// Numbas version: finer_feedback_settings {"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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For each prefix listed on the left click the button that corresponds to the correct abbreviation listed above.

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Metric prefixes in everyday use
PrefixSymbolMultiple of the base unitMultiple of the base unit (as a fraction)Multiple of the base unit (as a power)Also known as
teraT$1\\,000\\,000\\,000\\,000$$1\\,000\\,000\\,000\\,000$$10^{12}$trillion
gigaG$1\\,000\\,000\\,000$$1\\,000\\,000\\,000$$10^9$billion
megaM$1\\,000\\,000$$1\\,000\\,000$$10^6$million
kilok$1\\,000$$1\\,000$$10^3$thousand
hectoh$100$$100$$10^2$hundred
decada$10$$10$$10^1$ten
decid$0.1$$\\frac{1}{10}$$10^{-1}$tenth
centic$0.01$$\\frac{1}{100}$$10^{−2}$hundredth
millim$0.001$$\\frac{1}{1\\,000}$$10^{−3}$thousandth
micromc or $\\mu$$0.000\\,001$$\\frac{1}{1\\,000\\,000}$$10^{−6}$millionth
nanon$0.000\\,000\\,001$$\\frac{1}{1\\,000\\,000\\,000}$$10^{−9}$billionth
picop$0.000\\,000\\,000\\,001$$\\frac{1}{1\\,000\\,000\\,000\\,000}$$10^{−12}$trillionth
\n

Of these, mega, kilo, centi and milli are probably the most common in everyday use.

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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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Metric prefixes in everyday use
PrefixSymbolMultiple of the base unitMultiple of the base unit (as a fraction)Multiple of the base unit (as a power)Also known as
teraT$1\\,000\\,000\\,000\\,000$$1\\,000\\,000\\,000\\,000$$10^{12}$trillion
gigaG$1\\,000\\,000\\,000$$1\\,000\\,000\\,000$$10^9$billion
megaM$1\\,000\\,000$$1\\,000\\,000$$10^6$million
kilok$1\\,000$$1\\,000$$10^3$thousand
hectoh$100$$100$$10^2$hundred
decada$10$$10$$10^1$ten
decid$0.1$$\\frac{1}{10}$$10^{-1}$tenth
centic$0.01$$\\frac{1}{100}$$10^{−2}$hundredth
millim$0.001$$\\frac{1}{1\\,000}$$10^{−3}$thousandth
micromc or $\\mu$$0.000\\,001$$\\frac{1}{1\\,000\\,000}$$10^{−6}$millionth
nanon$0.000\\,000\\,001$$\\frac{1}{1\\,000\\,000\\,000}$$10^{−9}$billionth
picop$0.000\\,000\\,000\\,001$$\\frac{1}{1\\,000\\,000\\,000\\,000}$$10^{−12}$trillionth
\n

Of these, mega, kilo, centi and milli are probably the most common in everyday use.

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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.

\n

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

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[1,2,3] etc signifies which dot points are needed

\n

[1000,100,10] are the factors needed for the multiplication in the order that it is explained in (which is the opposite of the dot points)

", "templateType": "anything", "can_override": false}, "capacity": {"name": "capacity", "group": "Ungrouped variables", "definition": "random(\n [\"millilitres\",\"kilolitre\",1000000,[1,2],[1000,1000]],\n [\"millilitres\",\"kilolitre\",1000000,[1,2],[1000,1000]],\n [\"millilitres\",\"megalitre\",1000000000,[1,2,3],[1000,1000,1000]]\n)", "description": "", "templateType": "anything", "can_override": false}, "mass": {"name": "mass", "group": "Ungrouped variables", "definition": "random(\n [\"milligrams\",\"tonne\",1000000000,[1,2,3],[1000,1000,1000]],\n [\"grams\",\"tonne\",1000000,[2,3],[1000,1000]],\n [\"milligrams\",\"kilogram\",1000000,[1,2],[1000,1000]],\n [\"grams\",\"tonne\",1000000,[2,3],[1000,1000]],\n [\"milligrams\",\"kilogram\",1000000,[1,2],[1000,1000]],\n)", "description": "", "templateType": "anything", "can_override": false}, "time": {"name": "time", "group": "Ungrouped variables", "definition": "random(\n [\"milliseconds\",\"day\",24*3600000,[1,2,3,4],[24,60,60,1000]],\n [\"milliseconds\",\"hour\",3600000,[1,2,3],[60,60,1000]],\n [\"minutes\",\"day\",1440,[3,4],[24,60]],\n [\"seconds\",\"day\",86400,[2,3,4],[24,60,60]],\n [\"milliseconds\",\"hour\",3600000,[1,2,3],[60,60,1000]],\n [\"minutes\",\"day\",1440,[3,4],[24,60]],\n [\"seconds\",\"day\",86400,[2,3,4],[24,60,60]]\n)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["distance", "time", "mass", "capacity"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": "/* This is to get right parenthesis in the list so it doesn't look like a decimal and I can still refer to the facts by number */\nol {\n counter-reset: list;\n}\nol > li {\n list-style: none;\n}\nol > li:before {\n content: counter(list) \") \";\n counter-increment: list;\n}"}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

There are 

\n

[[0]] millimetres in a centimetre,

\n

[[1]] centimetres in a metre, and 

\n

[[2]] metres in a kilometre.

\n

From some or all of this, we can calculate that there are [[3]] {distance[0]} in a {distance[1]}.

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There are 

\n
    \n
  1. $10$ millimetres in $1$ centimetre,
  2. \n
  3. $100$ centimetres in $1$ metre, and 
  4. \n
  5. $1000$ metres in $1$ kilometre.
  6. \n
\n

Since we want to know the number of {distance[0]} in a {distance[1]}, we can use the facts labelled {join(distance[3],\", \")} above.

\n

In each kilometre, there are $1000$ metres, and in each metre there are $100$ centimetres, and in each centimetre there are $10$ millimetres, therefore there are $\\var{latex(join(distance[4],'\\\\times'))}=\\var{distance[2]}$ {distance[0]} in a {distance[1]}.

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There are 

\n

[[0]] milliseconds in a second, 

\n

[[1]] seconds in a minute, 

\n

[[2]] minutes in an hour, and 

\n

[[3]] hours in a day.

\n

From some or all of this, we can calculate that there are [[4]] {time[0]} in a {time[1]}.

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There are 

\n
    \n
  1. $1000$ milliseconds in a second, 
  2. \n
  3. $60$ seconds in a minute, 
  4. \n
  5. $60$ minutes in an hour, and 
  6. \n
  7. $24$ hours in a day.
  8. \n
\n

Since we want to know the number of {time[0]} in a {time[1]}, we can use the facts labelled {join(time[3],\", \")} above.

\n

In each day, there are $24$ hours, and in each hour, there are $60$ minutes, and in each minute there are $60$ seconds, and in each second there are $1000$ milliseconds, therefore there are $\\var{latex(join(time[4],'\\\\times'))}=\\var{time[2]}$ {time[0]} in a {time[1]}.

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There are 

\n

[[0]] milligrams in a gram,

\n

[[1]] grams in a kilogram, and

\n

[[2]] kilograms in a tonne. 

\n

From some or all of this, we can calculate that there are [[3]] {mass[0]} in a {mass[1]}.

\n

", "stepsPenalty": "2.5", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

There are 

\n
    \n
  1. $1000$ milligrams in $1$ gram,
  2. \n
  3. $1000$ grams in $1$ kilogram, and 
  4. \n
  5. $1000$ kilograms in $1$ tonne.
  6. \n
\n

Since we want to know the number of {mass[0]} in a {mass[1]}, we can use the facts labelled {join(mass[3],\", \")} above.

\n

In each tonne, there are $1000$ kilograms, and in each kilogram there are $1000$ grams, and in each gram there are $1000$ milligrams, therefore there are $\\var{latex(join(mass[4],'\\\\times'))}=\\var{mass[2]}$ {mass[0]} in a {mass[1]}.

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There are

\n

[[0]] millilitres in a litre,

\n

[[1]] litres in a kilolitre, and

\n

[[2]] litres in a megalitre.

\n

From some or all of this, we can calculate that there are [[3]] {capacity[0]} in a {capacity[1]}.

", "stepsPenalty": "2.5", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

There are

\n
    \n
  1. $1000$ millilitres in a litre,
  2. \n
  3. $1000$ litres in a kilolitre, and
  4. \n
  5. $1000$ kilolitres in a megalitre.
  6. \n
\n

Since we want to know the number of {capacity[0]} in a {capacity[1]}, we can use the facts labelled {join(capacity[3],\", \")} above.

\n

In each megalitre, there are $1000$ kilolitres, and in each kilolitre there are $1000$ litres, and in each litre there are $1000$ millilitres, therefore there are $\\var{latex(join(capacity[4],'\\\\times'))}=\\var{capacity[2]}$ {capacity[0]} in a {capacity[1]}.

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$\\var{cm1}$ cm $=$[[0]]m

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There are $100$ cm in $1$ metre.

\n

So if we have a number of centimetres, and want to know how many metres we have, we need to determine how many hundreds of centimetres we have. To determine the number of hundreds, we divide by $100$. Recall that dividing by $100$ means we are making the number one hundred times smaller by moving the decimal place twice. 

\n

That is, $\\var{cm1}\\div 100 = \\var{m1}$, and so we have $\\var{m1}$ m.

\n

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$\\var{m2}$ m $=$[[0]]cm

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$1$ metre is $100$ cm.

\n

So if we have a number of metres, and want to know how many centimetres we have, we need to multiply the number of metres we have by one hundred. Recall that multiplying by $100$ means we are making the number one hundred times larger by moving the decimal place twice. 

\n

That is, $\\var{m2}\\times 100 = \\var{cm2}$, and so we have $\\var{cm2}$ cm.

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

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

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.

", "advice": "

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 

", "templateType": "anything", "can_override": false}, "bpowerdiff": {"name": "bpowerdiff", "group": "Ungrouped variables", "definition": "P2[3]-P3[3]", "description": "", "templateType": "anything", "can_override": false}, "P0": {"name": "P0", "group": "Ungrouped variables", "definition": "prefix[RP1[0]]", "description": "", "templateType": "anything", "can_override": false}, "thousandsb": {"name": "thousandsb", "group": "Ungrouped variables", "definition": "(P2[3]-P3[3])/3", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["prefix", "units", "mult", "random_order", "RP1", "RP2", "P0", "P1", "P2", "P3", "ansa", "apowerdiff", "ansb", "bpowerdiff", "thousandsa", "thousandsb"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

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

\n

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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 $10^3=1000$. 

\n

giga
mega
kilo
base unit
milli
micro
nano

\n

\n

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

\n

$(1000)^\\var{thousandsa}=$$10^\\var{apowerdiff}=\\simplify{10^{apowerdiff}}$.

\n

\n

When converting from {P0[2]+units[0][1]} to {P1[2]+units[0][1]} the units are getting larger and so 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

\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

If this doesn't make sense to you consider a simpler example:

\n

$5$ one-dollar coins is the same amount of money as $1$ five-dollar note.

\n

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 $5$.

\n

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. 

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

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

\n

giga
mega
kilo
base unit
milli
micro
nano

\n

\n

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

\n

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

\n

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.

\n

\n

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}

\n

\n

If this doesn't make sense to you consider a simpler example:

\n

$5$ one-dollar coins is the same amount of money as $1$ five-dollar note.

\n

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 $5$.

\n

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. 

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