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A certain sum amounts to Rs. 320 at 6% p...

A certain sum amounts to Rs. 320 at 6% per anuum simple interest and to Rs. 360 at 8% per annum simple interest. Find the principal.

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To solve the problem step by step, we will find the principal amount using the information given about the amounts at different interest rates. ### Step 1: Set up the equations for both interest scenarios. Let the principal amount be \( P \). 1. At 6% per annum, the amount is Rs. 320. \[ \text{Simple Interest} = \text{Amount} - \text{Principal} = 320 - P \] The formula for Simple Interest (SI) is: \[ SI = \frac{P \times R \times T}{100} \] Here, \( R = 6\% \) and \( T = t \) (time in years). Thus, we can write: \[ 320 - P = \frac{P \times 6 \times t}{100} \tag{1} \] 2. At 8% per annum, the amount is Rs. 360. \[ \text{Simple Interest} = 360 - P \] Using the same formula for Simple Interest: \[ 360 - P = \frac{P \times 8 \times t}{100} \tag{2} \] ### Step 2: Rearranging both equations. From equation (1): \[ 320 - P = \frac{6Pt}{100} \] Rearranging gives: \[ 320 - P = \frac{6Pt}{100} \implies 320 = P + \frac{6Pt}{100} \implies 320 = P\left(1 + \frac{6t}{100}\right) \tag{3} \] From equation (2): \[ 360 - P = \frac{8Pt}{100} \] Rearranging gives: \[ 360 - P = \frac{8Pt}{100} \implies 360 = P + \frac{8Pt}{100} \implies 360 = P\left(1 + \frac{8t}{100}\right) \tag{4} \] ### Step 3: Set equations (3) and (4) equal to each other. From (3) and (4): \[ P\left(1 + \frac{6t}{100}\right) = 320 \] \[ P\left(1 + \frac{8t}{100}\right) = 360 \] ### Step 4: Divide the two equations. Dividing equation (3) by equation (4): \[ \frac{320}{360} = \frac{P\left(1 + \frac{6t}{100}\right)}{P\left(1 + \frac{8t}{100}\right)} \] This simplifies to: \[ \frac{8}{9} = \frac{1 + \frac{6t}{100}}{1 + \frac{8t}{100}} \] ### Step 5: Cross-multiply to solve for \( P \). Cross-multiplying gives: \[ 8\left(1 + \frac{8t}{100}\right) = 9\left(1 + \frac{6t}{100}\right) \] Expanding both sides: \[ 8 + \frac{64t}{100} = 9 + \frac{54t}{100} \] Rearranging gives: \[ 8 - 9 = \frac{54t}{100} - \frac{64t}{100} \] \[ -1 = -\frac{10t}{100} \implies t = 10 \text{ years} \] ### Step 6: Substitute \( t \) back to find \( P \). Substituting \( t = 10 \) into equation (3): \[ 320 = P\left(1 + \frac{6 \times 10}{100}\right) \] \[ 320 = P\left(1 + 0.6\right) = P(1.6) \] \[ P = \frac{320}{1.6} = 200 \] ### Final Answer: The principal amount is Rs. 200. ---
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