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The enrollments at College X and College Y both grew by 8 percent from [#permalink]
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01 Feb 2018, 23:24
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Re: The enrollments at College X and College Y both grew by 8 percent from [#permalink]
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02 Feb 2018, 00:20
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% (YX) = 40 i.e. YX = 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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02 Feb 2018, 00:37
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 = YX = 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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02 Feb 2018, 10:32
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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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03 Feb 2018, 17:26
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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) 500Thanks 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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03 Feb 2018, 18:06
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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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03 Feb 2018, 18:39
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 (yx) have found the difference of 500 for year 1980... But in year 1985, an increase of further 840800=40 is done, so answer there too should be 500+840800=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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03 Feb 2018, 20:47
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 generisyour 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 \(=>yx=\frac{100(yx)}{108}+40\) \(=>yx\frac{100(yx)}{108}=40=>\frac{8(yx)}{108}=40\) \(=>yx=540\)



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The enrollments at College X and College Y both grew by 8 percent from [#permalink]
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04 Feb 2018, 13:38
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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05 Feb 2018, 10:25
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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