Question

Find the rate of compound interest per annum at which ₹25000 will amount to

₹ 31360 in 2 years.​

Answer 1

Answer :

• The rate of the compound Interest per annum at which Rs. 25000 will amount to Rs. 31360 , R = 12 % p.a.

Explanation :

Given :

• Sum of money , Principal (P) = Rs. 25000

• Amount, A = Rs. 31360

• Time period, n = 2 years

To find :

• Rate of interest p.a. , R = ?

Solution :

We know the formula for amount of a transaction (in case of compound interest) ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀A = P(1 + R/100)^n

Where :

• A = Amount
• P = Principal
• R = Rate of interest
• n = Time period

Now using the above formula and substituting the values in it, we get :

==> A = P(1 + R/100)^n

==> 31360 = 25000(1 + R/100)²

==> 31360/25000 = (1 + R/100)²

==> 3136/2500 = [(100 + R)/100]²

==> 784/625 = [(100 + R)/100]²

==> (28/25)² = [(100 + R)/100]²

==> 28/25 = (100 + R)/100

==> 28/25 × 100 = 100 + R

==> 28 × 4 = 100 + R

==> 112 = 100 + R

==> 112 - 100 = R

==> 12 = R

∴ R = 12% p.a.

Therefore,

• The rate of interest per annum is 12% p.a.

Answer 2

Solution

G I V E N :

• principal ( p) = ₹ 25000
• Amount ( A) = Rs.31360
• Time period ( T ) = 2 years

T O F I N D :

• Rate of interest per annum = ?

A N S W E R :

To find the rate per annum we have to use the formula of amount which is : A = p ( 1 + r/100)^ t

W H E R E ,

• A = Amount
• P = Principal
• R = Rate per annum
• T = Time period

N O W ,

• Putting the values in the equation we get ;

$\small{ = > a \: = \: p(1 + \frac{r}{100})} ^{2}$

$= > 31360 = 25000(1 + \frac{r}{100} ) ^{2}$

$= > \frac{31360}{25000} = (1 + \frac{r}{100} ) ^{2}$

$= > ( \frac{28}{25} ) = (100 + \frac{r}{100} ) ^{2}$

$= > (\frac{28}{25} ) ^{2} = (100 + \frac{r}{100} ) ^{2}$

$= > (\frac{28}{25} )^{ \cancel2} = (100 + \frac{r}{100} ) ^{ \cancel{2}}$

$= > \frac{28}{25} = 100 + \frac{r}{100}$

$= > \frac{28}{25} \times 100 = r + 100$

$= > 28 \times 4 = 100 + r$

$= > 112 = 100 + r$

$= > r \: = 112 - 100$

$= > r = 12 \%$

T H E R E F O R E ,

• Rate per annum is 12 % .