Kamran129239 wrote:
A car travels from town A to town B at constant speed. If it increases speed by 20%, it will arrive 1 hour ahead of schedule.If it increased by 25% after travelling the first 120 km at usual speed,it will arrive 36 minutes ahead.Find the distance from A to B.
a.205
b.210
c.230
d.220
e.250
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When distance travelled is the same, ratio of speeds is inverse of ratio of time taken.
Case 1: When car increases its speed by 20% over the entire distance.
Ratio of speeds = 5:6 (20% increase in speed), ratio of time taken = 6:5
The actual difference in time taken is 1 hr so in usual case, the car takes 6 hrs but with increased speed, it will take 5 hrs.
Case 2: When car increases its speed by 25% over second part of the distance (ignore the first 120 kms for the time being)
Over the distance that the car increases its speed by 25%, ratio of speeds = 4:5 (increase of 25%)
Then ratio of time taken = 5:4 (a difference of 1 on ratio scale)
The actual difference in time taken is 36 mins = 36/60 hrs = 3/5 hrs. So normally, the car takes 5*(3/5) hrs = 3 hrs over this distance.
Now note that for the entire distance, the car usually takes 6 hrs. For the second part distance, the car usually takes 3 hrs. So this second part distance must be exactly 1/2 of total distance.
Hence total distance must be 2*120 km = 240 km
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