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CAT 2019 Slot 1 — QA Question 20

Linear RaceEasy

In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?

Answer & solution

Answer: 880

Solution

Easy

All three start together, so in any matchup the speeds (hence distances covered in the same time) are in a fixed ratio. Write the distances when the first finishes (LL), then the second-vs-third ratio when the second finishes, and equate the two expressions for the speed ratio of horse 2 to horse 3.

1

When horse 1 finishes the course of length LL. It beats the 2nd by 11 m and the 3rd by 90 m.

d2=L−11,d3=L−90⇒ v2v3=L−11L−90(same time)\begin{aligned} &d_2=L-11,\qquad d_3=L-90\\ &\Rightarrow\ \frac{v_2}{v_3}=\frac{L-11}{L-90}\quad\text{(same time)} \end{aligned}
2

When horse 2 finishes the course. It beats the 3rd by 80 m.

v2v3=LL−80(same time)\begin{aligned} &\frac{v_2}{v_3}=\frac{L}{L-80}\quad\text{(same time)} \end{aligned}
3

Equate the speed ratios. The ratio v2:v3v_2:v_3 is constant.

L−11L−90=LL−80[step 1 = step 2]⇒ (L−11)(L−80)=L(L−90)⇒ L2−91L+880=L2−90L\begin{aligned} &\frac{L-11}{L-90}=\frac{L}{L-80}\quad\text{[step 1 = step 2]}\\ &\Rightarrow\ (L-11)(L-80)=L(L-90)\\ &\Rightarrow\ L^2-91L+880=L^2-90L \end{aligned}
4

Solve for LL. The L2L^2 terms cancel.

−91L+880=−90L⇒ 880=L\begin{aligned} &-91L+880=-90L\\ &\Rightarrow\ 880=L \end{aligned}
Length of racecourse=880 m\text{Length of racecourse}=880\text{ m}
CAT 2019 Slot 1 QA Q20: In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second be — Solution | TheCATExam