On the campaign trail

In question 1, the students need to calculate the number of pupils out of 30 that are represented by the various fractions. This can be done in several ways, but probably the easiest is to divide by the reciprocals. It is important that the students understand that finding, for example, 1/6 of 30 is the same as calculating 1/6 x 30, which is 30/6. This could be written as:

• 1/6 x 30 = 30 x 1/6 = 30 ÷ 6.

In other words, these are multiplication problems that can be turned into division problems by using the reciprocal of the fractions.

Before the students begin this activity, it may be helpful to develop their confidence in working with reciprocals. The reciprocal of a number is the number that results in 1 when it is multiplied by the original number. For instance, the reciprocal of 1/6 is 6 because 6 x 1/6 is 1. The reciprocal of a fraction can be found by swapping the numerator and the denominator. For example,

• the reciprocal of 2/3 is 3/2.

This also works for whole numbers where the denominator is 1 in fraction form.

For example:

• because 6 = 6/1, the reciprocal is 1/6, and the reciprocal of 1/7 is 7/1 or 7.

Now the next step can be taken, that is, establishing that finding a fraction of any number is the same as dividing the number by the reciprocal of the fraction. You can lead the students to this understanding by considering a simple case, such as finding 1/2 of some number. After asking them to work out half of various numbers, including some larger numbers such as 86 or 124, they can be asked how they did this. They will most likely say that they divided the 86 or 124 by 2, which is equivalent to solving 86/2 and 124/2 respectively.

It is then a matter of establishing that this same principle works for other numbers.

For example:

• finding 1/4 of 12 is the same as dividing 12 by 4.

With these insights, the students can now return to the activity and divide 30 by 6, 10, 3, and 10 respectively. Finding 3/10 of 30 consists of two steps: dividing by 10 and then multiplying the result by 3. This is equivalent to solving 3/10 x 30 = 90/10. Once the proportions are calculated, it is a relatively straightforward matter of arranging them in order from smallest to largest.

In this activity, the use of a calculator is not recommended for two reasons. Firstly, the calculations are simple enough to be done mentally, and secondly, the thirds and sixths could cause some confusion as the decimal equivalents are “messy”.

In question 2, the students need to consider the candidates’ policies in the light of Jed’s survey results. The answers provide some indication of how students may tackle this. The key to the election may be the three pupils who didn’t express any opinion. In short, there is no one right answer to question 2. You could conclude the activity by having the students justify their predictions during a whole-class discussion and reflect on the fact that candidates’ policies are only one of the factors that influence voter choice.

1.

3 students want a shorter school day. 5 students want girls to be able to wear make-up to school. 9 students want a longer lunch break. 10 students want to wear jackets in class.

2.

Predictions will vary, but Awatea should win. Explanations should relate to numerical data. For example, 10 students want to be able to wear jackets in class. This would give Awatea more votes than the other two candidates. However, 3 of the 30 students have not expressed an opinion. If they voted for Bram, he would win.