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Intern  Joined: 28 May 2010
Posts: 26
In a local school district, the high school and middle  [#permalink]

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10 00:00

Difficulty:   45% (medium)

Question Stats: 63% (02:06) correct 37% (02:25) wrong based on 237 sessions

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In a local school district, the high school and middle school each received r dollars toward funding for the student arts program. The high school enrolled 300 students and the middle school enrolled 200 students. Later, the middle school transferred s dollars to the high school so that they would have received the same funding per student. Which of the following is equivalent to s?

A. r/2
B. r/3
C. r/4
D. r/5
E. r/6

-------------
How I originally did it:

High school: \$r, 300 students
Middle: \$r, 200 students
X= amount transfer per student

let R = 600
H: 600/300 + X = 600/300 -X
2X = \$3/student - \$2/student
2X = \$1/student
X = \$0.5/student

therefore, Middle school transfers: S=(0.5)(200) = 100 dollars to High School.
Check:
H:700/300students = \$2.5/student
M:500/200students = \$2.5/student

Manager  Joined: 04 Feb 2010
Posts: 121
Re: 700-800 PS...simple algebra actually  [#permalink]

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1
2
I set up the equation as such.

(r - s) / 200 = (r + s) / 300

s leaving one school, s entering one school, hence the equation. Dividing the amount of money after the transfer by the number of students.

This gives us:

r / 200 - r / 300 = s / 300 + s / 200

100 r / 600 = 500 s / 600

1/6 r = 5/6 s

s = 1/5 r

Hence D.
Manager  Joined: 04 Feb 2010
Posts: 121
Re: 700-800 PS...simple algebra actually  [#permalink]

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Oooops, sorry, algebra mode. Check the solution. 700/500 <> 2.5 700/500 = 2.33333.

You forgot to divide the X - the amount transferred - by the number of students. The amount transferred must be 120 - for \$2.4 per student.
Intern  Joined: 19 Jul 2009
Posts: 34
Location: baltimore, md
Schools: kellogg, booth, stern, ann arbor
Re: 700-800 PS...simple algebra actually  [#permalink]

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yeah this problem can be difficult if you don't understand what the problem is asking. it took me a few minutes just to figure out exactly what the problem wanted. once i understood the problem, the algebra was very easy.
Intern  Joined: 28 May 2010
Posts: 26
Re: 700-800 PS...simple algebra actually  [#permalink]

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RGM wrote:
Oooops, sorry, algebra mode. Check the solution. 700/500 <> 2.5 700/500 = 2.33333.

You forgot to divide the X - the amount transferred - by the number of students. The amount transferred must be 120 - for \$2.4 per student.

RGM, I'm so sorry, I can't believe I still don't understand...

why am I dividing 700/500? where is 700 from?
X is the amount transferred per student, so it's = \$0.5/student, so Middle should transfer \$0.5/student to High. Since Middle has 200 students, it transfers \$100=s to High....

but again, that's wrong, and again, I don't see why it's wrong.
Manager  Joined: 04 Feb 2010
Posts: 121
Re: 700-800 PS...simple algebra actually  [#permalink]

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2
No worries.

Your calculations used this: ( r / 200 ) - s = ( r / 300 ) + s

when it should have used this: ( r - s ) / 200 = ( r + s ) / 300

See the difference. You set R as 600. 600/300 = \$2 and 600/200 = \$3

So you reasoned that amount to be transferred per student should be \$3 - \$2 = \$1. Divide by 2, and get \$0.5. \$0.5 per student. That's wrong. It is because the transfer isn't divided by the number of students - as such, it's lopsided again, and the amount of money per student isn't equal for both schools. Also the \$3 - \$2 = \$1 approach is wrong since you're combining per student dollar amounts that don't have the same number of students (mixing 200 and 300 students).

Case in point. You got the \$100 from \$0.5*200. Why not \$0.5*300? to get the amount of transfer?

You got the 700 by adding the 100 to the 600 original amount. 700 is to be shared by 300 students. the 500 is from 600 minus 100. 500 to be shared by 200 students. The amounts are not equal.

Hope this helps.
Intern  Joined: 28 May 2010
Posts: 26
Re: 700-800 PS...simple algebra actually  [#permalink]

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RGM wrote:
No worries.

Your calculations used this: ( r / 200 ) - s = ( r / 300 ) + s

when it should have used this: ( r - s ) / 200 = ( r + s ) / 300

See the difference. You set R as 600. 600/300 = \$2 and 600/200 = \$3

So you reasoned that amount to be transferred per student should be \$3 - \$2 = \$1. Divide by 2, and get \$0.5. \$0.5 per student. That's wrong. It is because the transfer isn't divided by the number of students - as such, it's lopsided again, and the amount of money per student isn't equal for both schools. Also the \$3 - \$2 = \$1 approach is wrong since you're combining per student dollar amounts that don't have the same number of students (mixing 200 and 300 students).

Case in point. You got the \$100 from \$0.5*200. Why not \$0.5*300? to get the amount of transfer?

You got the 700 by adding the 100 to the 600 original amount. 700 is to be shared by 300 students. the 500 is from 600 minus 100. 500 to be shared by 200 students. The amounts are not equal.

Hope this helps.

THANK YOU SO MUCH RGM. lol, that was stupid. 700/300 doesn't equal 2.5, it equals 2.333, so the result doesn't even equal. you're the best.
Manager  Joined: 04 Feb 2010
Posts: 121
Re: 700-800 PS...simple algebra actually  [#permalink]

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knabi wrote:
RGM wrote:
No worries.

Your calculations used this: ( r / 200 ) - s = ( r / 300 ) + s

when it should have used this: ( r - s ) / 200 = ( r + s ) / 300

See the difference. You set R as 600. 600/300 = \$2 and 600/200 = \$3

So you reasoned that amount to be transferred per student should be \$3 - \$2 = \$1. Divide by 2, and get \$0.5. \$0.5 per student. That's wrong. It is because the transfer isn't divided by the number of students - as such, it's lopsided again, and the amount of money per student isn't equal for both schools. Also the \$3 - \$2 = \$1 approach is wrong since you're combining per student dollar amounts that don't have the same number of students (mixing 200 and 300 students).

Case in point. You got the \$100 from \$0.5*200. Why not \$0.5*300? to get the amount of transfer?

You got the 700 by adding the 100 to the 600 original amount. 700 is to be shared by 300 students. the 500 is from 600 minus 100. 500 to be shared by 200 students. The amounts are not equal.

Hope this helps.

THANK YOU SO MUCH RGM. lol, that was stupid. 700/300 doesn't equal 2.5, it equals 2.333, so the result doesn't even equal. you're the best.

Sure thing. Nah, it's not stupid - we simply just don't see our mistakes from time to time, happens to the best of us.
Intern  Joined: 06 Jul 2010
Posts: 31
WE 1: Corporate Strategy
Re: 700-800 PS...simple algebra actually  [#permalink]

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1
I did this the same way as RGM at first, but after reading knabi's attempted solution, I actually think you were on track for a simpler method. You just didn't quite nail it.

if r = \$600, then the total cash given is \$1,200 (\$600 to school A, \$600 to school B) and the total students are 500 (200 at A, 300 at B). So for \$/student to be the same, it has to be \$1200 / 500 students = \$2.40/student. School A has \$600/200 = \$3/student, so they need to give away s = \$0.60 for each of their 200 students, or \$3/5 * 200.

Thus s = 3/5 * 200 = 600/5 = r/5

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Re: 700-800 PS...simple algebra actually  [#permalink]

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1
The way i solved it:

After s transfer both schools have total for head.
So at High School it will be:

r+s = (300)* (2r/500) = 6r/5
i.e. s = r/5
Manager  Joined: 12 Jun 2007
Posts: 97
Re: 700-800 PS...simple algebra actually  [#permalink]

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how i go about is let R =600 (because of 300, 200 i choose R=600) :

Lets S is fiven by middle school : 600+s /300 = 600-s /200
5s = 600
s = 120 (600/5)

This show S= r/5

Intern  Joined: 19 Jul 2010
Posts: 13
Re: 700-800 PS...simple algebra actually  [#permalink]

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I think you may be overcomplicating it.

If they get the same amount of funding per student in the end, then the first school will have 60% and the 2nd school will have 40% of the total funding (300 vs 200).

They start with 50% each. So, 2nd school has to transfer 10% of the total to the first school...which is 1/5 of the initial amount they got.
Intern  Joined: 04 Sep 2010
Posts: 24
Re: 700-800 PS...simple algebra actually  [#permalink]

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this is simpl..just use basic maths..
(r-s)/200 = (r+s)/300
s=r/5
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Re: 700-800 PS...simple algebra actually  [#permalink]

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300(r-s) = 200(r+s)
r = 5s
Intern  Joined: 09 Nov 2013
Posts: 8
Re: In a local school district, the high school and middle  [#permalink]

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plug smart numbers:
H got \$250 (r)
M got \$250 (r)
S= \$50
H+S=\$250+\$50=\$300
M-S=\$250-\$50=\$200
now H+S has \$1 per 1 student, and H-S has \$1 per 1 student. S=\$50, r=\$250 Check with answers: r/5=\$250/5=\$50
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Re: In a local school district, the high school and middle  [#permalink]

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Hi All,

This is a great question to TEST VALUES. The prompt tells us there are 500 total students. Later on, we're told that money has to move transferred, so that the end result is the same funding PER student. Let's TEST \$1 per student....

500 students
\$1 each
\$500 total

The question further states that the two schools each received \$R each, so each school received the same amount. That means…

R = \$250

So, the High School got \$250 for 300 students
and the Middle School got \$250 for 200 students

Later, the middle school transferred \$S to the High School to make the average \$1 per student, so the Middle School would have transferred \$50

S = \$50

Now, the High School has \$300 for 300 students
and the Middle School has \$200 for 200 students

Thus, we're looking for an answer that equals 50 when R=250. There's only one answer that matches..

GMAT assassins aren't born, they're made,
Rich
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Re: In a local school district, the high school and middle  [#permalink]

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knabi wrote:
In a local school district, the high school and middle school each received r dollars toward funding for the student arts program. The high school enrolled 300 students and the middle school enrolled 200 students. Later, the middle school transferred s dollars to the high school so that they would have received the same funding per student. Which of the following is equivalent to s?

A. r/2
B. r/3
C. r/4
D. r/5
E. r/6

After the transfer, the middle school has r - s and the high school has r + s dollars. The funding per student for the middle school and the high school are (r - s)/200 and (r + s)/300, respectively. Since we are told that the funding per student is the same for both schools, we have:

(r - s)/200 = (r + s)/300

300r - 300s = 200r + 200s

100r = 500s

r = 5s

s = r/5

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Re: In a local school district, the high school and middle  [#permalink]

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EMPOWERgmatRichC I tried the testing values method and keep getting the wrong answer e) I attached my step by step. Any idea why I getting this wrong?

Attachment: IMG_1761.jpg [ 1.84 MiB | Viewed 615 times ]
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Re: In a local school district, the high school and middle  [#permalink]

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1
Hi whollymoses,

TESTing VALUES works nicely on this question, but you have to be careful with your note-taking. Based on the work that you showed, you're doing repeated calculations at the END of your work (instead of just doing 1 calculation at the beginning) - so that's a somewhat inefficient way to approach the work.

R = 6,000 meaning that EACH School received \$6,000 dollars - and the TOTAL money = \$12,000

The prompt tells us that the Middle School will transfer \$S to the High School so that the Schools will have the same money PER STUDENT.

Between the 2 Schools, there are 300+200 = 500 students, so \$12,000/500 = \$24 per student.
Right now, the High School has \$6,000/300 = \$20 per student.... but we need it to be \$24/student.
Thus, the Middle School has to transfer (\$4)(300) = \$1200 to the High School.

We're looking for an answer that equals 1200 when R = 6000. There's only one answer that matches.

GMAT assassins aren't born, they're made,
Rich
_________________ Re: In a local school district, the high school and middle   [#permalink] 13 Nov 2019, 19:09

# In a local school district, the high school and middle  