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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? A. 1% B. 3.5% C. 6.5% D. 7% E. 13.75%
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Re: 700 level Ps [#permalink]
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20 Aug 2012, 07:04
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 = ??
SouthAverageNorth 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: 700 level Ps [#permalink]
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16 Dec 2011, 05:32
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 \(105.562.7 =42.8\) Hence the percentage increase in north is \([(42.840)/40 ]*100 = 7%\)
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Re: 700 level Ps [#permalink]
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16 Dec 2011, 22:45
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 D
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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



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Re: At the beginning of 2010, 60% of the population of Town X [#permalink]
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27 Jan 2013, 05:25
south = 60* 9/2 *1/100=2.7 5.52.7=2.8 (2.8/40)=0.07 0.07*100%=7%
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Re: 700 level Ps [#permalink]
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26 May 2014, 14:57
macjas 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?
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 = ??
SouthAverageNorth 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% Hi I'm not sure if this is right  judging by your setup, wouldn't your second equation have to be: North % growth + 3(n) = 5.5% This yield 4, not 7?



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Re: At the beginning of 2010, 60% of the population of Town X [#permalink]
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12 Mar 2016, 05:06
got complicated way
0.6x*1.045=0.627x, so 1.055x0.627x=0.428x
(0.428x0.4x/0.4x)*100=7%
D



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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.



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Re: At the beginning of 2010, 60% of the population of Town X [#permalink]
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14 Nov 2016, 08:54
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% suppose we have 10,000 total in south we have 6,000 n north we have 4,000 population grew by 5.5% 1% = 100 5% = 500 0.5% = 50 total population increased by 550. In south, population increased by 4.5% 6,000 = 100% 60 = 1% 240 = 4% 30 = 0.5% 270 = 4.5% so total in south now is 6270. now..we have in north: 10,550  6,270 = 4,280 people 280 increase. 280/4000 = 28/400 = 7/100 or 7% increase. the answer is D.



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Re: At the beginning of 2010, 60% of the population of Town X [#permalink]
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14 Nov 2016, 10:55
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% Population in 2010 = 1000 Population of South = 600 Population of North = 400 Population in 2011 = 1055 Population of South = 627 Population of North = 428 ( ie, 1055 627) So, Population growth in north is 28/400*100 = 7% Hence, answer will be (D)
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Re: At the beginning of 2010, 60% of the population of Town X [#permalink]
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28 May 2017, 20:51
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% We can use weighted average to solve the problem or smart numbers. Weighted average3/5*(4.5) + 2/5*(x) = 5.5 3*0.9+2/5*x = 5.5 2/5*x = 5.52.7 2/5*x = 2.8 x/5 = 1.4 x = 5+2 = 7 Smart numbersLet the population be 100 It grows to 105.5. The south grows from 60 to 62.7 since 60*1.045 = 62.7. 105.5  62.7 = 42.8 The increase in the north is thus 2.8 2.8/40 = 28/400 = 7/100 or 7%.



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Re: At the beginning of 2010, 60% of the population of Town X [#permalink]
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04 Sep 2017, 17:49
i set up an algebraic expression. kind of like a weighted averages q. SEE: (\(\frac{3}{5}\))in South, so (\(\frac{2}{5}\)) in North. Total up 5.5% > South up 4.5%. Since overall up by 5.5%, North needs to have gone up by more than 4.5%, so immediately eliminate A&BExpress in fractions (easier to calculate): (\(\frac{3}{5}\))*(4.5) + (\(\frac{2}{5}\))*(x) = 5.5 > Simplify: 2.7 + (\(\frac{2}{5}\))x=5.5 > (\(\frac{2}{5}\))x=2.8 > x=2.8*(\(\frac{5}{2}\)) > 1.4*5=7 Ans: D) 7Kudos please if helpful



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Re: At the beginning of 2010, 60% of the population of Town X [#permalink]
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06 Dec 2017, 12:17
Tried this. Came upon once I did some hit and trial.
Case I> 2010 (in the beginning) > 60 (S) and 40 (N). Total 100. This grew by 5.5% > which is say approximately 106 persons now [as 5.5/100*100 + 100]
Second part > S grew by 4.5 % > 60 + 4.5/100(60 approximately now 63 people. Balance amount of people in (N) to make up 106 persons is 106  63 = 43 people. So N grew by > 4340/40*100 = 15/2 = 7.5% Now, as I approximated, the values below it but around 7% makes sense which is D.



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At the beginning of 2010, 60% of the population of Town X [#permalink]
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04 Feb 2018, 07:09
Question is test of weighted average concept.
Weight1(north)/weight2(south) = 40/60 = 2/3
4.5 (south weight = 3) <2>5.5<3> ? (north whose weight is 2) so=> weight average rule, (5.5  4.5)/ x = 2/5 => 1/x = 2/5 => x = 2.5 (where x is difference between north population increase and south population increase) => which means population increase in north is 4.5 + x = 7




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