Banking Class 10
Banking
Exercise
1. Archana deposited ₹ 400 per month for 3 years in a bank's recurring deposit account. If the bank pays interest at the rate of 10% p. a., find the amount she gets on maturity.
Solution :
MI = ₹ 400
n = 3 years = 36 month
r = 10%
MV = ?
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
= 400 * 36 + [ 400 * 36 ( 36 +1) 10 / 2 * 12 * 100]
= 14400 + [400 * 36 * 37 *10 / 2 * 12 * 100]
= 14400 + [60 * 37]
= 14400 + 2220
= 16620
2. Mr. Antao has a two year deposit account in a bank where he deposits ₹ 900 per month. Find the amount received by him at the time of maturity, if the rate of interest is 5% p.a.
Solution :
n = 2 years = 24 months
MI = ₹ 900
r = 5%
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
= 900 * 24 + [ 900 * 24 ( 24 +1) 5 / 2 * 12 * 100]
= 21600 + [900 * 24 * 25 *5 / 2 * 12 * 100]
= 21600 + [9 * 25 * 5]
= 21600 + 1125
= ₹ 22725
3. Joseph has an account in recurring deposit scheme for 2 years. He deposits ₹1500 per month. If the rate of interest is 8% p. a., calculate the amount he would receive at the time of maturity.
Solution :
n = 2 years = 24 months
MI = ₹ 1500
r = 8 %
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
= 1500 * 24 + [ 1500 * 24 ( 24 +1) 8 / 2 * 12 * 100]
= 36000 + [1500 * 24 * 25 * 8 / 2 * 12 * 100]
= 36000 + [15 * 25 * 8]
= 36000 + 3000
= ₹ 39000
4. Mrs. Ramani has a three-year recurrent deposit account in the State Bank. She deposits ₹ 600 per month. Calculate the amount she would receive at the time of maturity if the rate of interest in 9% p. a.
Solution :
n = 3 years = 36 months
MI = ₹ 600
r = 9 %
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
= 600 * 36 + [ 600 * 36 ( 36 +1) 9 / 2 * 12 * 100]
= 21600 + [600 * 36 * 37 * 9 / 2 * 12 * 100]
= 21600 + [9 * 37 * 9]
= 21600 + 2997
= ₹ 24597
5. Rajesh deposit ₹ 1000 every month in a recurring deposit account for three years. Calculate the rate of interest if the matured value is ₹ 40,440.
Solution :
MI = ₹ 1000
n = 3 years = 36 months
r = ?
MV = ₹ 40440
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
40440 = 1000 * 36 + [ 1000 * 36 ( 36 +1) r / 2 * 12 * 100]
40440 = 36000 + [1000 * 36 * 37 * r / 2 * 12 * 100]
40440 = 36000 + [5 * 3 * 37 * r]
40440 = 36000 + 555r
555r = 40440 - 36000
555r = 4440
r = 4440 / 555
r = 8 %
6. Ravina deposits ₹ 600 per month in a recurring deposit scheme for two years. If she receives rupees 15,450 at the time of maturity. Calculate the rate of interest per annum.
Solution :
MI = ₹ 600
n = 2 years = 24 months
MV = ₹ 15450
r = ?
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
15450 = 600 * 24 + [ 600 * 24 ( 24 +1) r / 2 * 12 * 100]
15450 = 14400 + [600 * 24 * 25 * r / 2 * 12 * 100]
15450 = 14400 + [3 * 2 * 25 * r]
15450 = 14400 + 150r
150r = 15450 - 14400
150r = 1050
r = 1050 / 150
r = 7 %
7. Vineeta deposits ₹ 800 per month in a cumulative deposit account for 3 years. If the account payable at the time of maturity is ₹ 31,464. Calculate the rate of interest.
Solution :
MI = ₹ 800
MV = ₹ 31464
n = 3 years = 36 months
r = ?
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
31464 = 800 * 36 + [ 800 * 36 ( 36 +1) r / 2 * 12 * 100]
31464 = 28800 + [800 * 36 * 37 * r / 2 * 12 * 100]
31464 = 28800 + [4 * 3 * 37 * r]
31464 = 28800 + 444r
444r = 31464 - 28800
444r = 2664
r = 2664 / 444
r = 6 %
8. Mr. Madhav Rao gets ₹ 6455 at the end of 1 year when he deposits ₹ 500 per month in a recurring deposit scheme. Find the rate of interest.
Solution :
MI = ₹ 500
MV = ₹ 6455
n = 1 years = 12 months
r = ?
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
6455 = 500 * 12 + [ 500 * 12 ( 12 +1) r / 2 * 12 * 100]
6455 = 6000 + [500 * 12 * 13 * r / 2 * 12 * 100]
6455 = 6000 + [5 * 13r/2]
6455 = 6000 + 65r/2
65r/2 = 6455 - 6000
65r/2 = 455
r = 455 * 2 / 65
r = 910 /65
r = 14 %
9. Zaheeda deposits a certain sum of money, every month in a recurring deposit account for 2 years. if she receives ₹ 37875 at the time of maturity and the rate of interest is 5%, find a monthly deposit.
Solution :
MV = ₹ 37875
n = 2 years = 24 months
r = 5 %
MI = ?
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
37875 = MI * 24 + [MI * 24 ( 24 +1) 5 / 2 * 12 * 100]
37875 = 24MI + [MI * 24 * 25 * 5 / 2 * 12 * 100]
37875 = 24MI + 5MI/4
37875 = 96MI + 5MI/4
37875 = 101MI/4
37875 * 4 = 101MI
MI = 37875 *4/101
MI = ₹ 1500
10. Srinidhee deposits a certain sum of money every month in a recurring deposit scheme for 2 years at 6% p.a. If the amount payable to her at the time of maturity of the account is ₹ 20,400, find the monthly installment.
Solution :
MV = ₹ 20400
n = 2 years = 24 months
r = 6 %
MI = ?
MV = MI * n + [ MI * n(n+1) r / 2 * 12 * 100]
20400 = MI * 24 + [MI * 24 ( 24 +1) 6 / 2 * 12 * 100]
20400 = 24MI + [MI * 24 * 25 * 6 / 2 * 12 * 100]
20400 = 24MI + 3MI/2
20400 = 48MI + 3MI/2
20400 = 51MI/2
20400 * 2 = 51MI
MI = 20400 * 2/51
MI = ₹ 800
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