[4 marks sum]

Question 14 Marks

Samita has a recurring deposit account in a bank of Rs 2000 per month at the rate of 10% p.a. If she gets Rs 83100 at the time of maturity. Find the total time for which the account was held.

Answer

Deposit per month $=$ Rs 2000,
Rate of interest $=10 \%$, Let period $=n$ months
$
\begin{aligned}
& =2000 \times \frac{n(n+1)}{2} \\
& =100 n(n+1) \text { and interest } \\
& =\frac{1000 n(n+1) \times 10 \times 1}{100 \times 12} \\
& =\frac{100 n(n+1)}{12}
\end{aligned}
$
$\therefore$ Maturity value
$
\begin{aligned}
& =2000 \times n+\frac{100 n(n+1)}{12} \\
& \therefore 2000 n+\frac{100 n(n+1)}{12}=83100 \\
& \Rightarrow 24000 n+100 n^2+100 m=83100 \times 12 \\
& \Rightarrow 240 n+n^2+n=831 \times 12 \\
& \Rightarrow n^2+241 n-9972=0 \\
& \Rightarrow n^2+27 n-36 n-9972=0 \\
& \Rightarrow 2(n+277)-36(n+277)=0 \\
& \Rightarrow(n+277)(n-36)=0
\end{aligned}
$
Either $n+277=0$, then $n=-277$, which is not possible.
or $n-36=0$, then $x=36$
$\therefore$ Period $=36$ months or 3years.

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Question 24 Marks

Mr. Chaturvedi has a recurring deposit account in Grindlay's Bank for $4 \frac{1}{2}$ years at $11 \%$ p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly installment.

Answer

Let each monthly instalment $=$ Rs $x$
Rate of interest $=11 \%$
Period $(n)=4 \frac{1}{2}$ years or 54 months,
Total principal for one month
$
\begin{aligned}
& =\text { ₹ } x \times \frac{ n ( n +1)}{2} \\
& =\text { ₹ }x \times \frac{54(54+1)}{2} \\
& =x \times \frac{54 \times 55}{2} \\
& =1485 x
\end{aligned}
$
Interest
$
\begin{aligned}
& =\frac{1485 x \times 11 \times 1}{100 \times 12} \\
& =13.6125 x
\end{aligned}
$
$\therefore$ Total amount of maturity
$
=54 x +13.6125 x
$
$
=67.6125 x
$
$\therefore 67.6125 x=101418.75$
$
\begin{aligned}
& x=\frac{101418.75}{67.6125} \\
& =\text { ₹ } 1500
\end{aligned}
$
$\therefore$ Deposit per month $=\text { ₹ } 1500$.

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Question 34 Marks

Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and deposits Rs 800 per month. If she gets Rs 48200 at the time of maturity, find
(i) the rate of simple interest,
(ii) the total interest earned by Shilpa

Answer

Deposit per month $(P)=R s 800$
Amount of maturity $=$ Rs48200
Period $(n)=4$ years $=48$ months
Let rate of interest be R\%p.a.
Total principal for one month
$
\begin{aligned}
& =\frac{ P ( n )( n +1)}{2} \\
& =\frac{800 \times 48 \times(48+1)}{2} \\
& =\text { ₹ } \frac{800 \times 48 \times 49}{2} \\
& =\text { ₹ } 940800 \\
& \text { Total deposit } \\
& =\text { ₹ } 800 \times 48 \\
& =\text { ₹ } 38400
\end{aligned}
$
Total deposit
$
\begin{aligned}
& =\text { ₹ } 800 \times 48 \\
& =\text { ₹ }38400
\end{aligned}
$
and amount of maturity
$
=\text { ₹ } 48200
$
$\therefore$ Interest earned
$
\begin{aligned}
& =\text { ₹ }48200 \text { -\text { ₹ }38400 } \\
& =\text { ₹ } 9800
\end{aligned}
$
$
\text { (i) } \therefore \text { Rate of interest }
$
$
\begin{aligned}
& =\frac{ S . I \times 100}{ P \times T } \\
& =\frac{9800 \times 100 \times 12}{940800} \\
& =12.5 \%
\end{aligned}
$
(ii) Total interest earned by Shilpa
$
=\text { ₹ }9800 \text {. }
$

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Question 44 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 54 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 [4 marks sum]

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