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

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sandy
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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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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.
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A
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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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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 ATown B
Now160,00080,000
Year 1160,000(1.2) = 192,00080,000 + 40,000 = 120,000
Year 2192,000(1.2) = 230,400120,000 + 60,000 = 180,000
Year 3230,400(1.2) = 276,480180,000 + 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.
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post   13 Feb 2021, 04:11
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Yes, there is no shortcut. However, since you don't need to come up with an accurate anwser and it's just quantity comparison as aforementioned by one of the fellows above, you can simply ignore the zeros and use the calculator to multiply each town's population by its growth rate three times. So, we have:

Town's B population after 3 years: 8 x 1.5 x 1.5 x 1.5 = 27
Town's A population after 3 years: 16 x 1.2 x 1.2 x 1.2 = 27.648

We can see that the population in town A is greater than town B. Hence the answer is A.
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post   13 Feb 2021, 06:01
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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.


Function for Town A = \(160,000 (1.20)^t\)

Function for Town B = \(80,000 (1.50)^t\)

Col. A: Population of B > Population of A
i.e. \(80,000 (1.50)^t\) > \(160,000 (1.20)^t\)
\(\frac{(1.50)^t}{(1.20)^t}\) > 2
\((1.25)^t\) > 2
t > 3

Col. B: 3

Col. A > Col. B

Hence, option B
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink] New post   10 Jun 2022, 09:07
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