Question

The compound interest on rs.1800 at the rate of 10% per annum for a certain period of time is rs.378. find the time in years

Answer 1

A = P(1+r/100)^t
A = 1800+378 = 2178
2178 = 1800(1+10/100)^t
2178/1800 = (1+0.1)^t
1.21 = 1.1^t
1.1^t = 1.1^2
Therefore t = 2 because the base 1.1 is same on both sides.

Answer 2

Answer:

Time (T) = 2 years

Explanation:

Given

Principal (P) = Rs 1800

Rate of interest (r) = 10% p.a

Let Time = T years

Number of times interest paid = n

Compound interest (C.I) = Rs378

Amount (A) = P+C.I

= Rs 1800 + Rs 378

= Rs 2178

We know that,

$\boxed { A = P \left(1+\frac{r}{100}\right)^{n}}$

Now , substitute the values in the above formula, we get

$\implies 2178 = 1800\times \left( 1+ \frac{10}{100}\right)^{n}$

$\implies \frac{2178}{1800} = \left( 1+ \frac{1}{10}\right)^{n}$

$\implies \frac{121}{100} = \left(\frac{10+1}{10}\right)^{n}$

$\implies \left(\frac{11}{10}\right)^{2} = \left(\frac{11}{10}\right)^{n}$

$n=2$

/* we know the Exponential Law:

$\boxed { If \: a^{m} = a^{n} \implies m = n }$

*/

Therefore,

Number of times interest paid (n) = 2

Time (T) = 2 years

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