Dr Martin the chemist is weighing out some pills.

He has some 5g weights and some 7g weights.

Can he weigh exactly 38g of pills?

This problem gives students the chance to do some number investigations using combinations of 5 and 7. This helps extend their experience and knowledge of number, and can also extend their experience of weights and the use of balance scales.

Copymaster of the problem (Māori)

Balance scales and weights (5g and 7g)

### Problem

Dr Martin the chemist is weighing out some pills. He has some 5g weights and some 7g weights. Can he weigh exactly 38g of pills?

### Teaching sequence

- Ask the students to find an object that they estimate weighs 20g. Check estimates on the (balance) scales.
- Discuss students' ideas about how they made their estimates of 20g (eg, weight of small chip packet = 18g, flake bar = 30g).
*What object in your desk would weigh close to 38g? How did you decide that?*

How do you use weights on a balance scale? /How do you use these kitchen scales? - Pose the problem.
- As the students work on the problem in pairs, ask questions that focus their understanding of the size of grams.
- Focus their thinking on working systematically by asking questions about the way that they are keeping track of their work.
*What are you doing?*

How will you share what you have done with others in the class.

How do you know that you are on the right track? - Share solutions

#### Extension to the problem

Can the chemist weigh out 52g ? Can this be done in more than one way?

#### Solution

38 is not exactly divisible by 5 or 7. Hence both 5g and 7g weights are needed. 38 – 7 = 31, 38 – 2 x 7 = 24, and 38 – 3 x 7 = 17 are not divisible by 5. However, 38 – 4 x 7 = 10 = 2 x 5. So Dr Martin can use four 7g weights and two 5g weights.

#### Extension:

( 9 x 5, 1 x 7 ) or (6 x 7, 2 x 5)