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A
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Difficulty:
75%
(hard)
Question Stats:
49% (01:56) correct
51%
(02:00)
wrong
based on 470
sessions
History
Date
Time
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Not Attempted Yet
If t and u are positive integers, what is the value of \(t^{-3}*u^{-2}\)?
(1) \(t^{-2}*u^{-3} = \frac{1}{36}\).
(2) \(u*(t^{-1}) = \frac{1}{6}\).
(1) \(t^{-2}*u^{-3} = \frac{1}{36}\).
(2) \(u*(t^{-1}) = \frac{1}{6}\).
24 Aug 2012, 10:42
Kudos
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The question should read:
If t and u are positive integers, what is the value of \(t^{-3}*u^{-2}\)?
(1) \(t^{-2}*u^{-3} = \frac{1}{36}\) --> \(\frac{1}{t^{2}*u^{3}}=\frac{1}{36}\) --> \(t^{2}*u^{3}=36\) --> since \(t\) and \(u\) are positive integers, then only possible case is \(t^{2}*u^{3}=6^2*1^3\) (\(u\) cannot be any other positive integer but 1, since 36 doesn't have a prime factor in power of 3) --> \(t=6\) and \(u=1\). Sufficient.
(2) \(u*(t^{-1}) = \frac{1}{6}\) --> \(\frac{u}{t}=\frac{1}{6}\) --> infinite number of values are possible for \(u\) and \(t\) (1, and 6, 2 and 12, 3 and 18, ...), thus infinite number of values are possible for \(t^{-2}*u^{-3}\). Not sufficient.
Answer: A.
Hope it's clear.
If t and u are positive integers, what is the value of \(t^{-3}*u^{-2}\)?
(1) \(t^{-2}*u^{-3} = \frac{1}{36}\) --> \(\frac{1}{t^{2}*u^{3}}=\frac{1}{36}\) --> \(t^{2}*u^{3}=36\) --> since \(t\) and \(u\) are positive integers, then only possible case is \(t^{2}*u^{3}=6^2*1^3\) (\(u\) cannot be any other positive integer but 1, since 36 doesn't have a prime factor in power of 3) --> \(t=6\) and \(u=1\). Sufficient.
(2) \(u*(t^{-1}) = \frac{1}{6}\) --> \(\frac{u}{t}=\frac{1}{6}\) --> infinite number of values are possible for \(u\) and \(t\) (1, and 6, 2 and 12, 3 and 18, ...), thus infinite number of values are possible for \(t^{-2}*u^{-3}\). Not sufficient.
Answer: A.
Hope it's clear.
General Discussion
25 Aug 2012, 00:34
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Bunuel
You have corrected the error. Thanks once again.
The question now makes sense.
In the original question the answer I got was C
PS: I have edited the question again, as question stem and A have become same.
