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

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The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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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
[Reveal] Spoiler: OA

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Re: The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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Bunuel wrote:
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


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
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Re: The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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New post 01 Feb 2018, 23:37
Bunuel wrote:
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



(108 * X)/ 100 = 800 + X

X= 10,000

(108 * Y)/100 = 840 + Y

Y = 10,500

Ans = Y-X = 500

C
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The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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Bunuel wrote:
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



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
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Re: The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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Abhishek009 wrote:
Bunuel wrote:
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



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


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?
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The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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New post 03 Feb 2018, 17:06
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Bunuel wrote:
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

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
Quote:
the enrollment at College Y was how much greater than the enrollment at College Xin 1985?

11,340 - 10,800 = 540

Answer D

niks18 , chetan2u - am I missing something?
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Re: The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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generis wrote:
Bunuel wrote:
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

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
Quote:
the enrollment at College Y was how much greater than the enrollment at College Xin 1985?

11,340 - 10,800 = 540

Answer D

niks18 , chetan2u - am I missing something?


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%
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Re: The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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New post 03 Feb 2018, 19:47
generis wrote:
Bunuel wrote:
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

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
Quote:
the enrollment at College Y was how much greater than the enrollment at College Xin 1985?

11,340 - 10,800 = 540

Answer D

niks18 , chetan2u - am I missing something?


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

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

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

\(=>y-x=540\)
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The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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New post 04 Feb 2018, 12:38
Bunuel wrote:
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


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.
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Re: The enrollments at College X and College Y both grew by 8 percent from [#permalink]

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New post 05 Feb 2018, 09:25
Bunuel wrote:
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


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
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Re: The enrollments at College X and College Y both grew by 8 percent from   [#permalink] 05 Feb 2018, 09:25
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