Susan invests in two schemes A (8% p.a.) and B (9% p.a.). She receives Rs 1860 interest in total. If interchanged, the interest would be Rs 20 more. Find the amounts invested in each scheme.
Scheme A: Rs 12,000; Scheme B: Rs 10,000.
Step 1: Assume the amounts.
Let the amount invested in Scheme A = \(x\).
Let the amount invested in Scheme B = \(y\).
Step 2: Write the first condition.
Interest from A = \(8\% \times x = 0.08x\).
Interest from B = \(9\% \times y = 0.09y\).
Total interest = 1860.
So, \(0.08x + 0.09y = 1860\). … (1)
Step 3: Write the second condition.
If the amounts were interchanged:
Interest from A = \(9\% \times x = 0.09x\).
Interest from B = \(8\% \times y = 0.08y\).
Total interest now = 1880 (20 more).
So, \(0.09x + 0.08y = 1880\). … (2)
Step 4: Simplify the equations.
Multiply both equations by 100 to remove decimals:
(1) becomes \(8x + 9y = 186000\).
(2) becomes \(9x + 8y = 188000\).
Step 5: Subtract the equations.
Subtract (1) from (2):
\((9x - 8x) + (8y - 9y) = 188000 - 186000\).
\(x - y = 2000\). … (3)
Step 6: Solve for one variable.
From (3): \(x = y + 2000\).
Step 7: Substitute back.
Put \(x = y + 2000\) in equation (1):
\(8(y + 2000) + 9y = 186000\).
\(8y + 16000 + 9y = 186000\).
\(17y + 16000 = 186000\).
\(17y = 170000\).
\(y = 10000\).
Step 8: Find \(x\).
\(x = y + 2000 = 10000 + 2000 = 12000\).
Final Answer:
Scheme A = Rs 12,000 and Scheme B = Rs 10,000.