Question

The amount of a certain principal is ₹6655 in 3 years, compounded annually at the rate of 10 p.c.p.a. Find the principal.

Answer 1

Let principal is P
interest rate , r = 10 % per annual
time period , n = 3
Amount , A = 6655

use formula , A = P(1 + r/100)ⁿ
6655 = P(1 + 10/100)³
6655 = P(1 + 1/10)³ = P(11/10)³
6655 = P(1331/1000)
6655 × 1000/1331 = P
P = 6655 × 1000/1331 = 5000

hence, principal is ₹ 5000

Answer 2

GIVEN :-

amount ( a ) = 7986 rupees

time ( n ) = 3 years

rate ( r ) = 10 %

TO FIND :-

principal of the intrest

SOLUTION :-

as we know that the formula of compound intrest :-

$\implies \boxed{\rm{a = p (1 + \dfrac{r}{100}) {}^{n} } }$

now put the values in formula

$\implies \rm{7986 = p (1 + \dfrac{10}{100}) {}^{3} }$

$\implies \rm{7986 = p (1 + \dfrac{1}{10}) {}^{3} }$

$\implies \rm{7986 = p ( \dfrac{11}{10}) {}^{3} }$

$\implies \rm{7986 = p ( \dfrac{1331}{1000}) }$

$\implies \rm{7986 = p ( 1.331) }$

$\implies \rm{p = \dfrac{7986}{1.331} }$

$\implies \boxed { \boxed{ \rm{p = 6000 \: rupees }}}$

OTHER INFORMATION :-

Compound Interest Definition:

• Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is different from the simple interest where interest is not added to the principal while calculating the interest during the next period. Compound interest finds its usage in most of the transactions in the banking and finance sectors and also in other areas as well. Some of its applications are:

• Increase or decrease in population.

• The growth of bacteria.

• Rise or Depreciation in the value of an item.

The compound interest formula is given below:

• Compound Interest = Amount – Principal

Where,

• A= amount

• P= principal

• R= rate of interest

• n= number of times interest is compounded per year

It is to be noted that the above formula is the general formula for the number of times the principal is compounded in a year. If the amount is compounded annually, the amount is given as:

$\implies \rm{ \bf{a = p (1 + \dfrac{r}{100}) {}^{n} } }$