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Practice
Questions
international competitions and assessments for schools
ICAS Mathematics Practice Questions Paper I © EAA 2
1. The diagram below represents the products of (x + 5) and (3x + 2).
NOT TO SCALE
3x 2
5
x
3x + 2
x + 5
What product is represented by the shaded rectangle?
(A) 2x(B) 6x(C) x2
(D) 3x2
2. Jules has a package gift-wrapped, as shown.
10 cm
10 cm
30 cm
What is the volume, in cm3, of the package?
(A) 50(B) 300(C) 1400(D) 3000
3. During 2001, Australia exported goods to a total value of $123 000 million.The graph shows the percentage of these goods exported to different parts of the world.
41%
12%
10%
19% Japan
Key
USA
ASEAN
EU
China
Other12% 6%
What was the value, in millions of dollars, of the goods exported to China?
(A) 7 380(B) 12 300(C) 14 760(D) 23 370
3 ICASMathematicsPracticeQuestionsPaperI©EAA
4. Janewastossingacoin,butonesideofthecoinwasweightedmoreheavilythantheother.
Herearetheresultssheobtained.
Basedonherresults,whichoftheseisthebestestimateoftheprobabilityofgettingaheadinasingletossofJane’scoin?
(A) 0.4(B) 0.5(C) 0.6(D) 0.7
5. ThispictureisbasedonthestyleoftheDutchartistPietMondrian(1872–1944).
2
NOT TO SCALE
26
6
4
4
Whichexpressiongivesthetotalareaofthethreecolouredrectanglesinthepicture?
(A) 6a2 + 48ab – 32b2
(B) 20a2 + 24b2
(C) 4a2 + 36ab – 24b2
(D) 6a2 + 16b2
6.
Y = 5 F = 10 M =100 000 L = 120 000
L = M 1 – FX
Y
WhatisthevalueofXcorrecttotwodecimalplaces?
(A) –0.09(B) –0.02(C) 0.02(D) 0.09
7. Theareaofoneoftheseisoscelestrianglesis60squareunits.
base
Thelengthofthebaseofthetriangleis10units.
Thegridbelowismadeupoftrianglesexactlythesamesizeasthetriangleabove.
base
Whatistheperimeteroftheshadedshape? (A) 105units (B) 150units (C) 160units (D) 162units
ICAS Mathematics Practice Questions Paper I © EAA 4
8. Anna forgot the code of a 3-digit lock on her case. The code was made of 3 digits ranging from 0 to 9. She remembered that the first digit was less than 5, the second digit was an odd number, and the third one was either 7 or 8. There were no identical digits in the code.
How many different combinations could possibly open her lock?
(A) 25(B) 36(C) 41(D) 50
9. The diagram represents a regular octagon of area 500 cm2.
How long, to the nearest centimetre, is one side of the octagon?
10. In the diagram H represents the position of a hawk hovering above the ground, and M the position of a mouse on the ground.
M
H
500
050100 100150 150200 200250
250
200
150
100
50
ALL MEASUREMENTS IN CENTIMETRES
The mouse moves to a new position N, which is 50 cm from position M.
What is the maximum possible distance, in cm, from H to the new position N correct to the nearest whole number?
END OF PAPER
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5 ICAS Mathematics Practice Questions Paper I © EAA
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FIRST NAME to appear on certificate LAST NAME to appear on certificate
Are you male or female? Male Female
Does anyone in your home usually speak a language other than English? Yes No
School name:
Town / suburb:
Today’s date: Postcode:
CLASSDATE OF BIRTHDay Month Year
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HOW TO FILL OUT THIS SHEET:
• Ruboutallmistakescompletely.• Printyourdetailsclearly intheboxesprovided.• Makesureyoufillinonly oneovalineachcolumn.
EXAMPLE 1: Debbie BachFIRST NAME LAST NAME
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M THE UNIVERSITY OF NEW SOUTH WALES
International Competit ions and Assessments for Schools
PRACTICE QUESTIO
NS
*045911*
PaPer
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Your privacy is assured as EAA fully complies with appropriate Australian privacy legislation. Visit www.eaa.unsw.edu.au for more details.
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International Competit ions and Assessments for SchoolsM
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MULTIPLE CHOICE FREE RESPONSE
Example: 4+6= Example:6+6=
(A) 2 ●Theansweris12,soWRITEyour (B) 9 answerintheboxes. (C) 10 ●WriteonlyONEdigitineachbox, (D) 24 asshown,andfillinthecorrectoval, asshown.Theansweris10,sofillintheoval,asshown.C
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ICAS Mathematics Practice Questions Paper I © EAA
QUESTION KEY SOLUTION STRAND LEVEL OF DIFFICULTY
1 AThe shaded rectangle has a side of 2 and a side of x. Therefore, the product of these two sides is 2x.
Algebra and Pattern Easy
2 DVolume of a box = length × width × heightV = 30 × 10 × 10V = 3000 cm3
Measurement Easy
3 A6% of the total products are exported to China.6% of 123 000 million = 0.06 × 123 000 million = $7380 million
Chance and Data Easy
4 C
Experimental Probability equals:
Number of times an event has occurredNumber of trials
67 + 286 + 581 + 2989100 + 500 + 1000 + 5000
Applying this formula:
Number of times an event has occurredNumber of trials
67 + 286 + 581 + 2989100 + 500 + 1000 + 5000
= 0.59
Therefore, the best estimate is 0.6.
Chance and Data Medium
5 A
The green area is 2a × a = 2a2
The orange area is 4a × a = 4a2
The blue area is = length × width = (6b + 2b) × (6a − 4b)= 8b × (6a − 4b)= 48ab − 32b2
Therefore the total area that is coloured is = green + orange + blue= 2a2 + 4a2 + 48ab − 32b2
= 6a2 + 48ab − 32b2
Algebra and Pattern Medium
6 A
Substitute all values in the given expression:
120000 = 100000(1− )10X5
6 = 5(1− )10X5
= (1− )10X5
65
1− =10X5
65
X = 5(1−10 )65
By simplifying you will get:120000 = 100000(1− )10X5
6 = 5(1− )10X5
= (1− )10X5
65
1− =10X5
65
X = 5(1−10 )65
, then divide by 5 to get:
120000 = 100000(1− )10X5
6 = 5(1− )10X5
= (1− )10X5
65
1− =10X5
65
X = 5(1−10 )65
By applying the tenth root to the two sides of the equation:
120000 = 100000(1− )10X5
6 = 5(1− )10X5
= (1− )10X5
65
1− =10X5
65
X = 5(1−10 )65
Make X the subject of the equation:
120000 = 100000(1− )10X5
6 = 5(1− )10X5
= (1− )10X5
65
1− =10X5
65
X = 5(1−10 )65
X = −0.09
Algebra and Pattern Medium/Hard
ICAS Mathematics Practice Questions Paper I © EAA
7 C
The perimeter of the shaded shape is made of 11 bases of the given triangle + 2 heights + 2 sides.You need to calculate the height and the side of the triangle first.Calculating the height: the area is given (60 square units), therefore you can find the height using the formula: Area = ½ × height
× base, h × 10
260 =
s = 122 + 52
which gives h = 12 unitsCalculating the side by cutting the triangle down the middle, a right-angled triangle is formed with base = 5 cm and height = 12 cm. The unknown side is called s. Using Pythagoras’ Theorem, h2 + 52 =s2. Substitute h = 12 and find s:
h × 10 260 =
s = 122 + 52 . Therefore, s = 13 units.Now add all the values in the original calculation: 11 bases of the given triangle + 2 heights + 2 sides.11 × 10 + 2 × 12 + 2 × 13 = 160 units.
Space and Geometry Hard
8 C
The first two digits filled 5 × 5 ways, the last digit 2 ways BUT must delete117, 118, 337,338, 077,177,277,377 and 477.5 × 5 × 2 – 9 = 41.
Chance and Data Hard
9 10
This item requires a lot of work and some knowledge of Pythagoras’ Theorem and the internal angles of an octagon.
There are several approaches, but they all work by finding the area of triangles, rectangles or trapeziums. From there, an equation that relates the area of the octagon to the length of the side can be written.One method is to think of the octagon as a rectangle between two trapeziums. A different way is to think of it as a square with triangles cut off.
y
y
y
x
x
x
x
45°
22xx
2)x2 500 = 2(1+
2x
Space and Geometry Hard
ICAS Mathematics Practice Questions Paper I © EAA
y
y
y
x
x
x
x
45°
22xx
2)x2 500 = 2(1+
2x
Either method results in the same answer.
Let’s approach it using the second way:We need to calculate y first.using Phythagoras’ Theorem:
y y x
y x
2 2 2
2
+ =
=
The side of the square equals x + 2y
that is x x+ 22
The area of the square is x x+
22
2
The area of each of the four triangles on the
corners is 12
12 2 2 4
2
y y x x x× = ×× × =
Therefore,area of square – area of 4 triangles = 500
x x x
x x x x x
x
− =
×
− =
=
22
500
22 22
500
242
500
2
2
2 2 2
2
Solving for x gives an answer of 10.176, cm.Which is 10 to the nearest centimetre.
ICAS Mathematics Practice Questions Paper I © EAA
10 190
Apart from reading 3D coordinates the main mathematics in this question is Pythagoras’ Theorem.If we look at the mouse and the hawk from above we would see this:
MH
NH
1002 + 502
161.82 + 1002
N
H
161.8 cm
100 cm
The line shows the hawk’s path. The distance along the ground of this path (the horizontalcomponent) is
MH
NH
1002 + 502
161.82 + 1002
N
H
161.8 cm
100 cm
. This is about 111.8 cm. The mouse runs 50 cm away from the hawk to a new position ‘N’. The mouse can run in any direction but wants to maximise the distance from the hawk. This means he should run in the same direction as the line NH in the diagram below.
MH
NH
1002 + 502
161.82 + 1002
N
H
161.8 cm
100 cm
Measurement HardAlong the ground this gives a distance of 111.8+50=161.8 cmThis is just the horizontal distance. Fortunately for the mouse, the hawk is further away than that because it is hovering above the ground at a height of 100 cm.We can show this on a new diagram from a different point of view.
MH
NH
1002 + 502
161.82 + 1002
N
H
161.8 cm
100 cm
We can now use Pythagoras’ Theorem again to find the distance from the hawk to the mouse.
MH
NH
1002 + 502
161.82 + 1002
N
H
161.8 cm
100 cm
This gives an answer of 190.2 cm. To the nearest whole number this is 190.CommentThe underlying mathematics in this problem is not very difficult and boils down to two instancesof Pythagoras’ Theorem. As a problem, though, the question is more difficult. Students have to realise that Pythagoras’ Theorem is the appropriate piece of mathematics to use and have to extract information presented in an unusual way. Also some insight is required to understand in what direction the mouse should run.
ICAS Mathematics Practice Questions Paper I © EAA
Level of difficulty refers to the expected level of difficulty for the question.
Easy more than 70% of candidates will choose the correct option
Medium about 50–70% of candidates will choose the correct option
Medium/Hard about 30–50% of candidates will choose the correct option
Hard less than 30% of candidates will choose the correct option
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