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Lyn Gardner

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4 years, 7 months ago

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Lyn Gardner 4 years, 7 months ago

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Lyn Gardner 4 years, 7 months ago

Gave some feedback: Ready to use

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Lyn Gardner 7 years, 5 months ago

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Question statement

Subtraction.

Work out the following without using a calculator.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	576
x	integer	5
y	integer	7
z	integer	6
u	integer	3
v	integer	1
t	integer	4
b	integer	431
c	integer	567
d	integer	413
f	integer	754
g	integer	436
h	integer	531
l	integer	76
a1	rational	288/5
a2	rational	144/25
b1	rational	431/10
b2	rational	431/100
g1	rational	218/5
f2	rational	377/50

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
100*x+10*y+z

JME function reference

Description

xyz

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

576

← Depends on:

→ Used by:

Advice
"Part a)" - prompt
"Part a)" → "Gap 0." - Minimum accepted value
"Part a)" → "Gap 0." - Maximum accepted value

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
Part c) 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.

i) $\var{a}-\var{b}$ =Gap 0.

ii) $\var{c} -\var{d}$ =Gap 1.

Advice

Subtraction by hand

To carry out subtraction by hand the easiest way is to line the numbers up one above the other. When lining them up make sure that each place value is lined up correctly. Remember that the first number in the subtraction is placed above the second. The order of subtraction is important and it must be remembered throughout the calculation that it is the top number minus the bottom number.

Part a)

i)

Lining these up gives

$\begin{align}\var{a}&\\\underline{-{\hspace 0.5cm}\var{b}}&\\\end{align}$

Carrying out subtraction down each column in turn starting with the furthest right and placing the answer of these two digits underneath each column gives

$\var{z}-\var{v}=\var{z-v}$ ,

$\begin{align}\var{a}&\\\underline{-{\hspace 0.5cm}\var{b}}&\\\var{z-v}&\\\end{align}$

And for the other columns $\var{y}-\var{u}=\var{y-u}$ and $\var{x}-\var{t}=\var{x-t}$

$\begin{align}\var{a}&\\\underline{-{\hspace 0.5cm}\var{b}}&\\\var{a-b}&\\\end{align}$

ii)

Lining these up gives

$\begin{align}\var{c}&\\\underline{-{\hspace 0.5cm}\var{d}}&\\\end{align}$

Carrying out subtraction down each column in turn starting with the furthest right and placing the answer of these two digits underneath each column gives

$\var{y}-\var{u}=\var{y-u}$ ,

$\begin{align}\var{c}&\\\underline{-{\hspace 0.5cm}\var{d}}&\\\var{y-u}&\\\end{align}$

And for the other columns $\var{z}-\var{v}=\var{z-v}$ and $\var{x}-\var{t}=\var{x-t}$

$\begin{align}\var{c}&\\\underline{-{\hspace 0.5cm}\var{d}}&\\\var{c-d}&\\\end{align}$

Part b)

In part a for each column the number at the top was equal to or greater than the number below it. In this case the subtraction did not need any rearranging or consideration of surrounding columns. This is not often the case with subtraction and when the top number of a column is less than the number beneath it some rearranging may be required. At this point remembering that the column is part of a bigger calculation enables the digits to be changed so that the top number is bigger in each column. It can be helpful to think of it in terms of money and changing a larger domination coin into smaller ones, 10p = 10 x 1p. If the digit at the top of a column is less than the digit below it then this can be changed by looking to the column to the left. If this is not 0 at the top you can change one from this column into 10 in your column.

i)

Lining up the numbers in the same way gives

$\begin{align}\var{f}&\\\underline{-{\hspace 0.5cm}\var{g}}\\\end{align}$

Looking at the furthest right column the calculation we want to perform is $\var{t}-\var{z}$ , but the number at the top is smaller than the number at the bottom. We need to change this by reducing the number at the top in the column to the left of it by 1 and adding 10 to the top of this column. So the top row now becomes $\var{y},\var{x-1},\var{t+10}$ and the subtraction is performed on the following numbers.

$\begin{align}\var{y}^{\;\var{x-1}}\rlap{/}\var{x}^{\;1}\var{t}&\\\underline{-{\hspace 0.5cm}\var{t}\ \;\;\var{u}\ \;\; \var{z}}&\\\end{align}$

So now carrying out the subtraction down each column gives

$\var{t+10}-\var{z}=\var{t+10-z}$ , $\var{x-1}-\var{u}=\var{x-1-u}$ , $\var{y}-\var{t}=\var{y-t}$

$\begin{align}\var{y}^{\;\var{x-1}}\rlap{/}\var{x}^{\;1}\var{t}&\\\underline{-{\hspace 0.5cm}\var{t}\ \;\;\var{u}\ \;\;\var{z}}&\\\var{y-t}\ \;\;\var{x-1-u}\ \;\;\var{t+10-z}&\\\end{align}$

ii)

Again lining up the numbers gives

$\begin{align}\var{h}&\\\underline{-{\hspace 0.5cm}\var{l}}\\\end{align}$

Looking at each column and exchanging along the top where the bottom number is larger the subtraction becomes

$\begin{align}\var{x}^{\;\var{u-1}}\rlap{/}\var{u}^{\;1}\var{v}&\\\underline{-{\hspace 1cm}\var{y}\ \;\; \var{z}}&\\\end{align}$

There is still a column with the top number smaller than the bottom, so the process is repeated, carried out in all positions from right to left where the top number is smaller than the bottom.

$\begin{align}^{\;\var{x-1}}\rlap{/}\var{x}^{\;\var{u-1+10}}\rlap{/}\var{u}^{\;1}\var{v}&\\\underline{-{\hspace 1cm}\var{y}\ \;\; \var{z}}&\\\end{align}$

Carrying out subtraction down each column in turn starting with the furthest right and placing the answer of these two digits underneath each column gives

$\begin{align}^{\;\var{x-1}}\rlap{/}\var{x}^{\;\var{u-1+10}}\rlap{/}\var{u}^{\;1}\var{v}&\\\underline{-{\hspace 1cm}\var{y}\ \; \var{z}}&\\\var{x-1}\ \;\;\; \var{u-1+10-y}\ \; \var{v+10-z}&\\\end{align}$

Part c)

With decimals the key point is to line up the numbers by the decimal point and it can be made easier by adding 0's in to any gaps.

i)

So lining up the numbers in this case gives

$\begin{align}\var{x}\var{y}.\var{z}0&\\\underline{-{\hspace 0.5cm}\var{t}.\var{u}\var{v}}&\\\end{align}$

The same method is used as before, looking at each column in turn and if the top number is smaller than the bottom then change the top numbers by taking one from the column to the left and adding 10 to the current column so the top number becomes more than the bottom in that column.

$\begin{align}\var{x}\ \;\; \var{y}.^{\var{z-1}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{t}.\ \; \var{u}\ \;\; \var{v}}&\\\end{align}$

Carrying out the calculation the answer is found. Make sure that the decimal point is placed in the correct place in the answer

$\begin{align}\var{x}\ \;\; \var{y}.^{\var{z-1}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{t}.\ \; \var{u}\ \;\; \var{v}}&\\\var{x}\ \;\; \var{y-t}.\ \; \var{z-1-u}\ \;\; \var{10-v}&\\\end{align}$

ii)

$\begin{align}\var{t}\var{u}.\var{z}0&\\\underline{-{\hspace 0.5cm}\var{y}.\var{x}\var{t}}&\\\end{align}$

$\begin{align}\var{t}\ \;\; \var{u}.^{\var{z-1}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{y}.\ \; \var{x}\ \;\; \var{t}}&\\\end{align}$

Carying out the calculation the answer is found . Make sure that the decimal point is placed in the correct place in the answer.

$\begin{align}\var{t}\ \;\;^{\var{u-1}} \rlap{/}\var{u}.^{\var{z-1+10}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{y}.\;\ \; \var{x}\ \;\; \var{t}}&\\\end{align}$

$\begin{align}^{\var{t-1}}\rlap{/}\var{t}\ \;\;^{\var{u-1+10}} \rlap{/}\var{u}.^{\var{z-1+10}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{y}.\;\ \; \var{x}\ \;\; \var{t}}&\\\end{align}$

Now carrying out the calculations in each column gives

$\begin{align}^{\var{t-1}}\rlap{/}\var{t}\ \;\;^{\var{1}} \var{u}.^{\var{z-1}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{y}.\ \; \var{x}\ \;\; \var{t}}&\\\end{align}$

Now carrying out the calculations in each column gives

$\begin{align}^{\var{t-1}}\rlap{/}\var{t}\ \;\;^{\var{1}} \var{u}.^{\var{z-1}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{y}.\ \; \var{x}\ \;\; \var{t}}&\\\var{t-1}\ \;\; \var{u+10-y}.\ \; \var{z-1-x}\ \;\; \var{10-t}&\\\end{align}$

$\begin{align}^{\var{t-1}}\rlap{/}\var{t}\ \;\;^{\var{1}} \var{u}.^{\var{z-1}}\rlap{/}\var{z}^{\ \;1}0&\\\underline{-{\hspace 0.5cm}\var{y}.\ \; \var{x}\ \;\; \var{t}}&\\\end{align}$

Now carrying out the calculations in each column gives

Additional resources

For further examples and questions

Multi Digit Subtraction - Khan Academy

Subtracting Decimals - Khan Academy

Give a worked solution to the whole question.

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Part	Test	Passed?
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Gap-fill		Hasn't run yet
Number entry
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Number entry
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Gap-fill
Gap-fill		Hasn't run yet
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Number entry
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Gap-fill
Gap-fill		Hasn't run yet
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This question is used in the following exams:

Addition and Subtraction by Lyn Gardner in Lyn's workspace.
ABID trial test by Rebecca Smith in Rebecca's workspace.
Lauren's CHANGED exam by Lauren O in Lauren's workspace.

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Error

Subtraction

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Lyn Gardner 4 years, 7 months ago

Lyn Gardner 4 years, 7 months ago

Lyn Gardner 7 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

Subtraction by hand

Part a)

i)

ii)

Part b)

i)

ii)

Part c)

i)

ii)

Additional resources

Loading...

Error

Subtraction

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Lyn Gardner 4 years, 7 months ago

Lyn Gardner 4 years, 7 months ago

Lyn Gardner 7 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Subtraction by hand

Part a)

i)

ii)

Part b)

i)

ii)

Part c)

i)

ii)

Additional resources

Generated value: `integer`