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