# CGL & CPO QUESTIONS ON FRUSTUM

#### QUESTIONS ON FRUSTUM

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Frustum of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

4. Grade-II DASS Exam conducted by Delhi Staff Subordinate Services (DSSSB)

#### QUERY 1

**Plz tell me Lateral Side Area and total surface area of the frustum.**

#### QUERY 2

**The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum of the cone is 44 cm³. The radius of the upper circular surface of the frustum is?**

A) 13

B) ∛13

C) ∛27

D) 27

**AMIT JHA**

Volume of the entire cone = (1/3)π3²*9 = 27π

Volume of the Frustum = 44 cm³ = 14*22/7 = 14π (given)

We know that if a cone is cut by a plane parallel to its base; we get two part i.e. the upper part is a cone itself in shape and the lower part is a frustum (like a bucket).

Thus volume of the upper cone = 27π – 14π= 13π

Heights and radii of both the cones got in this way are always proportional;

so letting h & h1 and r & r1 being the heights and radii of the new and the original cones respectively h : h1 = r : r1

Here r1 = 3 cm and h1 = 9 cm (given)

So h : 9 = r : 3

=> h = 3r

Now Volume of the upper cone = (1/3)πr²h = 13π (shown above)

=> r²*3r = 39 (by putting h = 3r and solving)

=> r³ = 13

=> r = ∛13 (option ‘B’)

#### QUERY 3

**If the radii of the circular ends of a conical bucket are 28 cm and 7 cm, whose height is 45 cm. Find the capacity of the bucket.**

A) 48510 cm³

B) 84510 cm³

C) 45005 cm³

D) 48000 cm³

**MAHA GUPTA
**You see, the bucket forms a frustum of a cone such that the radii of its circular ends are R = 28 cm, r = 7 cm and height is = 45 cm.

Therefore, capacity of the bucket (volume of the frustum)

= ^{1}⁄_{3}πh (R² + Rr + r²)

= ^{1}⁄_{3 }π*45(28² + 7² + 28 *7)

= 22*15*(28*4 + 7 + 28)

= 330*147 cm³

= 48510 cm³ (option ‘A’)

#### QUERY 4

**The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm respectively. Find the lateral surface area and total surface area of the frustum.**

A) 620.57 cm², 1300 cm²

B) 628.57 cm², 1357.71 cm²

C) 528.56 cm², 1300.75 cm²

D) 600.50 cm², 1400.15 cm²

**MAHA GUPTA
**Here, R = 14 cm, r = 6 cm and h = 6 cm

Now, let l be the slant height of the frustum;

then, l = √[(R – r)² + h²]

=> l = √(64 + 36)

=> l = 10 cm

Now, lateral surface area of the frustum = π(R + r)l

= π(14 + 6)10

= 628.57 cm² (option ‘B’)

And, total surface area of the frustum = π[(R + r)l + R² + r²]

= π(200 + 196 + 36)

= π432

= 1357.71 cm² (option ‘B’)

#### QUERY 5

**A friction clutch is in the form of a frustum of a cone, the diameters of the ends being 32 cm and 20 cm and length 8 cm. Find its bearing surface and volume.**

A) 817.14 cm², 4324.57 cm³

B) 917.14 cm², 4342.57 cm³

C) 800.24 cm², 4224.75 cm³

D) 850.24 cm², 4304.75 cm³

**MAHA GUPTA
**

Let ABB’A’ be the friction clutch of slant height l cm.

We have, R = 16 cm, r = 10 cm and h = 8 cm.

Therefore, l² = h² + (R – r)²

=> l² = 64 + (16 – 10)²

=> l = 10 cm

Bearing surface of the clutch (lateral surface of the frustum)

= π(R + r)l

= π(16 + 10)10

= 817.14 cm² (option ‘A’)

Now, the volume = ^{1}⁄_{3}π(R² + Rr + r²)h

= ^{1}⁄_{3}π*8(16² + 16*10 + 10²)

= ^{1}⁄_{3}π*8(256 + 160 + 100)

= 4324.57 cm³ (option ‘A’)

#### QUERY 6

**A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.**

A) 12 cm

B) 10 cm

C) 15 cm

D) 14 cm

**MAHA GUPTA
**Let the height of the bucket be h cm

We have, R = 28 cm, r = 21 cm and Volume of the bucket = 28.490 litres = 28.490 × 1000 cm³ = 28490 cm³

V (volume) = 28490 cm³

=> ^{1}⁄_{3}π(R² + Rr + r²)h = 28490

=> ^{22}⁄_{21 }× h(784 +588 + 441) = 28490

=> h = 15 cm (option ‘C’)