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A builder borrows ₹ 2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be ?
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- ₹ 1352
- ₹ 1377
- ₹ 1275
- ₹ 1283
Correct Option: A
Here , A = ₹ 2550 , R = 4% per annum , n = 2 years
Let each of the two equal instalments be y .
| Present worth = | ||||
![]() | 1 + | |||
![]() | n | |||
| 100 | ||||
| P1 = | ||||
![]() | 1 + | |||
![]() | 1 | |||
| 100 | ||||
| P1 = | ||||
| 1 + | ||||
| 25 | ||||
| = | ||||
| 25 | ||||
| or P1 = | y | |
| 26 |
Similarly,
| P2 = | ![]() | ![]() | 2 | y = | y | ||
| 26 | 676 |
P1 + P2 = A
| ∴ | y + | y = 2550 | ||
| 26 | 676 |
| ⇒ | = 2550 | |
| 676 |
| ⇒ | y = 2550 | |
| 676 |
| ⇒ y = 2550 × | = ₹ 1352 | |
| 1275 |
Second Method to solve this question :
Here, P = ₹ 2550, n = 2, r = 4%
| Each instalment = | |||||||
![]() | |||||||
![]() | + | ![]() | ![]() | 2 | |||
| 100 + r | 100 + r | ||||||
| = | |||||||
![]() | |||||||
![]() | + | ![]() | ![]() | 2 | |||
| 100 + 4 | 100 + 4 | ||||||
| = | |||||
| + | ![]() | ![]() | 2 | ||
| 104 | 104 | ||||
| Each instalment = | ||||||
![]() | 1 + | ![]() | ||||
| 104 | 104 | |||||
| Each instalment = | ||||||
![]() | ![]() | |||||
| 104 | 104 | |||||
| Each instalment = | = ₹ 1352 | |
| 20400 |


