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Town A has a population of 160,000 and is growing at a rate [#permalink]
30 Jul 2018, 10:31
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Question Stats:
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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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Re: Town A has a population of 160,000 and is growing at a rate [#permalink]
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)³TOWN 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)³Now 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)³There 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 populationcase c) (160)(1.2)³/ (80)(1.5)³ > 1, in which case, Town A's population is GREATER THAN Town B's population160/80 = 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 populationSo, 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]
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]
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= 4.141*1.5= 1.5*1.5= 2.25*1.5= 3.37*1.5= 5.06Consider 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]
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]
07 Aug 2018, 14:13
Use this Regards Attachment:
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Re: Town A has a population of 160,000 and is growing at a rate [#permalink]
12 Aug 2018, 06:04
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ExplanationSet 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]
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]
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





