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At the beginning of 2010, 60% of the population of Town X [#permalink]

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16 Dec 2011, 04:51

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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?

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 = ??

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

D _________________

Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.

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

\([(42.8-40)/40 ]*100 = 7%\) _________________

Regards, Rajesh Helping hands are anytime better than praying hearts

Re: At the beginning of 2010, 60% of the population of Town X [#permalink]

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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

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 = ??

Re: At the beginning of 2010, 60% of the population of Town X [#permalink]

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13 Mar 2016, 08:33

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%

Supposing that the population of town X is 100; 60 are in the South and 40 are in the North. During 2010 population grew to 105.5 population in the South grew to 62.7 The remaining population in the North is 42.8 For 40, the growth is 2.8 Therefore for 100 the growth is 7.

gmatclubot

Re: At the beginning of 2010, 60% of the population of Town X
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13 Mar 2016, 08:33

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