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CAT 2018 Slot 1 — QA Question 2

Basics of TSD/ProportinalityEasy

Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

Answer & solution

Correct answer: 12

Answer: 12

Solution

Easy

P divides AB with BP=3 APBP = 3\,AP. Let AP=xAP = x, so PB=3xPB = 3x. Car 1 covers xx to reach P; car 2 covers 3x3x to reach P and arrives 1 hour (60 min) later, with car 2's speed half of car 1's. Translate "same distance ratio at given speeds" into one equation in the unknown time tt.

A P B x 3x
1

Name the times and speeds. Let car 1 reach P in tt minutes, so car 2 reaches P in (t+60)(t+60) minutes.

speed of car 1=xt,speed of car 2=3xt+60\begin{aligned} &\text{speed of car 1} = \frac{x}{t}, \qquad \text{speed of car 2} = \frac{3x}{t+60} \end{aligned}
2

Use the speed relation. Car 2's speed is half of car 1's, so 3xt+60=12⋅xt\dfrac{3x}{t+60} = \dfrac12\cdot\dfrac{x}{t}.

3xt+60=x2t⇒ 3t+60=12t(cancel x)⇒ 6t=t+60(cross-multiply)\begin{aligned} &\frac{3x}{t+60} = \frac{x}{2t}\\ &\Rightarrow\ \frac{3}{t+60} = \frac{1}{2t} \quad\text{(cancel }x\text{)}\\ &\Rightarrow\ 6t = t + 60 \quad\text{(cross-multiply)} \end{aligned}
3

Solve for tt.

5t=60⇒ t=12 minutes\begin{aligned} &5t = 60\\ &\Rightarrow\ t = 12 \text{ minutes} \end{aligned}
12 minutes\boxed{12 \text{ minutes}}

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CAT 2018 Slot 1 QA Q2: Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and m — Solution | TheCATExam