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Town A has a population of 160,000 and is growing at a rate

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Town A has a population of 160,000 and is growing at a rate [#permalink] New post 30 Jul 2018, 10:31
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Question Stats:

68% (01:31) correct 31% (01:39) wrong based on 159 sessions
Town A has a population of 160,000 and is growing at a rate of 20% annually. Town B has a population of 80,000 and is growing at a rate of 50% annually.

Quantity A
Quantity B
The number of years until town B’s population is greater than that of town A
3


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 04 Aug 2018, 06:02
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sandy wrote:
Town A has a population of 160,000 and is growing at a rate of 20% annually. Town B has a population of 80,000 and is growing at a rate of 50% annually.

Quantity A
Quantity B
The number of years until town B’s population is greater than that of town A
3


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Let's keep track of each population for 3 years

TOWN A
Present Population: 160 (we'll drop the 000's at the end for both populations)
Population in 1 year: (160)(1.2)
Population in 2 years: (160)(1.2)(1.2)
Population in 3 years: (160)(1.2)(1.2)(1.2) = (160)(1.2)³


TOWN B
Present Population: 80
Population in 1 year: (80)(1.5)
Population in 2 years: (80)(1.5)(1.5)
Population in 3 years: (80)(1.5)(1.5)(1.5) = (80)(1.5)³


Now that we know the populations after 3 years, we can check whether or not 3 years is enough time for Town B's population to exceed Town A's population.
So, let's examine the fraction: (160)(1.2)³/(80)(1.5)³

There are 3 possible cases:
case a) (160)(1.2)³/(80)(1.5)³ = 1, in which case, the 2 populations are EQUAL
case b) (160)(1.2)³/(80)(1.5)³ < 1, in which case, Town A's population is LESS THAN Town B's population
case c) (160)(1.2)³/(80)(1.5)³ > 1, in which case, Town A's population is GREATER THAN Town B's population

160/80 = 2 and (1.2)³/(1.5)³ = (1.2/1.5)³ = (12/15)³ = (4/5)³ = 64/125
So, (160)(1.2)³/(80)(1.5)³ = (2)(64/125) = 128/125
Since 128/125 > 1, we have a case c situation.
So, after 3 years, Town A's population is still GREATER THAN Town B's population
So, it will take MORE THAN 3 YEARS for Town B's population to exceed Town A's population

We get:
Quantity A: MORE THAN 3 YEARS
Quantity B: 3 years

Answer: A

Cheers,
Brent
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 06 Aug 2018, 00:26
I solved this question on paper, and it took me 05 minutes 20 seconds to get the answer. I went through the traditional percentile method year by year. The answer was right, but I am worried about taking a long time to solve it.
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 06 Aug 2018, 01:38
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In that time you should solve almost 3 questions not one only.

The Brent's approach above is fine and i do not think there is a shortcut. Sometimes you have to go through the question step by step moving fast.

Notice how B is 1/2 of A. So A could be written as 2 for simplicity and B = 1

2*1.2=2.4*1.2=2.8*1.2=3.4*1.2=4.14

1*1.5=1.5*1.5=2.25*1.5=3.37*1.5=5.06

Consider that you do have the calculator and the question to solve will end up in 30 seconds.

Hope this helps
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 07 Aug 2018, 03:48
Carcass wrote:
In that time you should solve almost 3 questions not one only.

Consider that you do have the calculator and the question to solve will end up in 30 seconds.



This was helpful, and I am working on to improve my skills. I lack of mathematical tricks, though.
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 07 Aug 2018, 14:13
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Use this

Regards

Attachment:
GMAT Club Math Book v3 - Jan-2-2013.pdf [2.83 MiB]
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To download please login or register as a user


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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 12 Aug 2018, 06:04
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Explanation

Set up a table and calculate the population of each town after every year; use the calculator to calculate town A’s population. If you feel comfortable multiplying by 1.5 yourself, you do not need to use the calculator for town B. Instead, add 50% each time (e.g., from 80,000 add 50%, or 40,000, to get 120,000).




Town A Town B
Now 160,000 80,000
Year 1 160,000(1.2) = 192,000 80,000 + 40,000 = 120,000
Year 2 192,000(1.2) = 230,400 120,000 + 60,000 = 180,000
Year 3 230,400(1.2) = 276,480 180,000 + 90,000 = 270,000


Note that, after three years, town A still has more people than town B. It will take longer than 3 years, then, for town B to surpass town A, so Quantity A is greater.
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 08 Oct 2019, 00:46
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Well I combined the two approaches-

1) Drop all the zeros
2) Use the table

.............. Town A.....................Town B
Now......... 16 ..........................8
Year......... 16(1.2) = 19.2 ............8 + 4 = 12
Year 2....... 19.2(1.2) = 23.04 ........12 + 6= 18
Year 3....... 23.04(1.2) = 27.6.........18+ 9 = 27

So, at the end of year 3, the population of town A is > town B. Therefore, Quantity A is greater.
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post 22 Nov 2019, 04:35
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Treat this question as a compound interest question.

You will get below relationship

160,000* (1+0.2)exponential variable a< 80, 000 * (1+0.5) exponential variable a
2* (1.2) exponential variable a < 1.5 * exponential variable a

put a= 3 first

LHS RHS

3.456 3.375

so to increase value of RHS we need to put a value greater than 3 . It means option A is greater.
Re: Town A has a population of 160,000 and is growing at a rate   [#permalink] 22 Nov 2019, 04:35
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