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# If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990

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Math Expert
Joined: 02 Sep 2009
Posts: 58402
If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990  [#permalink]

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01 Mar 2018, 00:13
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5% (low)

Question Stats:

86% (00:35) correct 14% (00:27) wrong based on 34 sessions

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If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990, what was the percent of increase in enrollment?

(A) 3%
(B) 25%
(C) 125%
(D) 300%
(E) 400%

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If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990  [#permalink]

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01 Mar 2018, 00:19
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Bunuel wrote:
If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990, what was the percent of increase in enrollment?

(A) 3%
(B) 25%
(C) 125%
(D) 300%
(E) 400%

The enrollment saw an increase of $$9000(12000-3000)$$ from 3000 in 1965 to 12000 in 1990.

Therefore, the percentage increase in enrollment is $$\frac{9000}{3000}*100 = 300$$% (Option D)
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Re: If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990  [#permalink]

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03 Mar 2018, 00:03
Bunuel wrote:
If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990, what was the percent of increase in enrollment?

(A) 3%
(B) 25%
(C) 125%
(D) 300%
(E) 400%

IMO D
12000-3000*100/3000
9000*100/3000= 3*100
=300%
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Re: If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990  [#permalink]

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03 Mar 2018, 20:49
% increase= $$\frac{Change}{Initial Value}$$*100

= $$\frac{(12000-3000)}{3000}$$*100

= $$\frac{9000}{3000}$$*100

= 300%

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If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990  [#permalink]

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03 Mar 2018, 23:03
In general, if the final value is K times the initial value, then the increase is always (K-1)*100 %.

In this particular case K=4.
If the enrollment at State U. was 3,000 in 1965 and 12,000 in 1990   [#permalink] 03 Mar 2018, 23:03
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