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# Revolutions

**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

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Age 11 to 14

ShortChallenge Level

- Problem
- Solutions

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

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

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.

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

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.

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