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

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

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

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

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

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.

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.

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}

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

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

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.

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.

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}

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