M and N start at the same time from two ends S and T, at speeds of

Question

M and N start at the same time from two ends S and T, at speeds of 3 m/sec and 4 m/sec towards each other, ST being 7 m. After reaching the opposite ends, they retrace their path and start moving towards each other. What is the total distance travelled by M when he meets N for the second time?

A. 26 m
B. 20 m
C. 15 m
D. 9 m
E. 8 m

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illumunde · Answer

M moves at  3 m/s , N at  4 m/s , distance ST =  7 m . 
 After  1 second :
 
 M has travelled  3 m . 
 N has travelled  4 m . 
 Together, they’ve covered  7 m , so they meet for the  first time .  
 
 After  2 seconds :
 
 M has travelled  6 m  (just 1 m short of reaching T). 
 N has travelled  8 m , which means N has already reached S (at 7 m) and started coming back.  
 
 After  3 seconds :
 
 M has travelled  9 m total  (he reached T at 7 m, then went 2 m back). 
 N has travelled  12 m total  (7 m forward to S, then 5 m back). 
 At this point, they meet for the  second time .  
  
So, the total distance travelled by M by the second meeting is  9 metres .

Ayeka · Answer

Distance ST= 7 m 
Speed M= 3 and N=4 
Relative speed of M+N=3+4=7 m/s 
They meet first time after = 7/7 =1 sec 
After 2 sec, M has covered 6 m and N has covered 8m 
M has covered 1 m less and N has covered 1 m more then 7m.
After 3 sec, M has covered 9m and N has covered 12 m and they meet for the second time.
M total distance = (3+3+3=9m 
N total distance = (4+4+4=12) 
For first time they meet after 1 sec . Then, meetings repeat every 2 sec because of reflections, so second meeting is at 3 sec, and 3rd meeting at 5 sec, 4th at 7th sec and so on.

Distance covered by M by second meeting = 9 m 
D

Souradeep17 · Answer

M starts at S speed is 3m/s
N starts at T speed is 4m/s
 distance between ST is 7m

They start travelling in opp direction.

So after 1second they travel 3m and 4m respectively where they meet first time

After 2second M reaches a point which 1M away from T 
N travels 8 , 7m to S and then 1m back towards T

After 3seconds
M travels 1m to T and then 2m towards S
N travels 4m towards to T

So the position after 3 seconds for M is a point 2m away from T
For N the same point as he travels (4m and previous 1m=5m towards T from S)

So the total distance travelled during this time by M 3*3=9m Option D

Dereno · Answer

The distance ST = 7m

Speed of M and N are 3m/sec and 4m/sec.

They are separated by a distance of 7 m.

The speed ratios of M and N = 3:4

The distance ratios of M and N = 3:4 ( speed is proportional to distance ).

The Time ratio of M and N = 4:3.

when they meet for the FIRST time, the distance travelled by both = 7m.

When they meet for the SECOND time, the distance travelled by both = 2*7m = 14m.

The total distance travelled by both during the second meet = 7+ 14 = 21 m.

The distance ratio of M : N = 3:4

Distance travelled by M = (3/7) *21 = 3*3 =  9 m.

Option D

paragw · Answer

There is a simple concept
When the ratio of given speeds is less than or equal to 2
then Total distance covered by both of them will go like this

Now for 0-2 meeting points, overall distance covered will be D+2D=3D=3*7=21
Now the distance of 21 will be divided into the ratio of their respective speed i.e. 3:4
Hence, distance covered by M=9 & N=12 till they meet for the second time.

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KevoK · Answer

My thought process is like this:

First, we look at the total distance travelled by both M and N: Because they walk the entirety of ST and the question ask the total distance by the time they meet each other for the 2nd time, the total travelled by both is:

7 (Since N reached the end) + 7 (Since M reached the end) + 7 (sum of the distance M and N travelled in order to meet each other again) = 21m

Next, we're finding the time it took both M and N to travel a total 21m

3n + 4n = 21 
 7n = 21 
 n = 3

Now, plugging back, we can find the distance M has travelled, which is 3*3 = 9

IMO D

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