Math, asked by Class8student, 10 months ago

Compound interest on rupees 1800 at 10% per annum for a certain period of time in Rupees 378 find the time in years

Answers

Answered by nishapragu
1

Step-by-step explanation:

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}}

A=P(1+

100

r

)

n

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

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

100

10

)

n

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

1800

2178

=(1+

10

1

)

n

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

100

121

=(

10

10+1

)

n

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

10

11

)

2

=(

10

11

)

n

n=2n=2

/* we know the Exponential Law:

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

Ifa

m

=a

n

⟹m=n

*/

Therefore,

Number of times interest paid (n) = 2

Time (T) = 2 years

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