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If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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08 Jul 2012, 18:20
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If a, b, c, and d, are positive numbers, is \(\frac{a}{b} < \frac{c}{d}\)? (1) \(0 < \frac{(ca)}{(db)}\) (2) \((\frac{ad}{bc})^2 < \frac{(ad)}{(bc)}\)
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Last edited by Bunuel on 12 May 2017, 01:29, edited 2 times in total.
Edited the question.



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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08 Jul 2012, 18:46
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catty2004 wrote: 92. If a, b, c, and d, are positive numbers, is a/b < c/d?
1) 0 < (ca) / (db)
2) (ad/bc)^2 < (ad)/(bc) Hi catty, We're looking for whether a/b < c/d. Fortunately, we're told a useful bit of info in the question stem. All four terms are positive. That's very important with inequalities, because it means that we can multiply and divide without having to worry about the direction of the inequality signs. In this case, we could rephrase the question to whether ad < bc by crossmultiplying. This will be useful laters. Statement 1) is not useful, however. (ca) and (db) could both be positive or negative; that means that when me multiply to get rid of a term, we might or might not have to flip the terms. Since any of the variables could be greater or less than any of the other variables, this statement is insufficient. Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get ad/bc < 1, and can crossmultiply to get ad < bc. That answers our question with a definite yes, so it's sufficient and the answer is (B)
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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19 Sep 2012, 15:56
First of all, I have rephrased the statement. I have become "a/b < c/d" to "ad < cb".
Then, I have answered (B) due to the fact that I know that the result of a proper fraction to the power of 2 is always less than the result of the proper fraction. Thus, in this case (ad/bc)^2 < (ad)/(bc), the fraction ad/bc must be a proper fraction and therefore it must be true that ad<bc.
Is it right this reasoning?



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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19 Sep 2012, 21:02
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catty2004 wrote: If a, b, c, and d, are positive numbers, is a/b < c/d?
(1) 0 < (ca) / (db)
(2) (ad/bc)^2 < (ad)/(bc) We know that a,b,c and d are positive numbers. This is a Yes/No DS question type  is a/b < c/d. Since we are certain that we have no negative values, we can manipulate the inequality question to  is ad < bc? It's much easier to look at. (1) (ca)/(db)  a positive fraction or whole number Say c=d=5 and a=2 and b=1 for 3/4, then ad < bc is false Say c=d=5 and a=1 and b=2 for 4/3, then ad < bc is true thus (1) is INSUFFICIENT (2) Thus, YES! SUFFICIENT. See attachment. Answer: B
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Last edited by mbaiseasy on 20 Sep 2012, 07:21, edited 2 times in total.



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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20 Sep 2012, 01:51
I agree with your approach to the problem. But is not it easier to realize about the rule of proper fraction to the power of 2 instead of manipulate the ecuation in the stem 2?



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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20 Sep 2012, 07:13
racingip wrote: I agree with your approach to the problem. But is not it easier to realize about the rule of proper fraction to the power of 2 instead of manipulate the ecuation in the stem 2? You are right. That's what I did. I did cancelling of of the powers of . Sorry my explanation is not clear. haha! I just summarized that when you start cancelling out, it's like multiplying that fraction I put up.
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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18 Jun 2014, 07:11
KapTeacherEli wrote: catty2004 wrote: 92. If a, b, c, and d, are positive numbers, is a/b < c/d?
Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get \(ad/bc\) < 1, and can crossmultiply to get \(ad < bc\). That answers our question with a definite yes, so it's sufficient and the answer is (B) Hi could you please explain the part on cross multiplication? I am getting \(a/b\) > \(b/c\).



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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18 Jun 2014, 08:48
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pretzel wrote: KapTeacherEli wrote: catty2004 wrote: 92. If a, b, c, and d, are positive numbers, is a/b < c/d?
Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get \(ad/bc\) < 1, and can crossmultiply to get \(ad < bc\). That answers our question with a definite yes, so it's sufficient and the answer is (B) Hi could you please explain the part on cross multiplication? I am getting \(a/b\) > \(b/c\). \((\frac{ad}{bc})^2 < \frac{ad}{bc}\) > reduce by ad/bc: \(\frac{ad}{bc} <1\) > multiply by bc: \(ad<bc\). Hope it's clear.
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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18 Jun 2014, 18:20
Bunuel wrote: \((\frac{ad}{bc})^2 < \frac{ad}{bc}\) > reduce by ad/bc: \(\frac{ad}{bc} <1\) > multiply by bc: \(ad<bc\).
Hope it's clear.
If \(ad<bc\), then \(\frac{a}{b}\) < \((\frac{c}{d})\)?



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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11 Sep 2014, 21:29
we can conceive logically that (ad/bc)^2<ad/bc only when ad/bc<1, so ad<bc
B



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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21 May 2015, 09:05
Is a/b<c/d or (because all the values are >0) ad<bc ? Prethinking: c/d>a/b if casame and d<b or db  same and c>a... 1) (ca)/(db) > c>a and d>b Not suff. yes and no.. see explanation above... 2) (ad/bc)^2 < ad/bc this means we have a proper fraction here (1/2^2 < 1/2) this also means that ad<bc Sufficient > see underlined part above
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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09 Nov 2015, 08:46
Bunuel, 2) This quite clearly sufficient 1) However, it was not very clear to me how this one is insufficient. Is there any way we know this choice is not sufficient w/o resorting to number picking? Thanks!
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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10 Nov 2015, 11:49
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If a, b, c, and d, are positive numbers, is a/b < c/d? (1) 0 < (ca) / (db) (2) (ad/bc)^2 < (ad)/(bc) If we modify the question, the sign of the inequality does not change as a,b,c,d are positive integers. Hence, we want to know whether a/b < c/d?, or ad<bc. From condition 1, 0 < (ca) / (db), and if we multiply (db)^2 on both sides, 0<(ca)(db). We cannot know whether ad<bc, so this is insufficient. From condition 2, we can divide both sides by (ad)/(bc), which gives us (ad/bc)^2 < (ad)/(bc), or (ad/bc)<1, and when we multiply bc on both sides, (ad/bc)<1, or ad<bc. This answers the question 'yes' so this is sufficient, and the answer becomes (B). Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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10 Dec 2015, 07:07
Please correct me if I'm wrong.
1) not sufficient
2) sufficient
Let ab=4 and bc=2 so (ab/bc)^2=4, which cannot be less than (ab/bc)=2. We can try any values such as ab=4 and bc=3.
Therefore we can conclude that ab has to be less than bc.
So 2) is sufficient.
My only problem is that how would I be able to think of such an approach in the main exam , which requires that every question be solved in less than 2 mins ?
Pls help! :p



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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NoHalfMeasures wrote: Bunuel,
2) This quite clearly sufficient 1) However, it was not very clear to me how this one is insufficient. Is there any way we know this choice is not sufficient w/o resorting to number picking?
Thanks! I know your question is to Bunuel, but I am just sharing how I eliminated the first statement: Statement 1 tells us that \(\frac{ca}{db}>0\), i.e. \(ca\) and \(db\) have same signs (either both are +ve or both are ve). However that doesn't tells us anything about their individual values, which makes this statement insufficient. Hope it helps.



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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21 Apr 2017, 03:03
hey, for statement 1: can't we just multiply the equation by db ? 1. 0 < ca/db > multiply by db > 0 < ca > a < c so a < c but we don't know whether a/b < c/d Is that correct ?
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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21 Apr 2017, 03:15
daviddaviddavid wrote: hey,
for statement 1:
can't we just multiply the equation by db ?
1. 0 < ca/db > multiply by db > 0 < ca > a < c so a < c but we don't know whether a/b < c/d
Is that correct ? We cannot multiply the equation by db because we do not know if db is less or greater than zero. For instance, d=3 and b=4, then 34 = 1. Rule of thumb: we cannot manipulate inequalities without knowing the signs. (and as I mentioned before we don't know the sign of db). Hope it helps



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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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21 Apr 2017, 03:26
The question is asking whether ad<bc ? 1) Not sufficient 2) (ad/bc)^2 <ad/bc That means denominator > numerator bc>ac Sufficient.
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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21 Apr 2017, 03:36
racingip wrote: daviddaviddavid wrote: hey,
for statement 1:
can't we just multiply the equation by db ?
1. 0 < ca/db > multiply by db > 0 < ca > a < c so a < c but we don't know whether a/b < c/d
Is that correct ? We cannot multiply the equation by db because we do not know if db is less or greater than zero. For instance, d=3 and b=4, then 34 = 1. Rule of thumb: we cannot manipulate inequalities without knowing the signs. (and as I mentioned before we don't know the sign of db). Hope it helps ok thanks so but we could still change the inequality and consider both cases, right ? 1a. 0 < ca/db > multiply by db > 0 < ca > a < c so a < c but we don't know whether a/b < c/d 1b. 0 < ca/db > multiply by db > 0 > ca > a > c so a > c but we don't know whether a/b > c/d
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