. Point out the mistake or ambiguity in the following statement, “A person goes X to Y on cycle at 20 km/h and returns at 24 km/h his average speed was 22 km/h”
(Ans:21.82km/h)
Answers
Answer:
If a motorist travels from a to b at a speed of 40km/hr and returns at speed of 60km/hr, what is his average speed?
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This is a very easy question to answer, but I am going to answer it at length, so as to explain the derivation rather than just give it. This is very similar to some of the math used on a naval vessel. We had to work things like this out for calculating things like time, fuel load, exposure (a different tale altogether), and so forth.
The most basic representation of this question in its culmination is V = D/T
You did not supply 2 of the variables, leaving one to insert his own.
The equation is extended beyond the basic by the given variation in speed for each leg of a fixed-distance round trip.
The distance between points A & B must be stated; since unsupplied, assume 100 Km.
Elapsed time is crucial, as with differing speed given for each leg, elapsed time will vary on each leg, and total elapsed time--another unsupplied variable--must be known to derive average speed over a known, fixed distance.
Defining terms: V, speed in kph; D, distance in kilometers; T, time = hours. Again, basic equation is V = D/T; or,
D/V = T; solve for T.
100 km / 40 kph = 2.5 hr. This is leg 1, or A.
100 km / 60 kph = 1.6667 hr. This is keg 2, or B.
Now we can average speed between the two legs. Average is the mean between the combination of in this case 2 elements, A and B, legs of travel along a fixed distance. Therefore, since the distance is 100 km, the final equation becomes:
2(D) / (TA + TB) = V
200 / 2.5 + 1.6667 = V
200 / 4.1667 = V
47.9996 = V
On any speedometer, this would register as 48 kph.