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Expert replies
by logitech » Fri Nov 28, 2008 1:18 am
If t and u are positive integers, what is the value of t^(-2)u^(-3) ?

(1) t^(-3)u^(-2) = 1/36

(2) t(u^(-1)) = 1/6
Last edited by logitech on Fri Nov 28, 2008 10:39 am, edited 1 time in total.
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Source: — Data Sufficiency |

by jimmiejaz » Fri Nov 28, 2008 4:57 am
IMO ans is E.

given t and u are both positive integers, there sum or differences can never be fractions.
both equations are just given to confuse.
I hope am correct.
What if i have not yet beat the beast, I know i will beat it!!!!!!!!
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by sudhir3127 » Fri Nov 28, 2008 5:08 am
jimmiejaz wrote:IMO ans is E.

given t and u are both positive integers, there sum or differences can never be fractions.
both equations are just given to confuse.
I hope am correct.
i go with E as well.. As the question says "INTEGERS" there is no way that their addition/subtraction will result in a Non integer..
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by logitech » Fri Nov 28, 2008 10:40 am
I do apologize for the typo. I corrected the question.
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by cramya » Fri Nov 28, 2008 10:58 am
I am getting C)

VALUE = 1/6*1/36

Made a mistake. THinking more about it only values possible for t and u ar 1 and 6 hence A)
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by logitech » Fri Nov 28, 2008 11:52 am
cramya wrote:I am getting C)

VALUE = 1/6*1/36

Made a mistake. THinking more about it only values possible for t and u ar 1 and 6 hence A)
That was the trap Cramya. I am glad you are still alive!! ;-)
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by vittalgmat » Fri Nov 28, 2008 1:49 pm
I get C.

Here is my approach.

Stmt 1:
1/ (t^3 u^2) = 1/36
insufficient coz we need to know 1 / (t^2 u^3)

Stmt 2:
t/u = 1/6.
Insufficient.

Combining:

Multiply stmt 1 by stmt 2.

we get 1/ (t^3 u^2) * t/u = 1/36 * 1/6
1/(t^2 u^3) = 1/6^3 a definite answer
So C is sufficient.

What is the OA.

thanks
-V
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by cramya » Fri Nov 28, 2008 1:57 pm
Vittal,
From Logitech' reply it looks like the OA is A)
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by vittalgmat » Fri Nov 28, 2008 2:06 pm
Damn!!! Fell into the abyss!!!!!.. someone save me!!!! :(
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Re: TUTUTUTUTUT

by Stuart@KaplanGMAT » Fri Nov 28, 2008 3:20 pm
logitech wrote:If t and u are positive integers, what is the value of t^(-2)u^(-3) ?

(1) t^(-3)u^(-2) = 1/36

(2) t(u^(-1)) = 1/6
(1) let's start by making this less complicated.

t^(-3) is just 1/t^3... u^(-2) is just 1/(u^2).. so, we have:

(1/t^3)(1/u^2) = 1/36

taking the reciprocal of both sides:

(t^3)(u^2) = 36

Since we know that t and u are positive integers, we should be suspicious that there are limited possibilities. Let's break 36 down to primes:

36 = 2*2*3*3

Well, there's no way we're getting an integer cubed by using primes. Therefore, the only possible value for t is 1. Once we know that t has to be 1, u must be 6.

Since we know the values of t and u, we can answer any question about them: sufficient.

(2) simplifying, we get:

t/u = 1/6

Well, there's an infinite number of positive integers (e.g 1/6, 2/12, 3/18, ...) that fit that ratio and, since t and u appear in a different ratio in the question, each set of numbers will give us a different answer: insufficient.

(1) is sufficient, (2) isn't: choose (A).
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by cramya » Fri Nov 28, 2008 5:17 pm
Thanks Stuart!
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by vittalgmat » Fri Nov 28, 2008 5:59 pm
Excellent explanation Stuart!!!
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