ML Aggarwal Solutions for Class 10 Maths Chapter 2 Banking can be used by the students to understand the topics covered in an effective way. The solutions contain brief stepwise explanations, which are purely based on the latest syllabus of the ICSE board. To know more about these concepts, students can access ML Aggarwal Solutions for Class 10 Maths Banking PDF, from the links which are given below.
ML Aggarwal Solutions for Class 10 Maths Chapter 2 Banking has problems based on determining the maturity value, rate of interest, deposit, and important concepts, as per the prescribed textbook. By solving the problems on a regular basis, students can improve their conceptual knowledge, which is important from the exam perspective. The ML Aggarwal Solutions can be referred to as cross-check answers and understand the other ways of solving problems effortlessly.
ML Aggarwal Solutions for Class 10 Maths Chapter 2 Banking :-
Exercise 2
1. Mrs. Goswami deposits ₹ 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.
Solution:
It is given that
Amount deposited by Mrs. Goswami = ₹ 1000
Rate of interest = 8% p.a.
Period (x) = 3 years = 36 months
We know that
Total principal for one month = 1000 × [x (x + 1)]/ 2
Substituting the value of x
= 1000 × (36 × 37)/ 2
By further calculation
= ₹ 666000
Interest = PRT/ 100
Substituting the values
= (666000 × 8 × 1)/ (100 × 12)
So we get
= ₹ 4440
So the amount of maturity = P × x + SI
= 1000 × 36 + 4440
= 36000 + 4440
= ₹ 40440
2. Sonia had a recurring deposit account in a bank and deposited ₹ 600 per month for 2 ½ years. If the rate of interest was 10% p.a., find the maturity value of this account.
Solution:
Its given that,
Amount deposited by Sonia per month = ₹ 600
Rate of interest (r) = 10% p.a.
Period (n) = 2 ½ years = 30 months
The interest earned during this period is calculated using the formula:
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
Hence, the maturity value of Sonia’s account will be ₹ 20325.
3. Kiran deposited ₹ 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?
Solution:
It is given that
Amount deposited by Kiran = ₹ 200
Rate of interest = 11% p.a.
Period (x) = 36 months
So the amount deposited in 36 months = 200 × 36 = ₹ 7200
We know that
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
Substituting the values
= (133200 × 11 × 1)/ (100 × 12)
So we get
= ₹ 1221
So the amount of maturity = P × x + SI
= 7200 + 1221
= ₹ 8421
4. Haneef has a cumulative bank account and deposits ₹ 600 per month for a period of 4 years. If he gets ₹ 5590 as interest at the time of maturity, find the rate of interest per annum.
Solution:
Interest at the time of maturity = ₹ 5880
Amount deposited by Haneef = ₹ 600
Period (x) = 4 years = 48 months
We know that
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)
So we get
5880 = 588r
By further calculation
r = 5880/588 = 10
Hence, the rate of interest = 10% p.a.
5. David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7725 at the time of maturity, find the rate of interest per annum.
Solution:
It is given that
Amount deposited per month = ₹ 300
Period (x) = 2 years = 24 months
Amount received at the time of maturity = ₹ 7725
Consider R as the rate percent
We know that
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)
So we get
= 75R
So we get
300 × 24 + 75R = 7725
By further calculation
7200 + 75R = 7725
75R = 7725 – 7200 = 525
R = 525/75 = 7
Hence, the rate of interest is 7% p.a.
6. Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2500 per month for two years. At the time of maturity he got ₹ 67500. Find:
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.
Solution:
It is given that
Amount deposited by Mr. Gupta per month = ₹ 2500
Period (x) = 2 years = 24 months
Amount got at the time of maturity = ₹ 67500
We know that
Total principal for one month = 2500 × [x (x + 1)]/ 2
Substituting the value of x
= 2500 × (24 × 25)/ 2
By further calculation
= ₹ 750000
Interest = Maturity value – x × deposit per month
Substituting the values
= 67500 – 24 × 2500
= 67500 – 60000
= ₹ 7500
We know that
Period = 1 month = 1/12 year
So the rate of interest = (SI × 100)/ (P × T)
Substituting the values
= (7500 × 100 × 12)/ (750000 × 1)
= 12%
7. Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 1 ½ years. If he received ₹ 15084 at the time of maturity, find the rate of interest per annum.
Solution:
Amount deposited by Shahrukh per month = ₹ 800
We know that
No. of months (n) = 1 ½ = 3/2 × 12 = 18 months
We know that
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
Substituting the values
= (136800 × r × 1)/ (100 × 12)
So we get
= 114r
So the amount of maturity = P × x + SI
15084 = 800 × 18 + 114r
By further calculation
114r = 15084 – 14400
114r = 684
r = 684/114 = 6%
Hence, the rate of interest per annum is 6%.
8. Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives ₹ 441 as interest at the time of maturity. Find the amount Rekha deposited each month.
Solution:
Here,
The number of months Rekha deposited (n) = 20
Rate of interest per annum (r) = 9%
Let the amount deposited by Rekha each month be ₹ x, then P = ₹ x
Now, we know that
I = P × [n(n + 1)/ (2 × 12)] × r/100
= x × [20(20 + 1)/ (2 × 12)] × 20/100
= x × [20(21)/ (2 × 12)] × 1/5
= x × [(5 × 21)/6)] × 1/5
According to given,
441 = x × [(5 × 21)/6)] × 1/5
2205 = x × [(5 × 21)/6)]
x = (2205 × 6)/ (5 × 21)
x = 126
Hence, the amount deposited by Rekha each month is ₹ 126.
9. Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find
(i) the monthly installment.
(ii) the amount of maturity.
Solution:
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
We know that
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
10. Mr. R. K. Nair gets ₹ 6455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly installment.
Solution:
Consider ₹ P as the monthly installment
Period (x) = 1 year = 12 months
We know that
Total principal for one month = P × [x (x + 1)]/ 2
Substituting the value of x
= P × (12 × 13)/ 2
By further calculation
= 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
By further calculation
P = 6455/12.91 = ₹ 500
11. Samita has a recurring deposit account in a bank of ₹ 2000 per month at the rate of 10% p.a. If she gets ₹ 83100 at the time of maturity, find the total time for which the account was held.
Solution:
Amount deposited in the account per month = ₹ 2000
Rate of interest = 10%
Consider period = n months
We know that
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
Substituting the values
2000n + [100 n (n + 1)]/ 12 = 83100
By further calculation
24000n + 100n2 + 100n = 83100 × 12
Dividing by 100
240n + n2 + n = 831 × 12
n2 + 241n – 9972 = 0
We can write it as
n2 + 277n – 36n – 9972 = 0
n (n + 277) – 36 (n + 277) = 0
(n + 277) (n – 36) = 0
Here n + 277 = 0
So we get
n = – 277 which is not possible
Similarly
n – 36 = 0 where x = 36
So the period = 36 months or 3 years
Hence, the total time for which the account was held is 3 years.
Chapter Test
1. Mr. Dhruv deposits ₹ 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.
Solution:
It is given that
Amount deposited by Mr. Dhruv = ₹ 600
Rate of interest = 10% p.a.
Period (n) = 5 years = 60 months
We know that
Total principal for one month = 600 × n (n + 1)/ 2
Substituting the value of n
= 600 × (60 × 61)/ 2
So we get
= ₹ 1098000
Here Interest = PRT/ 100
Substituting the values
= (1098000 × 10 × 1)/ (100 × 12)
= ₹ 9150
So the amount of maturity = 600 × 60 + 9150
= 36000 + 9150
= ₹ 45150
2. Ankita started paying ₹ 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying ₹ 500 per month in a 2 ½ years recurring deposit. The bank paid 10% p.a. simple interest for both. At maturity who will get more money and by how much?
Solution:
Case 1 – Ankita
Amount deposited per month = ₹ 400
Period (n) = 3 years = 36 months
Rate of interest = 10%
We know that
Total principal for one month = 400 × n (n + 1)/ 2
Substituting the value of n
= 400 × (36 × 37)/ 2
So we get
= ₹ 266400
Here Interest = PRT/ 100
Substituting the values
= (266400 × 10 × 1)/ (100 × 12)
= ₹ 2220
So the amount of maturity = 400 × 36 + 2220
= 14400 + 2220
= ₹ 16620
Case 2 – Anshul
Amount deposited per month = ₹ 500
Period (n) = 2 ½ years = 30 months
Rate of interest = 10%
We know that
Total principal for one month = 500 × n (n + 1)/ 2
Substituting the value of n
= 500 × (30 × 31)/ 2
So we get
= ₹ 232500
Here Interest = PRT/ 100
Substituting the values
= (232500 × 10 × 1)/ (100 × 12)
= ₹ 1937.50
So the amount of maturity = 500 × 30 + 1937.50
= 15000 + 1937.50
= ₹ 16937.50
We know that at maturity Anshul will get more amount
So the difference = 16937.50 – 16620 = ₹ 317.50
3. Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and deposits ₹ 800 per month. If she gets ₹ 48200 at the time of maturity, find
(i) the rate of simple interest,
(ii) the total interest earned by Shilpa
Solution:
It is given that
Amount deposited per month (P) = ₹ 800
Amount of maturity = ₹ 48200
Period (n) = 4 years = 48 months
Consider R% p.a. as the rate of interest
We know that
Total principal for one month = 800 × n (n + 1)/ 2
Substituting the value of n
= 800 × (48 × 49)/ 2
So we get
= ₹ 940800
Here the total deposit = 800 × 48 = ₹ 38400
Amount of maturity = ₹ 48200
So the interest earned = 48200 – 38400 = ₹ 9800
(i) Rate of interest = (SI × 100)/ (P × T)
Substituting the values
= (9800 × 100 × 12)/ (940800 × 1)
= 12.5%
(ii) Total interest earned by Shilpa = ₹ 9800
4. Mr. Chaturvedi has a recurring deposit account in Grindlay’s Bank for 4 ½ years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly installment.
Solution:
Consider ₹ x as the each monthly installment
Rate of interest = 11%
Period (n) = 4 ½ years = 54 months
We know that
Total principal for one month = x × n (n + 1)/ 2
Substituting the value of n
= x × (54 × 55)/ 2
So we get
= 1485x
Here Interest = PRT/ 100
Substituting the values
= (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.
5. Rajiv Bhardwaj has a recurring deposit account in a bank of ₹ 600 per month. If the bank pays simple interest of 7% p.a. and he gets ₹ 15450 as maturity amount, find the total time for which the account was held.
Solution:
It is given that
Amount deposited per month (P) = ₹ 600
Rate of interest = 7% p.a.
Amount of maturity = ₹ 15450
Consider n months as the period
We know that
Total principal for one month = 600 × n (n + 1)/ 2
By further calculation
= 600 (n2 + n)/ 2
= 300 (n2 + n)
Here Interest = PRT/ 100
Substituting the values
= (300 (n2 + 1) × 7 × 1)/ (100 × 12)
= 7/4 (n2 + n)
Amount of maturity = 600n + 7/4 (n2 + n)
Substituting the values
600n + 7/4 (n2 + n) = 15450
By further calculation
2400 + 7n2 + 7n = 61800
7n2 + 2407n – 61800 = 0
We can write it as
7n2 – 168n + 2575n – 61800 = 0
7n (n – 24) + 2575 (n – 24) = 0
(n – 4) (7n + 2575) = 0
Here n – 24 = 0 where n = 24
Similarly
7n + 2575 = 0
Where 7n = -2575
n = -2575/7 which is not possible as it is negative
Period (n) = 24 months or 2 years
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