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Enrollment in City College in 1980 was 83 1/3 percent of enrollment in

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Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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Enrollment in City College in 1980 was 83 1/3 percent of enrollment in 1990. What was the percent increase in the college’s enrollment from 1980 to 1990?

A. 10%
B. 16 2/3 %
C. 20%
D. 25%
E. 183 1/3%
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Jul 2015, 01:44, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 05 Aug 2006, 09:08
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C =>20%

Let no is 1990 = x, then 1980 = 5/6x

Hence difference = (x-5x/6)/5x/6 = x/6 * 6/5x = 1/5 = 20%

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 05 Aug 2006, 09:24
It is C.

Let 100 is the enrollment in 1990.
Enrollment in 1980= 83.3


% increase= (Increase from 1980 to 1990)/Enrollment in 1983

=>16.7*100/83.3 =20%

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 05 Aug 2006, 11:12
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1990 -> x
1980 --> 0.833(x)

% increase from 80 to 90 is (New - old/old) * 100
x - 0.833x/0.833x * 100
0.167x/0.833x * 100
(1/6)/(83 1/3) * 100
(1/6) * (3/250) * 100
= approx 20%

C

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 20 Jul 2015, 02:18
from 90 - 80 83.33% = 5/6

from 80 - 90 6/5 = 1.2 = 20%

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 27 May 2016, 08:29
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83.33% --> 83%

1980 --> 83 students
1990 --> 100 students

\((New-Old)/(Old)\)

\((100-83)/83\)

\(17/83\)

\(1/5\) --> 20%

Answer C

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Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 24 Jan 2017, 06:12
An easier method that I adopted.

\(X_{1980}\) = 83\(\frac{1}{3}\) % x \(X_{1990}\)

\(X_{1980}\) = \(\frac{250}{300}\)\(\) x \(X_{1990}\) - Upon expanding 83\(\frac{1}{3}\)%

So now let us assume that \(X_{1980}\) = 250 and \(X_{1990}\) = 300.

% increase = \(\frac{300-250}{250}\) x 100
= \(\frac{1}{5}\) x 100
= 20%

I think plugging in easy values will save considerable time rather than working with decimals. That it the trap the GMAT exam wants us to fall into.

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 26 Oct 2017, 18:36
1990: X studens;
1980: 83 1/3% * X = 250X/300 = 5/6x;

Hence 5/6X * K = X -> K = 6/5=1,2 or 20%.

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]

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New post 05 Dec 2017, 03:35
Let Enrollment in City College in 1990 -- x
Therefore, Enrollment in City College in 1980 -- 83 1/3 % of x

PERCENT INCREASE in enrollment FROM 1980 To 1990 =

= (x - 83 1/3 % of x)
_____________________
83 1/3 % of x

= 16 2/3 % of x
__________________
83 1/3 % of x

= 50/3 % of x
__________________
250/3 % of x

= 1/5 = 20%

Ans : C

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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in   [#permalink] 05 Dec 2017, 03:35
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