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S95-09

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The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?


A. 480
B. 1600
C. 1920
D. 2080
E. 2400
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 01:49
Official Solution:


The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?


A. 480
B. 1600
C. 1920
D. 2080
E. 2400


We must determine the number of students enrolled at school \(Y\) last year.

We can set up two equations to describe this situation.

First: last year, the number of students enrolled at school \(X\) (which we will call \(x\)) and the number of students enrolled at school \(Y\) (which we will call \(y\)) together were equal to 4000. So \(x + y = 4000\).

Second: the number of additional students at school \(X\) this year (equal to 7 percent of last year's student body of \(x\), or \(0.07x\)) is 40 more than the number of additional students at school \(Y\) this year (equal to 3 percent of last year's student body of \(y\), or \(0.03y\)). Therefore, \(0.07x - 0.03y = 40\).

Now we use substitution to solve for \(y\). Rearrange the first equation to express \(x\) in terms of \(y\): \(x = 4000 - y\). Then, substitute for \(x\) in the second equation: \(0.07(4000 - y) - 0.03y = 40\).

Distribute the 0.07 to get: \(0.07(4000) - 0.07y - 0.03y = 40\)

Simplify and combine like terms: \(280 - 40 = (0.07 + 0.03)y\), or \(240 = 0.1y\).

Multiply both sides by 10: \(2400 = y\).


Answer: E
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New post 13 Feb 2015, 03:41
Bunuel wrote:
Official Solution:


The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?


A. 480
B. 1600
C. 1920
D. 2080
E. 2400


We must determine the number of students enrolled at school \(Y\) last year.

We can set up two equations to describe this situation.

First: last year, the number of students enrolled at school \(X\) (which we will call \(x\)) and the number of students enrolled at school \(Y\) (which we will call \(y\)) together were equal to 4000. So \(x + y = 4000\).

Second: the number of additional students at school \(X\) this year (equal to 7 percent of last year's student body of \(x\), or \(0.07x\)) is 40 more than the number of additional students at school \(Y\) this year (equal to 3 percent of last year's student body of \(y\), or \(0.03y\)). Therefore, \(0.07x - 0.03y = 40\).

Now we use substitution to solve for \(y\). Rearrange the first equation to express \(x\) in terms of \(y\): \(x = 4000 - y\). Then, substitute for \(x\) in the second equation: \(0.07(4000 - y) - 0.03y = 40\).

Distribute the 0.07 to get: \(0.07(4000) - 0.07y - 0.03y = 40\)

Simplify and combine like terms: \(280 - 40 = (0.07 + 0.03)y\), or \(240 = 0.1y\).

Multiply both sides by 10: \(2400 = y\).


Answer: E



X + Y =4000
Difference in no. of stds joined in current year = 40
Let us assume Y as 2400 and X 1600

X increased by 7% and Y by 3%.

X = 1712, Y - 2472
Diff in no. of students joined in current year is 40.


Hi Bunel ,

I was not clear on this equation. .07X - .03 Y = 40

It should be 1.07X -1.03Y = 40. The questions states that 7% more than last year.

can u please clarify

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S95-09 [#permalink]

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New post 03 Apr 2015, 10:33
It states that X grew by.......
therefore,check "Grew" word...which means the difference between final state and original state.
Thus,
0.07X....
Similarly for Y.....

Hope this helps!! :-D

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Re: S95-09 [#permalink]

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New post 07 Jul 2015, 03:34
So, I have a question. The stem says that for school X the number of students this year was 7% more than that of ast year. So, if x is the number of students last year, shouldn't the number of students this year be x + 0.07x = 1.07x (the total students this year for school X).

Similarly for Y, this year there would be 1.03Y.

The stem also says that this year X grew by 40 more students than Y did. Does not this mean that the number of students X has this year (1.07X) is 40 more than the number of students school Y has this year (1.03Y)? Because this translates into 1.07X = 1.03Y+40.

Then, using the same relationship X+Y = 4000, from where we result in X = 4000 - Y, and substituting into the relationship above (1.07X = 1.03Y+40) we result in a different number for Z.

So, what I do not understand is why 40 relates only to the % of the increase (0.07X or 0.03Y), and not to the increased student number itself (1.07X and 1.03Y)?

Thanks!

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Re: S95-09 [#permalink]

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New post 01 Oct 2016, 03:08
I have the same doubt as pacifist85.On increasing from x by 7% it becomes 1.07x-1.03y=40

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New post 01 Oct 2016, 03:10

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New post 23 Aug 2017, 11:16
For what it's worth, (A) (C) and (D) can all be eliminated right away because they cannot be increased by 3% without creating partial students (gross). This leaves (B) and (E). From here it doesn't take long to figure which is correct by plugging each into the stem.

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Re: S95-09   [#permalink] 23 Aug 2017, 11:16
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