population of bees

[rahul.s]

4 years from now, the population of a colony of bees will reach 1.6 * 10^8. if the population of the colony doubles every 2 years, what was the population 4 years ago?

sorry folks, but there are no answer options given for this problem.

OA: [spoiler]1 * 10^7[/spoiler]

[Stuart@KaplanGMAT]

4 years from now, the population of a colony of bees will reach 1.6 * 10^8. if the population of the colony doubles every 2 years, what was the population 4 years ago?

sorry folks, but there are no answer options given for this problem.

If we see that this is a power of 2 question, we can rewrite it to make life much easier.

For powers of 2 questions, it's easiest to use numbers that are also powers of 2. So, let's rewrite:

1.6 * 10^8

as

16 * 10^7.

Next, let's sketch out a timeline:

4 years ago ---- 2 years ago ---- now ---- 2 years from now ---- 4 years from now

to see that to get from 4 years from now to 4 years ago, we need to "undouble" 4 times; of course, "undoubling" is the same as dividing by 2.

So:

4 years from now: 16 * 10^7
2 years from now: 8 * 10^7
now: 4 * 10^7
2 years ago: 2 * 10^7
4 years ago: 1 * 10^7

Of course we could have done the same thing starting with 1.6 * 10^8; our final answer will be .1 * 10^8, which is 1 * 10^7.

[shashank.ism]

4 years from now, the population of a colony of bees will reach 1.6 * 10^8. if the population of the colony doubles every 2 years, what was the population 4 years ago?

sorry folks, but there are no answer options given for this problem.

OA: [spoiler]1 * 10^7[/spoiler]

Let the population four years ago was X and since it doubles every two year.
so present population = X . 2^2
population after 4 years = X. 2^4 = 1.6* 10^8

--> X= 1.6*10^8/16 = 10^7 Ans.

[thephoenix]

4 years from now, the population of a colony of bees will reach 1.6 * 10^8. if the population of the colony doubles every 2 years, what was the population 4 years ago?

sorry folks, but there are no answer options given for this problem.

OA: [spoiler]1 * 10^7[/spoiler]

there is a formula for such problems
P=S*N^t/I

where
P=population
S=starting value
N=mutiplication amnt
time=time
I=int for change of value
now here
let S=x
N=since it doubles therefore=2
t=as per situation i.e for 4 yrs from now=+4 and for 4 yrs ago=-4
I=2( every two year)

now for t=+4
p=1.6*10^8
therefore
1.6*10^8=x*2^4/2=4x--->x=.4*10^8

now for t=-4
p=.1*10^8=1*10^7