The enrollments at College X and College Y both grew by 8 percent from

Question

The enrollments at College X and College Y both grew by 8 percent from 1980 to 1985. If the enrollment at College Xgrew by 800 and the enrollment at College Y grew by 840, the enrollment at College Y was how much greater than the enrollment at College Xin 1985?

(A) 400

(B) 460

(C) 500

(D) 540

(E) 580

(A) 400

(B) 460

(C) 500

(D) 540

(E) 580

GMATinsight · Answer

8% of X = 800

and 8% of Y = 840

i.e. 8% of Y - 8% of X = 840 - 800

i.e. 8% (Y-X) = 40

i.e. Y-X = 500

Answer: Option C

rocko911 · Answer

(108 * X)/ 100 = 800 + X

X= 10,000

(108 * Y)/100 = 840 + Y

Y = 10,500

Ans = Y-X = 500

C

Abhishek009 · Answer

Enrolment in 1980 in College X is  \frac{800}{8}*100 = 10000 
Enrolment in 1985 in College X is  1000*\frac{108}{100} = 10800

Enrolment in 1980 in College Y is  \frac{840}{8}*100 = 10500 
Enrolment in 1985 in College Y is  10500*\frac{108}{100} = 11340

So, enrollment at College Y > enrollment at College X in the year by 500, Answer must be (C) 500

anoushhashemi · Answer

Thanks for the clear answer, however, if the 1985 enrollments for X and Y are 10800 and 11340, respectively, then wouldn't the enrollment difference for year 1985 be 540?

generis · Answer

EDIT:  anoushhashemi , I didn't see your post. I have the same question.

I seem to be missing something. The posters above me are rarely (never?) wrong, let alone at the same time.

Enrollment grows by 8 percent (OF 1980 figure) from 1980 to 1985
X increases by 800 students
Y increases by 840 students

.08x = 800 
 x=\frac{800}{.08}=\frac{80,000}{8}= 10,000   students in 1980
10,000 + 800 =  10,800  students in 1985 at X

.08y = 840 
 y=\frac{840}{.08}=\frac{84,000}{8}10,500   students in 1980
10,500 + 840 =  11,340  students in 1985 at Y
 
11,340 - 10,800 = 540

Answer D

niks18  ,  chetan2u  - am I missing something?

chetan2u · Answer

Hi,
You are not wrong...
The members who have choosen the method of 8% of (y-x) have found the difference of 500 for year 1980...
But in year 1985, an increase of further 840-800=40 is done, so answer there too should be 500+840-800=540..
Your approach and answer both are correct 100%

niks18 · Answer

Hi  generis

your answer is correct. :thumbup:

We can also use below alternate approach. Let  y  &  x  be the new enrollments after increase so we have

y=\frac{100y}{108}+840  (new enrollment=old enrollment + increase)

x=\frac{100x}{108}+800 . subtract both equation to get increased enrollment

=>y-x=\frac{100(y-x)}{108}+40

=>y-x-\frac{100(y-x)}{108}=40=>\frac{8(y-x)}{108}=40

=>y-x=540

rahul16singh28 · Answer

Given,  1.08x = x + 800 ,  X = 10,000 
 1.08y = y + 800 ,  Y = 10,500 
 Y - X = 500 .. This is in 1980
 (Y+840)- (X+800) = 540 ... This is in 1985.

ScottTargetTestPrep · Answer

Let’s let X = the enrollment at College X in 1980 and Y = the enrollment at College Y in 1980. We can create the following equations:

1.08X = X + 800

0.08X = 800

X = 10,000

and

1.08Y = Y + 840

0.08Y = 840

Y = 10,500

Thus, in 1985, the enrollment in College Y was 10,500 + 840 = 11,340 and the enrollment in College X was 10,000 + 800 = 10,800, and so, in 1985, the enrollment in College Y was 11,340 - 10,800 = 540 greater than that of College X.

Answer: D

GmatPoint · Answer

Let the total enrollments at college X be X in 1980, and that of Y be Y in 1980.

Given that 8% of Y - 8% of X = 840 - 800

Y*8/100 - X*8/100 = 40

Y - X = 500.

Since the difference in 1980 was 500, difference in 1985 = 500 + (840 - 800) = 540

Thus, the correct option is D.

bumpbot · Answer

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The enrollments at College X and College Y both grew by 8 percent from

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