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Question
Mr. Banerjee opens a recurring deposit account for Rs 3,000 per month at 9% simple interest pa. On maturity, he gets Rs. 1,70,460. Find the period for which he continued with the account.
Solution
Given that cumulative deposit per month =Rs 3000, Period= t Months, R=9°/o, Maturity amount= Rs. 1,70,460
Money deposited= Monthly value x No ofMonths=3000 x t= Rs 3000t
Total Principal for 1 Month = `(3000 xx ("t")("t" + 1))/2`
= 1500t2 + 1500 t
Interest= Principal for One month x RI ( 12 x 100)
..... ( 1)
Putting Values in (1), we get
170460 - 3000 t = `((1500 "t"^2 + 1500 "t") xx 9)/1200`
170460 - 3000 t = `(45 "t"^2 + 45 "t")/4`
45t2 + 12045 t - 681840 = 0
45t2 - 2160 t + 14205 t - 681840 = 0
45t (t- 48) + 14205 (t- 48) = 0
(t - 48)(45t + 14205) = 0
t = 48 , t = `- 14205/45`
The number of months cannot be negative.
Hence, t = 48 months = 4 years
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