| A man covers a certain distance on scooter. Had he moved 3 km/h faster, he would have taken 40 min less. If he had moved 2 km/h slower, he would have taken 40 min more. The distance (in km) is |
A) 42.5
B) 36
C) 37.5
D) 40
Correct Answer: D
Solution :
| Let distance and original speed of the man be d km and s km/h. |
| Then, \[\frac{d}{s}-\frac{d}{s+3}=\frac{2}{3}\] |
| \[\Rightarrow \] \[\frac{d\,(s+3-s)}{s\,(s+3)}=\frac{2}{3}\] |
| \[\Rightarrow \] \[9d=2s\,(s+3)\] ... (i) |
| and \[\frac{d}{s-2}-\frac{d}{s}=\frac{2}{3}\] |
| \[\Rightarrow \] \[\frac{d\,(s-s+2)}{s\,(s-2)}=\frac{2}{3}\] |
| \[\Rightarrow \] \[3d=s(s-2)\] (ii) |
| From Eqs. (i) and (ii), we get |
| \[3s\,(s-2)=2s\,(s+3)\] |
| \[\Rightarrow \] \[3{{s}^{2}}-6s=2{{s}^{2}}+6s\]\[\Rightarrow \]\[{{s}^{2}}=12s\] |
| \[\therefore \] \[s=12\] |
| From Eq. (ii), we get |
| \[3d=12\,(12-2)\] |
| \[\therefore \] \[d=40\,km\] |
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