Compound Interest-Chapter 9- Exercise 9(A)-Class 8

byAmlesh Mahato •June 11, 2020

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Exercise-9 (A)

Find the compound interest on the following:

Principal	Rate % p.a	Time (In years)
1. Rs 10000	12 %	2
2. Rs 5000	10 %	2

1.Solution:

⇨Given

⇨p = Rs 10000

⇨r = 12 %

⇨n = 2

⇨Compound interest will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( 1+\frac{12}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( \frac{100+12}{100} \right )^{2}-1 \right \}$ $\large \Rightarrow C.I=10000\left \{ \left ( \frac{112}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \frac{12544}{10000}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \frac{12544-10000}{10000} \right \}$

$\large \Rightarrow C.I=10000\times \frac{2544}{10000}$

$\large \Rightarrow C.I=2544$ Rs

2.Solution:

⇨Given

⇨p = Rs 5000

⇨r = 10 %

⇨n = 2

⇨Compound interest will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=5000\left \{ \left ( 1+\frac{10}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=5000\left \{ \left ( \frac{100+10}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=5000\left \{ \left ( \frac{110}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=5000\left \{ \frac{12100}{10000}-1 \right \}$

$\large \Rightarrow C.I=5000\left \{ \frac{12100-10000}{10000} \right \}$

$\large \Rightarrow C.I=5000\times \frac{2100}{10000}$

$\large \Rightarrow C.I=\frac{10500000}{10000}$

$\large \Rightarrow C.I=1050$ Rs

3. Find the amount and the compound interest on a sum of Rs 15625 at 4% per annum for 3 years compounded annually.

Solution:

⇨Given

⇨p= 15625

⇨r= 4%

⇨n= 3

⇨We have , C.I = A - P

$\large \Rightarrow C.I=p(1+\frac{r}{100})^{n}-p$

$\large \Rightarrow C.I=p\left \{ (1+\frac{r}{100})^{n}-1 \right \}$

$\large \Rightarrow C.I=15625\left \{ \left (1+\frac{4}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=15625\left \{ \left (\frac{100+4}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=15625\left \{ \left (\frac{104}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=15625 \left (\frac{1124864}{1000000}-1 \right )$

$\large \Rightarrow C.I=15625 \left (\frac{1124864-1000000}{1000000} \right )$

$\large \Rightarrow C.I=15625 \left (\frac{124864}{1000000} \right )$

$\large \Rightarrow C.I=15625 \times \frac{124864}{1000000}$

$\large \Rightarrow C.I= \frac{1951000000}{1000000}$

$\large \Rightarrow C.I=1951$

⇨Then Amount = P + C.I.=15625 + 1951=17576 Rs

4. Shankar takes a loan of Rs 10000 at a compound interest rate of 10% per annum (p.a.)

i. Find the compound interest after one year.

Solution:

⇨Given

⇨p =10000

⇨ r =10%

⇨n =1

⇨Compound interest after 1 year will be

$\large C.I= A-P$

$\large C.I= p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large C.I= p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large C.I= 10000\left \{ \left ( 1+\frac{10}{100} \right )^{1}-1 \right \}$

$\large C.I= 10000\left \{ \frac{100+10}{100}-1 \right \}$

$\large C.I= 10000\left \{ \frac{110}{100}-1 \right \}$

$\large C.I= 10000\left \{ \frac{110-100}{100} \right \}$

$\large C.I= \frac{100000}{100}$

$\large C.I=1000$ Rs

ii. Find the compound interest for 2 years.

Solution:

⇨Given
⇨p =10000
⇨r =10%
⇨n=2

⇨ Compound interest after 2 year will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( 1+\frac{10}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( \frac{100+10}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( \frac{110}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \frac{12100}{10000}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \frac{12100-10000}{10000} \right \}$

$\large \Rightarrow C.I=10000\times \frac{2100}{10000}$

$\large \Rightarrow C.I=2100$ Rs

iii. Find the sum of money required to clean the debit at the end of 2 years.

Solution:

⇨From Above, We have

⇨p =10000 Rs

⇨C.I. of two years=2100 Rs

⇨Therefore ,

⇨The sum of money required to clean the debit at the end of 2 years Will be

= p + C.I. of two years

= 10000 + 2100

=12100 Rs

iv. Find the differences between the compound interest and the simple interest at the same rate for 2 years.

Solution:

⇨From Above C.I. of two years = 2100 Rs

⇨Given

⇨p=10000

⇨r=10

⇨n=2

⇨For Simple interest

$\large S.I=\frac{p\times r\times n}{100}$

$\large \Rightarrow S.I=\frac{10000\times 10\times 2}{100}$

$\large \Rightarrow S.I=\frac{200000}{100}$

$\large \Rightarrow S.I=2000$ Rs

⇨Therefore the differences between the compound interest and the simple interest at the same rate for 2 years will be

= C.I.- S.I

=2100 - 2000

=100 Rs

Multiple Choice Questions

5. If the simple interest on a sum at 4% per annum for 2 years is Rs 80, then the compound interest on the same sum for the same period is-

Solution:

⇨Given
⇨r = 4%
⇨ n= 2
⇨S.I.= 80
⇨Using Simple Interest formula

$\large S.I=\frac{p\times r\times n}{100}$

$\large \Rightarrow 80=\frac{p\times 4\times 2}{100}$

$\large \Rightarrow 8000=8p$

$\large \Rightarrow p=\frac{8000}{8}$

$\large \Rightarrow p=1000$ Rs

⇨Then Compound interest will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=1000\left \{ \left ( 1+\frac{4}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=1000\left \{ \left ( \frac{104}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=1000\left \{ \frac{10816}{10000}-1 \right \}$

$\large \Rightarrow C.I=1000\left \{ \frac{10816-10000}{10000} \right \}$

$\large \Rightarrow C.I=1000\times \frac{816}{10000}$

$\large \Rightarrow C.I= \frac{816000}{10000}$

$\large \Rightarrow C.I=81.6$ Rs

6. What is the difference between the compound interest and simple interest on Rs 8000 at 15% per annum for 2 years?

Solution:

⇨Given

⇨p = 8000

⇨r = 15%

⇨n = 2

⇨Simple interest will be

$\large S.I=\frac{p\times r\times n}{100}$

$\large \Rightarrow S.I=\frac{8000\times 15\times 2}{100}$

$\large \Rightarrow S.I=80\times 15\times 2$

$\large \Rightarrow S.I=2400$ Rs

⇨And Compound interest will be

$\large C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=8000\left \{ \left ( 1+\frac{15}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=8000\left \{ \left ( \frac{100+15}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=8000\left \{ \left ( \frac{115}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=8000\left \{ \frac{13225}{100}-1 \right \}$

$\large \Rightarrow C.I=8000\left \{ \frac{13225}{10000}-1 \right \}$

$\large \Rightarrow C.I=8000\left \{ \frac{13225-10000}{10000} \right \}$

$\large \Rightarrow C.I=8000\times \frac{3225}{10000}$

$\large \Rightarrow C.I=\frac{25800000}{10000}$

$\large \Rightarrow C.I=2580$ Rs

⇨Therefore the difference between the compound interest and simple interest will be

= C.I.- S.I.

=2580 - 2400

=180 Rs

Tags: Math-Class-8

Compound Interest-Chapter 9- Exercise 9(A)-Class 8

Principal

Rate % p.a

Time (In years)

1. Rs 10000

12 %

2

2. Rs 5000

10 %

2

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