SYDNEY
BOYS
HIGH
SCHOOL
2025
YEAR 8 STAGE 4
TERM 2
HALF YEARLY
Sample Solutions
NOTE:
This process of checking your mark is about reading the solutions and the comments.
Before putting in an appeal re marking, first consider that the mark is not linked to the
amount of writing you have done.
Just because you have shown some ‘working’ does not justify that your solution is worth
any marks.
Students who used pencil, an erasable pen and/or whiteout, may NOT be able to
appeal.
Question 1 (22 marks)
1
Is 1.21 rational or irrational?
1
Rational, since 1.21 = 1.1 which is terminating decimal.
1 mark for correctly indicating “rational”. No half marks.
Very well done. Almost all students got this one correct (most with reasoning, which is nice).
2
a3
Which of the following is equivalent to −4 ?
a
Circle the correct answer.
A.
a7
Solution:
B.
a–7
1
1
a
C.
D.
−1
a3
a3 a 4
7
A since −4 = −4 × 4 = a .
a
a
a
1 mark for selecting the correct option. No half marks.
Mostly well done.
3
What is the value of
17.1 − 1.89
, correct to 3 decimal places?
10.4 + 8.36
3
1
Using a calculator gives 0.1864118168 which is equal to 0.186 when rounded to 3 decimal places.
1 mark for the correct answer, rounded correctly.
Mostly well done. A common error was the result 3.130 , which comes from entering the fraction
using a division symbol without using brackets to group the denominator as a single expression.
i.e. 3 17.1 − 1.89 ÷ 10.4 + 8.36 is incorrect.
(
)
Grouping correctly is 3 17.1 − 1.89 ÷ 10.4 + 8.36 .
a S (17.1p1.89) R 10.4+ s8.36=
To do S you need to press q s
When entering a complicated fraction on your calculator, just use the fraction button!
4
p is the largest prime number between 50 and 100.
q is the smallest prime number between 50 and 100.
2
Calculate the value of p − q .
By Eratosthenes’ Sieve, we only need to check for divisibility up to 100 in order to find primes.
We can find that: p = 97 and q = 53 . Therefore, p − q =
44 .
1 mark for finding either p or q.
1 mark for correctly calculating the difference.
Mostly well done.
Almost all students found at least one of the prime numbers.
A common error was having q = 51 , but 51 is divisible by 3!
5
The diagram below shows the sector of a circle.
Calculate the area of this sector, correct to 2 decimal places.
……………………
2
This sector has a radius of 4.8 cm and has a subtending angle of 45° .
We can calculate the area of the sector to be:
45°
2
= 9.05
A= π ( 4.8 ) ×
(2 decimal places)
360°
1 mark for recognising that the area of the sector is an eighth of the circle.
1 mark for the correct answer, rounded correctly.
Generally done ok.
Most students recognised that the sector was an eighth of the corresponding circle.
Most errors came from incorrect circle area formulas and using π = 3.14 instead of the exact value.
–2–
6
The diagram below shows a triangle with sides of length 2 x + 3 , 3x and 11 − x .
(i)
1
Explain why x must be less than 11.
Since 11 − x is a distance, it must be positive. Therefore, x must be less than 11.
1 mark for recognising that a distance must be positive. No half marks.
Generally well done.
Many students incorrectly assumed that this was a right triangle and attempted to use
Pythagoras’ theorem in their response.
Many students also assumed that x must be an integer (this did not lose marks but indicates
a limited outlook on numbers).
(ii)
Write down an expression, in terms of x, for the perimeter of the triangle.
Leave your answer in simplest possible form.
( 2 x + 3) + ( 3x ) + (11 − x )
= ( 2 x + 3 x − x ) + ( 3 + 11)
P=
= 4 x + 14
1 mark for the correct expression in simplest possible form.
Mostly well done.
Aside from algebraic errors, a common error was not fully simplifying.
–3–
1
7
(i)
1
Expand 2 x( x 2 − 2 y )
2 x( x 2 − 2 y=
) 2 x ( x 2 ) + 2 x ( −2 y )
= 2 x 3 − 4 xy
1 mark for the fully expanded (and simplified) answer.
Very well done.
(ii)
Expand and simplify 3(3w − 2) − 2(2 w − 3)
2
3(3w − 2) − 2(2 w =
− 3) 3 ( 3w ) + 3 ( −2 ) + ( −2 )( 2 w ) + ( −2 )( −3)
= 9w − 6 − 4w + 6
= 5w
1 mark for correctly fully expanding.
1 mark for correctly fully simplifying.
Generally done ok.
A common error was dropping negative signs when expanding, resulting in the expression
9w − 6 − 4w − 6 .
Most students correctly simplified after expanding.
–4–
8
At 6:09 am Mr Wangers bought 987 fish at a fish supermarket for $5.79 each.
Exactly 96 minutes later, he sold them all to a supermarket for $7.95 each.
(i)
1
What time did he sell the fish?
We can convert 96 minutes into 1 hour and 36 minutes.
6:09 + 1:36 =
7:45 am.
Alternate method: we can use the calculator to do 6°9° + 0°96° and convert to DMS.
1 mark for the correct time. No half marks.
Very well done.
(ii)
2
Calculate his total profit.
Method 1
Earnings:
Cost:
Total profit:
7.95 × 987 =
7846.65
5.79 × 987 =
5714.73
7846.65 − 5714.73 =
2131.92
Method 2
Profit per fish: 7.95 − 5.79 =
2.16
2131.92
Total profit: 2.16 × 987 =
1 mark for correctly calculating earnings or cost or profit per fish.
1 mark for the correct total profit.
Answers without working could not earn full marks.
Mostly well done.
The majority of errors came from silly mistakes.
–5–
9
−2
3
3
.
Show that 8 + 2−3 =
8
Write down all the steps of your working.
3
−2
LHS = 8 3 + 2−3
1 1
= 2+ 3
83 2
1
1
=
+ 3
2
3
2
8
1 1
+
2 2 23
1 1
=
+
4 8
3
=
8
= RHS
=
1 mark for converting the negative index form into a fraction.
1 mark for converting the fractional index form into a surd.
1 mark for the correct solution, via full and correct working.
Generally done poorly.
Many students failed to show ALL the required working (if it is a 3-mark question, calculator
answers will not suffice!).
Only a couple of students used LHS and RHS to structure their response- this is the expected format
of a response to any “show that” problem for an equation.
–6–
10
(i)
Calculate the volume of the prism below.
3
Various approaches were possible, breaking up the composite solid into
some valid sum or difference of rectangular prisms. It was also possible
to calculate the base as a composite shape, then calculate the prism.
The following calculations are the most common approaches:
210
( 3 × ( 3 + 5) × 7 ) + ( 2 × 3 × 7 ) =
( ( 3 + 2 ) × 3 × 7 ) + ( 3 × 5 × 7 ) =210
210
( ( 3 + 2 ) × ( 5 + 3) × 7 ) − ( 5 × 2 × 7 ) =
30 × 7 210
( 2 × 3) 30, =
( 3 × ( 5 + 3) ) +=
1 mark for a valid decomposition of the solid.
1 mark for correct calculations of the parts making up the composite solid/shape.
1 mark for the correct answer.
Mostly well done.
Some responses left out part of the volume during decomposition.
(ii)
How much liquid can the prism hold, in litres?
Since 1 cm3 = 1 mL and 1 000 mL = 1 L, therefore 210 cm3 = 210 mL = 0.21 L.
1 mark for the correct answer.
Very well done.
End of Question 1 Solutions
–7–
1
Question 2 (18 marks)
1
Which choice is equivalent to
%&! '"
?
%&# '%&$ '
'%" (#
%"!# %$%
=
=
=$
'%! ('%& ( %!+& %$%
1
Circle the correct response.
A.
B.
Comment:
2
C.
1
D.
4
A majority of students got this correct (C) with (A) being the second most popular.
Answer (A) resulted from the incorrect use of indices in the numerator.
Given that y = 3x, express ! ! in terms of y.
1
" ! = ( #! ) = ( #! ) = " !
!
Comment:
3
!
This was poorly done, showing again the lack of knowledge of power to a power and
multiples of powers in indices.
The most common incorrect answer was ! ! = "#"! $
Factorise 5g2h + 10hj.
Task out the HCF:
Comment:
4
1
"! ( " ! + # )
This question was very well done showing strong grasp of simple factorisation.
The most common error was to incorrectly factorise the 10hj term.
Simplify
.
1
!"!!
! "! " ( "! )
=
#
$ =
#!
$#
% # &
!
!
Comment:
5
This was reasonably well done, with the main error resulting from not squaring all of the
terms: 5, m and 6
Simplify
.
( ! ) "(" ) = ! " " = !
# !!
# "
! # " !"
Comment:
!$
! # " !"
%
2
!$! #
" %!' !"( = ! !)* " & +,
"&
! )*
This question caused a lot of problems.
The errors stemmed from performing multiple operations in one step.
( ! ) produced the most mistakes.
" !!
– Q2 P1 –
6
Using your knowledge of surds, explain why
lies between 9 and 10 on the number line.
Your answer should not rely on the use of approximations, calculator or otherwise.
1
First ! = "# , ! " = " ! #$ = $% , and !" = !"" .
As
Comment:
!" < #! < "$$ then ! < " # < $%
Very poorly done.
Most students did not interpret the word “explain” in the question properly.
Students generally wrote a lot of unrelated numbers
7
Perimeter = 3 × 9 + 6 × 2 + 3 × 2 + 3 = 48 cm
Comment:
8
Reasonably well done.
Most mistakes were simple errors rather concept errors.
A rectangular sheet of paper ABCD is made into an open cylinder with the edge AB meeting
the edge CD. AD = 28 cm and AB = 20 cm.
(i)
Show that the radius of the cylinder is 4.46 cm, correct to 2 decimal places.
2
!! ! = !" " ! ! = #$
#$
# ! = $ $%$&'((" = $%$' )! *%+%,
Comment:
(ii)
!
Very well done.
Although the progression of logic was lacking in many cases, most got the information
required.
Hence calculate the volume of the cylinder.
1
! = "# = ! A # = ! " ( "#"$) " !% # &!"'#() *! +#,#!
Comment:
!
A variety of answers were given depending on the values used, such as π on the calculator as
well as the value of the radius. The ‘Hence’ meant the correct answer is using the 4.46 from
part (i)
– Q2 P2 –
9
A pattern is made using identical rectangular tiles.
2
Find the total area of the pattern.
!! = " ! + "
#= !+ "
"" = ! ! + #$ ! !%
"" ! $ = ! ! ! !
" ! = &' " = (
" # = & # & # ( = !& )* !
Comment:
10
Reasonably well done.
Some solved by simultaneous equations and some by inspection.
Answers were well communicated
Sydney is 10 hours ahead of London. Texas is 6 hours behind London.
2
If it is Tuesday 6th May 12 pm in Sydney, what time and day is it in Texas?
Sydney:
London:
Texas:
Tuesday 6th May
12 pm
Tuesday 6th May
12 – 10 = 2 am
2 am – 9 hours = 5 pm previous day
8 pm Monday 5th May
In one go: Texas is 16 hours behind.
Rather than subtracting 16 this is the same as adding 8
(but previous day) i.e. Texas 5th May 12 + 8 = 20 = 10 pm
Comment:
Very well done.
The main mistake was not going back a day when going back from London to Texas time.
– Q2 P3 –
11
A parallelogram has base (2x – 1) m and height (4x – 7) m.
2
The area of the parallelogram is 1 m2.
Show that " ! ! ! # ! + $ = % .
( ! ! ! ")( # ! ! $ ) = "
" % ! ! ! "% ! + $ = "
" % ! ! ! "% ! + & = ' # ! ( # ! ! ! ( ! + )) = '
" # !! ! ( ! + ) = '
Comment:
Poorly done.
Many students attempted methods which made it clear they did not understand the question.
End of Question 2 Solutions
– Q2 P4 –
2x°
NOT TO
SCALE
Question 3 (18 marks)
x°
Question 3 (18 marks)
1
!
(x – 10)°
"
Simplify # ! $# !% .
1
Find the value of x.
Circle the correct response.
A.
2
1
B.
x3
2x3
C.
D.
(2x)0 = 1
Express
4x3
Answer(a) x =
2
Marking Scheme
Marker’s Comments
!!
!!
!"
+
"
!"
#
1 mark – Correct answer.
• Majority
of candidates
answered
(b) (i) Write the four missing terms
in the table
for sequences
A, B, Cthis
and D.
question
correctly.
as a single fraction with no negative indices
Term
1
2
3
4
5
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Sequence A
–4
2
5
8
…………………………………………………………………………………………………………
Sequence B
1
4
9
16
25
…………………………………………………………………………………………………………
Sequence C
5
10
15
20
25
…………………………………………………………………………………………………………
Sequence D
3
6
14
24
36
50
(i)
Expand and simplify (2x – 1)(x + 4).
1
(ii) Which term in sequence D is equal to 500?
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Answer(b)(ii)
(ii)
Hence simplify
2
2
x − 16
.
Marking
Marker’s Comments
(c) Scheme
Simplify
2
1 mark – Correctly converting both terms
into
•
Common
mistakes included:
2x + 7x − 4
fractions.
!" ! + !
- Not simplifying
"
…………………………………………………………………………………………………………
!" !
1 mark – Correct final answer.
!
- Incorrectly stating !" !! =
…………………………………………………………………………………………………………
!"
- Incorrectly simplifying
…………………………………………………………………………………………………………
!!
" !!
+
=
"! "! "!
…………………………………………………………………………………………………………
!
=
…………………………………………………………………………………………………………
"
Answer(c)
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
© UCLES 2012
0580/42/M/J/12
…………………………………………………………………………………………………………
– Q3 P1 –
– 13 –
3
Marking Scheme
1 mark – Correct answer.
Marking Scheme
1 mark – Correct factorisation of numerator.
•
•
Marker’s Comments
Generally, well done by candidates.
Common silly error was stating
! ! ! ! = !" !#
•
Marker’s Comments
Many candidates did not use the result from
part (i) despite the prompt “Hence.”
1 mark – Correct final answer.
•
– Q3 P2 –
Incorrect cancellation seen, i.e. cancelling
! ! or cancelling –16 and –4 in
! ! ! "#
$ which is invalid as these are
! !! + % ! ! &
not common factors.
Marking Scheme
1 mark – Factorising one pair of terms
correctly.
•
Marker’s Comments
Poorly done overall.
•
Many missed the difference of two squares
factorisation.
•
Silly errors in factorising by grouping in
pairs were also common.
1 mark – Correct grouping in pairs
factorisation.
1 mark – Correct final answer.
– Q3 P3 –
Marking Scheme
1 mark – Correct area of circle.
•
1 mark – Correct side length of square or area of
one triangle (depending on the method
used.)
•
1 mark – Correct final answer in exact form.
– Q3 P4 –
Marker’s Comments
Many candidates struggled to calculate the
area of the square correctly.
Some candidates were penalised for not
leaving their final answer in exact form, as
required.
Marking Scheme
1 mark – Correctly expanding the LHS.
•
Marker’s Comments
Many candidates struggled to answer this
question correctly.
1 mark – Finding both correct values for y.
•
1 mark –Determining both correct values for f.
– Q3 P5 –
Some candidates were making the mistake
in only giving one value for y2 = 25 when
there should be 2 values.
Marking Scheme
1 mark – Correct lowest common denominator
(LCD) for all 3 fractions.
•
1 mark – Correct expansion of both
!" ! # !$! ! + %$ &'()*!+ !$! ! + %$,
Marker’s Comments
Many candidates made silly arithmetic
errors and struggled to add algebraic
fractions with different denominators
correctly.
Some also incorrectly cancelled
denominators, which is not valid.
1 mark – Correct final simplified answer.
– Q3 P6 –
Question 4 (12 marks)
1
Let n be an integer.
3
By factorising, or otherwise, prove that
35 + (3n + 1)2 – 2n(4n – 3)
is always a perfect square number.
…………………………………………………………………………………………………………
36 n 12h
35 9m on 1 8m on
…………………………………………………………………………………………………………
Iiii
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
which is always a perfect
square
…………………………………………………………………………………………………………
Common mistakes
9m 1
…………………………………………………………………………………………………………
3h 1
Recall a b
…………………………………………………………………………………………………………
a Lab b
8m on
2n 4h 3
…………………………………………………………………………………………………………
becomes
…………………………………………………………………………………………………………
Recall
minusminus
plus
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Studentstried to testvalues of n this does not prove that it
…………………………………………………………………………………………………………
is always a perfect
You would have to testinfinitely
square
…………………………………………………………………………………………………………
numbers
many
…………………………………………………………………………………………………………
e
1 2 3 4,5 6 Can we assume60 is always
60 is divisible
g
by
…………………………………………………………………………………………………………
number
divisible
by any
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
students need to revise how tofactorise quadratics
– 18 –
Problem C
Diagonal Distance
2
Square ABCD
hasABCD
K on has
BC,KL on
onBC,
DC,LMonon
AD,
N onand
ABNsuch
thatsuch
KLMN
Square
DC,
Mand
on AD,
on AB
that
forms a rectangle.
KLM N forms a rectangle, 4AM N and 4LKC are congruent isosceles
As well, Δtriangles,
AMN and
LKC4M
areDL
identical
isosceles
triangles,isosceles
and also
Δ MDLIf and
andΔalso
and 4BN
K are congruent
triangles.
the Δ BNK are
total area triangles.
of the four triangles is 50 cm2 , what is the length of M K?
identical isosceles
If the total area of the four triangles is 50 cm2 , what is the length of MK?
3
y
y
let
…………………………………………………………………………………………………………
MD DL X and KC LC
y
…………………………………………………………………………………………………………
X
50
Atriangles
y
…………………………………………………………………………………………………………
Using Pythagoras Theorem ML Jx x
J
2
2
…………………………………………………………………………………………………………
Fy
…………………………………………………………………………………………………………
Similarly LK
…………………………………………………………………………………………………………
KLMN is a
rectangle
…………………………………………………………………………………………………………
LMLK 90
J
…………………………………………………………………………………………………………
J 2 2 242
mainly x 1 and y 7
…………………………………………………………………………………………………………
J2 X y
whatabout x JT4
…………………………………………………………………………………………………………
ML
LK
Using Pythagoras Theorem MK
100
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
10CM
…………………………………………………………………………………………………………
students tried to let x and be numerical values You havemade
y
…………………………………………………………………………………………………………
the question easier and haven'taccounted for all possible x and
y values Usepronumerals instead
– 19 –
4
Problem E
Stained Glass
3
A stained
glass window
is in the
shapeisofina the
rectangle
length ofwith
8 cma and
a width
A stained
glass window
hanging
shape with
of a arectangle
length
of
3
8 cm and a width of 6 cm.
of 6 cm.
Rectangle
represents
the window
AB
Rectangle
ABCDABCD
represents
the window
with ABhanging
= 8 andwith
BC =
6. = 8 and BC = 6.
The points E, F, G, and H are the midpoints of sides AB, BC, CD, and AD,
The points E, F, G, and H are the midpoints of sides AB, BC, CD, and AD, respectively.
respectively. The point J is the midpoint of line segment EH. Triangle F GJ is
The point
J is the
midpoint of line segment EH.
coloured
blue.
3
3
1
4
5
3
5
5
3
4
4
What is the area of triangle FGJ ?
Determine the area of the blue triangle.
…………………………………………………………………………………………………………
Method 1
…………………………………………………………………………………………………………
Using
Pythagoras Theorem HEMeEF FGto at 132 42 5
…………………………………………………………………………………………………………
EFAH is a rhombus four sides
equal
Ex
diagonal I diagonal 2
…………………………………………………………………………………………………………
EG HF
AEFan
recall Arnombus
6 8
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
24
…………………………………………………………………………………………………………
since J is the midpoint of HE
24
…………………………………………………………………………………………………………
AAFaJ
…………………………………………………………………………………………………………
12cm
…………………………………………………………………………………………………………
See Extra Working Page for Method 2
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Poorly done
…………………………………………………………………………………………………………
HE 5 which was a good
Many students
got
startingpoint
…………………………………………………………………………………………………………
assumed LJHA or LH A 90
Manystudents
incorrectly
Some students incorrectly assumed EFGH is a square
– 20 –
4
In the diagram below, points D, B and F are collinear (i.e. lie on a straight line).
3
Points C, A and F are also collinear.
Find x, showing all necessary working. Leave your answer in simplest surd form.
F
I
1
LBDA 90 converse of Pythagoras Theorem
…………………………………………………………………………………………………………
J
…………………………………………………………………………………………………………
82 42 80 455
Using Pythagoras Theorem AB
…………………………………………………………………………………………………………
FE bisects AB altitude in isoscelestriangle bisectsbase
LAB
…………………………………………………………………………………………………………
AE
120
or
255
…………………………………………………………………………………………………………
J
…………………………………………………………………………………………………………
52_ 255
x
…………………………………………………………………………………………………………
25 20
55
…………………………………………………………………………………………………………
units
…………………………………………………………………………………………………………
Some students wrote AB 182 42 458
…………………………………………………………………………………………………………
students realised CBDA 90 but didn't
justifywhy
Many
…………………………………………………………………………………………………………
sinceit's not labelled in the diagram students mustexplain
why
…………………………………………………………………………………………………………
and explicitly mention converse
…………………………………………………………………………………………………………
It is disappointing to see students are either not
the
reading
…………………………………………………………………………………………………………
feedback properly or have not learnt from their mistakes CCT1
in
Note
5802
540 similarly 1
End of paper
– 21 –
54
Extra Working Page for Question 4 only
If you use this space, clearly indicate which question you are answering.
Method
6
Az
1.5
As
A
6
45
As
4
4
AAFas
Ay
3
J is the midpoint of HE
i X is the midpoint of DG
Y is the midpoint of BF
s
12cm
t
Ii.tt o 2x3x4 Ex4.5x2
– 22 –