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CAT 2020 Slot 2 — QA Question 5

Linear RaceEasy

In a car race, car A beats car B by 45 km, car B beats car C by 50 km, and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is

Answer & solution

  • A

    550

  • 450

  • C

    475

  • D

    500

Solution

Easy

In equal-time races, each "beats by" statement is a ratio of speeds over the same race length DD. Since all cars run for the same time in any pairwise comparison, the speed ratios multiply: (a/b)(b/c)=a/c(a/b)(b/c)=a/c. This gives one equation in DD.

1

Translate the three statements. When the winner finishes DD, the loser has covered D−(margin)D-(\text{margin}) in the same time, so the speed ratio equals the distance ratio.

ab=DD−45(A beats B by 45)(1)bc=DD−50(B beats C by 50)(2)ac=DD−90(A beats C by 90)(3)\begin{aligned} &\frac{a}{b} = \frac{D}{D-45} \quad\text{(A beats B by 45)}\quad (1)\\ &\frac{b}{c} = \frac{D}{D-50} \quad\text{(B beats C by 50)}\quad (2)\\ &\frac{a}{c} = \frac{D}{D-90} \quad\text{(A beats C by 90)}\quad (3) \end{aligned}
2

Combine. Multiplying (1) and (2) gives a/ca/c, which must equal (3).

DD−45⋅DD−50=DD−90[(1)×(2)=(3)]⇒ D(D−45)(D−50)=1D−90\begin{aligned} &\frac{D}{D-45}\cdot\frac{D}{D-50} = \frac{D}{D-90} \quad\text{[(1)×(2)=(3)]}\\ &\Rightarrow\ \frac{D}{(D-45)(D-50)} = \frac{1}{D-90} \end{aligned}
3

Solve for DD. Cross-multiply and simplify.

D(D−90)=(D−45)(D−50)⇒ D2−90D=D2−95D+2250⇒ 5D=2250⇒ D=450 km\begin{aligned} &D(D-90) = (D-45)(D-50)\\ &\Rightarrow\ D^2 - 90D = D^2 - 95D + 2250\\ &\Rightarrow\ 5D = 2250\\ &\Rightarrow\ D = 450\ \text{km} \end{aligned}
D=450 kmD = 450\ \text{km}
CAT 2020 Slot 2 QA Q5: In a car race, car A beats car B by 45 km, car B beats car C by 50 km, and car A beats car C by 90 km. The dis — Solution | TheCATExam