Lakshman Patel RJIT
asked
in Quantitative Aptitude
Mar 31, 2020
retagged
Sep 13, 2020
by Lakshman Patel RJIT

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$\text{Amount(A) = Principal(P)}\left(1+\frac{r}{100}\right)^{n};$ where $r = $ rate of interest and $n = $ time.

$A = 8000\left(1+\frac{5}{100}\right)^{2} = 8000\left(\frac{21}{20}\right)^{2} = ₹8,820.$

$\text{PS:}$ We can do using simple interest also, while calculating the interest, we calculate for every one year.

So, the correct answer is $(C).$

$A = 8000\left(1+\frac{5}{100}\right)^{2} = 8000\left(\frac{21}{20}\right)^{2} = ₹8,820.$

$\text{PS:}$ We can do using simple interest also, while calculating the interest, we calculate for every one year.

So, the correct answer is $(C).$