A) 40
B) 50
C) 60
D) 70
Correct Answer: B
Solution :
Let bus starts with x number of passengers. After 1st stoppage, number of passengers = \[x-\frac{x}{5}+40=\frac{5x-x+200}{5}\] = \[\frac{4x+200}{5}\] After 2nd stoppage, number of passengers = \[\frac{4x+200}{5}-\frac{4x+200}{5\times 2}+30\] = \[\frac{4x+200}{5}-\frac{4x+200}{10}+30\] \[\Rightarrow \,\,\,\,\,\,\,\frac{4x+200}{5}\left( 1-\frac{1}{2} \right)+30=70\,\,(given)\] \[\Rightarrow \,\,\,\,\,\,\,\frac{4x+200}{10}+30=70\] \[\Rightarrow \,\,\,\,\,\,\,4x+200=400\Rightarrow 4x=200\] \[\therefore \,\,\,\,\,\,x=\mathbf{50}\]You need to login to perform this action.
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