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CAT 2018 Slot 1 — QA Question 20

Basics of Mensuration/PrismEasy

A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is

Answer & solution

Correct answer: 198

Answer: 198

Solution

Easy

The remaining solid is the original cone minus the small cone sliced off the top. Use similar triangles to get the small cone's base radius, then subtract volumes.

A r=1 R=4 3 ft 9 ft
1

Dimensions. Original cone: height 1212, base radius 82=4\tfrac{8}{2}=4. The cut is 99 ft from the base, so the small top cone has height 12−9=312-9=3.

big cone: h=12, R=4small cone: h′=3\begin{aligned} &\text{big cone: } h=12,\ R=4\\ &\text{small cone: } h'=3 \end{aligned}
2

Small cone's radius by similarity. The small cone is similar to the big one (cut parallel to the base):

r′R=h′h=312=14⇒ r′=14×4=1\begin{aligned} &\frac{r'}{R} = \frac{h'}{h} = \frac{3}{12} = \frac14\\ &\Rightarrow\ r' = \tfrac14\times 4 = 1 \end{aligned}
3

Subtract the volumes. Remaining volume == big cone −- small cone, with π=227\pi=\tfrac{22}{7}:

V=13πR2h−13πr′2h′⇒ V=13π(42⋅12−12⋅3)(factor 13π)⇒ V=13π(192−3)=13π⋅189⇒ V=13⋅227⋅189=227⋅63=22⋅9=198\begin{aligned} &V = \tfrac13\pi R^2 h - \tfrac13\pi r'^2 h'\\ &\Rightarrow\ V = \tfrac13\pi\big(4^2\cdot 12 - 1^2\cdot 3\big) \quad\text{(factor }\tfrac13\pi)\\ &\Rightarrow\ V = \tfrac13\pi(192-3) = \tfrac13\pi\cdot 189\\ &\Rightarrow\ V = \tfrac13\cdot\tfrac{22}{7}\cdot 189 = \tfrac{22}{7}\cdot 63 = 22\cdot 9 = 198 \end{aligned}
Remaining volume=198 cubic ft\text{Remaining volume} = \mathbf{198}\ \text{cubic ft}

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CAT 2018 Slot 1 QA Q20: A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut — Solution | TheCATExam