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These kind of questions, are best done when we go backwards.

1280000 when halved gives 640000(4 hours ago)
4 more hours ago the population of the colony of bacterial colony would be 320000
4 more hours ago, the population will be 160000
4 more hours ago, the population will be 80000 which is 8 * \(10 ^ 4\)(Option B)
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It is possible to do using powers to simplify:
(future) 1.28 x 10^6
Going back 4 times from +4y to -12y: multiply by 1/(2^4)
1.28 x 10^6 = 128 x 10^4 = 2^7 x 10^4 == you could stop here looking to the possible answers
(2^7 x 10^4) / 2^4 = 2^3 x 10^4 = 8 x 10^4
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Four hours from now, the population of a colony of bacteria will reach 1.28 * \(10 ^ 6\). If the population of the colony doubles every 4 hours, what was the population 12 hours ago?

A. 6.4 * \(10 ^ 2\)
B. 8.0 * \(10 ^ 4\)
C. 1.6 * \(10 ^ 5\)
D. 3.2 * \(10 ^ 5\)
E. 8.0 * \(10 ^ 6\)


Since the population doubles every 4 hours and the population will be 1.28 x 10^6 four hours from now, the current population must be half of 1.28 x 10^6, or ½(1.28 x 10^6) = 0.64 x 10^6.

Thus:

4 hours ago, the population was half the current population: ½(0.64 x 10^6) = 0.32 x 10^6.

8 hours ago, the population was half the population 4 hours ago: ½(0.32 x 10^6) = 0.16 x 10^6.

Finally, 12 hours ago the population was half the population 8 hours ago: ½(0.16 x 10^6) = 0.08 x 10^6 = 8 x 10^-2 x 10^6 = 8 x 10^4.

Answer: B
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It will double every 4 hour => if 4 hours from now the population is 1.28 * 10^6
=> now the population is 1.28 * 10^6 / 2

It will double (x2) every 4 hour => 12 hours ago, the population is 1/ 2^ 3 = 1/8 the population now (3 = 12/4)
=> The population 12 hours ago = 1.28 * 10^6 /2 * 1/8 = 8 * 10^4

=> B
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carcass
Four hours from now, the population of a colony of bacteria will reach 1.28 * \(10 ^ 6\). If the population of the colony doubles every 4 hours, what was the population 12 hours ago?

A. 6.4 * \(10 ^ 2\)
B. 8.0 * \(10 ^ 4\)
C. 1.6 * \(10 ^ 5\)
D. 3.2 * \(10 ^ 5\)
E. 8.0 * \(10 ^ 6\)


pushpitkc, is my approach / reasoning correct ? :)

1.28 * \(10 ^ 6\) means we need to move decimal point 6 points to the right so it is 1,280,000, so i just leave out zeros and work with 128

if in 4 hours the population will be 128, then now it is 64

4 hours ago it was 32
8 hours ago 16
12 hrs ago 8

so all i need is to add 4 zeros and express it like this \(8*10^4\) :)

one thing I didn't get why in answer option B. 8.0 * \(10 ^ 4\) \(8\) is expressed as \(8.0\) and not just \(8\) I mean with one zero after decimal point... :?
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dave13
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Four hours from now, the population of a colony of bacteria will reach 1.28 * \(10 ^ 6\). If the population of the colony doubles every 4 hours, what was the population 12 hours ago?

A. 6.4 * \(10 ^ 2\)
B. 8.0 * \(10 ^ 4\)
C. 1.6 * \(10 ^ 5\)
D. 3.2 * \(10 ^ 5\)
E. 8.0 * \(10 ^ 6\)


pushpitkc, is my approach / reasoning correct ? :)

1.28 * \(10 ^ 6\) means we need to move decimal point 6 points to the right so it is 1,280,000, so i just leave out zeros and work with 128

if in 4 hours the population will be 128, then now it is 64

4 hours ago it was 32
8 hours ago 16
12 hrs ago 8

so all i need is to add 4 zeros and express it like this \(8*10^4\) :)

one thing I didn't get why in answer option B. 8.0 * \(10 ^ 4\) \(8\) is expressed as \(8.0\) and not just \(8\) I mean with one zero after decimal point... :?

Hi dave13

This problem is best solved when we go backwards!

8 *10^4 = 80000
Now see the sequence when we double the population of the colony of bacteria
80000 * 2 = 160000 | 160000 * 2 = 320000 | 320000 * 2 = 640000 | 640000 * 2 = 1280000
(You can see that 12 hours from the start, the population has increased
from 8 *10^4 to 6.4 *10^5 and then to 1.28 * 10^6)

Your approach is perfect otherwise. All you need to understand is this part of why the ZEROES increase!

Hope this helps you!
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1.28E6 = 128E4. 12 hours before now = 2^4 --> 128E4 / 2^4 = 128E4 / 16 = 128/16 = 8E4.
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backwards (1.28*10^6)/2 ---present time
(1.28*10^6)/4 ---4 hrs before
(1.28*10^6)/8--- 8 hrs before
(1.28*10^6)/16 --- 12 hrs before
ans 8*10^4
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Upsetting this took me over 5 minutes just to try and understand the doubling, etc.

4 hours from now the population doubles to 1.28 * 10^6

This means the CURRENT population is half of this. Then we have to go back 3 doublings so 2^4

1.28 * 10^6 / 2^4 = 128 x 10^4 / 2^4 = 2^7 x 10^5 / 2^4 = 8 x 10^4

B
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As GC2808 says, first you need to speed up this problem by switching from 1,28*10^6 to 128*10^4.

Then you need to observe that are 4 periods in which the population doubles (16 hours total, 4 hours each interval) that need to take into account for "discounting" the final value of the population.

So you basically do (128*10^4)/2^4

(128*10^4)/16 results in 8*10^4.

There are no casual numbers here (and probably in none of the GMAT problems)

The key here is the first step. Doing that leads you to solve the problem in 2 min or less
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carcass
Four hours from now, the population of a colony of bacteria will reach 1.28 * \(10 ^ 6\). If the population of the colony doubles every 4 hours, what was the population 12 hours ago?

A. 6.4 * \(10 ^ 2\)
B. 8.0 * \(10 ^ 4\)
C. 1.6 * \(10 ^ 5\)
D. 3.2 * \(10 ^ 5\)
E. 8.0 * \(10 ^ 6\)

Why on earth would we deal with scientific notation?

4 hours from now: 1280000
Now: 64000
4 hours ago: 32000
8 hours ago: 16000
12 hours ago: 8000

Answer choice B.
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