Question : If a certain sum becomes 3 times in 6 years at compound interest, then in how many years, it will become 81 times?
Option 1: 81 years
Option 2: 162 years
Option 3: 27 years
Option 4: 24 years
Correct Answer: 24 years
Solution : By applying the formula: Amount = P[$(1+\frac{r}{100})^n$] where P is principal, $r$ is the rate of interest compounded annually for $n$ years. Let the sum be $x$, then: Amount = $x(1+\frac{r}{100})^6 = 3x$ ⇒ $(1+\frac{r}{100})^6 = \frac{3x}{x}=3$ ⇒ $(1+\frac{r}{100}) =3^\frac{1}{6}$ Now, the sum becomes 81 times after $n$ years, then: $x(1+\frac{r}{100})^n = 81x$ ⇒ $(1+\frac{r}{100})^n = \frac{81x}{x}=81=3^4$ ⇒ $3^\frac{n}{6} = 3^4$ Thus, $n$ = 6 × 4 = 24 years Hence, the correct answer is 24 years.
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Question : If a certain sum becomes two times in seven years at Compound Interest, then in how many years will it become eight times?
Option 1: 14 years
Option 2: 21 years
Option 3: 28 years
Option 4: 35 years
Question : A certain sum becomes 5 times in 3 years, at Simple Interest, then in how many years it will become 13 times?
Option 1: 6
Option 2: 15
Option 3: 9
Option 4: 12
Question : A certain sum becomes 7 times in 8 years at Simple Interest, then in how many years will it become 19 times of itself?
Option 1: 15
Option 2: 18
Option 3: 28
Option 4: 24
Question : A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself?
Option 1: 4 years
Option 2: 6 years
Option 3: 6.4 years
Option 4: 7.5 years
Question : A certain sum doubles in 7 years at Simple Interest. The same sum under the same interest will become 4 times in how many years?
Option 2: 28 years
Option 3: 21 years
Option 4: 10 years
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