NIMCET 2017 Question Paper with solutions

Let number of breads available be V.
According to question,
Number of breads robbed by lot thief,
= $\frac{\mathit{x}}{2}+\frac{1}{2}$ = $\frac{\mathit{x}+1}{2}$
Now, number of breads left after 1st robbery, = x - $\frac{\mathit{x}+1}{2}$ = $\frac{\mathit{x}-1}{2}$
Again, number of breads robbed by 2nd thief,
= $\frac{1}{2}\left(\frac{\mathit{x}-1}{2}\right)+\frac{1}{2}$
= $\frac{\mathit{x}-1}{4}+\frac{1}{2}$ = $\frac{\mathit{x}+1}{2}$
So, number of bread left after 2nd robbery,
= $\left(\frac{\mathit{x}-1}{2}\right)-\left(\frac{\mathit{x}+1}{4}\right)$
= $\frac{2\mathit{x}-2-\mathit{x}-1}{4}$ = $\frac{\mathit{x}-3}{4}$
Now, number of breads robbed by 3rd thief,
= $\frac{1}{2}\left(\frac{\mathit{x}-3}{4}\right)+\frac{1}{2}$
= $\frac{\mathit{x}-8}{8}+\frac{1}{2}$ = $\frac{\mathit{x}+1}{8}$
Hence, number of breads left after 3rd robbery = $\left(\frac{\mathit{x}-3}{4}\right)-\left(\frac{\mathit{x}-1}{8}\right)$ = $\frac{\mathit{x}-7}{8}$
Now , $\frac{\mathit{x}-7}{8}$ = 3
⇒ x = 31
Therefore, total number of breads present initially is 31.