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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 |