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CAT 2017 Slot 2 — QA Question 12

Relative SpeedEasy

A motorbike leaves point A at 1 pm and moves towards point B at a uniform speed. A car leaves point B at 2 pm and moves towards point A at a uniform speed which is double that of the motorbike. They meet at 3:40 pm at a point which is 168 km away from A. What is the distance, in km, between A and B?

Answer & solution

Correct answer: 378

  • A

    364

  • 378

  • C

    380

  • D

    388

Solution

Easy

At the meeting point the two distances add to ABAB. Use the bike's known distance and travel time to get its speed, double it for the car, then find how far the car travelled in its own time window.

1

Bike's speed. It runs from 1:001{:}00 to 3:403{:}40, i.e. 22 h 4040 min =83=\tfrac83 h, covering 168168 km to the meeting point.

vbike=1688/3=168×38=63 km/h\begin{aligned} &v_{\text{bike}}=\frac{168}{8/3}=168\times\frac{3}{8}=63\ \text{km/h} \end{aligned}
2

Car's speed and distance. The car is twice as fast and runs from 2:002{:}00 to 3:403{:}40, i.e. 11 h 4040 min =53=\tfrac53 h.

vcar=2×63=126 km/hdcar=126×53=210 km\begin{aligned} &v_{\text{car}}=2\times 63=126\ \text{km/h}\\ &d_{\text{car}}=126\times\frac{5}{3}=210\ \text{km} \end{aligned}
3

Total distance ABAB. The bike's 168168 km (from AA) plus the car's 210210 km (from BB) span the whole route.

AB=168+210=378 km\begin{aligned} &AB=168+210=378\ \text{km} \end{aligned}
AB=378 kmAB=378\ \text{km}

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CAT 2017 Slot 2 QA Q12: A motorbike leaves point A at 1 pm and moves towards point B at a uniform speed. A car leaves point B at 2 pm — Solution | TheCATExam