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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]

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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%
[Reveal] Spoiler: OA

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Re: 700 level Ps [#permalink]

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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]

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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

D
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Re: 700 level Ps [#permalink]

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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]

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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

5.5-2.7=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 = ??

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%

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 1.055x-0.627x=0.428x

(0.428x-0.4x/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 average

3/5*(4.5) + 2/5*(x) = 5.5

3*0.9+2/5*x = 5.5

2/5*x = 5.5-2.7

2/5*x = 2.8

x/5 = 1.4

x = 5+2 = 7

Smart numbers

Let 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&B

Express 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) 7

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Re: At the beginning of 2010, 60% of the population of Town X   [#permalink] 04 Sep 2017, 17:49
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