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At the beginning of 2010, 60% of the population of Town X [#permalink]
16 Dec 2011, 03:51

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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? 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

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?

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]
22 Sep 2012, 20:25

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 = ??

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...