Q. Peter borrows Rs. 12000 for 2 years at 10% p.a. compound interest. He repays Rs. 8000 at the end of first year. Find:
(i) the amount at the end of first year, before making the repayment.
(ii) the amount at the end of first year, after making the repayment.
(iii) the principal for the second year.
(iv) the amount to be paid at the end of second year, to clear the account.
A money lender borrows INR 40000 at 3% interest rate and lends the same amount at 9% interest rate what is the interest earned by the money lender after 3 years in this transaction if the interest is compounded yearly. (Approximate values for easy calculation)
- 8000
- 8320
- 9360
- 9500
Answer (Detailed Solution Below)
Option 2 : 8320
Formula used:
(i) Compouded Amount A = P(1 + R) T
(ii) A = P (1 + R) T
Given:
(i) P = 40000,
(ii) R = 3%,
(iii) T = 3 Years
Calculations:
⇒ A = 40000 (1 + 0.03) 3 ––––((1.03)3 = 1.092727 ≈ 1.09)
⇒ A = 40000 (1.09)
⇒ A = 43680
CI = 43680 – 40000 = 3680
For P = 40000, R = 9%, T = 3 Years
⇒ A = 40000(1 + 0.09) 3 ------ ((1.09)3 = 1.2950 ≈ 1.30)
⇒ A = 40000(1.30)
⇒ A = 52000
⇒ CI = 52000 – 40000 = 12000
Interest earned = 12000 – 3680
Interest earned = 8320
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What is the interest earned if Rs. 40000 is compounded annually at 12.5% p.a. for 2 years?
- Rs. 10000
- Rs. 5000
- Rs. 10025
- Rs. 5625
- Rs. 10625
Answer (Detailed Solution Below)
Option 5 : Rs. 10625
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Given:
Principal = Rs. 40000
Rate% = 12.5%
Time = 2 years
Formula used:
\(CI = P × \left[ {{{\left( {1 + \frac{R}{{100}}} \right)}^n}-1} \right]\)
Calculation:
\(CI = P × \left[ {{{\left( {1 + \frac{R}{{100}}} \right)}^n}-1} \right]\)
\(⇒ CI = 40000 × \left[ { {{\left( {1 + \frac{12.5}{{100}}} \right)}^2}-1} \right]\)
⇒ CI = 40000 × [(1 + 1/8)2 – 1 ]
⇒ CI = 40000 × (81/64 – 1)
⇒ CI = 40000 × (17/64)
∴ CI is Rs. 10625
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