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Volume of original cone

CAT 2020 Slot 1 · Quantitative Ability · Easy · Geometry

This is an easy Quantitative Ability question from the CAT 2020 Slot 1 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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A solid right circular cone of height 27 cm is cut into 2 pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in the volume of the two pieces is 225 cc, the volume, in cc, of the original cone is

  1. A.

    264

  2. B.

    232

  3. C.

    243

  4. D.

    256

Answer

C

Explanation

Total height of cone H=27H = 27 cm. Height of the cut from base =18= 18 cm, so height of the smaller top cone h=27−18=9h = 27 - 18 = 9 cm.

Ratio of heights hH=927=13\frac{h}{H} = \frac{9}{27} = \frac{1}{3}.

Since volume is proportional to the cube of height: VtopVtotal=(13)3=127\frac{V_{\text{top}}}{V_{\text{total}}} = \left(\frac{1}{3}\right)^3 = \frac{1}{27}

So Vtop=127VV_{\text{top}} = \frac{1}{27}V and Vfrustum=V−127V=2627VV_{\text{frustum}} = V - \frac{1}{27}V = \frac{26}{27}V.

Difference in volume between the two pieces: Vfrustum−Vtop=2627V−127V=2527V=225V_{\text{frustum}} - V_{\text{top}} = \frac{26}{27}V - \frac{1}{27}V = \frac{25}{27}V = 225 V=225×2725=9×27=243 ccV = 225 \times \frac{27}{25} = 9 \times 27 = 243\text{ cc}

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