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1. A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical ball. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is :

A. 2.5 cm

B. 2.66 cm

C. 3 cm

D. 3.5 cm

Solution:

Let the radius of the third ball be R cm
Then,

\frac{4}{3} π \times {(\frac{3}{4})}^{3} + \frac{4}{3} π \times {(1)}^{3}

+ \frac{4}{3} π \times R^{3}

= \frac{4}{3} π \times {(\frac{3}{2})}^{3}

\begin{aligned} \Rightarrow \frac{27}{64} + 1 + R^{3} = \frac{27}{8} \\ \Rightarrow R^{3} = \frac{125}{64} = \frac{{(5)}^{3}}{{(4)}^{3}} \\ \Rightarrow R = \frac{5}{4} \end{aligned}

∴ Diameter of the third ball :

= 2 R = \frac{5}{2} c m = 2.5 c m

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