| A person travels a certain distance at 3 km/h and reaches 15 min late. If he travels at 4 km/h, he reaches 15 min earlier. The distance he has to travel is |
A) 4.5 km
B) 6 km
C) 7.2 km
D) 12 km
Correct Answer: B
Solution :
| Let the certain distance bed and timer. |
| Now, by given condition, |
| \[\frac{d}{3}=(t+15)\min =\frac{(t+15)}{60}h\] |
| \[\Rightarrow \]\[20d=t+15\]\[\Rightarrow \]\[t=20\,\,d-15\] (i) |
| and \[\frac{d}{4}=(t-15)\min =\frac{(t-15)}{60}h\] |
| \[\Rightarrow \]\[15\,\,d=t-15\]\[\Rightarrow \]\[t=15\,\,d+15\] (ii) |
| From Eqs. (i) and (ii), we get |
| \[20\,\,d-15=15\,\,d+15\]\[\Rightarrow \]\[5d=30\] |
| \[\therefore \] \[d=6\,\,km\] |
You need to login to perform this action.
You will be redirected in
3 sec
