Let the total amount of money be \(x\) rupees and the rate of interest be \(y\%\). So, using the simple interest formula: \[ x + \frac{x \times 3 \times y}{100} = 944 \] Multiplying through by 100: \[ 100x + 3xy = 94400 \Rightarrow 3xy = 94400 - 100x \Rightarrow xy = \frac{94400 - 100x}{3} \quad \text{—— (i)} \] Now, if the rate of interest is increased by 25%, the new rate becomes: \[ y + \frac{25y}{100} = \frac{5y}{4} \] So, the new amount becomes: \[ x + \frac{x \times 3 \times \frac{5y}{4}}{100} = 980 \Rightarrow x + \frac{15xy}{400} = 980 \Rightarrow x + \frac{3xy}{80} = 980 \Rightarrow 80x + 3xy = 78400 \Rightarrow 3xy = 78400 - 80x \Rightarrow xy = \frac{78400 - 80x}{3} \quad \text{—— (ii)} \] Equating equations (i) and (ii): \[ \frac{94400 - 100x}{3} = \frac{78400 - 80x}{3} \Rightarrow 94400 - 100x = 78400 - 80x \Rightarrow -20x = -16000 \Rightarrow x = 800 \] Now substituting \(x = 800\) into equation (i): \[ 800y = \frac{94400 - 80000}{3} = \frac{14400}{3} \Rightarrow y = \frac{14400}{3 \times 800} = 6 \] ∴ The principal amount is ₹80% and the rate of interest is 6%