Frank Solutions for Class 10 Maths Chapter 3 Banking

Frank Solutions for Class 10 Maths Chapter 3 Banking covers all the exercise questions in the Frank textbook. We, at BYJU’S, ensure that you get the best Frank Solutions study material that can help you excel in your studies with a well-organised way of learning. These Frank Solutions for Class 10 Maths are prepared by our well-trained and qualified subject experts and are provided here to assist students in learning the concepts quickly and with precision.

Chapter 3 – Banking is a world of numbers and Mathematics is used in the way accounts are handled for calculating interest rates and for determining credits scores. In the Frank Solutions for Class 10 Maths Chapter 3, students are going to study more about parameters in the passbook and also different types of problems.

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1. Mr. Burman open a saving back account with Bank of India on 3rd April 2007 with a cash deposit of Rs 5,000/-. Subsequently, he deposited Rs 16,500/- by cheque on 11th April 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.

Solution:-

(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.

Date

Particular

Withdrawals

Deposits

Balance

03.04.2007

By Cash

5,000.00

5,000.00

11.04.2007

By Cheque

16,500.00

21,500.00

10.05.2007

To Self

4,000.00

17,500.00

07.07.2007

By Cheque

3,500.00

14,000.00

09.08.2007

By Cash

6,000.00

20,000.00

12.10.2007

To Self

1,500.00

18,500.00

(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.

Months

Minimum balance between 10th day and the last day

April

5,000

May

17,500

June

17,500

July

14,000

August

20,000

September

20,000

October

18,500

November

18,500

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

Solution:-

(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.

Date

Particular

Withdrawals

Deposits

Balance

05.04.2007

By Cheque

11,000.00

11,000.00

12.05.2007

To Self

3,200.00

7,800.00

03.06.2007

By Cheque

8,800.00

16,600.00

18.06.2007

To Cheque

2,000.00

14,000.00

(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.

Months

Minimum balance between 10th day and the last day

Minimum balance in nearest multiple of 10

2007, April

11,000

11,000

May

7,800

7,800

June

14,600

14,600

July

14,600

14,600

August

14,600

14,600

September

14,600

14,600

October

14,600 + 322 = 14,922

14,920

November

14,922

14,920

December

14,922

14,920

2008, January

14,922

14,920

February

14,922

14,920

March

14,922

14,920

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.

Date

Particular

Withdrawals

Deposits

Balance

05.01.2008

By Cash

15,500.00

15,000.00

10.01.2008

To Cheque

4,800.00

10,700.00

15.02.2008

To Cheque

5,300.00

5,400.00

08.03.2008

By Cash

19,200.00

24,600.00

17.03.2008

To Cheque

7,400.00

32,000.00

Solution:-

(a) As per the condition given in the question, the principal amount in January, February and March which will be considered for interest for interest calculation

Date

Particular

Withdrawals

Deposits

Balance

Qualifying amount

05.01.2008

By Cash

15,500.00

15,000.00

15,500.00

10.01.2008

To Cheque

4,800.00

10,700.00

10,700.00

15.02.2008

To Cheque

5,300.00

5,400.00

5,400.00

08.03.2008

By Cash

19,200.00

24,600.00

24,600.00

17.03.2008

To Cheque

7,400.00

32,000.00

32,000.00

So,

In January ₹ 10,700.00 as this is minimum of 10th and 31st January

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,

Months

Minimum balance between 10th day and the last day

January

10,700.00

February

5,400.00

March

24,600.00

Therefore, total principal for at the end of March = ₹ 40,700

Interest = (40,700 × 6 × 1)/(100 × 12)

= ₹ 203.50

Hence interest = ₹ 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.

Date

Particular

Withdrawals

Deposits

Balance

08.02.2008

By Cash

12,000.00

15.03.2008

To Cash

3,000.00

08.04.2008

To Cheque

2,500.00

18.04.2008

By Cash

16,000.00

10.06.2008

By Cash

8,000.00

Solution:-

From the given table,

Date

Particular

Withdrawals

Deposits

Balance

08.02.2008

By Cash

12,000.00

12,000.00

15.03.2008

To Cash

3,000.00

9,000.00

08.04.2008

To Cheque

2,500.00

6,500.00

18.04.2008

By Cash

16,000.00

22,500.00

10.06.2008

By Cash

8,000.00

30,500.00

Then,

Months

Minimum balance between 10th day and the last day

February

12,000

March

9,000

April

6,500

May

22,500

June

30,500

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.

Date

Particular

Withdrawals

Deposits

Balance

01.04.2007

By B/F

16,500.00

15.04.2007

By Cash

2,500.00

09.06.2007

To Cheque

6,500.00

04.07.2007

By Cash

9,000.00

12.07.2007

To Cash

3,500.00

05.09.2007

To Cash

4,000.00

10.11.2007

By Cheque

12,000.00

Solution:-

We know that, Balance = Previous Balance + Deposit – Withdrawal

Date

Particular

Withdrawals

Deposits

Balance

01.04.2007

By B/F

16,500.00

15.04.2007

By Cash

2,500.00

19,000.00

09.06.2007

To Cheque

6,500.00

12,500.00

04.07.2007

By Cash

9,000.00

21,500.00

12.07.2007

To Cash

3,500.00

18,000.00

05.09.2007

To Cash

4,000.00

14,000.00

10.11.2007

By Cheque

12,000.00

26,000.00

Then, interest earned by account holder in the month of November

Months

Minimum balance between 10th day and the last day

April

16,500

May

19,000

June

12,500

July

18,000

August

18,000

September

14,000

October

14,000

November

26,000

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.

Date

Particular

Withdrawals

Deposits

Balance

01.01.2007

By B/F

6,500.00

05.02.2007

By Cheque

7,500.00

14,000.00

09.02.2007

To Cash

1,500.00

12,500.00

06.06.2007

By Cash

1,725.00

14,225.00

08.09.2007

By Cheque

375.00

14,600.00

06.11.2007

By Cash

20,600.00

10.12.2007

To Cheque

2,500.00

6,000.00

18,100.00

Solution:-

Minimum balance between 10th day and the last day is mentioned in the table.

Months

Minimum balance between 10th day and the last day

January

6,500

February

12,500

March

12,500

April

12,500

May

12,500

June

14,225

July

14,225

August

14,225

September

14,600

October

14,600

November

20,600

December

18,100

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.

Date

Particular

Withdrawals

Deposits

Balance

03.01.2006

By B/F

17,900.00

09.01.2006

To Cash

3,700.00

14,200.00

06.02.2006

To Cheque

2,450.00

11,750.00

21.02.2006

By Cash

15,600.00

27,350.00

17.03.2006

By Cash

9,850.00

37,200.00

31.03.2006

By Interest

06.06.2006

To Cheque

4,100.00

22.08.2006

To Cash

1,500.00

05.09.2006

By Cheque

17,300.00

09.09.2006

To Cash

6,300.00

30.09.2006

By Interest

04.12.2006

To Cash

3,000.00

11.12.2006

By Cheque

11,760.00

Solution:-

Minimum balance between 10th day and the last day is mentioned in the table.

Months

Minimum balance between 10th day and the last day

January

14,200

February

11,750

March

27,350

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,

Date

Particular

Withdrawals

Deposits

Balance

03.01.2006

By B/F

17,900.00

09.01.2006

To Cash

3,700.00

14,200.00

06.02.2006

To Cheque

2,450.00

11,750.00

21.02.2006

By Cash

15,600.00

27,350.00

17.03.2006

By Cash

9,850.00

37,200.00

31.03.2006

By Interest

200.00

37,400.00

06.06.2006

To Cheque

4,100.00

33,300.00

22.08.2006

To Cash

1,500.00

31,800.00

05.09.2006

By Cheque

17,300.00

49,100.00

09.09.2006

To Cash

6,300.00

42,800.00

30.09.2006

By Interest

810.00

43,610.00

04.12.2006

To Cash

3,000.00

40,610.00

11.12.2006

By Cheque

11,760.00

52,370.00

Now, interest calculating at the end of September.

Months

Minimum balance between 10th day and the last day

April

37,400

May

37,400

June

33,300

July

33,300

August

31,800

September

42,800

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 25th July 08.

Date

Particular

Withdrawals

Deposits

Balance

05.01.2008

By B/F

24,650.00

09.01.2008

By Cash

14,390.00

39,040.00

15.02.2008

To Cheque

7,600.00

31,440.00

21.02.2008

By Cheque

8,350.00

39,790.00

07.03.2008

To Cash

4,000.00

35,790.00

31.03.2008

By Interest

08.04.2008

By Cheque

13,670.00

12.04.2008

To Cash

6,000.00

01.05.2008

By Cheque

17,350.00

16.06.2008

By Cash

9,000.00

27.06.2008

To Cash

4,370.00

04.07.2008

By Cheque

21,320.00

11.07.2008

To Cheque

9,460.00

Solution:-

Months

Minimum balance between 10th day and the last day

January

39,040

February

31,440

March

35,790

Total principal for at the end of March = ₹ 1,06,270

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,

Date

Particular

Withdrawals

Deposits

Balance

05.01.2008

By B/F

24,650.00

09.01.2008

By Cash

14,390.00

39,040.00

15.02.2008

To Cheque

7,600.00

31,440.00

21.02.2008

By Cheque

8,350.00

39,790.00

07.03.2008

To Cash

4,000.00

35,790.00

31.03.2008

By Interest

399.00

36,189.00

08.04.2008

By Cheque

13,670.00

49,859.00

12.04.2008

To Cash

6,000.00

43,859.00

01.05.2008

By Cheque

17,350.00

61,209.00

16.06.2008

By Cash

9,000.00

70,209.00

27.06.2008

To Cash

4,370.00

65,839.00

04.07.2008

By Cheque

21,320.00

87,159.00

11.07.2008

To Cheque

9,460.00

77,699.00

Therefore, Net money that Mr. Punjwani will get ₹ 77,699

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

Solution:-

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.

Solution:-

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.

Solution:-

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.

Solution:-

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.

Solution:-

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.

Solution:-

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.

Solution:-

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.

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