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In a local school district, the high school and middle [#permalink]
05 Jul 2010, 12:42

3

This post was BOOKMARKED

00:00

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Difficulty:

45% (medium)

Question Stats:

62% (02:25) correct
38% (01:29) wrong based on 172 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?

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. _________________

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.

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.

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.

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.

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.

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.

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

Re: In a local school district, the high school and middle [#permalink]
03 Nov 2015, 16:51

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