Frank Solutions for Chapter 3 Banking Class 10 ICSE Mathematics
Exercise 3.1
1. Mr. Burman open a saving back account with Bank of India on 3rdApril 2007 with a cash deposit of Rs 5,000/-. Subsequently, he deposited Rs 16,500/- by cheque on 11thApril 2007, withdraw Rs 4,000/- on 10th May, paid Rs 3,500 for insurance by cheque on 7th July 2007, deposited Rs. 6,000/- in cash on 9th August 2007 and withdrew Rs 1,500/- on 12th Oct 2007.
(a) Make the entries in his passbook
(b) If he closed the account on 14th December and if the rate of simple interest is 4% pa, then find the amount he received on closing the account.
Answer
(a) From data given in the question,
We have to make the entries in passbook,
So, the table contains 5 columns. The data in 5 columns are, (i) Date (ii) Particulars (iii) Withdrawals (iv) Deposits (v) Balance.
Where, Date is the date of transaction, Particular is the details of transaction primarily the name, Withdrawal is the amount that has been taken out from account, Deposit is the amount that has been given to the account, balance is net amount remaining in the account after subtracting or adding the amount as applicable.
(b) As per the condition given in the question,
If he closed the account on 14th December and if the rate of simple interest is 4% pa. Then we have to find the amount he received on closing the account.
Given, June, September and November month where no transaction were made but the bank will give interest based on the amount which is reflected in the last month.
Therefore, total principal for at the end of November = ₹ 1,31,000
Interest = (1,31,000 × 4 × 1)/(100 × 12)
= ₹ 437
So, while closing the account,
Mr. Burman will get = principal + interest which amounts to
= ₹ 18,500 + ₹ 437
= ₹ 18,937
2. Ms. Chitra opened a saving bank account with SBI on 05.04.2007 with a cheque deposit of Rs 11,000/. Subsequently, she took out Rs 3,200/- on 12.05.2007; deposited a cheque of Rs. 8,800/- on 03.06.2007 and paid Rs 2,000/- by cheque on 18.06.2007.
(a) Make the entries in her passbook
(b) If the rate of simple interest was 5% pa compounded at the end of March and September, find her balance on 1.04.2008
Answer
(a) From data given in the question,
We have to make the entries in passbook,
So, the table contains 5 columns. The data in 5 columns are, (i) Date (ii) Particulars (iii) Withdrawals (iv) Deposits (v) Balance.
Where, Date is the date of transaction, Particular is the details of transaction primarily the name, Withdrawal is the amount that has been taken out from account, Deposit is the amount that has been given to the account, balance is net amount remaining in the account after subtracting or adding the amount as applicable.
(b) As per the condition given in the question,
If the rate of simple interest was 5% pa compounded at the end of March and September, find her balance on 1.04.2008
Then we have to find her balance on 1.04.2008.
Therefore, total principal for at the end of September 2007,
= ₹ 11,000 + ₹ 7,800 + ₹(14,600 × 4)
= ₹ 77,200
Interest at the end of September 2007 = (77,200 × 5 × 1)/(100 × 12)
= ₹ 321.66
So, interest = ₹ 322
Then, again principal at the end of March 2008 = 14920 × 6
= ₹ 89520
Interest at the end of March 2008 = (89520 × 5 × 1)/(100 × 12)
= ₹ 373
Therefore, Account balance as on 01.04.2008 = ₹ 14920 + ₹ 373
= ₹ 15,293
3. Given below is a page from the passbook of a saving bank account that Mr. Sharma has with SBI. If the bank gives interest at 6%pa, find
(a) The principal amount in January, February and March which will be considered for interest for interest calculation.
(b) The interest she gets at the end of March.
In February ₹ 5,400.00 as this is minimum of 10th and 28th February
In March ₹ 24,600.00 as this is minimum of 10th and 31st March (including 17th March)
(b) The interest she gets at the end of March given below,
Total principal at the end of March = 40700
Interest = (40700 × 6 × 1)/(100 × 12) = 203.50
Thus interest = Rs. 204
4. Given below is a page from the passbook of the savings bank account of Mr. Rajesh. Complete the entries in the passbook and calculate the interest paid to him by the bank at 6% pa in the end of June.
From the given table,
Then,
Therefore, total principal for at the end of June= ₹ 80,500
So, Interest = (80,500 × 6 × 1)/(100 × 12)
= ₹ 402.50
= ₹ 403
Hence, the interest paid to Mr. Rajesh by the bank at 6% pa in the end of June is ₹ 403
5. Given below is a page from the passbook of the savings bank account of Dolly Majumdar. Complete the entries in the passbook and find the interest earned by the account holder in the month of November if the rate of simple interest is 5% pa.
We know that, Balance = Previous Balance + Deposit – Withdrawal
Then, interest earned by account holder in the month of November
Therefore, total principal for at the end of November = ₹ 1,38,000
So, Interest = (138000 × 5 × 1)/(100 × 12)
= ₹ 575
6. The following are the entries in the passbook of a saving account of Ananya during the year 2007. If interest is calculated at 5% pa, find the interest earned by Ananya during the year.
Minimum balance between 10th day and the last day is mentioned in the table.
Therefore, total principal for at the end of December = ₹ 1,67,075
So, Interest = (1,67,075 × 5 × 1)/(100 × 12)
= ₹ 696.14
Hence, interest is ₹ 696
7. The following are from the saving bank account passbook of Mr. Ramesh. If the rate of interest paid by the bank is 4.5% pa calculated at the end of March and September, find the balance in his account at the end of the year.
Minimum balance between 10th day and the last day is mentioned in the table.
Total principal for at the end of March = ₹ 53,300
So, Interest at the end of March = (53,300 × 4.5 × 1)/(100 × 12)
= ₹ 199.87
Hence, interest is ₹ 200
Then, entering the interest in the passbook we get,
Now, interest calculating at the end of September.
Total principal for at the end of September = ₹ 2,16,000
So, Interest at the end of September = (2,16,000 × 4.5 × 1)/(100 × 12)
= ₹ 810
Therefore, by entering the interest in the passbook above we get the balance ₹ 52,370 at the end of year.
8. Mr. Punjwanis saving account passbook had the following entries, The bank pays interest at 4.5% on all SB accounts. Find the amount received by Mr. Punjwani when he closed the account on 25thJuly 08.
So, Interest at the end of March = (1,06,270 × 4.5 × 1)/(100 × 12)
= ₹ 398.51
Hence, interest is ₹ 399
Then, entering the interest in the passbook we get,
9. Mrs. Chhabra deposits Rs 500 per month in a recurring deposit account for 4 years at a simple interest rate of 6% pa.
(a) Find the maturity value of deposit.
(b) Find the total interest she will earn after 2 years
Answer
From the question it is given that,
Mrs. Chhabra deposits ₹ 500 per month in a recurring deposit
Period = 4 years
We know that, 1 year = 12 Months
So, 4 years = 4 × 12 = 48 Months
Rate = 6 % pa
Then, Money deposited = Monthly value × Number of Months
= ₹ 500 × 48
= ₹ 24,000 …[equation i]
So, total principal for 1 month = [500 × 48(48 + 1)]/2
= ₹ 5,88,000
Now, interest = (6 × 5,88,000)/(12 × 100)
= ₹ 2,940 …[equation ii]
For getting maturity amount we have to add both equation (i) and equation (ii)
= ₹ 24,000 + ₹ 2,940
= ₹ 26,940
Therefore, Maturity amount is ₹ 26,940
10. Mrs. Khandelkar invests Rs 900 every month in a recurring deposit account for a period of 3 years at a simple interest rate of 8% pa.
(a) Find the total interest she will earn at the end of the period.
(b) Find the maturity value of her deposits.
Answer
From the question it is given that,
Mrs. Khandelkar invests ₹ 900 every month in a recurring deposit account.
Period = 3 years
We know that, 1 year = 12 Months
So, 3 years = 3 × 12 = 36 Months
Rate = 8 % pa
Then, Money deposited = Monthly value × Number of Months
= ₹ 900 × 36
= ₹ 32,400 …[equation i]
So, total principal for 1 month = [900 × (36(36 + 1))]/2
= ₹ 5,99,400
Now, interest = (8 × 5,99,400)/(12 × 100)
= ₹ 3,996 …[equation ii]
For getting maturity amount we have to add both equation (i) and equation (ii)
= ₹ 32,400 + ₹ 3,996
= ₹ 36,396
Therefore, Maturity amount is ₹ 36,396
11. Mr. Patel deposit Rs 2,250 per month in a recurring deposit account for a period of 3 years. At the time of maturity, he gets Rs 90,990.
(a) Find the rate of simple interest per annum.
(b) Find the total interest earned by Mr. Patel.
Answer
From the question it is given that,
Mr. Patel deposit ₹ 2,250 per month in a recurring deposit account.
Period = 3 years
We know that, 1 year = 12 Months
So, 3 years = 3 × 12 = 36 Months
Maturity = ₹ 90,990
Rate = R % pa
Then, Money deposited = Monthly value × Number of Months
= ₹ 2,250 × 36
= ₹ 81,000
Interest get for this period = Maturity amount – Amount deposited
= 90,990 – 81,000
= ₹ 9,990
So, total principal for 1 month = [2,250 × (36(36 + 1))]/2
= ₹ 14,98,500
Now, interest = (R × 14,98,500)/(12 × 100)
⇒ 9,990 = (R × 14,98,500)/(12 × 100)
⇒ R = (9,990 × 12 × 100)/(14,98,500)
⇒ R = 8%
12. Mr. Menon deposit Rs 1,200 per month in a cumulative deposit account for a period of 5 years. After the end of the period, he will receive Rs 88,470.
(a) Find the rate of the interest per annum.
(b) Find the total interest that Mr. Menon will earn.
Answer
From the question it is given that,
Mr. Menon deposit Rs 1,200 per month in a cumulative deposit account.
Period = 5 years
We know that, 1 year = 12 Months
So, 5 years = 5 × 12 = 60 Months
Maturity = ₹ 88,470
Rate = R % pa
Then, Money deposited = Monthly value × Number of Months
= ₹ 1,200 × 60
= ₹ 72,000
Interest get for this period = Maturity amount – Amount deposited
= 88,470 – 72,000
= ₹ 16,470
So, total principal for 1 month = [1,200 × (60(60 + 1))]/2
= ₹ 21,96,000
Now, interest = (R × 21,96,000)/(12 × 100)
⇒ 16,470 = (R × 21,96,000)/(12 × 100)
⇒ R = (16,470 × 12 × 100)/(21,96,000)
⇒ R = 9%
13. Aarushi has a recurring deposit account for 2 years at 6% pa. She receives Rs 1,125 as interest on maturity.
(a) Find the monthly instalment amount.
(b) Find the maturity amount.
Answer
From the question it is given that,
Aarushi has a recurring deposit account for 2 years.
Period = 2 years
We know that, 1 year = 12 Months
So, 2 years = 2 × 12 = 24 Months
Rate = 6 % pa
Then, Money deposited = Monthly value × Number of Months
= P × 24
= ₹ 24P
So, total principal for 1 month = [P × (24(24 + 1))]/2
= ₹ 300P
Now, interest = (Total principal for 1 month × R)/(12 × 100)
₹ 1,125 = (300P × 6)/(12 × 100)
⇒ P = (1,125 × 12 × 100)/(300 × 6)
⇒ P = ₹ 750
For getting maturity amount = (P × 24) + Interest
= (750 × 24) + 1125
Therefore, Maturity amount is ₹ 19,125
14. Mr. Mohan has a cumulative deposit account for 3 years at 7% interest pa. She receives Rs 8,547 as a maturity amount after 3 years.
(a) Find the monthly deposit.
(b) Find the total interest receivable after maturity.
Answer
From the question it is given that,
Mr. Mohan has a cumulative deposit account for 3 years.
Period = 3 years
We know that, 1 year = 12 Months
So, 3 years = 3 × 12 = 36 Months
Rate = 7 % pa
Maturity amount = ₹ 8,547
Then, Money deposited = Monthly value × Number of Months
= P × 36
= ₹ 36P
So, total principal for 1 month = [P × (36(36 + 1))]/2
= ₹ 666P
Now, interest = (Total principal for 1 month × R)/(12 × 100)
₹ 8,547 – 36P= (666P × 7)/(12 × 100)
⇒ ₹ 8,547 – 36P = 3.885P
⇒ ₹ 8,547 = 3.885P + 36P
⇒ ₹ 8,547 = 39.885p
⇒ P = ₹ 8,547/39.885
⇒ P = ₹ 214.3
Interest amount = 8547 – 36P
Substitute the value of P we get,
Interest amount = 8547 – (36 × 214.3) = ₹ 832
15. Mr. Banerjee opens a recurring deposit account for Rs 3,000 per month at 9% simple interest pa. On maturity, he gets Rs. 1,70,460. Find the period for which he continued with the account.
Answer
From the question it is given that,
Mr. Banerjee opens a recurring deposit account for ₹ 3,000 per month.
Period = t
Rate = 9% pa
Maturity amount = ₹ 1,70,460
Then, Money deposited = Monthly value × Number of Months
= 300 × t
= ₹ 3000t
So, total principal for 1 month = [3000 × (t(t + 1))]/2
= 1500 (t2 + t)
= 1500t2 + 1500t
Now, interest = (Total principal for 1 month × R)/(12 × 100)
₹ 1,70,460 – 3000t= ((1500t2 + 1500t) × 9)/(12 × 100)
⇒ ₹ 1,70,460 – 3000t = (45t2 + 45t)/4
By cross multiplication we get,
681840 – 12000t = 45t2 + 45t
Then transposing we get,
45t2 + 45t + 12000t – 681840 = 0
⇒ 45t2 + 12045t – 681840 = 0
⇒ 45t2 – 2160t + 14205t – 681840 = 0
By taking out common we get,
45t(t – 48) + 14205 (t – 48) = 0
⇒ (t – 48) (45t + 14205) = 0
Then,
t – 48 = 0, 45t + 14205 = 0
⇒ t = 48, t = -14205/45
So, the number of months cannot be negative,
Therefore, t = 48 months i.e. 4 years.
Exercise 3.2
1. Mr. Chhabra deposits Rs 500 per month in a recurring deposit account for 4 years at a simple interest rate of 6% p.a.
(a) Find the maturity value of deposit.
(b) Find the total interest she will earn after 2 years
Answer
Given that recurring deposit per month = Rs 500, Period = 4 years = 48 months,
R = 6%
2. Mrs Khandekar invests Rs 900 every month in a recurring deposit account for a period of 3 years at a simple interest rate of 8% p.a.
(a) Find the total interest she will earn at the end of the period.
(b) Find the maturity value of her deposits.
Answer
Given that Recurring deposit per month = Rs 900, Period = 3 years = 36 months,
R = 8%
3. Mr. Patel deposits Rs 2,250 per month in a recurring deposit account for a period of 3 years. At the time of maturity, he gets Rs 90,990.
(a) Find the rate of simple interest per annum.
(b) Find the total interest earned by Mr. Patel.
Answer
Given that Recurring deposit per month = Rs 2,250, Period = 3 years = 36
4. Mr. Menon deposit Rs 1,200 per month in a cumulative deposit account for a period of 5 years. After the end of period, he will receive Rs 88,470.
(a) Find the rate of the interest per annum.
(b) Find the total interest that Mr. Menon will earn.
Answer
Given that Recurring deposit per month = Rs 1,200, Period = 5 years = 60 months, R = R%, Maturity value = Rs 88,470
5. Aarushi has a recurring deposit account for 2 years at 6% p.a. She receive Rs 1,125 as interest on maturity.
(a) Find the monthly instalment amount.
(b) Find the maturity amount.
Answer
Given that Recurring deposit per month = P, Period = 2 years = 24 months,
6. Mr. Mohan has a cumulative deposit account for 3 years 7% interest p.a. She receives Rs 8,547 as a maturity amount after 3 years.
(a) Find the monthly deposit.
(b) Find the total interest receivable after maturity.
Answer
Given that cumulative deposit per month = P, Period = 3 years = 36 Months,
R = 7%, Maturity amount = Rs 8,547
7. Mr. Banerjee opens a recurring deposit account for Rs 3,000 per month at 9% simple interest p.a. On maturity, he gets Rs 1,70, 460. Find the period for which he continued with the account.
Answer
Given that cumulative deposit per month = Rs 3,000, Period = t Months, R = 9%
Maturity amount = Rs 1,70, 460
8. Mr. Saha opened a cumulative deposit account of monthly installment of Rs 1,200 at 9% p.a. simple interest. She earned a total interest of Rs 5,328. How much installments did she pay?
Answer
Given that cumulative deposit per month = Rs 1200, Period = t Months, Rs = 9%
9. What should be minimum monthly deposit in a recurring deposit for 3 years to get Rs 20,220 on maturity after 3 years at an interest rate of 8%. P.a.
Answer
Given that Recurring deposit per month = P, Period = 3 years = 36 months
10. Mr. Pradip deposited Rs 2,400 pm in bank for One year 6 months under the recurring deposit scheme. If the maturity value of his deposit is Rs 47,304 find the rate of interest per annum.
Answer
Given that Recurring deposit per month = Rs 2400, Period = 3 years = 18
Months, R = R%, Maturity value = Rs. 47304
Money deposited = Monthly value × No. of Months = 2400 ×18 = Rs 43,200
⇒ Interest = Maturity Value - Amount deposited
= Rs. (47,304 - 43,200) = Rs. 4,104
Total Principal for 1 Month = 2400× (18)(18+1)/2 = Rs. 4,10,400
⇒ 4104 = 4,10,400 × R/1200
⇒ R= 12%