RRB NTPC Prelims Model Paper 5 with solutions

(1) After reducing 20% in rate,20 more orange are found.
So, reduced cost of 20 orange$=20\%\mathit{o}\mathit{f}120=120\times \frac{20}{100}=24$
∴ reduced cost of 1 orange$=\frac{24}{20}=\mathit{₹}1.2$
∴ 20% reduced in cost.
So, Reduced cost of orange is the 80% of basic cost.
∴ 80% of (basic cost) = 1.2
$\Rightarrow \frac{80}{100}\times \left(\mathit{b}\mathit{a}\mathit{s}\mathit{i}\mathit{c}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\right)=1.2$
∴ basic cost =$\frac{12}{10}\times \frac{100}{80}=\mathit{₹}1.5$