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CAT 2018 Slot 1 — QA Question 1

Relative SpeedEasy

Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?

Answer & solution

Answer: 15

Solution

Easy

Let the full distance be dd and TT's speed be ss, so SS's speed is 34s\tfrac34 s. Use the head-start (TT travels alone for 1 hour) and the meeting condition (X to Z is 35\tfrac35 of XY) to pin down dd in terms of ss. The travel time XY is simply d/sd/s.

1

Set up the head start. TT starts at 3 pm; SS starts at 4 pm. In that 1 hour TT covers ss km, so when SS begins the gap between them is (d−s)(d-s).

speed of T=s,speed of S=34s⇒ gap at 4 pm=d−s(T already moved s km)\begin{aligned} &\text{speed of }T = s,\quad \text{speed of }S = \tfrac34 s\\ &\Rightarrow\ \text{gap at 4 pm} = d - s \quad\text{(}T\text{ already moved }s\text{ km)} \end{aligned}
2

Time until they meet (after 4 pm). They close the gap at their combined speed.

t=d−s s+34s =d−s74s⇒ t=4(d−s)7s(combined speed =74s)\begin{aligned} &t = \frac{d-s}{\,s + \tfrac34 s\,} = \frac{d-s}{\tfrac74 s}\\ &\Rightarrow\ t = \frac{4(d-s)}{7s} \quad\text{(combined speed }=\tfrac74 s\text{)} \end{aligned}
3

Distance TT has covered at the meeting point Z. Add the head-start ss to what TT does in time tt.

XZ=s+s⋅t=s+s⋅4(d−s)7s⇒ XZ=s+4(d−s)7=3s+4d7(from step 2)\begin{aligned} &XZ = s + s\cdot t = s + s\cdot\frac{4(d-s)}{7s}\\ &\Rightarrow\ XZ = s + \frac{4(d-s)}{7} = \frac{3s + 4d}{7} \quad\text{(from step 2)} \end{aligned}
4

Apply the position of Z. Given XZ=35dXZ = \tfrac35 d, equate and solve for dd.

3s+4d7=35d⇒ 5(3s+4d)=21d(cross-multiply)⇒ 15s+20d=21d⇒ d=15s\begin{aligned} &\frac{3s + 4d}{7} = \frac{3}{5}d\\ &\Rightarrow\ 5(3s+4d) = 21d \quad\text{(cross-multiply)}\\ &\Rightarrow\ 15s + 20d = 21d\\ &\Rightarrow\ d = 15s \end{aligned}
5

Travel time of TT for XY. Divide distance by TT's speed.

time=ds=15ss=15 hours\begin{aligned} &\text{time} = \frac{d}{s} = \frac{15s}{s} = 15 \text{ hours} \end{aligned}
15 hours\boxed{15 \text{ hours}}
CAT 2018 Slot 1 QA Q1: Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves — Solution | TheCATExam