Question

Each of A and B both opened recurring deposit accounts in a bank. If A deposited ₹1,200 per month for 3 years and B deposited ₹1,500 per month for 2.5 years; find, on maturity, who will get more amount and by how much? The rate of interest paid by the bank is 10% per annum.

Options:

  • A, ₹812.50
  • B, ₹952.50
  • B, ₹860.50
  • Both of them get the same amount

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Answer


The correct option is B $$ \text { B, ₹ 952.50} $$

Calculation for A:

  • Installment per month (P): ₹ 1,200
  • Number of months (n): $3 \times 12 = 36$ months
  • Total Amount Deposited: $$ 36 \times 1200 = ₹ 43,200 $$
  • Rate of interest (r): 10% per annum

Simple Interest (S.I.) Calculation:

$$ \begin{array}{rl} \text{S.I.} & = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \ & = 1200 \times \frac{36(36+1)}{2 \times 12} \times \frac{10}{100} \ & = 1200 \times \frac{1332}{24} \times \frac{10}{100} \ & = ₹ 6660 \end{array} $$

Maturity Value for A:

$$ \begin{array}{r} \text{Total Amount Deposited} + \text{Interest} \ = ₹ 43,200 + ₹ 6,660 \ = ₹ 49,860 \end{array} $$

Calculation for B:

  • Installment per month (P): ₹ 1,500
  • Number of months (n): $2.5 \times 12 = 30$ months
  • Total Amount Deposited: $$ 30 \times 1500 = ₹ 45,000 $$
  • Rate of interest (r): 10% per annum

Simple Interest (S.I.) Calculation:

$$ \begin{array}{rl} \text{S.I.} & = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \ & = 1500 \times \frac{30(30+1)}{2 \times 12} \times \frac{10}{100} \ & = 1500 \times \frac{930}{24} \times \frac{10}{100} \ & = ₹ 5812.50 \end{array} $$

Maturity Value for B:

$$ \begin{array}{r} \text{Total Amount Deposited} + \text{Interest} \ = ₹ 45,000 + ₹ 5812.50 \ = ₹ 50,812.50 \end{array} $$

Comparison:

  • B gets more amount than A.
  • Difference in Maturity Amount: $$ ₹ 50,812.50 - ₹ 49,860 = ₹ 952.50 $$

Therefore, B gets ₹ \mathbf{952.50} more than A.


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