[5 marks sum]

Question 15 Marks

Rajiv Bhardwaj has a recurring deposit account in a bank of Rs 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs 15450 as maturity amount, find the total time for which the account was held.

Answer

Deposit during the month $(P)=R s 600$
Rate of interest $=7 \%$ p.a.
Amount of maturity $=$ Rs15450
Let time $= n$ months
$\therefore$ Total principal
$
\begin{aligned}
& =\frac{P(n)(n+1)}{2} \\
& =\frac{600 \times n(n+1)}{2} \\
& =\frac{600\left(n^2+n\right)}{2} \\
& =300\left(n^2+n\right)
\end{aligned}
$
$\therefore$ Interest
$
\begin{aligned}
& =\frac{\text { PRT }}{100} \\
& =\frac{300\left(n^2+n\right) \times 7 \times 1}{100 \times 12} \\
& =\frac{7}{4}\left(n^2+n\right) \\
& \therefore 600 n+\frac{7}{4}\left(n^2+n\right)=15450
\end{aligned}$
$
\begin{aligned}
& \Rightarrow 2400 n+7 n^2+7 n=61800 \\
& \Rightarrow 7 n^2+2407 n-61800=0 \\
& \Rightarrow 7 n^2-168 n+257 n-61800=0 \\
& \Rightarrow 7 n(n-24)+2575(n-24)=0 \\
& \Rightarrow(n-24)(7 n+2575)=0
\end{aligned}$
Either $n-24=0$, then $n=24$
or
$7 n+2575=0$, then
$7 n=-2575$
$
\Rightarrow n =\frac{-2575}{7}
$
Which is not possible being negative.
$
\therefore n =24
$
$\therefore$ Period $=24$ months or 2 years.

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Question 25 Marks

Ankita started paying Rs 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs 500 per month in a $2 \frac{1}{2}$ years recurring deposit. The bank paid $10 \%$ p.a. simple interest for both. At maturity who will get more money and by how much?

Answer

In case of Ankita,
Deposit per month $=$ Rs 400
Period $(n)=3$ years $=36$ months
Rate of interest $=10 \%$
Total principal for one month
$
\begin{aligned}
& =400 \times \frac{ n ( n +1)}{2} \\
& =400 \times \frac{36(36+1)}{2} \\
& =\text { ₹ } \frac{400 \times 36 \times 37}{2} \\
& =\text { ₹ }266400
\end{aligned}
$
Interset
$
=\frac{ prt }{100}
$
$
\begin{aligned}
& =\frac{266400 \times 10 \times 1}{100 \times 12} \\
& =\text { ₹ }2220
\end{aligned}
$
$\therefore$ Amount of maturity
$
\begin{aligned}
& =\text { ₹ } 400 \times 36+\text { ₹ } 2220 \\
& =\text { ₹ } 14400+\text { ₹ }2220 \\
& =\text { ₹ }16620
\end{aligned}
$
In case of Anshul,
Deposit P.m. =\text { ₹ }500
Rate of interest $=10 \%$
Period $(n)=2 \frac{1}{2}$ year $=30$ months
$\therefore$ Total principal for one month
$
\begin{aligned}
& =\text { ₹ } 500 \times \frac{n(n+1)}{2} \\
& =500 \times \frac{30(30+1)}{2} \\
& =\text { ₹ }\frac{500 \times 30 \times 31}{2} \\
& =\text { ₹ } 232500
\end{aligned}
$
Interest
$
\begin{aligned}
& =\frac{232500 \times 10 \times 1}{100 \times 12} \\
& =\text { ₹ }1937.50
\end{aligned}
$
Amount of maturity
$
\begin{aligned}
& =\text { ₹ }500 \times 30+\text { ₹ }1937.50 \\
& =\text { ₹ } 15000+\text { ₹ }1937.50 \\
& =\text { ₹ }16937.50
\end{aligned}
$
At maturity Anshul will get more amount
Difference
$
\begin{aligned}
& =\text { ₹ } 16937.50-\text { ₹ }16620.00 \\
& =\text { ₹ }317.50 .
\end{aligned}
$

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Mathematics [5 marks sum]

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