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