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Updated on: 10 Jul 2019, 03:41
Originally posted by Bunuel on 30 May 2017, 10:05.
Last edited by Sajjad1994 on 10 Jul 2019, 03:41, edited 1 time in total.
Last edited by Sajjad1994 on 10 Jul 2019, 03:41, edited 1 time in total.
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Difficulty:
5%
(low)
Question Stats:
91% (00:42) correct
9%
(00:43)
wrong
based on 103
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A town has a population growth rate of 10% per year. The population in 1990 was 2000. What was the population in 1992?
(A) 1600
(B) 2200
(C) 2400
(D) 2420
(E) 4000
(A) 1600
(B) 2200
(C) 2400
(D) 2420
(E) 4000
Source: Nova GMAT
Difficulty Level: 550
Difficulty Level: 550
30 May 2017, 12:43
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Bunuel
Method I - calculate each "step" in the percent increase, using (original #) + (percentage of that #) = new total
1. 1990, population is 2,000
2. 1991, population increases by 10% of 2,000.
Ten percent of 2000 is 200.
Increase = 200 + 2,000 = 2,200 (new total)
3. 1992, population increases by another 10%, this time of 2,200.
Ten percent of 2,200 is 220.
Increase of 220 + 2,200 = 2,240
By 1992 the population has grown to 2,240
Method II - multiply the multipliers, then multiply by original value
1. A 10% increase has a multiplier of 1.1. That percent increase happens twice.
2. (1.1)*(1.1) = 1.21
3. 1.21 * 2,000 = 2,240
[Obviously, you can also simply multiply 1.1 * 1.1 * 2000.]
