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A single question which asks you to round to the nearest ten, hundred or thousand. 

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$\\var{number}$ rounded to the nearest {seedtext} is [[0]].

\n

\n

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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 $\\var{number}$ to the nearest {seedtext}, we look to the right of the {seedtext}s column and see the digit $\\var{digit_to_the_right}$. 

\n

Since $\\var{digit_to_the_right}$ is {less_greater} $5$ we round {direction} to $\\var{ans}$. 

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

\n

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

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

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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.

\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.

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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.

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

\n

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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]] $\\text{cm}$

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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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