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CAT 2017 Slot 1 — QA Question 4

Basics of TSD/ProportinalityEasy

A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is:

Answer & solution

Correct answer: 20

Answer: 20

Solution

Easy

Same distance both times, so time is inversely proportional to speed. Find the time gap between the two journeys (it equals 10+10=2010+10=20 minutes), express it as a fraction of the slower time, then get the distance.

1

Ratio of times. Speeds are 1212 and 1515, i.e. the second is 54\tfrac54 of the first, so its time is 45\tfrac45 of the first.

t2t1=1215=45(time∝1/speed)⇒ t2=t1−15t1\begin{aligned} &\frac{t_2}{t_1}=\frac{12}{15}=\frac{4}{5} \quad\text{(time}\propto 1/\text{speed)}\\ &\Rightarrow\ t_2 = t_1 - \tfrac15 t_1 \end{aligned}
2

The time gap. One arrival is 1010 min late, the other 1010 min early, so the difference is 2020 minutes =15t1=\tfrac15 t_1.

15t1=20 min⇒ t1=100 min=10060=53 hours\begin{aligned} &\tfrac15 t_1 = 20 \text{ min}\\ &\Rightarrow\ t_1 = 100 \text{ min} = \frac{100}{60}=\frac{5}{3}\text{ hours} \end{aligned}
3

Distance. Use the slow journey: speed 1212 km/h for 53\tfrac53 h.

d=12×53⇒ d=20 km\begin{aligned} &d = 12\times\frac{5}{3}\\ &\Rightarrow\ d = 20 \text{ km} \end{aligned}
20 km20 \text{ km}

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CAT 2017 Slot 1 QA Q4: A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after th — Solution | TheCATExam