Exercise : Chapter 2 Banking Recurring Deposit Accounts Exercise 2A MCQ Questions and Answers
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.
Solution :
Given:
Installment per month(P) = ₹ 600
Number of months(n) = 20
Rate of interest(r) = 10% p.a.
We know that:
S.I. = P x [n(n+1)/2x12] x (r/100)
= 600 x [20(20+1)/2x12] x (10/100)
= 600 x (420/24) x (10/100)
= ₹ 1,050
The amount that Manish will get as the maturity value
=(P x n)+S.I
= ₹ (600×20) + ₹ 1,050
= ₹ 12,000 + ₹ 1,050
= ₹ 13,050
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2.
Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited ₹ 640 per month for 4 ½ years Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.
Solution :
Given:
Installment per month(P) = ₹ 640
Number of months(n) = 54
Rate of interest(r)= 12% p.a.
We know that:
S.I. = P x [n(n+1)/2x12] x (r/100)
= 640 x [54(54+1)/2x12] x (12/100)
= 640 x (2970/24) x (12/100)
= ₹ 9,504
The amount that Manish will get at the time of maturity
=(P x n)+S.I
= ₹ (640×54)+ ₹ 9,504
= ₹ 34,560 + ₹ 9,504
= ₹ 44,064
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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 2 ½ 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 :
Given:
For A
Installment per month(P) = ₹ 1,200
Number of months(n) = 36
Rate of interest(r) = 10% p.a.
We know that:
S.I. = P x [n(n+1)/2x12] x (r/100)
= 1200 x [36(36+1)/2x12] x (10/100)
= 1200 x (1332/24) x (10/100)
= ₹ 6,660
The amount that A will get at the time of maturity
=(P x n)+S.I
= ₹ (1,200×36) + ₹ 6,660
= ₹ 43,200 + ₹ 6,660
= ₹ 49,860
Given:
For B
Instalment per month(P) = ₹ 1,500
Number of months(n) = 30
Rate of interest(r) = 10% p.a.
We know that:
S.I. = P x [n(n+1)/2x12] x (r/100)
= 1500 x [30(30+1)/2x12] x (10/100)
= 1500 x (930/24) x (10/100)
= ₹ 5,812.50
The amount that B will get at the time of maturity
=(P x n)+S.I
= ₹ (1,500×30) + ₹ 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
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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 :
Let Installment per month(P) = ₹ y
Number of months(n) = 12
Rate of interest(r) = 11% p.a.
S.I. = P x [n(n+1)/2x12] x (r/100)
= y x [12(12+1)/2x12] x (11/100)
= y x (156/24) x (11/100)
= ₹ 0.715y
Maturity value = ₹ (y × 12) + ₹ 0.715y = ₹ 12.715y
Given maturity value = ₹ 12,715
Then ₹ 12.715y = ₹ 12,715
⇒ y = 12715/12.715 = ₹ 1,000
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5.
A man has a Recurring Deposit Account in a bank for 3 ½ 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 :
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)/2x12] x (r/100)
= y x [42(42+1)/2x12] x (12/100)
= y x (1806/24) x (12/100)
= ₹ 9.03y
Maturity value= ₹ (y × 42) + ₹ 9.03y= ₹ 51.03y
Given maturity value = ₹ 10,206
Then ₹ 51.03y = ₹ 10206
⇒ y = 10206/51.03 = ₹ 200
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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 x [n(n+1)/2x12] x (r/100)
= 140 x [48(48+1)/2x12] x (r/100)
= y x (2352/24) x (r/100)
= ₹ 137.20 r
Maturity value= ₹ (140 × 48) + ₹ (137.20)r
Given maturity value = ₹ 8,092
Then ₹ (140 × 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 x [n(n+1)/2x12] x (r/100)
= 300 x [24(24+1)/2x12] x (r/100)
= 300 x (600/24) x (r/100)
= ₹ 75 r
Maturity value = ₹ (300 × 24) + ₹ (75)r
Given maturity value = ₹ 7,725
Then ₹ (300 × 24) + ₹ (75)r = ₹ 7,725
⇒ 75 r = ₹ 7,725 – ₹ 7,200
⇒ r = 525/75 = 7%
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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 :
Installment per month(P) = ₹ 150
Number of months(n) = 8
Rate of interest(r) = 8% p.a.
S.I. = P x [n(n+1)/2x12] x (r/100)
= 150 x [8(8+1)/2x12] x (8/100)
= 150 x (72/24) x (8/100)
= ₹ 36
The amount that Manish will get at the time of maturity
= ₹ (150 × 8) + ₹ 36
= ₹ 1,200 + ₹ 36
= ₹ 1,236
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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 :
Installment per month(P) = ₹ 350
Number of months(n) = 15
Let rate of interest(r)= r% p.a.
S.I. = P x [n(n+1)/2x12] x (r/100)
= 350 x [15(15+1)/2x12] x (r/100)
= 350 x (240/24) x (r/100)
= ₹ 35 r
Maturity value= ₹ (350 × 15) + ₹ (35)r
Given maturity value = ₹ 5,565
Then ₹ (350 × 15) + ₹ (35)r = ₹ 5,565
⇒ 35r = ₹ 5,565 – ₹ 5,250
⇒ r = 315/35 = 9%
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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 :
Installment per month(P) = ₹ 1,200
Number of months(n) = n
Let rate of interest(r) = 8% p.a.
S.I. = P x [n(n+1)/2x12] x (r/100)
= 1200 x [n(n+1)/2x12] x (8/100)
= 1200 x [n(n+1)/24] x (8/100)
= ₹ 4n(n+1)
Maturity value = ₹ (1,200 × n) + ₹ 4n(n+1) = ₹ (1200n+4n^2 +4n)
Given maturity value= ₹ 12,440
Then 1200n+4n^2 +4n = 12,440
⇒ 4n^2 + 1204n - 12440 = 0
⇒ n^2 + 301n - 3110 = 0
⇒ (n + 311)(n-10) = 0
⇒ n = -311 or n = 10
Then number of months = 10
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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 :
Installment per month(P) = ₹ 300
Number of months(n) = n
Let rate of interest(r)= 12% p.a.
S.I. = P x [n(n+1)/2x12] x (r/100)
= 300 x [n(n+1)/2x12] x (12/100)
= 300 x [n(n+1)/24] x (12/100)
= ₹ 1.5n(n+1)
Maturity value= ₹ (300 × n)+ ₹ 1.5n(n+1)
= ₹ (300n+1.5n^2 +1.5n)
Given maturity value= ₹ 8,100
Then 300n+1.5n^2 +1.5n = 8,100
⇒ 1.5n^2 + 301.5n - 8100 = 0
⇒ n^2 + 201n - 5400 = 0
⇒ (n + 225)(n - 24)
⇒ n = -225 or n = 24 months
Then time = 2 years
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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.
Solution :
(i)
Maturity value = ₹ 67,500
Money deposited = ₹ 2,500 × 24= ₹ 60,000
Then total interest earned = ₹ 67,500 – ₹ 60,000 = ₹ 7,500 Ans.
(ii)
Installment per month(P) = ₹ 2,500
Number of months(n) = 24
Let rate of interest(r)= r% p.a.
S.I. = P x [n(n+1)/2x12] x (r/100)
= 2500 x [24(24+1)/2x12] x (r/100)
= 2500 x (600/24) x (r/100)
= ₹ 625 r
then 625 r = 7500
⇒ r = 7500/625 = 12%
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