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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%
संबंधित प्रश्न
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| Jan 9, 2008 | By Cash | - | 10000 | 10000 |
| Feb 12, 2008 | By Cash | - | 15500 | 25500 |
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| April 30, 2008 | To Self | 2000 | - | 20000 |
| July 16, 2008 | By Cheque | - | 6500 | 26500 |
| August 4, 2008 | To Self | 5500 | - | 21000 |
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| Dec. 12, 2008 | By Cash | - | 1700 | 21500 |
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