Jack and John simultaneously start running in the same direction aroun

Question

Jack and John simultaneously start running in the same direction around a circular track. Jack travels at 6 m/s and John runs at b m/s. If they cross each other at exactly two points on the circular track and b is a natural number less than 30, how many values can b take?

A. 3
B. 4
C. 7
D. 5
E. 10

A. 3
B. 4
C. 7
D. 5
E. 10

ShivamoggaGaganhs · Answer

You can get sol here: https://www.careerbless.com/qna/discuss ... ionid=2742

unflinchingSubhs · Answer

If a:b is the ratio of speeds of the persons running in the same direction then no of meetings point will be modulus of (a-b). Since meetings points given is 2 a:b or b:a should be 3:1. So posdibles values are 2 and 18. Not sure how they got 10.

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himssss29 · Answer

Answer: A. 3

Key formula:    For two runners moving in the  same direction  on a circular track, if their speed ratio (in lowest/reduced terms) is $p:q, they meet at exactly|p-q| distinct points.

We need:    ∣  p−q  ∣  =2

Check all valid b values (b < 30, b ≠ 6):

b   
 
    gcd(6,b)   
 
    Reduced ratio (p:q)   
 
    |p−q|

2  
  2  
  3:1  
  2   ✓  
 
   10  
  2  
  3:5  
  2   ✓  
 
   18  
  6  
  1:3  
  2   ✓

Valid values: b = 2, 10, 18 → 3 values

onlyPlanA · Answer

How are we going to approach this question if they are moving in opposite directions? They will surely meet at one or more points, so what will be the distinct number of meeting points in that case?

egmat · Answer

Hi onlyPlanA,

Great question, and it's the natural next step once you've got the same-direction rule from himssss29's table.

The formula you saw ( meeting points = |p - q| , using the reduced speed ratio p:q) is built specifically for runners going the  same way . The reason it's a  difference  is that when both run the same direction, only the  relative speed  |v1 - v2| matters - the faster runner has to gain a full lap on the slower one to catch up again.

What changes when they run opposite ways

Now the runners close the gap from both sides, so the relative speed becomes the  sum  v1 + v2. That single change flips the formula:

-  Same direction:  distinct meeting points =  |p - q| 
-  Opposite directions:  distinct meeting points =  p + q

where p:q is the speed ratio in lowest terms. Same setup, just add instead of subtract.

Quick check with real numbers

Take Jack at  6  and John at  2  m/s. Reduced ratio is  3 : 1 .

- Same direction - | 3  -  1 | =  2  points.
- Opposite directions -  3  +  1  =  4  points.

You can feel why: heading toward each other, they cross much more often, so more distinct crossing spots.

One thing to be careful about

Always reduce the ratio first. For b =  10 , the ratio  6:10  reduces to  3:5 , so opposite-direction points would be  3  +  5  =  8 , not  6  +  10 . Reducing is exactly what kept the same-direction counts honest in the table above - the same discipline applies here.

So if this question had said "opposite directions," you'd solve  p + q = 2 , which forces p:q =  1:1  - meaning equal speeds, and that's a different (degenerate) situation entirely. That's a nice illustration of how much the direction changes the whole problem.

Answer: A

Jack and John simultaneously start running in the same direction aroun

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b	gcd(6,b)	Reduced ratio (p:q)	\|p−q\|
2	2	3:1	2 ✓
10	2	3:5	2 ✓
18	6	1:3	2 ✓

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Jack and John simultaneously start running in the same direction aroun

Prep Toolkit

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