Mrs Rao deposited ₹250 per month in a recurring deposit account for 3 years. She received ₹10,110 at maturity. Find:
(a) the rate of interest.
(b) how much more interest she would receive if she deposited ₹50 more per month at the same rate and for the same time.
(a) Monthly deposit, P = ₹250
Time = 3 years = 36 months
Total deposit = ₹250 × 36 = ₹9,000
Interest = Maturity value − Total deposit
⇒ Interest = ₹10,110 − ₹9,000
⇒ Interest = ₹1,110
For a recurring deposit:
I = \mathrm{P}\times\frac{\mathrm{n}(\mathrm{n}+1)}{2}\times\frac{\mathrm{r}}{12\times100}
1110=250\times\frac{36\times37}{2}\times\frac{\mathrm{r}}{1200}⇒ 1110=\frac{333000\mathrm{r}}{2400}
⇒ r = \frac{1110\times2400}{333000}
⇒ r = 8%
(b) New monthly deposit = ₹250 + ₹50 = ₹300
\mathrm{I}=300\times\frac{36\times37}{2}\times\frac{8}{1200}⇒ I = ₹1,332
Additional interest = ₹1,332 − ₹1,110
⇒ Additional interest = ₹222
Therefore, the rate is 8% and the additional interest is ₹222.
More ICSE Hard level questions
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