A certain sum was invested in a high-interest bond for which : GMAT Data Sufficiency (DS)
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A certain sum was invested in a high-interest bond for which

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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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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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New post 16 Mar 2011, 08:24
thnx so much fluke
regards
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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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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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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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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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New post 18 Mar 2011, 19:05
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: A certain sum was invested in a high-interest bond for which [#permalink]

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New post 19 Mar 2011, 10:54
Not a Gmat question. Do not waste your time.
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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.

can someone please have a look into this and let me know how to solve this? :cry:
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New post 15 Nov 2011, 15:47
So we now there was an investment in a bond that compounds monthly for X months. X is a whole number so no fractional months, and that the value of the bond doubled in our X months of investment.

Statement 1 - gives us not info about value - granted we can back into a range of X but a dollar amount is impossible - insuff.
Statement 2- gives us a value at X+1, but its impossible to determine the starting value as we do not know either the interest rate or the starting value - insuff.

Together you have a range of rates and a X-1 ending value. Since the starting value doesn't have to be an integer and the interest rate doesn't have to be an integer either their are a number of possible answers for this.


I would say it should be E.
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New post 15 Nov 2011, 20:54
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Answer is C.
You can find a good explanation here: http://www.urch.com/forums/gmat-math/20537-interest-rates.html
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New post 16 Nov 2011, 09:05
I decided on C after three mins, but it was more of a guess off of what I know rather than confirmation. Just hope you don't get one of these on the test
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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: A certain sum was invested in a high-interest bond for which [#permalink]

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New post 11 Sep 2013, 02:42
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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