[3 marks sum]

Question 13 Marks

Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs. 1200 as interest at the time of maturity find:
(i) the monthly instalment
(ii) the amount of maturity.

Answer

Interest = Rs 1200
Period $(n)=2$ years $=24$ months
Rate $(r)=6 \%$ p.a
Let monthly deposit = \text { ₹ }P p.m.
$\therefore$ Interest
$
\begin{aligned}
& =\frac{ P \times n ( n +1)}{2 \times 12} \times \frac{ r }{100} \\
& 1200=\frac{ P \times 24 \times 25}{24} \times \frac{6}{100}
\end{aligned}
$
$
\Rightarrow 1200=\frac{6}{4} p
$
$
\therefore P=\frac{1200 \times 4}{6}=800
$
$\therefore$ Monthly deposit $=\text { ₹ }800$
and maturity value $= Pxn +$ Interest
$
\begin{aligned}
& =\text { ₹ } 800 \times 24+\text { ₹ }1200 \\
& =\text { ₹ } 19200+\text { ₹ } 1200 \\
& =\text { ₹ }20400 .
\end{aligned}
$

View full question & answer→

Question 23 Marks

Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs 800 per month for years. If he received Rs 15084 at the time of maturity, find the rate of interest per annum.

Answer

Money deposited by Shahrukh per month $(P)=$ Rs 800
$
r=\text { ? }
$
No. of months(n)
$
=1 \frac{1}{2}
$
$
\begin{aligned}
& =\frac{3}{2} \times 12 \\
& =18 \text { months }
\end{aligned}
$
$\therefore$ Interest
$
= P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100}
$
$
\text { = ₹ } 800 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}
$
$
\text { = ₹ } 800 \times \frac{18 \times 19}{2 \times 12} \times \frac{r}{100}
$
$
=114 r
$
$\therefore$ Maturity amount
$
\begin{aligned}
& =114 r+800 \times 18 \\
& \text { ₹ } 15084=114 r+\text { ₹ } 14400 \\
& \Rightarrow \text { ₹ }15084-\text { ₹ } 14400=114 r \\
& \Rightarrow 684=114 r
\end{aligned}
$
$
\begin{aligned}
& r=\frac{684}{114} \\
& =6 \% .
\end{aligned}
$

View full question & answer→

Question 33 Marks

Mr. Gupta-opened a recurring deposit account in a bank. He deposited Rs. 2500 per month for two years. At the time of maturity he got Rs. 67500. Find :
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.

Answer

Deposit per month $=$ Rs. 2500
Period $=2$ years $=24$ months
Maturity value $=$ Rs. 67500
$\therefore$ Total principal for 1 month
$
\begin{aligned}
& =\frac{ P \times n ( n +1)}{2} \\
& =\text { ₹ }\frac{2500 \times 24 \times 25}{2} \\
& =\text { ₹ }750000 \\
& \therefore \text { Interest } \\
& =\text { ₹ }67500-24 \times 2500 \\
& =\text { ₹ } 67500-60000 \\
& =\text { ₹ } 7500
\end{aligned}
$
Period
$
\begin{aligned}
& =1 \text { month } \\
& =\frac{1}{12} \text { year }
\end{aligned}
$
$\therefore$ Rate of interest
$
\begin{aligned}
& =\frac{\text { S.I. } \times 100}{ P \times T } \\
& =\frac{7500 \times 100 \times 12}{750000 \times 1} \\
& =12 \% .
\end{aligned}
$

View full question & answer→

Question 43 Marks

David opened a Recurring Deposit Account in a bank and deposited Rs 300 per month for two years. If he received Rs 7,725 at the time of maturity, find the rate of interest per annum.

Answer

Instalment per month $(P)=$ Rs 300
Number of months( $n)=24$
Let rate of interest $(r)=r \%$ p.a
$
\begin{aligned}
& \therefore \text { S.I }= P \times \frac{ n ( n +1)}{2 \times 12} \times \frac{ r }{100} \\
& =300 \times \frac{24(24+1)}{2 \times 12} \times \frac{ r }{100} \\
& =300 \times \frac{600}{24} \times \frac{ r }{100}=\operatorname{Rs}(75) r
\end{aligned}
$
Maturity value $=\operatorname{Rs}(300 \times 24)+\operatorname{Rs}(75) r$
Given maturity value $=\operatorname{Rs~7,725}$
Then $\text { ₹ }(300 \times 24)+\operatorname{Rs}(75) r=\operatorname{Rs} 7,725$
$\Rightarrow 75 r =$ Rs $7,725-$ Rs 7,200
$
\Rightarrow r =\frac{525}{75}=7 \%
$

View full question & answer→

Question 53 Marks

Haneef has a cumulative bank account and deposits Rs. 600 per month for a period of 4 years. If he gets Rs. 5880 as interest at the time of maturity, find the rate of interest.

Answer

Interest = Rs. 58800
Monthly deposit (P) = Rs. 600
Period $(n)=4$ years or 48 months
$\therefore$ Deposit for 1 month
$
\begin{aligned}
& =\frac{ P ( n )( n +1)}{2} \\
& =\frac{600 \times 48 \times 49}{2} \\
& =\text { Rs. } 705600
\end{aligned}
$
Let, rate of interest $=r \%$ p.a.
Interest
$
=\frac{\text { Prt }}{100}
$
$\Rightarrow 5880=\frac{705600 \times r \times 1}{100 \times 12}$
$5880=588 r$
$\therefore r=\frac{5880}{588}=10$
$\therefore$ Rate of interest $=10 \%$ p.a.

View full question & answer→

Question 63 Marks

Mrs. Goswami deposits Rs. 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value. (2009)

Answer

Deposit per month $(P)=$ Rs. 1000
Period $=3$ years $=36$ months
Rate $=8 \%$
Total principal
$
=\frac{36(36+1)}{2} \times 1000
$
Interest
$
=\frac{ PRT }{100}
$
$
=\frac{36 \times 37 \times 1000 x 8}{2 \times 12 \times 100}
$
$
=12 \times 37 \times 10
$
$
=4440
$ Matured value
$
\begin{aligned}
& =\text { P } \times n+\text { S.I. } \\
& =1000 \times 36+4440 \\
& =36000+4440 \\
& =\text { Rs. } 40440 .
\end{aligned}
$

View full question & answer→

Question 73 Marks

Salom deposited Rs 150 per month in a bank for 8 months under the Recurring Deposit Scheme. ‘What will be the maturity value of his deposit if the rate of interest is 8% per annum?

Answer

Deposit per month $=$ Rs. 150
Rate of interest $=8 \%$ per
Period $(x)=8$ month
$\therefore$ Total principal for one month
$
\begin{aligned}
& =150 \times \frac{x(x+1)}{2} \\
& =\text { Rs. } 150 \times \frac{8(8 \times 1)}{2} \\
& =\frac{150 \times 8 \times 9}{2} \\
& =\text { Rs.5400. } \\
& \therefore \text { Interest }
\end{aligned}
$
$\therefore$ Interest
$
=\frac{\text { prt }}{100}
$
$
\begin{aligned}
& =\frac{5400 \times 8 \times 1}{100 \times 12} \\
& =\text { Rs. } 36
\end{aligned}
$
$\therefore$ Amount of Maturity
$=$ Rs. $150 \times 8+$ Rs. 36
$=$ Rs. $1200+36$
$=$ Rs. 1236.

View full question & answer→

Question 83 Marks

Shweta deposits Rs. 350 per month in a recurring deposit account for one year at the rate of 8% p.a. Find the amount she will receive at the time of maturity.

Answer

Deposit per month $=$ Rs 350 ,
Rate of interest $=8 \%$ p.a.
Period $(x)=1$ year
$=12$ months
$\therefore$ Total principal for one month
$
\begin{aligned}
& =350 \times \frac{x(x+1)}{2} \\
& =\text { Rs. } 350 \times \frac{12 \times 13}{2} \\
& =\text { Rs. } 350 \times 78 \\
& =\text { Rs. } 27300 \\
& \therefore \text { Interest }
\end{aligned}
$
$\therefore$ Interest
$
=\frac{ prt }{100}
$
$
\begin{aligned}
& =\frac{27300 \times 8 \times 1}{100 \times 12} \\
& =\text { Rs. } 182
\end{aligned}
$
$\therefore$ Amount of Maturity
$
\begin{aligned}
& =\text { Rs. } 350 \times 12+\text { Rs. } 182 \\
& =\text { Rs. } 4200+182 \\
& =\text { Rs. } 4382 .
\end{aligned}
$

View full question & answer→

Question 93 Marks

Mr. Dhruv deposits Rs 600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.

Answer

Deposit per month $=$ Rs 600
Rate of interest $=10 \%$ p.a.
Period $(n)=5$ years 60 months.
Total principal for one month
$
\begin{aligned}
& =\text { ₹ }600 \times \frac{ n ( n +1)}{2} \\
& =\text { ₹ } 600 \times \frac{60(60+1)}{2} \\
& =\text { ₹ } \frac{600 \times 60 \times 61}{2} \\
& =\text { ₹ }1098000
\end{aligned}
$
Interest
$
\begin{aligned}
& =\frac{\text { prt }}{100} \\
& =\frac{1098000 \times 10 \times 1}{100 \times 12} \\
& =\text { ₹ } 9150
\end{aligned}
$
$\therefore$ Amount of maturity
$=\text { ₹ } 600 \times 60+\text { ₹ }9150$
$=\text { ₹ }36000+\text { ₹ }9150$
$=\text { ₹ } 45150$.

View full question & answer→

Mathematics [3 marks sum]

Test yourself on this topic