A car after traveling 18 km from a point A developed some problem in the engine and the speed became $$\frac{4}{5}$$ th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:
A. 25 kmph
B. 30 kmph
C. 20 kmph
D. 35 kmph
E. None of these
Answer: Option C
Solution(By Examveda Team)
He proceeds at $$\frac{4}{5}$$ S where S is his usual speed means $$\frac{1}{5}$$ decrease in speed which will lead to $$\frac{1}{4}$$ increase in time.Now the main difference comes in those 12km (30 - 18) and the change in difference of time = (45 - 36) min = 9 min Thus, $$\frac{1}{4}$$ × T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km T = 36 min = $$\frac{36}{60}$$ hours = 0.6 hours. Speed of the car = $$\frac{12}{0.6}$$ = 20 kmph.
This Question Belongs to Arithmetic Ability >> Speed Time And Distance
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Related Questions on Speed Time and Distance
A. 48 min.
B. 60 min.
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D. 62 min.
E. 66 min.
A. 262.4 km
B. 260 km
C. 283.33 km
D. 275 km
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A. 4 hours
B. 4 hours 30 min.
C. 4 hours 45 min.
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45 mins -36 mins =9 mins . Ye 9 mins ka diff is coming because of change in km nd speed of 12km /9mins = 80 km / hr speed nhi hogi ??
Late,
Before problem speed of car = 5x km/hr
After problem it is = 4x km/hr
So,
12/4x - 12/5x = 9/60
1/20x = 3/20*12
1/x = 1/4
x = 4
speed = 20 km/hr
Please explain this line
T = 36 min = 36 /60 hours = 0.6 hours.
Travelling extra 12 km decreases the delay time by 9 min. Therefore for every 1 extra km delay time decreases by 0.75 min. Thus travelling 60 km more , decreases delay time by 45 min.Thus total distance covered is 18 +60 =78 km.
Also , 12 km is covered in 36 min thus, original speed is 20 km per hour.