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# GRE Math Challenge #104- x > 0

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GRE Math Challenge #104- x > 0 [#permalink]  09 May 2015, 11:22
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Question Stats:

91% (01:01) correct 8% (00:34) wrong based on 71 sessions
$$x > 0$$

 Quantity A Quantity B $$\frac{(590+x)}{800}$$ $$\frac{(600+x)}{790}$$

• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
[Reveal] Spoiler: OA

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Sandy
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Re: GRE Math Challenge #104 [#permalink]  12 May 2015, 08:30
Expert's post
sandy wrote:
x > 0

Quantity A: (590+x)/800
Quantity B: (600+x)/790

Two nice rules concerning fractions with positive numerators and denominators:
1) Making a numerator bigger makes the fraction bigger.
2) Making a denominator smaller makes the fraction bigger.

So, from rule number 1, we can see that (600+x)/800 is GREATER THAN (590+x)/800
In other words, (600+x)/800 > (590+x)/800

From rule number 2, we can see that (600+x)/790 is GREATER THAN (600+x)/800
In other words, (600+x)/790 > (600+x)/800

Combine the inequalities to get: (600+x)/790 > (600+x)/800 > (590+x)/800
So, we can be certain that (600+x)/790 > (590+x)/800

[Reveal] Spoiler:
B

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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Re: GRE Math Challenge #104 [#permalink]  17 Aug 2019, 16:49
GreenlightTestPrep wrote:
sandy wrote:
x > 0

Quantity A: (590+x)/800
Quantity B: (600+x)/790

Two nice rules concerning fractions with positive numerators and denominators:
1) Making a numerator bigger makes the fraction bigger.
2) Making a denominator smaller makes the fraction bigger.

So, from rule number 1, we can see that (600+x)/800 is GREATER THAN (590+x)/800
In other words, (600+x)/800 > (590+x)/800

From rule number 2, we can see that (600+x)/790 is GREATER THAN (600+x)/800
In other words, (600+x)/790 > (600+x)/800

Combine the inequalities to get: (600+x)/790 > (600+x)/800 > (590+x)/800
So, we can be certain that (600+x)/790 > (590+x)/800

[Reveal] Spoiler:
B

Cheers,
Brent

Hi Brent,

I got the correct answer independently but I think it was just dumb luck. Can you please help steer me in the right direction if there are flaws in my logic? I simplified both QA and QB by dividing both by 10 to get QA: (59+x)/80 and QB: (60+x)/79. Then I decided that the overall impact of x was unimportant because it would be the same on both sides so I gave x a value of 1 for both quantities. I was left to compare QA: 60/80 and QB: 61/79. I decided that QB was greater because the denominator was smaller and the differences in the numerator was too minuscule to matter.
Re: GRE Math Challenge #104   [#permalink] 17 Aug 2019, 16:49
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# GRE Math Challenge #104- x > 0

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