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A certain sum was invested in a high-interest bond for which

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A certain sum was invested in a high-interest bond for which [#permalink] New post 16 Mar 2011, 06:49
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A certain sum was invested in a high-interest bond for which the interest is compounded monthly. The bond was sold x number of months later, where x is an integer. If the value of the original investment doubled during this period, what was the approximate amount of the original investment in dollars?

(1) The interest rate during the period of investment was greater than 39 percent but less than 45 percent.
(2) If the period of investment had been one month longer, the final sale value of the bond would have been approximately $2,744.
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Re: compound intrest [#permalink] New post 16 Mar 2011, 07:40
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punyadeep wrote:
Q A certain sum was invested in a high-interest bond for which the interest is compounded monthly. The bond
was sold x number of months later, where x is an integer. If the value of the original investment doubled
during this period, what was the approximate amount of the original investment in dollars?
(1) The interest rate during the period of investment was greater than 39 percent but less than 45 percent.
(2) If the period of investment had been one month longer, the final sale value of the bond would have
been approximately $2,744.


Investment = $P
time: x months = x/12 years
Periods = n = 12

Return after application of the compound Interest for x months;

P(1+\frac{r}{n})^{nt}
P(1+\frac{r}{12})^{12*x/12}
P(1+\frac{r}{12})^x
It is given that the investment doubles after x months;
P(1+\frac{r}{12})^x=2P
(1+\frac{r}{12})^x=2

r and x are unknown

1. 0.4<=r<=0.44
(1+\frac{0.4}{12})^x=(1.033)^x = 2
to
(1+\frac{0.44}{12})^x=(1.036)^x=2

x can be found; but we don't know P.
Not Sufficient.

2.
P(1+\frac{r}{12})^{(x+1)}=2744
P(1+\frac{r}{12})^x*(1+\frac{r}{12})=2744
2P*(1+\frac{r}{12})=2744
We still have two unknowns.
Not Sufficient.

Combining both;

r= 0.4
2P*(1+\frac{0.4}{12})=2744

Now, we can get approx value of P.

Ans: "C"
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Re: compound intrest [#permalink] New post 16 Mar 2011, 08:24
thnx so much fluke
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Re: compound intrest [#permalink] New post 16 Mar 2011, 21:19
Where is this question from? It makes no sense, in DS, to ask for the 'approximate value' of something -- how could you possibly know what information would be sufficient? If I ask the following question:

What is the approximate value of x?

1. 3 < x < 5
2. 4 < x < 4.5

Is Statement 1 sufficient? Statement 2? You can't possibly know. It's a nonsensical question to ask in DS, so I wouldn't use other questions from the same source.
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Re: compound intrest [#permalink] New post 17 Mar 2011, 06:10
Good point Ian, based on (2) I could easily say "initial investment is between $0 and $1372, approximately"
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Re: compound intrest [#permalink] New post 18 Mar 2011, 09:37
Fluke - how did u go from the 2nd stage to the 3rd?!?!

2.
P(1+\frac{r}{12})^{(x+1)}=2744
P(1+\frac{r}{12})^x*(1+\frac{r}{12})=2744
2P*(1+\frac{r}{12})=2744
We still have two unknowns.
Not Sufficient.
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Re: compound intrest [#permalink] New post 18 Mar 2011, 10:00
144144 wrote:
Fluke - how did u go from the 2nd stage to the 3rd?!?!

2.
P(1+\frac{r}{12})^{(x+1)}=2744
P(1+\frac{r}{12})^x*(1+\frac{r}{12})=2744
2P*(1+\frac{r}{12})=2744
We still have two unknowns.
Not Sufficient.


From the stem, please see the [highlight]highlighted[/highlight] part.

fluke wrote:
Investment = $P
time: x months = x/12 years
Periods = n = 12

Return after application of the compound Interest for x months;

P(1+\frac{r}{n})^{nt}
P(1+\frac{r}{12})^{12*x/12}
P(1+\frac{r}{12})^x
It is given that the investment doubles after x months;
P(1+\frac{r}{12})^x=2P [highlight]--------------A[/highlight]
(1+\frac{r}{12})^x=2

r and x are unknown

1. 0.4<=r<=0.44
(1+\frac{0.4}{12})^x=(1.033)^x = 2
to
(1+\frac{0.44}{12})^x=(1.036)^x=2

x can be found; but we don't know P.
Not Sufficient.

2.
P(1+\frac{r}{12})^{(x+1)}=2744
P(1+\frac{r}{12})^x*(1+\frac{r}{12})=2744
[highlight]Substitute from equation A[/highlight]
2P*(1+\frac{r}{12})=2744
We still have two unknowns.
Not Sufficient.

Combining both;

r= 0.4
2P*(1+\frac{0.4}{12})=2744

Now, we can get approx value of P.

Ans: "C"

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Re: compound intrest [#permalink] New post 18 Mar 2011, 19:05
Expert's post
This question reminds me:

Sometimes, it's useful to know that in compound interest, the principal doubles approximately every 72/r years where r is the rate of interest.

i.e. if rate of interest is 10, the principal doubles in approximately 72/10 = 7.2 years.
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Re: compound intrest [#permalink] New post 19 Mar 2011, 10:54
Not a Gmat question. Do not waste your time.
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Re: compound intrest [#permalink] New post 10 Sep 2013, 17:19
VeritasPrepKarishma wrote:
This question reminds me:

Sometimes, it's useful to know that in compound interest, the principal doubles approximately every 72/r years where r is the rate of interest.

i.e. if rate of interest is 10, the principal doubles in approximately 72/10 = 7.2 years.


Couldn't the following give multiple values of x (may be with a minor difference each)? So, we wouldn't have a single answer to the question & hence 1 & 2 aren't sufficient together. SO, the answer shd be E. Please correct me if I'm wrong. Thank you.
**********************************
(1+\frac{0.4}{12})^x=(1.033)^x = 2
to
(1+\frac{0.44}{12})^x=(1.036)^x=2

x can be found; but we don't know P.
**********************************
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Re: compound intrest [#permalink] New post 11 Sep 2013, 02:42
Expert's post
divineacclivity wrote:
VeritasPrepKarishma wrote:
This question reminds me:

Sometimes, it's useful to know that in compound interest, the principal doubles approximately every 72/r years where r is the rate of interest.

i.e. if rate of interest is 10, the principal doubles in approximately 72/10 = 7.2 years.


Couldn't the following give multiple values of x (may be with a minor difference each)? So, we wouldn't have a single answer to the question & hence 1 & 2 aren't sufficient together. SO, the answer shd be E. Please correct me if I'm wrong. Thank you.
**********************************
(1+\frac{0.4}{12})^x=(1.033)^x = 2
to
(1+\frac{0.44}{12})^x=(1.036)^x=2

x can be found; but we don't know P.
**********************************


This is a poor quality question. Check here: a-certain-sum-was-invested-in-a-high-interest-bond-for-which-110991.html#p893606
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Re: compound intrest   [#permalink] 11 Sep 2013, 02:42
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