// Numbas version: exam_results_page_options {"name": "Skills Audit for Maths and Stats - Biomedical Science (BIS114)", "metadata": {"description": "

Skills Audit for Maths and Stats for module BIS114 students.

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Whole number division in a context of number of tablets per day based on a tablet size and a daily prescribed amount.

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

You prescribe a patient {p} milligrams (mg) per day of a particular drug. The drug is available in tablets of {t} milligrams (mg). How many tablets should you instruct the patient to take each day?

", "advice": "\n\n\n\n\n\n

Since each tablet is {t} mg of the necessary drug, we must prescribe enough tablets to add up to {p} milligrams. By dividing {p} by {t} we can find out how many {t} mg tablet make up {p} mg, which gives {answer}.

Use this link to find some resources which will help you revise this topic.

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The daily number of tablets prescribed should be: [[0]]

Division resulting in decimals in the context of a dose per hour.

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

You are required to give an infusion of a drug of {d}ml over {t}hours, what is the rate per hour?

(answer to 1 decimal place)

", "advice": "\n\n\n\n\n\n

If there is {d}ml in {t} hours, then {d} should be divided by {t} to get {answert}ml per hour.

Make use of the following resource(s) to revise if necessary:

Use this link to find some resources which will help you revise this topic

If you were unsure how to round your answer then look at the following resource(s):

Use this link to find some resources which will help you revise this topic

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[[0]] ml per hour

Simple unit conversion with metric units.

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

Express {liquid} litres ($l$) in millilitres ($ml$).

", "advice": "

There are $1000ml$ in $1l$. To work out the conversion: $\\var{liquid}*1000 = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

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[[0]]$ml$

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Using prefixes (milli) in this case.

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

Express {x} milligrams ($mg$) in grams ($g$). Give your answer to 3 decimal places.

", "advice": "

There are $1000mg$ in $1g$. To work out the conversion: $\\frac{\\var{x}}{1000} = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

[[0]]$g$

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Using prefixes - milli and micro.

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

Express {x} milligrams ($mg$) in micrograms ($\\mu g$).

", "advice": "

There are $1000\\mu g$ in $1mg$. To work out the conversion: $\\var{x}*1000 = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

[[0]]$\\mu g$

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Multiples, factors, lowest common multiples and highest common factors.

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

Factors multiply together in pairs to give the original number (e.g., the factors of 10 are 1, 2, 5 and 10, because 1x10=10 and 2x5=10.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"ca2": {"name": "ca2", "group": "Ungrouped variables", "definition": "random(answers except ca1)", "description": "", "templateType": "anything", "can_override": false}, "ca1": {"name": "ca1", "group": "Ungrouped variables", "definition": "random(answers)", "description": "", "templateType": "anything", "can_override": false}, "hcf": {"name": "hcf", "group": "Ungrouped variables", "definition": "gcd(k,l)", "description": "", "templateType": "anything", "can_override": false}, "lcm4": {"name": "lcm4", "group": "Ungrouped variables", "definition": "lcm(p,q)", "description": "", "templateType": "anything", "can_override": false}, "lcm2": {"name": "lcm2", "group": "Ungrouped variables", "definition": "lcm(n,m)", "description": "", "templateType": "anything", "can_override": false}, "product": {"name": "product", "group": "Ungrouped variables", "definition": "a*b*c", "description": "", "templateType": "anything", "can_override": false}, "answers": {"name": "answers", "group": "Ungrouped variables", "definition": "[a,b,c,a*b,a*c,b*c,1,product]", "description": "", "templateType": "anything", "can_override": false}, "hcf3": {"name": "hcf3", "group": "Ungrouped variables", "definition": "gcd(p,q)", "description": "", "templateType": "anything", "can_override": false}, "hcf2": {"name": "hcf2", "group": "Ungrouped variables", "definition": "gcd(n,m)+1", "description": "", "templateType": "anything", "can_override": false}, "lcm": {"name": "lcm", "group": "Ungrouped variables", "definition": "lcm(k,l)", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(1,2,3)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(5,7)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(2,3)", "description": "", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "random(1..12)", "description": "", "templateType": "anything", "can_override": false}, "g": {"name": "g", "group": "Ungrouped variables", "definition": "random(2..10 except b)", "description": "", "templateType": "anything", "can_override": false}, "f": {"name": "f", "group": "Ungrouped variables", "definition": "random(2..10)", "description": "", "templateType": "anything", "can_override": false}, "h": {"name": "h", "group": "Ungrouped variables", "definition": "random(1..100 except [d*1,d*2,d*3,d*4,d*5,d*6,d*7,d*8,d*9,d*10,f*1,f*2,f*3,f*4,f*5,f*6,f*7,f*8,f*9,f*10,g*1,g*2,g*3,g*4,g*5,g*6,g*7,g*8,g*9,g*10])", "description": "", "templateType": "anything", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "random(4..12)", "description": "", "templateType": "anything", "can_override": false}, "j": {"name": "j", "group": "Ungrouped variables", "definition": "random(1..50 except [d*1,d*2,d*3,d*4,d*5,d*6,d*7,d*8,d*9,d*10,f*1,f*2,f*3,f*4,f*5,f*6,f*7,f*8,f*9,f*10,g*1,g*2,g*3,g*4,g*5,g*6,g*7,g*8,g*9,g*10,h])", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "random(2..15 except [k,l,lcm])", "description": "", "templateType": "anything", "can_override": false}, "l": {"name": "l", "group": "Ungrouped variables", "definition": "random(4..15 except k)", "description": "", "templateType": "anything", "can_override": false}, "n": {"name": "n", "group": "Ungrouped variables", "definition": "random(6..15 except [k,l,lcm,m])", "description": "", "templateType": "anything", "can_override": false}, "q": {"name": "q", "group": "Ungrouped variables", "definition": "random(4..36 except p)", "description": "", "templateType": "anything", "can_override": false}, "p": {"name": "p", "group": "Ungrouped variables", "definition": "random(6..24)", "description": "", "templateType": "anything", "can_override": false}, "wa2": {"name": "wa2", "group": "Ungrouped variables", "definition": "random(1..product except [a,b,c,a*b,a*c,b*c,1,product,wa1])", "description": "", "templateType": "anything", "can_override": false}, "wa1": {"name": "wa1", "group": "Ungrouped variables", "definition": "random(1..product except answers)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["hcf2", "lcm2", "a", "b", "c", "product", "answers", "wa1", "wa2", "ca1", "ca2", "d", "f", "g", "h", "j", "k", "l", "lcm", "m", "n", "hcf", "p", "q", "hcf3", "lcm4"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "m_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, 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Choose all of the factors of $\\var{product}$.

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$\\var{ca1}$

", "

$\\var{ca2}$

", "

$\\var{wa1}$

", "

$\\var{wa2}$

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Questions testing understanding of the precedence of operators using BIDMAS, applied to integers. These questions only test DMAS. That is, only Division/Multiplcation and Addition/Subtraction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following expression:

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

And is a way for us to remember guidance about the order in which calculations are carried out to ensure that everyone doing teh same sum gets the same answer. In this case the first thing that is in the question is Multiplication.

First work through the expression from left to right, evaluating any multiplication as you come to them. You should be left with an expression involving only pluses and minuses. Evaluate this expression, again working from left to right. Thus:

\\[\\var{a}-\\var{b} \\times \\var{c}\\]

\\[=\\var{a}-\\var{b*c}\\]

\\[=\\var{a-b*c}\\]

Use this link to find some resources which will help you revise this topic.

Calculate

$\\var{a}-\\var{b} \\times\\var{c}$

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Questions testing understanding of the precedence of operators using BIDMAS, applied to integers. These questions only test DMAS. That is, only Division/Multiplcation and Addition/Subtraction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following expression:

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

First work through the expression from left to right, evaluating any division as you come to it. You should be left with an expression involving only pluses and minuses. Evaluate this expression, again working from left to right. Thus:

\\[\\var{h}-\\var{a2*b2} \\div \\var{b2}\\]

\\[=\\var{h}-\\var{a2}\\]

\\[=\\var{h-a2}\\]

Use this link to find some resources which will help you revise this topic.

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$\\var{h}-\\var{a2*b2} \\div \\var{b2}$

", "minValue": "{h-a2}", "maxValue": "{h-a2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NC3 BIDMAS with a division 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Applying the order of operators.

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

To calculate the following expression you press a sequence of buttons on your calculator.

\\begin{align}\\frac{\\var{num}}{\\var{a}\\times\\var{b}}\\end{align}

Which of the following would give the WRONG answer?

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

This is the standardized order of operations that we carry out and is part of how the calculator is designed to work. The most effective way to use most modern calculators is to use either the fraction button (on scientific calculators) or as is hinted at in this question, use brackets.

Use this link to find some resources which will help you revise this topic.

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Round numbers to a given number of decimal places.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

We can approximate numbers by rounding.

Round $\\var{c1}$ to a given number of decimal places.

", "advice": "

The first thing to do when we are rounding numbers is to identify the last digit we are keeping.

When you're asked to round your answer to a number of decimal places, you need to decide whether to keep the last digit the same (rounding down) or increase it by 1 (rounding up). If the following digit is less than 5 we round down and we round up when the next digit is 5 or more.

To write it down in steps:

Identify the last digit we need to keep.
Look at the following digit.
If it's 5 or more, increase the previous digit by one.
If it's 4 or less, keep the previous digit the same.
Fill any spaces to the right of the digit with zeros if needed.

It is important to keep in mind that if the digit we are increasing is 9, it becomes zero and we increase the previous digit instead. If this digit is 9 as well, we move along to the left side until we find a digit less than 9.

To round a number to a given number $n$ of decimal places, we look at the $n$th digit after the decimal point.

We have $\\var{c1}$.

We look at the first digit after the decimal point. This is $\\var{cdig[4]}$ and the following digit is $\\var{cdig[3]}$ so we round updown to get $\\var{precround(c1, 1)}$.

ii)

The second digit after the decimal point is $\\var{cdig[3]}$. It is followed by $\\var{cdig[2]}$ so we round updown to get $\\var{precround(c1, 2)}$.

iii)

The 3rd decimal place is $\\var{cdig[2]}$, followed by $\\var{cdig[1]}$. We get $\\var{precround(c1, 3)}$. The 4th decimal place is $\\var{cdig[1]}$, followed by $\\var{cdig[0]}$. We get $\\var{precround(c1, 4)}$.

Use this link to find some resources which will help you revise this topic

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"c1": {"name": "c1", "group": "Ungrouped variables", "definition": "n_from_digits(cdig)*10^(-5) + random(1..5)", "description": "

Random number with 5 decimal places.

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Number of decimal places to round.

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i) $\\var{c1}$ rounded to 1 decimal place is: [[0]]

ii) $\\var{c1}$ rounded to 2 decimal places is: [[1]]

iii) $\\var{c1}$ rounded to {dp} decimal places is: [[2]]

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Round numbers to a given number of significant figures.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

The first thing to do when we are rounding numbers is to identify the last digit we are keeping.

When you're asked to round your answer to a number of significant figures, you need to decide whether to keep the last digit same (rounding down) or increase it by 1 (rounding up). If the following digit is less than 5 we round down and we round up when the next digit is 5 or more.

To write it down in steps:

Identify the last digit we need to keep.
Look at the following digit.
If it's 5 or more, increase the previous digit by one.
If it's 4 or less, keep the previous digit the same.
Fill any spaces to the right of the digit with zeros if needed.

The last digit we need to keep will depend on how many zeros there are. We don't consider leading zeros to be significant,
i.e. 0.03 and 0.3 both have 1 significant figure (but 0.30 has two significant figures, since the second zero isn't a 'leading' zero).

We round $\\var{d1}$ to 1 significant figure. The first non-zero digit is $\\var{ddig[5]}$. The following digit is $\\var{ddig[4]}$ so we round updown to get $\\var{dpformat(siground(d1, 1), 0)}$.

ii)

We round $\\var{d1}$ to {sf} significant figures. The first non-zero digit is $\\var{ddig[5]}$. The second following digit is $\\var{ddig[4]}$, the third following digit is $\\var{ddig[3]}$ and the fourth following digit is $\\var{ddig[2]}$. The digit following the last digit we are keeping is $\\var{ddig[3]}$$\\var{ddig[2]}$$\\var{ddig[1]}$, so we round to get $\\var{sigformat(d1, sf)}$. These are our {sf} significant figures.

Use this link to find some resources which will help you revise this topic.

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Random integer.

", "templateType": "anything", "can_override": false}, "e1": {"name": "e1", "group": "Ungrouped variables", "definition": "n_from_digits(edig)*10^(random(-6,-7,-8))", "description": "

Random number with 7 decimal places.

", "templateType": "anything", "can_override": false}, "ddig": {"name": "ddig", "group": "Ungrouped variables", "definition": "repeat(random(1..9), 6)", "description": "", "templateType": "anything", "can_override": false}, "sf": {"name": "sf", "group": "Ungrouped variables", "definition": "3", "description": "

Number of significant figures to round.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "100"}, "ungrouped_variables": ["sf", "ddig", "edig", "d1", "e1"], "variable_groups": [], "functions": {"n_from_digits": {"parameters": [["digits", "list"]], "type": "number", "language": "jme", "definition": "if(\n len(digits)=0,\n 0,\n digits[0]+10*n_from_digits(digits[1..len(digits)])\n)"}}, "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": "

Round $\\var{d1}$

i) $\\var{d1}$ rounded to 1 significant figure is: [[0]]

ii) $\\var{d1}$ rounded to {sf} significant figures is: [[1]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "precround(siground(d1, 1),0)", "maxValue": "precround(siground(d1, 1),0)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "siground(d1, sf)", "maxValue": "siground(d1, sf)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "ND3 Rounding SF (decimal)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": ["rounding"], "metadata": {"description": "

Round numbers to a given number of significant figures.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

The first thing to do when we are rounding numbers is to identify the last digit we are keeping.

To write it down in steps:

Identify the last digit we need to keep.
Look at the following digit.
If it's 5 or more, increase the previous digit by one.
If it's 4 or less, keep the previous digit the same.
Fill any spaces to the right of the digit with zeros if needed.

We round $\\var{e1}$ to 1 significant figure. The first non-zero digit is $\\var{edig[4]}$, followed by $\\var{edig[3]}$. This is lower than 5 so we round downmore than 5 so we round up to get $\\var{sigformat(e1,1)}$.

ii)

We round $\\var{e1}$ to {sf} significant figures. The first non-zero digit is $\\var{edig[4]}$. The second following digit is $\\var{edig[3]}$, the third following digit is $\\var{edig[2]}$ and the fourth following digit is $\\var{edig[1]}$. The digit following the last digit we are keeping is $\\var{edig[2]}$$\\var{edig[1]}$$\\var{edig[0]}$, so we round to get $\\var{sigformat(e1, sf)}$. These are our {sf} significant figures.

Use this link to find some resources which will help you revise this topic.

Random integer.

", "templateType": "anything", "can_override": false}, "e1": {"name": "e1", "group": "Ungrouped variables", "definition": "n_from_digits(edig)*10^(random(-6,-7,-8))", "description": "

Random number with 7 decimal places.

Number of significant figures to round.

Round $\\var{e1}$

iii) $\\var{e1}$ rounded to 1 significant figure is: [[0]]

iv) $\\var{e1}$ rounded to {sf} significant figures is: [[1]]

State the Upper and lower bound of a distance that has been rounded to either the nearest 10 or 100 miles.

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

The distance between two towns had been rounded to the nearest {x} miles in an aticle in the newspaper. If they reported that the distance was {y} miles, what are the upper and lower bound for the reported number?

", "advice": "

If a number like {y} has been rounded to the nearest {x} then {y} would have been rounded down if it was less than {y+x/2} because {y} is the nearest multiple of {x}.

Similarly {y} would have been rounded up if it was larger than or equal to {y-x/2}. This means the lower bound is {y-x/2} and the upper bound is {y+x/2}.

Use this link to find some resources which will help you revise this topic.

Upper bound:

[[0]]

Lower bound:

[[1]]

Calculate the percentage increase (as a percentage) given a number and the size of the increase.

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

What is the percentage increase in a class of {total} if {additional} more are added to it?

Give your answer to 2 decimal places.

", "advice": "

To calculate a percentage increase you need to find how much the increase is as a percentage of the original number. In this question the increase is {additional} and the original number is {total} so the percentage is

\\[ \\frac{\\var{additional}}{\\var{total}}\\times100\\%=\\var{dpformat(additional/total,4)}\\times 100\\%=\\var{dpformat(percentage,2)}\\%\\]

Use this link to find some resources which will help you revise this topic.

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[[0]]%

Given a student discount, calculate a discounted price.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{pname} is buying a new {item}. The price of the model he picked is £{price}. On a website with discounts for students, he found a voucher for a discount of {percentage}%.

", "advice": "

There are multiple methods to approach this problem. The first method involves working out the discounted price as a percentage of the original, while the second method calculates the value of the discount and subtracts that from the listed price.

Method 1

There is a {percentage}% decrease in price. This means that the new price will be {100-percentage}% of the old price.

\\[\\begin{align} \\frac{\\var{100-percentage}}{100} \\times \\var{price} &= \\var{dpformat((100-percentage)/100*price,4)} \\\\&= \\var{dpformat((100-percentage)/100*price, 2)}\\text{.} \\end{align}\\]

Or, using the multiplier method,

\\[\\begin{align} \\var{(100-percentage)/100} \\times \\var{price} &= \\var{dpformat((100-percentage)/100*price,4)}\\\\&= \\var{dpformat((100-percentage)/100*price, 2)}\\text{.} \\end{align}\\]

When we are talking about money, it is usually assumed that we will round the answer to 2 decimal places.

Method 2

We find the discount first. This is

\\[\\frac{\\var{percentage}}{100} \\times \\var{price} = \\var{dpformat((percentage)/100*price,4)}\\text{.}\\]

Or using a decimal multiplier,

\\[\\var{(percentage)/100} \\times \\var{price} = \\var{dpformat((percentage)/100*price,4)}\\text{.}\\]

Then we subtract the discount from the original price to get the new price:

\\[ \\var{price} - \\var{dpformat(discount,2)} = \\var{dpformat(price - discount, 2)}\\text{.} \\]

Use this link to find some resources which will help you revise this topic.

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Names.

", "templateType": "anything", "can_override": false}, "discountrounded": {"name": "discountrounded", "group": "Ungrouped variables", "definition": "precround(discount,2)", "description": "", "templateType": "anything", "can_override": false}, "price": {"name": "price", "group": "Ungrouped variables", "definition": "switch(\n item = \"TV\", random(170.99..1199.99), \n item = \"laptop\", random(200.99..799.99),\n item = \"smartphone\", random(100.99..799.99),\n item = \"PC\", random(200.99..969.99),\n item = \"gaming console\", random(80.99..349.99),\n random(110.99..649.99))\n", "description": "

Price of an item.

", "templateType": "anything", "can_override": false}, "item": {"name": "item", "group": "Ungrouped variables", "definition": "random(\"TV\", \"laptop\", \"smartphone\", \"PC\", \"gaming console\", \"fridge\")", "description": "

The bought item.

", "templateType": "anything", "can_override": false}, "percentage": {"name": "percentage", "group": "Ungrouped variables", "definition": "random(5..40 #5)", "description": "

Discount percentage.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "1000"}, "ungrouped_variables": ["item", "pname", "price", "percentage", "discount", "discountrounded"], "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": "

What will the discounted price of the {item} be?

Round your answer to the nearest penny.

£ [[0]]

Your answer does not make sense in real life, we cannot divide a penny any further. Shops always round their prices for items. That is why you should have rounded your answer to $\\var{precround((100-percentage)/100*price, 2)}$.

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Compound percentage change.

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

The value of a car is initially {StartingPrice}. If the value decreases by {dec}%, and then increases by {inc}%, what is the final value?

Give your answer correct to two decimal places.

", "advice": "

There is a {dec}% decrease in price. This means that price after the decrease will be {100-dec}% of the old price.

\\[\\frac{\\var{100-dec}}{100} \\times \\var{StartingPrice} = \\var{(100-dec)/100*StartingPrice}\\]

Then there is a {inc}% increase in price. This means the final price will be {100+inc}% of the price after the decrease.

\\[\\frac{\\var{100+inc}}{100} \\times \\var{(100-dec)/100*StartingPrice} = £\\var{dpformat((100+inc)/100*(100-dec)/100*StartingPrice,2)}\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"dec": {"name": "dec", "group": "Ungrouped variables", "definition": "random(1..50)", "description": "", "templateType": "anything", "can_override": false}, "inc": {"name": "inc", "group": "Ungrouped variables", "definition": "random(1..50)", "description": "", "templateType": "anything", "can_override": false}, "FinalPrice": {"name": "FinalPrice", "group": "Ungrouped variables", "definition": "StartingPrice*(1-dec/100)*(1+inc/100)", "description": "", "templateType": "anything", "can_override": false}, "StartingPrice": {"name": "StartingPrice", "group": "Ungrouped variables", "definition": "random(600..8000 # 10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["dec", "inc", "FinalPrice", "StartingPrice"], "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": "

£[[0]]

Find the original price before a discount by dividing the new price by the percentage discount.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{name1} and {name2} are friends. {name1} noticed {name2}'s new {item} when he came over to visit her house. He immediately knew he wanted to buy the same model. When he got home, he bought the {item} online for £{newprice}.

", "advice": "

We need to find the original price paid by {name2}. This value represents 100%.

By the time {name1} bought the {item}, the price had decreased by {percentage}%.

{name1} therefore paid {100-percentage}% of the price {name2} paid.

We use the unitary method to find the original price. We know the price paid by {name1}.

\\[\\var{100-percentage}\\text{%} = \\var{newprice} \\text{.}\\]

Divide both sides by {100-percentage} to get

\\[\\begin{align} 1\\text{%} &= \\var{newprice} \\div \\var{100-percentage} \\\\&= \\var{newprice/(100-percentage)} \\text{.} \\end{align}\\]

Multiply both sides by 100 to get

\\[\\begin{align} 100\\text{%} &= \\var{newprice/(100-percentage)} \\times 100 \\\\&= \\var{newprice/(100-percentage)*100} \\\\&= \\var{oldprice}\\text{.} \\end{align}\\]

This is the original price paid by {name2} before the {percentage}% decrease.

We can check our answer with a different method.

\\[\\begin{align} \\var{100-percentage}\\text{% of } \\var{oldprice} &= \\var{(100-percentage)/100} \\times \\var{oldprice} \\\\&= \\var{(100-percentage)/100*oldprice} \\\\&= \\var{precround((100-percentage)/100*oldprice, 2)} \\text{.} \\end{align}\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"newprice": {"name": "newprice", "group": "Ungrouped variables", "definition": "precround(oldprice*(100-percentage)/100,2)", "description": "", "templateType": "anything", "can_override": false}, "name2": {"name": "name2", "group": "Ungrouped variables", "definition": "random(\"Kaden\",\"Ola\",\"Pat\",\"Skylar\",\"Wren\",\"Zendaya\")", "description": "", "templateType": "anything", "can_override": false}, "name1": {"name": "name1", "group": "Ungrouped variables", "definition": "random(\"Adair\",\"Aya\",\"Bergen\",\"Dua\",\"Fadhili\",\"Harper\")", "description": "", "templateType": "anything", "can_override": false}, "oldprice": {"name": "oldprice", "group": "Ungrouped variables", "definition": "switch(\n item = \"TV\", random(179.99..1199.99 #10), \n item = \"laptop\", random(209.99..799.99 #10),\n item = \"smartphone\", random(109.99..799.99 #10),\n item = \"PC\", random(209.99..969.99 #10),\n item = \"gaming console\", random(89.99..349.99 #10),\n 399.99)", "description": "", "templateType": "anything", "can_override": false}, "percentage": {"name": "percentage", "group": "Ungrouped variables", "definition": "random(5..30)", "description": "

Discount percentage.

The bought item.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "precround(precround(oldprice*(100-percentage)/100,2)*100/(100-percentage),2) = oldprice", "maxRuns": "1000"}, "ungrouped_variables": ["item", "name1", "percentage", "name2", "oldprice", "newprice"], "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": "

When {name1} told {name2} how much he had paid for the {item}, {name2} said the price had decreased by {percentage}% since she bought it.

How much did {name2} pay for the {item}?

£ [[0]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "oldprice", "maxValue": "oldprice", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF5 - Percentage of amount 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Find a percentage of an amount.

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

What is {x}% of {y}?

", "advice": "

Taking {x}% of {y} is calculated by multiplying,

\\[\\frac{\\var{x}}{100}\\times\\var{y}.\\]

For this question we can calculate this by noticing that 10% of {y} is {y*0.1} and then since $\\var{x}\\%=\\var{x/10}\\times10\\%$ we can calculate {x}% of {y} as,
\\[\\var{x/10}\\times10\\%\\times \\var{y}=\\var{x/10}\\times\\var{y*0.1}=\\var{x/100*y}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(10..90 #10)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(10.. 100 # 10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "y*(x/100)", "maxValue": "y*(x/100)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF6 - Percentage of amount 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Calculate one number as percentage of another.

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

To find the percentage of a number we can use the formula:

\\[ \\text{New value } = \\text{Original value } \\times \\text{Percentage in decimal form} \\]

Firstly, to convert a percentage into decimal form we need to divide by $100$:

\\[ \\var{p} \\% = \\var{p/100} \\]

Therefore,

\\[ \\begin{split} \\var{p} \\% \\,\\text{ of } \\var{og} &\\,= \\var{og} \\times \\var{p/100} \\\\ &\\,= \\var{ans} \\end{split} \\]

Use this link to find resources to help you revise how to work out percentages.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"p": {"name": "p", "group": "Ungrouped variables", "definition": "random(1..99)", "description": "", "templateType": "anything", "can_override": false}, "og": {"name": "og", "group": "Ungrouped variables", "definition": "random(101..999)", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "og*p*0.01", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["p", "og", "ans"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find {p}% of {og}

", "minValue": "ans", "maxValue": "ans", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF7 One number as a percentage of another", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Don Shearman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/680/"}, {"name": "Adelle Colbourn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2083/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Given the number of international students enrolled on a course of $n$ students, calculate the percentage of 'home' students.

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

{num_students} of the {class_size} students enrolled on a course are international students. What percentage are 'home' students?

", "advice": "

First work out the number of students who are not international. In this case it is {class_size} - {num_students} = {class_size-num_students} students.

Then write this as a fraction out of {class_size}. $ \\frac{\\var{class_size-num_students}} {\\var{class_size}} $

Then convert this to a percentage. You should put this fraction into your calculator and then multiply by 100:

$ \\frac{\\var{class_size-num_students}} {\\var{class_size}} \\times 100 = \\var{(class_size-num_students)/class_size*100}\\%$

Use this link to find resources to help you revise how to calculate percentages.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"num_students": {"name": "num_students", "group": "Ungrouped variables", "definition": "per*class_size/100\n", "description": "

The number of students in the class who do speak a language other than English.

", "templateType": "anything", "can_override": false}, "class_size": {"name": "class_size", "group": "Ungrouped variables", "definition": "random(80..300)", "description": "", "templateType": "anything", "can_override": false}, "per": {"name": "per", "group": "Ungrouped variables", "definition": "random(5..90 except 50)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "num_students = precround(num_students,0)", "maxRuns": 100}, "ungrouped_variables": ["num_students", "class_size", "per"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "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": "

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["$\\var{class_size-num_students}\\%$", "$\\var{num_students*100/class_size}\\%$", "$\\var{num_students}\\%$", "$\\var{(class_size-num_students)/class_size*100}\\%$"], "matrix": [0, 0, 0, "1"], "distractors": ["Have you converted this to a percentage? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Then convert this to a percentage. Click on Reveal Answer and scroll down for Advice regarding this question.", "Well done!"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NJ3 - Dividing amounts in ratios", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Dividing amounts in ratios

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

The ratio of ethanol to water is {a}:{b} for an experiment. If I have {volWater}ml of water, how much ethanol do I need?

", "advice": "

If there is a ratio of {a}:{b} for ethanol:water then that means for every {b}ml of water we need {a}ml of ethanol.

In our experiment there is {volwater}ml of water so to find the amount of ethanol we divide by {b} and then multiply by {a}.

\\[\\var{volwater}\\text{ml}\\times\\frac{\\var{a}}{\\var{b}}=\\var{volwater*a/b}\\text{ml}\\]

Use this link to find some resources which will help you revise this topic.

[[0]]ml

Convert numbers greater than 1 into standard form/scientific notation.

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

Write the following numbers in scientific notation.

", "advice": "

Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. The decimal point is currently to the right of the last digit in $\\var{q2}$ and needs to be between the first and second digits, i.e $\\var{dec2}$. Count the places that the digits must move and you get $\\var{pow2}$ places. That is,

\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]

We have a positive $\\var{pow2}$ as the power because we need to make the number $\\var{dec2}$ bigger to get to $\\var{q2}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"pow2": {"name": "pow2", "group": "Ungrouped variables", "definition": "random(4..8)", "description": "", "templateType": "anything", "can_override": false}, "q2": {"name": "q2", "group": "Ungrouped variables", "definition": "precround(dec2*10^pow2,0)", "description": "", "templateType": "anything", "can_override": false}, "dec2": {"name": "dec2", "group": "Ungrouped variables", "definition": "random(1.1..9.9#0.001)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["dec2", "pow2", "q2"], "variable_groups": [], "functions": {"spacenumber": {"parameters": [["n", "number"]], "type": "string", "language": "javascript", "definition": "var parts=n.toString().split(\".\");\n if(parts[1] && parts[1].length<2) {\n parts[1]+='0';\n }\n return parts[0].replace(/\\B(?=(\\d{3})+(?!\\d))/g, \" \") + (parts[1] ? \", \" + parts[1] : \"\");"}}, "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": "

$\\var{q2} =$ [[0]]$\\times 10$ ^[[1]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{dec2}", "maxValue": "{dec2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{pow2}", "maxValue": "{pow2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NK2 standard form (small)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": ["converting", "scientific notation", "standard form"], "metadata": {"description": "

Convert numbers between 0 and 1 intro standard form/scientific notation.

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

Write the following numbers in scientific notation.

", "advice": "

Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. Count the places that the digits must move and you get $\\var{-pow2}$ places to the right. That is,

\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]

We have a negative $\\var{-pow2}$ as the power because we need to make the number $\\var{dec2}$ smaller to get to $\\var{q2}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"pow2": {"name": "pow2", "group": "Ungrouped variables", "definition": "random(list(-6..-1))", "description": "", "templateType": "anything", "can_override": false}, "dec2": {"name": "dec2", "group": "Ungrouped variables", "definition": "random(1.1..9.9#0.001)", "description": "", "templateType": "anything", "can_override": false}, "q2": {"name": "q2", "group": "Ungrouped variables", "definition": "precround(dec2*10^pow2,adjpow)", "description": "", "templateType": "anything", "can_override": false}, "adjpow": {"name": "adjpow", "group": "Ungrouped variables", "definition": "If(round(mod(dec2*1000,10))<>0,3-pow2,If(round(mod(dec2*1000,100))<>0,2-pow2,If(round(mod(dec2*1000,1000))<>0,1-pow2,0-pow2)))", "description": "", "templateType": "anything", "can_override": false}, "test": {"name": "test", "group": "Ungrouped variables", "definition": "mod(1000*dec2,10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["dec2", "pow2", "q2", "adjpow", "test"], "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": "

$\\var{q2}$ = [[0]]$\\times 10$ ^[[1]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{dec2}", "maxValue": "{dec2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{pow2}", "maxValue": "{pow2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AA4 Indices - Fractional 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": ["category: Indices"], "metadata": {"description": "

Using indices rules to rewrite an expression from $a^\\frac{m}{n}$ to $b$, for integers $a$, $b$, $m$ and $n$.

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

Evaluate the following expression:

\\[\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}\\]

", "advice": "

To find $\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}$, we want to make use of the following rule:

\\[\\left(a^n\\right)^m = a^{n\\times m}\\]

By rewriting the power $\\frac{\\var{m}}{\\var{n}}$ as a product of $\\var{m} \\times \\frac{1}{\\var{n}}$, we can apply this rule:

\\[ \\begin{split} \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} &\\,= \\var{a^n}^{\\left(\\var{m} \\times \\frac{1}{\\var{n}}\\right)} \\\\ &\\,= \\left(\\var{a^n}^\\frac{1}{\\var{n}}\\right)^\\var{m} \\\\ &\\,= \\var{a}^\\var{m}\\end{split} \\]

Then calculating what is left:

\\[ \\begin{split} \\var{a}^\\var{m} &\\,=\\var{a^(m)} \\end{split} \\]

Therefore,

\\[ \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} =\\var{a^(m)}. \\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"m": {"name": "m", "group": "Ungrouped variables", "definition": "random(2,3)", "description": "", "templateType": "anything", "can_override": false}, "n": {"name": "n", "group": "Ungrouped variables", "definition": "random(2..3 except m)", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(2,3,4)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["m", "n", "a"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{a^m}", "maxValue": "{a^m}", "correctAnswerFraction": false, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AA5 - Indices - negative", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

perform a calculation involving negative indices.

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

Evaluate and simplify the following expression:

\\[\\frac{\\var{x}^\\var{n}}{\\var{y}^\\var{m}}\\]

", "advice": "

To simplify this expression we use the rule $a^{-n}=\\frac1{a^n}$.

\\[\\frac{\\var{x}^\\var{n}}{\\var{y}^\\var{m}}=\\frac{\\var{y}^\\var{-m}}{\\var{x}^\\var{-n}}=\\frac{\\var{y^-m}}{\\var{x^-n}}=\\simplify{{y^-m}/{x^-n}}\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"n": {"name": "n", "group": "Ungrouped variables", "definition": "random(-3..-1)", "description": "", "templateType": "anything", "can_override": false}, "x": {"name": "x", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "random(-3..-1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["n", "x", "y", "m"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{x^n/y^m}", "maxValue": "{x^n/y^m}", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": true, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AB1 - Collecting terms", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Simple exercise in collecting terms in different powers of \$x\$

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following expression by combining \"like\" terms.

", "advice": "

The idea is to collect together and combine any terms that are the same kind of term so:

$\\var{b}$ and $\\var{f}$ are ordinary numbers. We can combine them to get $\\var{b+f}$

We can combine $\\var{a}x$ and $\\var{d}x$ to get $\\var{a+d}x$.

We combine $\\var{c}y$ and $\\var{e}y$ to get $\\var{c+e}y$. So our answer is:

$\\simplify{{c+e}y+{a+d}x+{b+f}}$

Use this link to find some resources which will help you revise this topic.

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$\\simplify[!collectNumbers]{{a}x+{b}+{c}y+{d}x+{f}+{e}y}$

", "answer": "({c}+{e})y+({a}+{d})x+({b}+{f})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": "`+-$n`?*y+`+-$n`?*x+`+-$n`?", "partialCredit": 0, "message": "", "nameToCompare": ""}, "valuegenerators": [{"name": "x", "value": ""}, {"name": "y", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AB3 - Collecting terms (higher powers)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Simple exercise in collecting terms in different powers of \$x\$

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following expression by combining \"like\" terms.

", "advice": "

First we expand the minus sign in the bracket.

\\[\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4-({f}x+{e}x^3)}=\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4+{-f}x+{-e}x^3}\\]

The idea is to collect together and combine any terms that are the same kind of term so:

$\\var{b}x$ and $\\var{-f}x$ both have an $x$ term. We can combine them to get $\\var{b-f}x$

We can combine $\\var{a}x^4$ and $\\var{d}x^4$ to get $\\var{a+d}x^4$.

We combine $\\var{c}x^3$ and $\\var{-e}x^3$ to get $\\var{c-e}x^3$. So our answer is:

$\\simplify{{a+d}x^4+{c+e}x^3+{b+f}}$

Use this link to find some resources which will help you revise this topic.

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$\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4-({f}x+{e}x^3)}$

", "answer": "({a}+{d})x^4+({c}-{e})x^3+({b}-{f})x", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": "`+-$n`?*x^4+`+-$n`?*x^3+`+-$n`?*x", "partialCredit": 0, "message": "", "nameToCompare": ""}, "valuegenerators": [{"name": "x", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AB5 Expand single brackets", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Bradley Bush", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1521/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": ["brackets", "expanding brackets", "expansion of brackets", "simplifying algebraic expressions", "simplifying expressions", "taxonomy"], "metadata": {"description": "

This question is made up of 10 exercises to practice the multiplication of brackets by a single term.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Expand the expression below by multiplying each of the terms inside the brackets by the term outside. Give the answer in its simplest form.

", "advice": "

Expand brackets using the general formula $\\displaystyle a(x+c)=ax+ac$. This means we multiply each term inside the brackets by the term outside the brackets.

It is easy to forget that the sign outside the brackets also needs to be involved in the multiplication so remember that when two of the same sign are multiplied, the resultant term is positive and when opposite signs are multiplied, the result is negative.

\\[
\\begin{align}
\\simplify[terms]{{a[7]}x({a[8]}x^2+{a[9]}x)}&=
\\simplify[!collectNumbers]{{a[7]}x{a[8]}x^2+{a[7]}x{a[9]}x}\\\\&=
\\simplify{{a[7]}*{a[8]}x^3+{a[7]}*{a[9]}x^2}\\text{.}
\\end{align}
\\]

Use this link to find resources to help you revise how to expand single brackets

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$\\simplify{{a[7]}x({a[8]}x^2+{a[9]}x)}=$ [[0]]

Factorise an expression of 2 or 3 terms where the gcd is a letter times a number. Part of HELM Book 1.3.4

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

Factorise $\\var{q2expr}$

", "advice": "

The two terms share a common factor of $\\var{q2gcd}\\var{latex(q2v[0])}$ which can be factored out.

So $\\var{q2expr} = \\var{q2ans}$

Use this link to find some resources which will help you revise this topic.

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Expanding two linear brackets multiplied together.

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

Expand the brackets and simplify

", "advice": "

To expand the brackets $\\simplify{({a[1]}x+{a[2]})({a[3]}x+{a[4]})}$ We first multiply all the terms in the left bracket by all the terms in the right bracket. This gives us

\\[\\var{a[1]}\\times\\var{a[3]}x^2+\\var{a[1]}x\\times\\var{a[4]}+\\var{a[2]}\\times\\var{a[3]}x+\\var{a[2]}\\times\\var{a[4]}=\\var{a[1]*a[3]}x^2+\\var{a[1]*a[4]}x+\\var{a[2]*a[3]}x+\\var{a[2]*a[4]}.\\]

We can then collect the terms to give us the final answer of

\\[\\var{a[1]*a[3]}x^2+\\var{a[1]*a[4]+a[2]*a[3]}x+\\var{a[2]*a[4]}.\\]

Use this link to find some resources which will help you revise this topic.

$\\simplify{({a[1]}x+{a[2]})({a[3]}x+{a[4]})}=$[[0]]

Expand two brackets involving powers of $x$.

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

Expand the brackets and simplify

", "advice": "

To expand the brackets $\\simplify{({a[1]}x^{b[1]}+{a[2]}x^{b[2]})({a[3]}x^{b[3]}+{c[1]}x^{b[4]})}$ We first multiply all the terms in the left bracket by all the terms in the right bracket. This gives us

\\[\\var{a[1]}x^\\var{b[1]}\\times\\var{a[3]}x^\\var{b[3]}+\\var{a[1]}x^\\var{b[1]}\\times\\var{c[1]}x^\\var{b[4]}+\\var{a[2]}x^\\var{b[2]}\\times\\var{a[3]}x^\\var{b[3]}+\\var{a[2]}x^\\var{b[2]}\\times\\var{c[1]}x^\\var{b[4]}\\]

We can then simplify to give us the final answer of

$\\simplify{{a[1]*a[3]}*x^{b[1]+b[3]}+{a[1]*c[1]}*x^{b[1]+b[4]}+{a[2]*a[3]}*x^{b[2]+b[3]}+{a[2]*c[1]}*x^{b[2]+b[4]}}.$

Use this link to find some resources which will help you revise this topic.

$\\simplify{({a[1]}x^{b[1]}+{a[2]}x^{b[2]})({a[3]}x^{b[3]}+{c[1]}x^{b[4]})}=$[[0]]

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Solve linear equations with unkowns on both sides. Including brackets and fractions.

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

To solve an equation like

$\\displaystyle{\\frac{x+\\var{num1}}{\\var{num2}}+\\frac{x}{\\var{num3}}=\\var{num4}},$

the first thing to deal with is the denominators of the fractions. In order to do that you multiply both sides of the equation by both denominators $\\var{num2}$ and $\\var{num3}$ (or their lowest common multiple to be slightly more efficient). This will give something equivalent to:

$\\displaystyle{\\var{num3 + num2} x+\\var{num3*num1} = \\var{num2*num3*num4}.}$

Then proceeding by subtracting $\\var{num3*num1} from both sides:

$\\displaystyle{\\var{num3 + num2} x = \\var{num2*num3*num4-num3*num1}.}$

And finally dividing by $\\var{num2+num3}$:

$\\displaystyle{x = \\frac{\\var{num2*num3*num4-num3*num1}}{\\var{num2+num3}}.}$

Use this link to find resources to help you revise how to solve linear equations

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Solve $\\displaystyle{\\frac{x+\\var{num1}}{\\var{num2}}+\\frac{x}{\\var{num3}}=\\var{num4}}$.

$x=$ [[0]]

Substitute values into an algebraic expression and calculate the result.

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

Evaluate the following expression,

\\[\\simplify{p^{n}+{a}*r*t+{c}},\\]

when $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$.

", "advice": "

In order to evaluate $\\simplify{p^{n}+{a}*r*t+{c}},$ with the given values, $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$, we replace each instance of that letter with its corresponding value and then apply the rules of BIDMAS:

\\[\\var{pval}^\\var{n}+\\var{a}\\times \\var{rval} \\times \\var{tval} + \\var{c}\\]

Which gives the answer $\\var{ans}$.

Follow this link for more help on tackling these kind of questions.

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Solving a pair of linear simultaneous equations, giving answers as integers or fractions.

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

Solve the simultaneous equations for x and y, giving your answers as integers or fractions, but not decimals.

\\[ \\begin{split} \\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\,=\\var{c} \\\\ \\simplify[!noLeadingminus,unitFactor]{{a1}x +{b1}y} &\\,=\\var{c1} \\end{split}\\]

", "advice": "

\\[\\begin{split}\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\,=\\var{c} \\qquad\\qquad&(1)\\\\ \\simplify[!noLeadingminus,unitFactor]{{a1}x +{b1}y} &\\,=\\var{c1} \\qquad\\qquad&(2)\\end{split}\\]

{advice1}

Use this link to find some resources which will help you revise this topic.

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For these equations, it is easiest to get a solution for $y$ first, due to the $x$-terms having {eqoroppa} coefficients.

\\n

If we {aorsa} equation (2) {torfa} equation (1) this eliminates the $x$-terms leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({b}+{sgna*(b1)})y} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c}+{sgna*(c1)}}\\\\\\\\ \\\\simplify{{b+sgna*(b1)}y} &\\\\,= \\\\simplify{{c+sgna*(c1)}} \\\\\\\\ y &\\\\,= \\\\simplify[all, fractionNumbers]{{c+sgna*(c1)}/{b+sgna*(b1)}} \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}x + \\\\var{b} \\\\times \\\\simplify[all, fractionNumbers]{{c+sgna*(c1)}/{b+sgna*(b1)}} &\\\\,= \\\\var{c} \\\\\\\\ \\\\var{a}x &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c} - {c*b+b*sgna*(c1)}/{b+sgna*(b1)}} \\\\\\\\ x &\\\\,= \\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "eqoroppb": {"name": "eqoroppb", "group": "Ungrouped variables", "definition": "if(abs(b)*b1=abs(b1)*b,'equal','equal and opposite')", "description": "", "templateType": "anything", "can_override": false}, "eqoroppa": {"name": "eqoroppa", "group": "Ungrouped variables", "definition": "if(abs(a)*a1=abs(a1)*a,'equal','equal and opposite')", "description": "", "templateType": "anything", "can_override": false}, "samey": {"name": "samey", "group": "Ungrouped variables", "definition": "\"

For these equations, it is easiest to get a solution for $x$ first, due to the $y$-terms having {eqoroppb} coefficients.

\\n

If we {aorsb} equation (2) {torfb} equation (1) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({a}+{sgn*(a1)})x} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c}+{sgn*(c1)}}\\\\\\\\ \\\\simplify{{a+sgn*(a1)}x} &\\\\,= \\\\simplify{{c+sgn*(c1)}} \\\\\\\\ x &\\\\,= \\\\simplify[all, fractionNumbers]{{c+sgn*(c1)}/{a+sgn*(a1)}} \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a} \\\\times\\\\simplify[fractionNumbers]{{c+sgn*(c1)}/{a+sgn*(a1)}} + \\\\var{b}y &\\\\,= \\\\var{c} \\\\\\\\ \\\\var{b}y &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c} - {c*a+a*sgn*(c1)}/{a+sgn*(a1)}} \\\\\\\\ y &\\\\,= \\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcmb": {"name": "lcmb", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (2) by $\\\\simplify{{abs(b/b1)}}$ we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\\\,=\\\\var{c} \\\\qquad\\\\qquad&(3)\\\\\\\\ \\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(b/b1)}x +{b1*abs(b/b1)}y} &\\\\,=\\\\var{c1*abs(b/b1)} \\\\qquad\\\\qquad&(4)\\\\end{split}\\\\]

\\n

If we {aorsb} equation (4) {torfb} equation (3) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({a}+{sgn*(a1*abs(b/b1))})x} &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c}+{sgn*(c1*abs(b/b1))}}\\\\\\\\ \\\\simplify{{a+sgn*(a1*abs(b/b1))}x} &\\\\,= \\\\simplify{{c+sgn*(c1*abs(b/b1))}} \\\\\\\\ x &\\\\,= \\\\simplify[all,fractionNumbers]{{c+sgn*(c1*abs(b/b1))}/{a+sgn*(a1*abs(b/b1))}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}\\\\times\\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c+sgn*c1*abs(b/b1)}/{(a)+sgn*a1*abs(b/b1)}) + {b}y} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(a*c)+a*sgn*c1*abs(b/b1)}/{(a)+sgn*a1*abs(b/b1)})} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(a*c+a*sgn*c1*abs(b/b1))/(a+sgn*a1*abs(b/b1))}} \\\\\\\\ y &\\\\,=\\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcmb1": {"name": "lcmb1", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (1) by $\\\\simplify{{abs(b1/b)}}$ we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(b1/b)}x +{b*abs(b1/b)}y} &\\\\,=\\\\var{c*abs(b1/b)} \\\\qquad\\\\qquad&(3) \\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1}x+{b1}y} &\\\\,=\\\\var{c1} \\\\qquad\\\\qquad&(4)\\\\\\\\ \\\\end{split} \\\\]

\\n

If we {aorsb} equation (4) {torfb} equation (3) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({(a*abs(b1/b))}+{sgn*a1})x} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{(c*abs(b1/b))}+{sgn*c1}}\\\\\\\\ \\\\simplify{{(a*abs(b1/b))+sgn*a1}x} &\\\\,= \\\\simplify{{(c*abs(b1/b))+sgn*c1}} \\\\\\\\ x &\\\\,= \\\\simplify[all, fractionNumbers]{{(c*abs(b1/b))+sgn*c1}/{(a*abs(b1/b))+sgn*a1}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}\\\\times\\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({(c*abs(b1/b))+sgn*c1}/{(a*abs(b1/b))+sgn*a1}) + {b}y} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(a*c*abs(b1/b))+a*sgn*c1}/{(a*abs(b1/b))+sgn*a1})} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(a*c*abs(b1/b)+a*sgn*c1)/(a*abs(b1/b)+sgn*a1)}} \\\\\\\\ y &\\\\,=\\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "full": {"name": "full", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (1) by $\\\\var{abs(b1)}$ and equation (2) by $\\\\var{abs(b)}$, we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(b1)}x+{b*abs(b1)}y} &\\\\,=\\\\var{c*abs(b1)} \\\\qquad\\\\qquad&(3)\\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(b)}x +{b1*abs(b)}y} &\\\\,=\\\\var{c1*abs(b)} \\\\qquad\\\\qquad&(4) \\\\end{split}\\\\]

\\n

Now, {aorsb} equation (4) {torfb} equation (3) to eliminate the $y$ terms:

\\n

\\\\[ \\\\begin{split} (\\\\simplify[!collectNumbers]{{a*abs(b1)} +{sgn*a1*abs(b)}}) x &\\\\,= \\\\simplify[!collectNumbers]{{c*abs(b1)}+{sgn*c1*abs(b)}} \\\\\\\\ \\\\simplify{{a*abs(b1)+sgn*a1*abs(b)}} x &\\\\,= \\\\simplify{{c*abs(b1)+sgn*c1*abs(b)}} .\\\\end{split} \\\\]

\\n

So the solution for $x$ is \\\\[ x=\\\\simplify{{c*abs(b1)+sgn*c1*abs(b)}/{a*abs(b1)+sgn*a1*abs(b)}}.\\\\]

\\n

To obtain a solution for $y$ we can substitute this value of $x$ into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[noLeadingminus,fractionNumbers,unitFactor]{{a} {xsimp} + {b}y} &\\\\,=\\\\var{c} \\\\\\\\ \\\\var{b}y &\\\\,= \\\\simplify[!collectNumbers,fractionNumbers]{{c}-{a*xsimp}} \\\\\\\\\\\\var{b}y &\\\\,= \\\\simplify[fractionNumbers]{{c-a*xsimp}} \\\\\\\\y &\\\\,= \\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}} \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "aorsa": {"name": "aorsa", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,'subtract','add')", "description": "", "templateType": "anything", "can_override": false}, "torfa": {"name": "torfa", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,'from','to')", "description": "", "templateType": "anything", "can_override": false}, "sgna": {"name": "sgna", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,-1,1)", "description": "", "templateType": "anything", "can_override": false}, "lcma": {"name": "lcma", "group": "Ungrouped variables", "definition": "\"

To get a solution for $y$, if we multiply equation (2) by $\\\\simplify{{abs(a/a1)}}$ we will have two equations with {eqoroppa} $x$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\\\,=\\\\var{c} \\\\qquad\\\\qquad&(3)\\\\\\\\ \\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(a/a1)}x +{b1*abs(a/a1)}y} &\\\\,=\\\\var{c1*abs(a/a1)} \\\\qquad\\\\qquad&(4)\\\\end{split}\\\\]

\\n

If we {aorsa} equation (4) {torfa} equation (3) this eliminates the $x$-terms, leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({b}+{sgna*(b1*abs(a/a1))})y} &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c}+{sgna*(c1*abs(a/a1))}}\\\\\\\\ \\\\simplify{{b+sgna*(b1*abs(a/a1))}y} &\\\\,= \\\\simplify{{c+sgna*(c1*abs(a/a1))}} \\\\\\\\ y &\\\\,= \\\\simplify[all,fractionNumbers]{{c+sgna*(c1*abs(a/a1))}/{b+sgna*(b1*abs(a/a1))}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{{a}x + {b}}\\\\times \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c+sgna*c1*abs(a/a1)}/{(b)+sgna*b1*abs(a/a1)})} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(b*c)+b*sgna*c1*abs(a/a1)}/{(b)+sgna*b1*abs(a/a1)})} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(b*c+b*sgna*c1*abs(a/a1))/(b+sgna*b1*abs(a/a1))}} \\\\\\\\ x &\\\\,=\\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcma1": {"name": "lcma1", "group": "Ungrouped variables", "definition": "\"

To get a solution for $y$, if we multiply equation (1) by $\\\\simplify{{abs(a1/a)}}$ we will have two equations with {eqoroppa} $x$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(a1/a)}x +{b*abs(a1/a)}y} &\\\\,=\\\\var{c*abs(a1/a)} \\\\qquad\\\\qquad&(3) \\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1}x+{b1}y} &\\\\,=\\\\var{c1} \\\\qquad\\\\qquad&(4) \\\\end{split}\\\\]

\\n

If we {aorsa} equation (4) {torfa} equation (3) this eliminates the $x$-terms, leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({(b*abs(a1/a))}+{sgna*b1})y} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{(c*abs(a1/a))}+{sgna*c1}}\\\\\\\\ \\\\simplify{{(b*abs(a1/a))+sgna*b1}y} &\\\\,= \\\\simplify{{(c*abs(a1/a))+sgna*c1}} \\\\\\\\ y &\\\\,= \\\\simplify[all, fractionNumbers]{{(c*abs(a1/a))+sgna*c1}/{(b*abs(a1/a))+sgna*b1}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{{a}x + {b}}\\\\times \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c*abs(a1/a)+sgna*c1}/{(b*abs(a1/a))+sgna*b1})} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(b*c*abs(a1/a))+b*sgna*c1}/{(b*abs(a1/a))+sgna*b1})} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(b*c*abs(a1/a)+b*sgna*c1)/(b*abs(a1/a)+sgna*b1)}} \\\\\\\\ x &\\\\,=\\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

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$x=$ [[0]]

$y=$ [[1]]

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Rearrange a specific formula. No randomisation.

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

Rearrange the following equation, to make $y$ the subject:

\\[{cy -b = 3x}\\]

", "advice": "

In order to rearrange the equation so that it is in terms of $y$, we must first add $b$ to both sides, and then divide both sides of the equation by $c$:

\\begin{split} cy-b &= 3x \\\\ cy &= 3x + b \\\\ y &=\\frac{3x+b}{c} \\end{split}

Use this link to find some resources which will help you revise this topic.

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$y=$ [[0]]

Calculations involving Standard form.

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

To divide two numbers in standard form we can calculate the division of each part of the standard form number separately. In general we have,

\\[\\frac{x\\times10^j}{y\\times10^k}=\\frac xy\\times\\frac{10^j}{10^k}=\\frac xy\\times 10^{j-k}\\]

In this question we therefore have,

\\[\\frac{\\var{a}\\times10^{\\var{n}}}{\\var{b}\\times10^{\\var{m}}}=\\frac{\\var{a}}{\\var{b}}\\times\\frac{10^{\\var{n}}}{10^{\\var{m}}}=\\var{aDivBRound}\\times10^\\var{n-m}.\\]

Since {aDivBRound} is less than 1 then our answer isn't in standard form. In this case we need to reduce the exponent by 1 so the final answer is

\\[\\var{MantAnsRound}\\times10^{\\var{ExponentAns}}.\\]

Use this link to find some resources which will help you revise this topic.

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For the equation

\\[\\frac{\\var{a}\\times10^{\\var{n}}}{\\var{b}\\times10^{\\var{m}}}=a\\times10^n\\]

find the values of $a$ and $n$ which keep the answer in standard form.

Give $a$ to two decimal places.

$a=$[[0]]

$n=$[[1]]

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Draws a triangle based on 3 side lengths and randomises asking for hypotenuse or not.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{statement}

Find $x$.

", "advice": "

Only round your final answer to 1 decimal place.

{advice}

Use this link to find some resources to help you revise how to use pythagoras' theorem.

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Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = \\\\var{sh1}$, $b =\\\\var{sh2}$ and $c = x$

\\n

So

\\\\[\\\\var{sh1}^2 + \\\\var{sh2}^2 = x^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{sh1^2} + \\\\var{sh2^2}\\\\]

\\\\[x = \\\\sqrt{\\\\var{sh1^2} + \\\\var{sh2^2}}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{sh1^2+sh2^2}}\\\\]

$x = \\\\var{hyp}$ to 1 d.p.

\"", "description": "", "templateType": "long string", "can_override": false}, "advice1": {"name": "advice1", "group": "Varying q and advice", "definition": "\"

Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = x$, $b =\\\\var{sh2}$ and $c = \\\\var{hyp}$

\\n

\\\\[x^2 + \\\\var{sh2}^2 = \\\\var{hyp}^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{hyp^2} - \\\\var{sh2^2}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{hyp^2-sh2^2}}\\\\]

$x = \\\\var{sh1}$ to 1 d.p.

\"", "description": "", "templateType": "long string", "can_override": false}, "statement1": {"name": "statement1", "group": "Varying q and advice", "definition": "{geogebra_applet('https://www.geogebra.org/m/zhu32wjq',[sh1: sh1, sh2: sh2])}", "description": "

", "templateType": "anything", "can_override": false}, "statement2": {"name": "statement2", "group": "Varying q and advice", "definition": "{geogebra_applet('https://www.geogebra.org/m/b8wz2vn9',[sh1: sh1, sh2: sh2])}", "description": "", "templateType": "anything", "can_override": true}, "statement": {"name": "statement", "group": "Varying q and advice", "definition": "if(setup=1,statement1,statement2)", "description": "", "templateType": "anything", "can_override": false}, "hyp": {"name": "hyp", "group": "Unnamed group", "definition": "precround(sqrt(sh1^2+sh2^2),1)", "description": "", "templateType": "anything", "can_override": false}, "sh1_gen": {"name": "sh1_gen", "group": "Unnamed group", "definition": "random(5 .. 10#0.1)", "description": "

one of two shortest sides for calculations.

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$x=$[[0]] to 1 d.p.

Finding the area of a circle when given the diameter of the circle.

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

Find the area of a circle with diameter $\\var{d}$ cm giving your answer to 1 decimal place.

{geogebra_applet('https://www.geogebra.org/m/ngcchpcj',[d: d])}

", "advice": "

To calculate the area of a circle we want to use the formula \\[ A = \\pi r^2, \\]

where $r$ is the radius of the circle.

So, if the diameter, d, is $\\var{d}$ cm, then the radius is, $r=\\frac{d}{2}=\\var{{d}/2}$ cm, then

\\[ \\begin{split} Area &\\,=\\var{{d}/2}^2 \\times \\pi \\text{ cm}^2 \\\\ &\\,= \\simplify[all, fractionNumbers]{{{{d}^2/4}}pi} \\text{ cm}^2 \\\\ &\\,= \\var{precround({d}^2/4*pi,1)} \\text{ cm}^2. \\end{split} \\]

Use this link to find some resources to help you revise how to calculate the area of a circle.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"d": {"name": "d", "group": "Ungrouped variables", "definition": "random(6,8,10,12,14,16,18,20)", "description": "", "templateType": "anything", "can_override": true}, "t": {"name": "t", "group": "Ungrouped variables", "definition": "random(2..6)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["d", "t"], "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": "

$Area=$ [[0]] $\\text{ cm}^2$

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Calculate the volume of different 3D shapes, given the units and measurements required. The formulae for the volume of each shape are available as steps if required.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

For a triangular prism, we first need to find the area of one of the faces then multiply this area by the depth of the prism.
In this example the easiest way to calculate the volume is to take the area of the triangular face first with $\\mathrm{base} = \\var{w6}m$ and $\\mathrm{height} = \\var{h6}m\\thinspace$.

\\begin{align}
\\mathrm{Area\\thinspace_\\triangle} &= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\\\
&= \\frac{\\var{w6} \\times \\var{h6}}{2} \\\\
&= \\var{0.5*w6*h6}\\, \\mathrm{m}^2\\,.
\\end{align}

Now that we have the area of the triangular face ($\\mathrm{Area\\thinspace_\\triangle}$) we can multiply this by the $\\mathrm{depth} = \\var{d6}m\\thinspace$.

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\var{0.5*w6*h6} \\times \\var{d6} \\\\
&= \\var{0.5*w6*h6*d6}\\, \\mathrm{m}^2\\,.
\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a triangular prism.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"d4": {"name": "d4", "group": "Cuboid ", "definition": "random(2..5#1)", "description": "

Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w6": {"name": "w6", "group": "Triangular prism", "definition": "random(5..9#1)", "description": "

Creates base of triangle.

", "templateType": "anything", "can_override": false}, "d8": {"name": "d8", "group": "Square based pyramid", "definition": "random(3..6#0.1)", "description": "

One side of square base.

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Height of pyramid.

", "templateType": "anything", "can_override": false}, "w7": {"name": "w7", "group": "Cylinder", "definition": "random(7..15#0.1)", "description": "

Depth of cylinder.

", "templateType": "anything", "can_override": false}, "d6": {"name": "d6", "group": "Triangular prism", "definition": "random(9..15#0.1)", "description": "

Depth of triangular prism.

", "templateType": "anything", "can_override": false}, "r7": {"name": "r7", "group": "Cylinder", "definition": "random(2..6#1)", "description": "

Radius of the cylinder.

", "templateType": "anything", "can_override": false}, "h4": {"name": "h4", "group": "Cuboid ", "definition": "random(2..5#1 except d4)", "description": "

Side of square in cuboid.

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Width of cuboid.

", "templateType": "anything", "can_override": false}, "w8": {"name": "w8", "group": "Square based pyramid", "definition": "random(3..7#1)", "description": "

One side of square base.

", "templateType": "anything", "can_override": false}, "h6": {"name": "h6", "group": "Triangular prism", "definition": "random(2..5#1)", "description": "

Height of traingle.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "Cuboid ", "variables": ["w4", "d4", "h4"]}, {"name": "Triangular prism", "variables": ["w6", "h6", "d6"]}, {"name": "Cylinder", "variables": ["r7", "w7"]}, {"name": "Square based pyramid", "variables": ["h8", "w8", "d8"]}], "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": "

Calculate the $\\mathrm{Volume}$ of the following triangular prism.

$\\mathrm{Volume} =$[[0]]$\\mathrm{m}^3$.

", "stepsPenalty": "1", "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": "

Volume of a triangular prism:

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\times \\mathrm{depth}
\\end{align}

"}], "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "0.5{w6}{h6}{d6}", "maxValue": "0.5{w6}{h6}{d6}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GA5 Volume of cylinder", "extensions": [], "custom_part_types": [], "resources": [["question-resources/sqbasedpyramid_sEpkGzO.svg", "/srv/numbas/media/question-resources/sqbasedpyramid_sEpkGzO.svg"], ["question-resources/triangularprism.svg", "/srv/numbas/media/question-resources/triangularprism.svg"], ["question-resources/cylinder.svg", "/srv/numbas/media/question-resources/cylinder.svg"], ["question-resources/cuboid.svg", "/srv/numbas/media/question-resources/cuboid.svg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Aiden McCall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1592/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": ["3D shapes", "cuboid", "Cylinder", "cylinder", "pyramid", "taxonomy", "triangular prism", "volume", "Volume", "volume of a cuboid", "volume of a cylinder", "volume of a pyramid", "volume of a triangular prism"], "metadata": {"description": "

Calculate the volume of different 3D shapes, given the units and measurements required. The formulae for the volume of each shape are available as steps if required.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

For a cylinder, we first need to find the area of the circular face then multiply this area by the depth of the cylinder.
In this example the radius of the circular face is $\\mathrm{radius} = \\var{r7}m$ which can be used to calculate the area of the circular face.

\\begin{align}
\\mathrm{Area\\thinspace_\\bigcirc} &= \\pi \\times \\mathrm{radius}^2 \\\\
&= \\pi \\times \\var{r7}^2 \\\\
&= \\var{pi * (r7)^2}\\, \\mathrm{m}^2 \\,.
\\end{align}

Now that we have the area of the circular face ($\\mathrm{Area\\thinspace_\\bigcirc}$) we can multiply this by the $\\mathrm{depth} =\\var{w7}m\\thinspace$.

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\bigcirc} \\times \\mathrm{depth} \\\\
&= \\var{pi*(r7)^2} \\times \\var{w7} \\\\
&= \\var{dpformat(pi*w7*(r7)^2, 5)} \\\\
&= \\var{dpformat(pi*w7*(r7)^2, 1)}\\, \\mathrm{m}^2\\,. \\quad \\text{1 d.p.}
\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a cylinder.

Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w6": {"name": "w6", "group": "Triangular prism", "definition": "random(5..9#1)", "description": "

Creates base of triangle.

", "templateType": "anything", "can_override": false}, "d8": {"name": "d8", "group": "Square based pyramid", "definition": "random(3..6#0.1)", "description": "

One side of square base.

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Height of pyramid.

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Depth of cylinder.

", "templateType": "anything", "can_override": false}, "d6": {"name": "d6", "group": "Triangular prism", "definition": "random(9..15#0.1)", "description": "

Depth of triangular prism.

", "templateType": "anything", "can_override": false}, "r7": {"name": "r7", "group": "Cylinder", "definition": "random(2..6#1)", "description": "

Radius of the cylinder.

", "templateType": "anything", "can_override": false}, "h4": {"name": "h4", "group": "Cuboid ", "definition": "random(2..5#1 except d4)", "description": "

Side of square in cuboid.

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Width of cuboid.

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One side of square base.

", "templateType": "anything", "can_override": false}, "h6": {"name": "h6", "group": "Triangular prism", "definition": "random(2..5#1)", "description": "

Height of traingle.

Calculate the $\\mathrm{Volume}$ of the following cylinder.

$\\mathrm{Volume} =$[[0]] $\\mathrm{m}^3$. Round your answer to 1 decimal place.

Volume of a cylinder:

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\bigcirc} \\times \\mathrm{depth} \\\\
&= \\pi \\times \\mathrm{r}^2 \\times \\mathrm{depth}
\\end{align}

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Calculating gradient and finding intercept from a geogebra graph.

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

{app}
Find the gradient of the line.

", "advice": "

Firstly draw a right angled 'step' from left to right. This triangle can be anywhere, but it is more helpful for it to have corners on the vertices (whole number points) of the graph and it is easier to calculate with postive numbers.

{app_advice}

Before we start to calculate, notice that the line is {uod}, so the gradient will be {pon} and the line is {sos}, so the absolute value of the number will be {mol}.

Now find the coordinates of the places your triangle meets the line

$(x_1,y_1)=(\\var{ax},\\var{ay})$ and $(x_2,y_2)=(\\var{bx},\\var{by})$

We need to compare the 'rise on the y-axis' to the 'run across the x-axis', we can say that:

$\\text{gradient} = \\frac{\\text{rise}}{\\text{run}}$

This is equivalent to using the formula:

$ m = \\frac{y_2 - y_1}{x_2 - x_1} $

and substitute the coordinates of the vertices of the triangle:

$\\begin{split} &\\, m = \\frac{\\var{by} - \\var{ay}}{\\var{bx} - \\var{ax}} \\\\
&\\, = \\frac{\\var{by-ay}}{\\var{bx-ax}} \\\\
&\\, = \\var[fractionNumbers]{m} \\\\
\\end{split} $

Use this link to find resources to help you revise straight line graphs and how to find the gradient of them.

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if(m=abs(m),'positive','negative')

", "templateType": "anything", "can_override": false}, "ax": {"name": "ax", "group": "Ungrouped variables", "definition": "random(0,1)", "description": "", "templateType": "anything", "can_override": false}, "ay": {"name": "ay", "group": "Ungrouped variables", "definition": "random(0,1,2,3)", "description": "", "templateType": "anything", "can_override": false}, "bx": {"name": "bx", "group": "Ungrouped variables", "definition": "random(ax+1..3) \n", "description": "", "templateType": "anything", "can_override": false}, "by": {"name": "by", "group": "Ungrouped variables", "definition": "random(0..4 except ay)\n", "description": "", "templateType": "anything", "can_override": false}, "app_advice": {"name": "app_advice", "group": "Ungrouped variables", "definition": "geogebra_applet(\n 800,500,\n [\n A: [\n definition: p1,\n label_visible: false,\n visible: true\n ],\n B: [\n definition: p2,\n label_visible: false,\n visible: true \n ],\n \n C: [\n definition: p3,\n label_visible: false,\n visible: false \n ],\n \n line1: [\n definition: \"Line(A,B)\",\n label_visible: false,\n visible: true\n ],\n \n line2: [\n definition: \"Segment(A,C)\",\n label_visible: false,\n visible: true\n ],\n \n \n \n line3: [\n definition: \"Segment(C,B)\",\n label_visible: false,\n visible: true\n ]\n ]\n)", "description": "", "templateType": "anything", "can_override": false}, "p3": {"name": "p3", "group": "Ungrouped variables", "definition": "vector(bx,ay)", "description": "", "templateType": "anything", "can_override": false}, "pon": {"name": "pon", "group": "Ungrouped variables", "definition": "if(m=0,'zero',if(m=abs(m),'a positive number','a negative number'))", "description": "", "templateType": "anything", "can_override": false}, "sos": {"name": "sos", "group": "Ungrouped variables", "definition": "if(m=0,'horizontal',if(abs(m)<1,'shallow','steep'))", "description": "", "templateType": "anything", "can_override": false}, "mol": {"name": "mol", "group": "Ungrouped variables", "definition": "if(m=0,'zero',if(abs(m)<1,'less than 1','greater than or equal to 1'))", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "m<>1", "maxRuns": 100}, "ungrouped_variables": ["app", "m", "c", "P1", "P2", "uod", "ax", "ay", "bx", "by", "app_advice", "p3", "pon", "sos", "mol"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you have incorrectly rounded this answer. You might want to look at some resources on rounded decimals. You can also leave your answer in fraction form as
$\\var[fractionNumbers]{m}$

", "useAlternativeFeedback": false, "answer": "{m}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{m}", "showPreview": true, "checkingType": "dp", "checkingAccuracy": "1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SA6 Calculate Range", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": ["mean", "measures of average and spread", "median", "mode", "range", "taxonomy"], "metadata": {"description": "

This question provides a list of data to the student. They are asked to find the \"range\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\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

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

Range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a2": {"name": "a2", "group": "Ungrouped variables", "definition": "repeat(random(1..9), 20)", "description": "

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

", "templateType": "anything", "can_override": false}, "modea2": {"name": "modea2", "group": "Ungrouped variables", "definition": "mode(a2)", "description": "", "templateType": "anything", "can_override": false}, "a3": {"name": "a3", "group": "Ungrouped variables", "definition": "shuffle([ random(0..1),\n 2, \n random(4..6),\n random(0..3 except 2), \n random(0..3 except 2),\n random(4..6),\n 2,\n 2,\n random(4..6),\n random(7..8),\n random(0..3 except 2 except 1), \n random(4..6),\n 2,\n random(1..3 except 2), \n random(7..8),\n 2,\n random(7..8),\n random(4..6), \n random(0..3 except 2), \n 2\n])", "description": "

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

", "templateType": "anything", "can_override": false}, "range": {"name": "range", "group": "final list", "definition": "max(a) - min(a)", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "final list", "definition": "if(len(modea1) = 1, a1, if(len(modea2) = 1, a2, a3))", "description": "

The final list.

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Find the range.

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Calculating the Mean from a basic list of integers.

", "licence": "None specified"}, "statement": "

Calculate the Mean from a list

", "advice": "

The MEAN is the sum, divided by the number of values summed i.e.

$\\frac{\\var{list[0]} + \\var{list[1]} + \\var{list[2]} + \\var{list[3]} + \\var{list[4]}}{5}$

use your calculator to find

mean = $\\var{mean}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"list": {"name": "list", "group": "Ungrouped variables", "definition": "repeat(random(0..20), 5)", "description": "", "templateType": "anything", "can_override": false}, "mean": {"name": "mean", "group": "Ungrouped variables", "definition": "mean(list)", "description": "", "templateType": "anything", "can_override": false}, "median": {"name": "median", "group": "Ungrouped variables", "definition": "median(list)", "description": "", "templateType": "anything", "can_override": false}, "order": {"name": "order", "group": "Ungrouped variables", "definition": "sort(list)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["list", "mean", "median", "order"], "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": "

Given a list of numbers:

{list}

Calculate the mean: [[0]]

This question provides a list of data to the student. They are asked to find the \"mode\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\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

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

The mode is the value that occurs the most often in the data.

To find a mode, we can look at our sorted list:

$\\var{a_s[0]}, \\var{a_s[1]}, \\var{a_s[2]}, \\var{a_s[3]}, \\var{a_s[4]}, \\var{a_s[5]}, \\var{a_s[6]}, \\var{a_s[7]}, \\var{a_s[8]}, \\var{a_s[9]}, \\var{a_s[10]}, \\var{a_s[11]}, \\var{a_s[12]}, \\var{a_s[13]}, \\var{a_s[14]}, \\var{a_s[15]}, \\var{a_s[16]}, \\var{a_s[17]}, \\var{a_s[18]}, \\var{a_s[19]}$.

We notice that $\\var{mode1}$ occurs the most ($\\var{modetimes[mode1]}$ times) so $\\var{mode1}$ is the mode.

Use this link to find some resources which will help you revise this topic.

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

", "templateType": "anything", "can_override": false}, "mode1": {"name": "mode1", "group": "final list", "definition": "mode[0]", "description": "

Mode as a value.

", "templateType": "anything", "can_override": false}, "mode": {"name": "mode", "group": "final list", "definition": "mode(a)", "description": "

Mode as a vector.

", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "final list", "definition": "if(len(modea1) = 1, a1, if(len(modea2) = 1, a2, a3))", "description": "

The final list.

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Find the mode.

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This question provides a list of data to the student. They are asked to find the \"median\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\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

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

The median is the middle value. We need to sort the list in order:

\\[ \\var{a_s[0]}, \\quad \\var{a_s[1]}, \\quad \\var{a_s[2]}, \\quad \\var{a_s[3]}, \\quad \\var{a_s[4]}, \\quad \\var{a_s[5]}, \\quad \\var{a_s[6]}, \\quad \\var{a_s[7]}, \\quad \\var{a_s[8]}, \\quad \\var{a_s[9]}, \\quad \\var{a_s[10]}, \\quad \\var{a_s[11]}, \\quad \\var{a_s[12]}, \\quad \\var{a_s[13]}, \\quad \\var{a_s[14]}, \\quad \\var{a_s[15]}, \\quad \\var{a_s[16]}, \\quad \\var{a_s[17]}, \\quad \\var{a_s[18]}, \\quad \\var{a_s[19]} \\]

There is an even number of responses, so there are two numbers in the middle (10^th and 11^th place). To find the median, we need to find the mean of these two numbers $\\var{a_s[9]}$ and $\\var{a_s[10]}$:

\\begin{align}
\\frac{\\var{a_s[9]} + \\var{a_s[10]}}{2} &= \\frac{\\var{a_s[9] + a_s[10]}}{2} \\\\
&= \\var{median} \\text{.}
\\end{align}

Use this link to find some resources which will help you revise this topic.

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

", "templateType": "anything", "can_override": false}, "modea1": {"name": "modea1", "group": "Ungrouped variables", "definition": "mode(a1)", "description": "", "templateType": "anything", "can_override": false}, "median": {"name": "median", "group": "final list", "definition": "median(a)", "description": "", "templateType": "anything", "can_override": false}, "a1": {"name": "a1", "group": "Ungrouped variables", "definition": "repeat(random(0..8), 20)", "description": "

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "mean": {"name": "mean", "group": "final list", "definition": "mean(a)", "description": "", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

", "templateType": "anything", "can_override": false}, "range": {"name": "range", "group": "final list", "definition": "max(a) - min(a)", "description": "", "templateType": "anything", "can_override": false}, "mode1": {"name": "mode1", "group": "final list", "definition": "mode[0]", "description": "

Mode as a value.

", "templateType": "anything", "can_override": false}, "mode": {"name": "mode", "group": "final list", "definition": "mode(a)", "description": "

Mode as a vector.

", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "final list", "definition": "if(len(modea1) = 1, a1, if(len(modea2) = 1, a2, a3))", "description": "

The final list.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["modea1", "modea2", "a1", "a2", "a3"], "variable_groups": [{"name": "final list", "variables": ["a", "a_s", "mean", "median", "mode", "mode1", "range", "modetimes"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find the median.

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This is a tool for you! It is here to help you diagnose whether there are any maths or statistics pre-requisites for your course that you may want to brush up on. If at any point you are struggling with any question you should find a link at the end of the \"reveal answer\" section that will take you to some recommended online resources on that subject area. You can also always contact the Maths and Stats Help team (MaSH) to arrange a one to one appointment or check out our workshop timetable to see if you can access the support you need that way. Find all this information via our website here!

", "end_message": "

Thanks for completing the Skills Audit. You can attempt this as many times as you need. Remember the score is not what matters - this is in no way assessed work - this is simply a tool for working out whether you may need to brush up on anything to ensure that you can access all the material on your course and get off to the best possible start.

Don't forget to look up what support is available to you through our web pages here!

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