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When we are rounding we look at the first digit that we might discard. If it is $5$ or greater we round up. If it is less than $5$ we round down.

\n

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a) To round $\\var{number1}$ to the nearest ten, we look at the units column and see $\\var{lastdigit1}$.

\n

Since $\\var{lastdigit1}$ is less greater than $5$ we round down up to $\\var{ans1}$.

\n

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b) To round $\\var{number2}$ to the nearest hundred, we look at the tens column and see $\\var{tens2}$.

\n

Since $\\var{tens2}$ is less greater than $5$ we round down up to $\\var{ans2}$.

\n

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c) To round $\\var{number3}$ to the nearest thousand, we look at the hundreds column and see $\\var{hundreds3}$.

\n

Since $\\var{hundreds3}$ is less greater than $5$ we round down up to $\\var{ans3}$.

Round random numbers to the nearest ten, hundred and thousand.

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If you are unsure of how to do a question, click on Show steps to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again.

\n

Round the following numbers to the indicated precision.

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$\\var{number1}$ to the nearest ten is [[0]].

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When we are rounding we look at the first digit that we might discard. If it is $5$ or greater we round up. If it is less than $5$ we round down.

\n

\n

To round $\\var{number1}$ to the nearest ten, we look at the units column and see $\\var{lastdigit1}$.

", "showCorrectAnswer": true, "unitTests": []}]}, {"scripts": {}, "unitTests": [], "type": "gapfill", "sortAnswers": false, "variableReplacements": [], "prompt": "

$\\var{number2}$ to the nearest hundred is [[0]].

", "marks": 0, "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "gaps": [{"scripts": {}, "allowFractions": false, "unitTests": [], "type": "numberentry", "showCorrectAnswer": true, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "minValue": "{ans2}", "correctAnswerStyle": "plain", "marks": 1, "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "showFeedbackIcon": true, "maxValue": "{ans2}", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "correctAnswerFraction": false}], "stepsPenalty": "1", "showFeedbackIcon": true, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "steps": [{"scripts": {}, "marks": 0, "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "type": "information", "showFeedbackIcon": true, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "prompt": "

When we are rounding we look at the first digit that we might discard. If it is $5$ or greater we round up. If it is less than $5$ we round down.

\n

\n

To round $\\var{number2}$ to the nearest hundred, we look at the tens column and see $\\var{tens2}$.

\n

Since $\\var{tens2}$ is less greater than $5$ we round down up to $\\var{ans2}$.

\n

", "showCorrectAnswer": true, "unitTests": []}]}, {"scripts": {}, "unitTests": [], "type": "gapfill", "sortAnswers": false, "variableReplacements": [], "prompt": "

$\\var{number3}$ to the nearest thousand is [[0]].

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When we are rounding we look at the first digit that we might discard. If it is $5$ or greater we round up. If it is less than $5$ we round down.

\n

\n

To round $\\var{number3}$ to the nearest thousand, we look at the hundreds column and see $\\var{hundreds3}$.

\n

Since $\\var{hundreds3}$ is less greater than $5$ we round down up to $\\var{ans3}$.

", "showCorrectAnswer": true, "unitTests": []}]}], "rulesets": {}, "type": "question"}, {"name": "Times tables (multiplication)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

This creates a 4x4 times table grid using numbers from 2 to 9. The step gives some advice on how to work out times tables (products, multiplications) if you can't remember them.

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\large\\times$ $\\large\\var{c[0]}$ $\\large\\var{c[1]}$ $\\large\\var{c[2]}$ $\\large\\var{c[3]}$ $\\large\\var{r[0]}$ [[0]] [[1]] [[2]] [[3]] $\\large\\var{r[1]}$ [[4]] [[5]] [[6]] [[7]] $\\large\\var{r[2]}$ [[8]] [[9]] [[10]] [[11]] $\\large\\var{r[3]}$ [[12]] [[13]] [[14]] [[15]]

Memorising as many of the times tables as possible is really helpful for lots of things in mathematics.

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If you can't memorise them, for some of the times tables there are little tricks you can use to work them out (sometimes quickly, sometimes slowly).

\n
\n
• Recall that the order of multiplication is irrelevant. For example, $8\\times 9$ is the same as $9\\times 8$ so sometimes thinking about the multiplication in the reverse order makes it easier to recall or work out.
• \n
• If you don't know a times table but you know one 'near' it, you can use that to work the unknown one out! For example, say you don't know $6\\times 7$, but you do know $6\\times 6=36$, then $6\\times 7$ is just one more lot of $6$, and so $6 \\times 7=36+6=42$.
• \n
• As a last resort you can always just use repeated addition. For example, for $8\\times 3$ you could just work out $8+8+8=16+8=24$, note this is less work than $8$ lots of $3$.
• \n
• Multiply by $\\bf 2$. Most people are comfortable with doubling a number. If not, just add the number to itself.
• \n
• Multiplying by $\\bf 3$? You could just double it and add on another lot. For example, for $3\\times 7$ you could double $7$ then add another $7$ on, $3\\times 7=14+7=21$.
• \n
• Multiplying by $\\bf 4$? You could just double it and then double that. For example, for $4\\times 8$ you could double $8$ and get $16$ and then double $16$ to get $32$.
• \n
• Multiplying by $\\bf 5$? You could just multiply by $10$ (that is add on a zero) and then halve that. For example, for $5\\times 6$ you multiply $6$ by $10$ to get $60$ and then you halve $60$ to get $30$.
• \n
• Multiplying by $\\bf 6$? You could multiply by $5$ and then add one more lot on. For example, for $6\\times 8$ you could work out that $5\\times 8=40$ and then add on another $8$ to get $48$.
• \n
• Multiplying by $\\bf 7$? If you couldn't work it out another way, then you could multiply the number by $5$ and then add on the number doubled. For example, $7\\times 8=5\\times 8+2\\times 8=40+16=56$.
• \n
• Multiplying by $\\bf 8$? Here are three not-so-great ways (you can normally find a better way using the above tricks):\n
\n
• Double, then double, then double. For example, for $8\\times 7$, double $7$ to get $14$, double that to get $28$, then double that to get $56$.
• \n
• Multiply the number by $5$ and $3$ separately and then add them up. For example, for $8\\times 7$ you could do $5\\times 7=35$ and $3\\times 7=21$ and so $8\\times 7=35+21=56$.
• \n
• Multiply the number by $10$ and then subtract the number doubled. For example, $8\\times 7$ will be $10\\times 7-2\\times 7=70-14=56$.
• \n
\n
• \n
• Multiplying by $\\bf 9$? Here are two ways. \n
\n
• You could multiply by $10$ (that is add on a zero) and then subtract one lot. For example, $9\\times 8=10\\times 8-8=80-8=72$.
• \n
• If you are multiplying $9$ by a number from $2$ to $10$ then take the number you are multiply by and subtract one from it, this is the start of your answer, then tack on whatever number makes the digits add up to $9$. For example, the answer to $9\\times 7$ starts the digit $6$ (one less than seven) and since $6+3=9$ the answer must be $9\\times 7=63$.
• \n
\n
• \n
• Multiply by $\\bf 10$? Just add a zero to the end of the number. For example, $10\\times 7=70$.
• \n
\n

If you are stuck on a times table, just for practice see if you can work it out in more than one way.

", "unitTests": [], "scripts": {}}]}], "statement": "

Complete this random selection of times tables. If you cannot recall the answer, try to work it out!

\n

For more practice, click 'Try another question like this one'.

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Write the following questions down on paper and evaluate them without using a calculator.

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If you are unsure of how to do a question, click on Show steps to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again.

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random pair 1 - designed to be increasing

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random pair 2 - designed to be decreasing

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Write the following questions down on paper and evaluate them without using a calculator.

\n

If you are unsure of how to do a question, click on Show steps to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again.

", "parts": [{"variableReplacements": [], "prompt": "

$\\var{mult[0]}\\, \\text{cm} =$ [[0]] $\\text{m}$

\n

", "stepsPenalty": "1", "showCorrectAnswer": true, "steps": [{"variableReplacements": [], "prompt": "

The scaling factor between {P0[2]} and {P1[2]} is  $10^\\var{apowerdiff}=\\simplify{10^{apowerdiff}}$.

\n

When converting from {P0[2]+units[0][1]} to {P1[2]+units[0][1]} the units are getting larger and so the the number will have to get smaller. That is, we will divide by $\\simplify{10^{apowerdiff}}$ to convert from {P0[2]+units[0][1]} to {P1[2]+units[0][1]}. Recall dividing by $\\simplify{10^{apowerdiff}}$ is simply moving the decimal place to the left $\\var{apowerdiff}$ places in order to make the number smaller.

\n

If this doesn't make sense to you consider a simpler example. $5$ one-dollar coins is the same amount of money as $1$ five-dollar note. As the units get bigger you need less of them to have the same amount of money. In this example the scaling factor would be $5$ since the size of the units increases by a factor of $5$ and the number of units decreases by a factor of $5$. Now, say we had $20$ one-dollar coins and we wanted to convert this to five-dollar notes, we would calculate $20\\div 5$ which equals $4$ and therefore we would get $4$ five-dollar notes.

\n

Therefore, \\begin{align}\\var{mult[0]} \\,\\var{P0[0]}\\var{units[0][0]}&=\\left(\\var{mult[0]} \\div\\simplify{10^{P1[3]-P0[3]}} \\right)\\,\\var{P1[0]}\\var{units[0][0]}\\\\&=\\var{ansa}\\,\\var{P1[0]}\\var{units[0][0]}.\\end{align}

\n

\n

\n

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$\\var{mult[1]} \\,\\text{m} =$ [[0]] $\\text{cm}$

", "stepsPenalty": "1", "showCorrectAnswer": true, "steps": [{"variableReplacements": [], "prompt": "

The scaling factor between {P2[2]} and {P3[2]} is  $10^\\var{bpowerdiff}=\\simplify{10^{bpowerdiff}}$.

\n

When converting from {P2[2]+units[0][1]} to {P3[2]+units[0][1]} the units are getting smaller and so the the number will have to get bigger. That is, we will multiply by $\\simplify{10^{bpowerdiff}}$ to convert from {P2[2]+units[0][1]} to {P3[2]+units[0][1]}. Recall multiplying by $\\simplify{10^{bpowerdiff}}$ is simply moving the decimal place to the right $\\var{bpowerdiff}$ places in order to make the number larger.

\n

If this doesn't make sense to you consider a simpler example. $1$ five-dollar note is the same amount of money as $5$ one-dollar coins. As the units get smaller you need more of them to have the same amount of money. In this example the scaling factor would be $5$ since the size of the units decreases by a factor of $5$ and the number of units increases by a factor of $5$. Now, say we had $4$ five-dollar notes and we wanted to convert this to one-dollar coins, we would calculate $4\\times 5$ which equals $20$ and therefore we would get $20$ one-dollar coins.

\n

Therefore, \\begin{align}\\var{mult[1]} \\,\\var{P2[0]}\\var{units[0][0]}&=\\left(\\var{mult[1]} \\times\\simplify{10^{bpowerdiff}} \\right)\\,\\var{P3[0]}\\var{units[0][0]}\\\\&=\\var{ansb}\\,\\var{P3[0]}\\var{units[0][0]}.\\end{align}

\n

\n

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\n

Converting between metres (m) and centimetres (cm).

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