ICSE Solutions for Chapter 2 Banking (Recurring Deposit Account) Class 10 Mathematics
Solution 1:
Installment per month(P) = ₹ 600
Number of months(n) = 20
Rate of interest(r) = 10% p.a.
∴ S.I. = P × n(n + 1)/(2 × 12) × (r/100)
= 600 × {20(20 + 1)}(2 × 12) × (10/100)
= 600 × (420/24) × (10/100)
= Rs 1,050
The amount that Manish will get at the time of maturity
= ₹ (600 x 20) + ₹ 1,050
= ₹ 12,000 + ₹ 1,050
= ₹ 13,050
Question 2. Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited ₹ 640 per month for 4.1/2 years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.
Solution 2: Installment per month(P) = ₹ 640
Number of months(n) = 54
Rate of interest(r) = 12% p.a.
∴ S.I.= P x {n(n+1)}/(2 × 12) × (r/100)
= 640 × {54(54 + 1)}/(2 × 12) × (12/100)
= 640 × (2970/24) × (12/100)
= Rs 9,504
The amount that Manish will get at the time of maturity
= ₹ (640 x 54)+ ₹ 9,504
= ₹ 34,560 + ₹ 9,504
= ₹ 44,064
Question 3. Each of A and B both opened recurring deposit accounts in a bank. If A deposited ₹ 1,200 per month for 3 years and B deposited ₹ 1,500 per month for 21/2 years, find, on maturity, who will get more amount and by how much? The rate of interest paid by the bank is 10% per annum.
Solution 3: For A,
Installment per month(P) = 1,200
Number of months(n) = 36
Rate of interest(r) = 10% p.a.
∴ S.I. = P × {n(n + 1))/(2 × 12) × (r/100)
=1,200 × {36(36 + 1)}/(2 × 12) × (10/100)
= 1,200 × (1332/24) x (10/100)
= Rs 6,660
The amount that A will get at the time of maturity
= ₹ (1,200 x 36) + ₹ 6,660
= ₹ 43,200 + ₹ 6,660
= ₹ 49,860
For B,
Instalment per month(P) = ₹ 1,500
Number of months(n) = 30
Rate of interest(r) = 10% p.a.
∴ S.I. = P × {n(n+1)}/(2 × 12) × (r/100)
= 1.500 × {30(30 + 1)}/(2 × 12) × (10/100)
= 1,500 × (930/24) × (10/100)
= Rs 5,812.50
The amount that B will get at time of maturity
= ₹ (1500 × 30) + ₹ 5,812.50
= ₹ 45,000 + ₹ 5,812.50
= ₹ 50,812.50
Difference between both amounts = ₹ 50,812.50 - ₹ 49,860
= ₹ 952.50
Then B will get more money than A by ₹ 952.50
Question 4. Ashish deposits a certain sum of money every month is a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets ₹ 12,715 as the maturity value of this account, what sum of money did money did he pay every month?
Solution 4: Let Installment per month(P) = ₹ y
Number of months(n) = 12
Rate of interest(r) = 11%p.a.
∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)
= y × {12(12 + 1)/(2 × 12)} × (11/100)
= y × (156/24) × (11/100) = Rs 0.715y
Maturity value = ₹(y × 12) + ₹ 0.715y = ₹ 12.715y
Given Maturity value = ₹ 12,715
Then ₹ 12.715y = ₹ 12,715
⇒ y = 12,715/12.715 = Rs 1,000
Question 5. A man has a Recurring Deposit Account in a bank for 3½ years. If the rate of interest is 12% per annum and the man gets ₹ 10,206 on maturity, find the value of monthly instalments.
Solution 5: Let Installment per month(P) = ₹ y
Number of months(n) = 42
Rate of interest(r) = 12% p.a.
∴ S.I. = P x {n(n + 1)}/(2 × 12) × (r/100)
= y × {42(42 + 1)/(2 × 12)} × (12/100)
= y × (1806/24) x (12/100) = Rs 9.03y
Maturity value = ₹ (y x 42) + ₹ 9.03y = ₹ 51.03y
Given maturity value = ₹ 10,206
Then ₹ 51.03y = ₹ 10206
⇒ y = 10206/51.03 = Rs 200
Question 6.
(i) Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits ₹ 140 per month for 4 years. If he gets ₹ 8,092 on maturity, find the rate of interest given by the bank
(ii) David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7,725 at the time of maturity, find the rate of interest per annum.
Solution:
Instalment per month(P) = ₹ 1,500
Number of months(n) = 30
Rate of interest(r) = 10% p.a.
∴ S.I. = P × {n(n+1)}/(2 × 12) × (r/100)
= 1.500 × {30(30 + 1)}/(2 × 12) × (10/100)
= 1,500 × (930/24) × (10/100)
= Rs 5,812.50
The amount that B will get at time of maturity
= ₹ (1500 × 30) + ₹ 5,812.50
= ₹ 45,000 + ₹ 5,812.50
= ₹ 50,812.50
Difference between both amounts = ₹ 50,812.50 - ₹ 49,860
= ₹ 952.50
Then B will get more money than A by ₹ 952.50
Question 4. Ashish deposits a certain sum of money every month is a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets ₹ 12,715 as the maturity value of this account, what sum of money did money did he pay every month?
Solution 4: Let Installment per month(P) = ₹ y
Number of months(n) = 12
Rate of interest(r) = 11%p.a.
∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)
= y × {12(12 + 1)/(2 × 12)} × (11/100)
= y × (156/24) × (11/100) = Rs 0.715y
Maturity value = ₹(y × 12) + ₹ 0.715y = ₹ 12.715y
Given Maturity value = ₹ 12,715
Then ₹ 12.715y = ₹ 12,715
⇒ y = 12,715/12.715 = Rs 1,000
Question 5. A man has a Recurring Deposit Account in a bank for 3½ years. If the rate of interest is 12% per annum and the man gets ₹ 10,206 on maturity, find the value of monthly instalments.
Solution 5: Let Installment per month(P) = ₹ y
Number of months(n) = 42
Rate of interest(r) = 12% p.a.
∴ S.I. = P x {n(n + 1)}/(2 × 12) × (r/100)
= y × {42(42 + 1)/(2 × 12)} × (12/100)
= y × (1806/24) x (12/100) = Rs 9.03y
Maturity value = ₹ (y x 42) + ₹ 9.03y = ₹ 51.03y
Given maturity value = ₹ 10,206
Then ₹ 51.03y = ₹ 10206
⇒ y = 10206/51.03 = Rs 200
Question 6.
(i) Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits ₹ 140 per month for 4 years. If he gets ₹ 8,092 on maturity, find the rate of interest given by the bank
(ii) David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7,725 at the time of maturity, find the rate of interest per annum.
Solution:
(a) Installment per month(P) = ₹ 140
Number of months(n) = 48
Let rate of interest(r) = r % p.a.
∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)
= 140 ×{48(48 + 1)}/(2 × 12) × (r/100)
= 140 × (2352/24) × (r/100) = Rs(137.20)r
Maturity value = ₹ (140 x 48) + ₹ (137.20)r
Given maturity value = ₹ 8,092
Then ₹ (140 x 48) + ₹ (137.20)r = ₹ 8,092
⇒ 137.20r = ₹ 8,092 - ₹ 6,720
⇒ r = (1,372/137.20) = 10%
Number of months(n) = 48
Let rate of interest(r) = r % p.a.
∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)
= 140 ×{48(48 + 1)}/(2 × 12) × (r/100)
= 140 × (2352/24) × (r/100) = Rs(137.20)r
Maturity value = ₹ (140 x 48) + ₹ (137.20)r
Given maturity value = ₹ 8,092
Then ₹ (140 x 48) + ₹ (137.20)r = ₹ 8,092
⇒ 137.20r = ₹ 8,092 - ₹ 6,720
⇒ r = (1,372/137.20) = 10%
(b) Instalment per month(P) = ₹ 300
Number of months(n) = 24
Let rate of interest(r) = r% p.a.
∴ S.I. = P × {n(n+1}/(2 × 12) × (r/100)
= 300 × {24(24 + 1)}/(2 × 12) × (r/100)
= 300 × (600/24) × (r/100) = Rs (75)r
Maturity value = ₹(300 × 24) + ₹(75)r
Given maturity value = ₹ 7,725
Then ₹(300 × 24) + ₹ (75)r = ₹ 7,725
⇒ 75r = ₹ 7,725 - ₹ 7,200
⇒ r = 525/75 = 7%
Question 7: Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month?
Solution 7: Installment per month(P) = ₹ 150
Number of months(n) = 8
Rate of interest(r) = 8% p.a.
∴ S.I. = P ×{n(n + 1)/(2 × 12)} × (r/100)
= 150 × {8(8 + 1)}/(2 × 12) × (8/100)
= 150 × (72/24) × (8/100) = Rs 36
The amount that Manish will get at the time of maturity
= ₹ (150 × 8) + ₹ 36
= ₹ 1,200 + ₹ 36
= ₹ 1,236
Question 8. Mrs. Geeta deposited ₹ 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ₹ 5,565, find the rate of interest per annum.
Solution 8: Installment per month(P) = ₹ 350
Number of months(n) = 15
Let rate of interest (r) = r % p.a.
∴ S.I.= P × {n(n + 1)}/(2 × 12) × (r/100)
= 350 × (15(15+1)}/(2 + 12) × (r/100)
= 350 × (240/24) × (r/100) = Rs (35)r
Maturity value= ₹ (350 x 15) + ₹ (35)r
Given maturity value = ₹ 5,565
Then ₹(350 × 15) + ₹ (35)r = ₹ 5,565
⇒ 35r = ₹5,565 - ₹5,250
⇒ r = 315/35 = 9%
Question 9. A recurring deposit account of ₹ 1,200 per month has a maturity value of ₹ 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time in months) of this Recurring Deposit Account
Solution 9: Installment per month(P) = ₹ 1,200
Number of months(n) = n
Let rate of interest(r) = 8% p.a.
∴ S.I.= P ×{n(n + 1)}/(2 × 12) × (r/100)
= 1,200 ×{n(n +1)} /(2 × 12) × (8/100)
= 1,200 ×{n(n + 1)}/24 × (8/100) = Rs 4n(n + 1)
Maturity value = ₹ (1,200 × n) + ₹ 4n(n + 1) = ₹ (1200n + 4n2 + 4n)
Given maturity value = ₹ 12,440
Then 1200n + 4n2 + 4n = 12,440
⇒ 4n2 + 1204n - 12440 = 0
⇒ n2 + 301n - 3110 = 0
⇒ (n + 311)(n - 10) = 0
⇒ n = - 311 or n =10 months
Then number of months = 10
Question 10. Mr. Gulati has a Recurring Deposit Account of ₹ 300 per month. If the rate of interest is 12% and the maturity value of this account is ₹ 8,100; find the time in years) of this Recurring Deposit Account
Solution 10: Installment per month(P) = ₹ 300
Number of months(n) = n
Let rate of interest(r) = 12% p.a.
∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)
= 300 × {n(n + 1)}/(2 × 12) × (12/100)
= 300 × {n(n +1)}/24 × (12/100) = Rs 1.5n(n +1)
Maturity value = ₹ (300 × n) + ₹ 1.5n(n+1)
= ₹ (300n + 1.5n2 + 1.5n)
Given maturity value= ₹ 8,100
Then 300n + 1.5n2 + 1.5n = 8,100
⇒ 1.5n2 + 301.5n - 8100 = 0
⇒ n2 + 201n - 5400 = 0
⇒ (n + 225)(n - 24) = 0
⇒ n = -225 or n = 24 months
Then time = 2 years
Number of months(n) = 24
Let rate of interest(r) = r% p.a.
∴ S.I. = P × {n(n+1}/(2 × 12) × (r/100)
= 300 × {24(24 + 1)}/(2 × 12) × (r/100)
= 300 × (600/24) × (r/100) = Rs (75)r
Maturity value = ₹(300 × 24) + ₹(75)r
Given maturity value = ₹ 7,725
Then ₹(300 × 24) + ₹ (75)r = ₹ 7,725
⇒ 75r = ₹ 7,725 - ₹ 7,200
⇒ r = 525/75 = 7%
Question 7: Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month?
Solution 7: Installment per month(P) = ₹ 150
Number of months(n) = 8
Rate of interest(r) = 8% p.a.
∴ S.I. = P ×{n(n + 1)/(2 × 12)} × (r/100)
= 150 × {8(8 + 1)}/(2 × 12) × (8/100)
= 150 × (72/24) × (8/100) = Rs 36
The amount that Manish will get at the time of maturity
= ₹ (150 × 8) + ₹ 36
= ₹ 1,200 + ₹ 36
= ₹ 1,236
Question 8. Mrs. Geeta deposited ₹ 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ₹ 5,565, find the rate of interest per annum.
Solution 8: Installment per month(P) = ₹ 350
Number of months(n) = 15
Let rate of interest (r) = r % p.a.
∴ S.I.= P × {n(n + 1)}/(2 × 12) × (r/100)
= 350 × (15(15+1)}/(2 + 12) × (r/100)
= 350 × (240/24) × (r/100) = Rs (35)r
Maturity value= ₹ (350 x 15) + ₹ (35)r
Given maturity value = ₹ 5,565
Then ₹(350 × 15) + ₹ (35)r = ₹ 5,565
⇒ 35r = ₹5,565 - ₹5,250
⇒ r = 315/35 = 9%
Question 9. A recurring deposit account of ₹ 1,200 per month has a maturity value of ₹ 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time in months) of this Recurring Deposit Account
Solution 9: Installment per month(P) = ₹ 1,200
Number of months(n) = n
Let rate of interest(r) = 8% p.a.
∴ S.I.= P ×{n(n + 1)}/(2 × 12) × (r/100)
= 1,200 ×{n(n +1)} /(2 × 12) × (8/100)
= 1,200 ×{n(n + 1)}/24 × (8/100) = Rs 4n(n + 1)
Maturity value = ₹ (1,200 × n) + ₹ 4n(n + 1) = ₹ (1200n + 4n2 + 4n)
Given maturity value = ₹ 12,440
Then 1200n + 4n2 + 4n = 12,440
⇒ 4n2 + 1204n - 12440 = 0
⇒ n2 + 301n - 3110 = 0
⇒ (n + 311)(n - 10) = 0
⇒ n = - 311 or n =10 months
Then number of months = 10
Question 10. Mr. Gulati has a Recurring Deposit Account of ₹ 300 per month. If the rate of interest is 12% and the maturity value of this account is ₹ 8,100; find the time in years) of this Recurring Deposit Account
Solution 10: Installment per month(P) = ₹ 300
Number of months(n) = n
Let rate of interest(r) = 12% p.a.
∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)
= 300 × {n(n + 1)}/(2 × 12) × (12/100)
= 300 × {n(n +1)}/24 × (12/100) = Rs 1.5n(n +1)
Maturity value = ₹ (300 × n) + ₹ 1.5n(n+1)
= ₹ (300n + 1.5n2 + 1.5n)
Given maturity value= ₹ 8,100
Then 300n + 1.5n2 + 1.5n = 8,100
⇒ 1.5n2 + 301.5n - 8100 = 0
⇒ n2 + 201n - 5400 = 0
⇒ (n + 225)(n - 24) = 0
⇒ n = -225 or n = 24 months
Then time = 2 years