[2 Mark Question Answer]

Question 12 Marks

Kiran deposited Rs. 200 per month for 36 months in a bank’s recurring deposit account. If the bank pays interest at the rate of 11% per annum, find the amount she gets on maturity.

Answer

$P=$ Rs. $200, n=36$ months and $r=11 \%$
$
\begin{aligned}
& I=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \\
& \Rightarrow I=R s 200 \times \frac{36 \times 37}{24} \times \frac{11}{100}=\text { Rs } 1221
\end{aligned}
$
Sum deposited Rs. $200 \times 36=$ Rs. 7200
$\therefore$ Amount that kiran gets on maturity $=$ Rs $7200+\operatorname{Rs} 1221=R s 8421$

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Question 22 Marks

Mr. R.K. Nair gets Rs 6,455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly installment.

Answer

Let monthly instalment is Rs $P$
here $n=1$ year $=12$ months
$n=12$
$
\begin{aligned}
& \because M . V .=\frac{ n ( n +1)}{2 \times 12} \times \frac{ P \times R }{100}+ P . n \\
& \Rightarrow \text { ₹ } 6455=\frac{12(12+1)}{2 \times 12} \times \frac{ P \times 14}{100}+ P .12 \\
& \text { ₹ } 6455=\frac{13 \times P \times 7}{100}+ P .12 \\
& \Rightarrow \text { ₹ } 6455=\frac{91 P +1200 P }{100} \\
& \Rightarrow \text { ₹ } 645500=1291 P \\
& \Rightarrow P =\frac{645500}{1291} \\
& =\text { ₹ }500 .
\end{aligned}
$

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Mathematics [2 Mark Question Answer]

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