[4 marks sum]

Question 14 Marks

A recurring deposit account of Rs 1,200 per month has a maturity value of Rs 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.

Answer

$ \text { Installment per month }(P)=\text { Rs } 1,200 $
Number of months( $n$ ) $=n$
Let rate of interest $(r)=8 \%$ p.a.
$ \begin{aligned} & \therefore S . I=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \\ & =1200 \times \frac{n(n+1)}{2 \times 12} \times \frac{8}{100} \\ & =1200 \times \frac{n(n+1)}{24} \times \frac{8}{100}=\operatorname{Rs} 4 n(n+1) \end{aligned} $
$\begin{aligned} & \text { Maturity value }=\text₹(1,200 \times n)+\text₹ 4 n(n+1)=\operatorname{Rs}\left(1200 n+4 n^2+4 n\right) \\ & \text { Given maturity value }=\text₹ 12,440 \\ & \text { Then } 1200 n+4 n^2+4 n=12,440 \\ & \Rightarrow 4 n^2+1204 n-12440=0 \\ & \Rightarrow n^2+301 n-3110=0 \\ & \Rightarrow(n+311)(n-10)=0 \\ & \Rightarrow n=-311 \text { or } n=10 \text { months }\end{aligned}$
Then number of months $=10$

View full question & answer→

Question 24 Marks

Each of A and B both opened recurring deposit accounts in a bank. If A deposited Rs 1,200 per month for 3 years and B deposited Rs 1,500 per month for 2 ½ years; find, on maturity, who will get more amount and by how much? The rate of interest paid by the bank is 10% per annum.

Answer

Installment per month $(P)=$ Rs 1,200
Number of months $(n)=36$
Rate of interest $(r)=10 \%$ p.a.
$ \begin{aligned} & \text { S.I }=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \\ & =1200 \times \frac{36(36+1)}{2 \times 12} \times \frac{10}{100} \\ & =1200 \times \frac{1332}{24} \times \frac{10}{100} \\ & =\text { Rs. } 6,660 \end{aligned} $
The amount that A will get at the time of maturity
$ \begin{aligned} & =(1,200 \times 36)+6,660 \\ & =43,200+6,660 \\ & =\text { Rs. } 49,860 \end{aligned} $
For B
$ \text { Instalment per month }(P)=\text { Rs. } 1,500 $
Number of months $(n)=30$
Rate of interest $(r)=10 \%$ p.a.
$ \begin{aligned} & \text { S.I }=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \\ & =1500 \times \frac{30(30+1)}{2 \times 12} \times \frac{10}{100} \\ & =1500 \times \frac{930}{24} \times \frac{10}{100} \\ & =\text { Rs. } 5812.50 \end{aligned} $
The amount that B will get at the time of maturity
$ \begin{aligned} & =(1500 \times 30)+5812.50 \\ & =45000+5812.50 \\ & =\text { Rs. } 50812.50 \end{aligned} $
Difference between both amounts $=50,812.50-49,860$
$ =\text { Rs. } 952.50 $
Then B will get more money than A by Rs. 952.50 .

View full question & answer→

Question 34 Marks

Mr Gulati has a Recurring Deposit Account of Rs 300 per month. If the rate of interest is 12% and the maturity value of this account is Rs 8,100; find the time (in years) of this Recurring Deposit Account.

Answer

Installment per month $(P)=$ Rs 300
Number of months $(n)=n$
Let rate of interest $(r)=12 \%$ p.a.
$ \begin{aligned} & \therefore S . I=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \\ & =300 \times \frac{n(n+1)}{2 \times 12} \times \frac{12}{100} \\ & =300 \times \frac{n(n+1)}{24} \times \frac{12}{100}=\text { Rs } 1.5 n(n+1) \end{aligned} $
$ \begin{aligned} & \text { Maturity value }=\operatorname{Rs}(300 \times n)+\operatorname{Rs} 1.5 n(n+1) \\ & =\operatorname{Rs}\left(300 n+1.5 n^2+1.5 n\right) \end{aligned} $
Given maturity value $=$ Rs 8,100
Then $300 n+1.5 n^2+1.5 n=8,100$
$ \begin{aligned} & \Rightarrow 1.5^2+301.5 n-8100=0 \\ & \Rightarrow n^2+201 n-5400=0 \\ & =>(n+225)(n-24)=0 \\ & \Rightarrow n=-225 \text { or } n=24 \text { months } \end{aligned} $
Then time $=2$ years.

View full question & answer→

Mathematics [4 marks sum]

Test yourself on this topic