AO2 / AO3 Questions for Boxing Up

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question asking
me?
What Maths
will I be
using?
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do I already have?
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working out do I need to
do?
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my answer is correct?
Two companies, Barry's Bricks and Bricks ArUs,
deliver bricks.
The graph shows the delivery costs of bricks from
both companies.
Prakash wants Bricks ArUs to deliver some bricks.
He lives 2 miles away from Bricks ArUs.
(a) Write down the delivery cost.
.................................................................................
.............................................................
John needs to have some bricks delivered.
He lives 4 miles from Barry's Bricks.
He lives 5 miles from Bricks ArUs.
(b) Work out the difference between the two
delivery costs.
.................................................................................
.............................................................
(Total for Question is 4 marks)
A03 Question
What is the
difference between
the two delivery
costs?
Reading
information
from a
graph.
Subtraction.
Barry’s Bricks £50
Bricks R Us £65
£65 - £50 = £15
£50 + £15 = £65
Solution
Question
Working
(a)
(b)
Barry's Bricks
£50
Bricks ArUs
£65
65 − 50
Answer
56
Mark
1
15
3
Notes
B1 for 56
(accept answer
in the range 55
to 57)
M1 for 50 or 65
(accept 64 –
66)
M1 for 65 – 50
(accept 64-66
for 65)
A1 for 15
(accept answer
in range 14 to
16)
What is the
question asking
me?
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using?
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working out do I need to
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A03 Question
* Barbara goes on holiday to Prague. The
currency in Prague is the Koruna (KC).
This graph can be used to convert between £
(pounds) and KC (Koruna).
The exchange rate is £1 = 30 KC.
Barbara bought some things in London.
She saw the same things on sale in Prague.
The table shows the cost in £ (pounds) and the
cost in KC (Koruna).
Barbara thinks the total cost of these things
was more in London than in Prague.
Is she correct?
Give a reason for your answer.
You must show all your working.
(Total for Question is 5 marks)
Is the total cost
more in London or
in Prague
Converting
between
currencies,
Addition,
Multiplication,
Division
London
£15 + £34 + £ 26 = £75
£1 = 30KC
£75 x 30 = 2250KC
Prague
450KC + 750KC +810KC = 2010KC
She is wrong, 2050KC is more than
2010KC so cheaper in Prague.
2010KC ÷ 30 = £67
£67 is less than £75
Solution
Question
Working
Answer
London:
Yes.
£15, £34, £26 (£75) Cheaper in
→ 450, 1020, 780
Prague
(2250) KC
(More in
London)
Prague:
450, 750, 810 KC
(2010KC)
→ £15, £25,
£27 (£67)
£ to KC is ×30;
KC to £ is ÷30.
Mark
5
Notes
M1 conversion method (× or ÷ as
appropriate) or evidence of use of
graph (seen, or implied, by at least
lines or evidence of conversion by
marks on axes) for at least one
figure.
M1 (dep) conversion applied to 3
figures or totals (converted figures
must be stated, marks on graph
insufficient)
A1 converted figures shown (all
three individual items or totals
converted correctly; NB: no
tolerance on graph)
M1 totalling converted amounts
C1 (dep on at least M1)
comparison of "totals" and correct
conclusion
Eg "2250KC">"2010KC",
"£75">"£67" so cheaper to buy in
Prague.
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The diagram shows a garden in the shape of a
rectangle.
A03 Question
All measurements are in metres.
The perimeter of the garden is 32 metres.
Work out the value of x
......................
(Total for Question is 4 marks)
Solve an Equation
To calculate perimeter
add the lengths of all
the sides.
Perimeter = 32cm
4 + 3x + x + 6 + 4 + 3x + x + 6
Write an
expression
Simplify
Use inverse
operations
Perimeter = 8x + 20
Length 4 + (3 x 1.5) = 8.5
Width 1.5 + 6 = 7.5
8x + 20 = 32
8.5 + 7.5 = 16
8x = 12
x = 1.5
16 x 2 = 32
Solution
Question
Working
Answer
1.5
Mark
4
Notes
M1 for correct expression for
perimeter
eg. 4 + 3x + x + 6 + 4 + 3x + x + 6 oe
M1 for forming correct equation
eg. 4 + 3x + x + 6 + 4 + 3x + x + 6=
32 oe
M1 for 8x = 12 or 12 ÷ 8
A1 for 1.5 oe OR M1 for correct
expression for semi-perimeter
eg. 4 + 3x + x + 6 oe
M1 for forming correct equation
eg. 4 + 3x + x + 6 = 16
M1 for 4x = 6 or 6 ÷ 4
A1 for 1.5 oe
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ABC is a triangle.
A03 Question
Angle ABC = angle BCA.
The length of side AB is (3x − 5) cm.
The length of side AC is (19 − x) cm.
The length of side BC is 2x cm.
Work out the perimeter of the triangle.
Give your answer as a number of centimetres.
(Total for Question is 5 marks)
Solve an Equation
Write an equation
Work out the Perimeter
To calculate perimeter add the
lengths of all the sides.
Isosceles triangles have two equal
sides.
3x – 5 = 19 – x
4x – 5 = 19
4x = 24
x=6
19 – 6 = 13
6 x 2 = 12
13 + 13 + 12 = 38cm
Solve an equation
Substitute the
value of x into the
equation
If x = 6
3x – 5 = 13
19 – x = 13
So x = 6
Solution
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question asking
me?
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using?
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* Bill uses his van to deliver parcels.
For each parcel Bill delivers there is a
fixed charge plus £1.00 for each mile.
You can use the graph to find the total
cost of having a parcel delivered by Bill.
(a) How much is the fixed charge?
£......................
(a) Ed uses a van to deliver parcels.
For each parcel Ed delivers it costs £1.50
for each mile.
There is no fixed charge.
(b) Compare the cost of having a parcel
delivered by Bill with the cost of having
a parcel delivered by Ed.
(Total for Question is 4 marks)
A03 Question
Compare two delivery
costs.
Plot information
onto a graph.
Miles
5
10
15
20
25
30
35
40
45
50
Cost
£7.50
£15
£22.50
£30
£37.50
£45.00
£52.50
£60
£67.50
£75.00
Ed is cheaper up to 20 miles.
Ed and Bill cost the same for 20 miles.
Bill is cheaper after 20 miles.
Plot the information from the table onto the graph.
The graphs cross at 20 miles.
Before 20 miles the graph for Bill is steeper.
After 20 miles the graph for Ed is steeper.
Solution
Questio
Answer
Notes
n
(a) (b) 10 Ed is cheaper up B1 cao M1 for correct line for Ed intersecting at (20,30) ±1 sq tolerance or 10 + x = 1.5x oe
to 20 miles, Bill is C2 (dep on M1) for a correct full statement ft from graph
cheaper for
eg. Ed cheaper up to 20 miles and Bill cheaper for more than 20 miles
more than 20 miles (C1 (dep on M1) for a correct conclusion ft from graph
eg. cheaper at 10 miles with Ed ; eg. cheaper at 50 miles with Bill eg. same cost at 20
miles; eg for £5 go further with Bill or A general statement covering short and long
distances eg. Ed is cheaper for shorter distances and Bill is cheaper for long distances)
OR M1 for correct method to work out Ed's delivery cost for at least 2 values of n miles
where 0 < n ≤ 50
or for correct method to work out Ed and Bill's delivery cost for n miles where 0 < n ≤ 50
C2 (dep on M1) for 20 miles linked with £30 for Ed and Bill with correct full statement
eg. Ed cheaper up to 20 miles and Bill cheaper for more than 20 miles
(C1 (dep on M1) for a correct conclusion eg. cheaper at 10 miles with Ed; eg. cheaper at
50 miles with Bill eg. same cost at 20 miles; eg for £5 go further with Bill or a general
statement covering short and long distances eg. Ed is cheaper for shorter distances and
Bill is cheaper for long distances)
SC: B1 for correct full statement seen with no working
eg. Ed cheaper up to 20 miles and Bill cheaper for more than 20 miles
QWC Decision and justification should be clear with working clearly presented and
attributable
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question asking
me?
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A03 Question
There are 300 ml of medicine in a bottle.
Mary has to take two 5 ml spoons full of medicine twice a
day. Mary has to take the medicine until the bottle is empty.
(a) How many days does Mary have to take the medicine for?
. . . . . . . . . . . . . . . . . . . . . . Days
You can work out the amount of medicine, c ml, to give to a
child by using the formula
c = ma⁄150
m is the age of the child, in months.
a is an adult dose, in ml.
A child is 30 months old.
An adult's dose is 40 ml.
(b) Work out the amount of medicine you can give to the
child
. . . . . . . . . . . . . . . . . . . . . . ml
(Total for Question is 5 marks)
How many days
does Mary take the
medicine for?
How much
medicine can you
give a child?
a) 5ml x 2 = 10ml
10ml x 2 = 20ml a day
300ml ÷ 20 = 15 days
Substitution
into a formula.
Multiplication
Division
a) 20ml a day x 15 days = 300ml
b) 8 x 150 = 1200
b) (Age of child x adult dose) ÷ 150
(30 x 40) ÷ 150
1200 ÷ 150 = 8 ml
Solution
Working
Question
(a)
(b)
2 × 5 × 2 = 20
300 ÷ 20 =
c=
Answer
Mark
15
3
8
2
Notes
M2 for
300 ÷ ( 2 × 5 × 2 ) oe
(M1 for
2 × 5 × 2 or 20 seen
or
300 ÷ (2 × 5) or 30
seen A1 cao
M1 for
or 1200 seen
A1 cao
What is the
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The diagram shows shape A.
All the measurements are in centimetres.
(a) Find an expression, in terms of x, for
the perimeter of shape A.
.....................
A square has the same perimeter as shape
A.
(b) Find an expression, in terms of x, for
the length of one side of this square.
.....................
(Total for Question is 4 marks)
A03 Question
Write an expression for
the perimeter.
Write an expression for
the missing sides.
To calculate perimeter
add all the sides
together.
Write an expression for
perimeter
Simplify
The missing sides are
2x + 1 and 3x + 3
Side of square = (16x + 8) ÷ 4
Side of square = 4x + 2
4x + 2
4x + 2
Perimeter of shape = 16x + 8
4x + 2
4x + 2
4(4x + 2) = 16x + 8
Solution
Question
(a)
(b)
Working
Missing sides are
2x + 1 and 3x + 3
Perimeter = 5x + 1 + x + 3x + 2x
+ 3 +2x + 1 + 3x + 3
OR
2(5x + 1) + 2(2x + 3 + x)
Answer
16x + 8
4x + 2
Mark
3
1
Notes
M1 for 5x + 1 –3x or 2x + 3 + x or
identifying a missing side as 2x + 1
or 3x + 3(maybe on the diagram)
M1 for adding 5 or 6 sides from x,
5x + 1, 3x, 2x + 3, '2x + 1', '3x + 3'
where the missing sides are in the
form ax ± b (a and
b ≠ 0)
or 2(5x + 1) + 2(2x + 3 + x) oe
A1 for 16x + 8 oe for unsimplified
expression B1 ft for ['2(5x + 1) +
2(2x + 3 + x)'] ÷ 4 or ("16x + 8") ÷ 4
oe where the answer is an
algebraic expression in x
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