IAS 2020 — CSAT Maths

Correct Option (2)

The average speed for the entire journey is determined by the ratio of the total distance covered to the total time taken.

Let 'd' represent the distance between place X and place Y. The car completes four segments of travel, each covering a distance 'd'.

Total distance covered = d + d + d + d = 4d.

The time taken for each segment is calculated as distance divided by speed:

• From X to Y at v km/hr: $\frac{d}{v}$vd​ hours
• From Y to X at 2v km/hr: $\frac{d}{2v}$2vd​ hours
• From X to Y at 3v km/hr: $\frac{d}{3v}$3vd​ hours
• From Y to X at 4v km/hr: $\frac{d}{4v}$4vd​ hours

Total time taken = $\frac{d}{v} + \frac{d}{2v} + \frac{d}{3v} + \frac{d}{4v}$vd​+2vd​+3vd​+4vd​

To sum these fractions, a common denominator of 12v is used:

Total time taken = $\frac{12d}{12v} + \frac{6d}{12v} + \frac{4d}{12v} + \frac{3d}{12v} = \frac{(12 + 6 + 4 + 3)d}{12v} = \frac{25d}{12v}$12v12d​+12v6d​+12v4d​+12v3d​=12v(12+6+4+3)d​=12v25d​

Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{4d}{\frac{25d}{12v}}$Total timeTotal distance​=12v25d​4d​

Average speed = $\frac{4d \times 12v}{25d} = \frac{4 \times 12v}{25} = \frac{48v}{25}$25d4d×12v​=254×12v​=2548v​

Average speed = 1.92v km/hr.

Since 1.92v km/hr is greater than v km/hr and less than 2v km/hr, the average speed for the entire journey lies between v and 2v km/hr.

Incorrect Options:

Options 1, 3, and 4 are incorrect because the calculated average speed of 1.92v km/hr does not fall within their respective ranges:

• Option 1 suggests the average speed is less than v km/hr, which is false as 1.92v > v.
• Option 3 suggests the average speed lies between 2v and 3v km/hr, which is false as 1.92v < 2v.
• Option 4 suggests the average speed lies between 3v and 4v km/hr, which is false as 1.92v < 3v.