Locust in a swarm: any short-cuts?

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Locust in a swarm: any short-cuts?

by euro » Sun Oct 24, 2010 5:07 am
The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?
a) 6
b) 8
c) 10
d) 12
e) 14

This is from a MGMAT test.
Is there any way, other than creating a population chart, to solve problems of the type mentioned above?

[spoiler]OA is d) 12[/spoiler]
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by GMATGuruNY » Sun Oct 24, 2010 5:22 am
euro wrote:The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?
a) 6
b) 8
c) 10
d) 12 =
e) 14

This is from a MGMAT test.
Is there any way, other than creating a population chart, to solve problems of the type mentioned above?

[spoiler]OA is d) 12[/spoiler]
The formula for exponential growth is:

Final amount = Original amount * (multiplier)^(number of changes)

In the problem above:
Final amount = 250,000
Original amount = 1000
Multiplier = 2 (because the amount keeps doubling)

We need to solve for the number of changes. Plugging the amounts above into the formula, we get:

250,000 = 1000*2^x
250 = 2^x
2^8 = 256, so x ≈ 8.

Thus, the population needs to double 8 times. It doubles every 2 hours, so the total hours needed is 2*8 = 16. Since the doubling started 4 hours ago, the process will be complete 16-4 = 12 hours from now.

The correct answer is D.
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by pesfunk » Sat Oct 30, 2010 10:14 pm
Hey Mitch,

Thanks for the reply...however...have a small question

the population was 1000, 4 hours ago....and the question is "after how many hours" (i assume from NOW)

in that case the counting should start from 4000 right ?

4 hours ago = 1000
2 hours ago = 2000
NOW = 4000

250000 = 4000 * 2 ^ number of changes(x)

2^x = 62

x = 4 ?
GMATGuruNY wrote:
euro wrote:The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?
a) 6
b) 8
c) 10
d) 12 =
e) 14

This is from a MGMAT test.
Is there any way, other than creating a population chart, to solve problems of the type mentioned above?

[spoiler]OA is d) 12[/spoiler]
The formula for exponential growth is:

Final amount = Original amount * (multiplier)^(number of changes)

In the problem above:
Final amount = 250,000
Original amount = 1000
Multiplier = 2 (because the amount keeps doubling)

We need to solve for the number of changes. Plugging the amounts above into the formula, we get:

250,000 = 1000*2^x
250 = 2^x
2^8 = 256, so x ≈ 8.

Thus, the population needs to double 8 times. It doubles every 2 hours, so the total hours needed is 2*8 = 16. Since the doubling started 4 hours ago, the process will be complete 16-4 = 12 hours from now.

The correct answer is D.

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by GMATGuruNY » Sun Oct 31, 2010 2:20 am
pesfunk wrote:Hey Mitch,

Thanks for the reply...however...have a small question

the population was 1000, 4 hours ago....and the question is "after how many hours" (i assume from NOW)

in that case the counting should start from 4000 right ?

4 hours ago = 1000
2 hours ago = 2000
NOW = 4000

250000 = 4000 * 2 ^ number of changes(x)

2^x = 62

x = 4 ?
GMATGuruNY wrote:
euro wrote:The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?
a) 6
b) 8
c) 10
d) 12 =
e) 14

This is from a MGMAT test.
Is there any way, other than creating a population chart, to solve problems of the type mentioned above?

[spoiler]OA is d) 12[/spoiler]
The formula for exponential growth is:

Final amount = Original amount * (multiplier)^(number of changes)

In the problem above:
Final amount = 250,000
Original amount = 1000
Multiplier = 2 (because the amount keeps doubling)

We need to solve for the number of changes. Plugging the amounts above into the formula, we get:

250,000 = 1000*2^x
250 = 2^x
2^8 = 256, so x ≈ 8.

Thus, the population needs to double 8 times. It doubles every 2 hours, so the total hours needed is 2*8 = 16. Since the doubling started 4 hours ago, the process will be complete 16-4 = 12 hours from now.

The correct answer is D.
You are correct: the question is indeed asking us to calculate the number of hours from now. If you reread my explanation, you'll see that I calculated that the total number of hours needed is 16, and since the process started 4 hours ago, it will be completed 16-4 = 12 hours from now.

Your approach will arrive at the same answer, but you miscalculated the value of x after you determined that 2^x = 62.

2^x = 62
Since 2^6 = 64, x ≈ 6. (x ≠ 4, as stated in your post)
Thus, starting from now, the population has to double 6 times. Since it doubles every 2 hours, the process will be completed 2*6 = 12 hours from now -- same answer as I determined above.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by tapanmittal » Sun Sep 13, 2015 5:19 am
GMATGuruNY wrote:
euro wrote:The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?
a) 6
b) 8
c) 10
d) 12 =
e) 14

This is from a MGMAT test.
Is there any way, other than creating a population chart, to solve problems of the type mentioned above?

[spoiler]OA is d) 12[/spoiler]
The formula for exponential growth is:

Final amount = Original amount * (multiplier)^(number of changes)

In the problem above:
Final amount = 250,000
Original amount = 1000
Multiplier = 2 (because the amount keeps doubling)

We need to solve for the number of changes. Plugging the amounts above into the formula, we get:

250,000 = 1000*2^x
250 = 2^x
2^8 = 256, so x ≈ 8.

Thus, the population needs to double 8 times. It doubles every 2 hours, so the total hours needed is 2*8 = 16. Since the doubling started 4 hours ago, the process will be complete 16-4 = 12 hours from now.

The correct answer is D.
Hi GMATGuru,

Can you help me solving this problem with this formula --> [final population] = [initial population] (1 + r)^t

Thanks

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by [email protected] » Sun Sep 13, 2015 8:57 am
Hi tapanmittal,

Mitch's original post essentially presents the solution using that formula. There are a couple of details worth noting though:

1) Doubling the population means that R is 100% (so R=1)
2) Each time period (T) is actually 2 hours
3) Since the prompt starts us off 4 hours AGO, we have to start counting '2-hour time periods' in the past.

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