# Banking Chapter Test ICSE Class 10 Maths ML Aggarwal Solutions

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

## Banking Chapter Test ICSE Class 10 Maths ML Aggarwal Solutions

 Board ICSE Class 10 Subject Mathematics Book ML Aggarwal Chapter-2 Banking Topics Solution of Ch-Test Questions Edition 2024-2025

### Solution of Ch-Test Questions

Banking Chapter Test ICSE Class 10 Maths ML Aggarwal Solutions

#### Que-1:

Sol:   Mr. Dhruv deposited  = ₹ 600 / month

Rate of interest = 10% p.a.

Period (n) = 5 years = 60 months

Total principal for one month = 600 × n (n + 1)/ 2

= 600 × (60 × 61)/ 2

= ₹ 1098000

Here Interest = PRT/ 100

= (1098000 × 10 × 1)/ (100 × 12)

= ₹ 9150

So the amount of maturity = 600 × 60 + 9150

= 36000 + 9150

= ₹ 45150

#### Que-2:

Sol:  Ankita deposited per month = ₹ 400

Period (n) = 3 years = 36 months

Rate of interest = 10%

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

Substituting the value of n

= 400 × (36 × 37)/ 2

= ₹ 266400

Here Interest = PRT/ 100

= (266400 × 10 × 1)/ (100 × 12)

= ₹ 2220

So the amount of maturity = 400 × 36 + 2220

= 14400 + 2220

= ₹ 16620

Anshul deposited per month = ₹ 500

Period (n) = 2 ½ years = 30 months

Rate of interest = 10%

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

= 500 × (30 × 31)/ 2

= ₹ 232500

Here Interest = PRT/ 100

= (232500 × 10 × 1)/ (100 × 12)

= ₹ 1937.50

So the amount at maturity = 500 × 30 + 1937.50

= 15000 + 1937.50

= ₹ 16937.50

at maturity, Anshul will get more amount

So the difference = 16937.50 – 16620 = ₹ 317.50

#### Que-3:

Sol:   Shilpa deposited per month (P) = ₹ 800

Amount of maturity = ₹ 48200

Period (n) = 4 years = 48 months

Let rate be  R% p.a.

Total principal for one month = 800 × n (n + 1)/ 2

= 800 × (48 × 49)/ 2

= ₹ 940800

Here the total deposit = 800 × 48 = ₹ 38400

Amount at maturity = ₹ 48200

So the interest earned = 48200 – 38400 = ₹ 9800

(i) Rate of interest = (SI × 100)/ (P × T)

= (9800 × 100 × 12)/ (940800 × 1)

= 12.5%

(ii) Total interest earned by Shilpa = ₹ 9800

#### Que-4:

Sol:  Let  Mr. Chaturvedi monthly instalment =  ₹ x

Rate of interest = 11%

Period (n) = 4 ½ years = 54 months

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

= x × (54 × 55)/ 2

= 1485x

Here Interest = PRT/ 100

= (1485x × 11 × 1)/ (100 × 12)

= 13.6125x

So the amount of maturity = 54x + 13.6125x

= 67.6125x

By equating the value

67.6125x = 101418.75

x = 101418.75/67.6125 = ₹ 1500

Hence, the deposit per month is ₹ 1500.

#### Que-5:

Sol:  Rajiv Bhardwaj deposited per month (P) = ₹ 600

Rate of interest = 7% p.a.

Amount of maturity = ₹ 15450

Let the period = n months

Total principal for one month = 600 × n (n + 1)/ 2

= 600 (n2 + n)/ 2

= 300 (n2 + n)

Here Interest = PRT/ 100

= (300 (n2 + 1) × 7 × 1)/ (100 × 12)

= 7/4 (n2 + n)

Amount of maturity = 600n + 7/4 (n2 + n)

600n + 7/4 (n2 + n) = 15450

2400 + 7n2 + 7n = 61800

7n2 + 2407n – 61800 = 0

7n2 – 168n + 2575n – 61800 = 0

7n(n – 24) + 2575 (n – 24) = 0

(n – 4)  (7n + 2575) = 0

applying zero product rule

either 7n + 2575 = 0

then  7n = -2575

so n = -2575/7, impossible (time never in negative)

or  n – 24 = 0 where n = 24

Period (n) = 24 months or 2 years

— : End of Banking Chapter Test ICSE Class 10 Maths ML Aggarwal solutions :–