A) \[=\frac{P}{{{\left( 1-\frac{R}{100} \right)}^{n}}}\]
B) \[\frac{a}{b}=K\]
C) \[{{a}_{1}}\]
D) 60%
Correct Answer: A
Solution :
Let first number be x and second number be y and let second number be reduced by k % so \[\left( x+y-\frac{xy}{100} \right)%\] \[\left( 15+4-\frac{15\times 4}{100} \right)=19-\frac{60}{100}=19-\frac{3}{5}=19-0.6=18.4%\] \[S.I.=\frac{P\times R\times T}{100}\]
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