A car travels from a place X to place Y at an average speed of v km/hr, from Y to X at an average speed of 2v km/hr, again from X to Y at an average speed of 3v km/hr and again from Y to X at an average speed of 4v km/hr. Then the average speed of the car for the entire journey

(a) is less than v km/hr
(b) lies between v and 2v km/hr ✓ UPSC's answer
(c) lies between 2v and 3v km/hr
(d) lies between 3v and 4v km/hr

Why the answer is (b)

• Let the distance between X and Y be $d$ km. The total distance for the four legs is $4d$ km.

• The time taken for each leg is calculated as distance divided by speed: $t_1 = d/v$, $t_2 = d/(2v)$, $t_3 = d/(3v)$, and $t_4 = d/(4v)$.

• The total time $T$ is the sum of these times: $T = \frac{d}{v} + \frac{d}{2v} + \frac{d}{3v} + \frac{d}{4v} = \frac{d}{v} \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

• Simplifying the fraction inside the parenthesis: $1 + 0.5 + 0.333... + 0.25 = \frac{12+6+4+3}{12} = \frac{25}{12}$.

• Thus, $T = \frac{25d}{12v}$.

• The average speed is Total Distance / Total Time: $\frac{4d}{\frac{25d}{12v}} = \frac{48v}{25} = 1.92v$.

• Since $1.92v$ is greater than $v$ and less than $2v$, the average speed lies between $v$ and $2v$ km/hr.

Why the other options are wrong

(a) is less than v km/hr: The calculated average speed is 1.92v, which is greater than v, so it cannot be less than v.
(c) lies between 2v and 3v km/hr: The calculated average speed is 1.92v, which is less than 2v, so it does not lie between 2v and 3v.
(d) lies between 3v and 4v km/hr: The calculated average speed is 1.92v, which is significantly less than 3v, so it does not lie between 3v and 4v.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2020, held on 4 October 2020. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.