A) 20
B) 35
C) \[36\frac{2}{3}\]
D) 40
Correct Answer: D
Solution :
Let the distance be x km and initial speed y km/h. |
According to question, \[\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60}\] ...(i) |
and \[\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60}\] ?(ii) |
From Eqs. (i) and (ii), we get |
\[\frac{x}{y}-\frac{x}{y+3}=\frac{x}{y-2}-\frac{x}{y}\] |
\[\Rightarrow \] \[\frac{1}{y}-\frac{1}{y+3}=\frac{1}{y-2}-\frac{1}{y}\] |
\[\Rightarrow \] \[\frac{y+3-y}{y\,\,(y+3)}=\frac{y-y+2}{y\,\,(y-2)}\] |
\[\Rightarrow \] \[3\,\,(y-2)=2(y+3)\] |
\[\Rightarrow \] \[3y-6=2y+6\] |
\[\Rightarrow \] \[y=12\] |
From Eq. (i) \[\frac{x}{12}-\frac{x}{15}=\frac{40}{60}\]\[\Rightarrow \]\[\frac{5x-4x}{60}=\frac{2}{3}\] |
\[\Rightarrow \] \[x=\frac{2}{3}\times 60=40\] |
\[\therefore \] Distance = 40 km |
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