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16 Sep 2021, 04:15
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D
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Difficulty:
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59% (02:41) correct
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At the start of an experiment, a population consisted of x organisms. At the end of each month, after the start of the experiment, the population size increased by twice of its size at the beginning of that month. If the total population at the end of five months is greater than 1,000, what is the minimum possible value of x?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Updated on: 26 Sep 2021, 07:54
Originally posted by BrentGMATPrepNow on 16 Sep 2021, 06:47.
Last edited by BrentGMATPrepNow on 26 Sep 2021, 07:54, edited 1 time in total.
Last edited by BrentGMATPrepNow on 26 Sep 2021, 07:54, edited 1 time in total.
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Bunuel
NOTE: Don't make the same mistake I made and ignore the words "increased by" in the question.
So, each month, the population doesn't double; the population INCREASES by double the population from last month.
In other words, the population TRIPLES each month.
Let's start by using the table method to list the population for each month.
# months elapsed | population
0 | x
1 | (x)(3)
2 | (x)(3)(3)
3 | (x)(3)(3)(3)
4 | (x)(3)(3)(3)(3)
5 | (x)(3)(3)(3)(3)(3) = (3⁵)(x) = 243x
So 243x represents the population of organisms after 5 months.
We're told that which population is greater than 1,000.
So we can write: 243x > 1,000
Divide both sides of the inequality by 243 to get: x > 4.11
Since x must be an INTEGER, the minimum possible of x is 5
Answer: D
16 Sep 2021, 08:33
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Lets start with population and month:-
X-Start of first month(1)
2X-end(1)
2X-start(2)
4X-end(2)
4X-start(3)
8X-end(3)
8X-start(4)
16X-end(4)
16X-start(5)
32X-end(5)
now 32X>1000
=> value has to be 4(least value)
Hope this makes sense!
X-Start of first month(1)
2X-end(1)
2X-start(2)
4X-end(2)
4X-start(3)
8X-end(3)
8X-start(4)
16X-end(4)
16X-start(5)
32X-end(5)
now 32X>1000
=> value has to be 4(least value)
Hope this makes sense!
