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Question
A recurring deposit account of Rs. 1,200 per month has a maturity value of Rs. 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.
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Solution
Installment per month (P) = Rs. 1,200
Number of months (n) = n
Let rate of interest (r) = 8% p.a.
∴ `S.I. = P xx (n(n +1))/(2 xx 12) xx r/100`
= `1200 xx (n(n + 1))/(2 xx 12) xx 8/100`
= `1200 xx (n(n + 1))/24 xx 8/100`
= Rs. 4n(n + 1)
Maturity value = Rs. (1,200 × n) + Rs. 4n(n + 1)
= Rs. (1200n + 4n2 + 4n)
Given maturity value = Rs. 12,440
Then 1200n + 4n2 + 4n = 12,440
⇒ 4n2 + 1204n – 12440 = 0
⇒ n2 + 301n – 3110 = 0
⇒ (n + 311)(n – 10) = 0
⇒ n = – 311 or n = 10 months
Then number of months = 10
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Amount deposited (Rs.) |
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| 03/04/2006 | B/F | 4000.00 | ||
| 05/04/2006 | By cash | 2000.00 | 6000.00 | |
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If the interest on 1700 rupees is 340 rupees for 2 years the rate of interest must be ______.
Fill in the blanks in the table.
| Principal | Rate of interest (p.c.p.a.) | Time | Interest | Amount |
| 4200 | 7% | 3 years | ...... | ...... |
Fill in the blanks in the table.
| Principal | Rate of interest (p.c.p.a.) | Time | Interest | Amount |
| ...... | 6% | 4 years | 1200 | ...... |
