cone: πr(l+r) where r=21 cm and l=√(r²+h²)=√(441+400)=29 cm. Using π=22/7: Area = (22/7)×21×(29+21) = 66×50 = 3300 cm². Option A is only the curved surface area, C is incorrect, and D includes an extra factor.

Multiple choice

The length and breadth of a rectangle are increased by 25% and 30% respectively. The increase in the area of the resulting rectangle will be- किसी आयत की लम्बाई और चौड़ाई में क्रमशः 25% और 30% की वृद्धि की जाती है। परिणामी आयत के क्षेत्रफल में वृद्धि होगी-

  1. 55%

  2. 60%

  3. 62.5%

  4. 75%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

New length = 1.25L, new breadth = 1.30B. New area = 1.25 × 1.30 × LB = 1.625 LB. Increase = 62.5% of original area. Alternatively: (25 + 30 + 7.5)% = 62.5%.

Multiple choice

The radius of a wire is decreased to one third. If volume remains the same, length will increase by. किसी तार की त्रिज्या एक-तिहाई कम कर दी जाती है तो लंबाई को कितना बढ़ा दें कि आयतन में कोई परिवर्तन न हो |

  1. 3 times/गुना

  2. 1 times/गुना

  3. 9 times/गुना

  4. 6 times/गुना
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume of a cylinder (wire) = πr²h. If volume remains constant and radius decreases to 1/3, we have: πr²h = π(r/3)² × h'. This gives r²h = (r²/9)h', so h' = 9h. The length increases by 9 times to maintain the same volume when radius is reduced to one-third.

Multiple choice

A cone is 14.4 m. high and the radius of its base is 3.6 m.It is melted and recast into a sphere. The diameter of the sphere is एक शंकु 14.4 मी ऊँचा है और इसके आधार की त्रिज्या 3.6 मी है | इसे पिघलाकर एक गोला बनाया जाता है | गोले का व्यास है

  1. 6 m./मी

  2. 6.4 m./मी

  3. 7 m./मी

  4. 7.2 m./मी
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of cone = (1/3)πr²h = (1/3) × π × 3.6² × 14.4. When melted and recast, volume of sphere = (4/3)πR³. Equating volumes: (1/3) × π × 3.6² × 14.4 = (4/3)πR³. Cancelling (1/3)π: 3.6² × 14.4 = 4R³. 12.96 × 14.4 = 4R³, so 186.624 = 4R³, giving R³ = 46.656 and R = 3.6 m. Diameter = 2R = 7.2 m.