Banking Class 10 RS Aggarwal Exe-2 Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2. Step by step solutions of Banking questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## Banking Class 10 RS Aggarwal Exe-2 Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2

Board | ICSE |

Publications | Goyal Brothers Prakashan |

Subject | Maths |

Class | 10th |

Chapter-2 | Banking |

Writer | RS Aggarwal |

Book Name | Foundation |

Topics | Solution of Banking Exercise Questions |

Edition | 2024-2025 |

**Page- 19,20**

### Banking Exercise Questions

Class 10 RS Aggarwal Exe-2 Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2

**Exercise-2**

**Que-1: Mrs. Goswami deposits Rs1000 per month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.**

**Solution- **P = 1000

T = 3 years

n = 12×T

= 12×3 = 36

R = 8%

M.V = Pn + [{P×n(n+1)xR}/{2×12×100}]

= (1000×36) + [{1000×36×37×8}/{24×100}]

= 3600 + 4440

= 40,440

Hence Matured value = **Rs.40440** **Ans**.

**Que-2: Inderjeet opened a cumulative time deposit account with Punjab National Bank. He deposited Rs360 per month for 2 years. If the rate of interest be 7% per annum, how much did he get at the time of maturity?**

**Solution- **P = Rs.360

R = 7% per annum.

Time = 2 years = 24 months = n

Maturity value = (P * n) + [P * {n(n+1)/2} * {R/(12 * 100)}]

Maturity value = (360 * 24) + [360 * (24 * 25/2) * {7/(12 * 100)}]

→ Maturity value = 8640 + 360 * 300 * {7/(12 * 100)}

→ Maturity value = 8640 + (360 * 7)/4

→ Maturity value = 8640 + 630

→ Maturity value = **Rs.9270**** ****Ans****.**

**Que-3: (i) Neema had a recurring deposit account in a bank and deposited Rs600 per month for 2*(1/2) years. If the rate of interest was 10% per annum, find the maturity value of this account.
**

**(ii) Sajal invested Rs600 per month for 2*(1/2) years in a recurring deposit scheme of Oriental Bank of Commerce. If the bank pays simple interest at 6*(2/3)% per annum, find the amount received by him on maturity.**

**Solution- **(i) Maturity value for the recurring deposits = Total Sum of Money deposited + Interest earned on it.

P = Amount deposited every month

n = number of months the deposits were made

r% = rate of interest

M.V = Pn + [{P×n(n+1)xR}/{2×12×100}]

Here, P = Rs.600, n = 30, r = 10%

Maturity Value = (600×30) + [{600×30(30+1)x10}/ (2×12×100)]

**= Rs.20325 Ans.**

(ii) Total Money Deposited = 600*30 months = Rs.18000

Interest = (600*30*31*20)/(3*2400)

= Rs.1550

Maturity value = Money Deposit + Interest

**= Rs.19550 Ans.**

**Que-4: Mr. Richard has a recurring deposit account in a bank for 3 years at 7.5% per annum simple interest. If he gets Rs8325 as interest at the time of maturity, find : (i) the monthly deposit (ii) the maturity value.**

**Solution- **(i) Let the deposit per month = Rs.P

Number of months (n) = 36

Rate of interest (r) = 7.5% p.a.

:. S.I = [𝑃×𝑛(𝑛+1)]/(2×12)×𝑟/100

⇒ 8325 = [𝑃×36×37]/(2×12)×7.5/100

⇒ 8325 = (𝑃×3×37)/2×7.5/100

⇒ 𝑃 = (8325×2×100)/(3×37×7.5)

**= 𝑅𝑠2000 Ans.**

(ii) Maturity value = P x n + S.I

= Rs. (2000 x 36 + 8325)

**= Rs. 80325 Ans.**

**Que-5: Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at 6% per annum and the monthly installment is Rs1000, find :(i) interest earned in 2 years (ii) matured value.**

**Solution-**Katrina deposits 1000 per month for 2 years

Total money deposited = 1000×24

Rate of interest = 6% p.a

Equivalent principal = [1000×24(24+1)]/2

(1000×24×25)/2

= Rs(1000×12×25)

(i) Interest = (P×R×T)/100

= (1000×12×25×6)/100×1/12 = **Rs1500 Ans.**

(ii) Matured value = 24000 + 1500

= Rs 25500 Ans.

**Que-6: Ahmed has a recurring deposit account in a bank. He deposits Rs2500 per month for 2 years. If he gets Rs66250 at the time of maturity find : (i) the interest paid by the bank (ii) the rate of interest.**

**Solution- **(i) P = Rs. 2500,

n = 2 years = (2 × 12) months= 24 months

Total Principal = Rs. 2,500 × 24 = Rs. 60,000,

Amount = Rs. 66,250

Interest = Amount – Principal

= Rs. 66,250 – Rs. 60,000 = Rs. 6,250

Thus, the interest paid by the bank is** Rs. 6,250 Ans.**

(ii) Let r be the rate of interest.

N = [𝑛(𝑛+1)]/(2×12)

= (24×25)/(2×12) = 25 year

This is equivalent to depositing Rs. 2,500 for 25 yrs.

Interest = (𝑃×𝑁×𝑅)/100

⇒ 6250 = (2500×25×𝑅)/100

⇒ 𝑅 = 10%

Thus, the rate of interest is **10% Ans.**

**Que-7: Mr. Gupta opened a recurring deposit account in a bank. ****He deposited Rs2500 per month for 2 years. At the time of maturity he got Rs67500. Find : (i) the total interest earned by Mr. Gupta**

(ii) the rate of interest per annum.

(ii) the rate of interest per annum.

**Solution- **(i) Monthly instalment = Rs. 2500

n = 24,

Amount deposited = 2500 x 24 = Rs. 60000

Maturity value = Rs. 67500

Interest on his deposit = Rs. (67500 – 60000) **= Rs. 7500 Ans.**

(ii) Now Interest = [n(n+1)]/2 × (Instalment×Rate)/(100×12)

7500 = (24×25)/2 × (2500×R)/(100×12)

R = (7500×100×24)/(24×25×2500) **= 12% p.a Ans.**

**Que-8: Mr. Thomas a 4 year cumulative time deposit account in Corporation Bank and deposits Rs650 per month. If he receives Rs36296 at the time of maturity, find : (i) the total interest earned by Mr. Thomas (ii) the rate of interest per annum.**

**Solution- **(i) P = Rs. 650,

n = 4 years = 48 months,

A = Rs. 36296

I = {Pn(n+1)}/(2×12) x r/100

= (650 × 48 × 49) × 8/2400

= Rs.5096

(ii) A = Pn + {Pn(n+1)}/(2×12) x r/100

36296 = (650 × 48) + (650 × 48 × 49) × r/2400

5096 = 637r

r = 8% p. a.

**Que-9: Tanvy has a recurring deposit account in a finance company for 1.5 years at 9% per annum. If she gets Rs15426 at the time of maturity, how much per month has been invested by her?**

**Solution- **n = 1.5 years = 18 months

r = 9%

MV = Rs15426

MV = {(𝑃×𝑛) + n(𝑛+1)}/(2×12) × 𝑟/100

15426 = {(𝑃×18) + (𝑃×9×18×19)/(2×12×100)

15426 = P (18 + 1.2825)

P = 15526/19.2825

**P = Rs800 Ans.**

**Que-10: Punam opened a recurring deposit account with Bank of Baroda for 1.5 years. If the rate of interest is 6% per annum and the bank pays Rs11313 on maturity, find how much Punam deposited each month ?**

**Solution- **n = 1.5 years = 18 month

r = 6% p.a.

MV = Rs11313

Let the principle be P

Interest = [P × n(n+1)]/2 × (r×t)/100

= [P×18(18+1)]/2 × (6/100) × (1/12)

= [P×18×19]/2 × 1/(2×100)

= (P×342)/400

I = 0.86P−−−−(i)

MV = Pn + I

11313 = Px18 + 0.86P

11313 = 18P + 0.86P

11313 = 18.86P

P = 11313/18.86

P = Rs600 Ans.

**Que-11: Kavita has a cumulative time deposit account in a bank. She deposits Rs800 per month and gets Rs6165 at the time of maturity. If the rate of interest be 6% per annum. Find the total time for which the account was held.**

**Solution- **Interest = [P × n(n+1)]/2 × r/100

SI = Principal amount – ( monthly deposit × number of months)

SI = 6165 – 600×n

6165 – 600×n = [600xn(n+1)x6]/(12x2x100)

6165 – 600×n = 3/2[n² + n]

6165 – 600×n = 1.5n² + 1.5n

1.5n² + 601.5n – 6165 = 0

the above quadratic equation,

n = -601.5 ± [-601.5 ± √{(601.5)² + 4 x 6165}]/(2 x 1.5)

n = 30/3

n = 10 months. Ans.

**Que-12: Kavita has a cumulative time deposit account in a bank. She deposits Rs800 per month and gets Rs16700 as maturity value. If the rate of interest be 5% per annum, find the total time for which the account has held.**

**Solution- **Let time be x months

SI = [400x(x+1)*5*1]/1200

= (1/3)x(x+1)*5

= (5/3)x(x+1)

Principal = 800x

Hence,

800x + (5/3)x(x+1) = 16700

2400x + 5x = 16700*3

x² + 481x – 10020 = 0

x² + 501x – 20x – 10020 = 0

x(x + 501) -20(x + 501)

x = 20 months Ans.

**Que-13: David opened a recurring deposit account in a bank and deposited Rs300 per month for two years. if he received Rs7725 at the time of maturity, find the rate of interest per annum.**

**Solution-**Amount deposited per month = Rs 300

No. of installments in 2 years = 24 months

∴ Equivalent Principal = Rs 300(24×252) = Rs 90,000.

Maturity value = (24×300) + (90000×r×1)/(100×12)

⇒ 7725 = 7200 + 75r

⇒r = (7725−7200)/72

= 525/75 = 7%

Hence, the rate of interest is 7% p.a.

**Que-14: Preeti has a recurring deposit account of Rs1000 per month at 10% per annum. If she gets Rs5550 as interest at the time of maturity, find the total time for which the account was held.**

**Solution- **Let the required time be n months.

Then, P=₹ 1000, R=10 % p.a.,

S I = ₹ 5550

SI = [P x n(n+1)]/2 x R/(12×100)

= 5550 = [1000 x n(n+1)]/2 x 10/(12×100)

= 555 = 12 x 5(n²+n)

= n² + n – 1332 = 0

= n² + 37n – 36n – 1332 = 0

= n(n+37) – 36(n+37) = 0

= (n + 37) (n – 36)

= n = -37,36 [negative value is not possible]

**n = 36 month or 3 years Ans.**

**Que15: Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha received Rs441 as interest at the maturity. Find the amount Rekha deposited each month.**

**Solution- **Let the monthly deposits be of = 𝑥

Here, 𝑛 = 20, 𝑅 = 9% p.a interest = 441

441 = 𝑥 × (20×21)/(2×12) × 9/100

[∵Interest = [𝑃×𝑛(𝑛+1)/(2×12)×𝑟/100]

⇒ 𝑥 = (441×100×24)/(20×21×9) = 280

Hence, the monthly deposits is 280.

**Que-16: Mr. Sonu has a recurring deposit account and deposits Rs750 per month for 2 years. If he gets Rs19125 at the time maturity, find the rate of interest.**

**Solution- **M.V.= 19125

P = 750

T = 24

We will now substitute the above values to compute the rate of interest using the above formula.

⇒ 19125 = [(750×24) + {750×24(24+1)/100}×R/100

⇒ 19125 = 18000 + 18750R/100

Subtracting the above equation by 18000 on both sides, we get

⇒ 19125−18000 = 18000 + (18750R/100) − 18000

⇒ 1125 = 18750R/100

Multiplying the above equation by 100/18750 on both sides, we get

⇒ (100/18750) (1125) = (100/18750) × (18750R/100)

⇒ 112500/18750 = R

**⇒ R = 6% Ans.**

**Que-17: Salman deposits Rs1000 every month in a recurring deposit account for 2 years. If he receives Rs26000 on maturity, find : (i) the total interest Salman earns (ii) the rate of interest.**

**Solution- **P = ₹ 1000 per month

n = 2 × 12 = 24 months

and Maturity value = ₹ 26000

(i) Total Interest = Maturity value – Deposited value

= ₹ 26000 – ₹ 1000 × 24

= ₹ 26000 – ₹ 24000

**= ₹ 2000 Ans.**

(ii) S.I = [𝑃×𝑛(𝑛 +1)]/2 × 1/12 × 𝑟/100

2000 = [1000×24×25]/2 × 1/12 × 𝑟/100

2000 = 250r

∴ r = 2000/250

**= 8% P.a. Ans.**

— : End of Banking Class 10 RS Aggarwal Exe-2 Goyal Brothers Prakashan ICSE Foundation Maths Solutions : —

Return to :– ** RS Aggarwal ICSE Class 10 Solutions Goyal Brothers**

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