# Revolutions

Jack and Jill run at different speeds in opposite directions around the maypole. How many times do they pass in the first minute?

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Jack dances clockwise around the Maypole, making one revolution every five seconds.

Starting from a point diametrically opposite Jack's starting point, Jill dances anticlockwise, making one revolution every six seconds.

How many times do they pass each other in the first minute?

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

**Answer**: 22 times

**Keeping Jack or Jill still**

Imagine that Jack doesn't move.

Jill makes 60$\div$6 = 10 revolutions in a minute, so passes Jack 10 times

But Jack does move - he makes 60$\div$5 = 12 revolutions, so he passes Jill 10 times

Total 10 + 12 = 22 passes

*Note: If Jack and Jill's times weren't factors of 60, then this method would require rounding and could easily lead to mistakes. Below is a similar method which avoids this issue:*

**Keeping Jack or Jill still and using the lowest common multiple**

Every 6 seconds, Jill makes a revolution, so passes Jack.

Every 5 seconds, Jack makes a revolution, so passes Jill.

At common multiples of 5 and 6, Jack and Jill have made whole numbers of revolutions

30 = lowest common multiple: Jill passes Jack 5 times and Jack passes Jill 6 times

Total in 30 seconds = 11

Total in 60 seconds = 22

**Using angles**

Jack makes a revolution in 5 seconds, that is 72 degrees a second.

Jill makes a revolution in 6 seconds, that is 60 degrees in one second.

So Jill is turning 132 degrees per second relative to Jack.

In one minute Jill completes 132$\times$60$\div$360 revolutions = 22 revolutions relative to Jack.

As Jack is half way around relative to Jill, this means they will pass 22 times in the first minute.

**Using the speed-distance-time relationship**

In $t$ seconds, Jack does $\frac t5$ revolutions

Jill $\frac t6$

First meet: $\frac t5 + \frac t6 = \frac12$

$\frac{11t}{30}=\frac12$

$ t = \frac{15}{11}$

Time between meetings: $\frac t5 + \frac t6=1$

$\frac{11t}{30}=1$

$t=\frac{30}{11}$

$\frac {15}{11} + n\times\frac{30}{11}=60$ where $n$ is the number of times they meet after the first time

$15+30n=660$

$30n=645$

$n=21.5$

$n=21$ since $n$ must be the nearest whole number below

So they meet $22$ times