X, Y and Z are three contestants in a race of 1000 m. Assume that all run with different uniform speeds. X gives Y a start of 40 m and X gives Z a start of 64 m. If Y and Z were to compete in a race of 1000 m, how many metres start will Y give to Z ?
- (a) 20
- (b) 25 ✓ UPSC's answer
- (c) 30
- (d) 35
Why the answer is (b)
• X gives Y a start of 40 m in a 1000 m race, meaning when X covers 1000 m, Y covers 960 m; hence the speed ratio X : Y = 1000 : 960 = 25 : 24.
• X gives Z a start of 64 m in a 1000 m race, meaning when X covers 1000 m, Z covers 936 m; hence the speed ratio X : Z = 1000 : 936 = 125 : 117.
• To find the ratio Y : Z, divide the two ratios: Y/Z = (24/25) ÷ (117/125) = (24 × 125)/(25 × 117) = 3000/2925 = 40/39.
• Therefore, when Y runs 1000 m, Z runs (39/40) × 1000 = 975 m.
• The start Y must give Z in a 1000 m race is 1000 − 975 = 25 m, which corresponds to option (b).
Why the other options are wrong
- (a) 20
- 20 m would require Y:Z = 1000:980 = 50:49, but the actual ratio derived from the given starts is 40:39, not 50:49.
- (c) 30
- 30 m would require Y:Z = 1000:970 = 100:97, which does not match the 40:39 ratio obtained from the two given starts.
- (d) 35
- 35 m would require Y:Z = 1000:965 = 200:193, which is inconsistent with the 40:39 ratio derived from the problem data.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2019, held on 2 June 2019. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.