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# At the beginning of 2010, 60% of the population of Town X lived in the

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Re: At the beginning of 2010, 60% of the population of Town X lived in the [#permalink]
Bunuel wrote:
At the beginning of 2010, 60% of the population of Town X lived in the south and the rest lived in the north. During 2010, the population of Town X grew by 5.5%. If the population in the south grew by 4.5%, by how much did the population in the north grow?

A. 1%
B. 3.5%
C. 6.5%
D. 7%
E. 13.75%

5.5a=0.6a4.5+0.4ax
11a/2=3a/5*9/2+2ax/5
11a/2-3a/5*9/2=2ax/5
55a-27a/10=2ax/5
55a-27a/2=2ax
28a=4ax
x=7

Ans (D)
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Re: At the beginning of 2010, 60% of the population of Town X lived in the [#permalink]
Let 1000 people live in 2010 in town X.
600 lived in the south, and 400 lived in the north.

Population increase = 1000*5.5/100 = 55

Population increase in south = 600*4.5/100 = 27.

Thus, population increase in the north = 55 - 27 = 28

Required percentage = (28/400)*100 = 7%.
Thus, the correct answer is D.
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Re: At the beginning of 2010, 60% of the population of Town X lived in the [#permalink]
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Re: At the beginning of 2010, 60% of the population of Town X lived in the [#permalink]
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