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If k > n, is kx - nx > p - m? (1) x = 7/3 (2) k - p > n - m

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If k > n, is kx - nx > p - m? (1) x = 7/3 (2) k - p > n - m  [#permalink]

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New post 26 Sep 2019, 04:41
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  35% (medium)

Question Stats:

73% (01:19) correct 27% (01:32) wrong based on 21 sessions

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

(1) \(x = \frac{7}{3}\)
(2) \(k - p > n - m\)
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Re: If k > n, is kx - nx > p - m? (1) x = 7/3 (2) k - p > n - m  [#permalink]

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New post 26 Sep 2019, 08:13
1
Bunuel wrote:
If \(k > n\), is \(kx - nx > p - m?\)

(1) \(x = \frac{7}{3}\)
(2) \(k - p > n - m\)


(1) \(x = \frac{7}{3}\)
Does not talk about p or m --> Insufficient

(2) \(k - p > n - m\)
--> \(k - n > p - m\)
Does not talk about 'x' --> Insufficient

Combining (1) & (2),
\(k - n > p - m\) & \(x = \frac{7}{3}\)

Note that, If a > b --> 2a is also greater than b --> 2a > b
Similarly If \(k - n > p - m\)
--> \((k - n)\frac{7}{3} > p - m\) --> Sufficient

IMO Option C

Pls Hit kudos if you like the solution
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If k > n, is kx - nx > p - m? (1) x = 7/3 (2) k - p > n - m  [#permalink]

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New post 26 Sep 2019, 08:32
Given: k > n ==> k-n > 0

Therefor, the question can be rephrased as x > \(\frac{p - m}{k - n}\)?


St 1: x = \(\frac{7}{3}\) ==> Insufficient since we do not know the value of \(\frac{p - m}{k - n}\)

St 2: k - p > n - m ==> k - n > p - m ==> \(\frac{p - m}{k - n}\) < 1 ==> Insufficient since we don't know the value of x

Sts 1 and 2 combined ==> x = \(\frac{7}{3}\) > 1 > \(\frac{p - m}{k - n}\) ==> x > \(\frac{p - m}{k - n}\) ==> Sufficient (C is the answer)
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If k > n, is kx - nx > p - m? (1) x = 7/3 (2) k - p > n - m   [#permalink] 26 Sep 2019, 08:32
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