calendarBack
Quant

/

Arithmetic II

/

Time & Speed
ALL MODULES

CAT 2025 Lesson : Time & Speed - Walking To & Fro

bookmarked

7. Crossing each other in a straight line

When two people are walking in a corridor at different speeds, they will meet each other over and over again. The relative speed is the sum of their individual speeds.

Distance covered is directly proportional to the speed. So, the ratio of the distances covered by two individuals in a given time will be the ratio of their respective speeds.

The following concepts apply only when the faster person's speed is less than twice the slower person's speed.

7.1 Walking towards each other from two ends

Let's say two people or objects are at either ends of a path, which is of length ll, and start moving towards each other. Every time they reach the end, they turn back and start walking.

The first time they meet, they together would have covered a distance of ll.

The second time they meet, they would have covered a further distance of 2l2l. So, in total they cover a distance of 3l3l when they meet for the second time.

With every incremental meeting, they together cover an incremental distance of 2l2l. So, when they meet for the nthn^{th} time, they would have covered a distance of l+2(n−1)l=(2n−1)ll + 2(n - 1)l = (2n - 1)l

For instance, when they meet for the 7th7^{th} time, they cover a distance of 13l13l and when they meet for the 18th18^{th} time, they cover a distance of 35l35l.

7.2 Walking from a point in the center in opposite directions

When they start walking from some point along a path in opposite directions and they continue to turn back and start walking every time they reach an end, then the first time they meet, they would have covered a distance of 2l.

With every incremental meeting, the distance covered will remain the same as above, which is 2l2l. So, in this case, when they meet for the nthn^{th} time, they would have covered a distance of 2nl2nl.

7.3 Walking from one of the ends in the same direction

The faster person who touches the other end will meet the slower person on the way back. So, distance covered for the first meet is 2l2l. Distance covered for every incremental meeting will remain the same as above, which is 2l2l.

So, distance covered the nthn^{th} time they meet is 2nl2nl.

Example 23

John and Matthew start walking from two ends of a 100 metre straight corridor at speeds of 15 m/s and 25 m/s. After how many seconds do they meet for the third time?

Solution

Relative speed =15+25=40= 15 + 25 = 40 m/s

Distance covered =l+2l+2l=100+200+200=500= l + 2l + 2l = 100 + 200 + 200 = 500m

Time taken =50040=12.5= \dfrac{500}{40} = 12.5 seconds

Answer: 12.512.5 seconds

Example 24

Guru and Prasad start walking in a straight line from point A towards point B, which is 150 metres away. Every time they reach a point, they turn back and start walking towards the other point in the same path. Guru travels at 12m/s and Prasad travels at 8 m/s. When they meet for the 10th time (their position at the start is not considered as a meeting), what is the distance covered by Prasad (in km)?

Solution

Distance covered =2×10×l=2×10×150=3000= 2 \times 10 \times l = 2 \times 10 \times 150 = 3000 metres

Ratio of the distance covered == ratio of their speeds =12:8=3:2= 12 : 8 = 3 : 2

Distance covered by Prasad =25×3000=1200= \dfrac{2}{5} \times 3000 = 1200 metres =1.2= 1.2 km

Answer: 1.21.2 km

Example 25

A and B are two ends of a straight road. Ram and Shyam start from points A and B respectively at constant speeds and at the same time. When either of them reaches an end, the person immediately turns and starts walking in the other direction. When they meet for the first time at a point that is 66 km from point B. They meet for the second time at a point that is 33 km from point A. What is the distance between A and B?

Solution

Let the first and second meeting points be named C and D respectively. CB and AD are given to be 66 km and 33 km respectively. Let CD = xx km.



As they start from the two ends, distance covered for the second meet is twice that of the first meet. Ratio of total distances covered for the first and second meets is 1:21 : 2.

As Ram and Shyam start walking at the same time and travel at constant speeds, the ratios of each of their distances covered for the first and second meets is also 1:21 : 2.

Distance covered by Ram for the first meet =3+x= 3 + x
Distance covered by Ram for the second meet =6+6+x=12+x= 6 + 6 + x = 12 + x

Applying the ratio,

3+x12+x=12 \dfrac{3 + x}{12 + x} = \dfrac{1}{2} ⇒ 6+2x=12+x 6 + 2x = 12 + x

⇒ x=6 x = 6 km

∴ \therefore Total Distance =3+6+6=15= 3 + 6 + 6 = 15 km

Answer: 1515 km

Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock