CAT 1996 — QA Question 23

In a mile race, Akshay can be given a start of 128 m by Bhairav. This means that Bhairav can afford to start after Akshay has travelled 128 m and still complete one mile with him. In other words, Bhairav can travel one mile, i.e. 1,600 m in the same time as Akshay can travel (1600 – 128) = 1,472 m. Hence, the ratio of the speeds of Bhairav and Akshay = Ratio of the distances travelled by them in the same time = $\frac{1600}{1472}$ = 25 : 32. Bhairav can give Chinmay a start of 4 miles. This means that in the time Bhairav runs 100 m, Chinmay only runs 96 m. So the ratio of the speeds of Bhairav and Chinmay = $\frac{100}{96}$ = 25 : 24.

Hence, we have B : A = 25 : 23 and B : C = 25 : 24. So A : B : C = 23 : 25 : 24. This means that in the time Chinmay covers 24 m, Akshay only covers 23 m. In other words, Chinmay is faster than Akshay. So if they race for $1\frac{1}{2}$ miles = 2,400 m, Chinmay will complete the race first and by this time Aksahy would only complete 2,300 m. In other words, Chinmay would beat Akshay by 100 m = $\frac{1}{16}$ mile.