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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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Joined: 02 Sep 2009
Posts: 58446
If k > n, is kx - nx > p - m? (1) x = 7/3 (2) k - p > n - m  [#permalink]

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26 Sep 2019, 04:41
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Difficulty:

45% (medium)

Question Stats:

72% (01:18) correct 28% (01:40) wrong based on 17 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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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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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)
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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