Which of the following cannot be the ratio of speeds of two joggers ru

First, a point that is diametrically opposite the starting point means that the runners will meet at the other side of the diameter of the circle (i.e., 180 degrees from where they started the race).

Now, let’s assume that the runners start at position A. We see that position B is diametrically opposite to their starting point , and it is equivalent to a half-circle (since it is 180 degrees from the starting point). Now, if both of them run at the same speed (a ratio of 1 : 1), then they will, of course start at point A and will also meet at point B (and actually, they will meet at all points on the circle, since they are traveling at the same speed). Let’s look at each answer choice.

Choice A: If they have a speed ratio of 3 : 5, then the first runner will run from point A to point B, then from B to A, and then from point A to point B again, having run 3 half-circles around the track. In the same amount of time, the second runner will have run 5 half-circles, from A to B, then B to A, then A to B, then B to A, and finally A to B. In this case, they will meet at point B. Thus, the ratio 3 : 5 is possible.

Choice B: If they have a speed ratio of 1 : 3, then we see that the first runner advances from point A to point B (one half-circle). In the same time, the second runner will have gone three half-circles (from A to B, then B to A, and then A to B). Again, they will meet at point B, and so a ratio of 1 : 3 is possible.

Choice C: If they have a speed ratio of 1 : 5, then we see that the first runner advances from point A to point B (one half-circle). In the same amount of time, the second runner will have gone five half-circles (from A to B, then B to A, then A to B, then B to A, and, finally, from A to B). The ratio 1 : 5 is possible.

Choice D: If they have a speed ratio of 2 : 5, then we see that the first runner will go from A to B and then from B to A, while the second runner will go A to B, then B to A, then A to B, then B to A and finally A to B. They will not meet at point B. In fact, they will be on opposite sides of the diameter. Thus, a ratio of 2 : 5 is not possible.

By this point, you may have noticed that, in order for the two runners to meet at a point that is diametrically opposite from their starting point, the ratio must consist of two odd numbers (or be reducible to lowest terms as a ratio of two odd numbers). This is why choice D’s ratio of 2 : 5 will not work. In this case, when one runner is at point B (an odd number of half-circles from the starting point), the other runner will always be at point A (an even number of half-circles from the starting point), and vice versa.

Thus, we see that choice E, a ratio of 5 : 3, consists of two odd numbers, and thus the runners will also meet at point B. Thus, a ratio of 5 : 3 is possible.

Answer: D
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The ratio of the speed of 2 runners in reduced form is x:y.
If the 2 runners meet at a diametrically opposite point to the point from where both of them started, then
\(180= n* (\frac{360}{x±y})\), where n is a positive integer or x±y can never be odd.

Option D- x±y= 5±2 is odd

nick1816
Can you provide a more detailed explanation?
I couldn't understand the logic here

Are they both jogging in the same direction (i.e. clockwise or counterclockwise) or is the answer the same regardless?

For 2 people running along a circular track (having started from the same point) with speeds S1 and S2, we have:

We take the ratio of S1 and S2 and reduce it to the lowest integer form, say p : q
For example, let the speeds be 16 & 6; thus, the reduced ratio is 8 : 3

#1. If they are running in the same direction, the number of meeting points is simply \((p - q)\) (assuming p > q). In this example, it is 8 - 3 = 5
#2. If they are running in the opposite direction, the number of meeting points is simply \((p + q)\). In this example, it is 8 + 3 = 11

Note that: To be able to meet at a point diametrically opposite their starting point, the number of meeting points must be even, say 2, 4, 6, 8, 10, etc.


Looking at the options:

(A) 3 : 5 => Same direction: 5 - 3 = 2; Opposite direction: 3 + 5 = 8 --- both are even, hence possible
(B) 1 : 3 => Same direction: 3 - 1 = 2; Opposite direction: 3 + 1 = 4 --- both are even, hence possible
(C) 1 : 5 => Same direction: 5 - 1 = 4; Opposite direction: 5 + 1 = 6 --- both are even, hence possible
(D) 2 : 5 => Same direction: 5 - 2 = 3; Opposite direction: 5 + 2 = 7 --- both are odd, hence NOT possible
(E) 5 : 3 => Same direction: 5 - 3 = 2; Opposite direction: 5 + 3 = 8 --- both are even, hence possible


Answer D
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sujoykrdatta

Could you please explain algebraically or otherwise why the following is true:

"Note that: To be able to meet at a point diametrically opposite their starting point, the number of meeting points must be even, say 2, 4, 6, 8, 10, etc."

Let the speeds of A and B be s1 & s2 after reducing to lowest form (say s1 > s2)
For same direction:
relative speed = s1 - s2
Let length of track = L
Time to meet for the first time (t)
= L/(s1 - s2)

Time to meet at starting point again (T)
= LCM (L/s1, L/s2)
= [LCM (L,L)]/[GCD (s1, s2)]
= LCM (L,L) = L

(Since s1 and s2 have been reduced to their lowest form, their GCD equals 1)


If we divide T/t, we will get the number of meeting points

=> # meeting points = T/t

= L/[L/(s1 - s2)]

= s1 - s2

If the above is even, then on distributing the points equally along the track, there will always be a point diametrically opposite the starting point. Infact, it will be true for all other points. Say there are 4 points:
1,2,3,4 in order; then 1 is opposite 3 and 2 is opposite 4.

Hope this helps!

Posted from my mobile device
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Thanks for the algebraic explanation of the 1st axiom - the number of meeting points is (s1-s2) or (s1+s2) depending on whether the joggers are moving in the same or opposite directions.
I have devised an algebraic approach to prove that in order for them to meet at a point diametrically opposite to their starting point (S), (s1+s2) or (s1-s2) must be even. We will consider the case when they are moving in opposite directions:
Let C meters be the circumference of the track. Then, the first time they meet, A will have covered {s1/(s1+s2)}*C meters. The next time they meet, A will have covered another {s1/(s1+s2)}*C meters, i.e. 2{s1/(s1+s2)}*C after they started. Let us assume that they meet 'p' number of times until they meet at a point that is diametrically opposite to S, i.e. at a distance of C/2 meters from S. Let us assume that , in the process (i.e. during the time they started and up to the point they meet half-way from S), A crosses point S 'q' times. Then:

p{s1/(s1+s2)}*C = q*C + C/2....> (p*s1)/(s1+s2) = q + 1/2....> p*s1 = q(s1+s2) + (s1+s2)/2
LHS of the above equation is an integer so the RHS must also be an integer and since q(s1+s2) is an integer (s1+s2)/2 must be an integer so (s1+s2) must be divisible by 2 and hence must be even.

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