Bounds with area

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Simon Thomas

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Simon Thomas 5 years, 2 months ago

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Simon Thomas 5 years, 2 months ago

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Question statement

Give all answers to 2 decimal places:

Give any introductory information the student needs.

			Name	Type	Generated Value

width	number	4
length	number	5.6
minarea	number	21.9225
maxarea	number	22.8825
width2	integer	27
area	integer	230
minlength	number	8.1818181818
maxlength	number	8.8679245283

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Name

Data type

Value

xxxxxxxxxx
 
random (2..4#0.1)

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Description

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Generated value: `number`

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Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
Part b) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

A rectangle has a width of $\var{width}$ cm and length of $\var{length}$ cm, each rounded to the nearest mm.

What is:

(i) the minimum possible area of the rectangle in cm $^2$ Gap 0.

(ii) the maximum possible area of the rectangle in cm $^2$ Gap 1.

Advice

(a)

The given width = $\var{width}$ cm and length = $\var{length}$ cm are each rounded to the nearest 0.1cm. To find the upper and lower bounds we must add/subtract half of this value.

Since half of 0.1cm is 0.05cm we obtain:

Lower bound for width $= \var{width}-0.05=\var{width-0.05}$

Upper bound for width $= \var{width}+0.05=\var{width+0.05}$

Lower bound for length $= \var{length}-0.05=\var{length-0.05}$

Upper bound for length $= \var{length}+0.05=\var{length+0.05}$

Since area = length $\times$ width, we have:

lower bound area = lower bound length $\times$ lower bound width $=\var{length-0.05} \times \var{width-0.05} = \var{minarea}$ cm $^2=\var{precround(minarea,2)}$ cm $^2$ to 2d.p.

and

upper bound area = upper bound length $\times$ upper bound width $=\var{length+0.05} \times\var{width+0.05} = \var{maxarea}$ cm $^2=\var{precround(maxarea,2)}$ cm $^2$ to 2d.p.

(b)

The given width = $\var{width2}$ cm is rounded to the nearest 1cm. To find the upper and lower bounds we must add/subtract half of this value.

Since half of 1cm is 0.5cm we obtain:

Lower bound for width $= \var{width2}-0.5=\var{width2-0.5}$

Upper bound for width $= \var{width2}+0.5=\var{width2+0.5}$

The given area = $\var{area}$ cm is rounded to the nearest 10cm $^2$ . To find the upper and lower bounds we must add/subtract half of this value.

Since half of 10cm $^2$ is 5cm $^2$ we obtain:

Lower bound for area $= \var{area}-5=\var{area-5}$

Upper bound for area $= \var{area}+5=\var{area+5}$

Since length = area $\div$ width, we have:

lower bound length= lower bound area $\div$ upper bound width (to get a small answer we want to be dividing a small number by a big number)

$=\var{area-5} \div \var{width2+0.5} = \var{minlength}$ cm $=\var{precround(minlength,2)}$ cm to 2d.p.

and

upper bound length = upper bound area $\div$ lower bound width (to get a large answer we want to be dividing a large number by a small number)

$=\var{area+5} \div \var{width2-0.5} = \var{maxlength}$ cm $=\var{precround(maxlength,2)}$ cm to 2d.p.

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Part	Test	Passed?
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This question is used in the following exam:

Working with Bounds by Simon Thomas in Maths support.

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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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History

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Checkpoint description

Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

Error in variable testing condition

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Generated value: `number`

Parts

Gap-fill

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Bounds with area

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`