UPSC Prelims 2020 · CSAT
Question 72: Basic Numeracy
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
AnswerUPSC’s answer is (b).
Each leg covers the same distance d. Time taken = d/v + d/2v + d/3v + d/4v = (d/v) × (1 + 1/2 + 1/3 + 1/4) = 25d/12v. Total distance is 4d, so average speed = 4d ÷ (25d/12v) = 48v/25 = 1.92v. That lies between v and 2v. With equal distances, the average is a harmonic-type mean, not the simple average of the speeds.
Why each option is right or wrong
- (a) 1.92v is above v, so it is not less than v.
- (b) Right: 48v/25 = 1.92v lies between v and 2v.
- (c) 2v to 3v is where the plain average, 2.5v, sits; that is not the right method here.
- (d) 3v to 4v is far above 1.92v.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2020, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.