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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
If equal number of students joined then the ratio remains the same thus, Both quantities are equal
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
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Couple of things happening here.

First is you have a implied integer restriction. You can't have half a person right? The second portion says both boys and girls are equal. If you add the smallest possible increase to both (1) then it becomes \(\frac{1}{2}\) which is greater that 1/3.

Any integer you add to both the top and bottom of the fraction will increase it therefore A.
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
Thank You
I got fooled by the wording of the question
"Today, an equal number of boys and girls joined the school"
I interpreted it as the number of boys and girls equal to the ratio
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
Quantity A is Larger
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
SherpaPrep wrote:
Couple of things happening here.

First is you have a implied integer restriction. You can't have half a person right? The second portion says both boys and girls are equal. If you add the smallest possible increase to both (1) then it becomes \(\frac{1}{2}\) which is greater that 1/3.

Any integer you add to both the top and bottom of the fraction will increase it therefore A.




Oh Okay, understood now. I did not really took note of the equal number of boys and girls added.

Thank you.
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
SherpaPrep wrote:
Couple of things happening here.

First is you have a implied integer restriction. You can't have half a person right? The second portion says both boys and girls are equal. If you add the smallest possible increase to both (1) then it becomes \(\frac{1}{2}\) which is greater that 1/3.

Any integer you add to both the top and bottom of the fraction will increase it therefore A.


thank you!... I understand now, I thought that the number of girls and boys joined the school were equal to the ones were already there.
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
soumya1989 wrote:
Yesterday, at a certain school, the ratio of boys to girls was 1 to 3. Today, an equal number of boys and girls joined the school. The number that joined was greater than zero and no students left.


Quantity A
Quantity B
Ratio of boys to girls now
\(\frac{1}{3}\)


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.



I know this is not an ETS problem based on the horrible wording lol.
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
B / G = 1/3 = .33
Now,
3B = G
or, 3B + 1 = G + 1
if B = 1 and G = 1; we get--
3.1 + 1 = 1 + 1
or, 4 = 2; that means Boys to girl ratio will be 2/4 or 1/2 = .5 which is greater than 0.33.
So adding an equal number of girls and boys the ratio will be higher.
Ans: A
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
a
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
1
For a ratio less than 1, addition of the same number in the numerator and denominator brings it closer to 1 viz. the original fractional value. Hence, Option A is the right answer.

Carcass it seems the answer is yet to reflect the right answer. Could you please modify it?
Thanks!
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
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Fixed as A the OA
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
Jaymani335 wrote:
If equal number of students joined then the ratio remains the same thus, Both quantities are equal


This is false. Please try plugging in numbers.
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink]
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