 N10 Written and calculator methods

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N10 Written and calculator methods
Maths Age 11-14 The aim of this unit is to teach pupils to: Make and justify estimates and approximations (of numbers and calculations). Use efficient column methods for addition and subtraction of whole numbers, and extend to decimals. Refine written methods of multiplication and division of whole numbers to ensure efficiency, and extend to decimals with two places. Carry out more complex calculations using the facilities on a calculator. Interpret the display on a calculator in different contexts (fractions, decimals, money, metric measures, time). Use checking procedures, including working the problem backwards and considering whether the result is the right order of magnitude. N10 Written and calculator methods

N10.1 Estimation and approximation
Contents N10 Written and calculator methods A1 N10.1 Estimation and approximation A1 N10.2 Addition and subtraction A1 N10.3 Multiplication A1 N10.4 Division A1 N10.5 Using a calculator A1 N10.6 Checking results

Estimation four-in-a-line
Start by dividing the class into two teams, red and blue. Each team must select a spokesperson whose role it is to select a target square each time. The object of the game is to get four squares in a line either horizontally, vertically or diagonally. Decide which team will start. The spokesperson must call out a target number from the grid, for example Members of the team must then use approximation to estimate which two numbers from the boxes on the left would multiply together to make the target number. Using approximation, pupils must choose a number from the yellow box and a number from the green box to make the required target number. If pupils estimate that 7 × 9.96 would give the required answer, select 7 from the yellow box and 9.96 from the green box. Press the equals button and the corresponding square will be coloured in the team colour. If the team choose the wrong numbers to multiply together to get their target square then another square will be coloured in the team colour (corresponding to the answer to their chosen calculation). If this square is already taken then a message stating that the square is unavailable will appear. Play then passes to the opposing team. The game continues until one team gets four in a line or every square in the grid is coloured.

Estimation Martin uses his calculator to work out 39 × 72.
The display shows an answer of 1053. How do you know this answer must be wrong? 39 × 72  40 × 70 = 2800 “is approximately equal to” Explain that there are times when there is no need to calculate an exact answer and an estimate is sufficient. For example, we might be doing some supermarket shopping and we might want to keep track of how much we are spending. In maths, it is often useful to estimate an answer before doing a calculation (as we did in the starter activity) or after we’ve done a calculation to make sure our answer is sensible. Calculators can be very useful for working things out, but only if we press the right buttons! We should always check that the answer given by our calculator is sensible. Introduce the notation . Introduce the rule that the product of two numbers must end in the same digit as the product of the end digits of the numbers. Ask pupils if they can work out what Martin did wrong when he calculated 39 × 72 (he typed 39 × 27 into his calculator by mistake). Also, if we multiply together the last digits of 39 and 72 we have 9 × 2 = 18. 9 × 2 = 18. The product of 39 and 72 must therefore end in an 8.

How could we estimate the answer to 3.5 × 17.5?
Estimation How could we estimate the answer to 3.5 × 17.5? 3.5 × 17.5 can be approximated to: Explain that there can be many ways to estimate the answer to a problem. For example we could estimate 3.5 × 17.5 to be approximately equal to: 4 × 20, if we only need a rough estimate it is best to choose numbers we can work with easily in our heads – it may not give the most accurate answer but we can at least have the correct order of magnitude. 3 × 18, round 3.5 down and 17.5 up. (Also, discuss mental strategies for calculating 3 × 18. For example, 3 × 20 is 60, subtract 3 × 2, 6, equals 54. Or 3 × 10 is 30, plus 3 x 8, 24, equals 54.) 4 × 17, round 3.5 up and 17.5 down. (Also, discuss mental strategies for calculating 4 × 17. For example, double 17 is 34, and double 34 is 68. Or 4 × 10 is 40, plus 4 × 7, 28, equals 68.) Discuss the last approximation. We know that 3 × 17 must be below the actual answer because we’ve rounded both numbers down and we know that 4 × 18 must be above the actual answer because we’ve rounded both answers up. Therefore the answer must lie between 51 and 72. Discuss the benefits of each approximation. 4 × 20 is the least accurate, but it is the easiest to work out. 3 × 18 and 4 × 17 are similarly accurate – it depends on which one you would rather work out in you head. Since we have a strategy for multiplying by 4 (doubling and then doubling again) it may be sensible to use that approximation. Lastly, how can we work out what number is half-way between 51 and 72? Establish that we can work out and divide by 2 to get 61.5. 3.5 × 17.5 is actually 61.25, so 61.5 is an excellent approximation. The disadvantage of this approximation is that there is a lot of mental maths to do. It is also worth pointing out that when multiplying two numbers it is preferable to either round both numbers up or round both numbers down. 4 × 20 = 80 3 × 18 = 54 4 × 17 = 68 or between 3 × 17 = 51 and 4 × 18 = 72

Using points on a scale to estimate answers
Jessica is trying to estimate which number multiplied by itself will give the answer 32. She knows that 5 × 5 = 25 and that 6 × 6 = 36. The number must therefore be between 5 and 6. She draws the following scales to help her find an approximate answer. Ask pupils to verify, using a calculator, that 5.64 × 5.64 is almost 32. Notice that since squaring is not a linear function this method will only give us an approximate answer. The arrow on the bottom scale is pointing to 5 and 7 elevenths. This is not the square root of 32, but is quite close to it. It may not be necessary to discuss this point unless it arises. 25 26 27 28 29 30 31 32 33 34 35 36 5 6 5.64

Using points on a scale to estimate answers
Use Jessica’s method to estimate which number multiplied by itself will give an answer of 40. We know that 6 × 6 = 36 and that 7 × 7 = 49. Draw a scale from 36 to 49. Underneath, draw a scale from 6 to 7. 36 37 38 39 40 41 42 43 44 45 46 47 48 49 Give pupils a time limit to use the illustrated method to find the square root of 40. Reveal parts of the solution as necessary. The scales should be drawn on cm squared paper if possible. The top line should be 13 cm long as should the bottom line. To divide the bottom line into 10 equal parts pupils will need to use their rulers to mark on the divisions every 13 mm. After revealing the solution, ask pupils to verify, using their calculators, that 6.31 is a good approximation. 6 7 6.31

N10.2 Addition and subtraction
Contents N10 Written and calculator methods A1 N10.1 Estimation and approximation A1 N10.2 Addition and subtraction A1 N10.3 Multiplication A1 N10.4 Division A1 N10.5 Using a calculator A1 N10.6 Checking results

Adding and subtracting decimals
Use this activity to demonstrate that the decimal points must be lined up when adding and subtracting decimals. Use the pen tool to fill in any zero place holders and to complete the calculation. Ask volunteers to come to the board and complete calculations.

Adding and subtracting decimals
Jack is doing some DIY. He buys a 3m length of wood. Jack needs to cut off two pieces of wood – one of length 0.7m and one of length 1.92m. What is the total length of wood which Jack needs to cut off? What is the length of the piece of wood which is left over? 2 9 b) a) 0.7 Jack needs to cut off 2.62m altogether. 3 .00 1 1 The left-over wood will measure 0.38m (or 38cm). Before going through the calculations, establish that for part a) we must add together 0.7 and 1.92 and for part b) we must subtract the answer to part a) from 3m. State that to work out 0.7 plus 1.92 we will use standard column method. Remember that we have to line up the units tenths and hundredths and, most importantly, the decimal point. It is sensible to add a zero place holder after the 7 in 0.7 to make Remind pupils that we can write as many zeros as we like after a decimal number without changing its value. The decimal point will be in the same position in the answer. Go through the method for addition. Click to reveal a dashed line to show the decimal points lined up. Ask how we could check this answer by approximating. + 1.92 2.62 2 . 6 2 . 3 8 1

Sums and differences Ask pupils how we could use the digit cards to make the largest sum possible. Discuss strategies and check the answer by clicking on the show answer key. Now, ask how we could make the smallest sum possible. Again, discuss strategies and check the answer by clicking on the show answer key. Repeat the activity to find the largest difference possible and the smallest difference possible.

N10 Written and calculator methods
Contents N10 Written and calculator methods N10.1 Estimation and approximation N10.2 Addition and subtraction N10.3 Multiplication N10.4 Division N10.5 Using a calculator N10.6 Checking results

The grid method for multiplying decimals
Demonstrate how we can use the grid method to multiply decimals. If pupils lack confidence in multiplying by decimals, an alternative would be to write an equivalent whole number calculation by multiplying the numbers by 10, 100 and 1000 and dividing by the appropriate amount at the end of the calculation.

Using the standard column method
What is 2.28 × 7? Start by finding an approximate answer: 2.28 × 7  2 × 7 = 14 2.28 × 7 is equivalent to 228 × 7 ÷ 100 Put this multiplication into context. For example, suppose you are at a supermarket and you want to buy 7 boxes of cola cans costing £2.28 a box. You have £15 in your wallet. Do you have enough money? Start by establishing that 2.28 × 7 is too difficult to work out mentally. (If necessary, we could work out 2.28 × 8 by doubling 3 times and then subtract 2.28 to get 2.28 × 7). We will have to use a written method. We could use the grid method or, as shown here, the column method. (Pupils may have met other written methods such as the Chinese or lattice method.) Emphasize that whatever method we use we should start by writing down an approximate answer. Explain that it is easier to multiply by whole numbers than it is to multiply by decimals. It is therefore sensible to find an equivalent whole number calculation, in this case 228 × 7 ÷ 100. Talk through the standard column method for multiplying 7 and 228. Remind pupils that we always write down the units and carry the tens to the next column. Explain that what we are actually doing when we perform this calculation is work out 7 × 8, then 7 × 20 and then 7 × 200. As we carry the tens across each time, we are adding these three together as we go along. (This method is therefore more efficient than the grid method.) To find 2.28 × 7 we need to divide by 100. Check this answer by comparing it to the approximation made at the beginning. 228 × 7 Answer 15 9 6 2.28 × 7 = 1596 ÷ 100 = 15.96 1 5

Using the standard column method
What is × 0.8? Again, start by finding an approximate answer: 392.7 × 0.8  400 × 1 = 400 392.7 × 0.8 is equivalent to 3927 × 8 ÷ 100 Again, explain that it is easier to multiply by whole numbers than it is to multiply by decimals. It is therefore sensible to find an equivalent whole number calculation in this case 3927 × 8 ÷ 10 ÷ 10 = 3927 × 8 ÷ 100 Talk through the standard column method for multiplying 8 and Remind pupils that we always write down the units and carry the tens to the next column. To find × 0.8 we need to divide by 100. Check this answer by comparing it to the approximation made at the beginning. 3927 Answer × 8 392.7 × 0.8 = ÷ 10 ÷ 10 31 4 1 6 = 7 2 5

Drag and drop multiplication problem
Ask pupils to fill in the remaining digits to make the sum correct. For example, 9. × 3 = .2 Possible solutions are: 91.4 × 3 = 274.2, 92.4 × 3 = and 93.4 × 3 =

Multiplying two-digit numbers
Calculate 57.4 × 24. Estimate: 60 × 25 = 1500 Equivalent calculation: 57.4 × 10 × 24 ÷ 10 = 574 × 24 ÷ 10 574 × 24 4 × 574 = 2296 Talk through possible estimates for this calculation. The next step is to form an equivalent calculation using whole numbers. For example, if we are multiplying together 57.4 and 24, we can multiply 57.4 by 10 to make a whole number. We must then divide by 10 to undo the effect of multiplying by 10. Therefore 57.4 × 24 = 57.4 × 10 × 24 ÷ 10. Explain that when we multiply by a two digit whole number we have to partition the number into tens and units. We multiply by the units first, in this case 4, and then by the tens, in this case 20. These two answers are then added together. Emphasize that the final answer should be compared with the estimate to check the order of magnitude. 20 × 574 = 11480 13776 Answer: ÷ 10 =

Multiplying two-digit numbers
Calculate 23.2 × 1.8. Estimate: 23 × 2 = 46 Equivalent calculation: 23.2 × 10 × 1.8 × 10 ÷ 100 = 232 × 18 ÷ 100 232 × 18 8 × 232 = 1856 Talk through the problem as before. 10 × 232 = 2320 4176 Answer: 4176 ÷ 100 = 41.76

Multiplying two-digit numbers
Calculate 394 × 0.47. Estimate: 400 × 0.5 = 200 Equivalent calculation: 394 × 0.47 × 100 ÷ 100 = 394 × 47 ÷ 100 394 × 47 7 × 394 = 2758 Modify the numbers in this slide to make more examples. 40 × 394 = 15760 18518 Answer: ÷ 100 =

N10 Written and calculator methods
Contents N10 Written and calculator methods A1 N10.1 Estimation and approximation A1 N10.2 Addition and subtraction A1 N10.3 Multiplication A1 N10.4 Division A1 N10.5 Using a calculator A1 N10.6 Checking results

Dividing decimals – Example 1
What is 259.2 ÷ 6? Dividend Divisor Put this division into context. For example, six children have a combined mass of kg. What is their average mass? Ask pupils to give other real-life examples. Start by establishing that ÷ 6 is too difficult to work out mentally. We will have to use a written method. We always start by finding an approximate answer. Discuss a sensible approximation for this calculation. Remember, we must choose numbers that can be divided exactly without leaving remainders. We should also choose numbers that can be easily calculated mentally. For example, round to the nearest multiple of ten that can be divided by 6: 240 ÷ 6 = 40.

Using repeated subtraction
Start by finding an approximate answer: 259.2 ÷ 6  240 ÷ 6 = 40 259.2 6 – 240.0 6 × 40 Emphasize that whatever method we use we should start by writing down an approximate answer. Talk through this method. Explain that just as we can think of multiplication as repeated addition, we can think of division as repeated subtraction. When we divide a number with a decimal by a whole number we will have a remainder. The remainder will also be divided by the number to give a decimal answer. It is not efficient to subtract 6 again and again, and so we start by subtracting the largest multiple of 6 that we can work out mentally and that is less than From our approximation we know that 40 × 6 is 240. We can subtract 240, forty 6’s, in one step. We write to ensure that the decimal points are lined up. Write 6 × 40 beside so that we can see how many 6’s we’ve subtracted so far. We then subtract from Click to reveal 19.2. Ask, How many 6s can we subtract from 19.2? 3 × 6 is 18, so we can subtract three 6’s. We write 18.0 under 19.2, ensuring that the decimal points are lined up, and write 6 × 3 beside We then subtract 18.0 from 19.2. Click to reveal 1.2. Now, we know that 12 can be divided by 6, so what is 1.2 divided by 6? Establish that if 12 ÷ 6 = 2 then 1.2 ÷ 6 = 0.2 and so 6 x 0.2 = 1.2. We write 6 × 0.2 beside the 1.2. We can now subtract 1.2 from 1.2 to get 0. To find the answer, we need to add together 40, 3 and 0.2 to get 43.2. Tell pupils that the result of a division, in this case 43.2, is called the quotient. 19.2 – 18.0 6 × 3 1.2 – 1.2 6 × 0.2 Answer: 43.2

Using short division Start by finding an approximate answer: 259.2 ÷ 6  240 ÷ 6 = 40 4 3 . 2 6 2 1 1 We can also work this out using short division. This method can be used when dividing by a single digit and is much quicker than subtracting multiples. Once again, we start by writing down an approximate answer. Explain that when dividing 2 by 6, it is not strictly necessary to write down the 0 at the front. However, doing this helps to keep the digits lined up in the correct position. Point out the position of the decimal point, which must be lined up with the decimal point beneath it. Ensure that pupils are able to divide mentally and can find remainders mentally by subtraction or otherwise. 2.59 ÷ 6 = 43.2

Dividing decimals – Example 2
What is ÷ 9? Dividend Divisor Put this division into context. For example, nine people in a lottery syndicate win a total of £ How much do they win each if the prize money is shared equally? Ask pupils to give other real-life examples. Discuss a sensible approximation for this calculation. Remember, we must choose numbers that can be divided exactly. We should also choose numbers that can be easily calculated mentally. For example, 720 ÷ 9 = 80 (rounding to the nearest multiple of 10 that can be divided by 9). We could also use 710 ÷ 10 = 71 (but not 710 ÷ 9). Repeat the definitions of the keywords, dividend and divisor is the dividend and 9 is the divisor. The answer to this division is called the quotient.

Using repeated subtraction
Start by finding an approximate answer: ÷ 9  720 ÷ 9 = 80 714.06 9 Again, we start by finding an approximate answer. Talk through this method again. We can divide by 9 by subtracting successive multiples of 9 from We know from our approximation that 9 × 80 is 720. This is too big, however. So, let’s start by subtracting 9 × 70, which is We write to ensure that the decimal points are lined up. Write 9 × 70 beside so that we can see how many 9s we’ve subtracted so far. We then subtract from Click to reveal How many 9s can we subtract from 84.06? 9 × 9 is 81, so we can subtract nine 9’s. We write under 84.06, ensuring that the decimal points are lined up, and write 9 × 9 beside We then subtract from Click to reveal 3.06. Now, 3.06 can’t be divided by a whole number of 9’s. If this was 30.6, how many 9’s would divide into it? Establish that if three 9’s divide into 30.6 then 0.3 9’s divide into Pupils may find this step difficult so spend some time explaining it before moving on. 9 × 0.3 is 2.7. We write 2.70 under 3.06, ensuring that the decimal points are lined up, and write 9 × 0.3 beside We then subtract 2.70 from Click to reveal 0.36. What is 0.36 ÷ 9? Establish that if 36 ÷ 9 is equal to 4 then 0.36 ÷ 9 is equal to 0.04 and we can therefore subtract 9 × 0.04 from 0.36. We can now subtract 0.36 from We write 9 × 0.04 beside it. To find the answer we need to add together 70, 9, 0.3 and 0.04 to get Explain that as the numbers we are dividing become bigger, the short division method becomes the more efficient method. Note, however, that we can only use short division when dividing by single digit numbers. 9 × 70 84.06 – 81.00 9 × 9 3.06 – 2.70 9 × 0.3 0.36 – 0.36 9 × Answer: 79.34

Using short division Start by finding an approximate answer: ÷ 9  720 ÷ 9 = 80 7 9 . 3 4 9 7 8 3 3 Talk through the method for short division and note that when dividing by a whole single digit number, this method is much quicker and easier. Again, tell pupils that the result of a division, in this case 79.34, is called the quotient. ÷ 9 = 79.34

Drag and drop division problem
Drag three digits into the short division sum (one digit for each number, say) and ask pupils to fill in the remaining digits to make the sum correct. Some examples may have more than one possible solution.

Writing an equivalent calculation
What is 36.8 ÷ 0.4? This will be easier to solve if we write an equivalent calculation with a whole number divisor. ×10 36.8 0.4 368 We can write 36.8 ÷ 0.4 as = Explain that when dividing by a decimal it is sensible to form an equivalent calculation with a whole number divisor before using the standard column method of repeated subtraction. Explain that any division can be written using a dividing line. We can then find an equivalent fraction by multiplying the the numerator and the denominator by the same number. In this example, multiplying by 10 gives us a whole-number divisor. Ask pupils to give further justification for 36.8 ÷ 0.4 being equal to 368 ÷ 4. For example, 36.8 ÷ 4 will be ten times less than 36.8 ÷ 0.4 (because we are dividing by a number that’s ten times bigger). To compensate for this we need to make 36.8 ten times bigger. 4 ×10 36.8 ÷ 0.4 is equivalent to 368 ÷ 4 = 92

Find the equivalent calculation
Establish that when dividing an equivalent calculation can be made by multiplying both the dividend and the divisor by the same number (or dividing both the dividend and the divisor by the same number).

Dividing by two-digit numbers
Calculate 75.4 ÷ 3.1. Estimate: 75 ÷ 3 = 25 Equivalent calculation: 75.4 ÷ 3.1 = 754 ÷ 31 754 31 – 620 20 × 31 134 – 124 4 × 31 10.0 – 9.3 0.3 × 31 Talk through the standard method of repeated subtraction for division. Emphasize that we should always start by estimating the answer. When dividing we should always round to numbers that divide exactly. When dividing by a decimal, we write an equivalent calculation with a whole number divisor. We then use repeated subtraction to divide. Emphasize the need to work systematically when using this method. If we wish to find the quotient to one decimal place, as shown in this example, then it is necessary to find the value of the number in the second decimal place so that we know whether to round up or down. 0.70 – 0.62 0.02 × 31 0.08 Answer: 75.4 ÷ 3.1 = R 0.08 = 24.3 to 1 d.p.

Dividing by two-digit numbers
Calculate 8.12 ÷ 0.46. Estimate: 8 ÷ 0.5 = 16 Equivalent calculation: 8.12 ÷ 0.46 = 812 ÷ 46 812 46 – 460 10 × 46 352 – 322 7 × 46 30.0 – 27.6 0.6 × 46 In this example, we multiply both numbers by 100 to make an equivalent whole number calculation. If required copy this slide and change the numbers to give more examples. Set pupils variety of divisions by two-digit decimals to practise. 2.40 – 2.30 0.05 × 46 0.10 Answer: 8.12 ÷ 0.43 = R 0.1 = 17.7 to 1 d.p.

N10 Written and calculator methods
Contents N10 Written and calculator methods A1 N10.1 Estimation and approximation A1 N10.2 Addition and subtraction A1 N10.3 Multiplication A1 N10.4 Division A1 N10.5 Using a calculator A1 N10.6 Checking results

Solving complex calculations mentally
What is ? 7.4 – 2.4 7.4 – 2.4 = 10 5 = 2 When we say that a calculation is complex it usually means that it uses several operations and that it needs to be broken down into simpler steps. Sometimes we need to use a calculator. It is important with these calculations that we observe the correct order of operations. Remind pupils of BIDMAS. We always evaluate the contents of any Brackets first, then Indices (Powers), Division, Multiplication, Addition and Subtraction. Ask pupils what the line between and 7.4 – 2.4 means. It means that we have to divide the answer to by the answer to 7.4 –2.4. Ask pupils to give you the steps needed to work out this division mentally. Establish that we would need to evaluate the top and the bottom separately before we can divide. BIDMAS tells us that division should be preformed before addition and subtraction. However, the dividing line acts like brackets. When we have the dividing line like this we don’t need to put brackets around the ( ) and (7.4 – 2.4). We can work this problem out mentally. Reveal the stages on the board. Ask pupils to use their calculators to work out ( ) ÷ (7.4 – 2.4). Get pupils with the correct answer, 2, to put their hands up. Ask a volunteer to explain how they got the answer. Discuss how we can do this calculation on the calculator using the bracket keys. We could also write this calculation as: ( ) ÷ (7.4 – 2.4). How could we work this out using a calculator?

Using bracket keys on the calculator
What is ? 3.7 – 2.1 We start by estimating the answer: 3.7 – 2.1 6 2 = 3 Tell pupils that this problem is difficult to solve mentally because the number on the bottom will not divide exactly into the number on the top. We will use a calculator and we should start by estimating the answer. Remind pupils that when estimating answers to division problems we should round to numbers that divide into each other exactly. Demonstrate the calculation using bracket keys on the calculator. Verify that this answer is close to our approximation. Using brackets we key in: ( ) ÷ (3.7 – 2.1) = 3.625

Interpreting the calculator display
Use this activity to discuss how to interpret the calculator display in different contexts. Point out that when working with money convention dictates that we cannot write £4.3, for example, but must write £4.30. Ensure that pupils know metric conversions such as, 1km = 1000m and 1m = 100 cm. Make sure, too, that pupils understand how to convert decimal fractions of an hour into minutes and decimal fractions of a minute into seconds. Explain that we need to multiply the decimal part of the number by 60 because there are 60 minutes in an hour (and 60 seconds in a minute). Establish that units that are based on 10s, 100s and 1000s, such as pounds and pence and metric units are easy to work with on a calculator. Units of time and imperial units are much more difficult. For example, 1.25 hours is 1 hour and 15 minutes (1 quarter of 60 minutes), 1.25 feet is 1 foot and 3 inches (1 quarter of 12 inches) and 1.25 pounds in weight is 1 pound and 4 ounces (1 quarter of 16 ounces).

Finding whole number remainders
Sometimes, when we divide, we need the remainder to be expressed as a whole number. For example, 236 eggs are packed into boxes of 12. Division is probably one of the most difficult operations to compute mentally. This is because when a number does not divide exactly into another number we have a remainder. Ask pupils: How does a calculator display remainders? On a calculator remainders are always displayed as decimals. Depending on the context of the problem we are working out, we may need to find a remainder as a whole number rather than a decimal. Remainders can also be expressed as fractions. Emphasize that the way we express the remainder depends on the context of the problem. Discuss the problem on the board. Establish that the number of boxes filled is found by dividing 236 by 12. The remainder will give us the number of eggs left over. Instruct pupils to use their calculators to key in 236 ÷ 12. The calculator displays an answer of This is read as 19.6 recurring. That means that the 6 repeats infinitely. Ask pupils if they know why there is a 7 on the end (because of rounding the last 6 up) and what 0.6 recurring is as a fraction. The answer is 2/3 because it is double 0.3 recurring, which is 1/3. We can show that a digit in a decimal recurs by putting a dot above it. The whole number part, 19, tells us how many times 12 divides into 237. The decimal part, 0.6 recurring, tells us the remainder as a decimal. (Note: because we are dividing by 12 this does not mean 2/3 of 1 egg but 2/3 of 12 eggs). To find the remainder as a whole number we must multiply the decimal remainder by the number we are dividing by. Leaving on the display of their calculators, instruct pupils to subtract the whole number part, 19, and press the = key. The display shows Keying in × 12 = gives us the number of eggs left over, 8. Demonstrate this calculation on the calculator on the board. Ask pupils if they can think of an alternative method to find the number of eggs left over. We could work out 19 × 12, that is 228, and then subtract this from 236 to get the remainder, 8. We could use either method but emphasize that the method on the board is more efficient (that is, quicker). How many boxes are filled? Using a calculator: 236 ÷ 12 = This is 19.6 recurring or 19.6 . Number of boxes filled = 19 How many eggs are left over? . Number of eggs left over = 0.6 × 12 = 8

Finding whole number remainders
My calculator display shows the following: Find the remainder if this answer was obtained by: a) Dividing 384 by 60 0.4 × 60 = 24 Use this example exercise to discuss the fact that in each case we can find the remainder by multiplying the decimal part of 6.4 (that is, 0.4) by the divisor. Put each example into context. For example, 384 minutes = 6 minutes and 24 seconds. 160 pencils grouped into packs of 25 = 6 pack of 25 and 10 left over. 2464 newspapers put into batches of 385 = 6 batches and 154 left over. Ask how we could use our calculator to check our answer for part c). We could work out 385 × This should give us an answer of 2464. b) Dividing 160 by 25 0.4 × 25 = 10 c) Dividing by 2464 by 385 0.4 × 385 = 154

Working with units of time
What is 248 days in weeks and days? Using a calculator we key in: 2 4 8 ÷ 7 = Which gives us an answer of weeks. We have 35 whole weeks. To find the number of days left over we key in: Ask pupils what calculation we need to perform to convert 248 days into weeks and days. We need to divide by 7 because there are 7 days in a week. To find the number of days left over we leave on the display and subtract the whole number part 35. Emphasize that we must press the = key before we × 7. (This is because multiplication takes precedence over subtraction and we need the subtraction to be performed first.) Ask pupils if they can deduce what is as a fraction (3/7). As before, an alternative would be to work out 7 × 35, 245, and subtract this from 248 to find the remainder, 3. 3 5 = × 7 This give us the answer 3. 248 days = 35 weeks and 3 days.

Converting units of time to decimals
When using a calculator to work with with units of time it can be helpful to enter these as decimals. For example: 7 15 60 minutes 7 minutes and 15 seconds = = 7 1 4 minutes = 7.25 minutes 4 18 24 days Explain that an alternative to using decimals would be to convert to the smaller unit by multiplying. For example, 7 minutes and 15 seconds = (7 × 60) + 15 seconds = 435 seconds. This, however, means working with larger numbers and we would still have to convert back to the larger units by dividing at the end of the calculation. 4 days and 18 hours = = 4 3 days = 4.75 days

Find the correct answer
Four people used their calculators to work out 9 + 30 15 – 7 Tracy gets the answer 4. Fiona gets the answer Andrew gets the answer –4.4. Sam gets the answer Conclude by discussing each of these answers and deciding how they were obtained. (9 + 30) ÷ (15 – 7) = 39 ÷ 8. Fiona has the correct answer. Tracy probably keyed in ÷ 15 – 7 without using any bracket keys. Fiona could have keyed in (9 + 30) ÷ (15 –7), 39 ÷ (15 – 7), (9 + 30) ÷ 8 or 39 ÷ 8 to get the correct answer, Andrew could have keyed in (9 + 30) ÷ 15 – 7 or 39 ÷ 15 –7 to get –4.4. Sam must have keyed in either ÷ (15 – 7) or ÷ 8 to get Ask pupils if they think it is possible to find a whole number remainder for this calculation. Establish that 39 divided by 8 equals 4 remainder 7. Ask pupils to use this fact to deduce the fractional equivalent of (7/8). Who is correct? What did the others do wrong?

N10 Written and calculator methods
Contents N10 Written and calculator methods A1 N10.1 Estimation and approximation A1 N10.2 Addition and subtraction A1 N10.3 Multiplication A1 N10.4 Division A1 N10.5 Using a calculator A1 N10.6 Checking results

Making sure answers are sensible
When we complete a calculation, whether using a calculator, a mental method or a written method we should always check that the answer is sensible. Make sure that the sum of two odd numbers is an even number. When you multiply two large numbers together check the last digit. For example, 329 × 842 must end in an 8 because 9 × 2 = 18. Point out that if the problem is given in context then we can also use this to help estimate answers. For example, if we worked out that the average distance that pupils travelled to school was 3.72 m, then we would know that this answer was incorrect because it is too small. Use checks for divisibility when you multiply by 2, 3, 4, 5, 6, 8 and 9. For example, if you multiply a number by 9 the sum of the digits should be a multiple of 9.

Using rounding and approximation
We can check that answers to calculations are of the right order of magnitude by rounding the numbers in the calculation to find an approximate answer. Sam calculates that × 0.45 is Could this be correct? 387.4 × 0.45 is approximately equal to 390 × 0.5 = 195 Emphasize that when multiplying it is usually best to round one number up and the other number down. When dividing it is best to round both numbers up or both numbers down and to choose numbers that are exactly divisible. This approximate answer is a little larger than the calculated answer but since both numbers were rounded up, there is a good chance that the answer is correct.

Using inverse operations
We can use a calculator to check answers using inverse operations. We can check the solution to 34.2 × 45.9 = by calculating ÷ 34.2 If the calculation is correct then the answer will be 45.9.

Using inverse operations
We can use a calculator to check answers using inverse operations. We can check the solution to 4 7 of 224 = 128 by calculating 128 × 7 ÷ 4 If the calculation is correct then the answer will be 224.

Using inverse operations
We can use a calculator to check answers using inverse operations. We can check the solution to 6 ÷ 13 = … by calculating Point out that when working this out on the calculator we should leave the decimal answer on the screen before multiplying by 13. Some calculators may display the answer as because of the error caused by rounding an infinitely recurring decimal. 13 × If the calculation is correct then the answer will be 6.

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