// Numbas version: exam_results_page_options {"name": "Units 1D", "metadata": {"description": "

Most of the standard prefixes (not hecto, deca though), abbreviations, powers of 10, converting between prefixes for distance, capacity, mass.

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Recall of common units, along with understanding multiplication.

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In the second sentence for each part you can enter multiplication using an asterisk, for example $60\\times 12$ can be written as 60*12.

Avoid using commas or spaces between zeroes, for example ten thousand would be entered 10000.

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There are [[0]] millimetres in a centimetre, [[1]] centimetres in a metre and [[2]] metres in a kilometre.

From this we can calculate that there are [[3]] {distance[0]} in a {distance[1]}.

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Suppose you knew there were 100 centimetres in a metre and 1000 metres in a kilometre.

To determine how many centimetres are in a kilometre think \"there are 1000 metres in a kilometre, and each metre has 100 centimetres in it, therefore there are $1000\\times 100$ centimetres in a kilometre\".

You would normally write this as 100 000 but here you need to remove the space or write it as a multiplication, for example, 1000*100.

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There are [[0]] milliseconds in a second, [[1]] seconds in a minute, [[2]] minutes in an hour, and [[3]] hours in a day.

From this we can calculate that there are [[4]] {time[0]} in a {time[1]}.

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Suppose you knew there were 1000 milliseconds in a second and 60 seconds in a minute.

To determine how many milliseconds are in a minute think \"there are 60 seconds in a minute, and each second has 1000 milliseconds in it, therefore there are $60\\times 1000$ milliseconds in a minute\".

You would normally write this as 60 000 but here you need to remove the space or write it as a multiplication, for example, 60*1000.

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There are [[0]] milligrams in a gram, [[1]] grams in a kilogram, [[2]] kilograms in a tonne.

From this we can calculate that there are [[3]] {mass[0]} in a {mass[1]}.

Suppose you knew there were 1000 milligrams in a gram and 1000 grams in a kilogram.

To determine how many milligrams are in a kilogram think \"there are 1000 grams in a kilogram, and each gram has 1000 milligrams in it, therefore there are $1000\\times 1000$ milligrams in a kilogram\".

You would normally write this as 1 000 000 but here you need to remove the space or write it as a multiplication, for example, 1000*1000.

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There are [[0]] millilitres in a litre, [[1]] litres in a kilolitre and [[2]] litres in a megalitre.

From this we can calculate that there are [[3]] {capacity[0]} in a {capacity[1]}.

Suppose you knew there were 1000 millilitres in a litre and 1 000 000 litres in a megalitre.

To determine how many millilitres are in a megalitre think \"there are 1 000 000 litres in a megalitre, and each litre has 1000 millilitres in it, therefore there are $1\\,000\\,000\\times 1000 $ millilitres in a megalitre\".

You would normally write this as 1 000 000 000 but here you need to remove the space or write it as a multiplication, for example, 1000000*1000.

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Metric prefixes in everyday use
Text	Symbol	Multiple of the base unit	Multiple of the base unit (as a power)
tera	T	1000000000000	10¹²
giga	G	1000000000	10⁹
mega	M	1000000	10⁶
kilo	k	1000	10³
hecto	h	100	10²
deca	da	10	10¹
deci	d	0.1	10⁻¹
centi	c	0.01	10⁻²
milli	m	0.001	10⁻³
micro	mc or $\\mu$	0.000001	10⁻⁶
nano	n	0.000000001	10⁻⁹
pico	p	0.000000000001	10⁻¹²

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For each prefix listed on the left click the button that corresponds to the correct abbreviation listed above.

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For each prefix listed on the left click the button that corresponds to how many of the base unit it represents.

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Converting between giga, mega, kilo, base, milli and micro, nano. Metres, grams and litres.

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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.

", "advice": "

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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$\\var{mult[0]}\\, \\var{P0[0]}\\var{units[0][0]} =$ [[0]] $\\var{P1[0]}\\var{units[0][0]}$

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The following are some of the common SI prefixes used listed in descending order. The scaling factor between each adjacent prefix is $10^3=1000$.

giga
mega
kilo
base unit
milli
micro
nano

The scaling factor between {P0[2]} and {P1[2]} is therefore

$(1000)^\\var{thousandsa}=$$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.

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}

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.

$\\var{mult[1]}\\, \\var{P2[0]}\\var{units[1][0]} =$ [[0]] $\\var{P3[0]}\\var{units[1][0]}$

The following are some of the common SI prefixes used list in decending order. The scaling factor between each adjacent prefix is $10^3=1000$.

giga
mega
kilo
base unit
milli
micro
nano

The scaling factor between {P2[2]} and {P3[2]} is therefore

$(10^3)^\\var{thousandsb}=$$10^\\var{bpowerdiff}=\\simplify{10^{bpowerdiff}}$.

When converting from {P2[2]+units[1][1]} to {P3[2]+units[1][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[1][1]} to {P3[2]+units[1][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.

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

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.