Question

A car covers four successive 6 km stretches at speeds of 25kmph, 50kmph, 75kmph and 150kmph respectively. Its average speed over this distance is:
(a) 25kmph
(b) 50kmph
(c) 75kmph
(d) 150kmph

Answer 1

Hint: In this question, the distance is the same in all the four cases. So, by using the formula for average speed if the distance is the same we can get the result for this question.
\[\text{Average speed}=\dfrac{4\times abcd}{\left[ abc+bcd+abd+acd \right]}\]

Complete step-by-step answer:

SPEED: Distance travelled by an object in unit time.
AVERAGE SPEED: Ratio of the total distance and the total time taken by the object to cover that distance.
If an object covers a distance at x kilometre per hour and he covers the same distance at y kilometre per hour, then average speed during whole of the journey is
\[\dfrac{2xy}{x+y}\]
Similarly, we get the formula
\[\text{Average speed}=\dfrac{4\times abcd}{\left[ abc+bcd+abd+acd \right]}\]
Where a, b, c, d are the different speeds at the same distance.
Now, on comparing the given values in the question with the above formula we get,
\[a=25,b=50,c=75,d=150\]
Now, by substituting these values in the above average speed formula we get,
Let us assume the average speed as A.
\[\Rightarrow A=\dfrac{4\times 25\times 50\times 75\times 150}{\left[ 25\times 50\times 75+50\times 75\times 150+25\times 50\times 150+25\times 75\times 150 \right]}\]
Now, on further simplifying the individual terms we get,
\[\Rightarrow A=\dfrac{4\times 25\times 50\times 75\times 150}{\left[ 25\times 25\times 2\times 25\times 3+25\times 6\times 25\times 2\times 25\times 3+25\times 25\times 2\times 25\times 6+25\times 25\times 3\times 25\times 6 \right]}\]
Let us now take the common terms out and then rewrite it.
\[\Rightarrow A=\dfrac{4\times 25\times 50\times 75\times 150}{25\times 25\times 25\left[ 2\times 3+6\times 2\times 3+2\times 6+3\times 6 \right]}\]
Now, by cancelling out the common terms in the numerator and denominator and on further simplification we get,
\[\begin{align}
  & \Rightarrow A=\dfrac{4\times 2\times 3\times 150}{\left[ 6+36+12+18 \right]} \\
 & \Rightarrow A=\dfrac{4\times 6\times 150}{72} \\
 & \therefore A=50kmph \\
\end{align}\]
Hence, the correct option is (b).

Note: Instead of using the formula to find the average speed over this distance by directly substituting the values in the formula of average speed for the same distance we can also do it by calculating the respective speeds over the distance and then find the average to get the average speed. Both the methods give the same result but this is a bit lengthy. While calculating the average speed by substituting the values in the formula there are a lot of chances for calculation mistakes. So, while calculating we should not neglect any of the terms because it results in the wrong final answer.