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In the formula F
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01 Aug 2021, 21:15
01 Aug 2021, 21:15
Expert Reply
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Question Stats:
85% (00:43) correct
15% (00:38) wrong based on 40 sessions
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In the formula \(F=\frac{1}{(3k)^3}\), if k is halved , the F is multiplied by
A. 27
B. 8
C. 1
D. 1/8
E. 1/27
A. 27
B. 8
C. 1
D. 1/8
E. 1/27
In the formula F
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01 Aug 2021, 23:08
01 Aug 2021, 23:08
2
Solution:
Substituting K= 1, will help us to easily solve this question
F= \(\frac{1}{{3k}^3}\)=\(\frac{1}{27}\)
Now, when K is halved, 1 will become 1/2
Thus, we get \(\frac{1}{{(3*1/2)}^3}\)=\(\frac{ 1}{27/8}\)= \(\frac{8}{27}\)
Thus for the original value of F this value has 8 in the numerator, which we get if we multiply F by 8
i.e. 8F\(= \frac{1}{{(3K/2)}^3}\)
IMO B
Hope this helps!
Substituting K= 1, will help us to easily solve this question
F= \(\frac{1}{{3k}^3}\)=\(\frac{1}{27}\)
Now, when K is halved, 1 will become 1/2
Thus, we get \(\frac{1}{{(3*1/2)}^3}\)=\(\frac{ 1}{27/8}\)= \(\frac{8}{27}\)
Thus for the original value of F this value has 8 in the numerator, which we get if we multiply F by 8
i.e. 8F\(= \frac{1}{{(3K/2)}^3}\)
IMO B
Hope this helps!
gmatclubot
In the formula F [#permalink]
01 Aug 2021, 23:08
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