Frank Solutions for Class 10 Maths Chapter 3 Banking covers all the exercise questions in the Frank textbook. We, at BYJUâ€™S, ensure that you get the best Frank Solutions study material that can help you excel in your studies with a wellorganised way of learning. These Frank Solutions for Class 10 Maths are prepared by our welltrained 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.
Download the PDF of Frank Solutions for Class 10 Maths Chapter 3 Banking
Access answers to Frank Solutions for Class 10 Maths Chapter 3 Banking
1. Mr. Burman open a saving back account with Bank of India on 3^{rd}Â April 2007 with a cash deposit of Rs 5,000/. Subsequently, he deposited Rs 16,500/ by cheque on 11^{th}Â April 2007, withdraw Rs 4,000/ on 10^{th}Â May, paid Rs 3,500 for insurance by cheque on 7^{th}Â July 2007, deposited Rs. 6,000/ in cash on 9^{th}Â August 2007 and withdrew Rs 1,500/ on 12^{th}Â Oct 2007.Â
(a) Make the entries in his passbook
(b) If he closed the account on 14^{th}Â 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 14^{th}Â 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 10^{th} 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 = â‚¹ 1,31,000
Interest = (1,31,000 Ã— 4 Ã— 1)/(100 Ã— 12)
= â‚¹ 437
So, while closing the account,
Mr. Burman will get = principal + interest which amounts to
= â‚¹ 18,500 + â‚¹ 437
= â‚¹ 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.Â
(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 10^{th} 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,
= â‚¹ 11,000 + â‚¹ 7,800 + â‚¹(14,600 Ã— 4)
= â‚¹ 77,200
Interest at the end of September 2007 = (77,200 Ã— 5 Ã— 1)/(100 Ã— 12)
= â‚¹ 321.66
So, interest = â‚¹ 322
Then, again principal at the end of March 2008 = 14920 Ã— 6
= â‚¹ 89520
Interest at the end of March 2008 = (89520 Ã— 5 Ã— 1)/(100 Ã— 12)
= â‚¹ 373
Therefore, Account balance as on 01.04.2008 = â‚¹ 14920 + â‚¹ 373
= â‚¹ 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, findÂ
(a) The principal amount in January, February and March which will be considered for interest for interest calculation.Â
(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 â‚¹ 10,700.00 as this is minimum of 10^{th} and 31^{st} January
In February â‚¹ 5,400.00 as this is minimum of 10^{th} and 28^{th} February
In March â‚¹ 24,600.00 as this is minimum of 10^{th} and 31^{st} March (including 17^{th} March)
(b) The interest she gets at the end of March given below,
Months 
Minimum balance between 10^{th} 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 = â‚¹ 40,700
Interest = (40,700 Ã— 6 Ã— 1)/(100 Ã— 12)
= â‚¹ 203.50
Hence interest = â‚¹ 204
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 10^{th} 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= â‚¹ 80,500
So, Interest = (80,500 Ã— 6 Ã— 1)/(100 Ã— 12)
= â‚¹ 402.50
= â‚¹ 403
Hence, the interest paid to Mr. Rajesh by the bank at 6% pa in the end of June is â‚¹ 403
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 â€“ 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 10^{th} 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 = â‚¹ 1,38,000
So, Interest = (138000 Ã— 5 Ã— 1)/(100 Ã— 12)
= â‚¹ 575
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 10^{th} day and the last day is mentioned in the table.
Months 
Minimum balance between 10^{th} 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 = â‚¹ 1,67,075
So, Interest = (1,67,075 Ã— 5 Ã— 1)/(100 Ã— 12)
= â‚¹ 696.14
Hence, interest is â‚¹ 696
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 10^{th} day and the last day is mentioned in the table.
Months 
Minimum balance between 10^{th} day and the last day 
January 
14,200 
February 
11,750 
March 
27,350 
Total principal for at the end of March = â‚¹ 53,300
So, Interest at the end of March = (53,300 Ã— 4.5 Ã— 1)/(100 Ã— 12)
= â‚¹ 199.87
Hence, interest is â‚¹ 200
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 10^{th} 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 = â‚¹ 2,16,000
So, Interest at the end of September = (2,16,000 Ã— 4.5 Ã— 1)/(100 Ã— 12)
= â‚¹ 810
Therefore, by entering the interest in the passbook above we get the balance â‚¹ 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 25^{th} 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 10^{th} day and the last day 
January 
39,040 
February 
31,440 
March 
35,790 
Total principal for at the end of March = â‚¹ 1,06,270
So, Interest at the end of March = (1,06,270 Ã— 4.5 Ã— 1)/(100 Ã— 12)
= â‚¹ 398.51
Hence, interest is â‚¹ 399
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 â‚¹ 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.Â
(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 â‚¹ 500 per month in a recurring deposit
Period = 4 years
We know that, 1 year = 12 Months
So, 4 years = 4 Ã— 12 = 48 Months
Rate = 6 % pa
Then, Money deposited = Monthly value Ã— Number of Months
= â‚¹ 500 Ã— 48
= â‚¹ 24,000 â€¦ [equation i]
So, total principal for 1 month = [500 Ã— (48(48 + 1))]/2
= â‚¹ 5,88,000
Now, interest = (6 Ã— 5,88,000)/(12 Ã— 100)
= â‚¹ 2,940 â€¦ [equation ii]
For getting maturity amount we have to add both equation (i) and equation (ii)
= â‚¹ 24,000 + â‚¹ 2,940
= â‚¹ 26,940
Therefore, Maturity amount is â‚¹ 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 â‚¹ 900 every month in a recurring deposit account.
Period = 3 years
We know that, 1 year = 12 Months
So, 3 years = 3 Ã— 12 = 36 Months
Rate = 8 % pa
Then, Money deposited = Monthly value Ã— Number of Months
= â‚¹ 900 Ã— 36
= â‚¹ 32,400 â€¦ [equation i]
So, total principal for 1 month = [900 Ã— (36(36 + 1))]/2
= â‚¹ 5,99,400
Now, interest = (8 Ã— 5,99,400)/(12 Ã— 100)
= â‚¹ 3,996 â€¦ [equation ii]
For getting maturity amount we have to add both equation (i) and equation (ii)
= â‚¹ 32,400 + â‚¹ 3,996
= â‚¹ 36,396
Therefore, Maturity amount is â‚¹ 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 â‚¹ 2,250 per month in a recurring deposit account.
Period = 3 years
We know that, 1 year = 12 Months
So, 3 years = 3 Ã— 12 = 36 Months
Maturity = â‚¹ 90,990
Rate = R % pa
Then, Money deposited = Monthly value Ã— Number of Months
= â‚¹ 2,250 Ã— 36
= â‚¹ 81,000
Interest get for this period = Maturity amount â€“ Amount deposited
= 90,990 â€“ 81,000
= â‚¹ 9,990
So, total principal for 1 month = [2,250 Ã— (36(36 + 1))]/2
= â‚¹ 14,98,500
Now, interest = (R Ã— 14,98,500)/(12 Ã— 100)
9,990 = (R Ã— 14,98,500)/(12 Ã— 100)
R = (9,990 Ã— 12 Ã— 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 Ã— 12 = 60 Months
Maturity = â‚¹ 88,470
Rate = R % pa
Then, Money deposited = Monthly value Ã— Number of Months
= â‚¹ 1,200 Ã— 60
= â‚¹ 72,000
Interest get for this period = Maturity amount â€“ Amount deposited
= 88,470 â€“ 72,000
= â‚¹ 16,470
So, total principal for 1 month = [1,200 Ã— (60(60 + 1))]/2
= â‚¹ 21,96,000
Now, interest = (R Ã— 21,96,000)/(12 Ã— 100)
16,470 = (R Ã— 21,96,000)/(12 Ã— 100)
R = (16,470 Ã— 12 Ã— 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.Â
(a) Find the monthly instalment amount.Â
(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 Ã— 12 = 24 Months
Rate = 6 % pa
Then, Money deposited = Monthly value Ã— Number of Months
= P Ã— 24
= â‚¹ 24P
So, total principal for 1 month = [P Ã— (24(24 + 1))]/2
= â‚¹ 300P
Now, interest = (Total principal for 1 month Ã— R)/(12 Ã— 100)
â‚¹ 1,125 = (300P Ã— 6)/(12 Ã— 100)
P = (1,125 Ã— 12 Ã— 100)/(300 Ã— 6)
P = â‚¹ 750
For getting maturity amount = (P Ã— 24) + Interest
= (750 Ã— 24) + 1125
Therefore, Maturity amount is â‚¹ 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.Â
(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 Ã— 12 = 36 Months
Rate = 7 % pa
Maturity amount = â‚¹ 8,547
Then, Money deposited = Monthly value Ã— Number of Months
= P Ã— 36
= â‚¹ 36P
So, total principal for 1 month = [P Ã— (36(36 + 1))]/2
= â‚¹ 666P
Now, interest = (Total principal for 1 month Ã— R)/(12 Ã— 100)
â‚¹ 8,547 â€“ 36P= (666P Ã— 7)/(12 Ã— 100)
â‚¹ 8,547 â€“ 36P = 3.885P
â‚¹ 8,547 = 3.885P + 36P
â‚¹ 8,547 = 39.885p
P = â‚¹ 8,547/39.885
P = â‚¹ 214.3
Interest amount = 8547 â€“ 36P
Substitute the value of P we get,
Interest amount = 8547 â€“ (36 Ã— 214.3)
= â‚¹ 832
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 â‚¹ 3,000 per month.
Period = t
Rate = 9% pa
Maturity amount = â‚¹ 1,70,460
Then, Money deposited = Monthly value Ã— Number of Months
= 300 Ã— t
= â‚¹ 3000t
So, total principal for 1 month = [3000 Ã— (t(t + 1))]/2
= 1500 (t^{2} + t)
= 1500t^{2} + 1500t
Now, interest = (Total principal for 1 month Ã— R)/(12 Ã— 100)
â‚¹ 1,70,460 â€“ 3000t= ((1500t^{2} + 1500t) Ã— 9)/(12 Ã— 100)
â‚¹ 1,70,460 â€“ 3000t = (45t^{2} + 45t)/4
By cross multiplication we get,
681840 â€“ 12000t = 45t^{2} + 45t
Then transposing we get,
45t^{2} + 45t + 12000t â€“ 681840 = 0
45t^{2} + 12045t â€“ 681840 = 0
45t^{2} â€“ 2160t + 14205t â€“ 681840 = 0
By taking out common we get,
45t(t – 48) + 14205 (t – 48) = 0
(t – 48) (45t + 14205) = 0
Then,
t â€“ 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.