Satvik
borrows Rs. 45000 from a bank at 10% compound interest. He repays it in three
annual installments that are in arithmetic progression. He ends up paying 54000
totally. How much did he pay in year 1?
A.Rs.
16,500
B.Rs.
19,500
C.Rs.
21,000
D.Rs. 18,000
Let the repayments be Rs "a – d", Rs "a" and Rs. "a + d"
ReplyDeletea – d + a + a + d = 54000
3a = 54000
a = 18000
The payment at the end of year 2 is Rs. 18,000.
Borrowed amount = Rs. 45,000
Amount outstanding at the end of Year 1 = (45000 × 1.1) – (18000 – d)
= 31500 + d
Amount outstanding at the end of Year 2 = ((31500 + d) × 1.1) – 18000
= 34650 + 1.1d – 18000 = 16650 + 1.1d
Amount outstanding at the end of Year 3 = ((16650 + 1.1d) × 1.1) = 18000 + d
18315 + 1.21d = 18000 + d
0.21d = – 315 d = –1500
The payments are Rs. 19500, Rs. 18000 and Rs. 16500
Correct Answer: Rs. Rs. 19,500