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Kamal took 6800 as a loan which along with interest is to be repaid in two equal
value of each instalment isannual instalments. If the rate of interest is 12 1 %, compounded annually, then the 2
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- ₹ 8100
- ₹ 4150
- ₹ 4050
- ₹ 4000
Correct Option: C
Let the annual instalment be y
Using the given formula ,
A = P | ![]() | 1 + | ![]() | T | |
T |
y = P1 | ![]() | 1 + | ![]() | |
200 |
⇒ y = P1 × | |
8 |
⇒ P1 = | y | |
9 |
Similarly,P2 = | y | |
81 |
P1 + P2 = 6800
⇒ | y + | y = 6800 | ||
9 | 81 |
⇒ | = 6800 | |
81 |
⇒ | = 6800 | |
81 |
⇒ y = | = ₹ 4050 | |
136 |
We can find required answer with the help of given formula :
Here, P = ₹ 6800, R = | % , n = 2 | |
2 |
Each instalment = | |||||||
![]() | |||||||
![]() | + | ![]() | ![]() | 2 | |||
100 + r | 100 + r |
Each instalment = | |||||||
![]() | |||||||
![]() | + | ![]() | ![]() | 2 | |||
100 + (25/2) | 100 + (25/2) |
Each instalment = | ||||||
+ | ![]() | ![]() | 2 | |||
225 | 225 |
Each instalment = | ||||||
![]() | 1 + | ![]() | ||||
225 | 225 |
Each instalment = | = ₹ 4050 | |
200 × 425 |