At the start of an experiment, a certain population consisted of x org

Question

At the start of an experiment, a certain 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 1000, 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

Bunuel · Answer

This is a modified question of the the following OG question: https://gmatclub.com/forum/at-the-start ... 07202.html

vvrkumar · Answer

At the end of each month, the population is 3 times than that of the beginning, since it is said that the population size increases by twice the initial amount.

At end of 5th month, population is 3^5 times the original, say x.

Thus, 243x > 1000

=> x = 5

Option D

arvind910619 · Answer

Imo C
Let population be x then 
At the end of 1st month population will be =2x
At the end of 2st month population will be=4x
At the end of 3st month population will be=8x
At the end of 4st month population will be=16x
At the end of 5st month population will be=32x

32X>1000 we have x>3 
So Minimum value of x is 4

ydmuley · Answer

x ====> 2x

2x ===> 4x

4x ===> 8x

8x ===> 16x

16x ===> 32x

Which means we are looking for  x^5

x^5 > 3,000

So we are looking for a smaller number

x^4 = 1024  which is slightly greater than 1000

Hence, the number should be  4

Answer is C

generis · Answer

The answer is D.

The devil is in this detail:

"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."

Say we have a size of 10.  Twice of the size  10 is 10*2 = 20. 
That 20 should be added to  original  10 to get 30. So 10 increased by twice its size = 30.
10 + 200% of 10 = (10 + 20) = 30

In other words, at the end of each month, the value is  three  times the original, not two times.

Try 4. The second number in the sum is the amount added  to an already existing start number  that yields a new total at the end of each month:

Month 1: 4 + 8 = 12
Month 2: 12 + 24 = 36
Month 3: 36 + 72 = 108
Month 4: 108 + 216 = 324
Month 5: 324 + 648 = 972

972 < 1,000. A little too small.

Try 5:
Month 1: 5 + 10 = 15
Month 2: 15 + 30 = 45
Month 3: 45 + 90 = 135
Month 4: 135 + 270 = 405
Month 5: 405 + 810 = 1,215

1,215 > 1,000

The answer is D.

rocko911 · Answer

D) FOR SURE

We dont need 4*4*4*4*4
We need 4*2*2*2*2

so 5*2*2*2*2 > 1000
and 4*2*2*2*2 <1000

Hence D

GMATinsight · Answer

Initial size = x
Increase = 2x
Final size after first increase =  3x

After 5 months Population =  3^5*x

Now,  3^5*x > 1000

i.e. 243x > 1000
i.e. x > 4.1

i.e.   x_{min} = 5

Answer: Option D

Abhishek009 · Answer

At the time of Start is x
Size after 1st Month is 3x (ie x + 2x)
Size after 2nd Month is 9x (ie 3x + 2*3x)
Size after 3rd Month is 27x (ie 9x + 2*9x)
Size after 4th Month is 81x (ie 27x + 2*27x)
Size after 5th Month is 243x (ie 81x + 2*81x)

Or,  243x > 1000

So,  x > 4.11 , Thus among the given optionthe s, answer must be (D) 5

ScottTargetTestPrep · Answer

When the population size increased by twice its size every month, it was tripling every month, as shown here:

End of Month 1: x * 3

End of Month 2: x * 3 * 3 = x * 3^2

End of Month 3: x * 3 * 3 * 3 = x * 3^3

End of Month 4: x * 3 * 3 * 3 * 3 = x * 3^4

End of Month 5: x * 3 * 3 * 3 * 3 * 3 = x * 3^5

Therefore, we can create the inequality

x * 3^5 > 1000

x > 1000/3^5 
x > 4.11

Since x is an integer, x is at least 5.

Answer: D

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At the start of an experiment, a certain population consisted of x org

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