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If n ≠ 1, is (2xn - 2x)/(n-1) > y ?

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macjas
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If n ≠ 1, is (2xn - 2x)/(n-1) > y ? [#permalink] New post   27 May 2012, 09:35
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If n ≠ 1, is (2xn - 2x)/(n-1) > y ?

(1) x > y/2
(2) n = 3
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A
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narangvaibhav
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Re: If n ≠ 1, is (2xn - 2x)/(n-1) > y ? [#permalink] New post   28 May 2012, 00:44
macjas wrote:
If n ≠ 1, is (2xn - 2x)/(n-1) > y ?

1) x > y/2

2) n = 3

easy Q
take 2x out=>
Is 2x(n-1)/(n-1)> y? or is 2x>y
statement 1 clearly says 2x>y hece sufficient
statement 2 says about n not x or y hence not sufficient
Thus Ans: A
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Bunuel
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Re: If n ≠ 1, is (2xn - 2x)/(n-1) > y ? [#permalink] New post   28 May 2012, 04:47
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If n ≠ 1, is (2xn - 2x)/(n-1) > y ?

Is \(\frac{2xn - 2x}{n-1}>y\)? Factor out \(2x\) and reduce by \(n-1\): is \(2x>y\)?

(1) x > y/2 --> multiply by 2: \(2x>y\). Sufficient.
(2) n = 3. Not sufficient.

Answer: A.
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Re: If n ≠ 1, is (2xn - 2x)/(n-1) > y ? [#permalink] New post   07 Mar 2016, 23:35
macjas wrote:
If n ≠ 1, is (2xn - 2x)/(n-1) > y ?

(1) x > y/2
(2) n = 3

The key to solving this question is => we must know that we can factor out any term unless its not zero on the same side of the inequality but we cant change the sign..
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Re: If n ≠ 1, is (2xn - 2x)/(n-1) > y ? [#permalink] New post   13 Mar 2016, 09:07
Here after we cancel out n-1 from numerator and Denominator => We are asked if 2x>y or not.
Clearly statement 1 is sufficient
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Re: If n ≠ 1, is (2xn - 2x)/(n-1) > y ? [#permalink]
13 Mar 2016, 09:07
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Bunuel
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