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

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Bunuel
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At the beginning of 2010, 60% of the population of Town X lived in the [#permalink] New post   20 Dec 2019, 03:34
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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%


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Re: At the beginning of 2010, 60% of the population of Town X lived in the [#permalink] New post   20 Dec 2019, 04:26
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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%


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Explanation:
let total population in town X: 1000
In South: 600
In North: 400

Population Grew by 5.5%: 1055 , total increase by - 55
Population of south grew by 4.5%, so new population:600 + 600* 0.045 = 27
Population grew in North: 55-27 = 28
%= 28/400 = 7%
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Re: At the beginning of 2010, 60% of the population of Town X lived in the [#permalink] New post   25 Dec 2019, 11:03
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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%

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First of all, a low rise in proportionally larger population means the smaller proportion population must rise higher than that of larger population to match the overall rise. Thus, A and B are out. E would have been only considerable had the smaller proportion been very small(~20.3%), so E is out.

Let percentage rise for rest(40%) of the population living in north = x
Now, Rise in North + Rise in South = Total rise
40% * x + 60% * 4.5 = 1 * 5.5
0.4x + 2.7 = 5.5
x = \(\frac{2.8}{0.4}\) = 7
7%

Answer D.
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Re: At the beginning of 2010, 60% of the population of Town X lived in the [#permalink] New post   17 Jan 2020, 10:28
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] New post   19 Jul 2022, 05:05
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]
19 Jul 2022, 05:05
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