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CAT 2018 Slot 1 — QA Question 14

Basics of TSD/ProportinalityEasy

The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is

Answer & solution

Correct answer: 5

  • A

    4

  • B

    3

  • C

    6

  • 5

Solution

Easy

Write everyone's times in terms of Narayan's full-trip time. The two "time-gap" facts pin down how long Partha needs for the second half of the journey, which gives his speed directly.

1

Name the times. Let Narayan take tt hours for the full 6060 km. Partha takes 44 hours more:

Partha’s full-trip time=t+4(4 hours more than Narayan)\begin{aligned} &\text{Partha's full-trip time} = t+4 \quad\text{(4 hours more than Narayan)} \end{aligned}
2

Use the mid-point fact. Partha reaches the mid-point 22 hours before Narayan finishes, i.e. at time t−2t-2. So Partha's time to cover the first 3030 km is t−2t-2.

Partha’s time for first 30 km=t−2(2 hours before Narayan’s t)\begin{aligned} &\text{Partha's time for first }30\text{ km} = t-2 \quad\text{(2 hours before Narayan's }t\text{)} \end{aligned}
3

Time for the second half. Subtract step 2 from step 1 to get the time Partha needs for the remaining 3030 km:

(t+4)−(t−2)=6[step 1 − step 2]⇒ Partha covers 30 km in 6 hours\begin{aligned} &(t+4)-(t-2) = 6 \quad\text{[step 1 } - \text{ step 2]}\\ &\Rightarrow\ \text{Partha covers }30\text{ km in }6\text{ hours} \end{aligned}
4

Speed. Partha travels at constant speed, so use the clean second-half leg:

speed=306=5 km/h\begin{aligned} &\text{speed} = \frac{30}{6} = 5\ \text{km/h} \end{aligned}
Speed of Partha=5 km/h\text{Speed of Partha} = \mathbf{5}\ \text{km/h}

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CAT 2018 Slot 1 QA Q14: The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha — Solution | TheCATExam