Teaching mathematics
Teaching mathematics

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Teaching mathematics

3.2 Percentage change: using multiplying factors

Consider the following problem.

A customer goes into a restaurant and pays for a meal. Because they had the ‘early bird’ menu, their bill is discounted by 25%. There is also a 10% service charge added to their bill. What difference does it make to the customer's bill if the waiter deducts the 25% and then adds the 10% service charge, or alternatively, adds the service charge and then makes the deduction?

Activity 10 Does the order matter?

Timing: Allow 15 minutes

Work out the solution to the above problem in both ways and see what the difference is. Don’t forget that the second percentage change is enacted on the result of the first percentage change, i.e. having deducted 25% from the bill, the 10% service charge is worked out on the resulting discounted bill.

You could choose to calculate the solution of the problem using different amounts for the bill. Or you could prove it once and for all using a general case, which requires you to use algebra, by using a letter to represent the amount of the bill.

Discussion

One form of the solution uses algebra to find the general case:

Let the cost of the meal be c.

For 25% deduction followed by 10% increase:

multiline equation line 1 25 percent of postfix times times c equals one divided by four postfix times c postfix times postfix times open find 25 percent of the cost of the meal close line 2 c minus one divided by four postfix times c equals three divided by four postfix times c postfix times open subtract 25 percent from the cost of the meal close line 3 10 percent normal o times normal f postfix times times three divided by four postfix times c equals one divided by 10 multiplication three divided by four postfix times c equals three divided by 40 postfix times c postfix times open find 10 percent of the amended bill so far close line 4 three divided by four postfix times c plus three divided by 40 postfix times c equals 33 divided by 40 postfix times c postfix times open finall cost of the meal close

For 10% increase followed by 25% deduction:

multiline equation line 1 10 percent of postfix times times c equals one divided by 10 times c postfix times open find times 10 times percent of the cost of the meal close line 2 equation left hand side c plus one divided by 10 times c equals right hand side 11 divided by 10 times c postfix times open add 10 percent to the cost of the meal close line 3 25 percent normal o times normal f postfix times equation sequence 11 divided by 10 times c equals one divided by four times cap x times 11 divided by 10 times c equals 11 divided by 40 times c postfix times open find times 25 times percent of the amended bill so far close line 4 equation sequence 11 divided by 10 times c minus 11 divided by 40 times c equals 44 divided by 40 times c minus 11 divided by 40 times c equals 33 divided by 40 times c postfix times open find cost of the meal close

Were you surprised at the result? It does not matter which way round the final bill is calculated because a 25% deduction followed by adding on 10% works out the same as adding 10% first followed by deducting 25%.

To see why this is, consider the use of multiplying factors for each of the percentage changes.

A 10% increase means that, having started with 100% of the amount and adding an extra 10%, you end up with 110% of the amount.

That is equivanet to 1.1 as a decimal or postfix times postfix times as a fraction

A 25% decrease means that, having started with 100% of the amount and subtracting 25%, you end up with 75% of the amount.

That is equivanet to 0.75 as a decimal or postfix times three divided by four postfix times as a fraction

Using decimals for the multiplying factors (although you can also use fractions as the multiplying factors), an elegant solution to the problem can be found as follows:

Subtracting 25% then adding 10% gives c prefix multiplication of 0.75 multiplication 1.1

Adding 10% then subtracting 25% gives cprefix multiplication of 1.1 multiplication 0.75

Both multiplications come to 0.825c because it does not matter which way round you multiply (this is known as commutativity).

In short, a percentage change can be calculated by using the appropriate multiplying factor, which is calculated by considering what has happened to the original 100%.

The use of multiplying factors for percentage change does make the calculations much simpler. However, it may be a difficult concept for younger learners to grasp. If this is the case, it is better to allow learners to work the long way round.

In mathematics it is always best to work with the conceptual understanding that learners have demonstrated and to build on this carefully. Learners can do, understand and succeed when they have grasped the conceptual underpinning for the mathematics they are doing. This was summarised by Richard Skemp (1976)as relational understanding. Otherwise, if there is no understanding, learners will resort to trying to learn rules which are easily forgotten. (Skemp referred to the learning of rules and procedures without a conceptual underpinning as instrumental understanding.)

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