Management Aptitude Test 2014 Paper

(a) Let the cost of components 3x, 4x, 5x and 6x, respectively of A, B, C and D.
Change of cost in component A
= 3x + $\frac{3\mathit{x}\times 10}{100}=\frac{33}{100}\mathit{x}$
Change of cost in component B
= 4x - $\frac{4\mathit{x}\times 20}{100}=\frac{16}{5}\mathit{x}$
Change of cost in component C
= 5x - $\frac{5\mathit{x}\times 30}{100}=\frac{7}{2}\mathit{x}$
Change of cost in component D
= 6x + $\frac{6\mathit{x}\times 40}{100}=\frac{42}{5}\mathit{x}$
Now, total cost before change
= 3x + 4x + 5x + 6x = 18x
Total cost after change
= $\frac{33}{10}\mathit{x}+\frac{16}{5}\mathit{x}+\frac{7}{2}\mathit{x}+\frac{42}{5}\mathit{x}$ = $\frac{184}{10}\mathit{x}$
∴ Total change in cost = $\frac{184}{10}\mathit{x}$ - 18x = $\frac{4}{10}\mathit{x}$ = 0.4x
∴ Percentage change = $\frac{0.4\mathit{x}}{18\mathit{x}}$ × 100 = $\frac{40}{18}=\frac{20}{9}$ = 2.22%