Banking Exercise 2A ICSE Class-10 Concise ICSE Mathematics Selina Solution Ch-2. We Provide Step by Step Solutions / Answer of Exe-2(A) Banking of Selina Concise Maths. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Banking Class 10 Selina Concise Maths ICSE Maths Solutions Ch-2
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-2 | Banking |
| Writer | R.K. Bansal |
| Exe-2A | Solved Questions on Recurring Deposit |
| Edition | 2025-2026 |
Exe-2A Recurring Deposit
Banking Exercise 2A ICSE Class-10 Concise ICSE Mathematics Selina Solution Ch-2
Que-1. Manish opens a Recurring Deposit Account with the Bank of Rajasthan and deposits 600 per month for 20 months. Calculate the maturity value of this account, if the bank pays interest at the rate of 10% per annum.
Ans- Instalment per month(P) = Rs 600
Number of months(n) = 20
Rate of interest(r)= 10%p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 600 x [20(20+1)/2 x 12] x 10/100
=> Rs 1050
The amount that Manish will get at the time of maturity
=> Rs (600×20)+ Rs 1,050
=> Rs 12,000+ Rs 1,050
=> Rs 13,050
Que-2. Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited 640 per month for 4½years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.
Ans- Instalment per month(P) = Rs 640
Number of months(n) = 54
Rate of interest(r)= 12%p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 640 x [54(54+1)/2 x 12] x 12/100
=> Rs 9504
The amount that Mathew will get at the time of maturity
=Rs (640×54)+ Rs 9,504
=Rs 34,560+ Rs 9,504
= Rs 44,064
Que-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 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.
Ans- For A
Instalment per month(P) = Rs 1,200
Number of months(n) = 36
Rate of interest(r)= 10%p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 1200 x [36(36+1)/2 x 12] x 10/100
=> Rs 6660
The amount that A will get at the time of maturity
=Rs (1,200×36)+ Rs 6,660
=Rs 43,200+ Rs 6,660
= Rs 49,860
For B
Instalment per month(P) = Rs 1,500
Number of months(n) = 30
Rate of interest(r)= 10%p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 1500 x [30(30+1)/2 x 12] x 10/100
=> Rs 5812.50
The amount that B will get at the time of maturity
=Rs (1,500×30)+ Rs 5,812.50
=Rs 45,000+ Rs 5,812.50
= Rs 50,812.50
Difference between both amounts= Rs 50,812.50 – Rs 49,860
= Rs 952.50
Then B will get more money than A by Rs 952.50 Ans.
Que-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?
Ans- Let Instalment per month(P) = Rs y
Number of months(n) = 12
Rate of interest(r)= 11%p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> y x [12(12+1)/2 x 12] x 11/100
=> Rs 0.715 y
Maturity value= Rs(y x 12)+Rs 0.715 y= Rs 12.715 y
Given maturity value= Rs 12,715
Then Rs 12.715 y = Rs 12,715
y = 12715/12.715 = Rs 1000
Que-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 installments.
Ans- Let Instalment per month(P) = Rs y
Number of months(n) = 42
Rate of interest(r)= 12%p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> y x [42(42+1)/2 x 12] x 12/100
=> Rs 9.03 y
Maturity value= Rs(y x 42)+Rs 9.03 y= Rs 51.03 y
Given maturity value= Rs 10,206
Then Rs 51.03 y = Rs 10,206
y = 10206/51.3 = Rs 200
Ques-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.
Ans-(i) Instalment per month(P) = Rs 140
Number of months(n) = 48
Let rate of interest(r)= r %p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 140 x [48(48+1)/2 x 12] x r/100
=> Rs 0.715 r
Maturity value= Rs (140 x 48)+Rs (137.20)r
Given maturity value= Rs 8,092
Then Rs(140 x 48)+Rs(137.20)r = Rs 8,092
137.20 r = Rs 8,092 -; Rs 6,720
r = 1372/137.2 = 10%
(ii) Instalment per month(P) = Rs 300
Number of months(n) = 24
Let rate of interest(r)= r %p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 300 x [24(24+1)/2 x 12] x r/100
=> Rs 75 r
Maturity value= Rs(300 x 24)+Rs(75)r
Given maturity value= Rs 7,725
Then Rs(300 x 24)+Rs(75)r = Rs 7,725
75 r = Rs 7,725 -; Rs 7,200
r = 525/75 = 7%
Que-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?
Ans- Instalment per month(P) = Rs 150
Number of months(n) = 8
Rate of interest(r)= 8%p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 150 x [8(8+1)/2 x 12] x 8/100
=> Rs 36
The amount that Amit will get at the time of maturity
=Rs(150×8)+ Rs 36
=Rs 1,200+ Rs 36
= Rs 1,236
Que-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.
Ans- Instalment per month(P) = Rs 350
Number of months(n) = 15
Let rate of interest(r)= r %p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 350 x [15(15+1)/2 x 12] x r/100
=> Rs 35r
Maturity value= Rs(350 x 15)+Rs(35)r
Given maturity value= Rs 5,565
Then Rs(350 x 15)+Rs(35)r = Rs 5,565
35 r = Rs 5,565 -; Rs 5,250
r = 315/35 = 9
Que-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.
Ans- Instalment per month(P) = Rs 1,200
Number of months(n) = n
Let rate of interest(r)= 8 %p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 1200 x [n(n+1)/2 x 12] x 8/100
=> Rs 4n(n+1)
Maturity value= Rs(1,200 x n)+Rs 4n(n+1)= Rs(1200n+4n²+4n)
Given maturity value= Rs 12,440
Then 1200 n+4n²+4n = 12,440
=> 4n² + 1204n – 12440 = 0
=> (n+311)(n-10) = 0
Then number of months = 10
Ques-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.
Ans- Instalment per month(P) = Rs 300
Number of months(n) = n
Let rate of interest(r)= 12 %p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 300 x [n(n+1)/2 x 12] x 12/100
=> Rs 1.5n(n+1)
Maturity value= Rs(300x n)+Rs1.5n(n+1)
= Rs(300n+1.5n2+1.5n)
Given maturity value= Rs8,100
Then 300n+1.5n2+1.5n = 8,100
301.5n+1.5n²-8100 = 0
=> 4n² + 1204n – 12440 = 0
=> (n+225)(n-24) = 0
Then time= 2 years
Que-11. Mr. Gupta opened a recurring deposit account in a bank. He deposited 2,500 per month for two years. At the time of maturity he got 67,500. Find:(i) the total interest earned by Mr. Gupta (ii) the rate of interest per annum.
Ans- (i) Maturity value = Rs 67,500
Money deposited= Rs 2,500 x 24= Rs 60,000
Then total interest earned= Rs 67,500 – Rs 60,000= Rs 7,500
(ii) Instalment per month(P) = Rs 2,500
Number of months(n) = 24
Let rate of interest(r)= r %p.a.
∴ SI = P x [n(n+1)/2 x 12] x r/100
=> 2500 x [24(24+1)/2 x 12] x r/100
=> Rs 625r
Then 625 r= 7500
r = 7500/625 = 12%
— : End of Banking Exercise 2A ICSE Class-10 Concise ICSE Mathematics Selina Ch-2 :–
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