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

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.

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"

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.
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(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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