ML Aggarwal Solutions for Chapter 2 Banking Class 10 Maths ICSE
Answer
Deposit per month = Rs 350,
Rate of interest = 8% p.a.
Period (x) = 1 year = 12 months
∴ Total principal for one month 350 = x(x + 1)
= Rs 350 × (12 × 13)/12
= Rs 350 × 78
= Rs 27300
∴ Interest = prt/100 = (27300 × 8 × 1)/(100 × 12)
= Rs 182
∴ Amount of Maturity = Rs 350 × 12 + Rs 182
= Rs 4200 + 182
= Rs 4382.
2. Saloni deposited Rs 150 per month ina bank for 8 months under the Recurring Deposit Scheme. ‘What will be the maturity value of his deposit if the rate of interest is 8% per annum?
Answer
Deposit per month = Rs 150
Rate of interest = 8% per
Period (x) = 8 month
∴ Total principal for one month = Rs 150 × (x + 1)/2
= Rs 150 × 8(8 +1)/2
= (150 × 8 × 9)
= Rs 5400
∴ Interest = prt/100 = (5400 × 8 × 1)/(100 × 12) = Rs 36
Amount of maturity = Rs 150 × 8 + Rs 36
= Rs 1200 + Rs 36
= Rs 1236
3. Mrs Goswami deposits Rs 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.
Answer
Deposit per month (P) = Rs 1000
Period = 3 years = 36 months
Rate = 8%
Total Principal = 36(36 + 1)/2 × 1000
Interest = PRT/100 = (36 × 37 × 1000 × 8)/(2 × 12 × 100)
= 12 × 37 × 10
= 4440
Matured price = P × n × S.I.
= 1000 × 36 + 4440
= 36000 + 4440
= Rs 40440
4. Sonia had a deposit account in a bank and deposited Rs600 per month for 21/2 years. If the rate of interest was 10% p.a., find the maturity value of this account.
Answer
Principal (n) = 2.5 years = 2.5×12 month = 30 months
Principal (P) = Rs 600
Rate = 10% p.a.
Maturity Value = ….
S.I. = P×n(n+1)×r×1(2×100×12)
S.I. = 600×30×31/2×100×12
Simple Interest = Rs 2335
Now Maturity Value = P×n + S.I.
Maturity Value = 600×30 + 2325
= Rs 20325
5. Kiran deposited Rs 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?
Answer
Amount deposited month (P) = Rs 200
Period (n) = 36 months,
Rate (R) = 11% p.a.
Now amount deposited in 36 months = Rs 200 × 36
= Rs 7200
Simple interest (S.I.) = P(n(n +1)/2) × 1/12 × R/100
= 200(36(36+1))/2 × 1/12 × 11/110
= (200 × 36 × 37 × 11)/(2 × 12 × 100)
= 1221
∴ Kiran will get maturity value
= Rs 7200 + 1221
= Rs 8421
6. Haneef has a cumulative bank account and deposits Rs 600 per month for a period of 4 years. If he gets Rs. 5880 as interest at the time of maturity, find the rate of interest.
Answer:
Interest = Rs 58800
Monthly deposit (P) = Rs 600
Period (n) = 4 years or 48 months
∴ Deposit for 1 month = {P(n)(n+1)}/2
= (600 × 48 × 49)/2
= Rs. 705600
Let, rate of interest = r% p.a.
Interest = Prt/100
⇒ 58800 = (705600 × r × 1)/(100 × 12)
5880 = 588r
∴ r = 5880/588 = 10
∴ Rate of interest = 10% p.a.
7. David opened a Recurring Deposit Account in a bank and deposited Rs 300 per month for two years. If he received Rs. 7725 at the time of maturity, find the rate of interest per annum.
Answer
Deposit during one month (P) = Rs 300
Period = 2 years = 24 months
Maturity value= Rs 7725
Let R be the rate percent, then
Now principal for 1 month = P ×n(n +1)/2
= (300 × 24)(24 +1)/2 = (300 × 24 × 25)/2 = Rs 90000
∴ Interest earned = PRT/100 = (9000 × R × 1)/(100 × 12)
= 75R
Now 300 × 24 + 75R = 7725
⇒ 7200 + 75R = 7725
⇒ 75R = 7725 – 7200 = 525
⇒ R = 525/75 = 7
∴ Rate of Interest = 7% p.a.
8. Mr. Gupta-opened a recurring deposit account in a bank . He deposited Rs 2500 per month for two years. At the time of maturity he got Rs. 67500. Find:
(i) the total interest earned by Mr. Gupta
(ii) the rate of interest per annum.
Answer
Deposit per month = Rs 2500
Period = 2 years = 24 months
Maturity value = Rs 67500
∴ Total principal for 1 month = (P × n(n +1))/2
= ₹ (2500 × 24 × 25)/2
= ₹ 750000
∴ Interest = ₹ 67500 – 24 × 2500
= ₹ 67500 – 60000
= ₹ 7500
Period = 1 month = 1/12 year
∴ Rate of interest = S.I. × 100)/(7500 × 100 × 12)/(75000 × 1)
= 12%
9. Shahrukh opened a Recurring Deposit Account in a Bank and deposited Rs 800 per month for 1.1/2 years. If he received Rs 15084 at the time of maturity, find the rate of interest per annum.
Answer
Money deposited by Shahrukh per month (P) = Rs 800
r = ?
No. of months (n) = 1.1/2 = 3/2 × 12
= 18 months
∴ Interest = P × n(n +1)/(2 × 12) × r/100
= ₹ 800 × 18(18 + 1)/(2 × 12) × r/100
= ₹ 800 × (18 × 19)/(2 × 12) × r/100
= 114r
∴ Maturity amount = 114r + 800 × 18
= ₹ 15084 = 114r + ₹ 14400
⇒ ₹ 15084 - ₹ 14400 = 114r
⇒ 684 = 114r
r = 684/114 = 6%
10. Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs 1200 as interest at the time of maturity, find:
(i) the monthly instalment
(ii) the amount of maturity
Answer
Interest = Rs 1200
Period (n) = 2 years = 24 months
Let monthly deposit = ₹ P p.m.
∴ Interest = P × n(n +1)/(2 × 12) × r/100
1200 = (P × 24 × 25)/24 × 6/100
⇒ 1200 = 6/4.P
∴ P = (1200 × 4)/6 = 800
∴ Monthly deposit = ₹ 800
And maturity value = P × n + Interest
= ₹ 800 × 24 + ₹ 1200
= ₹ 19200 + ₹ 1200
= ₹ 20400
11. Mr. R.K. Nair gets Rs 6,455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly instalment.
Answer
Let monthly instalment is Rs P
Here n = 1 year = 12 months
n = 12
∵ M.V. = n(n +1)/(2 × 12) × (P × R)/100 + P.n
⇒ ₹ 6455 = 12(12 + 1)/(2 × 12) × (P × 14)/100 + P.12
₹ 6455 = (13 × P × 7)/100 + P.12
⇒ ₹ 6455 = (91P + 1200P)/100
⇒ ₹ 645500 = 1291 P
⇒ P = 645500/1291 = ₹500
12. Samita has a recurring deposit account in a bank of Rs2000 per month at the rate of 10% p.a. If she gets Rs 83100 at the time of maturity. Find the total time for which the account was held.
Answer
Deposit per month = Rs 2000,
Rate of interest = 10%,
Let period = n months
Then principal for one month = 2000 × n(n + 1)/2 = 1000n(n + 1) and interest
= {1000n(n + 1)×10×1}/(100 × 12)
= 100n(n + 1)/12
∴ Maturity value = 2000 × n + 100n(n + 1)/12
∴ 2000n + 100n(n + 1)/12 = 83100
⇒ 24000n + 100n2 + 100n = 83100 × 2
⇒ 240n + n2 + n = 831 × 12
⇒ n2 + 241n – 9972 = 0
⇒ n2 + 277n - 36n – 9972 = 0
⇒ n(n + 277) – 36(n + 277) = 0
⇒ (n + 277)(n – 36) = 0
Either n + 277 = 0, then n = - 277, which is not possible.
Or n – 36 = 0, then x = 36
∴ Period = 36 months or 3 years
MCQs of Chapter 2 Banking
1. If Shahrukh opened a recurring account in a bank and deposited Rs 800 per month for 1.1/2 years, then the total money deposited in the account is
(a) Rs 11400
(b) Rs 14400
(c) Rs 13680
(d) none of these
Answer
(b) Rs 14400
Monthly deposit = Rs 800
Period (n) = 1.1/2 years = 18 months
∴ Total money deposit = Rs 800 × 18
= Rs 14400
2. Mrs. Mehta deposit Rs 250 per month for one year in a bank’s recurring deposit account. If the rate of (simple) interest is 8% per annum, then the interest earned by her on this account is
(a) Rs 65
(b) Rs 120
(c) Rs 130
(d) Rs 260
Answer
(c) Rs 130
Deposit per month (P) = Rs 250
Period (n) = 1 year = 12 months
Rate (r) = 8% p.a.
∴ Interest = (P × n × (n + 1))/(2 × 12) × r/100
= (250 × 12 + 13)/(2 × 12) × 8/100
= ₹ 130
3. Mr. Sharma deposited Rs 500 every month in a cumulative deposit account for 2 years. If the bank pays interest at the rate of 7%per annum, then the amount he gets on maturity is
(a) Rs 875
(b) Rs 6875
(c) Rs 10875
(d) Rs 12875
Answer
(d) Rs 12875
Deposit (P) = Rs 500 per month
Period (n) = 2 years = 24 months
Rate (r) = 7% p.a.
∴ Interest = P × n × (n + 1)/(2 × 12) × r/100
= (500 × 24 × 25 × 7)/(2 × 12 × 100)
= ₹ 875
∴ Maturity value = P × 24 + Interest
= ₹ 500 × 24 + 875
= ₹ 12000 × 875
= ₹ 12875
4. John deposited Rs 400 every month in a bank’s recurring deposit for 2.1/2 years. If he gets Rs 1085 as interest at the time of maturity, then the rate of interest per annum is
(a) 6%
(b) 7%
(c) 8%
(d) 9%
Answer
(b) 7%
Deposit (P) = Rs 400 per month
Period (n) = 2.1/2 years = 3 months
Interest = Rs 1085
Let r% be the rate of interest
∴ Interest = P × n × (n + 1)/(2 × 12) × r/100
1085 = ₹ (400 × 30 × 31 × r)/(2 × 12 × 100)
1085 = 155r ⇒ r = 1085/155 = 7
∴ Rate 7% p.a.
Chapter Test for Banking
1. Mr. Dhruv deposits Rs 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.
Answer
Deposit per month = Rs 600
Rate of interest= 10% p.a.
Period (n) = 5 years 60 months
Total principal for one month = ₹ 600 × n(n + 1)/2
= 600 × 60(60 + 1)/2
= ₹ (600 × 60 × 61)/2
= ₹ 109800
= ₹ (600 × 60 × 61)/2
= ₹ 1098000
Interest = prt/100
= (1098000 × 10 × 1)/(100 × 12)
= ₹ 9150
∴ Amount of maturity = ₹ 600 × 60 + ₹ 9150
= ₹ 36000 + ₹ 9150
= ₹ 45150
2. Ankita started paying Rs 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs 500 per month in a 2.1/2 years recurring deposit. The Bank paid 10% p.a. simple interest for both. At maturity who will get more and by how much?
Answer:
In case of Ankita,
Deposit per month = Rs 400
Period (n) = 3 years= 36 months
Rate of Interest = 10%
Total principal for one month = 400 × n(n + 1)/2
= 400 × 36(36 + 1)/2
= ₹ (400 × 36 × 37)/2
= ₹ 266400
Interest = prt/100
= (266400 × 10 × 1)/(100 × 12)
= ₹ 2220
∴ Amount of maturity = ₹ 400 × 36 + ₹ 2220
= ₹ 14400 + ₹ 2220
= ₹ 16620
In case of Anshul,
Deposit p.m. = ₹ 500
Rate of Interest = 10%
Period (n) = 2.1/2 years= 30months
∴ Total principal for one month = ₹ 500 × n(n + 1)/2
= 500 × 30(30 + 1)/2
= ₹ (500 × 30 × 31)/2
= ₹ 232500
Interest = (232500 × 10 × 1)/(100 × 12)
= ₹ 1937.50
Amount of maturity = ₹ 500 × 30 + ₹1937.50
= ₹ 15000 + ₹ 1937.50
= ₹ 16937.50
At maturity Anshul will get more amount
Difference = ₹ 16937.50 - ₹ 16620.00
= ₹ 317.50
3. Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and Deposits Rs 800 per month. If she gets Rs 48200 at the time of maturity, find
(i) the rate of simple interest,
(ii) the total interest earned by Shilpa
Answer
Deposit per month (P) = Rs 800
Amount of maturity = Rs 48200
Period (n) = 4 years= 48 months
Let rate of interest be R% p.a.
Total principal for one month = P(n)(n + 1)/2
= {(800 × 48 × (48 × 1)}/2
= ₹ (800 × 48 × 49)/2
= ₹ 940800
Total deposit = ₹ 800 × 48
= ₹ 38400
And amount of maturity = ₹ 48200
∴ Interest earned = ₹ 48200 - ₹ 38400
= ₹ 9800
4. Mr. Chaturvedi has a recurring deposit account in Gindlay’s Bank for 4.1/2 years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly instalment.
Answer
Let each monthly instalment = Rs x
Rate of interest = 11%
Period (n) = 4.1/2 years or 54 months,
Total principal for one month = ₹ x × n(n + 1)/2
= ₹ x × 54(54 + 1)/2
= ₹ x (54 × 55)/2
= 1485x
Interest = (1485x 11 × 1)/(100 × 12)
= 13.6125x
∴ Total amount of maturity = 54x + 13.6125x
= 67.6125x
∴ 67.6125x = 101418.75
x = 101418.75/67.6125
= ₹ 1500
∴ Deposit per month = ₹ 1500
5. Rajiv Bhardwaj has a recurring deposit account in a bank of Rs 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs 15450 as maturity amount, find the total time for which the account was held.
Answer
Deposit during the month (P) = Rs 600
Rate of interest = 7% p.a.
Amount of maturity = Rs 15450
Let time = n months
∴ Total principal = P(n)(n + 1)/2
= 600 × n(n + 1)/2
= 600(n2 + n)/2
= 300(n2 + n)
∴ Interest = PRT/100
= (300(n2 + n) × 7 × 1)/(100 × 12)
= 7/4(n2 + n)
∴ 600n + 7/4(n2+ n) = 15450
⇒ 2400n + 7n2 + 7n = 61800
⇒ 7n2 + 2407n – 61800 = 0
⇒ 7n2 – 168n + 2575n – 61800 = 0
⇒ 7n(n – 24) + 2575(n – 24) = 0
⇒ (n – 24)(7n + 2575) = 0
Either n – 24 = 0, then n = 24
Or 7n + 2575 = 0, then
7n = -2575
⇒ n = -2575/7
Which is not possible being negative.
∴ n = 24
∴ Period = 24 months or 2 years
The solutions provided for Chapter 2 Banking of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 2 Banking contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.
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