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Bunuel
A sum of money is invested under compound interest for a few years. After how many years will the sum of money become nine times its present value?
(1) The sum of money invested under compound interest at the same rate of interest per annum became thrice its value in 6 years.
(2) The sum of money, under compound interest, at the same rate of interest per annum, would become twenty-seven times its present value in 9 years.

Project DS Butler Data Sufficiency (DS3)


For DS butler Questions Click Here

Solution:

If we assume the number of years when the sum of money will become nine times be \(n\), we can say \(9P=P[(1+\frac{r}{100})^n-1]\)
\(⇒9=(1+\frac{r}{100})^n-1\)
\(⇒10=(1+\frac{r}{100})^n\)..........\((i)\)

Statement 1: The sum of money invested under compound interest at the same rate of interest per annum became thrice its value in 6 years

According to this statement, \(3P=P[(1+\frac{r}{100})^6-1]\)
\(⇒3=(1+\frac{r}{100})^6-1\)
\(⇒4=(1+\frac{r}{100})^6\)
\(⇒1+\frac{r}{100}=4^{\frac{1}{6}}\)

Now we can plug this value of \(1+\frac{r}{100}\) in equation \((i)\) to get the value of \(n\)

Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: The sum of money, under compound interest, at the same rate of interest per annum, would become twenty-seven times its present value in 9 years

According to this statement, \(27P=P[(1+\frac{r}{100})^9-1]\)

Similar to statement 1, statement 2 alone is also sufficient


Hence the right answer is Option D
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SaquibHGMATWhiz
Bunuel
A sum of money is invested under compound interest for a few years. After how many years will the sum of money become nine times its present value?
(1) The sum of money invested under compound interest at the same rate of interest per annum became thrice its value in 6 years.
(2) The sum of money, under compound interest, at the same rate of interest per annum, would become twenty-seven times its present value in 9 years.

Project DS Butler Data Sufficiency (DS3)


For DS butler Questions Click Here

Solution:

If we assume the number of years when the sum of money will become nine times be \(n\), we can say \(9P=P[(1+\frac{r}{100})^n-1]\)
\(⇒9=(1+\frac{r}{100})^n-1\)
\(⇒10=(1+\frac{r}{100})^n\)..........\((i)\)

Statement 1: The sum of money invested under compound interest at the same rate of interest per annum became thrice its value in 6 years

According to this statement, \(3P=P[(1+\frac{r}{100})^6-1]\)
\(⇒3=(1+\frac{r}{100})^6-1\)
\(⇒4=(1+\frac{r}{100})^6\)
\(⇒1+\frac{r}{100}=4^{\frac{1}{6}}\)

Now we can plug this value of \(1+\frac{r}{100}\) in equation \((i)\) to get the value of \(n\)

Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: The sum of money, under compound interest, at the same rate of interest per annum, would become twenty-seven times its present value in 9 years

According to this statement, \(27P=P[(1+\frac{r}{100})^9-1]\)

Similar to statement 1, statement 2 alone is also sufficient


Hence the right answer is Option D

Dear SaquibHGMATWhiz
How did you deduce that number of compound periods are equal in the stimulus and in the statements 1 and 2?
The formula is \(P[(1+\frac{r}{100*n})^t^n]\)

Compound interests quarterly and annually will give different sum per year.

Thanks beforehand.
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