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28 Dec 2019, 01:30
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E
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Difficulty:
95%
(hard)
Question Stats:
41% (02:34) correct
59%
(02:28)
wrong
based on 177
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Not Attempted Yet
The value of a bar of gold weighing 400 gm, varies inversely with the square root of the fraction of impurities in it. If the value of a bar containing 25 gms of impurities is Rs. 45000, how many gms of pure gold is there in a bar costing Rs. 90000?
A. 6.25
B. 12.5
C. 375
D. 387.5
E. 393.75
A. 6.25
B. 12.5
C. 375
D. 387.5
E. 393.75
28 Dec 2019, 16:00
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\(\frac{V_1}{V_2}= \sqrt{\frac{I_2}{I_1}}\)
\(45000/90000\)= \(\sqrt{\frac{I_2}{25}}\)
\(I_2=6.25\)
pure gold in a bar costing Rs. 90000= 400-6.25= 393.75
\(45000/90000\)= \(\sqrt{\frac{I_2}{25}}\)
\(I_2=6.25\)
pure gold in a bar costing Rs. 90000= 400-6.25= 393.75
Bunuel
01 Jan 2020, 12:26
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Bunuel
\(4500 = \frac{k}{\sqrt25}\) so k= 225000
\(90000 = \frac{225000}{\sqrt m}\) where m is the amount of impurities.
\(\sqrt m = \frac{225000}{90000} =\frac{5}{2}\hspace{2mm} or \hspace{2mm} m=\frac{25}{4}=6.25\)
Therefore amount of impurities in 400 gms of gold is 6.25 gms hence pure gold is 400-6.25 = 393.75
Ans- E
Hope it's clear.
