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AccipiterQ
GMAT 2: 730 Q49 V41
Posts: 144
Updated on: 06 Nov 2025, 13:23
Originally posted by AccipiterQ on 22 Oct 2013, 14:21.
Last edited by Bunuel on 06 Nov 2025, 13:23, edited 12 times in total.
Last edited by Bunuel on 06 Nov 2025, 13:23, edited 12 times in total.
Moved to new forum
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Growth
Factor: \(p^\frac{1}{4}\) Population
at t = 0: \(p^\frac{3}{4}\)
Factor: \(p^\frac{1}{4}\) Population
at t = 0: \(p^\frac{3}{4}\)
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
74% (02:19) correct
26%
(02:56)
wrong
based on 1177
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A certain species of bacteria grows at a constant rate, its numbers increasing by an unknown multiplier every hour. At the end of one hour (time t = 1), the bacteria colony’s population size is p; at the end of five hours (t = 5), the colony’s population has grown to \(p^2\).
In the first column, indicate the factor by which the number of bacteria grows each hour. In the second column, indicate the population at the start of the time period described (at time t = 0). Make only two selections, one in each column.
In the first column, indicate the factor by which the number of bacteria grows each hour. In the second column, indicate the population at the start of the time period described (at time t = 0). Make only two selections, one in each column.
|
Growth Factor |
Population at t = 0 |
|
| \(p^\frac{1}{4}\) | ||
| \(p^\frac{1}{3}\) | ||
| \(p^\frac{1}{2}\) | ||
| \(p^\frac{3}{4}\) | ||
| \(p\) | ||
| \(p^\frac{3}{2}\) |
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Official Answer
Growth
Factor: \(p^\frac{1}{4}\) Population
at t = 0: \(p^\frac{3}{4}\)
Factor: \(p^\frac{1}{4}\) Population
at t = 0: \(p^\frac{3}{4}\)
22 Oct 2013, 15:00
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Dear AccipiterQ,
I'm happy to help. I believe the OA given is correct, and does correctly account for exponential growth. First, here's a blog on exponents:
https://magoosh.com/gmat/2012/exponent-p ... -the-gmat/
Part of what's tricky is that the start time, with population p, is at t = 1, and the the later time, at t = 5, is FOUR hours later. That's the first tricky thing.
Also, keep in mind, for exponential growth, of course the size of the population doesn't increase linearly, but the exponent on the function does increase linearly. Thus, at time #1, the population is a P^A, and at time #2, it's at P^B (where B>A), then at a time exactly between those two times, the population will have an exponent exactly between those two values (a.k.a. the average), P^[(A+B)/2]
Part one: Let r be the unknown hourly growth ratio:
t = 1 --> p
t = 2 --> p*r
t = 3 --> p*r*r = p*(r^2)
t = 4 --> p*r*r*r = p*(r^3)
t = 5 --> p*r*r*r*r = p*(r^4)
Set that last value equal to p^2
p*(r^4) = p^2
r^4 = p
r = p^(1/4)
Column #1, choice = (A)
Given that
t = 1 --> p
t = 0 --> p/r = p/[p^(1/4)] = p^(3/4)
Column #2, choice = (D)
Does all this make sense?
Mike
I'm happy to help. I believe the OA given is correct, and does correctly account for exponential growth. First, here's a blog on exponents:
https://magoosh.com/gmat/2012/exponent-p ... -the-gmat/
Part of what's tricky is that the start time, with population p, is at t = 1, and the the later time, at t = 5, is FOUR hours later. That's the first tricky thing.
Also, keep in mind, for exponential growth, of course the size of the population doesn't increase linearly, but the exponent on the function does increase linearly. Thus, at time #1, the population is a P^A, and at time #2, it's at P^B (where B>A), then at a time exactly between those two times, the population will have an exponent exactly between those two values (a.k.a. the average), P^[(A+B)/2]
Part one: Let r be the unknown hourly growth ratio:
t = 1 --> p
t = 2 --> p*r
t = 3 --> p*r*r = p*(r^2)
t = 4 --> p*r*r*r = p*(r^3)
t = 5 --> p*r*r*r*r = p*(r^4)
Set that last value equal to p^2
p*(r^4) = p^2
r^4 = p
r = p^(1/4)
Column #1, choice = (A)
Given that
t = 1 --> p
t = 0 --> p/r = p/[p^(1/4)] = p^(3/4)
Column #2, choice = (D)
Does all this make sense?
Mike
General Discussion
19 Sep 2017, 03:52
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Let us consider r the factor.
At end of 1 hr......p
At end of 2 hr......pr
.
.
At end of 5 hr....pr^4
pr^4=p^2
r=p^1/4
At t=0 , P^(1-1/4)=p^3/4 ...Ans
At end of 1 hr......p
At end of 2 hr......pr
.
.
At end of 5 hr....pr^4
pr^4=p^2
r=p^1/4
At t=0 , P^(1-1/4)=p^3/4 ...Ans
