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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] New post   Updated on: 20 Jul 2015, 02:44
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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%
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C

Originally posted by ong on 05 Aug 2006, 08:25.
Last edited by Bunuel on 20 Jul 2015, 02:44, 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] New post   27 May 2016, 09:29
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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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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink] New post   05 Aug 2006, 10:08
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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] New post   05 Aug 2006, 10:24
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] New post   05 Aug 2006, 12:12
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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] New post   20 Jul 2015, 03:18
from 90 - 80 83.33% = 5/6

from 80 - 90 6/5 = 1.2 = 20%
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Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink] New post   24 Jan 2017, 07:12
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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] New post   26 Oct 2017, 19:36
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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] New post   05 Dec 2017, 04:35
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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
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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] New post   01 Dec 2020, 08:06
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ong wrote:
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%


A quick approach is to test values that satisfy the given information.

Given: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in 1990
Important: We can save ourselves some time if we recognize that 83 1/3% is equivalent to 5/6.

So we can rewrite the given information as follows:
Given: Enrollment in City College in 1980 was 5/6 the enrollment in 1990

So, it COULD be the case that the 1990 enrolment was 6 students, and the 1980 enrolment was 5

What was the percent increase in the college’s enrollment from 1980 to 1990?
Percent change = (100)(new - old)/old
= (100)(6 - 5)/5
= (100)(1/5)
= 20%

Answer: C

Cheers,
Brent
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Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink] New post   24 Aug 2021, 23:05
So what I did which made life slightly easier was identify how I could turn 83.33% to a fraction and I noticed -
33.33 + 50 = 83.33
This means 1/3 + 1/2 would add to give the same value => 83.33 can be represented as 5/6 in fractions

Now,
1980 ; 1990
5/6 x ; x

Thus,
Option 1 - Assume x=600
Percent change = 600 - 500/ 500 = 1/5 = 20%

Option 2 - (x - 5/6 x) / (5/6x) = 1/5 = 20%
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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink] New post   31 Aug 2021, 02:34
Expert Reply

Solution



Given
• Enrollment in City College in 1980 was 83 1/3 percent of enrollment in 1990
To find
• The percent increase in the college’s enrollment from 1980 to 1990?


Approach and Working out
• Let the enrollment in City College in 1990 be x.
    o Then, the enrollment in City College in 1980 = 83 1/3 percent of enrollment in 1990 = 5x/6
• The percent increase from 5x/6 to x = [(x - 5x/6)/(5x/6)] * 100
    o [(x/6)/(5x/6)] * 100
    o (1/5) * 100
    o 20%

Correct Answer: Option C
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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink] New post   04 Dec 2023, 07:38
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Re: Enrollment in City College in 1980 was 83 1/3 percent of enrollment in [#permalink]
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