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Is (a - k)/(b - k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0

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Is (a-k)/(b-k)>(a+k)/(b+k) ?  [#permalink]

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New post Updated on: 31 Jul 2013, 10:50
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Originally posted by bagdbmba on 31 Jul 2013, 06:35.
Last edited by Bunuel on 31 Jul 2013, 10:50, edited 3 times in total.
Edited the question.
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Re: Is (a - k)/(b - k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0  [#permalink]

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New post 18 Jun 2015, 22:15
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Bunuel wrote:
Is (a - k)/(b - k) > (a + k)/(b + k) ?

(1) a > b > k
(2) k > 0


Kudos for a correct solution.


The question is based on the concept of adding the same number to the numerator and denominator of a fraction. We have discussed it here: http://www.veritasprep.com/blog/2011/06 ... round-one/

This is the crux of the post: When we add the same positive integer to the numerator and the denominator of a positive fraction, the fraction increases if it is less than 1 (but remains less than 1) and decreases if it is more than 1 (but remains more than 1). That is, we can say, that the fraction is pulled toward 1 in both the cases.
So 2/3 < (2+1)/(3+1) But 2/3 > (2-1)/(3-1)
But 4/3 > (4+1)/(3+1) But 4/3 < (4-1)/(3-1)

So whether (a - k)/(b - k) is greater than or less than (a + k)/(b + k) depends on whether a is greater than b or smaller, whether k is positive or not and whether the fraction a/b is positive or not. Together, both statements give us all this information and hence answer will be (C).
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Re: Is (a-k/b-k)>(a+k/b+k) ?  [#permalink]

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New post 31 Jul 2013, 06:57
Is \(\frac{a-k}{b-k}>\frac{a+k}{b+k}\) ?

i. \(a>b>k\)
If a=2,b=1 and k=0 the answer is NO. \(2>2\)
If a=3,b=1 and k=-2 the answer is YES. \(\frac{5}{3}>-1\)
Not sufficient

ii. \(k>0\)
Clearly not sufficient, no info about a, b.

1+2) \(a>b>k>0\)
Rewrite the question as
\(\frac{a-k}{b-k}-\frac{a+k}{b+k}>0\) or \(\frac{(a-k)(b+k)-(a+k)(b-k)}{(b+k)(b-k)}>0\) or \(\frac{2k(a-b)}{(b+k)(b-k)>0}\). Now we know that \(a>b\) so \(a-b>0\) and the Numerator is positive; we know also that \(b>k\) so \(b-k>0\) and \((b+k)\) will be positive as well because it's the sum of two positive numbers.
\(\frac{+ve}{+ve}=+ve>0\) Sufficient C
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Re: Is (a-k/b-k)>(a+k/b+k) ?  [#permalink]

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New post 31 Jul 2013, 08:14
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Zarrolou wrote:
Is \(\frac{a-k}{b-k}>\frac{a+k}{b+k}\) ?

i. \(a>b>k\)
If a=2,b=1 and k=0 the answer is NO. \(2>2\)
If a=3,b=1 and k=-2 the answer is YES. \(\frac{5}{3}>-1\)
Not sufficient

ii. \(k>0\)
Clearly not sufficient, no info about a, b.

1+2) \(a>b>k>0\)
Rewrite the question as
\(\frac{a-k}{b-k}-\frac{a+k}{b+k}>0\) or \(\frac{(a-k)(b+k)-(a+k)(b-k)}{(b+k)(b-k)}>0\) or \(\frac{2k(a-b)}{(b+k)(b-k)>0}\). Now we know that \(a>b\) so \(a-b>0\) and the Numerator is positive; we know also that \(b>k\) so \(b-k>0\) and \((b+k)\) will be positive as well because it's the sum of two positive numbers.
\(\frac{+ve}{+ve}=+ve>0\) Sufficient C


Thanks Zarrolou - so here you're assuming that thge given question is true and in turn you;re proving that your assumption is correct...! Great.

Good luck for your GMAT.
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Re: Is (a-k/b-k)>(a+k/b+k) ?  [#permalink]

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New post 31 Jul 2013, 08:18
bagdbmba wrote:

Thanks Zarrolou - so here you're assuming that thge given question is true and in turn you;re proving that your assumption is correct...! Great.

Good luck for your GMAT.


Thank for your wishes first.

I am saying that the equation will be positive, hence will be greater than zero, so the answer (with 1 and 2) will be always YES (hence they combined are sufficient).

Hope it's clear.
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Re: Is (a-k/b-k)>(a+k/b+k) ?  [#permalink]

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New post 31 Jul 2013, 08:28
Zarrolou wrote:
bagdbmba wrote:

Thanks Zarrolou - so here you're assuming that the given question is true and in turn you're proving that your assumption is correct...! Great.

Good luck for your GMAT.


Thank for your wishes first.

I am saying that the equation will be positive, hence will be greater than zero, so the answer (with 1 and 2) will be always YES (hence they combined are sufficient).

Hope it's clear.


Yeah! that means you're assuming that the equation to be true (as you're writing \(\frac{a-k}{b-k}-\frac{a+k}{b+k}>0\)) and proving that your assumption is correct by showing both numerator and denominator are +ve..Right?
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Re: Is (a-k/b-k)>(a+k/b+k) ?  [#permalink]

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New post 31 Jul 2013, 08:31
bagdbmba wrote:

Yeah! that means you're assuming that the equation to be true (as you're writing \(\frac{a-k}{b-k}-\frac{a+k}{b+k}>0\)) and proving that your assumption is correct by showing both numerator and denominator are +ve..Right?


Nope. I am doing exactly the opposite. I am asking: is this equation positive?

Then I show how the numerator and the denominator are both positive, HENCE the resulting fraction will be positive (hence greater than 0).I am not assuming anything.

Hope I've explained myself well.
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Re: Is (a-k/b-k)>(a+k/b+k) ?  [#permalink]

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New post 31 Jul 2013, 09:44
bagdbmba wrote:
Is \(\frac{a-k}{b-k}>\frac{a+k}{b+k}\) ?

i. \(a>b>k\)

ii. \(k>0\)


The question asks : Is \(\frac{a-k}{b-k}>\frac{a+k}{b+k}\) \(\to\) \(\frac{a-k}{b-k}-1>\frac{a+k}{b+k}-1\) \(\to \frac{(a-k)-(b-k)}{b-k}>\frac{(a+k)-(b+k)}{b+k}\)

Or Is \(\frac{(a-b)}{b-k}>\frac{(a-b)}{b+k} \to (a-b)(\frac{1}{b-k} - \frac{1}{b+k})>0 = \frac{(a-b)(2k)}{(b+k)(b-k)}>0\)

From F.S 1, we know that (a-b) and (b-k) both are positive. Thus, the question boils down to : IS\(\frac{2k}{b+k}>0\). Clearly Insufficient.

From F.S 2, we know that k>0. Hence, the question boils down to : IS \(\frac{(a-b)}{(b+k)(b-k)}>0\). Again Insufficient.

Taking both statements together, the question : IS\(\frac{1}{(b+k)}>0\). As k>0 and b>k, thus (b+k)>0. Suffcient.

C.
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Is (a - k)/(b - k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0  [#permalink]

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Re: Is (a - k)/(b - k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0  [#permalink]

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New post 18 Jun 2015, 05:06
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Bunuel wrote:
Is (a - k)/(b - k) > (a + k)/(b + k) ?

(1) a > b > k
(2) k > 0


Kudos for a correct solution.


(a - k)/(b - k) > (a + k)/(b + k)

<=> 1+ (a - b)/(b - k) > 1 + (a + b)/(b + k)

<=> (a - b)/(b - k) > (a - b)/(b + k)

(1) a > b > k => a-b>0, b-k>0, b+k>0. Just when k>0: (a - b)/(b - k) > (a - b)/(b + k) => INSUFFICIENT
(2) k > 0: Insufficient

(1) (2) Sufficient

Ans: C
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Re: Is (a - k)/(b - k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0  [#permalink]

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New post 18 Jun 2015, 19:58
1
Simplifying one can get
Is \(\frac{(2k*(a-b))}{(b^2 -a^2)} > 0\)

A: a>b>k
a-b > 0 & \((b^2 - k^2) > 0\)
Condition flips if k >0 or k <0.

B: k > 0
Condition flips if a>b or a < b

If one uses both A and B it would be sufficient.

So, C
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Re: Is (a - k)/(b - k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0  [#permalink]

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New post 22 Jun 2015, 05:30
Bunuel wrote:
Is (a - k)/(b - k) > (a + k)/(b + k) ?

(1) a > b > k
(2) k > 0


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

In rephrasing this question, we should recall that we do not know the sign of b – k and b + k. Thus, after cross-multiplying, we should set up Flow Charts to evaluate two different questions: one for the case in which (b – k) and (b + k) have the same sign and another for the case in which (b – k) and (b + k) have different signs:
Image

Statement (1) tells us that a > b > k. This is not sufficient. We do not know whether (b – k)(b + k) is positive, so we do not know which question to answer. Even if we did, we could get different results. For example, if a and b are positive and k is negative, then (b – k)(b + k) could be positive. Thus the relevant question would be “Is ak > bk?” Because k is negative, ak < bk. By contrast, if a, b, and k are all positive, then (b – k)(b + k) is positive. Thus the relevant question would be “Is ak > bk?” Because a > b > k, we would know ak > bk. We get two different answers depending on whether k is positive. INSUFFICIENT.

Statement (2) tells us that k > 0. This is not sufficient, because the statement tells us nothing about a and b. INSUFFICIENT.

Statements (1) and (2) combined are sufficient, because if a > b > k > 0, then (b – k)(b + k) > 0, so the relevant question is “Is ak > bk?” We know that a > b, and k is positive, so ak > bk, and the answer to the question is a definite “YES.”

Notice the use of the Scenario Chart—specifically, Flow Charts—to handle the different versions of the question depending on the sign of (b – k)(b + k). Additionally, we were careful to Beware of Inequalities—the inequalities in this problem make it easy to make a mistake in rephrasing the question in the Flow Chart or in evaluating the statements.

Notice also that if the algebra and thought process became too complicated, we could Cross-Multiply Inequalities and guess between C and E. Because the combined statements tell us that a > b > k > 0, we would know a lot about the relative values of the variables in the problems, and it might be reasonable to choose C as the best answer.

The correct answer is C.

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