How I estimate: Multiplication
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There were 14 friends at a birthday party. They were each given a bag, and each bag had about 26 jellybeans in them.
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Estimate about how many jellybeans there were altogether and show how you did this in the box below.
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b) |
Show another way of estimating about how many jellybeans there were.
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The task is best done after students have had teaching on alternative strategies for estimation. This task can be administered to individual students orally, with the teacher recording responses or with small groups of students or a whole class with students recording their responses on their own worksheets. A suggested lesson sequence:
- Emphasise that this is a multiplication problem (14 × 26).
- Ask the student(s) to make an estimate of the number of jellybeans and show how they did it in the box in a).
- Ask if they can think of up to two more different ways of making an estimate for this.
- Have a discussion sharing the different methods used. Prompt students if they think that each estimate is too small, about right, or too large (see Next steps 4. and 5. at the end of this resource).
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a) |
Acceptable estimation strategies with an appropriate estimate. For example:
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Results based on a sample of nine Year 8 students in one-to-one interviews. |
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- Rounding each number to the nearest 10 (i.e., 10 × 30) was by far the preferred option.
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A group of nine Year 8 students were given this problem both before and after they had completed a unit emphasising the use of a wide range of estimation strategies. Before the unit most students could come up with just one estimation strategy with three giving two strategies. After the unit, all but one student could come up with at least two strategies, with two suggesting four different ways of estimating. A wider range of strategies were used overall.
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Strategy name |
Likely calculation |
Typical estimates |
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10 × 20 |
200 |
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Front-end and final compensation |
10 × 20 + 4 × 20 + 6 × 10 |
340 |
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10 × 30 or |
300 |
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Rounding and intermediate compensation |
10 × 36 or |
360 |
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20 × 30 |
600 |
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15 × 25 (only accept if it can easily be done mentally using facts about the 25 times table) |
375 |
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10 × 26 |
260 |
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Rounding one number and final compensation |
10 × 26 + 4 × 25 |
360 |
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7 × 52 |
350 |
7 × 50NOTE:
- Accept if students attempt a strategy, but do not necessarily come up with an appropriate estimate.
- Results are based on a sample of nine Year 8 students in one-to-one interviews.
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Strategy name |
Likely calculation |
Typical answer(s) |
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Exact computation |
14 × 26 |
364 |
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Exact computation and then rounds |
14 × 26 |
360 |
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Rounds but still cannot do calculation mentally (see Next steps 3) |
15 × 25 (but does vertical algorithm or other exact method) |
375 |
Prior knowledge needed
- A firm base of multiplicative facts.
- Good place value concepts.
- Ability to mentally add numbers in tens, hundreds, etc, using standard place value concepts.
- Expose students to a wide range of estimation strategies (refer to Computational estimation information). This could be through a discussion of the ways different students in the class estimated with these problems. Students may equate rounding and estimation.
- Discuss that estimation aims to eliminate exact computation.
- Question the student if they can easily do the mental computation after the rounding that they have specified. If they cannot, they need to be exposed to the idea that numbers should be rounded to ones that they can easily add mentally. Students may think that any rounding implies estimation.
- Discuss the concept of intermediate compensation. i.e. Reduce one number by a similar proportion as you increase the other one by.
- Students who see 10 × 30 = 300 as "About right (because I rounded down by 4 and up by 4)" can be challenged with the question "What about 20 × 20 = 400? (Is it about right?)". They should be perplexed that 300 and 400 are both "about right". One response was "The answer must be between them" (Interval estimation). A more sophisticated answer, based on proportional reasoning, was "Taking 14 down to 10 is taking off 4/14 so I need to add about 4/14 to 26, which is about 8, so its about 10 × 34 = 340".