Question

4) Find the compound interest on ₹ 40,000 for 2 years at the rate of 8% ?​

Answer 1

Answer:

ANSWER

It is given that

Principal (P) = 40000

Amount (A) = 48620.25

Period (n) = 2 years = 4 half years

Consider rate of interest = r% p.a. = r/2% half-yearly

We know that

A/P= (1+r/100)

n

Substituting the values

48620.25/40000=(1+r/200)

4

By further calculation

(1+r/200)

4

=4862025/(100×40000)=194481/160000

So we get

(1+r/200)

4

=(21/20)

4

It can be written as

1+r/200=21/20

r/200=21/20−1=1/20

By cross multiplication

r=200×1/20=10

Hence the rate of interest per annum is 10%.

Answer 2

Given:

• Principal (P) = ₹ 40000
• Rate of Interest (R) = 8 %
• Time (n) = 2 years

To find:

• The Compound Interest (C.I)

The formula to find the Compound Interest (C.I) is

$\boxed{C.I = P[(1+ \frac{R}{100})^{n}-1]}$

Now, putting the respective values, we get,

$C.I = 40000 [(1+ \frac{8}{100})^{2} -1]$

$C.I = 40000 [(1+ \frac{2}{25})^{2} -1]$

$C.I = 40000 [( \frac{25+2}{25})^{2} -1]$

$C.I = 40000 [( \frac{27}{25})^{2} -1]$

$C.I = 40000 [( \frac{279}{625}) -1]$

$C.I = 40000 [ \frac{729-625}{625} ]$

$C.I = 40000 \times [ \frac{104}{625} ]$

$C.I = \frac{4160000}{625}$

C.I = ₹ 6,656

Hence,

• The Compound Interest (C.I) is  ₹ 6,656