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The population of a small beach town on Australia's east coast grows b

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The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post Updated on: 06 Oct 2016, 01:28
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A
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  45% (medium)

Question Stats:

70% (01:55) correct 30% (02:15) wrong based on 165 sessions

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The population of a small beach town on Australia's east coast grows by 50% from May 1 to June 1 and then grows by another 50% from June 1 to July 1. The population then decreases by 11.11% from July 1 to August 1. By approximately what percentage has the population grown from May 1 to August 1?

a) 50%
b) 90%
c) 100%
d) 200%
e) 225%

Please assist with above problem.

Originally posted by Nanobotstv on 04 Oct 2016, 06:28.
Last edited by abhimahna on 06 Oct 2016, 01:28, edited 1 time in total.
Corrected the OA
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Re: The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 04 Oct 2016, 06:45
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alanforde800Maximus wrote:
The population of a small beach town on Australia's east coast grows by 50% from May 1 to June 1 and then grows by another 50% from June 1 to July 1. The population then decreases by 11.11% from July 1 to August 1. By approximately what percentage has the population grown from May 1 to August 1?

a) 50%
b) 90%
c) 100%
d) 200%
e) 225%

Please assist with above problem.


Are we sure OA marked is correct?

I solved this way :

Let the population in May = 2x
June = 3x
July = 4.5x

July = 8/9 * 4.5x = 4x(Note : 11.11% is 1/9, so left over would be 8/9)

Thus , change from May to Aug = 2x, which is 100% of the original. Hence, I marked C.

Please let me know where am I missing.
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Re: The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 04 Oct 2016, 17:11
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This is not right,

1.5 * 1.5 * 0.989 * x ~ 2,22x

2,22x is a 122% increase, so the best option is C
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Re: The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 05 Oct 2016, 00:34
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1
alanforde800Maximus wrote:
The population of a small beach town on Australia's east coast grows by 50% from May 1 to June 1 and then grows by another 50% from June 1 to July 1. The population then decreases by 11.11% from July 1 to August 1. By approximately what percentage has the population grown from May 1 to August 1?

a) 50%
b) 90%
c) 100%
d) 200%
e) 225%

Please assist with above problem.


PopAug1 = PopMay1 * (3/2) * (3/2) * (8/9) = PopMay1 * 2

Population of August is twice the population of May. So it has grown by 100%.

Answer (C)

Check this post for successive percentage changes: https://www.veritasprep.com/blog/2011/0 ... e-changes/
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Re: The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 05 Oct 2016, 00:59
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Say the population was 100 to begin with
From May 1 to June 1 it increased by 50% i.e., population now is 150.
From June 1 to July 1 again it increased by 50% i.e., population now is 150+75= 225
Population decreased by 11.11% from July 1 to august 1 i.e., 225*(8/9)=200
% change in population from April 1 to August 1 is {(200-100)/100}*100= 100%
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Re: The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 21 Oct 2016, 03:10
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consider the population of beach town on 1 st may be 100
on 1 st june it increased by 50 % so the population is 1.5 * 100 = 150
on 1 st july it increased by another 50 % = 1.5 * 150 = 225
on 1 st august it decresed by 11.11 % or it became \(\frac{8}{9}\) of its value

so \(\frac{8}{9}\) * 225 = 200

clearly its a 100 % increase
correct answer - C
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The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 21 Oct 2016, 13:40
We can do this using successive percentage change formula.

If there is a change of a% followed by b% the overall change = a + b + ab/100 (in case of decrease use -a etc)

First change 50% and then another 50% increase

50 + 50 + 50*50/100 = 125%


Now we can pair this 125% and -11.11%

125 - 11.11 - 125*11.11/100

which will be approx 100%

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Re: The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 14 Dec 2017, 17:52
alanforde800Maximus wrote:
The population of a small beach town on Australia's east coast grows by 50% from May 1 to June 1 and then grows by another 50% from June 1 to July 1. The population then decreases by 11.11% from July 1 to August 1. By approximately what percentage has the population grown from May 1 to August 1?

a) 50%
b) 90%
c) 100%
d) 200%
e) 225%


We can let the initial population = n.

After the town grows by 50%, then another 50%, and then decreases by 11.11%, we have:

n(3/2)(3/2)(8/9) = 72n/36 = 2n

Let’s determine the percentage increase:

(2n - n)/n x 100 = n/n x 100 = 100%

Answer: C
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Re: The population of a small beach town on Australia's east coast grows b  [#permalink]

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New post 27 Jan 2018, 12:57
Hi All,

This question can be solved either Algebraically or by TESTing VALUES. Here's how you can TEST VALUES to get to the solution:

Population on May 1 = 100
Population on June 1 = 100 + (.5)(100) = 150
Population on July 1 = 150 + (.5)(150) = 225

Population on August 1 = 225 - (1/9)(225) = 200

The question asks for the percent increase in population during that time.

Percentage Change = (new - old)/old = (200 - 100)/100 = 1/1 = 100%

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Re: The population of a small beach town on Australia's east coast grows b   [#permalink] 27 Jan 2018, 12:57
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