UPSC Prelims 2013 · CSAT
Question 36: Basic Numeracy
A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original speed of the train in km/hr ?
AnswerUPSC’s answer is (c).
Let the original speed be x km/h. The times add up to 3 hours: 63/x + 72/(x + 6) = 3. Multiply out: 63(x + 6) + 72x = 3x(x + 6), so 135x + 378 = 3x^2 + 18x, which simplifies to x^2 - 39x - 126 = 0, or (x - 42)(x + 3) = 0. Speed cannot be negative, so x = 42. Check: 63/42 = 1.5 hours and 72/48 = 1.5 hours, a total of 3 hours.
Why each option is right or wrong
- (a) Wrong. At 24 km/h the times are 63/24 + 72/30, about 5 hours.
- (b) Wrong. At 33 km/h the times are 63/33 + 72/39, about 3.8 hours.
- (c) Right. 63/42 + 72/48 = 1.5 + 1.5 = 3 hours.
- (d) Wrong. At 66 km/h the times are 63/66 + 72/72, about 1.95 hours.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2013, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.