## Difficult Math Problem #122 - Rates

##### This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 354
Joined: 27 Jun 2006
Thanked: 11 times
Followed by:5 members

### Difficult Math Problem #122 - Rates

by 800guy » Mon Apr 23, 2007 9:22 am
A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?

A. 750 seconds
B. 50 seconds
C. 250 seconds
D. 375 seconds
E. 75 seconds

oa coming when people respond with explanations, from diff math problems set

Junior | Next Rank: 30 Posts
Posts: 26
Joined: 09 Apr 2007
Thanked: 2 times
by rajeshvellanki » Mon Apr 23, 2007 3:52 pm
HCF or GCD of 3,6 and 18 is 3. This is the first time they meet.

t=750/3=250

Community Manager
Posts: 789
Joined: 28 Jan 2007
Location: Silicon valley, California
Thanked: 30 times
Followed by:1 members
by jayhawk2001 » Mon Apr 23, 2007 7:20 pm
A completes 1 lap in 750/3 = 250 s
B completes 1 lap in 750/6 = 125 s
C completes 1 lap in 750/18 = 125/3 s

Comparing time vs laps completed in increments of 125/3 sec,

t=1*125/3 sec, A = 1/6 lap, B = 1/3 lap, C = 1 lap
t=2*125/3 sec, A = 2/6 lap, B = 2/4 lap, C = 2 laps
...
t=6*125/3 sec, A = 1 lap, B = 2 laps, C = 6 laps

So, they meet every 6*125/3 = 250 sec.

Alternatively, using LCM logic --

A and B meet every 250 sec
A and C meet every 125 sec
B and C meet every 250 sec

So, LCM of 3 cases above = 250 sec.

Newbie | Next Rank: 10 Posts
Posts: 4
Joined: 07 Mar 2007
by rajatmehta » Mon Apr 23, 2007 10:33 pm

In 250 sec A will complete 750m
In 250 sec B will complete 1500m
In 250 sec C will complete 4500m

Since, they are all multiples of length of track so all the runners will at the same point on the track. Try with 50,75 sec as well to confirm whether 250 sec is the least time in which they will be together.

Master | Next Rank: 500 Posts
Posts: 354
Joined: 27 Jun 2006
Thanked: 11 times
Followed by:5 members

### OA:

by 800guy » Fri Apr 27, 2007 12:45 pm
OA:

When two people are running in the same direction the relative speed is a difference in speeds of the two people.

In this case A=3 B=6 C=18

So relative speed of B wrt A is 6-3 = 3m/s
Relative speed of A wrt to C is 18-3 =15m/s

Therefore relative distances will be:
B wrt A is 750/3 =250
C wrt to A 750/15 = 50

So they have to bridge this distance of 250 and 50 between them which is the LCM of 250 and 50 which is 250.

Another Method: Simply put, Runner A's time take to run one lap is 250
Runner B's time is 125s
and Runner C's time is 41.67s

We can notice that A=2B
and thet B=3C

So when 250 s elapse, they will be at their starting point. A will have completed one lap, B 2 laps, and C 6 laps

• Page 1 of 1