Banking ICSE Maths Class 10 ML Aggarwal Solutions Ch-2. Step by step solutions of Banking Chapter 2 questions of Exercise 2. Method applying during solutions of this chapter as given in council prescribed itextbook for ICSE curriculum. althogh you can solve applying other method also. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

## Banking ICSE Maths Class 10 ML Aggarwal Solutions Ch-2

Board | ICSE |

Class | 10 |

Subject | Mathematics |

Book | ML Aggarwal |

Chapter-2 | Banking |

Topics | Solution of Exe-2 Questions |

Edition | 2024-2025 |

### Solution of Exe-2 Banking Questions

ICSE Maths Class 10 ML Aggarwal Solutions Ch-2

Note : Question has not shown only solutions are given hence keep your latest (2024-25) textbook of ML Aggarwal maths to understand the solutions more clearly.

**Que-1: **

**Sol:** Mrs. Goswami deposited by = ₹ 1000

Rate of interest = 8% p.a.

Period (x) = 3 years = 36 months

Total principal for one month = 1000 × [x (x + 1)]/ 2

Substituting the value of x

= 1000 × (36 × 37)/ 2

= ₹ 666000

Interest = PRT/ 100

Substituting the values

= (666000 × 8 × 1)/ (100 × 12)

= ₹ 4440

So the amount of maturity = P × x + SI

= 1000 × 36 + 4440

= 36000 + 4440

= ₹ 40440

**Que-2: **

**Sol:** Sonia deposited per month = ₹ 600

Rate of interest (r) = 10% p.a.

Period (n) = 2 ½ years = 30 months

I = P × [n(n + 1)/ (2 × 12)] × r/100

I = 600 × [30(30 + 1)/ (2 × 12)] × 10/100

= 600 × [(30 × 31)/ (2 × 12)] × 1/10

= 60 × [(15 × 31)/ 12]

= 5 × 15 × 31

I = ₹ 2325

Maturity value (MV) = P × n + I

MV = ₹ (600 × 30 + 2325)

= ₹ (18000 + 2325)

= ₹ 20325

**Que-3:**

**Sol:** Kiran deposited = ₹ 200 /month

Rate of interest = 11% p.a.

Period (x) = 36 months

So the amount deposited in 36 months = 200 × 36 = ₹ 7200

Total principal for one month = 200 × [x (x + 1)]/ 2

Substituting the value of x

= 200 × (36 × 37)/ 2

By further calculation

= ₹ 133200

Interest = PRT/ 100

= (133200 × 11 × 1)/ (100 × 12)

= ₹ 1221

So the amount of maturity = P × x + SI

= 7200 + 1221

= ₹ 8421

**Que-4:**

**Sol:** Haneef deposited = ₹ 600 / month

Interest earned at the time of maturity = ₹ 5880

Period (x) = 4 years = 48 months

Total principal for one month = 600 × [x (x + 1)]/ 2

Substituting the value of x

= 600 × (48 × 49)/ 2

By further calculation

= ₹ 705600

Consider r% p.a. as the rate of interest

Interest = PRT/ 100

Substituting the values

5880 = (705600 × r × 1)/ (100 × 12)

5880 = 588r

By further calculation

r = 5880/588 = 10 % p.a.

**Que-5:**

**Sol:** David deposited per month = ₹ 300

Period (x) = 2 years = 24 months

Amount received at the time of maturity = ₹ 7725

Consider R as the rate percent

Total principal for one month = 300 × [x (x + 1)]/ 2

Substituting the value of x

= 300 × (24 × 25)/ 2

By further calculation

= ₹ 90000

Interest = PRT/ 100

Substituting the values

= (90000 × R × 1)/ (100 × 12)

= 75R

So we get

300 × 24 + 75R = 7725

By further calculation

7200 + 75R = 7725

75R = 7725 – 7200 = 525

R = 525/75 = 7% p.a.

**Que-6:**

**Sol:** Mr. Gupta deposited per month = ₹ 2500

Period (x) = 2 years = 24 months

Amount got at the time of maturity = ₹ 67500

Total principal for one month = 2500 × [x (x + 1)]/ 2

Substituting the value of x

= 2500 × (24 × 25)/ 2

= ₹ 750000

Interest = Maturity value – x × deposit per month

= 67500 – 24 × 2500

= 67500 – 60000

= ₹ 7500

Period = 1 month = 1/12 year

So the rate of interest = (SI × 100)/ (P × T)

= (7500 × 100 × 12)/ (750000 × 1)

= 12%

**Que-7:**

**Sol:** Shahrukh deposited per month = ₹ 800

No. of months (n) = 1 ½ = 3/2 × 12 = 18 months

Total principal for one month = 800 × [x (x + 1)]/ 2

Substituting the value of x

= 800 × (18 × 19)/ 2

By further calculation

= ₹ 136800

Interest = PRT/ 100

= (136800 × r × 1)/ (100 × 12)

So we get

= 114r

So the amount of maturity = P × x + SI

15084 = 800 × 18 + 114r

114r = 15084 – 14400

114r = 684

r = 684/114 = 6%

**Que-8:**

**Sol:** update soon

**Que-9:**

**Sol:** Rekha deposited for time (n) = 20 month

Rate of interest per annum (r) = 9%

Let the amount deposited by Rekha each month be ₹ x, then P = ₹ x

I = P × [n(n + 1)/ (2 × 12)] × r/100

= x × [20(20 + 1)/ (2 × 12)] × 9/100

441 = x × [20(20 + 1)/ (2 × 12)] × 9/100

x = (441 × 100 × 24)/(20 × 21 × 9) = 280

**Que-10:**

**Sol:** Mohan gets Interest at the time of maturity = ₹ 1200

Period (x) = 2 years = 24 months

Rate of interest = 6% p.a.

Consider ₹ P p.m. as the monthly deposit

Interest = P × [x (x + 1)]/ (2 × 12) × r/100

Substituting the value of x

1200 = (P × 24 × 25)/ 24 × 6/100

By further calculation

1200 = 6/4P

By cross multiplication

P = (1200 × 4)/ 6 = 800

Here monthly deposit = ₹ 800

So the amount of maturity = P × x + SI

= 800 × 24 + 1200

= 19200 + 1200

= ₹ 20400

**Que-11:**

**Sol:** Mr R.K. Nair gets amonts = 6455

Letr ₹ P as the monthly instalment

Period (x) = 1 year = 12 months

Rate = 14 % pa

Total principal for one month = P × [x (x + 1)]/ 2

Substituting the value of x

= P × (12 × 13)/ 2

= 78P

Interest = PRT/ 100

Substituting the values

= (78P × 14 × 1)/ (100 × 12)

So we get

= 0.91P

So the amount of maturity = P × x + SI

6455 = P × 12 + 0.91P

6455 = 12.91P

P = 6455/12.91 = ₹ 500

**Que-12:**

**Sol:** Samita deposited in the account per month = ₹ 2000

Rate of interest = 10%

Consider period = n months

Principal for one month = 2000 × n (n + 1)/ 2 = 1000 n (n + 1)

Interest = [1000n (n + 1) × 10 × 1]/ [100 × 12]

= [100 n (n + 1)]/ 12

So the maturity value = 2000 × n + [100 n (n + 1)]/ 12

2000n + [100 n (n + 1)]/ 12 = 83100

24000n + 100n^{2} + 100n = 83100 × 12

Dividing by 100

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

Here n + 277 = 0

n = – 277, impossible

n – 36 = 0 where x = 36

So the period = 36 months or 3 years

— : End of Banking ICSE Maths Class 10 ML Aggarwal Solutions of Ch-2 questions :–

Return to :- ML Aggarwal Solutions for ICSE Class-10

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