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M12-11

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Re: M12-11  [#permalink]

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New post 30 Aug 2017, 22:35
I am agreeing to the point that option A should be the OA.

Need clarification from experts on this.
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New post 09 Dec 2017, 18:21
arjun7dk wrote:
Took population as 100! for it to double it means it should become 200.
1.now percentage increase from 100 to 200 is 100%
2.now divide 100% by 20% to get 5 years
3.the question asks how many years will elapse before the population of the country doubles?
so 5-1= 4 years.
I dont know if my approach is correct!


Thats incorrect mate. You have assumed 100 across all five years.

1st year population will be 120.
2nd year You have to take 120% of 120 rather than 100. So it will be 144.
3rd year 120% of 144 and so on..
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New post 17 Jan 2018, 03:04
Hi Experts,
I did not get how 4 is the ans despite in 4th year the value crossed double but in question we need the year before value getting doubled ( i hope in ques they need year when value <=1.99999999)Is it due to approximation ans is 4 or any other reason.
Pls explain.
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New post 26 Jan 2018, 13:39
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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New post 22 Aug 2018, 12:27
I think this is a poor-quality question and I don't agree with the explanation. Correct answer should be 3 (not 4) because of word "before" in question. please help.
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New post 13 Sep 2018, 05:50
I think this is a poor-quality question and I don't agree with the explanation. I think it should be A
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New post 15 Jan 2019, 12:14
Bunuel

Is there any faster way of calculating the number of years in a compound interest formula - (1.2)^n=2

I assume it would be very time consuming to try trial and error method if some other numbers are involved.

Please suggest!
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Re: M12-11  [#permalink]

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New post 15 Jan 2019, 16:51
I see there is some uncertainty over the problem wording; the intention of the problem is asking:

\(1.2^n\geq 2\), (solving for n, where n is the lowest possible integer)

I haven't seen this wording on official questions, but on this question, "before" means "how many years will it take for the population to at least double"? If "before" is taken literally as "prior to doubling", then 0 or 1 or 2 would also be acceptable responses, which doesn't make sense.


2 solutions below:

1) A reliable way to approach percent change over a short period of years (essentially, compound interest) is a table, picking the number 100 as a starting point:

Year 0: 100
After 1 year: 120
After 2 years: 120*1.2 = 144
After 3 years: 144*1.2 = 172.8 (a quick way to add 20% is add 2*10% --> add 2*14.4 to 144 --> 172.8
After 4 years: 172.8*1.2 = 207.36
Here, we can save time by rounding and approximating -- we just need to know Yes/No whether it's greater than 200 (just like Data Sufficiency, it's best to avoid unnecessary calculations)
Round to 173 and add 2*10% --> 173 + 2*17.3 > 200? --> Clearly yes (173 + 34.6 = 207.6)

Answer is B. 4


2) Using fractions instead of decimals:

In many cases, it's a helpful habit to use fractions:
Translate "increase (grows) by 20%" to "multiply by \(\frac{6}{5}\)"

Glance at the answer choices and try 3 years first:

If you have your cubes memorized, \((\frac{6}{5})^3=\frac{216}{125}<2.\) Less than 2, so not the answer. \((2 = \frac{250}{125}\))


Next, try 4 years: \(\frac{216}{125}*\frac{6}{5}=\frac{1296}{625} > 2.\) Greater than 2, so our answer is B. 4 (\(2 = \frac{1250}{625}\))
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M12-11  [#permalink]

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New post 18 Jan 2019, 11:31
ShrutiSaboo wrote:
Is there any faster way of calculating the number of years in a compound interest formula - (1.2)^n=2

I assume it would be very time consuming to try trial and error method if some other numbers are involved.

In a math course, you would use logarithms on your calculator.

On the GMAT, we use trial and error for this. Fortunately, the calculations on the actual GMAT are generally quite simple, because you only have 2 minutes per question. I would not expect a more difficult calculation than this; in fact, most are easier.

For more details, see the solution I posted above.
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M12-11   [#permalink] 18 Jan 2019, 11:31

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