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92. If a, b, c, and d, are positive numbers, is a/b < c/d?

1) 0 < (c-a) / (d-b)

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 cross-multiplying. This will be useful laters.

Statement 1) is not useful, however. (c-a) and (d-b) 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 cross-multiply to get ad < bc. That answers our question with a definite yes, so it's sufficient and the answer is (B)

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
19 Sep 2012, 14:56

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.

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
19 Sep 2012, 20:02

3

This post received KUDOS

catty2004 wrote:

If a, b, c, and d, are positive numbers, is a/b < c/d?

(1) 0 < (c-a) / (d-b)

(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) (c-a)/(d-b) - 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

Attachments

photo.JPG [ 259.84 KiB | Viewed 4255 times ]

_________________

Impossible is nothing to God.

Last edited by mbaiseasy on 20 Sep 2012, 06:21, edited 2 times in total.

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
20 Sep 2012, 00:51

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?

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
20 Sep 2012, 06:13

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.

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
26 May 2014, 07:36

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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 cross-multiply 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.

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 cross-multiply 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.