A person travels from X to Y at a speed of 40 km/hour and returns by increasing speed by 50%. What is his average speed for both the trips ? (Ans: 48km/hr)
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Answers
Answer:
mack me as brainiest please
Explanation:
mack me as brainiest please
Explanation:
A person travels from A to B at a speed of 40 km/h and returns by increasing his speed by 50%. What is his average speed for both the trips?
S -average speed
D -distance from A to B
T1 -time it takes to go from A to B
T2 -time it takes to go from B to A
RT = D -formula relating speed (Rate), Time and Distance
S = 2D/(T1 + T2)
T1 = D/40
T2 = D/(40 * 1.5)
T1 + T2 = D/40 + D/(40 * 1.5)
T1 + T2 = 1.5D + D/(40 * 1.5)
T1 + T2 = 2.5D/60
T1 + T2 = D/24
S = 2D/(D/24)
S = 48km/h
Normal speed=40 kmph
Increased speed by 50%=40*150/100=60 kmph
Let the distance be 120 km
Time taken from A to B=120/40=3 hrs
Time taken from B to A=120/60=2 hrs
Total time taken=3+2=5 hrs
Total distance covered=120+120=240 km
Average speed=240/5=48 kmph