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CAT 2024 Slot 3 — QA Question 8

Compound InterestEasy

Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36, then the minimum number of years required for the value of the investment to exceed Rs 20000 is

Answer & solution

Answer: 9

Solution

Medium

The ratio of the value after 3 years to that after 5 years gives 1/(1+r)21/(1+r)^2. Solve for the growth factor, then find the smallest nn for which 4000(1+r)n>200004000(1+r)^n>20000.

1

Find the growth factor. Let R=1+rR=1+r. The 3-year value is 4000R34000R^3 and 5-year value 4000R54000R^5.

4000R34000R5=2536⇒ 1R2=2536(cancel)⇒ R2=3625⇒ R=65=1.2(r=20%)\begin{aligned} &\frac{4000R^3}{4000R^5}=\frac{25}{36}\\ &\Rightarrow\ \frac{1}{R^2}=\frac{25}{36}\quad\text{(cancel)}\\ &\Rightarrow\ R^2=\frac{36}{25}\\ &\Rightarrow\ R=\frac{6}{5}=1.2\quad\text{(}r=20\%\text{)} \end{aligned}
2

Condition to exceed 20000. We need 4000(1.2)n>200004000(1.2)^n>20000, i.e. (1.2)n>5(1.2)^n>5.

(1.2)8=4.2998…<5⇒ (1.2)9=5.1598…>5(×1.2 again)\begin{aligned} &(1.2)^8=4.2998\ldots<5\\ &\Rightarrow\ (1.2)^9=5.1598\ldots>5\quad\text{(}\times1.2\text{ again)} \end{aligned}
3

Conclusion. The value first exceeds 2000020000 at n=9n=9.

4000(1.2)9=20639…>20000⇒ nmin⁡=9\begin{aligned} &4000(1.2)^9=20639\ldots>20000\\ &\Rightarrow\ n_{\min}=9 \end{aligned}
9 years9\text{ years}
CAT 2024 Slot 3 QA Q8: Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value o — Solution | TheCATExam