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A man cover a distance in t hrs, if he met with an accident after 20km and he then run at his 3/5th of his normal speed, so he reach his destination 40minutes late. If he met with an accident at 30km then he reach only 30min late. Find his original speed?
60kmph
50kmph
40kmph
24kmph
Solution: After moving 10km more, man saves 10minutes. For saving of 40minutes he has to cover 40km. Now speed: after accident before accident . 3 5 Time 5 3 . (+2) 2=40minutes 1=20min (Normal time) 3= 60min So he covers 40km in 60min with normal speed = 40kmph
By: Pranav Gupta ProfileResourcesReport error
JEEVANANTHAM P
After moving 10km more, man saves 10minutes. For saving of 40minutes he has to cover 40km more. So the distance is 60km ?????
Time saved by moving the accident point depends on the reduced speed, not the original speed. When the accident occurs earlier, a larger portion of the journey is covered at the slower speed, leading to a greater delay. As the accident point moves further, the portion covered at the slower speed decreases, and the delay reduces, but not at a constant rate. So, you can not say "For saving of 40min he has to cover 40km more"..
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