This problem solving activity has a measurement focus.
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?
- Measure using grams.
- Apply addition to a measurement problem.
- Devise and use problem solving strategies to explore situations mathematically (be systematic, use equipment).
This problem gives students the chance to do some number investigations using combinations of 5 and 7. It also builds their experience making measurement calculations. 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 (English)
- Copymaster of the problem (Māori)
- Balance scales and weights (5g and 7g) (and/or a digital representative)
The 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 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?
What else could we use to measure these things? - Pose the problem.
- As the students work on the problem in pairs, ask questions that focus their understanding of the size of grams. Encourage them to write their measurements with the correct unit (g - 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 are you recording your measurements?
How do you know that you are on the right track? - Share solutions
Extension
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.
Solution to the Extension
(9 x 5, 1 x 7 ) or (6 x 7, 2 x 5)