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School X and School Y

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by j_shreyans » Thu Sep 11, 2014 2:42 am
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


Guys ,

where i am wrong , pls correct me.

Let last year no. of students in school X be x and in school Y be y

x+y=4000 ----------(given in ques)

this year 7% more than the last year so = x(1+7/100) will be 1.07x this year students in school X

and in School Y = y(1+3/100) will be 1.03y this year students in school Y

According to the question :

1.07x=1.03y+40

So now we have two equation with two variables .

Pls help me out
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by GMATGuruNY » Thu Sep 11, 2014 3:05 am
j_shreyans wrote: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 can PLUG IN THE ANSWERS, which represent the number of students in school Y last year.
Since the enrollment in Y grows by 3% = 3/100, the correct answer choice must be a multiple of 100.
Eliminate A, C and D.
When the correct answer choice is plugged in, X's enrollment will grow by 40 more students than Y's.

Answer choice B: 1600
Increase in Y this year = 3% of 1600 = 48.
Since last year there were 4000 students between the 2 schools, X's enrollment last year = 4000-1600 = 2400.
Increase in X this year = 7% of 2400 = 168.
X's increase - Y's increase = 168-48 = 120.
Doesn't work.

The correct answer is E.

Answer choice E: 2400
Increase in Y this year = 3% of 2400 = 72.
Since last year there were 4000 students between the 2 schools, X's enrollment last year = 4000-2400 = 1600.
Increase in X this year = 7% of 1600 = 112.
X's increase - Y's increase = 112-72 = 40.
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by KlavdiaTeratai » Thu Sep 11, 2014 7:53 am
Can you please solve it directly without plugging in the answer choices? I understand that on the test plugging in the answers for such questions would be the best option, however, I would like to see the best way to solve this problem directly for thorough understanding of the concepts tested in this question. I have tried different approaches of solving this problem directly, but there comes a point in the calculations where I have to round up, leading me to get the wrong answer. Thank you very much!
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by GMATGuruNY » Thu Sep 11, 2014 8:09 am
j_shreyans wrote: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
Algebraic solution:

There were 4000 total enrolled students last year at both schools.
x + y = 4000.

School X (whose enrollment increased by 7%) grew by 40 more students than school Y (whose enrollment increased by 3%).
Since X's 7% increase is 40 more than Y's 3% increase, we get:
0.07x = 0.03y + 40
7x = 3y + 4000.

The first equation can be rephrased as follows:
x = 4000 - y
7x = 28000 - 7y.

Since 7x = 3y + 4000 and 7x = 28000 - 7y, the expressions in red are equal:
3y + 4000 = 28000 - 7y
10y = 24000
y = 2400.

The correct answer is E.
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by j_shreyans » Thu Sep 11, 2014 7:57 pm
Hi ,

Shouldn't be like below:

7% more than it was last year = x(1+7/100) =1.07x

3% more than it was last year = y(1+3/100) = 1.03y

Pls help me on this.
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by [email protected] » Thu Sep 11, 2014 9:31 pm
Hi j_shreyans,

The math translations that you've done would represent the CURRENT number of students at School X and School Y.

The third sentence tells us that "School X GREW by 40 more students than School Y did", so we don't need the current number of students at the two schools - we need the GROWTH at School X and the GROWTH at School Y

GROWTH at School X = .07X

GROWTH at School Y = .03Y

School X grew by 40 more students than school Y did --> .07X = .03Y + 40

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by j_shreyans » Thu Sep 11, 2014 10:02 pm
ooohhhh Yes.

Thanks for correting me .:)
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