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A man borrows ₹ 21000 at 10% compound interest. How much he has to pay annually at the end of each year, to settle his loan in two years ?
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- ₹ 12000
- ₹ 12100
- ₹ 12200
- ₹ 12300
Correct Option: B
If each instalment be y, then
∴Present worth of first instalment = | = | |||
[ 1 + ( 11 / 100 ) ] | 11 |
Present worth of second instalment = | = | |||
[ 1 + ( 11 / 100 ) ]² | 121 |
∴ | y + | y = 21000 | ||
11 | 121 |
⇒ | = 21000 | |
121 |
⇒ 210y = 21000 × 121
⇒ y = | = ₹ 12100 | |
210 |
Second Method to solve this question :
Here, n = 2 , P = ₹ 21000 , r = 10%
Each annual instalment = | |||||||
![]() | ![]() | + | ![]() | ![]() | 2 | ||
100 + r | 100 + r |
Each annual instalment = | |||||||
+ | ![]() | ![]() | 2 | ||||
110 | 110 |
Each annual instalment = | |||||||
+ | |||||||
110 | 12100 |
Each annual instalment = | |||||||
+ | |||||||
11 | 121 |
Each annual instalment = | × 121 | |
110 + 100 |
Each annual instalment = | × 121 = 12100 | |
210 |