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Is (ak)/(bk)>(a+k)/(b+k) ?
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Updated on: 31 Jul 2013, 10:50
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Is \(\frac{ak}{bk}>\frac{a+k}{b+k}\) ? (1) \(a>b>k\) (2) \(k>0\)
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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
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18 Jun 2015, 22:15
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 ... roundone/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 > (21)/(31) But 4/3 > (4+1)/(3+1) But 4/3 < (41)/(31) 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 (ak/bk)>(a+k/b+k) ?
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31 Jul 2013, 06:57
Is \(\frac{ak}{bk}>\frac{a+k}{b+k}\) ? 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 \(\frac{ak}{bk}\frac{a+k}{b+k}>0\) or \(\frac{(ak)(b+k)(a+k)(bk)}{(b+k)(bk)}>0\) or \(\frac{2k(ab)}{(b+k)(bk)>0}\). Now we know that \(a>b\) so \(ab>0\) and the Numerator is positive; we know also that \(b>k\) so \(bk>0\) 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 (ak/bk)>(a+k/b+k) ?
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31 Jul 2013, 08:14
Zarrolou wrote: Is \(\frac{ak}{bk}>\frac{a+k}{b+k}\) ?
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 \(\frac{ak}{bk}\frac{a+k}{b+k}>0\) or \(\frac{(ak)(b+k)(a+k)(bk)}{(b+k)(bk)}>0\) or \(\frac{2k(ab)}{(b+k)(bk)>0}\). Now we know that \(a>b\) so \(ab>0\) and the Numerator is positive; we know also that \(b>k\) so \(bk>0\) 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 (ak/bk)>(a+k/b+k) ?
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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 (ak/bk)>(a+k/b+k) ?
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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 \(\frac{ak}{bk}\frac{a+k}{b+k}>0\)) and proving that your assumption is correct by showing both numerator and denominator are +ve..Right?
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Re: Is (ak/bk)>(a+k/b+k) ?
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31 Jul 2013, 08:31
bagdbmba wrote: Yeah! that means you're assuming that the equation to be true (as you're writing \(\frac{ak}{bk}\frac{a+k}{b+k}>0\)) 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 (ak/bk)>(a+k/b+k) ?
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31 Jul 2013, 09:44
bagdbmba wrote: Is \(\frac{ak}{bk}>\frac{a+k}{b+k}\) ?
i. \(a>b>k\)
ii. \(k>0\) The question asks : Is \(\frac{ak}{bk}>\frac{a+k}{b+k}\) \(\to\) \(\frac{ak}{bk}1>\frac{a+k}{b+k}1\) \(\to \frac{(ak)(bk)}{bk}>\frac{(a+k)(b+k)}{b+k}\) Or Is \(\frac{(ab)}{bk}>\frac{(ab)}{b+k} \to (ab)(\frac{1}{bk}  \frac{1}{b+k})>0 = \frac{(ab)(2k)}{(b+k)(bk)}>0\) From F.S 1, we know that (ab) and (bk) 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 \(\frac{(ab)}{(b+k)(bk)}>0\). 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
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18 Jun 2015, 03:21
9. Inequalities For more check Ultimate GMAT Quantitative Megathread
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Re: Is (a  k)/(b  k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0
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18 Jun 2015, 05:06
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 => ab>0, bk>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
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18 Jun 2015, 19:58
Simplifying one can get Is \(\frac{(2k*(ab))}{(b^2 a^2)} > 0\) A: a>b>k ab > 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
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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 crossmultiplying, 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: 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 CrossMultiply 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.Attachment:
20150622_1626.png [ 99.76 KiB  Viewed 1661 times ]
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Re: Is (a  k)/(b  k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0
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Re: Is (a  k)/(b  k) > (a + k)/(b + k) ? (1) a > b > k (2) k > 0
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