ML Aggarwal ICSE Solutions Class 10 Math 2nd Chapter Banking
Class 10 Chapter 2 BankingExercise:
(1) Mrs. Goswami deposits Rs. 1000 every month in a recurring deposit, account for 3 years at 8% interest per annum. Find the matured value.
Solution:
Here, given that Mrs. Goswami deposits Rs. 1000 every month in a recurring deposit account for 3 years at the rate of 8% interest per annum.
Then, deposit per month (P) = Rs. 1000
Duration of period = 3 years = 36 months
Interest rate per annum = 8%
Then, Total principal = 36(36+1)/2 × 1000 = (36×37×1000)/2
And Interest = PRT/100 = (36×37×1000×8)/(2×12×100) = 12×37×10 = 4440
Thus, matured price = P × n Simple Interest
=1000 × 36 + 4440
= 36000 + 4440
= 40440
Thus, the matured value is found to be Rs. 40440.
(2) Sonia had a recurring deposit account in a bank and deposited Rs 600 per month for 2 1⁄2 years. If the rate of interest was 10% per annum. Find the maturity value of this account.
Solution:
Here, given that Sonia deposited Rs. 600 per month for 2 1⁄2 years. And, the rate of interest was 10% per annum.
Period of duration = 2.5 years
Then, = 2.5×12 months
= 30 months
Rate of interest per annum = 10%.
Then, Simple Interest = P×(n)(n+1)×r/2×100×12
= 600×30×31/2×100×12
Simple Interest = 2335 Rs
Thus, maturity value = P × n + Simple interest
= 600×30+2325
= 20,325
Thus, the maturity value is found to be 20,325 Rs.
(3) Kiran deposited Rs. 200 per month for 36 months in a bank’s rearing deposit account. If the banks pays interest at the rate of 11% per annum, find the amount she gets on maturity?
Solution:
Here, given that Kiran deposited Rs. 200 per month for 36 months.
The rate of interest per annum is 11%.
Then, the amount deposited (P) = 200 Rs.
Duration of Period (n) = 36 months
Rate of interest (R) = 11% per annum
Then, the total amount deposited in 36 months is 200×36 = Rs 7200
Thus, Simple interest = P×n×(n+1)×R/2×12×100
= 200×36×37×11/2×12×100
= (200×36×37×11)/(2×12×100)
= (200×36×37×11)/(2×12×100)
= 12,21
Hence, the maturity value = 7200 + 1221
= 8421
Thus, Kiran will get maturity value as Rs. 8421.
(4) 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.
Solution:
Here, given that Haneef deposits Rs. 600 per month for a period of 4 years. And he got an interest of Rs. 5880 at the time of maturity.
Then, Monthly deposit (P) = Rs 600
Interest = Rs 58,800
Period of duration (n) = 4 years = 48 months
Thus, deposit for 1 month = P×n×(n+1)/2
= (600×48×49)/2
= Rs.70,5600
Let us consider the rate of interest is 1% per annum.
Then, Interest = P×r×t/100
= 58,800 = (705600×r×1)/(100×12)
5880 = 588 r
∴ r = 5880/588 = 10
r = 10
Thus, the rate of interest is found to be 10%.
(5) 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.
Solution:
Here, given that David deposited Rs. 300 per month for two years. And he received Rs. 7725 at the time of maturity.
Here, Deposit during one month (P) = Rs. 300 duration of period = 2 years = 24 months. And maturity value = Rs. 7725
Let us consider ‘R’ is the rate percent which we have to find.
Thus, Principal for 1 month = P×n×(n+1)/2
= 300×24×25/2
= 90,000
Hence, Interest got = PRT/100
= (90,000×R×1)/(100×12)
= 75R
Thus, 300×24+75R = 7725
7200 + 75R = 7725
75R = 7725 – 7200 = 525
R = 525/75 = 7
R = 7
Thus, the rate of interest is found to be 7% per annum.
(6) 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. 67,500 find
(i) The total interest earned by Mr. Gupta
(i) The rate of interest per annum.
Solution:
Here, given that
Mr. Gupta deposited Rs. 2500 per month for 2 years. And he got Rs. 67,500 at the time of maturity. Thus, Amount deposited per month = Rs. 2500
Period of duration = 2 years = 24 months
Maturity value = Rs.67500
Then, Total principal for 1 month = P×n×(n+1)/2
= (2500×24×25)/2
= 7,50,000
Thus, Interest: = 67,500 – 24 × 2500
= 67,500 – 60,000
Interest = 7500
Period = 1 month = 1/12 year
Then, Rate of interest = (Simple interest×100)/(7500×100×12)
= 12%
Thus, (i) the total interest earned by Mr. Gupta is.7500 Rs.
(ii) And the rate of interest per annum is 12%.
(7) Shahrukh opened a recurring deposit account in a bank and deposited
Rs. 800 per month for 11/2 years. If he received Rs.15,084 at the time of maturity, find the rate of interest per annum.
Solution:
Here, given that Shahrukh deposited Rs.800 per month for 11/2 years. And he received Rs. 15,084 at the time of maturity.
Then, Amount deposited per month (P) = Rs 800
Period of months (n) = 1 1/2 = 3/2 years
n = 3/2×12 months
n = 18 months
Thus, Interest = P×n×(n+1)×r/2×12×100
= 800×18×19×r/2×12×100
Interest = 114r
But, maturity value = 15,084
Hence, 15084 = 114r + 800×18
15084 = 114r +14,400
15084 – 14,400 = 114r
684 – 114r
R = 684/114 = 6%.
R = 6%
Thus, the rate of interest per annum is found to be 6%.
(8) Rekha opened a recurring deposited account for 20 months and the rate of interest per annum is 9%. And Rekha received 441 as interest at the time of maturity-find the rate of interest per annum.
Solution:
Here, given that Rekha deposited amount for 20 months with rate of interest per annum 9%.
Rekha received Rs. 441 as interest at the time of maturity.
Then, n = 20 months, r = 90%.
Let the monthly deposited amount (P) = y
Thus, Interest = P× n(n+1)/2×12 × r/100
= y×20×21×9/2×12×100
= y×5×7×9/2×100
Interest = 1.575y Rs.
But, given that, Rekha got interest at the time of maturity = 441 Rs.
∴ 441 = 1.575y
= y = 280 Rs.
Thus, the rate of interest per annum is found to be Rs 280.
(9) Mohan has a recurring deposit account in a bank for 2 years at 6% per annum simple interest. If he gets Rs 1200 as interest per at the time of maturity. Find –
(i) The monthly instalment.
(ii) The amount of maturity.
Solution:
Here, given that Mohan deposited account in a bank for 2 years at 6% per annum simple interest.
He got Rs 1200 interest at the time maturity.
Interest = Rs 1200
Then, Period of duration (n) = 2 years = 24 months
Rate per annum (r) = 6%
Let us consider monthly deposit is Rs. ‘P’.
Then, Interest = P×n×(n+1)×r/2×12×100
1200 = P×24×25×6/2×12×100
1200 = 6/4 P
∴ P = 1200×4/6 = 800
∴ Monthly deposit = Rs 800
Then, Maturity value = P × n + interest
= 800 × 24 + 1200
= 19,200 + 1200
Maturity value = 20,400
Thus, (i) the monthly instalment is found to be Rs.800.
(ii) The amount of maturity is found to be Rs. 20,400.
(10) Mr. R.K. Nair gets Rs. 6,455 of the ends of one year at the rate of 14% per annum in a recurring deposit account. find the monthly instalment.
Solution:
Here, given that Mr. RK. Nair gets Rs. 6,455 at the end of one year & at the rate of 14% per annum.
Then, period of duration (n) = 1 year = 12 months
= R = 12
Let us consider, the monthly instalment is Rs. “P”
Then, Maturity value = p×n(n+1)×R/2×12×100 + (P×n)
6455 = 12×13×14×P/2×12×100 + (P×12)
6455 = (13×P×7)/100 + (P×12)
6455 = (91P + 1200P/100
645500 = 1291P
P = 6,45,500/1291
P = 500
Thus, the monthly instalment found Rs.500
(11) Samita has a recurring deposit account in a bank of Rs 2000 per month at the rate of 10% per annum. If she gets Rs 83,100 at the time of maturity. Find the total time for which, the account was held.
Solution:
Here, given that Samita deposited Rs 2000 per month at the rate of 10% per annum. And she got Rs 83,100 at the time of maturity.
Then, deposit per month (P) = 2000
Rate of interest (r) = 10%
Let us consider the period of duration is (n) months.
Then, (principal for one month) = 2000×n×(n+1)/2
= 1000n (n+1)
And interest = 1000n (n+1)×10×1/100×12
Interest = 100n (n+1)/12
Thus, maturity value = 2000 × n + 100n(n+1)/12
83,100 = 2000n + 100n (n+1)/12
24000n + 100n2+ 100n = 83100×12
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
n + 277 = 0 or n = 36
n = -277 or n = 36
Thus, n= 36 months = 3 years
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