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Re: GRE Math Challenge #104- x > 0 [#permalink]
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Bookmarks
We need to compare \(\frac{(590+x)}{800}\) and \(\frac{(600+x)}{790}\)
and we know that x > 0

Now, lets look at the numerator of both the quantities

Numerator for Quantity A (590+x) < Numerator for Quantity B (600+x)

Now, lets look at the denominator of both the quantities

Denominator for Quantity A (800) > Denominator for Quantity B (790)

Since Numerator of Quantity A < Numerator of Quantity B and Denominator for Quantity A > Denominator for Quantity B
[numerical example 12/20 < 13/19]

So, Quantity A < Quantity B

So, Answer will be B
Hope it helps!
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GRE Math Challenge #104- x > 0 [#permalink]
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JNeen

Just adding few things in your solution
When you are dividing both numerator and denominator by 10 then even x will become \(\frac{x}{10}\)

Now, x/10 can be assumed as X, lets continue with the solution and we know that X will also be > 0

Now, We know that 59/80 < 60/79 so Quantity A < Quantity B
So we can ignore X/80 and X/79 as in any case X/80 < X/79 and it will add to Quantity A < Quantity B
So, your approach was more or less correct with these minor additions.
Good work!
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Re: GRE Math Challenge #104- x > 0 [#permalink]
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Re: GRE Math Challenge #104- x > 0 [#permalink]
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