myhelper: Exercise 2

Exercise 2

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Question 1

Mr. Rajiv Anand has opened a recurring deposit account of Rs. 400 per month for 20 month in a bank. Find the amount he will get at the tie of maturity, if the rate of interest is 8.5% p. a., if the interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 400, n = 20 months, r = 8.5%.

Putting these values in SI formula, we get

SI $= \frac{\text{Pn(n + 1)r}}{2400}$

⇒ SI $= 400 \times 20 \times (20 + 1) \times \frac{ 8.5 }{ 2400}$

⇒ SI = 595

Thus, the maturity value is MV = Pn + SI = (400×20) + 595 = Rs. 8595

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Question 2

Mr. Savita Khosla deposits Rs 900 per month in a recurring account for 2 years. If she gets Rs. 1800 as interest at the time of maturity, find the rate of interest if the interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 900, n = 24 months, SI = Rs. 1800

Putting these values in SI formula, we get

SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒$1800 = 900 \times 24 \times (24 + 1) \times \frac{ r}{2400}$

⇒ r = 8

Thus, the required rate of interest is 8% p.a.

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Question 3

Mr. Brown deposit Rs. 1100 per month in a cumulative time deposit account in a bank for 16 months. If at the end of maturity he gets Rs. 19096, find the rate of interest if interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 1100, n = 16 months, MV = 19096,

So, SI = MV – Pn = 19096 – (1100×16) = Rs. 1496

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒ $1496 = 1100 \times 16 \times (16 + 1) \times \frac{r}{2400}$

⇒ r = 12

Thus, the required rate of interest is 12% p.a.

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Question 4

Sandhya has a recurring deposit account in Vijya Bank and deposits Rs 400 per month for 3 years. If she gets Rs 16176 on maturity, find the rate of interest given by the bank.

Sol :

From the question, we have P = Rs. 400, n = 36 months, MV = 16176,

So, SI = MV – Pn = 16176 – (400×36) = Rs. 1777

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒$ 1777 = 400 \times 36 \times (36 + 1) \times \frac{r}{2400}$

⇒ r = 8

Thus, the required rate of interest is 8% p.a.

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Question 5

A man deposits Rs 600 per month in a bank for 12 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits if the rate of interest is 8 % p.a. and interest is calculated at the end of every month?

Sol :

From the question, we have P = Rs. 600, n = 12 months, r = 8%.

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒ SI $= 600 \times 12 \times (12 + 1) \times \frac{8}{2400}$

⇒ SI = 312

Thus, the maturity value is MV = Pn + SI = (600×12) + 312 = Rs. 7512

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Question 6

Anil deposits Rs 300 per month in a recurring deposit account for 2 years. If the rate of interest is 10% per year, calculate the amount that Anil will receive at the end of 2 years, i.e at the time of maturity.

Sol :

From the question, we have P = Rs. 300, n = 24 months, r = 10%.

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒ SI $= 300 \times 24 \times (24 + 1) \times \frac{10}{2400}$

⇒ SI = 750

Thus, the maturity value is MV = Pn + SI = (300×24) + 750 = Rs. 7950

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Question 7

Sudhir opened a recurring deposit account with a bank for 1 ½ years. If the rate of interest is 10% and the bank pays Rs 1554 on maturity, find how much did Sudhir deposit per month?

Sol :

From the question, we have n = 18 months, r = 10%, MV = Rs. 1554

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒ SI $= P\times 18 \times (18 + 1) \times \frac{10 }{2400}$

⇒ SI $= \frac{57P}{40}$

Since the maturity value is MV = Pn + SI

⇒ $1554 = 18P + \frac{57P}{40}$

⇒ P = Rs. 80 per month.

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Question 8

Renu has a cumulative deposit account of Rs. 200 per month at 10% per annum. If she gets Rs 6775 at the time of maturity, find the total time for which the account was held.

Sol :

From the question, we have P = Rs. 200, r = 10%, MV = Rs. 6775

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒ SI $= 200 \times n(n + 1) \times \frac{10}{2400}$

⇒ SI $= \frac{\text{5n(n + 1)}}{6}$

Since the maturity value is MV = Pn + SI

⇒$ 6775 = 200n + \frac{5n(n + 1)}{6}$

⇒ n² + 241n – 8130 = 0

⇒ (n – 30)(n + 271) = 0

⇒ n = 30 months $= 2\frac{1}{2}$ years.

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

SChand CLASS 10 Chapter 2 Banking Exercise 2

Exercise 2

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ML Aggarwal Solution Class 10 Chapter 2 Banking Exercise 2

Exercise 2

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