Latest Elitmus Aptitude Question SOLUTION: A and B started from the same point on a circular track. The ratio of their speed is n:1 (n is a whole number). If 5th and 17th time they meet at the same point, what can not be t

Answer is n=6.

If speed of two bodies moving in same direction in circular path are in the simplest ratio of a:b, then no:of pts they will meet = (a-b) provided a>b.
Let circumference of circle be 'L'.
Time for 1st meeting at any of the (a-b) pts = L/(a-b).
Time for 1st meeting at Starting Pt = LCM of (L/a, L/b).

Time for nth meeting at Starting Pt =n* LCM of (L/a, L/b).

When n = 2 :
a=2,b=1.
No:of meeting pts = (2-1)=1pt.
Only meeting pt is L.
Time for first meeting = L/1.
Time for first meeting at starting pt = LCM of (L/2, L/1) = L.
Time for 5th meeting = 5L.
Time for 17th meeting = 17L.

Both tyms, they meet at pt L.
So, n=2 is valid.

When n = 3 :
a=3,b=1.
No:of meeting pts = (3-1)=2pts.
Meeting pts are : L/2, L .
Time for first meeting = L/2.
Time for first meeting at starting pt = LCM of (L/3, L/1) = L.
Time for 5th meeting = 5*L/2.
Time for 17th meeting = 17*L/2.

Both tyms, they meet at pt L/2.
So, n=3 is valid.

When n = 4 :
a=4,b=1.
No:of meeting pts = (4-1)=3pts.
Meeting pts are : L/3, L/2, L .
Time for first meeting = L/3.
Time for first meeting at starting pt = LCM of (L/4, L/1) = L.
Time for 5th meeting = 5*L/3.
Time for 17th meeting = 17*L/3.

Both tyms, they meet at pt L/3.
So, n=4 is valid.

When n = 6 :
a=6, b=1.
No:of meeting pts = (6-1)=5pts.
Meeting pts are : L/5, L/4, L/3, L/2, L .
Time for first meeting = L/5.
Time for first meeting at starting pt = LCM of (L/5, L/1) = L.
Time for 5th meeting = 5*L/5 = L.
Time for 17th meeting = 17*L/5.

Both tyms, they meet at different pts namely L and L/5.
So, n=6 is not valid.

6
as the total meeting points(mp) in a circle will be n-1(from ratio of velocities even it is not mentioned,faster will have more velocity=nk)

wnen n=2 mp=1 5th and 17 coincide

when n=6 mp5 5th will coincide with 10th 15th 20th.........

and 17th will coincide with 2th 7th 12th ..............

ans = D) 6

pradeep ...we cant sure for 3 and 4 right ...can u explain some thing clearly

How do u remember these questions..?? Please submit the and also..

please explain clearly.....

plz explain clearly........

thankss a lot kaustav!!!!

kaustav how did you determine the meeting points in each case ?

Meeting points are always equidistant from each other.

@ KAUSTAV MITRA
why no. of pts they meet is = a-b ? and why 1st meeting at any of the (a-b) pts = L/(a-b)?

you are genius

Check the answer
http://elitmus-page.blogspot.in/2015/10/a-and-b-started-from-same-point-on.html