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Question
Mr. Gulati has a Recurring Deposit Account of Rs. 300 per month. If the rate of interest is 12% and the maturity value of this account is Rs. 8,100; find the time (in years) of this Recurring Deposit Account.
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Solution
Let n be the maturity period, P be the money deposited every month and R be the rate of interest per annum. The interest received at the end of the maturity is given by,
`I = P xx (n(n + 1))/24xxr/100`
In the question, we are given that the amount deposited every month is
Rs. 300,
P = Rs. 300
Rate of interest per annum, R = 12%
Now, Mr. Gulati receives Rs. 8,100 at the time of maturity, so
I = 16220 − 400n
On putting the known values in the above equation, we get-
`8100 - 300n = 300xx(n(n + 1))/24 xx 12/100`
`8100 - 300n = (3n(n + 1))/2`
16200 − 600n = 3n2 + 3n
3n2 + 603n − 16200 = 0
⇒ n2 + 201n − 5400 = 0
Factorizing the above equation, we get,
n2 + 225n − 24n − 5400 = 0
n(n + 225) − 24(n + 225) = 0
(n − 24)(n + 225)
⇒ n = 24, n = −225
Now, the time period cannot be negative; the total time the account was kept is 24 months or 2 years.
RELATED QUESTIONS
A page from the Savings Bank Account of Mr Prateek is given below:
| Date | Particulars | Withdrawal (In Rs.) |
Deposit (In Rs.) |
Balances (In Rs.) |
| January 1st, 2006 | B/F | - | - | 1270 |
| January 7th, 2006 | By Cheque | - | 2310 | 3580 |
| March 9th, 2006 | To Self | 2000 | - | 1580 |
| June 26th, 2006 | By Cash | - | 6200 | 7780 |
| June 10th 2006 | To Cheque | 4500 | - | 3280 |
| July 15th, 2006 | By Clearing | - | 2630 | 5910 |
| October 18th, 2006 | To Cheque | 530 | - | 5380 |
| October 27th, 2006 | To Self | 2690 | - | 2690 |
| November 3rd, 2006 | By Cash | - | 1500 | 4190 |
| December 6th, 2006 | To Cheque | 950 | - | 3240 |
| December 23rd, 2006 | By Transfer | - | 2920 | 6260 |
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