myhelper: RS Aggarwal solution class 8 chapter 11 Compound Interest Exercise 11C

Exercise 11C

Page-154

Q1 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 1:

Find the amount and the compound interest on Rs 8000 for 1 year at 10% per annum, compounded half-yearly.

Answer 1:

$Principal, P = Rs . 8000 Time, n = 1 year = 2 half years Rate of interest per annum = 10 % Rate of interest for half year, R = \frac{10 %}{2} = 5 % The amount with the compound interest is given by Amount = Rs . P \times {(1 + \frac{R}{100})}^{2} = Rs . 8000 \times {(1 + \frac{5}{100})}^{2} = Rs . 8000 \times {(\frac{105}{100})}^{2} = Rs . 8000 \times {(\frac{21}{20})}^{2} = Rs . 8000 \times (\frac{21}{20}) \times (\frac{21}{20}) = Rs . (20 \times 21 \times 21) = Rs . 8820 ∴ Compound interest = amount - principal = Rs . (8820 - 8000) = Rs . 820$

Q2 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 2:

Find the amount and the compound interest on Rs 31250 for $1 \frac{1}{2}$ years at 8% per annum, compounded half-yearly.

Sol :

Principal, P = Rs. 31250

Annual rate of interest, R = 8%

Time, (n)

$=1\frac{1}{2}=\frac{3}{2}$ years

Then the amount with the compound interest is given by

$A=P\left(\frac{1}{1}+\frac{R}{2 \times 100}\right)^{2 n}$

$=31250\left(\frac{1}{1}+\frac{8}{2 \times 100}\right)^{2 \times \frac{3}{2}}$

$=31250\left(\frac{100+4}{100}\right)^3$

$=31250 \times \frac{104}{100}\times \frac{104}{100} \times \frac{104}{100}$

$=31250 \times \frac{26}{25}\times \frac{26}{25}\times \frac{26}{25}$

A=35152

Therefore, compound interest = amount - principal=Rs (35152- 31250) = Rs 3902

Q3 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 3:

Find the amount and the compound interest on Rs 12800 for 1 year at $7 \frac{1}{2} %$ per annum, compounded semi-annually.

Answer 3:

$Principal, P = Rs . 12800 Annual rate of interest, R = \frac{15}{2} % Rate of interest for a half year = \frac{1}{2} (\frac{15}{2}) % = \frac{15}{4} % Time, n = 1 year = 2 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 12800 \times {(1 + \frac{\frac{15}{4}}{100})}^{2} = 12800 \times {(1 + \frac{15}{100 \times 4})}^{2} = 12800 \times {(\frac{400 + 15}{400})}^{2} = 12800 \times {(\frac{415}{400})}^{2} = 12800 \times (\frac{83}{80}) \times (\frac{83}{80}) = (2 \times 83 \times 83) = Rs 13778 Therefore, compound interest = amount - principal = Rs (13778 - 12800) = Rs 978$

Q4 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 4:

Find the amount and the compound interest on Rs 160000 for 2 years at 10% per annum, compounded half-yearly.

Answer 4:

$Principal, P = Rs . 160000 Annual rate of interest, R = 10 % Rate of interest for a half year = \frac{10}{2} % = 5 % Time, n = 2 years = 4 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 160000 \times {(1 + \frac{5}{100})}^{4} = 160000 \times {(\frac{100 + 5}{100})}^{4} = 160000 \times {(\frac{105}{100})}^{4} = 160000 \times (\frac{21}{20}) \times (\frac{21}{20}) \times (\frac{21}{20}) \times (\frac{21}{20}) = (21 \times 21 \times 21 \times 21) = Rs 194481 Therefore, compound interest = amount - principal = Rs (194481 - 160000) = Rs 34481$

Q5 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 5:

Swati borrowed Rs 40960 from a bank to buy a piece of land. If the bank charges $12 \frac{1}{2} %$ per annum, compounded half-yearly, what amount will she have to pay after $1 \frac{1}{2}$ years? Also, find the interest paid by her.

Answer 5:

$Principal, P = Rs . 40960 Annual rate of interest, R = \frac{25}{2} % Rate of interest for half year = \frac{25}{4} % Time, n = 1 \frac{1}{2} years = 3 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 40960 \times {(1 + \frac{25}{100 \times 4})}^{3} = 40960 \times {(\frac{400 + 25}{400})}^{3} = 40960 \times {(\frac{425}{400})}^{3} = 40960 \times (\frac{17}{16}) \times (\frac{17}{16}) \times (\frac{17}{16}) = (10 \times 17 \times 17 \times 17) = Rs 49130 Therefore, compound interest = amount - principal = Rs (49130 - 40960) = Rs 8170 Therefore, Swati has to pay Rs . 49130, which includes an interest of Rs . 8170, to the bank after 1 \frac{1}{2} years .$

Q6 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 6:

Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is Rs 125000 and the Parishad charges interest at 12% per annum compounded half-yearly, find the interest paid by Aslam after a year and a half.

Answer 6:

$Let the principal amount be P = Rs . 125000 . Annual rate of interest, R = 12 % Rate of interest for a half year = 6 % Time, n = 1 \frac{1}{2} years = 3 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 125000 \times {(1 + \frac{6}{100})}^{3} = Rs . 125000 \times {(\frac{100 + 6}{100})}^{3} = Rs . 125000 \times {(\frac{106}{100})}^{3} = Rs . 125000 \times (\frac{53}{50}) \times (\frac{53}{50}) \times (\frac{53}{50}) = Rs . (53 \times 53 \times 53) = Rs . 148877 Now, C I = A - P = Rs . (148877 - 125000) = Rs . 23877 Therefore, Aslam has to pay an interest of Rs . 23877 to the bank after 1 \frac{1}{2} years .$

Q7 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 7:

Sheela deposited Rs 20000 in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is 6% per annum, what amount will she get after 1 year?

Answer 7:

$Let the principal amount be P = Rs . 20000 . Annual rate of interest, R = 6 % Rate of interest for half year = 3 % Time, n = 1 year = 2 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 20000 \times {(1 + \frac{3}{100})}^{2} = Rs . 20000 \times {(\frac{100 + 3}{100})}^{2} = Rs . 20000 \times {(\frac{103}{100})}^{2} = Rs . 20000 \times (\frac{103}{100}) \times (\frac{103}{100}) = Rs . (2 \times 103 \times 103) = Rs . 21218 Therefore, Sheela gets Rs . 21218 after 1 year .$

Q8 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 8:

Neeraj lent Rs 65536 for 2 years at $12 \frac{1}{2} %$ per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly?

Answer 8:

$Let the principal amount be P = Rs . 65536 . Annual rate of interest, R = \frac{25}{2} % Rate of interest for a half year = \frac{25}{4} % Time, n = 2 years = 4 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 65536 \times {(1 + \frac{25}{100 \times 4})}^{4} = Rs . 65536 \times {(\frac{400 + 25}{400})}^{4} = Rs . 65536 \times {(\frac{425}{400})}^{4} = Rs . 65536 \times {(\frac{17}{16})}^{4} = Rs . 65536 \times (\frac{17}{16}) \times (\frac{17}{16}) \times (\frac{17}{16}) \times (\frac{17}{16}) = Rs . (17 \times 17 \times 17 \times 17) = Rs . 83521 Now, CI = A - P = Rs . (83521 - 65536) = Rs . 17985 Therefore, interest earned when compounded half yearly = Rs . 17985 Amount when the interest is compounded yearly is given by A = P \times {(1 + \frac{R}{100})}^{n} = R s . 65536 \times {(1 + \frac{25}{100 \times 2})}^{2} = Rs . 65536 \times {(\frac{200 + 25}{200})}^{2} = Rs . 65536 \times {(\frac{225}{200})}^{2} = Rs . 65536 \times {(\frac{9}{8})}^{2} = Rs . 65536 \times (\frac{9}{8}) \times (\frac{9}{8}) = Rs . 82944 Therefore, CI = A - P = Rs . (82944 - 65536) = Rs . 17408 ∴ Difference between the interests compounded half yearly and yearly = Rs . (17985 - 17408) = Rs . 577$

Q9 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 9:

Sudershan deposited Rs 32000 in a bank, where the interest is credited quarterly. If the rate of interest be 5% per annum, what amount will he receive after 6 months?

Answer 9:

$Let the principal amount be P = Rs 32000 . Annual rate of interest, R = 5 % Rate of interest for a quarter year = \frac{5}{4} % Time, n = 6 months = 2 quarter years Then the amount with the compound interest is given by A = Rs . P \times {(1 + \frac{R}{100})}^{n} = Rs . 32000 \times {(1 + \frac{5}{100 \times 4})}^{2} = Rs . 32000 \times {(\frac{400 + 5}{400})}^{2} = Rs . 32000 \times {(\frac{405}{400})}^{2} = Rs . 32000 \times {(\frac{81}{80})}^{2} = Rs . 32000 \times (\frac{81}{80}) \times (\frac{81}{80}) = Rs . (5 \times 81 \times 81) = Rs . 32805 Therefore, Sudershan will receive an amount of Rs . 32805 after 6 months .$

Q10 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 10:

Arun took a loan of Rs 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year?

Answer 10:

$Let the principal amount be P = Rs 390625 . Annual rate of interest, R = 16 % Rate of interest for a quarter year = \frac{16}{4} % = 4 % Time, n = 1 year = 4 quarter years Then the amount with the compound interest is given by A = Rs . P \times {(1 + \frac{R}{100})}^{n} = Rs . 390625 \times {(1 + \frac{4}{100})}^{4} = Rs . 390625 \times {(\frac{100 + 4}{100})}^{4} = Rs . 390625 \times {(\frac{104}{100})}^{4} = Rs . 390625 \times {(\frac{26}{25})}^{4} = Rs . 390625 \times (\frac{26}{25}) \times (\frac{26}{25}) \times (\frac{26}{25}) \times (\frac{26}{25}) = Rs . (26 \times 26 \times 26 \times 26) = Rs . 456976 Therefore, Arun has to pay Rs 456976 after 1 year .$

RS Aggarwal solution class 8 chapter 11 Compound Interest Exercise 11C

Exercise 11C

Page-154

Q1 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 1:

Answer 1:

Q2 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 2:

Sol :

Q3 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 3:

Answer 3:

Q4 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 4:

Answer 4:

Q5 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 5:

Answer 5:

Q6 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 6:

Answer 6:

Q7 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 7:

Answer 7:

Q8 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 8:

Answer 8:

Q9 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 9:

Answer 9:

Q10 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 10:

Answer 10:

No comments:

Post a Comment

Contact Form