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

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Intern
Joined: 28 May 2010
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In a local school district, the high school and middle [#permalink]  05 Jul 2010, 12:42
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63% (02:27) correct 37% (01:27) wrong based on 148 sessions
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

[Reveal] Spoiler:
-------------
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

so, S=100dollars=R=600/6=100. But that's not the answer!! ah, please help.
[Reveal] Spoiler: OA
Manager
Joined: 04 Feb 2010
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Re: 700-800 PS...simple algebra actually [#permalink]  05 Jul 2010, 13:34
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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
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Re: 700-800 PS...simple algebra actually [#permalink]  05 Jul 2010, 13:45
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. Manager Joined: 19 Jul 2009 Posts: 53 Location: baltimore, md Schools: kellogg, booth, stern, ann arbor Followers: 1 Kudos [?]: 27 [0], given: 3 Re: 700-800 PS...simple algebra actually [#permalink] 05 Jul 2010, 14:19 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. _________________ Paaaaayyy Meeeee!!!!! Intern Joined: 28 May 2010 Posts: 37 Followers: 0 Kudos [?]: 17 [0], given: 15 Re: 700-800 PS...simple algebra actually [#permalink] 05 Jul 2010, 14:45 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: 200 Followers: 1 Kudos [?]: 34 [1] , given: 8 Re: 700-800 PS...simple algebra actually [#permalink] 05 Jul 2010, 15:07 1 This post received KUDOS 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
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Re: 700-800 PS...simple algebra actually [#permalink]  05 Jul 2010, 16:37
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: 200 Followers: 1 Kudos [?]: 34 [0], given: 8 Re: 700-800 PS...simple algebra actually [#permalink] 05 Jul 2010, 20:01 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.
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Re: 700-800 PS...simple algebra actually [#permalink]  07 Jul 2010, 16:52
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 Either method is fine, but I thought I'd share this in case it's helpful. Manager Status: one more time Joined: 05 Jul 2010 Posts: 72 Location: United States Concentration: Strategy, Entrepreneurship GMAT 1: 700 Q49 V37 GMAT 2: 740 Q50 V40 GPA: 3.48 WE: Web Development (Computer Software) Followers: 0 Kudos [?]: 8 [0], given: 6 Re: 700-800 PS...simple algebra actually [#permalink] 10 Jul 2010, 08:09 The way i solved it: Total per head = 2r/500 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: 128 Followers: 5 Kudos [?]: 49 [0], given: 2 Re: 700-800 PS...simple algebra actually [#permalink] 16 Jul 2010, 01:32 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 Answer is D Intern Joined: 19 Jul 2010 Posts: 13 Followers: 0 Kudos [?]: 3 [0], given: 1 Re: 700-800 PS...simple algebra actually [#permalink] 27 Jul 2010, 14:44 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: 13 Jul 2010 Posts: 4 Location: India Followers: 0 Kudos [?]: 0 [0], given: 2 Re: 700-800 PS...simple algebra actually [#permalink] 29 Jul 2010, 02:56 yups... i got the answer r/5... btw what is the standard or this question? is it 700+ range? Manager Joined: 04 Apr 2009 Posts: 68 Location: United Kingdom Schools: Cornell Followers: 1 Kudos [?]: 24 [0], given: 6 Re: 700-800 PS...simple algebra actually [#permalink] 24 Aug 2010, 15:11 This one is surely a 49ish question of Quant (because I got this on one of my preps where I scored 49 ) Manager Joined: 04 Sep 2010 Posts: 51 Followers: 2 Kudos [?]: 0 [1] , given: 1 Re: 700-800 PS...simple algebra actually [#permalink] 15 Oct 2010, 08:01 1 This post received KUDOS this is simpl..just use basic maths.. (r-s)/200 = (r+s)/300 s=r/5 Manager Status: ISB, Hyderabad Joined: 25 Jul 2010 Posts: 176 WE 1: 4 years Software Product Development WE 2: 3 years ERP Consulting Followers: 6 Kudos [?]: 35 [0], given: 15 Re: 700-800 PS...simple algebra actually [#permalink] 17 Oct 2010, 18:44 D Solve for s using the eqn: (r-s)/200 = (r+s)/300 _________________ -AD VP Status: There is always something new !! Affiliations: PMI,QAI Global,eXampleCG Joined: 08 May 2009 Posts: 1356 Followers: 14 Kudos [?]: 177 [0], given: 10 Re: 700-800 PS...simple algebra actually [#permalink] 15 Jun 2011, 23:44 300(r-s) = 200(r+s) r = 5s _________________ Visit -- http://www.sustainable-sphere.com/ Promote Green Business,Sustainable Living and Green Earth !! Current Student Joined: 09 Nov 2013 Posts: 12 Followers: 2 Kudos [?]: 7 [0], given: 53 Re: In a local school district, the high school and middle [#permalink] 17 Nov 2013, 04:47 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]  22 Sep 2014, 08:24
Can you explain how we got this (r-s)/200 = (r+s)/300? Thanks in advance.
Re: In a local school district, the high school and middle   [#permalink] 22 Sep 2014, 08:24
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# In a local school district, the high school and middle

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