Class 10: Banking – Exercise 4(c) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: 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. [2013]

Answer:

Qualifying principal for various months:

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

$\displaystyle P = \text{ Rs. } 29690 R = 6\% \text{ and } T= \frac{1}{12}$

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

$\\$

Question 2: A page from a passbook of saving bank account is given below:

Date	Particulars	Withdrawals (Rs.)	Deposits (Rs.)	Balance (Rs.)
09.08.1999	By Cash	–	10,000	10,000
11.08.1999	By Cheque	–	5,000	15,000
05.10.1999	To Cheque	12,000	–	3,000
10.10.1999	By Cash	–	17,000	20,000
27.11.1999	By Cheque	5,000	–	15,000
29.11.1999	By Cash	–	3,000	18,000

The account is closed on 2^nd January 2000. Find the amount received, if the rate of interest is 5% p.a.

Answer:

Qualifying principal for various months:

Month	Principal (Rs.)
August	10000
September	15000
October	20000
November	15000
December	18000
January	0
Total	78000

$\displaystyle P = \text{ Rs. } 78000 R = 5\% \text{ and } T= \frac{1}{12}$

$\displaystyle I = P \times R \times T = 78000 \times \frac{5}{100} \times \frac{1}{12} = \text{ Rs. } \ 325$

Amount received $\displaystyle = 18000+325 = \text{ Rs. } \ 18325$

$\\$

Question 3: A person had a Account in Bank. His passbook had the following entries:

Date	Particulars	Withdrawals (Rs.)	Deposits (Rs.)	Balance (Rs.)
Jan. 1, 2000	By Balance	–	–	9,600
Jan. 8	By Cash	–	6,000	15,600
Feb. 18	To Cheque	10,500	–	5,100
May 19	By Cash	–	6,300	11,400
July 15	By Self	2,400	–	9,000
Oct. 7	By Cash	–	3,600	12,000

On October 30th, 2000, he closed the account. If the amount of interest he received on closing account on 30^th Oct. 2000 is Rs.310; Calculate the rate of interest per annum.

Answer:

Qualifying principal for various months:

Month	Principal (Rs.)
January	15600
February	5100
March	5100
April	5100
May	5100
June	11400
July	9000
August	9000
September	9000
October	12000
Total	86400

$\displaystyle P = \text{ Rs. } \ 86400 \text{, Rate } = r\% \ and \ T= \frac{1}{12} , \ Interest = \text{ Rs. } \ 310$

$\displaystyle I = P \times R \times T$

$\displaystyle 310= 86400 \times \frac{r}{100} \times \frac{1}{12} \Rightarrow r = 4.31 \%$

$\displaystyle \\$

Question 4: A person deposited 600 per month in a recurring deposit account for 4 years. If the rate of interest is 8% per year; calculate the maturity value of his account.

Answer:

$\displaystyle P = \text{ Rs. } \ 600$ , no of months $\displaystyle = 48, \ r = 8\%$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle =600 \times 48 +600 \times \frac{48(48+1)}{2 \times 12} \times \frac{8}{100} = \text{ Rs. } 33504$

$\displaystyle \\$

Question 5: A person has a recurring deposit account in a bank and deposits 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of this account is Rs.1,554.

Answer:

$\displaystyle P = \text{ Rs. } \ 80$ , no of months $\displaystyle = 18 \text{, Rate } = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 1554$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 1554 =80 \times 18 +80 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle r = \frac{(1554-80 \times 18) \times (2 \times 12) \times 100}{80 \times 18 \times 19} \Rightarrow r=10\%$

$\displaystyle \\$

Question 6: The maturity value of recurring deposit account is 16,176. If the monthly installment is Rs.400 and the rate of interest is 8% find the time of this Account.

Answer:

$\displaystyle P = \text{ Rs. } \ 400$ , no of months $\displaystyle = n \text{, Rate } = 8\%$ Maturity Amount $\displaystyle = \text{ Rs. } 16176$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 16176 =400 \times n +400 \times \frac{n(n+1)}{2 \times 12} \times \frac{8}{100}$

$\displaystyle 16176 = 400n + \frac{4}{3} n(n+1)$

$\displaystyle \text{or } n^2+301n-12132=0 \Rightarrow n = 36 \ or -337$

Hence n = 36 months.

Notes: Please refer to quadratic equations for solving this.

$\displaystyle \\$

Question 7: A person needs Rs.30,000 after 2 years. What least money (in multiple of Rs.5) must he deposit every month in Recurring Deposit account to get required money after 2 years, the rate of interest being 8% p.a.

Answer:

$\displaystyle P = \text{ Rs. } \ x$ , no of months $\displaystyle = 24 \text{, Rate } = 8\%$ Maturity Amount $\displaystyle = \text{ Rs. } 30000$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 30000 =x \times 24 +x \times \frac{24(24+1)}{2 \times 12} \times \frac{8}{100}$

$\displaystyle x = \frac{30000}{26} = \text{ Rs. } 1153.84$

He must deposit Rs. 1155 every month.

$\displaystyle \\$

Question 8: A person has a recurring deposit account in a bank for 3 years at 8% p.a. simple interest. If he gets 9,990 as interest at the time of maturity, find

The monthly installment.
The amount of maturity

Answer:

$\displaystyle P = \text{ Rs. } \ x$ , no of months $\displaystyle = 36 \text{, Rate } = 8\%$ Interest Amount $\displaystyle = \text{ Rs. } 9990$

$\displaystyle \text{Interest Value } = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 9990 =x \times \frac{36(36+1)}{2 \times 12} \times \frac{8}{100}$

$\displaystyle x = \frac{9990}{4.44} = \text{ Rs. } 2250$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle =2250 \times 36 +2250 \times \frac{36(36+1)}{2 \times 12} \times \frac{8}{100} = \text{ Rs. } 90990$

$\displaystyle \\$

Question 9: A person has cumulative recurring deposit account and deposits 900 per month for a period of 4 years, if he gets Rs.52,020 at the time of maturity, find the rate of interest.

Answer:

$\displaystyle P = \text{ Rs. } \ 900$ , no of months $\displaystyle = 48 \text{, Rate } = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 52020$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 52020 =900 \times 48 +900 \times \frac{48(48+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle r = \frac{(52020-900 \times 48) \times (2 \times 12) \times 100}{900 \times 48 \times 49} \Rightarrow r=10\%$

$\displaystyle \\$

Question 10: A person has a 4 year recurring deposit account in a bank and deposits 1,800 per month. If she gets Rs.1,08,450 at the time of maturity, find the rate of interest.

Answer:

$\displaystyle P = \text{ Rs. } \ 1800$ , no of months $\displaystyle = 48 \text{, Rate } = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 108450$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 108540 =1800 \times 48 +1800 \times \frac{48(48+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle r = \frac{(108450-1800 \times 48) \times (2 \times 12) \times 100}{1800 \times 48 \times 49} \Rightarrow r=12.5\%$

$\displaystyle \\$

Question 11: Chaudhary opened a saving bank account at State Bank of India on 1^st April 2007. The entries of one year as shown in his passbook are given below:

Date	Particulars	Withdrawals (Rs.)	Deposits (Rs.)	Balance (Rs.)
1^st April 2007	By Cash	–	8,550.00	8,550.00
12^th April 2007	To Self	1,200.00	–	7,350.00
24^th April 2007	By Cash	–	4,550.00	11,900.00
8^th July 2007	By Cheque	–	1,500.00	13,400.00
10^th Sept. 2007	By Cheque	–	3,500.00	16,900.00
17^th Sept. 2007	To Cheque	2,500.00	–	14,400.00
11^th Oct. 2007	By Cash	–	800.00	15,200.00
6^th Jan. 2008	To Self	2,000.00	–	13,200.00
9^th 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 1^st April 2008. Give your answer correct to nearest rupee. [2011]

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 12: 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. [2013]

Answer:

$\displaystyle P = Rs. \ x$ , no of months $\displaystyle = 36$ , Rate $\displaystyle = 8\%$ Maturity Amount $\displaystyle = Rs. 8008$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 8008 =x \times 36 +x \times \frac{36(36+1)}{2 \times 12} \times \frac{8}{100}$

$\displaystyle x = \frac{8008}{40.44} = Rs. 198.02$

He must deposit Rs. $\displaystyle 200$ every month.

$\displaystyle \\$

Question 13: 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. [2014]

Answer:

$\displaystyle P = Rs. \ 800$ , no of months $\displaystyle = 18$ , Rate $\displaystyle = r\%$ Maturity Amount $\displaystyle = Rs. 15084$

$\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle 15084 =800 \times 18 +800 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}$

$\displaystyle r = \frac{(15084-800 \times 18) \times (2 \times 12) \times 100}{800 \times 18 \times 19} \Rightarrow r=6\%$

$\displaystyle \\$

Question 14: 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. [2014]

Answer:

Qualifying principal for various months:

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

$\displaystyle P = \text{ Rs. } 42000 \text{ Rate } = r\% \text{ and } T= \frac{1}{12} \text{ , Interest } = Rs. 175$

$\displaystyle I = P \times R \times T$

$\displaystyle 175= 42000 \times \frac{r}{100} \times \frac{1}{12} \Rightarrow r = 5 \%$

$\\$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 10: Banking – Exercise 4(c)

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