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

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Math Expert
Joined: 02 Sep 2009
Posts: 62442
At the beginning of 2010, 60% of the population of Town X lived in the  [#permalink]

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20 Dec 2019, 02:34
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Difficulty:

45% (medium)

Question Stats:

69% (02:25) correct 31% (03:06) wrong based on 96 sessions

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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]

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20 Dec 2019, 03:26
1
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%

Are You Up For the Challenge: 700 Level Questions

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]

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25 Dec 2019, 10:03
1
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%

Are You Up For the Challenge: 700 Level Questions

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%

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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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17 Jan 2020, 09:28
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)
Re: At the beginning of 2010, 60% of the population of Town X lived in the   [#permalink] 17 Jan 2020, 09:28
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