Solving a growth rate/time problem: 332 people on October 2016 increasing by 19.5%...

Whitehartlane

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Feb 27, 2017
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Hi guys, I am just asking if there's any way I can solve the following problem with an equation (I did it the long way but it's too time consuming).

Data 1: 332 people on October 2016 increasing by 19.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

Data 2: 300 people on October 2016 increasing by 25.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

In what month and year does Data 2 eclipse Data 1?

Long answer I got July 2017. I tried to make it into an inequality but I got April 2018 which isn't right.
 

Ishuda

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Jul 30, 2014
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Hi guys, I am just asking if there's any way I can solve the following problem with an equation (I did it the long way but it's too time consuming).

Data 1: 332 people on October 2016 increasing by 19.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

Data 2: 300 people on October 2016 increasing by 25.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

In what month and year does Data 2 eclipse Data 1?

Long answer I got July 2017. I tried to make it into an inequality but I got April 2018 which isn't right.
What have you been taught about exponential growth problems? If you can't remember, you should be able to work it out: Take the following problem starting at time zero with population of p0 population increasing by rate r percent every t0 periods. For each period we need to multiply the initial population 1+r/100. In an amount of time t, there are t/t0 periods of length t0. So at the end of time t, the population P is p0 multiplied by 1+r/100 t/t0 times or
P = p0 (1+r/100)t/t0

For your problem, the first data set has p0 = 332 r = 19.5% and t0=3 if we measure time in months starting from the beginning of Oct.
P1(t) = 332 (1.195)t/3
So what is P2(t) and at what time t does P2(t) become larger than P1(t).
 

Whitehartlane

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Feb 27, 2017
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332*1.195^n = 300*1.255^n

Solve for n

That'll give you the number of "3month periods".
Convert that to months.
I found it very difficult to solve for n so I used Wolfram Alpha which gave me this figure: n = 0.243694

It doesn't seem right to me.

 

JeffM

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Sep 14, 2012
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3,258
Hi guys, I am just asking if there's any way I can solve the following problem with an equation (I did it the long way but it's too time consuming).Data 1: 332 people on October 2016 increasing by 19.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)Data 2: 300 people on October 2016 increasing by 25.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)In what month and year does Data 2 eclipse Data 1?Long answer I got July 2017. I tried to make it into an inequality but I got April 2018 which isn't right.
\(\displaystyle 300 * 1.255^n > 322 * 1.195^n \implies\)\(\displaystyle log(300) + n * log(1.255) > log(322) + n * log(1.195).\)

The least integer value of n possible is the number of months from the start date. So if you get an answer like 9.1, the least integer is 10.
 
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