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  1. 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 ?
    1. ₹ 12000
    2. ₹ 12100
    3. ₹ 12200
    4. ₹ 12300
Correct Option: B

If each instalment be y, then

∴Present worth of first instalment =
y
=
10y
[ 1 + ( 11 / 100 ) ]11

Present worth of second instalment =
y
=
100y
[ 1 + ( 11 / 100 ) ]²121

∴ 
10
y +
100
y = 21000
11121

⇒ 
110y + 100y
= 21000
121

⇒  210y = 21000 × 121
⇒  y =
21000 × 121
= ₹ 12100
210

Second Method to solve this question :
Here, n = 2 , P = ₹ 21000 , r = 10%
Each annual instalment =
P
100
+
100
2
100 + r100 + r

Each annual instalment =
21000
100
+
100
2
110110

Each annual instalment =
21000
100
+
10000
11012100

Each annual instalment =
21000
10
+
100
11121

Each annual instalment =
21000
× 121
110 + 100

Each annual instalment =
21000
× 121 = 12100
210



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