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CAT 2020 Slot 1 — QA Question 2

Relative SpeedEasy

A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is

Answer & solution

  • A

    100

  • 90

  • C

    70

  • D

    80

Solution

Easy

For two bodies starting at opposite ends and meeting, there is a clean fact: if after meeting they take times t1t_1 and t2t_2 to finish, then the ratio of speeds is v1v2=t2t1\tfrac{v_1}{v_2}=\sqrt{\tfrac{t_2}{t_1}}. Even without recalling it, equating the two parts of the road (covered before vs. after the meeting) cracks it.

1

Set up the meeting point. Let them meet tt minutes after start, let Car 2's speed be bb km/h, and let MM be the meeting point. Each car's distance before the meeting equals the other car's distance after it.

AM=60 t=b⋅20(Car 1 took 45, Car 2 covers AM in 20)BM=b t=60⋅45(Car 2 took 20, Car 1 covers BM in 45)\begin{aligned} &AM=60\,t = b\cdot 20 \quad\text{(Car 1 took 45, Car 2 covers }AM\text{ in 20)}\\ &BM=b\,t = 60\cdot 45 \quad\text{(Car 2 took 20, Car 1 covers }BM\text{ in 45)} \end{aligned}
2

Multiply the two equations to eliminate bb.

(60 t)(b t)=(20b)(60⋅45)[(1)×(2)]⇒ 60 b t2=54000 b⇒ t2=900⇒ t=30 minutes\begin{aligned} &(60\,t)(b\,t)=(20b)(60\cdot 45) \quad\text{[(1)×(2)]}\\ &\Rightarrow\ 60\,b\,t^{2}=54000\,b\\ &\Rightarrow\ t^{2}=900\\ &\Rightarrow\ t=30\ \text{minutes} \end{aligned}
3

Back-substitute to get bb. Use 60 t=20b60\,t=20b from step 1 with t=30t=30.

60×30=20 b⇒ b=180020=90 km/h\begin{aligned} &60\times 30 = 20\,b\\ &\Rightarrow\ b=\frac{1800}{20}=90\ \text{km/h} \end{aligned}

Directly: v2v1=t1t2=4520=32\dfrac{v_2}{v_1}=\sqrt{\dfrac{t_1}{t_2}}=\sqrt{\dfrac{45}{20}}=\dfrac{3}{2}, so v2=32×60=90v_2=\dfrac{3}{2}\times 60=90 km/h.

b=90 km/hb=90\ \text{km/h}