The number of students enrolled at school X this year is 7 percent

Let us assume, number of student enrolled at school X last year is x
number of student enrolled at school Y last year is y.

this year enrollment at school X increases by 7% i.e. 7x/100
this year enrollment at school Y increases by 3% i.e. 3y/100

As per question stem, school X grew by 40 more students than school Y did
this means 7x/100 = 40+ 3y/100 --- equation(1)

last year total student at school was 4000
i.e. x+y = 4000 --- equation(2)

Equate Equation (1) & (2), y will be 2400

Answer: E

Regards,
Ammu

......................... School X ................ School Y ................... Total

Last Yr ................. 4000-a ..................... a ............................ 4000 (Assumed "a" as number of students in last yr)


Difference ......... \((4000-a) * \frac{7}{100}\) ................. \(a * \frac{3}{100}\) ..............
(Growth)

Given that "Growth" of School X is 40 more than growth of School Y

Setting up the equation

\(\frac{3a}{100} + 40 = \frac{7}{100} * (4000-a)\)

\(10a = 7 * 4000 - 4000\)

a = 2400

Answer = E

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.

Hey - I'm just wondering why it isnt 1.07X and 1.03Y as the question states that its "7% more than last year"... same with Y

Since the problem states that:

" If school X grew by 40 more students than school Y did "

It talks about the individual growth and not the total number of students.

Let x be the number of students in X and y in Y.

As per the question,

x +y = 4000 ... (1)
0.07x - 0.03y = 40
=> 7x - 3y = 4000 ... (2)

Solving equation 1 and 2:

y = 2400

why cant we say that 1.07X=1.03Y+40 ?

can you elaborate more?

is it because it talks about the growth of students?

How would the question have to be if we were to write 1.07X=1.03Y+40 ? "at the end of the year the school X had 40 students more than Y" ? or this would be the same as the stem?

These differences in language are so subtle..

iliavko

Since the question stem says - 'If school X grew by 40 more students than school Y did' i.e. 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), hence .07x=.03y+40.

Hope this helps.

let x and y=number of students at X and Y respectively last year
x=4000-y
.07(4000-y)-.03y=40
y=2400 students
E

Let the number of students last year be X and Y. This year the number of students will be 1.07X and 1.03Y.

We have 2 equations -

a) 0.07X=0.03Y+40
b) X+Y=4000

Solve the two and get the value of Y as 2400 or option E.

Since no of increased no. of student can not be fraction,

so the 7 percent of y shoud be integer, only C and E can do that than 40 plus 0.07y shoud also be able to divided with .03.

only E is left

School	Last Year	This Year	Increase
School X	100x	107x	7x
School Y	100y	103y	3y

(1) 7x - 3y = 40

(2) 100x + 100y = 4000

x + y = 40

Multiplying (2) by 7 ->

(2) 7x + 7y = 280
(1) 7x - 3y = 40

=> 10y = 240
=> y = 24

Number of students enrolled in school Y last year = 100y = 100 x 24 = 2400. Choice E.

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Harshavardhan