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Here's a sample of the good reasoning from Maithra of Henrietta Barnett School, and others :


At their first crossing A has swum 30 and B has swum x - 30 , where x is the pool length.

At their second crossing A has now swum x + 20 in all and B has swum 2x - 20

There respective distances will keep in a fixed ratio because their speeds are in a fixed ratio.

This equation expresses that

and can be solved to find the length of the pool.

x is zero or 70, and a pool of length zero is not our interest here so 70 m is the pool's length.

Here's another line of reasoning :

At first crossing A and B have swum one length between them, and A swam 30m of that length.

At second crossing A and B have swum two more lengths between them, so A will have swum 60m of those two lengths.

A has swum 90m in all, and that is one whole length plus a further 20m . So the pool is 70m long.








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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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