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GRE Math Challenge #14-at a certain school the ratio of boys

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soumya1989
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GRE Math Challenge #14-at a certain school the ratio of boys [#permalink] New post   Updated on: 26 Aug 2022, 10:45
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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.
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A

Originally posted by soumya1989 on 02 Sep 2014, 02:08.
Last edited by Carcass on 26 Aug 2022, 10:45, edited 1 time in total.
Fixed OA
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wongpcla
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink] New post   16 Jan 2018, 17:41
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A is greater, i try plug the number in,
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink] New post   18 Jan 2018, 00:08
2
please explain why the answer is A

Suppose there is 1 boy and 3 girls at school
and same number of boys and girls are added the new ratio of B/G would be 2/6 = 1/3

Suppose there are 4 boys and 12 girls at school
and same number of boys and girls are added the new ratio of B/G would be 8/24 = 1/3
(c)
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Jaymani335
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink] New post   18 Jan 2018, 08:25
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] New post   18 Jan 2018, 15:45
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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] New post   18 Jan 2018, 20:18
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] New post   18 Jan 2018, 22:47
Quantity A is Larger
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink] New post   19 Jan 2018, 01:05
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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Linamrs
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink] New post   26 Jan 2018, 10:00
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] New post   02 Oct 2020, 11:32
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] New post   10 Dec 2020, 05:06
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] New post   03 Dec 2021, 14:54
a
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Re: GRE Math Challenge #14-at a certain school the ratio of boys [#permalink] New post   26 Aug 2022, 08:15
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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] New post   26 Aug 2022, 10:45
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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] New post   12 Oct 2023, 19:26
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] New post   15 Oct 2025, 21:16
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