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

byAmlesh Mahato •June 11, 2020

0

The Biotic World is the best free Educational Website for Students. The Biotic World Provides NCERT Solution PDF, CBSE Notes PDF, NCERT Books PDF for class 8, 9, 10, 11, 12. The Biotic World also provides GU TDC Syllabus, Previous Years Question Papers, TDC notes on different topics of Physics, Chemistry, Mathematics, Zoology and Botany any many more. So, Follow us to get tuned with latest updates.

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

Post a Comment

Contact form