CAT 2021 Set-3 | Quantitative Aptitude | Question: 5

Question

Bank $\text{A}$ offers $6 \%$ interest rate per annum compounded half yearly. Bank $\text{B}$ and Bank $\text{C}$ offer simple interest but the annual interest rate offered by Bank $\text{C}$ is twice that of Bank $\text{B}.$ Raju invests a certain amount in Bank $\text{B}$ for a certain period and Rupa invests $ ₹ \; 10,000$ in Bank $\text{C}$ for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank $\text{A}$ for one year. The interest accrued, in $\text{INR},$ to Rupa is 

• A. $3436$
• B. $2436$
• C. $2346$
• D. $1436$

Answer 1

Given that,

• Bank $\text{A:}$
• Interest rate $ = 6 \% \; \text{per annum (CI)}$
• Time $ = \text{half yearly}$

• Bank $\text{B:}$
• Interest rate $ = x \% \; \text{per annum (SI)}$

• Bank $\text{C:}$
• Interest rate $ = 2x \% \; \text{per annum (SI)}$

Let Raju invested $₹ \; P$ in bank $\text{B}$ for $t$ years, hence Rupa invested $₹ \; 10000$ in bank $\text{C}$ for $2t$ years.

we know that, $\boxed{A = P \left(1 + \frac{R}{100} \right)^{t}}$

$\qquad \qquad \boxed{\text{CI = A – P}}$

Now, $\text{SI = CI}$

$\Rightarrow \frac{P \times x \times t}{100} = P \left(1 + \frac{3}{100} \right)^{2} – P$

$\Rightarrow \frac{xt}{100} = \left(\frac{103}{100} \right)^{2} – 1$

$\Rightarrow \frac{xt}{100} = 1.0609 – 1$

$\Rightarrow \frac{xt}{100} = 0.0609$

$\Rightarrow \boxed{xt = 6.09}$

Now, we can calculate $\text{SI}.$

$\text{SI} = \frac{10000 \times 2t \times 2x}{100}$

$\Rightarrow \text{SI} = 400 xt$

$\Rightarrow \text{SI} = 400 \times 6.09 \quad [\because xt = 6.09]$

$\Rightarrow \boxed{\text{SI} = ₹ \; 2436}$

$\therefore$ The interest accrued, in $\text{INR},$ to Rupa is $₹ \; 2436.$

Correct Answer $: \text{B}$