*There are 60 children in a school, in two classes, A and B. The school moves 5 children from Class A to Class B. Then 3 children are moved from Class B to Class A. They then decide to make a new class, Class C. So, 12 children from Class A and some children from Class B are moved to Class C, so that there is now the same number of children in each of the three classes.*

*How many children were there in Class A and Class B to begin with?*

To solve this problem we start with the end result – three classes of 20 – and then work backwards, carrying out the inverse of each of the given operations in reverse order. Using inverses like this is an important mathematical process and problem-solving strategy.

So, start from Class A having 20 children, Class B having 20 and Class C having 20.

The 20 children in C are made up of 12 children taken away from from A (we're told that) and (therefore) 8 children taken away from B.

So prior to this move A must have had 20 + 12 (= 32) and B must have had 20 + 8 (= 28).

Before this 3 children were taken away from B and added to A. So B must have had 28 + 3 (= 31) and A must have had 32 – 3 (= 29).

Before this 5 were taken away from A and added to B. So at the start A must have had 29 + 5 (=34) and B must have had 31 – 5 (= 26).

So the original distribution of children was 34 in Class A and 26 in Class B.

For primary school children (and adults!) this problem is a good test of the ability to extract and process information from a complex situation.

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