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