Practicing OP Malhotra Class 10 Solutions Chapter 2 Banking Ex 2 is the ultimate need for students who intend to score good marks in examinations.

## S Chand Class 10 ICSE Maths Solutions Chapter 2 Banking Ex 2

Question 1.

Mr. Rajiv Anand has opened a recurring deposit account of 7400 per month for 20 months in a bank. Find the amount he will get at the time of maturity, if the rate of interest is 8.5% p.a., and if the interest is calculated at the end of each month.

Solution:

Deposit per month (P) = ₹ 400

Period (n) = 20 months

Rate (R) = 8.5% p.a.

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

= \(\frac{400 \times 20 \times(20+1)}{2}\)

= ₹ \(\frac{400 \times 20 \times 21}{2}\) = ₹ 84000

∴ Interest = \(\frac {PRT}{100}\) = \(\frac{84000 \times 8.5 \times 1}{100 \times 12}\)

= \(\frac {1}{2}\) = ₹ 595

∴ Maturity value = P x 20 + Interest

= ₹ 400 x 20 + 595

= ₹ 8000 + 595

= ₹ 8595

Question 2.

Mrs. Savita Khosla deposits 7900 per month in a recurring account for 2 years. If she gets 71800 as interest at the time of maturity, find the rate of interest if the interest is calculated at the end of each month.

Solution:

Monthly deposit (P) = ₹ 900

Let Rate = R%

Period (n) = 2 years = 24 months

Interest = ₹ 1800

Now principal for 1 month

= \(P \times \frac{n(n+1)}{2}=₹ 900 \times \frac{24 \times(24+1)}{2}\)

= ₹ \(\frac{900 \times 24 \times 25}{2}\) = ₹ 270000

Interest = ₹ 1800

∴ Rate = \(\frac{\text { Interest } \times 100}{\mathrm{P} \times t}=\frac{1800 \times 100 \times 12}{270000 \times 1}\)

= 2 x 4 = 8% p.a.

Question 3.

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

Solution:

Deposit per month (P) = ₹ 1100

Period (n) = 16 months

Maturity value = ₹ 19096

Let rate = R%

Interest = Maturity value – Principal

= ₹ 19096 – 1100 x 16

= ₹ 19096 – 17600

= ₹ 1496

Principal for one month

= P x \(\frac{n(n+1)}{2}\) = ₹ 1100 x \(\frac{16 \times(16+1)}{2}\)

= ₹ \(\frac{1100 \times 16 \times 17}{2}\) = ₹ 149600

∴ Rate = \(\frac{\text { Interest } \times 100}{P \times t}=\frac{1496 \times 100 \times 12}{149600 \times 1}\)

= 12% p.a.

Question 4.

Sandhya has a recurring deposit account in Vijaya Bank and deposits ₹ 400 per month for 3 years. If she gets ₹ 16,176 on maturity, find the rate of interest given by the bank.

Solution:

Monthly deposit (P) = ₹ 400

Period (n) = 3 years or 36 months

∴ Maturity value = ₹ 16176

Let rate = R% p.a.

Principal for 1 month = P x \(\frac{n(n+1)}{2}\)

= 400 x \(\frac{36 \times(36+1)}{2}\)

= ₹ \(\frac{400 \times 36 \times 37}{2}\) = ₹ 266400

Interest = M.V. – Deposit

= 16176 – 400 x 36

= 16176 – 14400 = ₹ 1776

∴ Rate = \(\frac{\text { Interest } \times 100}{\mathrm{P} \times t}=\frac{1776 \times 100 \times 12}{266400 \times 1}\)

= 8% p.a.

Question 5.

A man deposits ₹ 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?

Solution:

Deposit per month (P) = ₹ 600

Period (n) = 12 months

Rate (R) = 8% p.a.

Principal for 1 month = P x \(\frac{n(n+1)}{2}\)

= ₹ 600 x \(\frac{12 \times(12+1)}{2}\)

= ₹ \(\frac{600 \times 12 \times 13}{2}\) = ₹ 46800

∴ Interest = \(\frac {PRT}{100}\) = \(\frac{46800 \times 8 \times 1}{100 \times 12}\)

∴Maturity value = P x n + Interest

= ₹ 600 x 12 + 312

= ₹ 7200 + 312

= ₹ 7512

Question 6.

Anil deposits ₹ 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.

Solution:

Deposit per month (P) = ₹ 300

Period (n) = 2 years = 24 months

Rate (R) = 10% p.a.

Principal for one month = P x \(\frac{n(n+1)}{2}\)

= ₹ \(\frac{300 \times 24 \times(24+1)}{2}\)

= ₹ \(\frac{300 \times 24 \times 25}{2}\) = ₹ 90000

∴ Interest = \(\frac {PRT}{100}\) = \(\frac{90000 \times 10 \times 1}{100 \times 12}\) = ₹ 750

∴Maturity value = P x n + Interest

= ₹ 300 x 24 + 750

= ₹ 7200 + 750

= ₹ 7950

Question 7.

Sudhir opened a recurring deposit account with a bank for 1\(\frac {1}{2}\) years. If the rate of interest is 10% and the bank pays ₹ 1554 on maturity, find how much did Sudhir deposit per month ?

Solution:

Maturity value = ₹ 1554

Rate (R) = 10%

Period (n) = 1\(\frac {1}{2}\) years = 18 months

Let P be the monthly deposit, then

principal for one month = \(\frac{\mathrm{P}(n)(n+1)}{2}\)

= \(\frac{P \times 18 \times(18+1)}{2}=\frac{P \times 18 \times 19}{2}\) = 171 P

Interest = \(\frac{\mathrm{PRT}}{100}=\frac{171 \mathrm{P} \times 10 \times 1}{100 \times 12}=\frac{171}{120}\)P

Now Maturity value = P x n + Interest

1554 = P x 18 + \(\frac {171}{120}\)P

⇒ 1554 = \(\frac{2160 \mathrm{P}+171 \mathrm{P}}{120}=\frac{2331}{120}\)P

⇒ p = \(\frac{1554 \times 120}{2331}\) = 80

∴ Monthly deposit = ₹ 80

Question 8.

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

Solution:

Deposit per month (P) = ₹ 200

Rate (R) = 10% p.a.

Maturity value = ₹ 6775

Let period = n months

∴ Principal for 1 month = P x \(\frac{n(n+1)}{2}\)

= \(\frac{n(n+1)}{2}\) x 200

= ₹ 100 n(n + 1)

Interest = \(\frac{\mathrm{PRT}}{100}=\frac{100 n(n+1) \times 10 \times 1}{100 \times 12}\)

= \(\frac{5n(n+1)}{6}\) … (i)

But Interest = M.V. – Deposit

= 6775 – 200 x n

= 6775 – 200n

From (i) and (ii)

\(\frac{5n(n+1)}{6}\) = 6775 – 200n

⇒ 5n(n+ 1) = 40650 – 1200n

⇒ 5n² + 5n + 1200n – 40650 = 0

⇒ 5n² + 1205n – 40650 = 0

⇒ n² + 241n – 8130 = 0

⇒ n² – 30n + 271n – 8130 = 0 {∵ \(\begin{aligned}

& 8130=-30 \times 271 \\

& 241=-30+270

\end{aligned}\)

= n(n- 30) + 271 (n – 30)

⇒ (n – 30) (n + 271) = 0

Either n – 30 = 0, then n = 30

or n + 271 = 0, n = – 271 which is not possible

∴ Period = n = 30 months

= 2\(\frac { 1 }{ 2 }\) years

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.

Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and the interest is calculated at the end of every month. (ICSE 2001)

Solution:

Deposit per month (P) = ₹ 150

Rate (R) = 8%

Period (n) = 8 months

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

= \(\frac{150 \times 8 \times(8+1)}{2}=₹ \frac{150 \times 8 \times 9}{2}\)

= ₹ 5400

Interest = \(\frac{\text { PRT }}{100}=₹ \frac{5400 \times 8 \times 1}{100 \times 12}\) = ₹ 36

∴ Maturity value = Deposit + Interest

= ₹ 150 x 8 + 36

= ₹ 1200 + 36 = ₹ 1236

Question 2.

Mr. R.K. Nair gets ₹ 6455 at the end of one year at the rate of 14% per annum in a Recurring Deposit Account. Find the monthly instalment. (ICSE 2005)

Solution:

Let deposit per month = ₹ x

Rate (R) = 14% p.a.

Period (n) = 1 year =12 months

Maturity value = ₹ 6455

Principal for 1 month = P x \(\frac{n(n+1)}{2}\)

= x × \(\frac{12 \times(12+1)}{2}=\frac{x+12 \times 13}{2}\) = 78 x

and interest = \(\frac{\mathrm{PRT}}{100}=\frac{78 x \times 14 \times 1}{100 \times 12}\)

= \(\frac { 91x }{ 100 }\) … (i)

Interest = Maturity value – Total deposit

= ₹ 6455 – x × 12 = 6455 – 12x … (ii)

From (i) and (ii)

\(\frac { 91x }{100}\) = 6455 – 12x

⇒ 91x = 645500 – 1200x

⇒ 91x + 1200x = 645500

⇒ 1291x = 645500

⇒ x = \(\frac { 645500 }{ 1291 }\) = 500

∴ Deposit per month = ₹ 500

Question 3.

Mohan deposits ₹ 80 per month in a cumulative deposit account for six years. Find the amount payable to him on maturity, if the rate of interest is 6% per annum.

Solution:

Deposit per month (P) = ₹ 80

Rate (R) = 6% p.a.

Period (n) = 6 years = 72 months

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

= \(\frac{80 \times 72 \times(72+1)}{2}\)

= ₹ \(\frac{80 \times 72 \times 73}{2}\) = ₹ 210240

Interest = \(\frac{P R T}{100}=\frac{210240 \times 6 \times 1}{100 \times 12}\)

= \(\frac { 105120 }{ 100 }\) = ₹ 1051.20

∴ Maturity value = Deposit + Interest

= 72 x 80+ 1051.20

= ₹ 5760 + 1051.20 = ₹ 6811.20

Question 4.

Saloni deposited ₹ 150 per month in her bank for eight months under the Recurring Deposit Scheme. What will the maturity value of her deposit, if the rate of interest is 8% per annum and the interest is calculated at the end of every month.

Solution:

Deposit per month = ₹ 150

Rate (R) = 8%

Period (n) = 8 months

∴ Principal for 1 month = \(\frac{P×n(n+1)}{2}\)

= ₹\(\frac{150 \times 8 \times(8+1)}{2}\)

= \(\frac{150 \times 8 \times 9}{2}\)

= ₹ 5400

∴ Interest = \(\frac{\mathrm{PRT}}{100}=\frac{5400 \times 8 \times 1}{100 \times 12}\) = ₹ 36

∴ Maturity value = P x n + Interest

= ₹ 150 x 8 + 36

= ₹ 1200 + 36

= ₹ 1236

Question 5.

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

Solution:

Deposit per month (P) = ₹ 300

Period (n) = 2 years or 24 months

Let R be the rate % per annum

Maturity value = ₹ 7725

Total principal for 1 month = \(\frac{P×n(n+1)}{2}\)

= \(\frac{300 \times 24 \times(24+1)}{2}\)

= \(\frac{300 \times 24 \times 25}{2}\) = ₹ 90000

∴ Interest = \(\frac{\text { PRT }}{100}=\frac{90000 \times R \times 1}{100 \times 12}\) = ₹ 75R … (i)

Now maturity value = P x n + Interest

7725 = 300 x 24 + 75R

⇒ 7725 = 7200 + 75R

⇒ 75R = 7725 – 7200 = 525

⇒ R = \(\frac { 525 }{ 75 }\) = 7

Rate of interest = 7% p.a.

Question 6.

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

Solution:

P = \(\frac{36(36+1)}{2}\) x 100

∴ Interest = \(\frac{36 \times 37 \times 1000 \times 8}{2 \times 12 \times 100}\)

= 12 x 37 x 10 = 4440

Matured value = 36000 + 4440 = ₹ 40440

Question 7.

Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2500 per month for two years. At the time of maturity he got ₹ 67,500. Find :

(i) the total interest earned by Mr. Gupta.

(ii) the rate of interest per annum.

Solution:

Total amount deposited by Mr. Gupta in 24 months = ₹ 2500 x 24 = ₹ 60000

Total principal for 1 month

= ₹ \(\frac{2500 \times 24 \times(24+1)}{2}\)

= ₹ 2500 x 12 x 25

= ₹ 750000

Let rate of interest be r% p.a.

Now interest on ₹ 750000 for 1 month

= 750000 x \(\frac { 1 }{ 12 }\) x \(\frac { r }{ 100 }\) = 625 r

Maturity amount = ₹ 67500

⇒ 60000 + 625r = 67500

625r = 67500 – 60000 = 7500

r = \(\frac { 7500 }{ 625 }\) = 12 % p.a.

Question 8.

Ahmed has a recurring deposit account in a bank. He deposits ₹ 2,500 per month for 2 years. If he get ₹ 66,250 at the time of maturity, find

(i) The interest paid by the bank.

(ii) The rate of interest.

Solution:

Monthly deposit (M.D.) = ₹ 2500

n = 2 x 12 = 24 months

Total deposited amount = ₹ 2500 x 24 = ₹ 60000

Matured amount = ₹ 66250

(i) Interest paid by the bank

= Rs. (66250 – 60000) = ₹ 6250

(ii) Equivalent principle for 1 month

= \(\frac{\text { M.D. } \times n(n+1)}{2}\)

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

R = \(\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{6250 \times 100 \times 12}{750000 \times 1}\) = 10 % p.a.

Question 9.

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

Solution:

P (Principal) = ₹ 200, n (Time) = 36 months,

R (Rate) = 11% p.a.

Amount deposited in 36 months

= ₹ 200 x 36 = ₹ 7200

S.I. = P x \(\frac{n(n+1)}{2} \times \frac{1}{12} \times \frac{\mathrm{R}}{100}\)

= ₹ \(\frac{200 \times 36 \times 37 \times 11}{2 \times 12 \times 100}\) = ₹ 1221

Amount Kiran will get on maturity

= ₹ (7200 + ₹ 1221) = ₹ 8421

Question 10.

Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is 8% per annum and Mr. Britto gets ₹ 8088 from the bank after 3 years, find the value of his monthly instalment.

Solution:

Let monthly installment = ₹ x

n = 3 x 12 months = 36 months

Rate = 8% p.a.

∴ I = P x \(\frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\)

= x × \(\frac{36 \times 37}{2 \times 12} \times \frac{8}{100}=\frac{444}{100}\) x

Maturity value = ₹ 8088

According to the question,

I + 36 × x = 8088

⇒ \(\frac { 444 }{ 100 }\) + 36x = 8088

4.44x + 36x = 8088

40.44x = 8088

x = \(\frac{8088}{40.44}=\frac{8080 \times 100}{4044}\) = ₹ 200

∴ Value of monthly installment = ₹ 200

Question 11.

Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 11/2 years. If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum.

Solution:

Money deposited per month (P) = ₹ 800 r = ?

No.of months (n) = 1 \(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 2 }\) x 12 = 18

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

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

= 800 x \(\frac{18 \times 19}{2 \times 12} \times \frac{r}{100}\) = 114r

∴ Maturity amount = 114r + 800 x 18

According to the question,

15084 = WAr + 14400

⇒ 15084 – 14400 = 114r

⇒ 684 = WAr

r = \(\frac { 684 }{ 114 }\) = 6%

Question 12.

Katrina opened a recurring deposit account with a National Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly instalment is ₹ 1,000, find the:

(i) interest earned in 2 years.

(ii) matured value.

Solution:

Given,

p = ₹ 1000

n = 2 years = 24 months

r = 6%

(i) Interest = p x \(\frac{n(n+1)}{2} \times \frac{r}{12 \times 100}\)

= 1000 x \(\frac{24(25+1)}{2}=\frac{6}{12 \times 100}\)

= ₹ 1000 x \(\frac{24(25)}{2} \times \frac{6}{12 \times 100}\)

Thus, the interest earned in 2 years is ₹ 1500

(ii) Sum deposited in two years = 24 x ₹ 1000 = ₹ 24,000

Maturity value = Total sum deposited in two years + Interest

= ₹ 24,000 + ₹ 1,500 = ₹ 25,500

Thus, the maturity value is ₹ 25,500

Question 13.

Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find :

(i) the monthly instalment

(ii) the amount of maturity

Solution:

(i) I = ₹ 1200,

n = 2 x 12 = 24 months,

r = 6%

I = P x \(\frac{n(n+1)}{2} \times \frac{r}{12 \times 100}\)

⇒ 1200 = \(\mathrm{P} \frac{24(24+1)}{2} \times \frac{1}{12 \times 100}\)

⇒ 1200 = \(P \times \frac{24(25)}{2} \times \frac{6}{12 \times 100}\)

⇒ 1200 = P x \(\frac { 3 }{ 2 }\)

⇒ P = \(\frac { 1200×2 }{ 3 }\)

⇒ P = ₹ 800

So the monthly instalment is ₹ 800

(ii) Total sum deposited = P x n = ₹ 800 x 24 = ₹ 19200

∴ The amount that Mohan will get at the time of maturity

= Total sum deposited + Interest on it

= ₹ 19200 + ₹ 1200 = ₹ 20400

Hence, the amount of maturity is ₹ 20400