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On a certain date, Pat invested $10,000 at x percent annual

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On a certain date, Pat invested $10,000 at x percent annual [#permalink] New post 02 Sep 2008, 19:50
On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???
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Re: compound interest [#permalink] New post 02 Sep 2008, 23:17
why nobody is answering ???

To all the einsteins out there...pls give soln to help lesser mortals .....
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Re: compound interest [#permalink] New post 02 Sep 2008, 23:24
arjtryarjtry wrote:
On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???


your second post blocked my post.

i cannot do it without calculator or computer and also beleieve this is not real gmat type question cuz its very difficult to get the value without those machines.

if i were to choose during the test, would go for 18/ or 20. probably 18 cuz not it wont take too long to get the value doubled....
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Re: compound interest [#permalink] New post 03 Sep 2008, 01:22
given 40000=10000(1+x/100)^12
or 4 ^1/12 = (1+x/100)
or x/100 = 4^1/12-1........A

asked 80000=10000(1+x/100)^n..............b
n=?

substitute value of x/100 from A in B.
8 = (1+4^1/12-1)^n
or 8= 4^n/12
2^3=2^2n/12
n=18

ans = C
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Re: compound interest [#permalink] New post 03 Sep 2008, 06:31
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arjtryarjtry wrote:
On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???


40k = 10k(1+x/100)^12 --> 4=(1+x/100)^12
--> (1+x/100)^6=2
80k=40k(1+x/100)^n
--> (1+x/100)^n =2 =(1+x/100)^6
n=6

Total years = 12+6=18
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Re: compound interest [#permalink] New post 03 Sep 2008, 12:17
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$40,000 is 4 times the original amount.

Therefore it doubled twice in 12 years.

Therefore its doubling time is 6.

$80,000 is double $40,000 so another 6 years will get us to $80,000.

Therefore 12+6=18 is the necessary amount of time.
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Re: compound interest [#permalink] New post 03 Sep 2008, 12:22
lsmv479 wrote:
$40,000 is 4 times the original amount.

Therefore it doubled twice in 12 years.

Therefore its doubling time is 6.

$80,000 is double $40,000 so another 6 years will get us to $80,000.

Therefore 12+6=18 is the necessary amount of time.


good thinking..!!!
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Re: compound interest [#permalink] New post 03 Sep 2008, 12:47
arjtryarjtry wrote:
On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???


Let's simplify this question.

First equation; 40,000 = 10,000(1 + x)^12
Given that (1 + x) = a and rearrange the equation above we get
4 = a^12 = 2^2
So now we know that a^12 = (a^6)^2 = 2^2
Therefore, (a^6) = 2

Second equation; 80,000 = 10,000(1 + x)^n
Or 8 = a^n
2^3 = a^n
(a^6)^3 = a^n = a^18 ---- (since a^6 = 2)

therefore, n = 18. C is the answer
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Re: compound interest [#permalink] New post 03 Sep 2008, 13:42
Even though it looks like a tough question, it is actually not. 4 times in 12 yesrs = > 2 times in 6 years ( from 40K to 80K ).
So total year = 12 +6 =18
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Re: compound interest [#permalink] New post 03 Sep 2008, 22:50
lsmv479 wrote:
$40,000 is 4 times the original amount.

Therefore it doubled twice in 12 years.

Therefore its doubling time is 6.

$80,000 is double $40,000 so another 6 years will get us to $80,000.

Therefore 12+6=18 is the necessary amount of time.


+1 for out of the box thinking and more imp..
GMAT way of thinking.
will give you kudos. :wink:
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Re: compound interest [#permalink] New post 02 Feb 2009, 02:35
$40,000 is 4 times the original amount.

Therefore it doubled twice in 12 years.

Therefore its doubling time is 6.

$80,000 is double $40,000 so another 6 years will get us to $80,000.

Therefore 12+6=18 is the necessary amount of time.

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is this the right method???
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Re: compound interest [#permalink] New post 02 Feb 2009, 03:55
LoyalWater wrote:
is this the right method???


I think it is exactly how we should solve the problem in GMAT: fast and elegantly
lsmv479, +1
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Re: compound interest [#permalink] New post 02 Feb 2009, 08:51
GMAT TIGER wrote:
arjtryarjtry wrote:
On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???


your second post blocked my post.

i cannot do it without calculator or computer and also beleieve this is not real gmat type question cuz its very difficult to get the value without those machines.

if i were to choose during the test, would go for 18 or 20. probably 18 cuz it wont take too long to get the value doubled....


I found a method: Rule of 72.

Given an x% return, it takes 10,000 to quadralope 12 years.

So according to the rule: 72/x is the no of years 10,000.00 took to double 20,000.00.
Again, 20,000.00 took to double 40,000.00 same (72/x) no of years.

72/x+ 72/x = 12
x = 12% (though rate here is not very much required).

Again, 40,000.00 takes the same (72/x) no of years to double 80,000.00.

72/x = 6 years.

So altogather:
10,000 - 20,000 = 6 years
20,000 - 40,000 = 6 years
40,000 - 80,000 = 6 years
total 18 years.
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Re: compound interest   [#permalink] 02 Feb 2009, 08:51
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