Mo2men 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 Y.
Since the increase in Y = 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, a 7% increase in X will be 40 more than a 3% increase in Y.
B: Y=1600, implying that X = 4000-1600 = 2400
7% of X = (7/100)(2400) = 168.
3% of Y = (3/100)(1600) = 48.
Difference = 168-48 = 120.
Here, the difference is too great.
Eliminate B.
The correct answer is
E.
Algebraic solution:
A 7% increase in X is 40 more than a 3% increase in Y:
(7/100)(X) = (3/100)Y + 40
7X = 3Y + 4000
There are 4000 students in total:
X + Y = 4000
X = 4000-Y
Substituting X=4000-Y into 7X = 3Y + 4000, we get:
7(4000-Y) = 3Y + 4000
28000 - 7Y = 3Y + 4000
24000 = 10Y
Y = 2400.
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