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If k > n, is kx - nx > p - m?

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Intern
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Joined: 24 May 2018
Posts: 6
If k > n, is kx - nx > p - m?  [#permalink]

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New post 12 Sep 2018, 09:12
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Question Stats:

74% (01:28) correct 26% (01:29) wrong based on 39 sessions

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If k > n, is kx - nx > p - m?

(1) x = 7/3
(2) k - p > n - m
Director
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Joined: 20 Feb 2015
Posts: 792
Concentration: Strategy, General Management
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Re: If k > n, is kx - nx > p - m?  [#permalink]

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New post 12 Sep 2018, 09:31
jennysussna wrote:
if k > n, is kx - nx > p - m?
1) x=7/3
2) k - p > n - m


1) x=7/3
above equation can be written as
x(k-n) > (p-m)
we do not know any other value apart from x
insufficient

2) k - p > n - m
can be rearranged
k-n > p-m

may seem sufficient , but we do not know value of x (various combinations of +ve and -ve numerator and denominators are possible)
insufficient

using both
we have x
so we have a definite yes
x(k-n) > (p-m)
sufficient
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Re: If k > n, is kx - nx > p - m?  [#permalink]

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New post 13 Sep 2018, 08:34
Each statement alone does not give any information about few variables but together they give complete information.

It is a straight forward C.
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Re: If k > n, is kx - nx > p - m?  [#permalink]

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New post 13 Sep 2018, 09:41
jennysussna wrote:
If k > n, is kx - nx > p - m?

(1) x = 7/3
(2) k - p > n - m


OA:C

\(k>n, \quad\)Is \(x*(k - n) > p - m\) ?

\((1) \quad x= \frac{7}{3}\)

Let \(k-n = 3\), and \(p-m=1\) then L.H.S\(= \frac{7}{3}*3\);R.H.S \(=1\)

L.H.S>R.H.S
Is \(kx - nx > p - m?\) : Yes

Let \(k-n = 3\), and \(p-m=8\) then L.H.S\(= \frac{7}{3}*3\);R.H.S \(=8\)

L.H.S<R.H.S
Is \(kx - nx > p - m?\) : No

Statement \((1)\) alone is insufficient.

\((2)\quad k - p > n - m\)

\(k-n>p-m\)

\(x\) can be a fraction less than \(1\), negative or positive.

Statement \((2)\) alone is insufficient.

Combining \((1)\) and \((2)\), we get \(\frac{7}{3}(k-n)>p-m\).
So combining statement \((1)\) and \((2)\) will be sufficient
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Re: If k > n, is kx - nx > p - m?   [#permalink] 13 Sep 2018, 09:41
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