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

A large solid metallic cylinder whose radius and height are equal to each other is to be melted and 48 identical solid balls so formed from the obtained liquid metal. What is the ratio of the radius of a ball to the radius of the cylinder? एक बड़े धात्विक बेलन जिसकी त्रिज्या और ऊँचाई एक दूसरे के बराबर है, को गलाया जाता है और प्राप्त द्रव धातु से 48 एक समान ठोस गेंदे बनायी जाती है। गेंद की त्रिज्या का बेलन की त्रिज्या से अनुपात क्या है?

  1. 4 : 1

  2. 1 : 4

  3. 8 : 1

  4. 1 : 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let cylinder radius = R, ball radius = r. Cylinder volume = πR²h = πR³ (since R = h). 48 balls volume = 48 × (4/3)πr³. Equating: πR³ = 48 × (4/3)πr³, giving R³ = 64r³, so R/r = 4. Ratio of ball radius to cylinder radius = r:R = 1:4, which equals 4:1 when expressed as ball:cylinder ratio.

Multiple choice

The base of a right pyramid is a square of a side 16 m long. If its height be 15 m , then the area of the lateral surface is ? एक समकोणीय पिरामिड जिसका आधार एक वर्ग हैं जिसकी भुजा 16 मी हैं| यदि इसकी ऊँचाई 15 मी हो तो इसका तिर्यक क्षेत्रफल ज्ञात करें ?

  1. 520

  2. 544

  3. 540

  4. 534
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Square base side = 16 m, height = 15 m. Slant height of each triangular face = √(8² + 15²) = 17 m (apothem of base = 8). Lateral surface area = 4 × area of one triangular face = 4 × (1/2 × 16 × 17) = 544 m².

Multiple choice

If the slant height of a right pyramid with square base is 4m and the perimeter of the base of the pyramid is 12m, then the ratio of total slant surface and area of the base is? एक समकोणीय पिरामिड की तिर्यक उचाई जिसका आधार एक वर्ग हैं 4 मी हैं और इसके आधार का परिमाप 12 मी है , तो इसके कुल तिर्यक प्रष्ट क्षेत्रफल और आधार के क्षेत्रफल का अनुपात ज्ञात कीजिये ?

  1. 4 : 3

  2. 7 : 4

  3. 8 : 5

  4. 8 : 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a right pyramid with square base, perimeter 12m means each side is 3m, so base area = 9 m². Slant surface area = 4 × (1/2 × 3 × 4) = 24 m² (four triangular faces). Ratio = 24:9 = 8:3. Option D is correct.

Multiple choice

Inside a square plot, a circular garden is developed which exactly fits in the square plot and the diameter of the garden is equal to the side of the square plot which is 28 meter. What is the area of the space left out in the square plot after developing the garden ?
  1. $98$
  2. $146$
  3. $84$
  4. $168$
  5. $154$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

वर्गाकार प्लॉट का क्षेत्रफल = 28 × 28 = 784 वर्ग मीटर। वृत्ताकार बगीचे का व्यास = 28 मीटर, अतः त्रिज्या = 14 मीटर। वृत्त का क्षेत्रफल = πr² = (22/7) × 14² = (22/7) × 196 = 616 वर्ग मीटर। शेष स्थान = वर्ग का क्षेत्रफल - वृत्त का क्षेत्रफल = 784 - 616 = 168 वर्ग मीटर। यह समस्या वृत्त और वर्ग के क्षेत्रफलों के अंतर की गणना मांगती है।

Multiple choice

The radii of concentric circles are 42 cm. and 35 cm. What will the area between these circles./संकेन्द्रीय वृत्तों की त्रिज्यायें 42 सेमी. और 35 सेमी. है। इन वृत्तों के बीच वाले स्थान का क्षेत्रफल क्या होगा ?

  1. 3850 cm2/सेमी2

  2. 1694 cm2/सेमी2

  3. 1925 cm2/सेमी2

  4. 1500 cm2/सेमी2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area between concentric circles = π(R² - r²) where R = 42 cm (outer radius) and r = 35 cm (inner radius). Area = π(42² - 35²) = π(1764 - 1225) = π(539) = 1693.46 ≈ 1694 cm². Option A (3850) is close to π×35² which would be the inner circle area. Option C (1925) is close to π×42² which would be the outer circle area. Option D appears to be a miscalculation.

Multiple choice

ABCD is a square inscribed in a circle of radius 14 m. Then the total area (in square units) of the portions of the circle lying outside the square is-/14 मी. त्रिज्या के वृत्त में ABCD एक वर्ग खींचा गया है। वर्ग के बाहर स्थित वृत्त के क्षेत्रों का कुल क्षेत्रफल है।

  1. 220 m2/मी2

  2. 224 m2/मी2

  3. 225 m2/मी2

  4. 230 m2/मी2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When a square is inscribed in a circle, the diagonal of the square equals the diameter of the circle. Circle radius = 14 m, so diameter = 28 m = diagonal of square. Side of square = diagonal/√2 = 28/√2 = 14√2 m. Area of square = (14√2)² = 392 m². Area of circle = π×14² = 196π ≈ 616 m². Area outside square = 616 - 392 = 224 m². Option A (220 m²) and D (230 m²) are common approximation errors. Option C (225 m²) is unrelated.

Multiple choice

Volume of a metallic cylindrical pipe is 748 m3. Its internal radius is 8 m and thickness is 1 m. Find the length of the cylinder. (use ) एक धात्विक बेलनाकार पाइप का आयतन 748 मी.3 है। यदि आन्तरिक त्रिज्या 8 मी. है और मोटाई 1 मी. है। बेलन की लम्बाई ज्ञात कीजिए।

  1. 16 m/मी

  2. 15 m/मी

  3. 14 m/मी

  4. 12 m/मी
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a cylindrical pipe, volume = π(R² - r²)×h where R = outer radius = 8+1 = 9 m, r = inner radius = 8 m, and h = length. Cross-sectional area = π(9² - 8²) = π(81-64) = 17π. Given volume = 748 m³, so h = 748/(17π) = 748/53.41 ≈ 14 m. Option A (16 m) would require different radius values. Option D (12 m) is a common miscalculation error.

Multiple choice

Let each side of a square is 20cm. Four equal circle each of radius 10cm, are drawn about the four corners of the square so that each touches two of the others. Find the area enclosed between the circumferences of the circles. माना एक वर्ग की प्रत्येक भुजा 20 सेमी है। 10 सेमी त्रिज्या वाले चार बराबर वृत्त वर्ग के चारों कोनो पर इस प्रकार खीचे जाते हैं कि प्रत्येक दो अन्य को स्पर्श करता है। वृत्तों की परिधि से बीच के घिरे हुए क्षेत्र का क्षेत्रफल क्या है ?

  1. 86 cm2/ सेमी2

  2. 314 cm2/ सेमी2

  3. None of these / इनमें से कोई नही

  4. 78 cm2/ सेमी2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When four circles of radius 10cm are drawn at the corners of a 20cm square, each circle is centered at a corner. The area enclosed between the circles equals the square's area minus the area of four quarter-circles (which equals one full circle). Square area = 400 cm², circle area = π(10)² ≈ 314 cm². Enclosed area = 400 - 314 = 86 cm².

Multiple choice

A figure is formed by revolving a rectangular sheet of dimensions 7 cm × 4 cm about its length. What is the volume of the figure thus formed? 7 सेमी. × 4 सेमी विमा वाली एक आयताकार चादर को अपनी लम्बाई के परितः परिक्रमण कराने से एक आकृति बनती है। इस प्रकार निर्मित आकृति का आयतन क्या है?

  1. 352 cm2/सेमी2

  2. 296 cm2/सेमी2

  3. 176 cm2/सेमी2

  4. 616 cm2/सेमी2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a rectangle of dimensions 7cm × 4cm is revolved about its length (7cm), it forms a cylinder with height = 7cm and radius = 4cm (the width becomes the radius). Volume = πr²h = π(4)²(7) = 112π cm³. Using π = 22/7: 112 × 22/7 = 16 × 22 = 352 cm³. Note: Option A shows cm² but should be cm³ - this is a unit error.

Multiple choice

Three circle of radius 3.5 cm are placed in such away that each circle touches the other two the area of the portion enclosed by the circle is- 3.5 सेमी. त्रिज्या के वृत्तों को इस प्रकार रखा गया है, कि प्रत्येक वृत्त दूसरे दो वृत्तों को स्पर्श करता है। वृत्तों के द्वारा बन्द भाग का क्षेत्रफल है-

  1. 1 cm3/सेमी3

  2. 2 cm3/सेमी3

  3. 3 cm3/सेमी3

  4. 4 cm3/सेमी3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Three circles of radius 3.5 cm touch each other. Their centers form an equilateral triangle of side 7 cm. Area enclosed = area of equilateral triangle - 3 × area of 60° sector. Triangle area = (√3/4) × 7² ≈ 21 cm². Each sector area = (60/360) × π × 3.5² ≈ 6.4 cm². Enclosed area = 21 - 3 × 6.4 ≈ 2 cm². Note: units should be cm², not cm³ as written.

Multiple choice

The base of a right pyramid is a square of side 12 m. long. If its height be 800 cm. then the area of the lateral surface in square metre is- एक समकोण पिरामिड का आधार 12 मी. लम्बी भुजा वाला एक वर्ग है। यदि इसकी ऊँचाई 800 सेमी. है तो पार्श्विक पृष्ठ का क्षेत्रफल कितने वर्ग मीटर है?

  1. 800

  2. 240

  3. 136

  4. 308
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a right pyramid with square base, lateral surface area = perimeter of base × slant height / 2. Slant height l = √(h² + (a/2)²) = √(8² + 6²) = 10m. Lateral area = (4 × 12) × 10 / 2 = 240 m². Option B is correct.

Multiple choice

A metal pipe of negligible thickness has radius 28 cm and length 161 cm . The outer curved surface area of the pipe in square cm is - एक नगण्य मोटाई का धातु का पाइप जिसकी त्रिज्या 28 सेमी. और लम्बाई 161 सेमी. है। पाइप का बाहरी वक्रपृष्ठ का क्षेत्रफल वर्ग सेमी. में है-

  1. 24680 sq.cm./ वर्ग मी.

  2. 18384 sq.cm. / वर्ग मी.

  3. 32464 sq.cm./ वर्ग मी.

  4. 28336 sq.cm./ वर्ग मी.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Outer curved surface area of a pipe (cylinder) = 2πrh, where r = 28 cm, h = 161 cm. Area = 2 × (22/7) × 28 × 161 = 2 × 22 × 4 × 161 = 176 × 161 = 28336 sq. cm. Option D is correct.

Multiple choice

What is the area of the copper sheet required to prepare a cone of base radius 30 cm with the height 40cm? एक 30 सेमी आधार त्रिज्या और 40 सेमी ऊँचाई वाले शंकु को बनाने के लिए आवश्यक कापर की चादर का क्षेत्रफल ज्ञात कीजिए?

  1. 7543 cm2/सेमी2

  2. 5146 cm2/सेमी2

  3. 5432 cm2/सेमी2

  4. 7246 cm2/सेमी2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a cone with r = 30 cm and h = 40 cm, slant height l = √(r² + h²) = √(900 + 1600) = 50 cm. Total surface area (sheet needed) = πr(l + r) = π × 30 × 80 = 2400π. Using π = 22/7: 2400 × 22/7 = 7542.85 ≈ 7543 cm².

Multiple choice

If the base radius and the height of a right circular cone are increased, 40% then find the net % increase in volume. यदि शंकु के आधार की त्रिज्या तथा ऊँचाई 40% बढ़ जाए तो उसके आयतन में कितने: की वृद्धि हो जायेगी।

  1. 174.4 %

  2. 120.8 %

  3. 54 %

  4. 64.8 %
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Original volume = (1/3)πr²h. New radius = 1.4r, new height = 1.4h. New volume = (1/3)π(1.4r)²(1.4h) = (1/3)πr²h × 1.4³ = V × 2.744. Increase = 2.744 - 1 = 1.744 = 174.4%. Volume scales with the product of dimension changes.

Multiple choice

If a right circular cone of height 24 cm has a volume of 1232 cm3, then the area of its curved surface (Take π = 22/7 ) is यदि एक लम्बवृत्तीय शंकु की ऊँचाई 24 सेमी. तथा आयतन 1232 घन सेमी है तो उसके वक्रपृष्ठ का क्षेत्रफल है- (लिया गया है π = 22/7)

  1. 1254 cm2 /सेमी2

  2. 704 cm2 /सेमी2

  3. 550 cm2 /सेमी2

  4. 154 cm2 /सेमी2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Cone volume = (1/3)πr²h = 1232. With h = 24, π = 22/7: (1/3)(22/7)r²(24) = 1232. Solving: r² = 49, r = 7. Slant height l = √(h² + r²) = √(576 + 49) = 25. Curved surface = πrl = (22/7) × 7 × 25 = 550 cm².