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Re: Town A has a population of 160,000 and is growing at a rate [#permalink]
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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]
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]
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Use this

Regards

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Re: Town A has a population of 160,000 and is growing at a rate [#permalink]
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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]
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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]
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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]
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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]
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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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Town A has a population of 160,000 and is growing at a rate [#permalink]
KarunMendiratta wrote:
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

Can you plz provide more details of, how it is done !?
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink]
Expert Reply
This discussion has several great answers from different users

See for example above https://gre.myprepclub.com/forum/town-a ... tml#p22768 if it is more clear to you
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink]
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We can simply use the relative percent increase or decrease in population per year and take 30% increase in population of B per year keeping the population of A constant to find the Number of years it takes to overtake population A. We can Simply use the calculator to do this problem.
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