Algebra
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Q.

A saint has a magic pot. He puts one gold ball of radius 1mm daily inside it for 10 days. If the weight of the first ball is 1 g and if the radius of a ball inside the pot doubles every day, how much gold has the saint made due to his magic pot?

 A.

\(\left(\dfrac{2^{30}-69}{7}\right)\) gm

 B.

\(\left(\dfrac{2^{30}+69}{7}\right)\) gm

 C.

\(\left(\dfrac{2^{30}-71}{7}\right)\)gm

 D.

\(\left(\dfrac{2^{30}+71}{7}\right)\)gm

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Solution:
Option(C) is correct

Weight of a solid spherical ball is proportional to the cube of its radius. 

The radius and weight of the 10 balls on the day 10 are tabulated below :

Bull put on day-1 , Radius $=2^9$ , Weight(gm) $=8^9$

Bull put on day-2 , Radius $=2^8$ , Weight(gm) $=8^8$

Bull put on day-3 , Radius $=2^7$ , Weight(gm) $=8^7$
...
Bull put on day-9 , Radius $=2^1$ , Weight(gm) $=8^1$

Bull put on day-10 , Radius $=2^0$ , Weight(gm) $=8^0$

The weight of the 10 balls before they were put in the pot = 10 gm.

The weight of the gold made by saint (gm):

\(=\left(\dfrac{2^{30}}{7}\right)-10\)

\(=\left(\dfrac{2^{30}-71}{7}\right)\)


(1) Comment(s)


Pranshu
 ()

could you pls explain this in detail how is the radius becoming $2^9$ and decreasing every day instead of increasing....and radius should be $2^3$ on first day....pls explain.thanku