Is a/b < c/d?

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Is a/b < c/d?

by catty2004 » Sun Jul 08, 2012 5:19 pm

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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)

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by eagleeye » Sun Jul 08, 2012 5:39 pm

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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 are told that a,b,c,d are positive integers, we need to find if a/b < c/d
Now:
a/b < c/d
=> a/b - c/d < 0
=> (ad-bc)/(bd)<0
=> ad-bc<0 (b and d are positive, Hence multiplying both sides by bd doesn't change the sign)
=> ad<bc
So we need to find it ad < bc

With that in mind let's look at the options:
1) 0 < (c-a) / (d-b)
This means that both c-a and d-b have the same sign:
If both are positive c-a>0, and d-b>0 => c>a and d>b. So we can't say anything about ad>bc.
We need not check the both less than 0 case now, because the both are positive case already led us to Insufficient data condition. Hence Insufficient.

2) (ad/bc)^2 <(ad/bc)
Since ad/bc is positive, dividing both sides by ad/bc
ad/bc <1 => ad < bc. So we get that ad < bc. Hence ad is definitely not greater than bc. Sufficient.

Hence B is the correct answer. :)

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by KapTeacherEli » Sun Jul 08, 2012 5:46 pm

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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)
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)
Eli Meyer
Kaplan GMAT Teacher
Cambridge, MA
www.kaptest.com/gmat

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by Nina1987 » Sun Dec 27, 2015 7:05 am

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Is it possible to prove St1 insufficient w/o resorting to number picking? thanks

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by Matt@VeritasPrep » Sun Dec 27, 2015 5:16 pm

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There is, but it's a little time consuming. I think S1 is mostly there to test your understanding of fractions; i.e. whether you recognize that (c/d) - (a/b) is NOT the same thing as (c - a)/(d - b).

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by ceilidh.erickson » Mon Oct 09, 2017 7:45 am

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Nina1987 wrote:Is it possible to prove St1 insufficient w/o resorting to number picking? thanks
This problem can be solved without choosing numbers, if you rephrase the question:
a/b < c/d --> Because all of these variables are positive, we're allowed to cross-multiply here:
ad < bc

Target question: is ad < bc?
Often it's easier to deal with products than with ratios on DS questions.

Statement 1: 0 < (c-a)/(d-b)
Here, we can't cross-multiply, because we don't know if (d - b) is positive or negative. Our question was asking us about ratios/products... in other words, PROPORTIONAL relationships. This statement is giving us information about DIFFERENCES, which cannot answer a proportion question. (You can certainly test numbers to prove the point, though, as Mitch pointed out).
Insufficient

[color=]Statement 2: (ad/bc)² < (ad)/(bc) [/color]
Here, we're given a proportion, which is already more helpful. Let's rephrase:
(ad/bc)² < (ad)/(bc) Because everything is positive, we know that all products and ratios will be positive. If the square of the ratio (ad/bc) is less than the ratio itself, what does that mean? It means the ratio must be a POSITIVE FRACTION.
ad/bc < 1
Multiply both sides by bc:
[color=]ad < bc [/color]
This is our target question. Sufficient.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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Re: Is a/b < c/d?

by Scott@TargetTestPrep » Thu May 20, 2021 6:57 am

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catty2004 wrote:
Sun Jul 08, 2012 5:19 pm
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)
Solution:

Question Stem Analysis:


We need to determine whether a/b < c/d given that a, b, c and d are positive. Since all quantities are positive, we can cross-multiply, obtaining ad < bc. In other words, we can rewrite the question as: Is ad < bc ?

Statement One Alone:

This is not sufficient. For example, if a = b = 1, c = 3 and d = 2, then ad = 2 and bc = 3, and the answer to the question is yes.. However, if a = 3, b = 2, and c = d = 1, we see that ad = 3 and bc = 2, and the answer to the question is no..

Statement Two Alone:

The square of a quantity is less than itself if the quantity has a value between 0 and 1. Therefore, ad/bc < 1. Since bc is positive, multiplying both sides by bc, we have ad < bc. Statement two alone is sufficient.

Answer: B

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