Question 1:  Given the following details, calculate simple interest at the rate of 6% per annum up to June 30.      [ICSE2003]

Date Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
Jan 1 – 24000 24000
Jan 20 5000 – 19000
Jan 29 – 10000 29000
March 15 8000 37000
April 3 7653 44653
May 6 3040 – 41613
May 8 – 5087 46700

Answer:

Qualifying principal for various months:

Month Principal (Rs.)
January 19000
February 29000
March 29000
April 44653
May 46700
June 46700
Total 215060

\displaystyle P = Rs. \ 215060 \ \ R = 6\% \ and \  T= \frac{1}{12}
\displaystyle I = P \times R \times T = 215060 \times \frac{6}{100} \times \frac{1}{12} = Rs. \  1075.27
\\

Question 2: Mr. Ashok has an account in the Central Bank of India. The following entries are from his passbook:-

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
01.01.05 B/F 1200
07.01.05 By Cash 500 1700
17.01.05 To Cheque 4000 1300
10..02.05 By Cash 800 2100
25.02.05 To Cheque 500 1600
20.09.05 By Cash 700 2300
21.11.05 To Cheque 600 1700
05.12.05 By Cash 300 2000

If Mr. Ashok gets Rs. 83.75 as interest at the end of the year, where the interest rate is compounded annually, calculate the rate of interest paid by the bank in his Savings Bank Account on 31st December 2005.      [ICSE2006]

Answer:

Qualifying principal for various months:

Month Principal (Rs.)
January 1300
February 1600
March 1600
April 1600
May 1600
June 1600
July 1600
August 1600
September 1600
October 2300
November 1700
December 2000
Total 20100

\displaystyle P = Rs. \ 20100 \ \ Rate = r\% \ and \  T= \frac{1}{12} , \ Interest = Rs. \ 83.75
\displaystyle I = P \times R \times T
\displaystyle 83.75= 20100 \times \frac{r}{100} \times \frac{1}{12} \Rightarrow r = 5 \%

Question 3: Kiran deposited Rs. 200 per month for 36 months in a bank’s recurring deposit account. If the bank pays interest at the rate of 11% per annum, find the amount she gets on maturity.      [ICSE2012]

Answer:

\displaystyle P = Rs. \ 200, \ no \ of \ months = 36, \ r = 11\%
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle  =200 \times 36 +200 \times \frac{36(36+1)}{2 \times 12} \times \frac{11}{100} = Rs. 8421

Question 4: Mohan deposited Rs. 80 per month in a cumulative deposit account for 6 years. Find the amount payable to him on maturity, it the rate of interest is 6% per annum.      [ICSE2006]

Answer:

\displaystyle P = Rs. \ 80, \ no \ of \ months = 72, \ r = 6\%
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle =80 \times 72 +80 \times \frac{72(72+1)}{2 \times 12} \times \frac{6}{100} = Rs. 6811.20

Question 5: Mr. R. K. Nair gets Rs. 6455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly installment.      [ICSE2005]

Answer:

\displaystyle P = Rs. \ x, \ no \ of \ months = 12, \ rate = 14\%  \ Maturity Amount = Rs. 6455
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 6455 =x \times 12 +x \times \frac{12(12+1)}{2 \times 12} \times \frac{14}{100}  
\displaystyle x = \frac{6455}{12.91} = Rs. 500
He must deposit Rs. 500 every month.

Question 6: Ahmed has a recurring deposit account in a bank He deposits Rs. 2500 per month for 2 years. If he gets Rs. 66250 at the time of maturity, find: i) interest paid by the bank ii) rate of interest.       [ICSE2011]

Answer:

\displaystyle P = Rs. \ 2500, \ no \ of \ months = 24, \ rate = r\%  \ Maturity Amount = Rs. 66250
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 66250 =2500 \times 24 +2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100} \Rightarrow r=10\%
\displaystyle Interest = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle = 2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{10}{100} = 6250

Question 7: The entries in a Saving Bank passbook are given below:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
01.01.03 B/F     14,00.00
01.02.03 By Cash   11,500.00 25,500.00
12.02.03 To Cheque 5,000   20,500.00
05.04.03 By Cash   3,750.00 24,250.00
15.04.03 To Cheque 4,250.00   20,000.00
09.05.03 By Cash   1,500.00 21,500.00
04.06.03 By Cash   1,500.00 23,000.00

Calculate the interest for six months (January to June) at 4% per Annum on the minimum balance on or after the tenth day of each month.      [ICSE2004]

Answer:

Qualifying principal for various months:

Month Principal (Rs.)
January 14000
February 20500
March 20500
April 20000
May 21500
June 23000

Total = Rs. 119500
\displaystyle P = Rs. \ 119500 \ \ R = 4\% \ and \ T= \frac{1}{12}
\displaystyle I = P \times R \times T = 119500 \times \frac{4}{100} \times \frac{1}{12} = Rs. \  398.33

Question 8: A page from the passbook of Mrs. Rama Bhalla is given below:

Date Year 2004 Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
January 1 B/F     2,000.00
January 9 By Cash   200.00 2,200.00
February 10 To Cheque 500.00   1,700.00
February 24 By Cheque   300.00 2,000.00
July 29 To Cheque 200.00   1,800.00
November 7 By Cash   300.00 2,100.00
December 8 By Cash   200.00 2,300.00

Calculate the interest to Mrs. Rama Bhalla for the period of January 2004 to December 2004, at the rate of 5% per annum.      [ICSE2005]

Answer:

Qualifying principal for various months:

Month Principal (Rs.)
January 2200
February 1700
March 2000
April 2000
May 2000
June 2000
July 1800
August 1800
September 1800
October 1800
November 2100
December 2300
Total 23500

\displaystyle P = Rs. \ 23500 \ \ R = 5\% \ and \  T= \frac{1}{12}
\displaystyle I = P \times R \times T = 23500 \times \frac{5}{100} \times \frac{1}{12} = Rs. \  97.92

Question 9: A page from Saving Bank account of Mr. Prateek is given below:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
January 1st 2006 B/F – – 1,270
January 7th 2006 By Cheque – 2,310 3,580
March 9th 2006 To Self 2,000 – 1,580
March 26th 2006 By Cash – 6,200 7,780
June 10th 2006 To Cheque 4,500 – 3,280
July 15th 2006 By Clearing – 2,630 5,910
October 18th 2006 To Cheque 530 – 5,380
October 27th 2006 To Self 2,690 – 2,690
November 3rd 2006 By Cash – 1,500 4,190
December 6th 2006 To Cheque 950 – 3,240
December 23rd 2006 By Transfer – 2,920 6,160

If he receives Rs.198 as interest on 1st January 2007. Find the rate of interest paid by the bank.      [ICSE2012]

Answer:

Qualifying principal for various months:

Month Principal (Rs.)
January 3580
February 3580
March 1580
April 7780
May 7780
June 3280
July 3280
August 5910
September 5910
October 2690
November 4190
December 3240
Total 52800

\displaystyle P = Rs. \ 52800 \ \ R = x\% \ and \  T= \frac{1}{12} I = Rs. \ 198
\displaystyle I = P \times R \times T \Rightarrow 52800 \times \frac{x}{100} \times \frac{1}{12} = 198 \Rightarrow x = 4.5 \%

Question 10: Mrs. Kapoor opened a Saving Bank Account in State Bank of India on 9th January 2008. Her passbook entries for the year 2008 are given below:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
Jan. 9, 2008 By Cash – 10,000 10,000
Feb. 12, 2008 By Cash – 15,500 25,500
April 6, 2008 To Cheque 3,500 – 22,000
April 30, 2008 To Self 2,000 – 20,000
July 16, 2008 By Cheque – 6,500 26,500
Aug. 4, 2008 To Self 5,500 – 21,000
Aug. 20, 2008 To Cheque 1,200 – 19,800
Dec. 12, 2008 By Cash – 1,700 21,500

Mrs. Kapoor closed the account on 31st December 2008. If the bank pays interest at 4% per annum, find the interest he receives on closing the account. Give your answer correct to the nearest rupee.        [ICSE2010]

Answer:  Qualifying principal for various months: 

Month Principal (Rs.)
January 10000
February 10000
March 25500
April 20000
May 20000
June 20000
July 20000
August 19800
September 19800
October 19800
November 19800
Total 204700

\displaystyle P = Rs. \ 204700 \ \ R = 4.0\% \ and \  T= \frac{1}{12}
\displaystyle I = P \times R \times T = 204700 \times \frac{4}{100} \times \frac{1}{12} = Rs. \  682.33 or Rs. 682

Question 11: Explain the following:

i) Punnet has a recurring deposit account in Bank of Baroda and deposits Rs.140 per month for 4 years. If he gets Rs.8,092 on maturity, find the rate of interest given by the bank.

ii) David opened a recurring deposit account in a bank and deposited Rs.300 per month for two years. If he received Rs.7,725 at the time of maturity, find the rate of interest per annum.       [ICSE2008]

Answer:

i)
\displaystyle P = Rs. \ 140, \ no \ of \ months = 48, \ rate = r\%  \ Maturity Amount = Rs. 8092
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 8092 =140 \times 48 +140 \times \frac{48(48+1)}{2 \times 12} \times \frac{r}{100}  
\displaystyle r = \frac{(8092-140 \times 48) \times (2 \times 12) \times 100}{140 \times 48 \times 49} \Rightarrow r=10\%
ii)
\displaystyle P = Rs. \ 300, \ no \ of \ months = 24, \ rate = r\%  \ Maturity Amount = Rs. 7725
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 7725 =300 \times 24 +300 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100}  
\displaystyle r = \frac{(7725-300 \times 24) \times (2 \times 12) \times 100}{300 \times 24 \times 25} \Rightarrow r=7\%

Question 12: Amit deposited 150 per month in a bank for 8 month under the recurring deposit scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month?      [ICSE2001, 2007]

Answer:

\displaystyle P = Rs. \ 150, \ no \ of \ months = 8, \ r = 8\%
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle =150 \times 8 +150 \times \frac{8(8+1)}{2 \times 12} \times \frac{8}{100} = Rs. 1236

Question 13: Mr. Gupta opened a recurring deposit account in a bank. He deposited Rs. 2,500 per month for two years. At the time of maturity, he got Rs.67,500. Find:

  • The total interest earned by Mr. Gupta
  • The rate of interest per annum.      [ICSE2010]

Answer:

\displaystyle P = Rs. \ 2500, \ no \ of \ months = 24, \ rate = r\%  \ Maturity Amount = Rs. 67500
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 67500 =2500 \times 24 +2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100}  
\displaystyle r = \frac{(67500-2500 \times 24) \times (2 \times 12) \times 100}{2500 \times 24 \times 25} \Rightarrow r=12\%
\displaystyle Interest =2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{12}{100} = Rs. \ 7500

Question 14: Given below are the entries in a saving Bank A/C passbook:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
Feb. 8 B/F – – 8,500
Feb. 18 To Self 4,000 – 4,500
April 12 By Cash – 2,230 6,730
June 15 To Self 5,000 – 1,730
July 8 By Cash – 6,000 7,730

Calculate the interest for six months from February to July at 6% p.a.      [ICSE2013]

Answer:

Qualifying principal for various months:

Month Principal (Rs.)
February 4500
March 4500
April 4500
May 6730
June 1730
July 7730
Total 29690

\displaystyle P = Rs. \ 29690 \ \ R = 6\% \ and \  T= \frac{1}{12}
\displaystyle I = P \times R \times T = 29690 \times \frac{6}{100} \times \frac{1}{12} = Rs. \  148.45

Question 15: Chaudhary opened a saving bank account at State Bank of India on 1st April 2007. The entries of one year as shown in his passbook are given below:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
1st April 2007 By Cash – 8,550.00 8,550.00
12th April 2007 To Self 1,200.00 – 7,350.00
24th April 2007 By Cash – 4,550.00 11,900.00
8th July 2007 By Cheque – 1,500.00 13,400.00
10th Sept. 2007 By Cheque – 3,500.00 16,900.00
17th Sept. 2007 To Cheque 2,500.00 – 14,400.00
11th Oct. 2007 By Cash – 800.00 15,200.00
6th Jan. 2008 To Self 2,000.00 – 13,200.00
9th March 2008 By Cheque – 950.00 14,150.00

If the bank pays interest at the rate of 5% per annum, find the interest paid on 1st April 2008. Give your answer correct to nearest rupee.      [ICSE2011]

Answer:

Qualifying principal for various months: 

Month Principal (Rs.)
April 7350
May 11900
June 11900
July 13400
August 13400
September 14400
October 14400
November 15200
December 15200
January 13200
February 13200
March 14150
Total 157700

\displaystyle P = Rs. \ 157700 \ \ R = 5\% \ and \  T= \frac{1}{12}
\displaystyle I = P \times R \times T = 157700  \times \frac{5}{100} \times \frac{1}{12} = Rs. \  657.08 or Rs. \ 657

Question 16: Bitto deposits a certain sum of money in a recurring deposit account of a Bank. If the rate of interest of 8% per annum and Mr. Bitto gets Rs.8,008 from the bank after 3 years, find the value of his monthly installment.      [ICSE2013]

Answer:

\displaystyle P = Rs. \ x, \ no \ of \ months = 36, \ rate = 8\%  \ Maturity Amount = Rs. 8008
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 8008 =x \times 36 +x \times \frac{36(36+1)}{2 \times 12} \times \frac{8}{100}  
\displaystyle x = \frac{8008}{40.44} = Rs. 198.02
He must deposit Rs. 200 every month.

Question 17: Shahrukh opened a recurring deposit account in a bank and deposited 800 per month for 1 ½ years. If he received Rs.15,084 at the time of maturity. Find the interest rate per annum.      [ICSE2014]

Answer:

\displaystyle P = Rs. \ 800, \ no \ of \ months = 18, \ rate = r\%  \ Maturity Amount = Rs. 15084
\displaystyle Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 15084 =800 \times 18 +800 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}  
\displaystyle r = \frac{(15084-800 \times 18) \times (2 \times 12) \times 100}{800 \times 18 \times 19} \Rightarrow r=6\%

Question 18: A page from the saving account of Priyanka is given below:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
03/04/2006 B/F     4000.00
05/04/2006 By Cash   2000.00 6000.00
18/04/2006 By Cheque   6000.00 12000.00
25/05/2006 To Cheque 5000.00   7000.00
30/05/2006 By Cash   3000.00 10000.00
20/07/2006 By Self 4000.00   6000.00
10/09/2006 By Cash   2000.00 8000.00
19/09/2006 To Cheque 1000.00   7000.00

If the interest earned by Priyanka for the period ending September 2006 is Rs.175, find the rate of interest.      [ICSE2014]

Answer:

Qualifying principal for various months: 

Month Principal (Rs.)
April 6000
May 7000
June 10000
July 6000
August 6000
September 7000
Total 42000

\displaystyle P = Rs. \ 42000 \ \ Rate = r\% \ and \  T= \frac{1}{12} , \ Interest = Rs. \ 175
\displaystyle I = P \times R \times T
\displaystyle 175= 42000 \times \frac{r}{100} \times \frac{1}{12} \Rightarrow r = 5 \%

Question 19: Mr. Dhoni has an account in the Union Bank of India. The following entries are from his passbook:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
Jan 3, 07 B/F – – 2642.00
Jan 16, 07 To Self 640.00 – 2002.00
March 5, 07 By Cash – 850.00 2852.00
April 10, 07 To Self 1130.00 – 1722.00
April 25, 07 By Check – 650.00 2372.00
June 15, 07 By Cash 577.00 – 1795.00

Calculate the interest from January 2007 to June 2007 at the rate od 4% per annum.      [ICSE2008]

Answer:

Qualifying principal for various months:

Month Principal (Rs.)
January 2002
February 2002
March 2852
April 1722
May 2372
June 1795
Total 12745

\displaystyle P = Rs. \ 12745 \ \ R = 4\% \ and \  T= \frac{1}{12}
\displaystyle I = P \times R \times T = 12745 \times \frac{4}{100} \times \frac{1}{12} = Rs. \  42.48

Question 20: Given below are the entries in a Saving Bank A/c passbook:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
Feb 8 B/F – – 8500
Feb, 18 To Self 4000 –
April, 12 By Cash – 2238
June, 15 To Self 5000 –
June, 8 By Cash – 6000  

Calculate the interest for the six months, February to July, at 4.5% per annum on the minimum balance on or after the 10th day of each month.           [ICSE2000, 2007]

Answer:

Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.)
Feb 8 B/F – – 8500
Feb, 18 To Self 4000 – 4500
April, 12 By Cash – 2238 6738
June, 15 To Self 5000 – 1738
June, 8 By Cash – 6000 7738

Qualifying principal for various months:

Month Principal (Rs.)
February 4500
March 4500
April 4500
May 6738
June 1738
July 7738
Total 29714

\displaystyle P = Rs. \ 29714 \ \ R = 4.5\% \ and \  T= \frac{1}{12}
\displaystyle I = P \times R \times T = 29714 \times \frac{4.5}{100} \times \frac{1}{12} = Rs. \  111.43

Question 21: Naveen deposits \displaystyle \text{Rs }800 every month in a recurring deposit account for 6 months. If he receives \displaystyle \text{Rs }4884 at the time of maturity, then the interest he earns is:
(a) \displaystyle \text{Rs }84
(b) \displaystyle \text{Rs }42
(c) \displaystyle \text{Rs }24
(d) \displaystyle \text{Rs }284
[ICSE 2023]
\displaystyle \text{Answer:}
\displaystyle (a)\ \text{Total money deposited in 6 months by Naveen}
\displaystyle =800 \times 6=\text{Rs }4800
\displaystyle \text{Money received at the time of maturity}=\text{Rs }4884
\displaystyle \therefore\ \text{The interest}=\text{Total received money}-\text{Total deposited money}
\displaystyle =\text{Rs }(4884-4800)=\text{Rs }84

Question 22: Mohit opened a recurring deposit account in a bank for 2 yr. He deposits \displaystyle \text{Rs }1000 every month and receives \displaystyle \text{Rs }25500 on maturity. The interest he earned in 2 yr is:
(a) \displaystyle \text{Rs }13500
(b) \displaystyle \text{Rs }3000
(c) \displaystyle \text{Rs }24000
(d) \displaystyle \text{Rs }1500
[ICSE Semester I 2022]
\displaystyle \text{Answer:}
\displaystyle (d)\ \text{Given, }P=\text{Rs }1000\ \text{and }n=24\ \text{months}
\displaystyle \text{Maturity value (MV)}=\text{Rs }25500
\displaystyle \therefore\ MV=P \times n+I
\displaystyle \Rightarrow 25500=1000 \times 24+I
\displaystyle \Rightarrow I=25500-24000
\displaystyle \Rightarrow I=\text{Rs }1500

Question 23: A man deposited \displaystyle \text{Rs }500 per month for 6 months and received \displaystyle \text{Rs }3300 as the maturity value. The interest received by him is:
(a) \displaystyle \text{Rs }1950
(b) \displaystyle \text{Rs }300
(c) \displaystyle \text{Rs }2800
(d) None of these
[ICSE Semester I 2022]
\displaystyle \text{Answer:}
\displaystyle (b)\ \text{Man deposited per month}=\text{Rs }500
\displaystyle \text{So, total deposit in 6 months}=6 \times 500=\text{Rs }3000
\displaystyle \text{and money received}=\text{Rs }3300
\displaystyle \text{Hence, the interest received}=\text{Rs }(3300-3000)=\text{Rs }300

Question 24: A man deposited \displaystyle \text{Rs }1200 in a recurring deposit account for 1 yr at 5% per annum simple interest. The interest earned by him on maturity is:
(a) \displaystyle \text{Rs }14790
(b) \displaystyle \text{Rs }390
(c) \displaystyle \text{Rs }4680
(d) \displaystyle \text{Rs }780
[ICSE Semester I 2022]
\displaystyle \text{Answer:}
\displaystyle (b)\ \text{Given, principal (P)}=\text{Rs }1200
\displaystyle \text{Time period (n)}=1\ \text{yr}=12\ \text{months}
\displaystyle \text{and Rate of interest (r)}=5\%
\displaystyle \text{So, interest received by him}
\displaystyle =P \times \frac{n(n+1)}{2} \times \frac{r}{12 \times 100}
\displaystyle =1200 \times \frac{12 \times 13}{2} \times \frac{5}{12 \times 100}=\text{Rs }390

Question 25 : Suresh has a recurring deposit account in a bank. He deposits \displaystyle \text{Rs }2000 per month and the bank pays interest at the rate of 8% per annum. If he gets \displaystyle \text{Rs }1040 as interest at the time of maturity, find in years the total time for which the account was held.      [ICSE 2024]
\displaystyle \text{Answer:}
\displaystyle 5.\ \text{Given, amount deposited by Suresh}=\text{Rs }2000\ \text{per month}
\displaystyle \text{Rate of interest paid by bank}=8\%\ \text{per annum}
\displaystyle \text{Amount of interest}=\text{Rs }1040
\displaystyle \therefore\ \text{Interest }(I)=\frac{P n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle 1040=\frac{2000 \times n(n+1)}{24} \times \frac{8}{100}
\displaystyle \Rightarrow \frac{1040 \times 3}{20}=n(n+1)
\displaystyle \Rightarrow 52 \times 3=n(n+1)=156=n(n+1)
\displaystyle \Rightarrow 12(12+1)=n(n+1)
\displaystyle \text{On comparing, we get }n=12\ \text{months}
\displaystyle \text{Hence, the time is 1 yr.}

Question 26 : Mr. Sonu has a recurring deposit account and deposits \displaystyle \text{Rs }750 per month for 2 yr. If he gets \displaystyle \text{Rs }19125 at the time of maturity, then find the rate of interest.    [ICSE 2020]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, money deposit per month, }(P)=\text{Rs }750
\displaystyle \text{Time }(n)=2\ \text{yr}=2 \times 12=24\ \text{months}
\displaystyle \text{and maturity value (MV)}=\text{Rs }19125
\displaystyle \text{Let the rate of interest be }r\%\ \text{per annum.}
\displaystyle \therefore\ \text{Maturity value (MV)}=P n \left(1+\frac{(n+1)r}{2400}\right)
\displaystyle \Rightarrow 19125=750 \times 24 \left(1+\frac{25r}{2400}\right)
\displaystyle \Rightarrow 19125=18000\left(1+\frac{r}{96}\right)
\displaystyle \Rightarrow 1+\frac{r}{96}=\frac{19125}{18000}
\displaystyle \Rightarrow \frac{r}{96}=\frac{19125-18000}{18000}
\displaystyle \Rightarrow \frac{r}{96}=\frac{1125}{18000}
\displaystyle \Rightarrow r=\frac{1125}{18000} \times 96
\displaystyle \Rightarrow r=6
\displaystyle \text{Hence, the required rate of interest is }6\%\ \text{per annum.}

Question 27 : Amit deposits \displaystyle \text{Rs }1600 per month in a bank for 18 months in a recurring deposit account. If he gets \displaystyle \text{Rs }31080 at the time of maturity, then what is the rate of interest per annum?      [ICSE 2020]
\displaystyle \text{Answer:}
\displaystyle  \text{We have, Amount deposited per month, }P=\text{Rs }1600
\displaystyle \text{Total months, }n=18\ \text{months and maturity amount}=\text{Rs }31080
\displaystyle \text{Let rate of interest per annum}=r\%
\displaystyle \text{Now, we know that Maturity value (MV)}
\displaystyle =P \times n+\left(P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\right)
\displaystyle \Rightarrow 31080=1600 \times 18+\frac{1600 \times 18(18+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow 31080-28800=\frac{1600 \times 18 \times 19}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow 2280=\frac{1600 \times 18 \times 19}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow r=\frac{2280 \times 2 \times 12 \times 100}{1600 \times 18 \times 19}=10\%
\displaystyle \text{Hence, the rate of interest per annum is }10\%.

Question 28 : Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives \displaystyle \text{Rs }441 as interest at the time of maturity. Find the amount Rekha deposited each month.      [ICSE 2019]
\displaystyle \text{Answer:}
\displaystyle  \text{Let money deposit per month be }\text{Rs }P.
\displaystyle \text{Number of months }(n)=20,\ \text{Interest earned }(I)=\text{Rs }441
\displaystyle \text{and Rate of interest }(r)=9\%.
\displaystyle \therefore\ I=\frac{P \times n(n+1) \times r}{2 \times 12 \times 100}
\displaystyle \Rightarrow 441=\frac{P \times 20 \times 21 \times 9}{2400}
\displaystyle \Rightarrow P=\frac{2400 \times 441}{20 \times 21 \times 9}=\text{Rs }280
\displaystyle \text{Hence, the amount deposited each month is }\text{Rs }280.

Question 29 : Priyanka has a recurring deposit account of \displaystyle \text{Rs }1000 per month at 10% per annum. If she gets \displaystyle \text{Rs }5550 as interest at the time of maturity, then find the total time for which the account was held.      [ICSE 2018]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }P=\text{Rs }1000,\ r=10\%\ \text{and }I=\text{Rs }5550
\displaystyle \therefore\ I=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow 5550=1000 \times \frac{n(n+1)}{24} \times \frac{10}{100}
\displaystyle \Rightarrow 5550=100 \times \frac{n(n+1)}{24}
\displaystyle \Rightarrow 5550 \times 24=n(n+1)
\displaystyle \Rightarrow n^2+n-1332=0
\displaystyle \Rightarrow n^2+37n-36n-1332=0
\displaystyle \Rightarrow n(n+37)-36(n+37)=0
\displaystyle \Rightarrow (n+37)(n-36)=0
\displaystyle \Rightarrow n=-37\ \text{or}\ 36\ \Rightarrow n=36\ \text{(since month cannot be negative)}
\displaystyle \text{Hence, the total time }(n)\text{ is }36\ \text{months.}

Question 30 : Sonia had a recurring deposit account in a bank and deposited \displaystyle \text{Rs }600 per month for \displaystyle 2\frac{1}{2} yr. If the rate of interest was 10% per annum, then find the maturity value of this account.      [ICSE 2018]
\displaystyle \text{Answer:}
\displaystyle 10.\ \text{Here, }P=\text{Rs }600,\ r=10\%\ \text{and }n=30\ \text{months}
\displaystyle \therefore\ MV=P n \left(1+\frac{(n+1)r}{2400}\right)
\displaystyle \Rightarrow MV=18000\left(1+\frac{31 \times 10}{2400}\right)
\displaystyle \Rightarrow =18000\left(1+\frac{31}{240}\right)=18000 \times \frac{271}{240}=\text{Rs }20325

Question 31 : Mr. Jacob has a 2 yr recurring deposit account in State Bank of India and deposits \displaystyle \text{Rs }1500 per month. If he receives \displaystyle \text{Rs }37875 at the time of maturity, then find the rate of interest.      [ICSE 2017]
\displaystyle \text{Answer:}
\displaystyle  \text{Monthly deposit}=\text{Rs }1500
\displaystyle \text{Amount deposited in 2 yr}=1500 \times 24=\text{Rs }36000
\displaystyle \text{Amount received on maturity}=\text{Rs }37875
\displaystyle \text{So, interest}=\text{Rs }(37875-36000)=\text{Rs }1875
\displaystyle \text{Now, }n=24\ \text{and }P=\text{Rs }1500
\displaystyle \therefore\ I=\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow 1875=\frac{1500 \times 24(24+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow 1875=1500 \times 12 \times 25 \times \frac{r}{100 \times 12}
\displaystyle \Rightarrow r=\frac{1875}{15 \times 25}=5\%
\displaystyle \text{Hence, the rate of interest is }5\%.

Question 32 : Mohan has a recurring deposit account in a bank for 2 yr at 6% per annum simple interest. If he gets \displaystyle \text{Rs }1200 as interest at the time of maturity, then find:
(i) the monthly instalment.
(ii) the amount of maturity.
[ICSE 2016]
\displaystyle \text{Answer:}
\displaystyle  (i)\ \text{Given, rate of interest }(r)=6\%\ \text{per annum}
\displaystyle \text{Interest }(I)=\text{Rs }1200\ \text{and Time }(n)=2\ \text{yr}=24\ \text{months}
\displaystyle \therefore\ I=\frac{P \times n(n+1) \times r}{2 \times 12 \times 100}
\displaystyle \Rightarrow 1200=\frac{P \times 24 \times 25 \times 6}{2400}
\displaystyle \Rightarrow P=\frac{1200 \times 2400}{24 \times 25 \times 6}=\text{Rs }800
\displaystyle \text{Hence, the monthly instalment is }\text{Rs }800.
\displaystyle (ii)\ \text{Amount on maturity}=P \times n+\text{Interest}
\displaystyle =800 \times 24+1200=19200+1200=\text{Rs }20400

Question 33 : Katrina opened a recurring deposit account with a nationalised bank for a period of 2 yr. If the bank pays interest at the rate of 6% per annum and the monthly instalment is \displaystyle \text{Rs }1000, then find:
(i) interest earned in 2 yr.
(ii) matured value.
[ICSE 2015]
\displaystyle \text{Answer:}
\displaystyle  \text{Here, we have, }P=\text{Rs }1000,\ n=24\ \text{months and }r=6\%.
\displaystyle (i)\ \therefore\ I=\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \text{Interest earned in 2 yr }(I)=\frac{1000 \times 24 \times 25}{2 \times 12} \times \frac{6}{100}=\text{Rs }1500
\displaystyle (ii)\ \text{Maturity value }(MV)=P \times n+I
\displaystyle =1000 \times 24+1500=\text{Rs }25500

Question 34 : Shahrukh opened a recurring deposit account in a bank and deposited \displaystyle \text{Rs }800 per month for \displaystyle 1\frac{1}{2} yr. If he received \displaystyle \text{Rs }15084 at the time of maturity, then find the rate of interest per annum. [ICSE 2014]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, money deposited per month }(P)=\text{Rs }800
\displaystyle \text{Time }(n)=1\frac{1}{2}\ \text{yr}=18\ \text{months}
\displaystyle \text{Maturity value }(MV)=\text{Rs }15084
\displaystyle \text{Let the rate of interest be }r\%\ \text{per annum.}
\displaystyle \therefore\ I=\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle =\frac{800 \times 18(18+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle =\frac{144 \times 19}{24}r=114r
\displaystyle \text{Total money deposited in 18 months}=18 \times 800=\text{Rs }14400
\displaystyle \therefore\ MV=\text{Money deposited}+\text{Interest}
\displaystyle \Rightarrow 15084=14400+114r
\displaystyle \Rightarrow 15084-14400=684
\displaystyle \Rightarrow r=\frac{684}{114}=6\%
\displaystyle \text{Hence, the rate of interest is }6\%\ \text{per annum.}

Question 35 : Mr. Britto deposits a certain sum of money each month in a recurring deposit account of a bank. If the rate of interest is 8% per annum and Mr. Britto gets \displaystyle \text{Rs }8088 from the bank after 3 yr, then find the value of his monthly instalment.     [ICSE 2013]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, rate of interest }(r)=8\%\ \text{per annum}
\displaystyle \text{Maturity amount }(A)=\text{Rs }8088\ \text{and time }(n)=3\ \text{yr}=36\ \text{months}
\displaystyle \text{Let the monthly instalment }(P)\text{ be }\text{Rs }x.
\displaystyle \text{Then, by the formula }MV=P \times n+\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow 8088=36x+\frac{x \times 36 \times 37}{2 \times 12} \times \frac{8}{100}
\displaystyle \Rightarrow 8088=36x+\frac{10656x}{2400}
\displaystyle \Rightarrow 8088=\frac{86400x+10656x}{2400}
\displaystyle \Rightarrow 8088=\frac{97056x}{2400}
\displaystyle \Rightarrow x=\text{Rs }200
\displaystyle \text{Hence, the monthly instalment is }\text{Rs }200.

Question 36 : Kiran deposited \displaystyle \text{Rs }200 for 36 months in a bank recurring deposit account. If the bank pays interest at the rate of 11% per annum, then find the amount she gets on maturity.      [ICSE 2012]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, money deposited per month}=\text{Rs }200
\displaystyle \text{Number of months }(n)=36\ \text{months and rate }(r)=11\%
\displaystyle \therefore\ I=\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \text{Interest received in 36 months}=\frac{200 \times 36 \times 37}{2 \times 12} \times \frac{11}{100}=\text{Rs }1221
\displaystyle \text{Now, maturity amount}=\text{Total amount deposit in 36 months}+\text{Interest}
\displaystyle =200 \times 36+1221=7200+1221=\text{Rs }8421
\displaystyle \text{Hence, Kiran will get }\text{Rs }8421\ \text{at the time of maturity.}

Question 37 : Ahmed has a recurring deposit account in a bank. He deposits \displaystyle \text{Rs }2500 per month for 2 yr. If he gets \displaystyle \text{Rs }66250 at the time of maturity, then find:
(i) the interest paid by the bank.
(ii) the rate of interest.
[ICSE 2011]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, money deposited per month}=\text{Rs }2500
\displaystyle \text{Number of months }(n)=24\ \text{and maturity value}=\text{Rs }66250
\displaystyle (i)\ \text{Clearly, interest paid by the bank, }I=\text{Maturity value}-\text{Money deposited}
\displaystyle =66250-24 \times 2500=66250-60000=\text{Rs }6250
\displaystyle (ii)\ \text{Let }r\%\ \text{be the rate of interest}
\displaystyle \therefore\ I=\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow 6250=\frac{2500 \times 24 \times 25}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow r=10
\displaystyle \text{Hence, the rate of interest is }10\%\ \text{per annum.}

Question 38: David opened a recurring deposit account in a bank and deposited \displaystyle \text{Rs }300 per month for 2 yr. If he received \displaystyle \text{Rs }7725 at the time of maturity, then find the rate of interest per annum.      [ICSE 2008]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, money deposited per month }(P)=\text{Rs }300
\displaystyle \text{Time }(n)=2\ \text{yr}=24\ \text{months and Maturity value }(MV)=\text{Rs }7725
\displaystyle \text{Let rate of interest be }r\%\ \text{per annum.}
\displaystyle \therefore\ MV=P \times n+I=P \times n+\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle =P n\left(1+\frac{(n+1)r}{2400}\right)
\displaystyle \Rightarrow 7725=300 \times 24\left(1+\frac{25r}{2400}\right)
\displaystyle \Rightarrow 7725=300 \times 24\left(\frac{2400+25r}{2400}\right)
\displaystyle \Rightarrow 7725=7200\left(\frac{2400+25r}{2400}\right)
\displaystyle \Rightarrow 7725 \times 2400=7200(2400+25r)
\displaystyle \Rightarrow 7725=2400+25r
\displaystyle \Rightarrow 25r=175\Rightarrow r=7
\displaystyle \text{Hence, the rate of interest is }7\%\ \text{per annum.}

Question 39:  Saloni deposited \displaystyle \text{Rs }150 per month in her bank for 8 months under the recurring deposit scheme. What will be the maturity value of her deposit, if the rate of interest is 8% per annum and the interest is calculated at the end of every month?      [ICSE 2007, 01]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, money deposited per month}=\text{Rs }150
\displaystyle \text{Number of months }(n)=8\ \text{months and rate of interest }(r)=8\%
\displaystyle \text{We know, }MV=P n+I
\displaystyle =P n+\frac{P \times n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle =P n\left[1+\frac{(n+1)r}{2400}\right]
\displaystyle \therefore\ MV\ \text{for given data}=150 \times 8\left[1+\frac{9 \times 8}{2400}\right]
\displaystyle =1200\left[1+\frac{3}{100}\right]
\displaystyle =1200 \times \frac{103}{100}
\displaystyle =\text{Rs }1236

Question 40 : Salman deposits \displaystyle \text{Rs }1000 every month in a recurring deposit account for 2 yr. If he receives \displaystyle \text{Rs }26000 on maturity, then find:
(i) the total interest Salman earns.
(ii) the rate of interest.
[ICSE 2023]
\displaystyle \text{Answer:}
\displaystyle  \text{Total money deposited by Salman in 2 yr}
\displaystyle (P)=1000 \times 24=\text{Rs }24000
\displaystyle \text{and Total received money }(A)=\text{Rs }26000
\displaystyle (i)\ \text{Total interest earned by Salman }(I)
\displaystyle =\text{Total received}-\text{Total deposited}
\displaystyle I=A-P
\displaystyle =26000-24000=\text{Rs }2000
\displaystyle (ii)\ \text{Let rate of interest be }r\%.
\displaystyle \text{We have by formula, Interest }(I)=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
\displaystyle \text{where }n\ \text{is total number of months and }P\ \text{is monthly deposits money.}
\displaystyle \Rightarrow 2000=1000 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100}
\displaystyle \Rightarrow r=\frac{2000}{250}=8\%
\displaystyle \text{Hence, the rate of interest is }8\%\ \text{per annum.}

Question 41 : Salman deposits \displaystyle \text{Rs }1200 every month in a recurring deposit account for \displaystyle 2\frac{1}{2} yr. If the rate of interest is 6% per annum, then find the amount he will receive on maturity.      [ICSE 2023]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, money deposited per month, }(P)=\text{Rs }1200,
\displaystyle \text{number of month }(n)=30\ \text{and rate of interest }(r)=6\%\ \text{per annum}
\displaystyle \left[\begin{array}{l} 1\ \text{yr}=12\ \text{months} \\ 2\frac{1}{2}\ \text{yr}=30\ \text{months} \end{array}\right]
\displaystyle \therefore\ \text{Maturity amount}=P n+\frac{P \times n(n+1)}{2400}
\displaystyle =P n\left[1+\frac{(n+1)r}{2400}\right]=1200 \times 30\left[1+\frac{6(30+1)}{2400}\right]
\displaystyle =36000\left[1+\frac{31}{400}\right]=36000\left[\frac{400+31}{400}\right]
\displaystyle =90 \times 431=\text{Rs }38790
\displaystyle \text{Hence, Salman will get the amount }\text{Rs }38790\ \text{at the time of maturity.}

Question 42 : Mr. Richard has a recurring deposit account in a bank for 3 yr at 7.5% per annum simple interest. If he gets \displaystyle \text{Rs }8325 as
interest at the time of maturity, then find:
(i) the monthly deposit.
(ii) the maturity value.
[ICSE 2017]
\displaystyle \text{Answer:}
\displaystyle  \text{Given,}
\displaystyle \text{Time, }(n)=3\ \text{yr}=36\ \text{months}
\displaystyle \text{Rate of interest, }(r)=7.5\%\ \text{per annum and Interest }(I)=\text{Rs }8325
\displaystyle (i)\ \text{Let }P=\text{Monthly deposit}
\displaystyle \therefore\ \text{Maturity value }(MV)=nP+I\qquad (i)
\displaystyle \text{We know that Maturity value }(MV)=nP\left[1+\frac{(n+1)r}{2400}\right]\qquad (ii)
\displaystyle \text{From Eqs. (i) and (ii), we get}
\displaystyle \Rightarrow 36P+8325=\left[1+\frac{(36+1)\times 7.5}{2400}\right]\times 36 \times P
\displaystyle \Rightarrow 36P+8325=\left[\frac{320+37}{320}\right]\times 36P
\displaystyle \Rightarrow 36P\left[\frac{357}{320}-1\right]=8325\Rightarrow 36P\left[\frac{37}{320}\right]=8325
\displaystyle \Rightarrow P=\frac{8325 \times 320}{36 \times 37}=2000
\displaystyle \therefore\ \text{Monthly deposit is }\text{Rs }2000.
\displaystyle (ii)\ \text{Now, maturity value }(MV)=36P+8325\qquad \text{(from Eq. (i))}
\displaystyle =(36 \times 2000)+8325=72000+8325=\text{Rs }80325
\displaystyle \text{Hence, the maturity values is }\text{Rs }80325.

Question 43 : On a certain sum of money, the difference between the compound interest for a year, payable half-yearly and the simple interest for a year is \displaystyle \text{Rs }16. Find the sum lent out, if the rate of interest in both cases is 8%.       [ICSE 2017]
\displaystyle \text{Answer:}
\displaystyle  \text{Given, rate of interest, }r=8\%
\displaystyle \text{Now, simple interest,}
\displaystyle SI=\frac{P \times T \times r}{100}=\frac{P \times 8 \times 1}{100}\qquad [\because\ T=1\ \text{yr, given}]
\displaystyle =\frac{8P}{100}
\displaystyle \text{For compound interest}
\displaystyle \text{Rate of interest becomes, }R=\frac{8}{2}=4\%
\displaystyle \text{and time, }T=2 \times 1=2\ \text{yr}
\displaystyle \text{Now, compound interest,}
\displaystyle CI=P\left[\left(1+\frac{R}{100}\right)^T-1\right]=P\left[\left(1+\frac{4}{100}\right)^2-1\right]

\displaystyle =P\left[\left(1+\frac{1}{25}\right)^2-1\right]=P\left[\left(\frac{26}{25}\right)^2-1\right]
\displaystyle =P\left[\frac{676}{625}-1\right]=P\left[\frac{676-625}{625}\right]=\frac{51P}{625}
\displaystyle \text{According to the question, }CI-SI=16
\displaystyle \Rightarrow \frac{51P}{625}-\frac{8P}{100}=16=\frac{204P-200P}{2500}
\displaystyle \Rightarrow \frac{4P}{2500}=16
\displaystyle \Rightarrow 4P=16 \times 2500
\displaystyle \Rightarrow P=4 \times 2500=\text{Rs }10000
\displaystyle \text{Hence, the sum of money is }\text{Rs }10000.

Question 44 : Mr. A Ramchandra has an account with Central Bank of India. The following entries are from his passbook:
\displaystyle \text{Complete the passbook given below}
\displaystyle \begin{array}{|c|l|c|c|c|} \hline \text{Date} & \text{Particulars} & \text{Withdrawals (in Rs)} & \text{Deposits (in Rs)} & \text{Balance (in Rs)}\\ \hline 05\ \text{Jan 2009} & \text{B/F} & & & 8000\\ \hline 20\ \text{Jan 2009} & \text{To self} & 2500 & & 5500\\ \hline 04\ \text{Feb 2009} & \text{By cash} & & 9000 & 14500\\ \hline 20\ \text{Feb 2009} & \text{By cash} & & 3000 & 17500\\ \hline 04\ \text{Mar 2009} & \text{To self} & 1000 & & 16500\\ \hline 15\ \text{Apr 2009} & \text{By cash} & & 12000 & 28500\\ \hline \end{array}

Complete the above page of his passbook and calculate the interest accumulated in four months, January to April at the rate of 3.5% per annum.
If the interest is added on 30th April, then find his balance on that date.      [ICSE 2017]
\displaystyle \text{Answer:}
\displaystyle  \text{Complete the passbook given below}
\displaystyle \begin{array}{|c|l|c|c|c|} \hline \text{Date} & \text{Particulars} & \text{Withdrawals (in Rs)} & \text{Deposits (in Rs)} & \text{Balance (in Rs)}\\ \hline 05\ \text{Jan 2009} & \text{B/F} & & & 8000\\ \hline 20\ \text{Jan 2009} & \text{To self} & 2500 & & 5500\\ \hline 04\ \text{Feb 2009} & \text{By cash} & & 9000 & 14500\\ \hline 20\ \text{Feb 2009} & \text{By cash} & & 3000 & 17500\\ \hline 04\ \text{Mar 2009} & \text{To self} & 1000 & & 16500\\ \hline 15\ \text{Apr 2009} & \text{By cash} & & 12000 & 28500\\ \hline \end{array}
\displaystyle \text{Table for minimum balance for qualifying amount}
\displaystyle \begin{array}{|c|c|} \hline \text{Months} & \text{Minimum balance between 10th day and the last day}\\ \hline \text{Jan} & 5500\\ \hline \text{Feb} & 14500\\ \hline \text{March} & 16500\\ \hline \text{April} & 16500\\ \hline \text{Total} & 53000\\ \hline \end{array}
\displaystyle \text{Here, principal for 1 month}=\text{Rs }53000
\displaystyle \text{Given, rate of interest, }R=3.5\%\ \text{per annum}
\displaystyle \therefore\ \text{Interest}=\frac{P \times R \times T}{100}=\frac{53000 \times 3.5 \times 1}{100 \times 12}
\displaystyle =\frac{1855}{12}=\text{Rs }154.58
\displaystyle \therefore\ \text{Balance on date 30th April}=\text{Rs }(28500+154.58)
\displaystyle =\text{Rs }28654.58

Question 45 : Joseph has a recurring deposit account in a bank for 2 yr at the rate of 8% per annum simple interest.      [ICSE Semester–I 2022]

(i) If at the time of maturity Joseph receives \displaystyle \text{Rs }2000
as interest, then the monthly instalment is:
(a) \displaystyle \text{Rs }1200
(b) \displaystyle \text{Rs }600
(c) \displaystyle \text{Rs }1000
(d) \displaystyle \text{Rs }1600

(ii) The total amount deposited in the bank is:
(a) \displaystyle \text{Rs }25000
(b) \displaystyle \text{Rs }24000
(c) \displaystyle \text{Rs }26000
(d) \displaystyle \text{Rs }23000

(iii) The amount Joseph receives on maturity is:
(a) \displaystyle \text{Rs }27000
(b) \displaystyle \text{Rs }25000
(c) \displaystyle \text{Rs }26000
(d) \displaystyle \text{Rs }28000

(iv) If the monthly instalment is \displaystyle \text{Rs }100 and the rate
of interest is 8%, in how many months Joseph will receive
\displaystyle \text{Rs }52 as interest?
(a) 18
(b) 30
(c) 12
(d) 6
\displaystyle \text{Answer:}
\displaystyle  \text{Given, rate of interest }(r)=8\%\ \text{per annum}
\displaystyle \text{Number of month }(n)=24\ \text{months}
\displaystyle (i)\ (c)\ \text{Interest }(I)=\text{Rs }2000
\displaystyle \therefore\ I=\frac{P n(n+1) \times r}{2400}
\displaystyle \Rightarrow 2000=\frac{P \times 24 \times 25 \times 8}{2400}
\displaystyle \Rightarrow P=\frac{2000 \times 2400}{25 \times 8 \times 24}=\text{Rs }1000
\displaystyle (ii)\ (b)\ \text{Monthly instalment}=\text{Rs }1000
\displaystyle \text{Time period }2\ \text{yr or }24\ \text{months}
\displaystyle \therefore\ \text{Total amount deposit}=\text{Total month} \times \text{Monthly instalment}
\displaystyle =24 \times 1000=\text{Rs }24000
\displaystyle (iii)\ (c)\ \text{Maturity amount}=\text{Total amount deposit}+\text{Interest}
\displaystyle =\text{Rs }(24000+2000)=\text{Rs }26000
\displaystyle (iv)\ (c)\ \text{Given, monthly instalment }(P)=\text{Rs }100
\displaystyle \text{and rate of interest }(r)=8\%
\displaystyle \text{and Interest}=\text{Rs }52
\displaystyle \therefore\ I=\frac{P n(n+1) \times r}{2400}
\displaystyle \Rightarrow 52=\frac{100 \times 8 \times n(n+1)}{2400}
\displaystyle \Rightarrow 52 \times 3=n(n+1)
\displaystyle \Rightarrow n^2+n-156=0
\displaystyle \Rightarrow n^2+13n-12n-156=0
\displaystyle \Rightarrow n(n+13)-12(n+13)=0
\displaystyle \Rightarrow (n-12)(n+13)=0
\displaystyle \Rightarrow n=12,-13
\displaystyle \text{Month cannot be negative.}
\displaystyle \text{So, number of month is }12.

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