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Re: If a > 0, is t^a > w^a ? [#permalink]
06 Sep 2015, 23:32
This is a typical GMAT type question that you can expect...
For such questions remember to consider: 1. Values <-1 2. -1 < Values < 0 3. 0 < Values < 1 4. 1 < Values [color=#0000ff]And you will never get such questions wrong !![/color]
Option1: The values differ when T,W are +VE and when they are -VE
Option 2: This is a TRAP. All those who will substitute value of 't' and and cancel out the terms on LHS & RHS - WILL GET THIS WRONG. You can cancel the terms ONLY if they are positive.
Combinig 1 & 2, We are left with only +ve values. For which the given equation is always consistent. Hence C.
Re: If a > 0, is t^a > w^a ? [#permalink]
07 Sep 2015, 01:48
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Following property tested:
If x>0, a>b>0, then a^x>b^x
In this question : a = x, t= a, w=b
S1 - No idea about the sigh of t & w - Insufficient
S2 - Again no idea about the sigh of t & w - Insufficient
Combining - (and using another concept - NEVER cancel the variables on both sides of equality/ inequality) the statements - we know t>w and w>0 so, C is the answer. _________________
Re: If a > 0, is t^a > w^a ? [#permalink]
07 Sep 2015, 03:08
Solution:
Statement1 : t > w. We dont know if the numbers are positive or negative . Anyhow lets check both the cases. If, t = 4 and w = 2. Since a>0, here \(t^{a} > w^{a}\). But, if t = -2, w = -4. Then t > w.Then for a=3, \(t^{a} > w^{a}\) and for a=2, \(t^{a} < w^{a}\). Therefore, Insufficient.
Statement2 : t=2w. If t = -4, w = -2. Then t = 2w. Then for a=3, \(t^{a} < w^{a}\) and for a=2, \(t^{a} > w^{a}\). Therefore, Insufficient.
Combined : We know that t and w can only be positive. Therefore, \(t^{a} > w^{a}\) holds for all possible values. Sufficient.
If a > 0, is t^a > w^a ? [#permalink]
07 Sep 2015, 06:22
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Expert's post
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.
If a > 0, is t^a > w^a ?
(1) t > w (2) t = 2w
In the original condition there are 3 variables (a,t,w), and 1 equation (a>0) thus in order to match the number of variables and equations we need 2 equations more. Since there is 1 each i 1) and 2), C is likely the answer. In actual calculation, using both 1) & 2) we get t=2w>w--> w>0, t>0 thus the answer is yes and the conditions are sufficient. therefore the answer is C _________________
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