Answer: B
Let \(x\) be the number of years after which both principals will amount to the same total. First case: Principal \((p) = ₹1200\) Time \((t) = x\) years Rate of interest \((r) = 6\frac{1}{4}\% = \frac{25}{4}\%\) Therefore, Interest = \(\frac{1200 \times x \times 25}{100 \times 4} = 75x\) rupees Second case: Principal \((p) = ₹1000\) Time \((t) = x\) years Rate of interest \((r) = 8\frac{1}{3}\% = \frac{25}{3}\%\) Therefore, Interest = \(\frac{1000 \times x \times 25}{100 \times 3} = \frac{250x}{3}\) rupees According to the question: \[ 1200 + 75x = 1000 + \frac{250x}{3} \] Multiplying through by 3 to eliminate the denominator: \[ 3600 + 225x = 3000 + 250x \] \[ 225x - 250x = 3000 - 3600 \Rightarrow -25x = -600 \Rightarrow x = 24 \]
Let \(x\) be the number of years after which both principals will amount to the same total. First case: Principal \((p) = ₹1200\) Time \((t) = x\) years Rate of interest \((r) = 6\frac{1}{4}\% = \frac{25}{4}\%\) Therefore, Interest = \(\frac{1200 \times x \times 25}{100 \times 4} = 75x\) rupees Second case: Principal \((p) = ₹1000\) Time \((t) = x\) years Rate of interest \((r) = 8\frac{1}{3}\% = \frac{25}{3}\%\) Therefore, Interest = \(\frac{1000 \times x \times 25}{100 \times 3} = \frac{250x}{3}\) rupees According to the question: \[ 1200 + 75x = 1000 + \frac{250x}{3} \] Multiplying through by 3 to eliminate the denominator: \[ 3600 + 225x = 3000 + 250x \] \[ 225x - 250x = 3000 - 3600 \Rightarrow -25x = -600 \Rightarrow x = 24 \]