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3.10 Find unknown quantities

Lesson

Are you ready?

Do you remember how to find the  missing number in a number sentence,  in order to make it balanced?

Examples

Example 1

Complete the number sentence: 48 - ⬚ = 19 + 16

Worked Solution
Create a strategy

Find the right hand side of the equation first then use a number line to complete the number sentence.

Apply the idea

Add the numbers in the right hand side of the equation in a vertical algorithm. \begin{array}{c} & &1 &9 \\ &+ &1 &6 \\ \hline \\ \hline \end{array}

Add the ones column first: 9 + 6 = 15. Bring down the 5 and carry the 1 to the tens column. \begin{array}{c} & &\text{}^1 1 &9 \\ &+ &1 &6 \\ \hline & & &5 \\ \hline \end{array}

Add the tens column: 1 + 1 + 1 = 3. \begin{array}{c} & &\text{}^1 1 &9 \\ &+ &1 &6 \\ \hline & &3 &5 \\ \hline \end{array}

The simplified number sentence would be: 48 - ⬚ = 35

Locate 48 in the number line.

303132333435363738394041424344454647484950

Jump back from 48 and count the number of units until we reach 35.

303132333435363738394041424344454647484950

We have jumped back 13 units from 48. So the complete number sentence would be: 48 - 13 = 19 + 16

Idea summary

For number sentences to be equivalent, or balanced, both sides must equal the same amount.

Find unknown quantities in single step problems

This video looks at how to find unknowns values in number sentences that involve one step.

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Examples

Example 2

Complete the statement with the missing number: 9\times⬚=135

Worked Solution
Create a strategy

Use the number line below to count how many jumps of 9 there are between 0 and 135.

0918273645546372819099108117126135
Apply the idea

Locate where 0 and 135 is.

0918273645546372819099108117126135

We can see that there are 15 jumps of 9 from 0 to 135. So, the complete statement is 9 \times 15 = 135

Idea summary

We can use a number line to find an unknown quantity.

Find unknown quantities in multi-step problems

Now we look at problems that will need more than one step to find the answer.

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Examples

Example 3

Complete the number sentence: 40 = 8 + 8 \times ⬚

Worked Solution
Create a strategy

Simplify the number sentence and solve it to help with the more complicated number sentence.

Apply the idea

We can simplify the number sentence to be: 40 = 8 + ⬚

To find the unknown, we locate where 8 is then jump to the right to get 40.

010203040

We have jumped 32 spaces to the right of 8.40 = 8 + 32

To complete the original number sentence, we need to find the unknown for the following number sentence: 8 \times ⬚ = 32

We can count up by multiples of 8 to get: 8,\,16,\,24,\,32, so we get 32 after 4 counts. This means that: 8\times 4 = 32

The complete number sentence is: 40 = 8 + 8 \times 4

Idea summary

To find the missing number in a multi-step number sentence, we can make the number sentence simpler and solve that one first.

Simplify problems

Sometimes you can make the problem easier if you simplify the question first, this video will show you how.

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Examples

Example 4

Let's work out what must equal to make the following number sentence true.

⬚ + 16 = 3 \times 8

a

First, what does 3 \times 8 equal?

Worked Solution
Create a strategy

We can think of it as 3 groups of 8, like in the array below where there are 3 rows of 8 squares.

The image shows 3 rows of 8 squares.
Apply the idea

Count the number of squares in the array to get: 3\times 8=24

b

So now we have ⬚ + 16 =24. What number plus 16 equals 24?

Worked Solution
Create a strategy

Use the number line below to count how many jumps are between the starting number and the target number.

14161820222426
Apply the idea

Plot 16 and 24 on the number line and count the number of jumps.

14161820222426

We have jumped 8 spaces to the right of 16. So: 8 + 16 = 24

This means that the unknown was 8: 8 + 16 = 3 \times 8

Idea summary

To balance a number sentence, work out the value of one side first, and use the answer as the target value to work out the other side.

Outcomes

MA3-6NA

selects and applies appropriate strategies for multiplication and division, and applies the order of operations to calculations involving more than one operation

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