// Numbas version: exam_results_page_options {"name": "Scaling one number into another", "metadata": {"description": "

A quick practice set of problems for education students to take in preparation for their numeracy test.

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Written to help prepare pre-service teachers for the numeracy test in Australia called LANTITE.

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Based on the Australian Core Skills Framework - numeracy - level 3 - personal and community sample activity:Accurately measures a range of quantities to follow a recipe or operating instructions
incorporating making a product of a smaller or larger size than specified, e.g. follows a recipe
for six people and can adjust it to cater for 24 people

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Has an annoying plural issue

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The following are the ingredients from a recipe that makes 12 {food}.

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1 1/2 cups milk
1 egg
2 teaspoons vanilla extract
2 cups self-raising flour
1/4 teaspoon bicarbonate of soda
1/3 cup caster sugar
25g butter, melted

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2 1/4 cups self-raising flour
3/4 cup caster sugar
1 egg
1/2 cup vegetable oil
3/4 cup milk
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If you wanted to make {new_amount[0]} {food}, how many {worded_ingredient[0]} would you need? 

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The recipe tells us that for 12 {food} we need $\\simplify[fractionNumbers]{{worded_ingredient[1]}}$ {worded_ingredient[0]}. Since we were asked about making $\\var{new_amount[0]}$ {food} and $\\simplify[fractionNumbers]{{new_amount[0]}/12={scaling_factor}}$, we have been asked about making {new_amount[1]} the number contained in the recipe. So we will need $\\simplify[fractionNumbers]{{scaling_factor}}\\times \\simplify[fractionNumbers]{{worded_ingredient[1]}={ans}}$ {worded_ingredient[0]}.

"}], "minValue": "ans", "maxValue": "ans", "correctAnswerFraction": true, "allowFractions": true, "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": "Scaling one number to another", "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/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(5..50)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(a..100 except 0..100#a)", "description": "", "templateType": "anything", "can_override": false}, "nums": {"name": "nums", "group": "Ungrouped variables", "definition": "shuffle([a,b])", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "nums[0]", "description": "", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "nums[1]", "description": "", "templateType": "anything", "can_override": false}, "n": {"name": "n", "group": "Ungrouped variables", "definition": "nums2[0]", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "nums2[1]", "description": "", "templateType": "anything", "can_override": false}, "temp_m": {"name": "temp_m", "group": "Ungrouped variables", "definition": "random(temp_n..100 except 0..100#temp_n except a except b)", "description": "", "templateType": "anything", "can_override": false}, "temp_n": {"name": "temp_n", "group": "Ungrouped variables", "definition": "random(5..50 except a except b)", "description": "", "templateType": "anything", "can_override": false}, "nums2": {"name": "nums2", "group": "Ungrouped variables", "definition": "shuffle([temp_n,temp_m])", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "b", "nums", "c", "d", "temp_n", "temp_m", "nums2", "n", "m"], "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": "

We can turn the number $\\var{c}$ into the number $\\var{d}$ by multiplication and division as follows:

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    \n
  1. We shrink $\\var{c}$ down to the number $1$ by dividing by [[0]]
  2. \n
  3. We stretch $1$ up to the number $\\var{d}$ by multiplying by [[1]]
  4. \n
\n

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Instead of using that stepping-stone of going to $1$ first, we can multiply by a fraction equivalent to the above division and multiplication. That is, 

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$\\var{c}\\times $ [[2]] $= \\var{d}$.

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We want to know what number to multiply $\\var{c}$ by in order to get the number $\\var{d}$.

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We can think of this as stretching/scaling $\\var{c}$ by a scaling factor to get the number $\\var{d}$.

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We can find this scaling factor by using the unitary method:

\n
    \n
  1. We shrink $\\var{c}$ down to the number $1$ by dividing by $\\var{c}$
  2. \n
  3. We stretch $1$ up to the number $\\var{d}$ by multiplying by $\\var{d}$
  4. \n
\n

This two-step process called the unitary method is used in many places (such as rates, ratios and percentages).

\n

Instead of using that stepping-stone of going to $1$ first, we can multiply by a fraction equivalent to the above division and multiplication. That is, 

\n

$\\var{c}\\times \\frac{\\var{d}}{\\var{c}}= \\var{d}$.

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Notice how multiplying by a number less than $1$ results in a smaller number.

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We can turn the number $\\var{n}$ into the number $\\var{m}$ by multiplying (or scaling) by [[0]].

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    \n

    ", "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": "

    We want to know what number to multiply $\\var{n}$ by in order to get the number $\\var{m}$.

    \n

    We can think of this as stretching/scaling $\\var{n}$ by a scaling factor to get the number $\\var{m}$.

    \n

    We can find this scaling factor by using the unitary method:

    \n
      \n
    1. We shrink $\\var{n}$ down to the number $1$ by dividing by $\\var{n}$
    2. \n
    3. We stretch $1$ up to the number $\\var{m}$ by multiplying by $\\var{m}$
    4. \n
    \n

    This two-step process called the unitary method is used in many places (such as rates, ratios and percentages).

    \n

    Instead of using that stepping-stone of going to $1$ first, we can multiply by a fraction equivalent to the above division and multiplication. That is, 

    \n

    $\\var{n}\\times \\frac{\\var{m}}{\\var{n}}= \\var{m}$.

    \n
    \n

    Notice how multiplying by a number less than $1$ results in a smaller number.

    \n
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