// Numbas version: exam_results_page_options {"name": "Q4 (Calculating area using algebraic equations)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["lent1", "red1", "area1", "ans11", "ans12", "per2", "ratio2", "ans21", "ans22", "ans23", "lent3", "per3", "ans31", "ans32", "ratio4", "per4", "ans41", "ans42", "ans43"], "name": "Q4 (Calculating area using algebraic equations)", "tags": ["area using algebra", "Rebel", "REBEL", "rebel", "rebelmaths", "teame"], "preamble": {"css": "", "js": ""}, "advice": "

Have a look at this youtube video to understand simultaneous equations better: Simultaneous Equations

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Part 1:

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Width:

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$\\var{lent1} \\times (width - \\var{red1}) = \\var{area1}cm$

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$ width: \\frac{(\\var{area1}+(\\var{red1} \\times \\var{lent1}))}{\\var{lent1}} = \\var{ans11}$

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Area:

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$\\var{lent1} \\times \\var{ans11}= \\var{ans12}cm^2$

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Part 2:

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Width:

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$(2 \\times Length)+ (2 \\times width) = \\var{per2}cm$

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$Length = \\var{ratio2} \\times width$

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$ width: \\frac{\\frac{\\var{per2}}{2}}{1 + \\var{ratio2}} = \\var{ans21}cm$

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Length:

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$\\var{ratio2} \\times \\var{ans21} = \\var{ans22}cm$

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Area:

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$\\var{ans21} \\times \\var{ans22}= \\var{ans23}cm^2$

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Part 3:

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Similar to part 2

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Length:

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$\\frac{\\frac{\\var{per3}}{2}}{2 + \\frac{\\var{lent3}}{100}} = \\var{ans31}cm$

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Breath:

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$\\var{ans31} \\times (1 + \\frac{\\var{lent3}}{100}) = \\var{ans32}cm$

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Part 4:

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Similar to part 2

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Width:

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$\\frac{\\frac{\\var{per4}}{2}}{1 + \\var{ratio4}} = \\var{ans41}cm$

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Length:

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$\\var{ratio4} \\times \\var{ans41} = \\var{ans42}cm$

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Area:

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$\\var{ans41} \\times \\var{ans42}= \\var{ans43}cm^2$

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", "rulesets": {}, "parts": [{"stepsPenalty": 0, "prompt": "

A rectangle has a length of $\\var{lent1} cm$ and a width of $x cm$. If the width was reduced by $\\var{red1} cm$, the area would be $\\var{area1} cm^2$. Find the original width and area of this rectangle.

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Width:  [[0]]$cm$

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Area:    [[1]]$cm^2$

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First sketch the original rectangle with the information given. Next draw the second rectangle with the new width. How can you represent this new width using x? Now use the information given to find x.

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Find the width, length and area of a rectangler room which has a perimeter of $\\var{per2} m$, if its length is $\\var{ratio2}$ times its width.

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Width:   [[0]]$m$

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Length: [[1]]$m$

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Area:     [[2]]$m^2$

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "

Sketch the rectangle and represent the information given in the question.

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The length of a rectangle is $\\var{lent3}$% greater than its breadth. If the perimeter of the rectangle is $\\var{per3} m$, find the length and breadth.

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Length:   [[0]]$m$

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Breadth:  [[1]]$m$

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Convert the precentage to decimals. E.g. 30% bigger would mean the length was 1.3x.

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Find the width, length and area of a rectangle room which has a perimeter of $\\var{per4} m$, if its length is $\\var{ratio4}$ times its width.

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Width:   [[0]]$m$

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Length:  [[1]]$m$

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Area:     [[2]]$m^2$

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Solve the following correct to 1 decimal place:

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Algebra word problems using area and perimeter.

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rebelmaths

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