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CAT 2022 Slot 1 — QA Question 18

TrainsEasy

Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X  is

Answer & solution

  • A

    12

  • 15

  • C

    10

  • D

    6

Solution

Easy

Use the classic "meeting point" result for two bodies starting together towards each other: if they meet after time tt, then t=tA⋅tBt=\sqrt{t_A\cdot t_B}, where tA,tBt_A,t_B are the times each takes to finish the remaining stretch after meeting.

1

Name the times. Let them meet tt minutes after starting. Train A reaches Y in 1010 min total, so after meeting it still needs (10−t)(10-t) min; train B needs 99 min after meeting:

A after meeting =10−tB after meeting =9\begin{aligned} &\text{A after meeting }=10-t\\ &\text{B after meeting }=9 \end{aligned}
2

Apply the meeting relation t2=(A after)×(B after)t^2=(\text{A after})\times(\text{B after}) and solve:

t2=(10−t)×9⇒ t2+9t−90=0⇒ (t+15)(t−6)=0⇒ t=6(t=−15 rejected)\begin{aligned} &t^2=(10-t)\times 9\\ &\Rightarrow\ t^2+9t-90=0\\ &\Rightarrow\ (t+15)(t-6)=0\\ &\Rightarrow\ t=6 \quad(t=-15\text{ rejected}) \end{aligned}
3

Total time for B == time to meet ++ time after meeting:

t+9=6+9=15 min\begin{aligned} &t+9=6+9=15\text{ min} \end{aligned}

Train B takes 15\mathbf{15} minutes from Y to X. Option (b).

Ratio shortcut: A afterB after=t?\dfrac{\text{A after}}{\text{B after}}=\dfrac{t}{?} comes from t=(10−t)⋅9t=\sqrt{(10-t)\cdot 9}. Once t=6t=6, B's journey is simply 6+9=156+9=15.

CAT 2022 Slot 1 QA Q18: Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, — Solution | TheCATExam