11.
Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. The length of the race is
-
36 metres
-
48 metres
-
60 metres
-
60 metres
B.
48 metres
Let A finishes x m which is the required distance Now, according to question, B finishes in (x -12) m C finishes in (x - 18) mAgain, If B finishes in x m then c finishes in (x - 8)m
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Given:
Speed when bus goes A to B = 40 km/h
Speed when bus returns B from A = 60 km/h
Formula used:
Average speed = 2ab/(a + b)
a = Speed a object when it goes from A to B.
b = Speed of a object when it returns same place
Formula is only used when the distance travels by all the speed is equal.
Calculation:
According to the question,
Average speed = (2 × 40 × 60)/(40 + 60)
⇒ Average speed = (2 × 40 × 60)/100
⇒ Average speed = 48 km/h
∴ The average speed of the bus is 48 km/h.
1) When the two speed are given,
Average speed = 2xy/(x + y)
2) When three speeds are given,
Average speed = 3xyz/(xy + yz + zx)
Note: All the above formula is only used when the distance travels by all the speed is equal.
Let's discuss the concepts related to Speed Time and Distance and Average Speed. Explore more from Quantitative Aptitude here. Learn now!
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A motorist travels A to B at a speed at 40 km/h and returns at speed of 60 km/h. his average speed will be between 40 & 60. -------------- Use d = r*t t = d/r d = distance one way. Total time = d/40 + d/60
Round trip avg =
= = 2*40*60/(40+60) = 48 km/hr ---------------- Notice it's similar to parallel resistors and parallel flows. Avg = 2*r1*r2/(r1 + r2) ==============
Also notice the distance is not a factor.