Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
11 Jun 2025, 06:41
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
33% (03:03) correct
67%
(01:06)
wrong
based on 3
sessions
History
Date
Time
Result
Not Attempted Yet
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. Find the total distance travelled by M when he meets N for the second time.
(A) 20 m
(B) 15 m
(C) 9 m
(D) 26 m
(A) 20 m
(B) 15 m
(C) 9 m
(D) 26 m
11 Jun 2025, 09:47
Kudos
Bookmarks
M and N will meet 1st time after 1 sec
7/(3+4)=1 sec
They will reach to respective destination and again meet after 2 sec
So total 3 sec
Hence distance travelled by M in 3 sec = 3×3=9m
Option C
7/(3+4)=1 sec
They will reach to respective destination and again meet after 2 sec
So total 3 sec
Hence distance travelled by M in 3 sec = 3×3=9m
Option C
12 Jun 2025, 02:41
Kudos
Bookmarks
Given:
Calculating the time for their first meeting:
They are moving towards each other, so we add their speeds:
Relative speed=3+4=7 m/s
Distance between them = 7 m
Time taken to meet for the first time:
Time = 7/7 = 1 second
In this 1 second:
After meeting, they continue to the ends and turn back
After meeting:
They'll both have turned back after LCM(4/3, 3/4) seconds.
But we can continue without calculating that exactly, by simply tracking where they are and when they’ll meet again.
Second Meeting
They continue their motion. After turning, they start moving towards each other again.
Let’s calculate the total time passed when they meet the second time.
Let’s call the second meeting time as t seconds after the start.
We know:
At first meeting:
So:
So total time until 2nd meeting:
3t+4t = 7t = 21⇒ t=3 seconds
How much distance M has travelled in 3 seconds?
M’s speed = 3 m/s
So in 3 seconds:
Distance = 3×3 = 9 meters
Final Answer: C
- Two people M and N start from opposite ends of a straight path ST, which is 7 meters long.
- M’s speed = 3 m/s
- N’s speed = 4 m/s
- They start at the same time, move towards each other, and when they reach the opposite ends, they turn back and start moving toward each other again.
- We are to find the total distance M travels when he meets N for the second time.
Calculating the time for their first meeting:
They are moving towards each other, so we add their speeds:
Relative speed=3+4=7 m/s
Distance between them = 7 m
Time taken to meet for the first time:
Time = 7/7 = 1 second
In this 1 second:
- M travels = 3×1=3 m
- N travels = 4×1=4 m
After meeting, they continue to the ends and turn back
After meeting:
- M continues from his position (3 m from S) to T.
- Distance left for M to reach T = 7−3 = 4 m
- Time = 4/3 seconds
- N continues from his position (4 m from T) to S.
- Distance left for N to reach S = 7−4 = 3 m
- Time = 3/4 seconds
- M takes 4/3 sec to reach T
- N takes 3/4 sec to reach S
They'll both have turned back after LCM(4/3, 3/4) seconds.
But we can continue without calculating that exactly, by simply tracking where they are and when they’ll meet again.
Second Meeting
They continue their motion. After turning, they start moving towards each other again.
Let’s calculate the total time passed when they meet the second time.
Let’s call the second meeting time as t seconds after the start.
We know:
- M moves at 3 m/s, so in time t, he has traveled 3t meters.
- N moves at 4 m/s, so in time t, he has traveled 4t meters.
At first meeting:
- Total distance = 7 m → 3t+4t = 7t= 7 → t=1 sec
- The total distance they together travel = 21 meters
- They meet first after covering 7 m together.
- Then both continue to the end and turn back:
- M travels 4 m more to reach T
- N travels 3 m more to reach S
- Then they turn and start moving toward each other again.
So:
- First 7 m → 1st meeting
- Next 7 m (to ends) → turning
- Next 7 m → 2nd meeting
So total time until 2nd meeting:
3t+4t = 7t = 21⇒ t=3 seconds
How much distance M has travelled in 3 seconds?
M’s speed = 3 m/s
So in 3 seconds:
Distance = 3×3 = 9 meters
Final Answer: C
