Maths homework help needed.... we're useless parents and can't help ds!

[carriemumsnet]

Ds - year 6 - has a maths challenge homework
One of questions two people swim to and fro across a pool - they start to swim from the middle of adjacent sides at exactly the same time. They swim at the same rate, How far do they swim before they meet

There's then a diagram of a pool which is 70 m long and 30 m wide

Possible answers are a) they never meet b) 100m c) 105m d) 1050m e) 2100

We think 2100 but not sure how or why...

Any suggestions welcome!

THanks

I think the answer is 105.

Boy A is starting at the midpoint of the 70 M wall. So he has to swim 35 m to get to the middle.

Boy B is starting at the midpoint of the 30m wall, so he has to swim 15m to get to the middle.

So on the first length Boy B will be at the middle far earlier than Boy A.....

So the answer is 105 which is the lowest common multiple of 35 and 15. - ie the first time someone swimming 35 meters will intersect with someone swimming 15 meters.

Wow thanks
Dh now of course saying he was right and knew that all along but that does make sense to us now so thanks!

I thinks also that they would meet after 105m. But, I think it's not quite as straightforward as simply finding the lower common multiple. Doesn't it need also to be common multiple that is also an odd-multiple of both 15 and 35? As it happens, 105 is, so that works for this case. (3 half-lengths for the length-swimmer, which brings him back to the middle, and 7 half-widths for the width-swimmer, which brings him back to the middle).

If it had been 60m by 40m pool, then the distances each would swim to the middle would be 30m and 20m. The lowest common multiple of those numbers is 60m. But they wouldn't meet after 60m, because the first boy would have done an entire length and be at the side, and the second boy would have done 3 widths, and also be at the side.

(sorry, meant to say there that second boy would be at the middle, having done a full width and a half. But one would be at the side, and one at the middle, not both in the middle!)