# ML Aggarwal Solutions for Chapter 2 Banking Class 10 Maths ICSE

**1. Shweta deposits Rs 300 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

∴ Total principal for one month 350 = x(x + 1)

= Rs 350 × (12 × 13)/12

= Rs 350 × 78

= Rs 27300

∴ Interest = prt/100 = (27300 × 8 × 1)/(100 × 12)

= Rs 182

∴ Amount of Maturity = Rs 350 × 12 + Rs 182

= Rs 4200 + 182

= Rs 4382.

**2. Saloni deposited Rs 150 per month ina 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

∴ Total principal for one month = Rs 150 × (x + 1)/2

= Rs 150 × 8(8 +1)/2

= (150 × 8 × 9)

= Rs 5400

∴ Interest = prt/100 = (5400 × 8 × 1)/(100 × 12) = Rs 36

Amount of maturity = Rs 150 × 8 + Rs 36

= Rs 1200 + Rs 36

= Rs 1236

**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.**

**Answer**

Deposit per month (P) = Rs 1000

Period = 3 years = 36 months

Rate = 8%

Total Principal = 36(36 + 1)/2 × 1000

Interest = PRT/100 = (36 × 37 × 1000 × 8)/(2 × 12 × 100)

= 12 × 37 × 10

= 4440

Matured price = P × n × S.I.

= 1000 × 36 + 4440

= 36000 + 4440

= Rs 40440

**4. Sonia had a deposit account in a bank and deposited Rs600 per month for 21/2 years. If the rate of interest was 10% p.a., find the maturity value of this account.**

**Answer**

Principal (n) = 2.5 years = 2.5×12 month = 30 months

Principal (P) = Rs 600

Rate = 10% p.a.

Maturity Value = ….

S.I. = P×n(n+1)×r×1(2×100×12)

S.I. = 600×30×31/2×100×12

Simple Interest = Rs 2335

Now Maturity Value = P×n + S.I.

Maturity Value = 600×30 + 2325

= Rs 20325

**5. 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?**

**Answer**

Amount deposited month (P) = Rs 200

Period (n) = 36 months,

Rate (R) = 11% p.a.

Now amount deposited in 36 months = Rs 200 × 36

= Rs 7200

Simple interest (S.I.) = P(n(n +1)/2) × 1/12 × R/100

= 200(36(36+1))/2 × 1/12 × 11/110

= (200 × 36 × 37 × 11)/(2 × 12 × 100)

= 1221

∴ Kiran will get maturity value

= Rs 7200 + 1221

= Rs 8421

6. **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

∴ Deposit for 1 month = {P(n)(n+1)}/2

= (600 × 48 × 49)/2

= Rs. 705600

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

Interest = Prt/100

⇒ 58800 = (705600 × r × 1)/(100 × 12)

5880 = 588r

∴ r = 5880/588 = 10

∴ Rate of interest = 10% p.a.

**7. 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.**

**Answer**

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 = P ×n(n +1)/2

= (300 × 24)(24 +1)/2 = (300 × 24 × 25)/2 = Rs 90000

∴ Interest earned = PRT/100 = (9000 × R × 1)/(100 × 12)

= 75R

Now 300 × 24 + 75R = 7725

⇒ 7200 + 75R = 7725

⇒ 75R = 7725 – 7200 = 525

⇒ R = 525/75 = 7

∴ Rate of Interest = 7% p.a.

**8. 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

∴ Total principal for 1 month = (P × n(n +1))/2

= ₹ (2500 × 24 × 25)/2

= ₹ 750000

∴ Interest = ₹ 67500 – 24 × 2500

= ₹ 67500 – 60000

= ₹ 7500

Period = 1 month = 1/12 year

∴ Rate of interest = S.I. × 100)/(7500 × 100 × 12)/(75000 × 1)

= 12%

**9. Shahrukh opened a Recurring Deposit Account in a Bank and deposited Rs 800 per month for 1.1/2 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 = ?

No. of months (n) = 1.1/2 = 3/2 × 12

= 18 months

∴ Interest = P × n(n +1)/(2 × 12) × r/100

= ₹ 800 × 18(18 + 1)/(2 × 12) × r/100

= ₹ 800 × (18 × 19)/(2 × 12) × r/100

= 114r

∴ Maturity amount = 114r + 800 × 18

= ₹ 15084 = 114r + ₹ 14400

⇒ ₹ 15084 - ₹ 14400 = 114r

⇒ 684 = 114r

r = 684/114 = 6%

**10. 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

Let monthly deposit = ₹ P p.m.

∴ Interest = P × n(n +1)/(2 × 12) × r/100

1200 = (P × 24 × 25)/24 × 6/100

⇒ 1200 = 6/4.P

∴ P = (1200 × 4)/6 = 800

∴ Monthly deposit = ₹ 800

And maturity value = P × n + Interest

= ₹ 800 × 24 + ₹ 1200

= ₹ 19200 + ₹ 1200

= ₹ 20400

**11. 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.**

**Answer**

Let monthly instalment is Rs P

Here n = 1 year = 12 months

n = 12

∵ M.V. = n(n +1)/(2 × 12) × (P × R)/100 + P.n

⇒ ₹ 6455 = 12(12 + 1)/(2 × 12) × (P × 14)/100 + P.12

₹ 6455 = (13 × P × 7)/100 + P.12

⇒ ₹ 6455 = (91P + 1200P)/100

⇒ ₹ 645500 = 1291 P

⇒ P = 645500/1291 = ₹500

**12. Samita has a recurring deposit account in a bank of Rs2000 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.**

**Answer**

Deposit per month = Rs 2000,

Rate of interest = 10%,

Let period = n months

Then principal for one month = 2000 × n(n + 1)/2 = 1000n(n + 1) and interest

= {1000n(n + 1)×10×1}/(100 × 12)

= 100n(n + 1)/12

∴ Maturity value = 2000 × n + 100n(n + 1)/12

∴ 2000n + 100n(n + 1)/12 = 83100

⇒ 24000n + 100n^{2} + 100n = 83100 × 2

⇒ 240n + n^{2} + n = 831 × 12

⇒ n^{2} + 241n – 9972 = 0

⇒ n^{2} + 277n - 36n – 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

**MCQs of Chapter 2 Banking**

**1. If Shahrukh opened a recurring account in a bank and deposited Rs 800 per month for 1.1/2 years, then the total money deposited in the account is**

**(a) Rs 11400**

**(b) Rs 14400**

**(c) Rs 13680**

**(d) none of these**

**Answer**

(b) Rs 14400

Monthly deposit = Rs 800

Period (n) = 1.1/2 years = 18 months

∴ Total money deposit = Rs 800 × 18

= Rs 14400

**2. Mrs. Mehta deposit Rs 250 per month for one year in a bank’s recurring deposit account. If the rate of (simple) interest is 8% per annum, then the interest earned by her on this account is**

**(a) Rs 65**

**(b) Rs 120**

**(c) Rs 130**

**(d) Rs 260**

**Answer**

(c) Rs 130

Deposit per month (P) = Rs 250

Period (n) = 1 year = 12 months

Rate (r) = 8% p.a.

∴ Interest = (P × n × (n + 1))/(2 × 12) × r/100

= (250 × 12 + 13)/(2 × 12) × 8/100

= ₹ 130

**3. Mr. Sharma deposited Rs 500 every month in a cumulative deposit account for 2 years. If the bank pays interest at the rate of 7%per annum, then the amount he gets on maturity is**

**(a) Rs 875**

**(b) Rs 6875**

**(c) Rs 10875**

**(d) Rs 12875**

**Answer**

(d) Rs 12875

Deposit (P) = Rs 500 per month

Period (n) = 2 years = 24 months

Rate (r) = 7% p.a.

∴ Interest = P × n × (n + 1)/(2 × 12) × r/100

= (500 × 24 × 25 × 7)/(2 × 12 × 100)

= ₹ 875

∴ Maturity value = P × 24 + Interest

= ₹ 500 × 24 + 875

= ₹ 12000 × 875

= ₹ 12875

**4. John deposited Rs 400 every month in a bank’s recurring deposit for 2.1/2 years. If he gets Rs 1085 as interest at the time of maturity, then the rate of interest per annum is**

**(a) 6%**

**(b) 7%**

**(c) 8%**

**(d) 9%**

**Answer**

(b) 7%

Deposit (P) = Rs 400 per month

Period (n) = 2.1/2 years = 3 months

Interest = Rs 1085

Let r% be the rate of interest

∴ Interest = P × n × (n + 1)/(2 × 12) × r/100

1085 = ₹ (400 × 30 × 31 × r)/(2 × 12 × 100)

1085 = 155r ⇒ r = 1085/155 = 7

∴ Rate 7% p.a.

**Chapter Test for Banking**

**1. 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 = ₹ 600 × n(n + 1)/2

= 600 × 60(60 + 1)/2

= ₹ (600 × 60 × 61)/2

= ₹ 109800

= ₹ (600 × 60 × 61)/2

= ₹ 1098000

Interest = prt/100

= (1098000 × 10 × 1)/(100 × 12)

= ₹ 9150

∴ Amount of maturity = ₹ 600 × 60 + ₹ 9150

= ₹ 36000 + ₹ 9150

= ₹ 45150

**2. Ankita started paying Rs 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs 500 per month in a 2.1/2 years recurring deposit. The Bank paid 10% p.a. simple interest for both. At maturity who will get more and by how much?**

**Answer:**

In case of Ankita,

Deposit per month = Rs 400

Period (n) = 3 years= 36 months

Rate of Interest = 10%

Total principal for one month = 400 × n(n + 1)/2

= 400 × 36(36 + 1)/2

= ₹ (400 × 36 × 37)/2

= ₹ 266400

Interest = prt/100

= (266400 × 10 × 1)/(100 × 12)

= ₹ 2220

∴ Amount of maturity = ₹ 400 × 36 + ₹ 2220

= ₹ 14400 + ₹ 2220

= ₹ 16620

In case of Anshul,

Deposit p.m. = ₹ 500

Rate of Interest = 10%

Period (n) = 2.1/2 years= 30months

∴ Total principal for one month = ₹ 500 × n(n + 1)/2

= 500 × 30(30 + 1)/2

= ₹ (500 × 30 × 31)/2

= ₹ 232500

Interest = (232500 × 10 × 1)/(100 × 12)

= ₹ 1937.50

Amount of maturity = ₹ 500 × 30 + ₹1937.50

= ₹ 15000 + ₹ 1937.50

= ₹ 16937.50

At maturity Anshul will get more amount

Difference = ₹ 16937.50 - ₹ 16620.00

= ₹ 317.50

**3. Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and Deposits Rs 800 per month. If she gets Rs 48200 at the time of maturity, find**

**(i) the rate of simple interest,**

**(ii) the total interest earned by Shilpa**

**Answer**

Deposit per month (P) = Rs 800

Amount of maturity = Rs 48200

Period (n) = 4 years= 48 months

Let rate of interest be R% p.a.

Total principal for one month = P(n)(n + 1)/2

= {(800 × 48 × (48 × 1)}/2

= ₹ (800 × 48 × 49)/2

= ₹ 940800

Total deposit = ₹ 800 × 48

= ₹ 38400

And amount of maturity = ₹ 48200

∴ Interest earned = ₹ 48200 - ₹ 38400

= ₹ 9800

**4. Mr. Chaturvedi has a recurring deposit account in Gindlay’s Bank for 4.1/2 years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly instalment.**

**Answer**

Let each monthly instalment = Rs x

Rate of interest = 11%

Period (n) = 4.1/2 years or 54 months,

Total principal for one month = ₹ x × n(n + 1)/2

= ₹ x × 54(54 + 1)/2

= ₹ x (54 × 55)/2

= 1485x

Interest = (1485x 11 × 1)/(100 × 12)

= 13.6125x

∴ Total amount of maturity = 54x + 13.6125x

= 67.6125x

∴ 67.6125x = 101418.75

x = 101418.75/67.6125

= ₹ 1500

∴ Deposit per month = ₹ 1500

**5. Rajiv Bhardwaj has a recurring deposit account in a bank of Rs 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs 15450 as maturity amount, find the total time for which the account was held.**

**Answer**

Deposit during the month (P) = Rs 600

Rate of interest = 7% p.a.

Amount of maturity = Rs 15450

Let time = n months

∴ Total principal = P(n)(n + 1)/2

= 600 × n(n + 1)/2

= 600(n^{2 }+ n)/2

= 300(n^{2} + n)

∴ Interest = PRT/100

= (300(n^{2} + n) × 7 × 1)/(100 × 12)

= 7/4(n^{2} + n)

∴ 600n + 7/4(n^{2}+ n) = 15450

⇒ 2400n + 7n^{2} + 7n = 61800

⇒ 7n^{2 }+ 2407n – 61800 = 0

⇒ 7n^{2 }– 168n + 2575n – 61800 = 0

⇒ 7n(n – 24) + 2575(n – 24) = 0

⇒ (n – 24)(7n + 2575) = 0

Either n – 24 = 0, then n = 24

Or 7n + 2575 = 0, then

7n = -2575

⇒ n = -2575/7

Which is not possible being negative.

∴ n = 24

∴ Period = 24 months or 2 years

The solutions provided for Chapter 2 Banking of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 2 Banking contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.

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