Question

The Amount of a certain principal is 6655 rupees in 3 years, compounded annually At the rate of 10 p. c. p. a. Find the Principal.​

Answer 1

we know that c = p((1+i)^n  

6625= R((1+10%)^3  

6625=  R((1+0.1)^3  

6625 =  R((1.1)^3  

6625 =  R(1.331)

6625/1.331 =  R

4977 =R is the principal amount..

Answer 2

$\large{ \bold{ \green{ \: solution}}}$

GIVEN:

Amount (A) = 6655 rupees

Rate (R) = 10 p.c.p.a.

Time (N) = 3 years.

Solution:

$\large{ \: A = P × (1 + \frac{r}{100})^{3}}$

$\large{ \: 6655 = P × (1 + \frac{10}{100})^{3}}$

$\large{ \: = P × ( \frac{110}{100})^{3}}$

so,

$\huge{ \: P = \frac{6655 \times {10}^{3} }{11 \times 11 \times 11}}$

$\huge{ \: P = 5 \times {10}^{3}}$

$\huge{ \: = 5000}$

Hence,

The principal was 5000 rupees.

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THANKS.!! ;-)