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प्रश्न
Deepa has a 4-year recurring deposit account in a bank and deposits Rs 1,800 per month. If she gets Rs 1,08,450 at the time of maturity, find the rate of interest.
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उत्तर
Installment per month (P) = Rs. 1,800
Number of months (n) = 48
Let rate of interest (r) = r% p.a.
∴ `S.I. = P xx (n(n + 1))/(2 xx 12) xx r/100`
= `1800 xx (48(48 + 1))/(2 xx 12) xx r/100`
= `1800 xx 2352/24 xx r/100`
= Rs. (1,764)r
Maturity value = Rs. (1,800 × 48) + Rs. (1,764)r
Given maturity value = Rs. 1,08,450
Then Rs. (1,800 × 48) + Rs. (1764)r = Rs. 1,08,450
`\implies` 1764r = Rs. 1,08,450 – Rs. 86,400
`\implies` r = `22050/1764` = 12.5%
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संबंधित प्रश्न
A page from a savings bank account passbook is given below:
| Date | Particulars | Amount withdrawn (Rs.) |
Amount Deposited (Rs.) |
Balance (Rs.) |
| Jan 7, 2016 | B/F | 3,000.00 | ||
| Jan 10, 2016 | By Cheque | 2600.00 | 5600.00 | |
| Feb 8, 2016 | To Self | 1500.00 | 4100.00 | |
| Apr 6, 2016 | By Cheque | 2100.00 | 2000.00 | |
| May 4, 2016 | By Cash | 6500.00 | 8500.00 | |
| May 27, 2016 | By Cheque | 1500.00 | 10000.00 |
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