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# GRE Math Challenge #71- (a/b) or (a+3)/(b+3)

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GRE Math Challenge #71- (a/b) or (a+3)/(b+3) [#permalink]  03 May 2015, 01:03
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Question Stats:

73% (00:37) correct 26% (00:23) wrong based on 71 sessions

a and b are positive integers.

 Quantity A Quantity B $$\frac{a}{b}$$ $$\frac{a+3}{b+3}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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[Reveal] Spoiler: OA

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Sandy
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Re: GRE Math Challenge #71 [#permalink]  03 May 2015, 06:29
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sandy wrote:
a and b are positive integers.

Quantity A: $$a/b$$

Quantity B: $$\frac{(a+3)}{(b+3)}$$

One of my favorite QC strategies is called "Looking for Equality" (free video on this topic here: http://www.greenlighttestprep.com/modul ... on?id=1099).
In this strategy, I first look for values of x such that the two quantities are equal. This often involves checking some easy numbers like 0, 1, -1 etc.

Consider these two cases:
Case #2: a = 1 and b = 1
Quantity A: a/b = 1/1 = 1
Quantity B: (a+3)/(b+3) = (1+3)/(1+3) = 4/4 = 1
Since the two quantities are EQUAL, the correct answer is either C or D

Now try some other values of a and b:
Case #2: a = 1 and b = 2
Quantity A: a/b = 1/2
Quantity B: (a+3)/(b+3) = (1+3)/(2+3) = 4/5
Since the two quantities are NOT EQUAL, the correct answer is D

Cheers,
Brent
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Last edited by GreenlightTestPrep on 18 May 2015, 11:25, edited 1 time in total.
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Re: GRE Math Challenge #71 [#permalink]  03 May 2015, 21:52
#Sandy this is not a challenge, it can be a practice question.....
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Re: a and b are positive integers [#permalink]  14 Feb 2017, 16:07
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Explanation

The best strategy here is to assume values of a and b.

Let a = 1 and b =2 .

Quantity A: $$\frac{1}{2}$$
Quantity B: $$\frac{1+3}{2+3}=\frac{4}{5}$$

Quantity B is greater.

Let a = 6 and b=3
Quantity A: $$\frac{6}{3}=2$$
Quantity B: $$\frac{6+3}{3+3}=\frac{3}{2}$$

Quantity A is greater.

Hence a relationship cannot be determined. Thus option D is correct.
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Re: a and b are positive integers [#permalink]  18 Feb 2017, 10:27
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sandy wrote:
Explanation

The best strategy here is to assume values of a and b.

Let a = 0 and b =1 .

Quantity A: $$\frac{0}{1}=0$$
Quantity B: $$\frac{0+3}{1+3}=\frac{3}{4}$$

Quantity B is greater.

Let a = 6 and b=3
Quantity A: $$\frac{6}{3}=2$$
Quantity B: $$\frac{6+3}{3+3}=\frac{3}{2}$$

Quantity A is greater.

Hence a relationship cannot be determined. Thus option D is correct.

Just wanted to say that 0 is not a positive integer. 0 is neither positive nor negative,
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Re: a and b are positive integers [#permalink]  19 Feb 2017, 03:24
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You are right. Changed it!

Thanks
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Re: GRE Math Challenge #71- (a/b) or (a+3)/(b+3) [#permalink]  17 Jan 2018, 00:06
since there is no restriction for a / b (i mean like a>1 b>a , this kind of restriction)
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Re: GRE Math Challenge #71- (a/b) or (a+3)/(b+3) [#permalink]  26 Jan 2018, 10:24
D
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Re: GRE Math Challenge #71- (a/b) or (a+3)/(b+3) [#permalink]  09 Aug 2018, 09:59
Why would you not assume different numbers for the variables a & b. If you assume a & b are the number 1, would they not be the same variable since they are the same number?
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Re: GRE Math Challenge #71- (a/b) or (a+3)/(b+3) [#permalink]  25 May 2019, 09:25
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PTDom wrote:
Why would you not assume different numbers for the variables a & b. If you assume a & b are the number 1, would they not be the same variable since they are the same number?

It's a common misconception that two different variables (e.g., x and y) must represent different values, but this is not the case.

Cheers,
Brent
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Re: GRE Math Challenge #71- (a/b) or (a+3)/(b+3)   [#permalink] 25 May 2019, 09:25
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