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

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At the beginning of 2010, 60% of the population of Town X [#permalink] New post 16 Dec 2011, 04:51
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64% (02:33) correct 35% (00:43) wrong based on 76 sessions
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: 700 level Ps [#permalink] New post 16 Dec 2011, 05:32
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ruturajp 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?
1%
3.5%
6.5%
7%
13.75%


Step 1:
Asuume the initial population to be 100 out of which 60 in south and 40 in north.

Step 2:
During 2010 the population grew by 5.5% ..i.e the population now is 105.5.

Step 3:

Population in south grew by 4.5%.So,

60+ [(4.5/100)*60] = 62.7 in south

Step 4:

New population in north is 105.5-62.7 =42.8

Hence the percentage increase in north is

[(42.8-40)/40 ]*100 = 7%
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Re: 700 level Ps [#permalink] New post 16 Dec 2011, 22:45
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ruturajp 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?
1%
3.5%
6.5%
7%
13.75%


This problem can be solved using the concept of weighted averages.

5.5 % is the total increase in population
60% live in the south & hence 40% live in the north
let % increase in north be x

5.5=0.6*4.5+0.4*x
5.5=2.7+0.4x
2.8x=0.4x
x=7

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Re: 700 level Ps [#permalink] New post 20 Aug 2012, 07:04
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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?

I prefer the weighted average method over dealing with creating and solving the algebraic expression:

Ratio of population - South:North :: 60%:40% = 3:2
Mean growth = 5.5%
South Growth: 4.5%
North Growth = ??

South-------------Average--------------North
4.5% __(2n)_____5.5%_____(3n)_____??

Since:
4.5 + 2n = 5.5
n = 1/2 (this is the multiplier of the ratio)
Therefore:
North = 5.5 + 3n
North = 5.5 + 3(0.5) = 7%
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Re: At the beginning of 2010, 60% of the population of Town X [#permalink] New post 22 Sep 2012, 21:25
ruturajp 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%


Can we not do this problem using residual weights ....

S / N = North - median / South - Median = x - 5.5 / 5.5 - 4.5 = 3 / 2 on solving we get 2x = 14 and x = 7
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Re: At the beginning of 2010, 60% of the population of Town X [#permalink] New post 27 Jan 2013, 05:25
south = 60* 9/2 *1/100=2.7

5.5-2.7=2.8

(2.8/40)=0.07
0.07*100%=7%
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Re: At the beginning of 2010, 60% of the population of Town X   [#permalink] 27 Jan 2013, 05:25
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