Bunuel
Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:If two objects are moving around a circle at different speeds, the number of distinct points on the circle at which they will meet can be determined by the difference between the (most) reduced ratios of their speeds.
Since b can be either greater than 6 or less than 6, let’s first assume that b < 6. If b < 6, we can let the reduced ratio be 6/b = r/s such r - s = 2. Notice that after being reduced, r < 6. So (r, s) can be (3, 1), (5, 3).
If (r, s) = (3, 1), we have 6/b = 3/1 and this gives us b = 2.
If (r, s) = (5, 3), we have 6/b = 5/3 and this gives us b = 18/5.
However, since b is a positive integer, b can’t be 18/5. So we only have one value for b when b < 6. Next, let’s assume now b > 6. Again we can let the reduced ratio be 6/b = r/s such s - r = 2. Notice that after being reduced, r < 6. So (r, s) can be (1, 3), (3, 5).
If (r, s) = (1, 3), we have 6/b = 1/3 and this gives us b = 18.
If (r, s) = (3, 5), we have 6/b = 3/5 and this gives us b = 10.
Since both values of b are integers, we have 2 integer values for b when b > 6. Therefore, altogether, there are 3 integer values for b.
Answer: CIs it that the values of r must be factors of 6? That would eliminate (4,2) as an option but (2,4) is still viable.