// Numbas version: exam_results_page_options {"name": "Decimals", "metadata": {"description": "

Face, Place and actual value, adding, subtracting, multiplying, dividing and rounding decimals.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "duration": 0, "percentPass": "0", "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], []], "questions": [{"name": "Decimals: how to read", "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": "

If you are unsure of how to do a question, click on Show steps to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"place": {"name": "place", "group": "Ungrouped variables", "definition": "random([placedig*0.001, \"tenths\", \"hundredths\", \"thousandths\" ]\n,[placedig*0.01, \"tenths\", \"thousandths\", \"hundredths\"],\n [placedig*0.1, \"thousandths\", \"hundredths\", \"tenths\"])", "description": "", "templateType": "anything", "can_override": false}, "pron": {"name": "pron", "group": "Ungrouped variables", "definition": "shuffle([random(\n[dpformat(0.10, 2), \"one zero\", \"ten\"], \n[dpformat(0.11, 2), \"one one\", \"eleven\"], \n[dpformat(0.12, 2), \"one two\", \"twelve\"],\n[dpformat(0.13, 2), \"one three\", \"thirteen\"],\n[dpformat(0.14, 2), \"one four\", \"fourteen\"],\n[dpformat(0.15, 2), \"one five\", \"fifteen\"],\n[dpformat(0.16, 2), \"one six\", \"sixteen\"],\n[dpformat(0.17, 2), \"one seven\", \"seventeen\"],\n[dpformat(0.18, 2), \"one eight\", \"eighteen\"],\n[dpformat(0.19, 2), \"one nine\", \"ninteen\"], \n\n[dpformat(0.20, 2), \"two zero\", \"twenty\"], \n[dpformat(0.21, 2), \"two one\", \"twenty one\"], \n[dpformat(0.22, 2), \"two two\", \"twenty two\"],\n[dpformat(0.23, 2), \"two three\", \"twenty three\"],\n[dpformat(0.24, 2), \"two four\", \"twenty four\"], \n[dpformat(0.25, 2), \"two five\", \"twenty five\"],\n[dpformat(0.26, 2), \"two six\", \"twenty six\"],\n[dpformat(0.27, 2), \"two seven\", \"twenty seven\"],\n[dpformat(0.28, 2), \"two eight\", \"twenty eight\"],\n[dpformat(0.29, 2), \"two nine\", \"twenty nine\"], \n \n[dpformat(0.30, 2), \"three zero\", \"thirty\"], \n[dpformat(0.31, 2), \"three one\", \"thirty one\"], \n[dpformat(0.32, 2), \"three two\", \"thirty two\"],\n[dpformat(0.33, 2), \"three three\", \"thirty three\"],\n[dpformat(0.34, 2), \"three four\", \"thirty four\"], \n[dpformat(0.35, 2), \"three five\", \"thirty five\"],\n[dpformat(0.36, 2), \"three six\", \"thirty six\"],\n[dpformat(0.37, 2), \"three seven\", \"thirty seven\"],\n[dpformat(0.38, 2), \"three eight\", \"thirty eight\"],\n[dpformat(0.39, 2), \"three nine\", \"thirty nine\"], \n \n[dpformat(0.40, 2), \"four zero\", \"forty\"], \n[dpformat(0.41, 2), \"four one\", \"forty one\"], \n[dpformat(0.42, 2), \"four two\", \"forty two\"],\n[dpformat(0.43, 2), \"four three\", \"forty three\"],\n[dpformat(0.44, 2), \"four four\", \"forty four\"], \n[dpformat(0.45, 2), \"four five\", \"forty five\"],\n[dpformat(0.46, 2), \"four six\", \"forty six\"],\n[dpformat(0.47, 2), \"four seven\", \"forty seven\"],\n[dpformat(0.48, 2), \"four eight\", \"forty eight\"],\n[dpformat(0.49, 2), \"four nine\", \"forty nine\"], \n \n[dpformat(0.50, 2), \"five zero\", \"fifty\"], \n[dpformat(0.51, 2), \"five one\", \"fifty one\"], \n[dpformat(0.52, 2), \"five two\", \"fifty two\"],\n[dpformat(0.53, 2), \"five three\", \"fifty three\"],\n[dpformat(0.54, 2), \"five four\", \"fifty four\"], \n[dpformat(0.55, 2), \"five five\", \"fifty five\"],\n[dpformat(0.56, 2), \"five six\", \"fifty six\"],\n[dpformat(0.57, 2), \"five seven\", \"fifty seven\"],\n[dpformat(0.58, 2), \"five eight\", \"fifty eight\"],\n[dpformat(0.59, 2), \"five nine\", \"fifty nine\"], \n \n[dpformat(0.60, 2), \"six zero\", \"sixty\"], \n[dpformat(0.70, 2), \"seven zero\", \"seventy\"], \n[dpformat(0.80, 2), \"eight zero\", \"eighty\"], \n[dpformat(0.90, 2), \"nine zero\", \"ninety\"], \n)]+[random(\n[dpformat(0.100, 3), \"one zero zero\", \"one hundred\"], \n[dpformat(0.200, 3), \"two zero zero\", \"two hundred\"], \n[dpformat(0.300, 3), \"three zero zero\", \"three hundred\"],\n[dpformat(0.400, 3), \"four zero zero\", \"four hundred\"],\n[dpformat(0.500, 3), \"five zero zero\", \"five hundred\"], \n[dpformat(0.600, 3), \"six zero zero\", \"six hundred\"],\n[dpformat(0.700, 3), \"seven zero zero\", \"seven hundred\"],\n[dpformat(0.800, 3), \"eight zero zero\", \"eight hundred\"],\n[dpformat(0.900, 3), \"nine zero zero\", \"nine hundred\"],\n[dpformat(0.120, 3), \"one two zero\", \"one hundred and twenty\"], \n[dpformat(0.230, 3), \"two three zero\", \"two hundred and thirty\"], \n[dpformat(0.340, 3), \"three four zero\", \"three hundred and forty\"],\n[dpformat(0.450, 3), \"four five zero\", \"four hundred and fifty\"],\n[dpformat(0.501, 3), \"five zero one\", \"five hundred and one\"], \n[dpformat(0.602, 3), \"six zero two\", \"six hundred and two\"],\n[dpformat(0.703, 3), \"seven zero three\", \"seven hundred and three\"],\n[dpformat(0.804, 3), \"eight zero four\", \"eight hundred and four\"],\n[dpformat(0.905, 3), \"nine zero five\", \"nine hundred and five\"]\n )])\n ", "description": "", "templateType": "anything", "can_override": false}, "placedig": {"name": "placedig", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "templateType": "anything", "can_override": false}, "hundredths": {"name": "hundredths", "group": "Ungrouped variables", "definition": "if(len(pron[0][0])=4,dec(pron[0][0])*100,dec(pron[1][0])*100)", "description": "", "templateType": "anything", "can_override": false}, "pron2": {"name": "pron2", "group": "Ungrouped variables", "definition": "random(\n[dpformat(0.100, 3), \"zero point one zero zero\", \"zero point one hundred\"], \n[dpformat(0.200, 3), \"zero point two zero zero\", \"zero point two hundred\"], \n[dpformat(0.300, 3), \"zero point three zero zero\", \"zero point three hundred\"],\n[dpformat(0.400, 3), \"zero point four zero zero\", \"zero point four hundred\"],\n[dpformat(0.500, 3), \"zero point five zero zero\", \"zero point five hundred\"], \n[dpformat(0.600, 3), \"zero point six zero zero\", \"zero point six hundred\"],\n[dpformat(0.700, 3), \"zero point seven zero zero\", \"zero point seven hundred\"],\n[dpformat(0.800, 3), \"zero point eight zero zero\", \"zero point eight hundred\"],\n[dpformat(0.900, 3), \"zero point nine zero zero\", \"zero point nine hundred\"],\n[dpformat(0.120, 3), \"zero point one two zero\", \"zero point one hundred and twenty\"], \n[dpformat(0.230, 3), \"zero point two three zero\", \"zero point two hundred and thirty\"], \n[dpformat(0.340, 3), \"zero point three four zero\", \"zero point three hundred and forty\"],\n[dpformat(0.450, 3), \"zero point four five zero\", \"zero point four hundred and fifty\"],\n[dpformat(0.501, 3), \"zero point five zero one\", \"zero point five hundred and one\"], \n[dpformat(0.602, 3), \"zero point six zero two\", \"zero point six hundred and two\"],\n[dpformat(0.703, 3), \"zero point seven zero three\", \"zero point seven hundred and three\"],\n[dpformat(0.804, 3), \"zero point eight zero four\", \"zero point eight hundred and four\"],\n[dpformat(0.905, 3), \"zero point nine zero five\", \"zero point nine hundred and five\"]\n )", "description": "", "templateType": "anything", "can_override": false}, "thousandths": {"name": "thousandths", "group": "Ungrouped variables", "definition": "if(len(pron[0][0])=5,dec(pron[0][0])*1000,dec(pron[1][0])*1000)", "description": "", "templateType": "anything", "can_override": false}, "identify": {"name": "identify", "group": "Ungrouped variables", "definition": "if(len(pron[0][0])=5,[[1000,dec(pron[0][0])*1000],[100, dec(pron[1][0])*100]],[[100, dec(pron[0][0])*100],[1000,dec(pron[1][0])*1000]])", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["placedig", "place", "pron", "identify", "thousandths", "hundredths", "pron2"], "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": "

The digit $\\var{placedig}$ in the decimal $\\var{place[0]}$ represents [[0]]

", "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": "

The decimal 0.1 is also known as \"one tenth\" (notice you need ten of them to make a whole).

The decimal 0.01 is also known as \"one hundredth\" (notice you need a hundred of them to make a whole).

The decimal 0.001 is also known as \"one thousandth\" (notice you need a thousand of them to make a whole).

That is, the digit $\\var{placedig}$ in the decimal $\\var{place[0]}$ is in the {place[3]} column and so represents $\\var{placedig}$ {place[3]}.

"}], "gaps": [{"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, "minMarks": 0, "maxMarks": "1", "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["

$\\var{placedig}$ {place[1]}

", "

$\\var{placedig}$ {place[2]}

", "

$\\var{placedig}$ {place[3]}

"], "matrix": [0, 0, "1"], "distractors": ["", "", ""]}], "sortAnswers": false}, {"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": "

The decimal $\\var{pron[0][0]}$ should be read as [[0]]

or, as $\\var{identify[0][1]}$ [[1]].

", "stepsPenalty": "2", "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": "

Say each digit individually after the decimal point.

It makes no sense to call 0.500, \"zero point five hundred\" since that sounds a lot bigger than \"zero point five\", or \"zero point fifty\", but these are all equal to the same number! Pronouncing decimals like this is misleading and doesn't help with your intuition. However, this decimal is 500 of something, it is 500 thousandths! But be careful, even reading it that way can be ambiguous when it's read aloud.

That is, $\\var{pron[0][0]}$ is read as {pron[0][1]}.

Alternatively, since the last digit written is in the hundredthsthousandths column, we can think of this (and read it out) as $\\var{identify[0][1]}$ hundredths.thousandths.

"}], "gaps": [{"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, "minMarks": 0, "maxMarks": "1", "shuffleChoices": true, "displayType": "dropdownlist", "displayColumns": 0, "showCellAnswerState": true, "choices": ["

zero point {pron[0][1]}

", "

zero point {pron[0][2]}

The decimal $\\var{pron[1][0]}$ should be read as [[0]]

or, as $\\var{identify[1][1]}$ [[1]].

Say each digit individually after the decimal point.

That is, $\\var{pron[1][0]}$ is read as {pron[1][1]}.

Alternatively, since the last digit written is in the hundredthsthousandths column, we can think of this as $\\var{identify[1][1]}$ hundredths.thousandths.

"}], "gaps": [{"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, "minMarks": 0, "maxMarks": "1", "shuffleChoices": true, "displayType": "dropdownlist", "displayColumns": 0, "showCellAnswerState": true, "choices": ["

zero point {pron[1][1]}

", "

zero point {pron[1][2]}

"], "matrix": ["1", 0], "distractors": ["", "Please see the steps"]}, {"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, "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "dropdownlist", "displayColumns": 0, "showCellAnswerState": true, "choices": ["tenths", "hundredths", "thousandths"], "matrix": [0, "if(identify[1][0]=100,1,0)", "if(identify[1][0]=1000,1,0)"], "distractors": ["", "", ""]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Decimals: common misconceptions", "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/"}], "variable_groups": [], "variables": {"fnzdigbig": {"group": "Ungrouped variables", "description": "", "definition": "random(fnzdigsmall+2..9)", "templateType": "anything", "name": "fnzdigbig"}, "trailshort": {"group": "Ungrouped variables", "description": "", "definition": "random(0.1..0.9#0.1)", "templateType": "anything", "name": "trailshort"}, "fnz": {"group": "Ungrouped variables", "description": "", "definition": "random([[fnzdigsmall/10,fnzdigbig/100,1,0],[fnzdigbig/100,fnzdigsmall/10,0,1]])", "templateType": "anything", "name": "fnz"}, "trail": {"group": "Ungrouped variables", "description": "", "definition": "shuffle(['\\$\\\\var{trailshort}\\$','\\$\\\\var{trailshort}0\\$','\\$\\\\var{trailshort}00\\$','\\$\\\\var{trailshort}000\\$'])[0..2]", "templateType": "anything", "name": "trail"}, "fnzdigsmall": {"group": "Ungrouped variables", "description": "

first non-zero digit

", "definition": "random(1..6)", "templateType": "anything", "name": "fnzdigsmall"}}, "preamble": {"js": "", "css": ""}, "tags": [], "rulesets": {}, "statement": "

Use the drop-down menu to create the correct sentence.

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The trailing zeros do not change the value of a decimal. In the same way that $42$ is no different to $000042$ (regardless of how many zeros are placed at the front), $\\var{trailshort}$ is no different to $\\var{trailshort}0000$ (regardless of how many zeros are placed at the back). This is why it is important to read things such as $0.200$ as \"zero point two zero zero\" and not as \"zero point two hundred\".

In general, the length or number of digits in a decimal does not tell us anything about how big the decimal is. The only things that affect the actual value of a decimal are the non-zero digits and their placement relative to the decimal point (that is their face value and place value).

", "showCorrectAnswer": true, "scripts": {}, "customMarkingAlgorithm": "", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "unitTests": []}], "prompt": "

The number {trail[0]} is [[0]] {trail[1]}

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You may have suspected that $\\var{fnz[0]}$ was greater than $\\var{fnz[1]}$ simply because $\\var{fnzdigbig}$ was greater than $\\var{fnzdigsmall}$, however, $\\var{fnzdigbig}$ is in a column with a smaller place value!

You may have suspected that $\\var{fnz[0]}$ was less than $\\var{fnz[1]}$ simply because $\\var{fnzdigsmall}$ was less than $\\var{fnzdigbig}$, however, $\\var{fnzdigsmall}$ is in a column with a larger place value!

In general, the first non-zero digit does not tell us anything about how big the decimal is. The only things that affect the actual value of a decimal are the non-zero digits and their placement relative to the decimal point (that is their face value and place value).

You can add zeros so that the decimals have the same number of decimals places, and then, comparing them might be easier. That is, by appending a zero (which doesn't affect the value) onto the end of $\\var{fnzdigsmall/10}$ it might be clearer that $\\var{fnzdigsmall/10}0$ is greater than $\\var{fnzdigbig/100}$. Note that $\\var{fnzdigsmall/10}0$ is $\\var{fnzdigsmall}0$ hundredths whereas $\\var{fnzdigbig/100}$ is $\\var{fnzdigbig}$ hundredths.

", "showCorrectAnswer": true, "scripts": {}, "customMarkingAlgorithm": "", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "unitTests": []}], "prompt": "

The number $\\var{fnz[0]}$ is [[0]] $\\var{fnz[1]}$

", "extendBaseMarkingAlgorithm": true, "marks": 0, "scripts": {}, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "unitTests": []}], "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

Some students believe a decimal is larger if it is longer, some believe a decimal is larger if its first non-zero digit is larger.

"}, "advice": "

", "type": "question"}, {"name": "Decimals: Addition", "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": "

Decimals addition algorithm. 2 and 3 digit numbers. Carrying.

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

Write the following question down on paper and evaluate it without using a calculator.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"chunsumlastdigit": {"name": "chunsumlastdigit", "group": "Ungrouped variables", "definition": "mod(chunsum,10)", "description": "", "templateType": "anything", "can_override": false}, "threedigit1": {"name": "threedigit1", "group": "Ungrouped variables", "definition": "cdigs[0]/1000+cdigs[1]/100+cdigs[2]/10", "description": "", "templateType": "anything", "can_override": false}, "cunitsum": {"name": "cunitsum", "group": "Ungrouped variables", "definition": "cdigs[0]+cdigs[3]", "description": "", "templateType": "anything", "can_override": false}, "ctencarry": {"name": "ctencarry", "group": "Ungrouped variables", "definition": "floor(ctensum/10)", "description": "", "templateType": "anything", "can_override": false}, "cunitsumlastdigit": {"name": "cunitsumlastdigit", "group": "Ungrouped variables", "definition": "mod(cunitsum,10)", "description": "", "templateType": "anything", "can_override": false}, "cunitcarry": {"name": "cunitcarry", "group": "Ungrouped variables", "definition": "floor(cunitsum/10)", "description": "", "templateType": "anything", "can_override": false}, "chunsum": {"name": "chunsum", "group": "Ungrouped variables", "definition": "ctencarry+cdigs[2]+cdigs[5]", "description": "", "templateType": "anything", "can_override": false}, "ctensum": {"name": "ctensum", "group": "Ungrouped variables", "definition": "cdigs[1]+cdigs[4]+cunitcarry", "description": "", "templateType": "anything", "can_override": false}, "threedigit2": {"name": "threedigit2", "group": "Ungrouped variables", "definition": "cdigs[3]/1000+cdigs[4]/100+cdigs[5]/10", "description": "", "templateType": "anything", "can_override": false}, "ctensumlastdigit": {"name": "ctensumlastdigit", "group": "Ungrouped variables", "definition": "mod(ctensum,10)", "description": "", "templateType": "anything", "can_override": false}, "chuncarry": {"name": "chuncarry", "group": "Ungrouped variables", "definition": "floor(chunsum/10)", "description": "", "templateType": "anything", "can_override": false}, "cans": {"name": "cans", "group": "Ungrouped variables", "definition": "threedigit1+threedigit2", "description": "", "templateType": "anything", "can_override": false}, "cdigs": {"name": "cdigs", "group": "Ungrouped variables", "definition": "[0]+shuffle(3..9)", "description": "", "templateType": "anything", "can_override": false}, "cunitsumtensdigit": {"name": "cunitsumtensdigit", "group": "Ungrouped variables", "definition": "0.1*(cunitsum-cunitsumlastdigit)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["cdigs", "threedigit1", "threedigit2", "cans", "cunitsum", "cunitsumlastdigit", "cunitcarry", "ctensum", "ctensumlastdigit", "ctencarry", "chunsum", "chunsumlastdigit", "chuncarry", "cunitsumtensdigit"], "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{threedigit1}+\\var{threedigit2} = $ [[0]]

Generally, we set up $\\var{threedigit1}+\\var{threedigit2}$ with the decimal points lined up vertically so that the columns with the same place value are also lined up vertically:

\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

$0$	.	$\\var{cdigs[2]}$	$\\var{cdigs[1]}$		$+$
$0$	.	$\\var{cdigs[5]}$	$\\var{cdigs[4]}$	$\\var{cdigs[3]}$
			$\\phantom{0}$

Note that we can pad out the decimal with zeros if we prefer:

\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

$0$	.	$\\var{cdigs[2]}$	$\\var{cdigs[1]}$	$\\color{red}{\\var{cdigs[0]}}$	$+$
$0$	.	$\\var{cdigs[5]}$	$\\var{cdigs[4]}$	$\\var{cdigs[3]}$
			$\\phantom{0}$

Now we add the digits in the column to the far right (in this case, the thousandths column).

This results in $\\var{cunitsum}$ and so we place $\\var{cunitsumlastdigit}$ under the line in this column.

This results in $\\var{cunitsum}$ and so we place $\\var{cunitsumlastdigit}$ under the line in this column and carry the $1$ into the next column to the left.

\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

$\\overset{\\phantom{1}}{0}$	.	$\\overset{\\phantom{1}}{\\var{cdigs[2]}}$	$\\overset{\\color{red}1}{\\var{cdigs[1]}}$ $\\overset{\\phantom{0}}{\\var{cdigs[1]}}$	$\\color{green}{\\overset{\\phantom{1}}{\\var{cdigs[0]}}}$	$+$
$0$	.	$\\var{cdigs[5]}$	$\\var{cdigs[4]}$	$\\color{green}{\\var{cdigs[3]}}$
				$\\color{red}{\\var{cunitSumLastDigit}}$

Now we add the digits in the next column to the left (in this case, the hundredths column).

This results in $\\var{ctenSum}$ and so we place $\\var{ctenSumlastdigit}$ under the line in this column.

This results in $\\var{ctenSum}$ and so we place $\\var{ctenSumlastdigit}$ under the line in this column and carry the $1$ into the next column to the left.

\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

$\\overset{\\phantom{1}}{0}$	.	$\\overset{\\color{red}{1}}{\\var{cdigs[2]}}$ $\\overset{\\phantom{1}}{\\var{cdigs[2]}}$	$\\color{green}{\\overset{1}{\\var{cdigs[1]}}}$ $\\color{green}{\\overset{\\phantom{0}}{\\var{cdigs[1]}}}$	$\\overset{\\phantom{1}}{\\var{cdigs[0]}}$	$+$
$0$	.	$\\var{cdigs[5]}$	$\\color{green}{\\var{cdigs[4]}}$	$\\var{cdigs[3]}$
			$\\color{red}{\\var{ctenSumlastdigit}}$	${\\var{cunitSumLastDigit}}$

Now we add the digits in the next column to the left (in this case, the tenths column).

This is $\\var{chunsum}$ so we place $\\var{chunsum}$ under the line in this column.

This is $\\var{chunsum}$ so we place $\\var{chunsumlastdigit}$ under the line in this column and carry $\\var{chuncarry}$ into the next column to the left (which in this case is the ones column on the other side of the decimal point).

\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

$\\overset{\\color{red}1}{0}$ $\\overset{\\phantom{1}}{0}$	.	$\\color{green}{\\overset{1}{\\var{cdigs[2]}}}$ $\\color{green}{\\overset{\\phantom{1}}{\\var{cdigs[2]}}}$	$\\overset{1}{\\var{cdigs[1]}}$ $\\overset{\\phantom{0}}{\\var{cdigs[1]}}$	$\\overset{\\phantom{1}}{\\var{cdigs[0]}}$	$+$
$0$	.	$\\color{green}{\\var{cdigs[5]}}$	$\\var{cdigs[4]}$	$\\var{cdigs[3]}$
	.	$\\color{red}{\\var{chunsumlastdigit}}$	$\\var{ctenSumlastdigit}$	${\\var{cunitSumLastDigit}}$

Now we add the digits in the next column to the left (in this case, the ones column).

This is just $0$ so we place $0$ under the line in this column.

This is just $\\var{chuncarry}$ so we place $\\var{chuncarry}$ under the line in this column.

\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

$\\color{green}{\\overset{1}{0}}$ $\\color{green}{\\overset{\\phantom{1}}{0}}$	.	$\\overset{1}{\\var{cdigs[2]}}$ $\\overset{\\phantom{1}}{\\var{cdigs[2]}}$	$\\overset{1}{\\var{cdigs[1]}}$ $\\overset{\\phantom{0}}{\\var{cdigs[1]}}$	$\\overset{\\phantom{1}}{\\var{cdigs[0]}}$	$+$
$\\color{green}{0}$	.	$\\var{cdigs[5]}$	$\\var{cdigs[4]}$	$\\var{cdigs[3]}$
$\\color{red}{\\var{chuncarry}}$	.	$\\var{chunsumlastdigit}$	$\\var{ctenSumlastdigit}$	${\\var{cunitSumLastDigit}}$

The answer is therefore $\\var{cans}$.

Subtracting a decimal with 3 decimal places from a decimal with 2 or 3 decimal places. borrowing is necessary. This was modified from a subtraction question using integers with each number divided by 1000 so the variables have names referring to ones, tens, hundreds etc.

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

Write the following question down on paper and evaluate it without using a calculator.

", "advice": "

Write the following question down on paper and evaluate it without using a calculator.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"tendiff": {"name": "tendiff", "group": "c", "definition": "if(unitdiff>=0,top[1]-bot[1],top[1]-1-bot[1])", "description": "", "templateType": "anything", "can_override": false}, "hundiff": {"name": "hundiff", "group": "c", "definition": "if(tendiff>=0,top[2]-bot[2],top[2]-1-bot[2])", "description": "", "templateType": "anything", "can_override": false}, "newtopten": {"name": "newtopten", "group": "c", "definition": "if(unitdiff>=0,top[1],top[1]-1)", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "c", "definition": "topnum-botnum", "description": "", "templateType": "anything", "can_override": false}, "newtophun": {"name": "newtophun", "group": "c", "definition": "if(tendiff>=0,top[2],top[2]-1)", "description": "", "templateType": "anything", "can_override": false}, "anshun": {"name": "anshun", "group": "c", "definition": "mod(floor(ans*10),10)", "description": "", "templateType": "anything", "can_override": false}, "topnum": {"name": "topnum", "group": "c", "definition": "top[0]/1000+top[1]/100+top[2]/10", "description": "", "templateType": "anything", "can_override": false}, "ansunit": {"name": "ansunit", "group": "c", "definition": "mod(ans*1000,10)", "description": "", "templateType": "anything", "can_override": false}, "unitdiff": {"name": "unitdiff", "group": "c", "definition": "top[0]-bot[0]", "description": "", "templateType": "anything", "can_override": false}, "bot": {"name": "bot", "group": "c", "definition": "if(top[0]<>0, \n random(\n [0, random(top[1]+1..9), random(1..top[2]-1)],\n [random(top[0]+1..9), random(0..9), random(1..top[2]-1)]), \n [random(1..9), random(2..9), 1])\n\n//original \n//random(\n//if(top[0]<9,[random(top[0]+1..9), random(top[1]..9), random(1..top[2]-1)],if(top[1]<9,[random(top[0]..9), random(top[1]+1..9), random(1..top[2]-1)]),\"error\"),\n//if(top[1]<9,[random(0..top[0]), random(top[1]+1..9), random(1..top[2]-1)],if(top[1]=9,[random(top[0]..9), random(0..9), random(1..top[2]-1)]),\"error\")\n//)\n\n", "description": "

This should force some borrowing and paying back, and that the final answer is positive.

", "templateType": "anything", "can_override": false}, "top": {"name": "top", "group": "c", "definition": "random([random(1..8),random(1..8),random(2..9)],[0,random(1..9),random(2..9)])", "description": "

the digits of a 2 or 3 decimal place number

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$\\var{topnum}-\\var{botnum} = $ [[0]]

Generally we set up $\\var{topnum}-\\var{botnum}$ with the decimal points lined up vertically so that the columns with the same place value are also lined up vertically. We also pad out the decimals with zeros:

\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

$0$	.	$\\var{top[2]}$	$\\var{top[1]}$	$\\color{red}{\\var{top[0]}}$ $\\var{top[0]}$	$-$
$0$	.	$\\var{bot[2]}$	$\\var{bot[1]}$	$\\color{red}{\\var{bot[0]}}$ $\\var{bot[0]}$
	.		$\\phantom{0}$

Now we try to subtract the digits in the column to the far right (in this case, the thousandths column).

Since this is $\\var{ansunit}$ we write $\\var{ansunit}$ under the line in this column.

Since we can't take $\\var{bot[0]}$ away from $\\var{top[0]}$ (without using negative numbers) we borrow from the next column to the left (in this case, the hundredths column). This means we cross out the $\\var{top[1]}$ in the hundredths column and replace it with a $\\var{top[1]-1}$, and the $\\var{top[0]}$ becomes a $\\var{10+top[0]}$. Now we can do $\\var{10+top[0]}-\\var{bot[0]}$, and write the result, $\\var{ansunit}$, under the line in the thousandths column.

\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

$\\overset{\\phantom{1}}{0}$	.	$\\overset{\\phantom{1}}{\\var{top[2]}}$	$\\overset{\\color{red}{\\var{newtopten}}}{\\var{top[1]}\\mkern-7.5mu\\color{red}/}$ $\\overset{\\phantom{1}}{\\var{top[1]}}$	$\\color{red}{^1}\\overset{\\phantom{1}}{\\color{green}{\\var{top[0]}}}$ $\\overset{\\phantom{1}}{\\color{green}{\\var{top[0]}}}$	$-$
$0$	.	$\\var{bot[2]}$	$\\var{bot[1]}$	$\\color{green}{\\var{bot[0]}}$
	.			$\\color{red}{\\var{ansunit}}$

Now we try to subtract the digits in the hundredths column.

Since this is $\\var{ansten}$ we write $\\var{ansten}$ under the line in this column.

Since we can't take $\\var{bot[1]}$ away from $\\var{newtopten}$ (without using negative numbers) we borrow from the next column to the left (in this case, the tenths column). This means we cross out the $\\var{top[2]}$ in the tenths column and replace it with a $\\var{top[2]-1}$, and the $\\var{newtopten}$ in the hundredths column becomes a $\\var{10+newtopten}$. Now we can do $\\var{10+newtopten}-\\var{bot[1]}$, and write the result, $\\var{ansten}$, under the line in the hundredths.

\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

$\\overset{\\phantom{1}}{0}$	.	\n $\\overset{\\color{red}{\\var{newtophun}}}{\\var{top[2]}\\mkern-7.5mu\\color{red}{/}}$ $\\overset{\\phantom{1}}{\\var{top[2]}}$ \n	\n $\\overset{\\color{red}{1}\\color{green}{\\var{newtopten}}}{\\var{top[1]}\\mkern-7.5mu/}$ $\\overset{\\color{green}{\\var{newtopten}}}{\\var{top[1]}\\mkern-7.5mu/}$ $\\color{red}{^1}\\overset{\\phantom{1}}{\\color{green}{\\var{top[1]}}}$ $\\overset{\\phantom{1}}{\\color{green}{\\var{top[1]}}}$ \n	${^1}\\overset{\\phantom{1}}{\\var{top[0]}}$ $\\overset{\\phantom{1}}{\\var{top[0]}}$	$-$
$0$	.	$\\var{bot[2]}$	$\\color{green}{\\var{bot[1]}}$	$\\var{bot[0]}$
	.		$\\color{red}{\\var{ansten}}$	$\\var{ansunit}$

Now we try to subtract the digits in the tenths column and then subtract the digits in the ones column.

\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

$\\overset{\\phantom{1}}{\\color{green}{0}}$	.	\n $\\overset{\\color{green}{\\var{newtophun}}}{\\var{top[2]}\\mkern-7.5mu/}$ $\\overset{\\phantom{1}}{\\color{green}{\\var{top[2]}}}$ \n	\n $\\overset{{1}\\var{newtopten}}{\\var{top[1]}\\mkern-7.5mu/}$ $\\overset{\\var{newtopten}}{\\var{top[1]}\\mkern-7.5mu/}$ ${^1}\\overset{\\phantom{1}}{\\var{top[1]}}$ $\\overset{\\phantom{1}}{\\var{top[1]}}$ \n	${^1}\\overset{\\phantom{1}}{\\var{top[0]}}$ $\\overset{\\phantom{1}}{\\var{top[0]}}$	$-$
$\\color{green}0$	.	$\\color{green}{\\var{bot[2]}}$	$\\var{bot[1]}$	$\\var{bot[0]}$
$\\color{red}0$	.	$\\color{red}{\\var{anshun}}$	$\\var{ansten}$	$\\var{ansunit}$

The answer is therefore $\\var{ans}$.

Write the following question down on paper and evaluate it without using a calculator.

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$\\var{poweroften[0]}$ has {pronpower[0]} $0$s after the leading $1$. This means to evaluate $\\var{dec1}\\times \\var{poweroften[0]}$ we just move the decimal point in $\\var{dec1}$ {pronpower[0]} decimal places to the right (to make the decimal $\\var{poweroften[0]}$ times bigger) and get $\\var{ans1}$.

", "scripts": {}, "variableReplacementStrategy": "originalfirst", "type": "information", "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "marks": 0, "variableReplacements": []}], "scripts": {}, "unitTests": [], "sortAnswers": false, "prompt": "

$\\var{dec1}\\times \\var{poweroften[0]}=$ [[0]]

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$\\var{poweroften[1]}$ has {pronpower[1]} $0$s after the leading $1$. This means to evaluate $\\var{dec2}\\times \\var{poweroften[1]}$ we just move the decimal point in $\\var{dec2}$ {pronpower[1]} decimal places to the right (to make the decimal $\\var{poweroften[1]}$ times bigger) and get $\\var{ans2}$.

$\\var{dec2}\\times\\var{poweroften[1]}=$ [[0]]

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$\\var{poweroften[2]}$ has {pronpower[2]} $0$s after the leading $1$. This means to evaluate $\\var{dec3}\\div \\var{poweroften[2]}$ we just move the decimal point in $\\var{dec3}$ {pronpower[2]} decimal places to the left (to make the decimal smaller) and get $\\var{ans3}$.

$\\var{dec3}\\div\\var{poweroften[2]}=$ [[0]]

", "extendBaseMarkingAlgorithm": true}, {"showFeedbackIcon": true, "gaps": [{"showFeedbackIcon": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerFraction": false, "variableReplacements": [], "type": "numberentry", "customMarkingAlgorithm": "", "showCorrectAnswer": true, "marks": 1, "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "allowFractions": false, "mustBeReduced": false, "scripts": {}, "correctAnswerStyle": "plain", "minValue": "ans4", "unitTests": [], "maxValue": "ans4", "extendBaseMarkingAlgorithm": true}], "variableReplacements": [], "type": "gapfill", "customMarkingAlgorithm": "", "showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "stepsPenalty": "1", "steps": [{"showFeedbackIcon": true, "unitTests": [], "customMarkingAlgorithm": "", "prompt": "

Recall, the fraction bar simply denotes division.

$\\var{poweroften[3]}$ has {pronpower[3]} $0$s after the leading $1$. This means to evaluate $\\var{dec4}\\div \\var{poweroften[3]}$ we just move the decimal point in $\\var{dec4}$ {pronpower[3]} decimal places to the left (to make the decimal smaller) and get $\\var{ans4}$.

$\\displaystyle \\frac{\\var{dec4}}{\\var{poweroften[3]}}=$ [[0]]

", "extendBaseMarkingAlgorithm": true}], "ungrouped_variables": ["dec1", "dec2", "dec3", "dec4", "power", "poweroften", "ans1", "ans2", "ans3", "ans4", "pronpower"], "rulesets": {}, "preamble": {"js": "", "css": ""}, "tags": [], "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

By powers of ten I mean a 1 followed by some 0s. The scientific notation questions will take care of the power of ten notation.

"}, "functions": {}, "advice": "

Multiplying or dividing by a power of ten (such as a $1$ followed by some $0$s) moves the decimal point. Multiplying moves the decimal point to make the number bigger (that is, to the right). Dividing moves the decimal to make the number smaller (that is, to the left). The number of $0$s indicates the number of places you should move the decimal place.

", "variables": {"ans4": {"name": "ans4", "description": "", "group": "Ungrouped variables", "definition": "dec4/poweroften[3]", "templateType": "anything"}, "dec4": {"name": "dec4", "description": "", "group": "Ungrouped variables", "definition": "random(list(0.111..0.499#0.001))", "templateType": "anything"}, "ans3": {"name": "ans3", "description": "

ans3

", "group": "Ungrouped variables", "definition": "dec3/poweroften[2]", "templateType": "anything"}, "dec1": {"name": "dec1", "description": "", "group": "Ungrouped variables", "definition": "random(list(0.111..0.999#0.001))*10", "templateType": "anything"}, "ans1": {"name": "ans1", "description": "", "group": "Ungrouped variables", "definition": "dec1*poweroften[0]", "templateType": "anything"}, "poweroften": {"name": "poweroften", "description": "", "group": "Ungrouped variables", "definition": "map(10^n,n,power)", "templateType": "anything"}, "power": {"name": "power", "description": "", "group": "Ungrouped variables", "definition": "shuffle([1,2,3,4])", "templateType": "anything"}, "pronpower": {"name": "pronpower", "description": "", "group": "Ungrouped variables", "definition": "[switch(power[0]=2,'two',power[0]=3,'three',power[0]=4,'four','one'),switch(power[1]=2,'two',power[1]=3,'three',power[1]=4,'four','one'),switch(power[2]=2,'two',power[2]=3,'three',power[2]=4,'four','one'),switch(power[3]=2,'two',power[3]=3,'three',power[3]=4,'four','one')]", "templateType": "anything"}, "dec2": {"name": "dec2", "description": "", "group": "Ungrouped variables", "definition": "random(list(0.011..0.099#0.0001))", "templateType": "anything"}, "dec3": {"name": "dec3", "description": "", "group": "Ungrouped variables", "definition": "random(list(0.500..0.999#0.001))*10", "templateType": "anything"}, "ans2": {"name": "ans2", "description": "", "group": "Ungrouped variables", "definition": "dec2*poweroften[1]", "templateType": "anything"}}, "type": "question"}, {"name": "Decimals: Multiplication", "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": "

a) Multiplying decimals with a single non-zero digit. Students are told to preserve the number of decimal places (from the question to the answer).

b) Multiplying decimals requiring the multiplication algorithm.

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

", "advice": "

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"ab0t0last": {"name": "ab0t0last", "group": "2digit", "definition": "mod(ab0t0,10)", "description": "", "templateType": "anything", "can_override": false}, "abot": {"name": "abot", "group": "2digit", "definition": "if(adigs[2]<>0,[adigs[3],adigs[2]],[adigs[2],adigs[3]])", "description": "

abot

", "templateType": "anything", "can_override": false}, "ab1t0last": {"name": "ab1t0last", "group": "2digit", "definition": "mod(ab1t0,10)", "description": "", "templateType": "anything", "can_override": false}, "ab0t1carry": {"name": "ab0t1carry", "group": "2digit", "definition": "(ab0t1pluscarry-ab0t1last)/10", "description": "", "templateType": "anything", "can_override": false}, "adps": {"name": "adps", "group": "2digit", "definition": "log(afactprod)", "description": "", "templateType": "anything", "can_override": false}, "adec1": {"name": "adec1", "group": "2digit", "definition": "atopnum/afact1", "description": "", "templateType": "anything", "can_override": false}, "afact1": {"name": "afact1", "group": "2digit", "definition": "random(10,100,1000,10000)", "description": "", "templateType": "anything", "can_override": false}, "adpsword": {"name": "adpsword", "group": "2digit", "definition": "switch(adps=6,\"six\", adps=5, \"five\", adps=4,\"four\", adps=3,\"three\", adps=2, \"two\", adps=1,\"one\",adps=7,\"seven\",adps)", "description": "", "templateType": "anything", "can_override": false}, "ab1t1carry": {"name": "ab1t1carry", "group": "2digit", "definition": "(ab1t1pluscarry-ab1t1last)/10", "description": "", "templateType": "anything", "can_override": false}, "adigs": {"name": "adigs", "group": "2digit", "definition": "shuffle(1..9)[0..4]", "description": "

we want distinct digits so it is easier to refer to digits unambiguously.

", "templateType": "anything", "can_override": false}, "ab0t1": {"name": "ab0t1", "group": "2digit", "definition": "abot[0]*atop[1]", "description": "", "templateType": "anything", "can_override": false}, "ab1t1": {"name": "ab1t1", "group": "2digit", "definition": "abot[1]*atop[1]", "description": "", "templateType": "anything", "can_override": false}, "ab0t1last": {"name": "ab0t1last", "group": "2digit", "definition": "mod(ab0t1pluscarry,10)", "description": "", "templateType": "anything", "can_override": false}, "ab0t1pluscarry": {"name": "ab0t1pluscarry", "group": "2digit", "definition": "ab0t1+ab0t0carry", "description": "

", "templateType": "anything", "can_override": false}, "aans": {"name": "aans", "group": "2digit", "definition": "atopnum*abotnum", "description": "", "templateType": "anything", "can_override": false}, "easydigprod": {"name": "easydigprod", "group": "Ungrouped variables", "definition": "easydig1*easydig2", "description": "", "templateType": "anything", "can_override": false}, "easyans": {"name": "easyans", "group": "Ungrouped variables", "definition": "easydigprod/(easyfactprod)", "description": "

eas

", "templateType": "anything", "can_override": false}, "easyfactprod": {"name": "easyfactprod", "group": "Ungrouped variables", "definition": "easyfact1*easyfact2", "description": "", "templateType": "anything", "can_override": false}, "easyfact2": {"name": "easyfact2", "group": "Ungrouped variables", "definition": "if(easyfact1=1,random(10,100,1000),random(1,10,100,1000))", "description": "", "templateType": "anything", "can_override": false}, "easy1": {"name": "easy1", "group": "Ungrouped variables", "definition": "easydig1/easyfact1", "description": "", "templateType": "anything", "can_override": false}, "afactprod": {"name": "afactprod", "group": "2digit", "definition": "afact1*afact2", "description": "", "templateType": "anything", "can_override": false}, "dps": {"name": "dps", "group": "Ungrouped variables", "definition": "log(easyfactprod)", "description": "", "templateType": "anything", "can_override": false}, "ab0t0carry": {"name": "ab0t0carry", "group": "2digit", "definition": "(ab0t0-ab0t0last)/10", "description": "", "templateType": "anything", "can_override": false}, "aanstho": {"name": "aanstho", "group": "2digit", "definition": "mod((aans-aansone-aansten*10-aanshun*100)/1000,10)", "description": "", "templateType": "anything", "can_override": false}, "adec2": {"name": "adec2", "group": "2digit", "definition": "abotnum/afact2", "description": "", "templateType": "anything", "can_override": false}, "atopnum": {"name": "atopnum", "group": "2digit", "definition": "atop[1]*10+atop[0]", "description": "", "templateType": "anything", "can_override": false}, "ab1t1pluscarry": {"name": "ab1t1pluscarry", "group": "2digit", "definition": "ab1t1+ab1t0carry", "description": "", "templateType": "anything", "can_override": false}, "asum1": {"name": "asum1", "group": "2digit", "definition": "abot[0]*atopnum", "description": "", "templateType": "anything", "can_override": false}, "ab0t0": {"name": "ab0t0", "group": "2digit", "definition": "atop[0]*abot[0]", "description": "", "templateType": "anything", "can_override": false}, "easy2": {"name": "easy2", "group": "Ungrouped variables", "definition": "easydig2/easyfact2", "description": "", "templateType": "anything", "can_override": false}, "asum2": {"name": "asum2", "group": "2digit", "definition": "10*abot[1]*atopnum", "description": "

sum2

", "templateType": "anything", "can_override": false}, "aansten": {"name": "aansten", "group": "2digit", "definition": "mod((aans-aansone)/10,10)", "description": "", "templateType": "anything", "can_override": false}, "aansone": {"name": "aansone", "group": "2digit", "definition": "mod(aans,10)", "description": "", "templateType": "anything", "can_override": false}, "ab1t0carry": {"name": "ab1t0carry", "group": "2digit", "definition": "(ab1t0-ab1t0last)/10", "description": "", "templateType": "anything", "can_override": false}, "abotnum": {"name": "abotnum", "group": "2digit", "definition": "abot[1]*10+abot[0]", "description": "

botnum

", "templateType": "anything", "can_override": false}, "atop": {"name": "atop", "group": "2digit", "definition": "if(adigs[0]<>0,[adigs[1],adigs[0]],[adigs[0],adigs[1]])", "description": "", "templateType": "anything", "can_override": false}, "ab1t0": {"name": "ab1t0", "group": "2digit", "definition": "abot[1]*atop[0]", "description": "", "templateType": "anything", "can_override": false}, "aanshun": {"name": "aanshun", "group": "2digit", "definition": "mod((aans-aansone-aansten*10)/100,10)", "description": "", "templateType": "anything", "can_override": false}, "adecans": {"name": "adecans", "group": "2digit", "definition": "aans/afactprod", "description": "", "templateType": "anything", "can_override": false}, "afact2": {"name": "afact2", "group": "2digit", "definition": "if(afact1=10000,random(10,100,1000),random(10,100,1000,10000))", "description": "", "templateType": "anything", "can_override": false}, "easyfact1": {"name": "easyfact1", "group": "Ungrouped variables", "definition": "random(1,10,100,1000)", "description": "", "templateType": "anything", "can_override": false}, "ab1t1last": {"name": "ab1t1last", "group": "2digit", "definition": "mod(ab1t1pluscarry,10)", "description": "", "templateType": "anything", "can_override": false}, "easydig1": {"name": "easydig1", "group": "Ungrouped variables", "definition": "random(3..9)", "description": "", "templateType": "anything", "can_override": false}, "easydig2": {"name": "easydig2", "group": "Ungrouped variables", "definition": "if(easydig1=3,random(4..9),random(3..9))", "description": "", "templateType": "anything", "can_override": false}, "dpsword": {"name": "dpsword", "group": "Ungrouped variables", "definition": "switch(dps=6,\"six\", dps=5, \"five\", dps=4,\"four\", dps=3,\"three\", dps=2, \"two\", dps=1,\"one\",dps)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["easydig1", "easydig2", "easyfact1", "easyfact2", "easydigprod", "easyfactprod", "easy1", "easy2", "easyans", "dps", "dpsword"], "variable_groups": [{"name": "2digit", "variables": ["adigs", "atop", "abot", "atopnum", "abotnum", "aans", "ab0t0", "ab0t0last", "ab0t0carry", "ab0t1", "ab0t1pluscarry", "ab0t1last", "ab0t1carry", "ab1t0", "ab1t0last", "ab1t0carry", "ab1t1", "ab1t1pluscarry", "ab1t1last", "ab1t1carry", "asum1", "asum2", "aansone", "aansten", "aanshun", "aanstho", "afact1", "afact2", "afactprod", "adps", "adpsword", "adec1", "adec2", "adecans"]}], "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{easy1}\\times \\var{easy2}= $ [[0]]

Remove the decimal points, do the multiplication of whole numbers, then put the decimal place in the answer so that the number of decimal places in the question and the answer are the same.

That is,

since $\\var{easydig1}\\times\\var{easydig2}=\\var{easydigprod}$, the answer will involve the digits $\\var{easydigprod}$,
since there are {dpsword} decimal places in the question, there will be {dpsword} decimal places in the answer, is only one decimal place in the question, there will be one decimal place in the question,

and therefore $\\var{easy1}\\times\\var{easy2}=\\var{easyans}0$. But note, we don't need to write the last zero so we could also write $\\var{easy1}\\times\\var{easy2}=\\var{easyans}$.

and therefore $\\var{easy1}\\times\\var{easy2}=\\var{easyans}$.

This procedure works because it is the following in disguise:

$\\begin{align}\\var{easy1}\\times\\var{easy2}&=\\frac{\\var{easydig1}}{\\var{easyfact1}}\\times\\frac{\\var{easydig2}}{\\var{easyfact2}}&&\\text{(convert the decimals to fractions)}\\\\&=\\frac{\\var{easydig1}\\times\\var{easydig2}}{\\var{easyfact1}\\times\\var{easyfact2}}&&\\text{(multiply the fractions)}\\\\&=\\frac{\\var{easydigprod}}{\\var{easyfactprod}}\\\\&=\\var{easyans}&&\\text{(convert back to a decimal)}\\end{align}$

$\\var{adec1}\\times\\var{adec2} = $ [[0]]

Remove the decimal points, do the multiplication of whole numbers, then put the decimal place in the answer so that the number of decimal places in the question and the answer are the same.

That is,

since $\\var{atopnum}\\times\\var{abotnum}=\\var{aans}$ (see the working below), the answer will involve the digits $\\var{aans}$,
since there are {adpsword} decimal places in the question, there will be {adpsword} decimal places in the answer, is only one decimal place in the question, there will be one decimal place in the question,

and therefore $\\var{adec1}\\times\\var{adec2}=\\var{adecans}0$. But note, we don't need to write the last zero so we could also write $\\var{adec1}\\times\\var{adec2}=\\var{adecans}$.

and therefore $\\var{adec1}\\times\\var{adec2}=\\var{adecans}00$. But note, we don't need to write the two trailing zeros so we could also write $\\var{adec1}\\times\\var{adec2}=\\var{adecans}$.

and therefore $\\var{adec1}\\times\\var{adec2}=\\var{adecans}$.

This procedure works because it is the following in disguise:

$\\begin{align}\\var{adec1}\\times\\var{adec2}&=\\frac{\\var{atopnum}}{\\var{afact1}}\\times\\frac{\\var{abotnum}}{\\var{afact2}}&&\\text{(convert the decimals to fractions)}\\\\&=\\frac{\\var{atopnum}\\times\\var{abotnum}}{\\var{afact1}\\times\\var{afact2}}&&\\text{(multiply the fractions)}\\\\&=\\frac{\\var{aans}}{\\var{afactprod}}\\\\&=\\var{adecans}&&\\text{(convert back to a decimal)}\\end{align}$

"}, {"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": "

How to calculate $\\var{atopnum}\\times\\var{abotnum}$

Generally we set up $\\var{atopnum}\\times\\var{abotnum}$ with the ones and tens columns lined up vertically:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{atop[1]}$	$\\var{atop[0]}$	$\\times$
$\\var{abot[1]}$	$\\var{abot[0]}$
$\\phantom{0}$

We need to multiply each digit in the bottom number by each digit in the top number whilst respecting their place values.

We multiply the digits in the ones column, that is, $\\color{green}{\\var{abot[0]}\\times \\var{atop[0]}}$.

Since this is just $\\var{ab0t0}$ we write $\\var{ab0t0}$ under the line in the ones column.

Since this is $\\var{ab0t0}$ we write the $\\var{ab0t0last}$ under the line in the ones column and carry the $\\var{ab0t0carry}$ into the tens column to be dealt with in the next step.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\overset{\\color{red}{\\var{ab0t0carry}}}{\\var{atop[1]}}$ $\\overset{\\phantom{1}}{\\var{atop[1]}}$	$\\color{green}{\\overset{\\phantom{1}}{\\var{atop[0]}}}$	$\\times$
$\\var{abot[1]}$	$\\color{green}{\\var{abot[0]}}$
	$\\color{red}{\\var{ab0t0last}}$

We now multiply diagonally, $\\color{green}{\\var{abot[0]}\\times \\var{atop[1]}}$.

This just gives us $\\var{ab0t1}$ so we write $\\var{ab0t1}$ under the line in the tens column.

This gives us $\\var{ab0t1}$ so we write this under the line with the $\\var{ab0t1last}$ in the tens column.

This gives us $\\var{ab0t1}$ but we have to add the $\\var{ab0t0carry}$ we carried earlier and so we write $\\var{ab0t1pluscarry}$ under the line with the $\\var{ab0t1last}$ in the tens column.

This gives us $\\var{ab0t1}$ but we have to add the $\\var{ab0t0carry}$ we carried earlier and so we write $\\var{ab0t1pluscarry}$ under the line in the tens column.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\color{green}{\\overset{{\\var{ab0t0carry}}}{\\var{atop[1]}}}$ $\\color{green}{\\overset{\\phantom{1}}{\\var{atop[1]}}}$	$\\overset{\\phantom{1}}{\\var{atop[0]}}$	$\\times$
	$\\var{abot[1]}$	$\\color{green}{\\var{abot[0]}}$
$\\color{red}{\\var{ab0t1carry}}$ $\\phantom{0}$	$\\color{red}{\\var{ab0t1last}}$	${\\var{ab0t0last}}$

We are now finished with the digit $\\var{abot[0]}$ and move on to work with the $\\var{abot[1]}$ in the tens column. Since this is really a $\\var{abot[1]*10}$ we place a zero in the ones column on the next line to pad our numbers out. We also crossout or erase any carry marks that we have used.

\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

	$\\overset{\\phantom{1}}{\\var{atop[1]}}$	$\\overset{\\phantom{1}}{\\var{atop[0]}}$	$\\times$
	$\\var{abot[1]}$	$\\var{abot[0]}$
${\\var{ab0t1carry}}$$\\phantom{0}$	${\\var{ab0t1last}}$	${\\var{ab0t0last}}$
		$\\color{red}0$

We now multiply along the other diagonal, that is, $\\color{green}{\\var{abot[1]}\\times\\var{atop[0]}}$.

Since this is just $\\var{ab1t0}$ we write $\\var{ab1t0}$ under the line in the tens column.

Since this is $\\var{ab1t0}$ we write the $\\var{ab1t0last}$ under the line in the tens column and carry the $\\var{ab1t0carry}$ into the tens column to be dealt with in the next step.

\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

	$\\overset{\\color{red}{\\var{ab1t0carry}}}{\\var{atop[1]}}$ $\\overset{\\phantom{1}}{\\var{atop[1]}}$	$\\color{green}{\\overset{\\phantom{1}}{\\var{atop[0]}}}$	$\\times$
	$\\color{green}{\\var{abot[1]}}$	$\\var{abot[0]}$
${\\var{ab0t1carry}}$$\\phantom{0}$	${\\var{ab0t1last}}$	${\\var{ab0t0last}}$
	$\\color{red}{\\var{ab1t0last}}$	${0}$

We now multiply the digits in the tens column, that is, $\\color{green}{\\var{abot[1]}\\times \\var{atop[1]}}$.

This just gives us $\\var{ab1t1}$ so we write $\\var{ab1t1}$ under the line in the hundreds column.

This gives us $\\var{ab1t1}$ so we write this under the line with the $\\var{ab1t1last}$ in the hundreds column.

This gives us $\\var{ab1t1}$ but we have to add the $\\var{ab1t0carry}$ we carried earlier and so we write $\\var{ab1t1pluscarry}$ under the line with the $\\var{ab1t1last}$ in the hundreds column.

This gives us $\\var{ab1t1}$ but we have to add the $\\var{ab1t0carry}$ we carried earlier and so we write $\\var{ab1t1pluscarry}$ under the line in the hundreds column.

\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\n\n\n\n

		$\\color{green}{\\overset{{\\var{ab1t0carry}}}{\\var{atop[1]}}}$ $\\color{green}{\\overset{\\phantom{1}}{\\var{atop[1]}}}$	$\\overset{\\phantom{1}}{\\var{atop[0]}}$	$\\times$
		$\\color{green}{\\var{abot[1]}}$	$\\var{abot[0]}$
	${\\var{ab0t1carry}}$$\\phantom{0}$	${\\var{ab0t1last}}$	${\\var{ab0t0last}}$
$\\color{red}{\\var{ab1t1carry}}$ $\\phantom{0}$	$\\color{red}{\\var{ab1t1last}}$	${\\var{ab1t0last}}$	${0}$

We now add the two results to get the total, that is, $\\color{green}{\\var{asum1}+\\var{asum2}}$.

\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\n\n\n\n\n\n\n\n\n\n\n

		$\\overset{{\\var{ab1t0carry}}}{\\var{atop[1]}}$ $\\overset{\\phantom{1}}{\\var{atop[1]}}$	$\\overset{\\phantom{1}}{\\var{atop[0]}}$	$\\times$
		$\\var{abot[1]}$	$\\var{abot[0]}$
	$\\color{green}{\\var{ab0t1carry}}$$\\phantom{0}$	$\\color{green}{\\var{ab0t1last}}$	$\\color{green}{\\var{ab0t0last}}$	$+$
$\\color{green}{\\var{ab1t1carry}}$ $\\phantom{0}$	$\\color{green}{\\var{ab1t1last}}$	$\\color{green}{\\var{ab1t0last}}$	$\\color{green}{0}$
$\\color{red}{\\var{aanstho}}$$\\phantom{0}$	$\\color{red}{\\var{aanshun}}$	$\\color{red}{\\var{aansten}}$	$\\color{red}{\\var{aansone}}$

The answer is therefore $\\var{aans}$.

Issues: alignment in columns in the working - not sure what to do about it

Decimal divided by a decimal. Multiply by a power of ten to get an integer divisor. Long and short division process. There is a remainder which we express as a decimal by continuing the division process. Rounding is required to some number of decimal places.

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

Write the following question down on paper and evaluate it without using a calculator.

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Was mod(floor(dividend1*100),10) but is now hard coded because there were rounding errors

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excluded 5 so that the decimal part is longer than 1 place.

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$\\var{givendividend}\\div\\var{givendivisor}=$[[0]] (2 decimal places)

We want to calculate $\\var{givendividend}\\div\\var{givendivisor}$. By the way, this is the same thing as $\\frac{\\var{givendividend}}{\\var{givendivisor}}$. Both of these expressions mean \"how many $\\var{givendivisor}$s go into $\\var{givendividend}$?\"

To work out such a thing we normally convert the division/fraction into an equivalent division/fraction where we are dividing by a whole number. We do this by multiplying both $\\var{givendividend}$ (the dividend) and $\\var{givendivisor}$ (the divisor) by $\\var{divisorscale}$ so that the decimal points are moved $\\var{divisorscaleorder}$ places to the right and so we get the division $\\var{dividend1}\\div\\var{divisor1}$.

Note we want the divisor to be a whole number but we don't need the dividend to be whole.

Why is this division equivalent? Think of division as sharing some amount equally amoungst some number of people. Now consider the scenario where you have $\\var{divisorscale}$ times the original amount to share but also $\\var{divisorscale}$ times the original number people, everyone will get the same amount as in the original scenario!

The following is the above working in terms of fractions:

\\begin{align}\\var{givendividend}\\div\\var{givendivisor}&=\\frac{\\var{givendividend}}{\\var{givendivisor}}\\\\[0.1cm]&=\\frac{\\var{givendividend}}{\\var{givendivisor}}\\times 1\\\\[0.1cm]&=\\frac{\\var{givendividend}}{\\var{givendivisor}}\\times\\frac{\\var{divisorscale}}{\\var{divisorscale}}\\\\[0.1cm]&=\\frac{\\var{givendividend}\\times \\var{divisorscale}}{\\var{givendivisor}\\times\\var{divisorscale}}\\\\[0.1cm]&=\\frac{\\var{dividend1}}{\\var{divisor1}}\\end{align}

Short Division

We want to calculate $\\var{dividend1}\\div\\var{divisor1}$. By the way, this is the same thing as $\\frac{\\var{dividend1}}{\\var{divisor1}}$. Both of these expressions mean \"how many $\\var{divisor1}$s go into $\\var{dividend1}$?\"

The short division algorithm allows you to work this out by working from the left to the right of $\\var{dividend1}$ whilst respecting place value. We normally set up the division in the following way:

$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dividend1}}$

Note the positions of the numbers!

Actually, since we want the answer to two decimal places we add as many zeroes after the decimal place to ensure we have three decimal places!

$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!\\var{dd2}\\var{dd1}\\var{dd0}}$

Why three? We use that extra digit to determine whether to round up or down.

We need to know the $\\var{divisor1}$ times tables or write the $\\var{divisor1}$ times tables out (by repeatedly adding $\\var{divisor1}$) so that we can refer to them.

\\[\\boxed{\\begin{align}1\\times\\var{divisor1}&=\\var{divisor1}\\\\2\\times\\var{divisor1}&=\\var{2*divisor1}\\\\3\\times\\var{divisor1}&=\\var{3*divisor1}\\\\4\\times\\var{divisor1}&=\\var{4*divisor1}\\\\5\\times\\var{divisor1}&=\\var{5*divisor1}\\\\6\\times\\var{divisor1}&=\\var{6*divisor1}\\\\7\\times\\var{divisor1}&=\\var{7*divisor1}\\\\8\\times\\var{divisor1}&=\\var{8*divisor1}\\\\9\\times\\var{divisor1}&=\\var{9*divisor1}\\end{align}}\\]

The ones column

Step 1: The first thing we ask ourselves is, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{dd3}}$?\"

Well, none! $\\var{divisor1}$ is too big to fit into $\\var{dd3}$. So we write $\\color{red}0$ above the $\\var{dd3}$ in the ones column: Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fit and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the ones column: Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fits and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the ones column:

$\\begin{array}{r} \\color{red}{\\var{qd3}\\,.\\phantom{^\\var{diff3}\\var{qd2}}\\,\\phantom{^{\\var{diff2}}\\var{qd1}^{\\var{diff1}}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}\\,.^\\phantom{\\var{diff3}}\\var{dd2}\\,^\\phantom{\\var{diff2}}\\var{dd1}^\\phantom{\\var{diff1}}\\var{dd0}} \\end{array}$

Step 2: We only made it to $\\var{prod3}$ but we were aiming for $\\var{dd3}$, so we were $\\var{diff3}$ away. This is the remainder (what remains to be divided) in the ones column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff3}}\\var{dd2}$ in the tenths column. We made it to exactly $\\var{dd3}$, so there is no remainder in the ones column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd2}$ in the tenths column. You do not need to do this but you can if you wish.

$\\begin{array}{r} {\\var{qd3}\\,.\\,\\phantom{^\\var{diff3}\\var{qd2}}\\,\\phantom{{^\\var{diff2}}\\var{qd1}\\,{^\\var{diff1}}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3}\\,. \\,^\\color{red}{\\var{diff3}}\\var{dd2}\\,\\phantom{^\\color{red}{\\var{diff2}}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

The tenths column

Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b2}}$?\" (note this $\\var{b2}$ actually represents $\\var{b2*0.1}$)

Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b2}$. So we write $\\color{red}0$ above the $\\var{dd2}$ in the tenths column: Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fit and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the tenths column: Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fits and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the tenths column:

$\\begin{array}{r} {\\var{qd3}\\,.\\,\\phantom{^\\var{diff3}}\\color{red}{\\var{qd2}}\\phantom{^\\var{diff2}\\var{qd1}\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,.\\,^\\color{green}{\\var{diff3}}\\color{green}{\\var{dd2}}\\phantom{^\\color{red}{\\var{diff2}}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

Step 2: We only made it to $\\var{prod2}$ but we were aiming for $\\var{b2}$, so we were $\\var{diff2}$ away. This is the remainder (what remains to be divided) in the tenths column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff2}}\\var{dd1}$ in the hundredths column. We made it to exactly $\\var{b2}$, so there is no remainder in the tenths column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd1}$ in the hundredths column. You do not need to do this but you can if you wish.

$\\begin{array}{r} {\\var{qd3}\\,.\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}\\var{qd1}\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3}\\,. \\,^{\\var{diff3}}{\\var{dd2}}\\,^\\color{red}{\\var{diff2}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

The hundredths column

Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b1}}$?\" (note this $\\var{b1}$ actually represents $\\var{b1*0.01}$)

Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b1}$. So we write $\\color{red}0$ above the $\\var{dd1}$ in the hundredths column: Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fit and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the hundredths column: Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fits and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the hundredths column:

$\\begin{array}{r} {\\var{qd3}\\,.\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}}\\color{red}{\\var{qd1}}\\phantom{\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,.\\,^{\\var{diff3}}{\\var{dd2}}\\,^\\color{green}{\\var{diff2}}\\color{green}{\\var{dd1}}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

Step 2: We only made it to $\\var{prod1}$ but we were aiming for $\\var{b1}$, so we were $\\var{diff1}$ away. This is the remainder (what remains to be divided) in the hundredths column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff1}}\\var{dd0}$ in the thousandths column. We made it to exactly $\\var{b1}$, so there is no remainder in the hundredths column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd0}$ in the thousandths column. You do not need to do this but you can if you wish.

$\\begin{array}{r} {\\var{qd3}\\,.\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}}{\\var{qd1}}\\phantom{\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,.\\,^{\\var{diff3}}{\\var{dd2}}\\,^{\\var{diff2}}{\\var{dd1}}\\,^\\color{red}{\\var{diff1}}\\var{dd0}}\\end{array}$

The thousandths column

Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b0}}$?\" (note this $\\var{b0}$ actually represents $\\var{b0*0.001}$)

Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b0}$. So we write $\\color{red}0$ above the $\\var{dd0}$ in the thousandths column: Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fit and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the thousandths column: Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fits and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the thousandths column:

$\\begin{array}{r} {\\var{qd3}\\,\\,.\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}}{\\var{qd1}}\\phantom{\\,^\\var{diff1}}\\color{red}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3}\\,. \\,^{\\var{diff3}}{\\var{dd2}}\\,^{\\var{diff2}}{\\var{dd1}}\\,^\\color{green}{\\var{diff1}}\\color{green}{\\var{dd0}}}\\end{array}$

Step 2: We could work out the remainder, we could keep adding zeros and continue the procedure but we only needed to determine the third decimal place in order to correctly round to the second decimal place and so we now stop the procedure.

Since the third decimal place was $\\var{qd0}$ we round updown to $\\var{ans}$. Therefore, $\\var{dividend1}\\div\\var{divisor1}=\\var{ans}$ (2 dec. pl.).

Long Division

We want to calculate $\\var{dividend1}\\div\\var{divisor1}$. Note, this is the same thing as $\\frac{\\var{dividend1}}{\\var{divisor1}}$ and both expressions mean \"how many $\\var{divisor1}$s go into $\\var{dividend1}$?\"

The long division algorithm allows you to work this out by working from the left to the right of $\\var{dividend1}$ whilst respecting place value. We normally set up the division in the following way:

$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dividend1}}$

Note the positions of the numbers!

Actually, since we want the answer to two decimal places we add as many zeroes after the decimal place to ensure we have three decimal places!

$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!\\var{dd2}\\var{dd1}\\var{dd0}}$

Why three? We use that extra digit to determine whether to round up or down.

The algorithm (or procedure) seems complicated at first but you might find a mnemonic helps to remember the steps. We work left to right doing the following steps

Divide
Multiply
Subtract
Bring down

and repeating until we run out of digits. The steps form the acronym DMSB. Popular mnemonics include \"Does McDonalds Sell Burgers?\", \"Dracula Must Suck Blood\" and \"Dead Mice Smell Bad\".

We need to know the $\\var{divisor1}$ times tables or write the $\\var{divisor1}$ times tables out (by repeatedly adding $\\var{divisor1}$) so that we can refer to them.

The ones column

D: The first thing we ask ourselves is, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{dd3}}$?\"

$\\begin{array}{r} \\color{red}{\\var{qd3}\\phantom{.\\!\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}.\\!\\var{dd2}\\var{dd1}\\var{dd0}} \\end{array}$

M: Now since $\\color{green}{\\var{qd3}}\\times \\color{green}{\\var{divisor1}}=\\var{prod3}$ we write $\\color{red}{\\var{prod3}}$ underneath in the ones column:

$\\begin{array}{r} \\color{green}{\\var{qd3}\\phantom{.\\!\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!\\var{dd2}\\var{dd1}\\var{dd0}} \\\\[-.5cm] \\color{red}{\\var{prod3}}\\phantom{.000}\\end{array}$

S: We now do the subtraction, $\\color{green}{\\var{dd3}-\\var{prod3}}$, to determine the remainder (what remains to be divided) in the ones column.

$\\begin{array}{r} {\\var{qd3}\\phantom{.\\!\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}.\\!\\var{dd2}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{\\color{green}{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]\\color{red}{\\var{diff3}}\\phantom{.000}\\end{array}$

B: Now we bring the $\\color{green}{\\var{dd2}}$ in the tenths column down next to the remainder so that it forms $\\var{diff3}\\var{dd2}$.

$\\begin{array}{r} {\\var{qd3}\\phantom{.\\!\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!\\color{green}{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}}\\phantom{.}\\color{red}{\\var{dd2}}\\phantom{00}\\end{array}$

The tenths column

D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b2}}$?\" (note this $\\var{b2}$ actually represents $\\var{b2/10}$ since it is in the tenths column)

$\\begin{array}{r} {\\var{qd3}.\\!\\color{red}{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]\\color{green}{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\end{array}$

M: Now since $\\color{green}{\\var{qd2}}\\times \\color{green}{\\var{divisor1}}=\\var{prod2}$ we write $\\color{red}{\\var{prod2}}$ underneath in the tenths column:

$\\begin{array}{r} {\\var{qd3}.\\!\\color{green}{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\color{red}{\\var{prod2}}\\phantom{00}\\end{array}$

S: We now do the subtraction, $\\color{green}{\\var{b2}-\\var{prod2}}$, to determine the remainder (what remains to be divided) in the tenths column.

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]\\color{green}{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]\\color{red}{\\var{diff2}}\\phantom{00}\\end{array}$

B: Now we bring the $\\color{green}{\\var{dd1}}$ in the hundredths column down next to the remainder so that it forms $\\var{diff2}\\var{dd1}$.

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}\\color{green}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]\\var{diff2}\\color{red}{\\var{dd1}}\\phantom{0}\\end{array}$

The hundredths column

D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b1}}$?\" (note this $\\var{b1}$ actually represents $\\var{b1/100}$ since it is in the hundredths column)

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}\\color{red}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]\\color{green}{\\var{diff2}\\var{dd1}}\\phantom{0}\\end{array}$

M: Now since $\\color{green}{\\var{qd1}}\\times \\color{green}{\\var{divisor1}}=\\var{prod1}$ we write $\\color{red}{\\var{prod1}}$ underneath in the hundredths column:

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}\\color{green}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{0}\\\\[-0.5cm]\\color{red}{\\var{prod1}}\\phantom{0}\\end{array}$

S: We now do the subtraction, $\\color{green}{\\var{b1}-\\var{prod1}}$, to determine the remainder (what remains to be divided) in the hundredths column.

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]\\color{green}{\\var{diff2}\\var{dd1}}\\phantom{0}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod1}}}\\phantom{0}\\\\[-0.5cm] \\color{red}{\\var{diff1}}\\phantom{0}\\end{array}$

B: Now we bring the $\\color{green}{\\var{dd0}}$ in the thousandths column down next to the remainder so that it forms $\\var{diff1}\\var{dd0}$.

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}{\\var{dd1}}\\color{green}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{0}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{0}\\\\[-0.5cm] {\\var{diff1}}\\color{red}{\\var{dd0}}\\end{array}$

The thousandths column

D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b0}}$?\" (note this $\\var{b0}$ actually represents $\\var{b0/1000}$ since it is in the thousandths column)

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}{\\var{qd1}}\\color{red}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{0}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{0}\\\\[-0.5cm] \\color{green}{\\var{diff1}\\var{dd0}}\\end{array}$

M: Now since $\\color{green}{\\var{qd0}}\\times \\color{green}{\\var{divisor1}}=\\var{prod0}$ we write $\\color{red}{\\var{prod0}}$ underneath in the thousandths column:

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}{\\var{qd1}}\\color{green}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{0}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{0}\\\\[-0.5cm]{\\var{diff1}\\var{dd0}}\\\\[-0.5cm]\\color{red}{\\var{prod0}}\\end{array}$

S: We now do the subtraction, $\\color{green}{\\var{b0}-\\var{prod0}}$, to determine the remainder (what remains to be divided) in the thousandths column.

$\\begin{array}{r} {\\var{qd3}.\\!{\\var{qd2}}{\\var{qd1}}{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}.\\!{\\var{dd2}}{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{.000}\\\\[-.7cm]{\\var{diff3}\\phantom{.}\\var{dd2}}\\phantom{00}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{00}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{0}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{0}\\\\[-0.5cm]\\color{green}{\\var{diff1}\\var{dd0}}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod0}}}\\\\[-0.5cm]\\color{red}{\\var{diff0}}\\end{array}$

Now we could keep adding zeros and continue the procedure but we only needed to determine the third decimal place in order to correctly round to two decimal places and so we now stop the procedure.

Since the third decimal place was $\\var{qd0}$ we round up down to $\\var{ans}$. Therefore, $\\var{dividend1}\\div\\var{divisor1}=\\var{ans}$ (2 dec. pl.).

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