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08 Dec 2020, 03:15
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A
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A wizard has a magic pot. In the morning of each day for 10 days, the wizard puts one gold spherical ball of radius 1mm and weighing 1 gram inside the pot. If the radius of a ball inside the pot doubles every 24 hours, how many grams of gold did the wizard gained by the end of the 10th day? (The volume of a sphere = \(\frac{4}{3}\pi r^3\))
A. \(\frac{2^{30}−71}{7}\) gm
B. \(\frac{2^{30}−69}{7}\) gm
C. \(\frac{2^{30}−1}{7}\) gm
D. \(\frac{2^{30}+69}{7}\) gm
E. \(\frac{2^{30}+71}{7}\) gm
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{2^{30}−71}{7}\) gm
B. \(\frac{2^{30}−69}{7}\) gm
C. \(\frac{2^{30}−1}{7}\) gm
D. \(\frac{2^{30}+69}{7}\) gm
E. \(\frac{2^{30}+71}{7}\) gm
Are You Up For the Challenge: 700 Level Questions
Updated on: 14 Dec 2020, 00:24
Originally posted by CrackverbalGMAT on 08 Dec 2020, 04:31.
Last edited by CrackverbalGMAT on 14 Dec 2020, 00:24, edited 1 time in total.
Last edited by CrackverbalGMAT on 14 Dec 2020, 00:24, edited 1 time in total.
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Bunuel
We do not need to worry about the constant (4/3) \(\pi\). We can take it as a constant k
When the radius is 1mm, the volume is 1 \(mm^3\) and the weight per 1 \(mm^3\) is 1 gm
So the sum of the volumes for the 10 days = Weight of the ball
The radius of the ball for the 10 days = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512
The volumes = 1, 8, 64, 512.... = 1 + [ \(2^3 + 2^6 + 2^9 + .... for \space 9 \space terms\)]
The series (in square brackets) is a GP, where \(a\) (1st term) = \(2^3 = 8\) and r (common ratio) = \(\frac{2^6}{2^3} = \frac{2^9}{2^6} = 8\) and n = 9
The sum of n terms of a GP, where r > 1 = \(\frac{a(r^n - 1)}{r - 1} = \frac{8(8^9 - 1)}{8 - 1}= \frac{8((2^3)^9 - 1)}{7} = \frac{(8 * 2^{27}) - 8}{7} = \frac{(2^3 * 2^{27}) - 8}{7} = \frac{2^{30} - 8}{7}\)
The total Sum = 1 + \(\frac{2^{30} - 8}{7} = \frac{7 + 2^{30} - 8}{7} = \frac{2^{30} - 1}{7}\)
The gain = \(\frac{2^{30} - 1}{7} - 10 = \frac{2^{30} - 71}{7}\)
Option A
Arun Kumar
