ML Aggarwal Solution Class 10 Chapter 2 Banking Exercise 2

Exercise 2

Question 1

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.

Sol :

Deposit per month = Rs 350,

Rate of interest = 8% p.a.

Period (x) = 1 year

= 12 months

∴Total principal for one month $350 \times \frac{x(x+1)}{2}=\operatorname{Rs} 350 \times \frac{12 \times 13}{2}$

= Rs. 350×78

= Rs. 27300

∴Interest $=\frac{PRT}{100}=\frac{27300 \times 8 \times 1}{100 \times 12}=$ Rs. 182

∴Amount of Maturity=Rs 350×12+Rs 182

=Rs 4200+182=Rs 4382 


Question 2

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 ?

Sol :

Deposit per month = Rs. 150

Rate of interest = 8% per

Period (x) = 8 month

∴Total principal for one month$=\operatorname{Rs.} 150 \times \frac{x(x+1)}{2}=\operatorname{Rs} 150 \times \frac{8(8+1)}{2}$ $=\frac{150 \times 8 \times 9}{2}=\operatorname{Rs} .5400 .$

∴Interest $=\frac{p r t}{100}=\frac{5400 \times 8 \times 1}{100 \times 12}=$ Rs. 36

∴Amount of Maturity=Rs 150×8+Rs 36

=Rs 1200+36=Rs 1236 


Question 3

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)

Sol :

Deposit per month (P) = Rs. 1000

Period = 3 years = 36 months

Rate = 8%

Total principal $=\frac{36(36+1)}{2} \times 1000$

Intercst $=\frac{\mathrm{PR} \mathrm{T}}{100}=\frac{36 \times 37 \times 1000 \times 8}{ 2 \times 12 \times 100}$

=12×37×10=4440

Matured value $=P \times n+S . I=1000 \times 36+4440=36000+4440=$ Rs. 40440


Question 4

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

Sol :

Amount deposited month (P) = Rs. 200

Period (n) = 36 months,

Rate (R) = 11% p.a.

Now amount deposited in 36 months = Rs. 200 x 36 = Rs 7200

Simple Interest $(\mathrm{S.I})=\mathrm{P}\left(\frac{n(n+1)}{2}\right) \times \frac{1}{12} \times \frac{R}{100}$

$=200\left(\frac{36(36+1)}{2}\right) \times \frac{1}{12} \times \frac{11}{100}$

$=\frac{200 \times 36 \times 37 \times 11}{2 \times 12 \times 100}=1221$

∴Kiran will get maturity value=7200+1210=8421


Question 5

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.

Sol :

Interest = Rs. 58800

Monthly deposit (P) = Rs. 600

Period (n)=4 years or 48 months

∴Deposit for 1 month $=\frac{P(n)(n+1)}{2}$

$=\frac{600 \times 48 \times 49}{2}=\mathrm{Rs} .705600$

Let, rate of interest =r % p.a.

Interest $=\frac{\operatorname{PRT}}{100}$

$5880=\frac{705600 \times r \times 1}{100 \times 12}$

5880=588r

∴$r=\frac{5880}{588}=10$

∴Rate of interest =10 % p.a.


Question 6

David opened a Recurring Deposit Account in a bank and deposited Rs. 300 per month for two years. If he received Rs. 7725 at the time of maturity, find the rate of interest per annum. (2008)

Sol :

Deposit during one month (P) = Rs. 300

Period = 2 years = 24 months.

Maturity value = Rs. 7725

Let R be the rate percent, then

Now principal for 1 month $=\frac{P \times n(n+1)}{2}$

$=\frac{300 \times 24(24+1)}{2}=\frac{300 \times 24 \times 25}{2}=$ Rs. 90000

∴Interest earned $=\frac{\mathrm{PRT}}{100}=\frac{90000 \times \mathrm{R} \times 1}{100 \times 12}$

=75R

Now 300×24+75R=7725

7200+75 R=7725

75 R=7725-7200=525

$\mathrm{R}=\frac{525}{75}=7$

∴Rate of Interest =7 % p.a.


Question 7

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.

Sol :
Deposit per month = Rs. 2500

Period = 2 years = 24 months

Maturity value = Rs. 67500

∴Total principal for 1 month $=\frac{\mathrm{P} \times n(n+1)}{2}$

$=₹ \frac{2500 \times 24 \times 25}{2}=₹ 750000$

∴Interest =₹ 67500-24×2500

=₹ 67500-60000=₹ 7500

Period $=1$ month $=\frac{1}{12}$ year

∴Rate of interest $=\frac{\mathrm{S.I.} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{7500 \times 100 \times 12}{750000 \times 1}$

=12%


Question 8

Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs 800 per month for $1 \frac{1}{2}$ yeas . If he received Rs 15084 at the time of maturity, find the rate of interest per annum.

Sol :

Money deposited by Shahrukh per month (P)= Rs 800

r = ?

No. of months $(n)=1 \frac{1}{2}=\frac{3}{2} \times 12$

=18 months

∴Interest $=\mathrm{P} \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$=₹ 800 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}$

$=₹ 800 \times \frac{18 \times 19}{2 \times 12} \times \frac{r}{100}=114 r$

∴Maturity amount =114 r+800×18

₹ 15084=114 r+₹ 14400

₹ 15084-₹ 14400=114 r

684=114 r

$r=\frac{684}{114}=6 \%$


Question 9

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. (2016)

Sol :

Interest = Rs 1200

Period (n) = 2 years = 24 months

Rate (r) = 6% p.a.

Let monthly deposit =₹ P p.m.

∴Interest $=\frac{\mathrm{P} \times n(n+1)}{2 \times 12} \times \frac{r}{100}$

$1200=\frac{P \times 24 \times 25}{24} \times \frac{6}{100}$

$1200=\frac{6}{4} \mathrm{P}$

∴$P=\frac{1200 \times 4}{6}=800$

∴Monthly deposit =₹ 800

and maturity value =P × n+ Interest

$=₹ 800 \times 24+₹ 1200$

$=₹ 19200+₹ 1200=₹ 20400$


Question 10

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 instalment.

Sol :

Let monthly instalment is Rs P

here n = 1 year = 12 months

n = 12

∴$\mathrm{M.V.}=\frac{n(n+1)}{2 \times 12} \times \frac{\mathrm{P} \times \mathrm{R}}{100}+\mathrm{P} . \mathrm{n}$

$₹ 6455=\frac{12(12+1)}{2 \times 12} \times \frac{P \times 14}{100}+P .12$

$₹ 6455=\frac{13 \times \mathrm{P} \times 7}{100}+\mathrm{P.} 12$

$₹ 6455=\frac{91 P+1200 P}{100}$

₹ 645500=1291 P

$P=\frac{645500}{1291}=₹ 500$


Question 11

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.

Sol :

Deposit per month = Rs 2000,

Rate of interest = 10%, Let period = n months

Then principal for one month

$=2000 \times \frac{n(n+1)}{2}=1000 n(n+1)$ and interest

$=\frac{1000 n(n+1) \times 10 \times 1}{100 \times 12}=\frac{100 n(n+1)}{12}$

∴$=2000 \times n+\frac{100 n(n+1)}{12}$

⇒24000 n+100n2+100 n=83100 \times 12

⇒240 n+n2+n=831 \times 12

⇒n2+241 n-9972=0

⇒n2+277 n-36 n-9972=0

⇒n(n+277)-36(n+277)=0

⇒(n+277)(n-36)=0

Either n+277=0, then n=-277, which is not possible.

or n-36=0, then x=36

∴Period=36 months or 3 years

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