Download 2011 Round 2 - NLCS Maths Department

HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 1 - STARTERS (INDIVIDUAL)
Marks: 2 marks to either or both competitors for the correct answer
Time: 30 seconds.
Year
13
1)
Write the algebraic fraction
12
2)
Find the median of the square numbers 1, 4, 9, …, 100 .
10-11 3)
in its simplest form.
Find the largest prime factor of 1650 .
7-9
4)
Find what time it is 900 minutes after 3.50 p.m.
13
5)
Find the coordinates of the minimum point on the graph of
.
12
6)
Find the coordinates of the point of intersection of the lines
and
.
10-11 7)
7-9
8)
Solve completely the equation
Evaluate 34 – 43 .
.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 2 - GEOMETRY AND TRIGONOMETRY (PAIRS)
Marks: 2 marks to either or both pairs for the correct answer
Time: 90 seconds.
Year
7-11
1)
12-13 2)
7-11
3)
Three regular polygons meet at a point as shown.
Polygon A has nine sides, polygon B is an
equilateral triangle.
Find how many sides polygon C has.
The sides of a triangle are cm, cm and
Find the sine of its largest angle.
A
C
cm.
D
Angles BAD and ACB are right
angles.
AB = 12 cm and AD = 5 cm.
Find the length of AC.
C
5 cm
A
12-13 4)
B
B
12 cm
A
Angle ADC is a right angle,
AD is 10 cm long,
sin ACD = and tan ABD = .
Find the length of BC.
10 cm
B
C
D
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 3 - MENTAL ARITHMETIC AND PROBABILITY (INDIVIDUAL)
Marks: 2 or 1 to opponent
Time: 60 seconds
Questions 1 - 4 are to be done mentally.
For Questions 1 - 4 you are given that 219 × 357 = 78183.
Year
7-9
1)
Evaluate 2.19 × 0.0357
2)
Evaluate 78.183 ÷ 0.219
10-11 3)
4)
Evaluate 78183 ÷ 73
Evaluate 657 × 119
Paper and pencil may be used for Questions 5 – 8.
In a board game a player rolls a fair six-sided die, numbered from 1 to 6
as normal.
If he rolls a 1, 2, 3 or 4 he moves forward that number of spaces.
If he rolls a 5 or 6 he moves forward 10 or 12 spaces respectively.
Giving your answer as a fraction in simplest form, find the probability of each of
the following outcomes after TWO rolls of the die.
12
13
5)
He moves forward 22 spaces.
6)
He moves forward 2 spaces.
7)
He moves forward 16 spaces.
8)
He moves forward 14 spaces.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 4 - TEAM QUESTION
Time: 5 minutes.
Use the numbers 2, 4, 6 and 8 precisely once each to form as many positive
integers as you can, starting from 1, with no omissions.
You may use the numbers in any order and the symbols
+ , – , ×, ÷, brackets and factorials (!).
[Note: 4! = factorial 4 = 4 × 3 × 2 × 1 = 24.]
Example (which you may not use): 1 = 6 ÷ 2 – 8 ÷ 4.
Marking will stop at the first error or omission.
The team captain is to produce a single list of answers on the sheet provided. An
extra minute after the 5 minutes is allowed to organise this.
Marks:
Stop at the first error or omission.
Winning team gets 5 points, losers get (5 – D) points, where D is the difference in
the numbers of correct answers produced by the teams, with minimum score zero.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 4 - TEAM QUESTION
Answer sheet
1=
11 =
21 =
2=
12 =
22 =
3=
13 =
23 =
4=
14 =
24 =
5=
15 =
25 =
6=
16 =
26 =
7=
17 =
27 =
8=
18 =
28 =
9=
19 =
29 =
10 =
20 =
30 =
etc.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 4 - TEAM QUESTION
Solutions (not unique)
1 = (8 + 2) ÷ (4 + 6)
11 = 8 + 6 ÷ (4 – 2)
21 = (8 – 4)! – 6 ÷ 2
2 = (6 – 8 ÷ 4) ÷ 2
12 = (8 – 4) × 6 ÷ 2
22 = 8 + 6 + 4 × 2
3 = (8 + 6 – 2) ÷ 4
13 = (8 × 4 – 6) ÷ 2
23 = 4! – (6 + 2) ÷ 8
4 = (8 + 6 + 2) ÷ 4
14 = 2 × 8 + 4 – 6
24 = 4! × (6 + 2) ÷ 8
5 = (8 + 2 × 6) ÷ 4
15 = 8 + 4 + 6 ÷ 2
25 = 4! + (6 + 2) ÷ 8
6=6+8÷2–4
16 = 8 + 6 + 4 – 2
26 = 4! + 6 – 8 ÷ 2
7 = (8 + 6) ÷ (4 – 2)
17 = 4! – (8 + 6) ÷ 2
27 = (8 – 4)! + 6 ÷ 2
8=8+6–4–2
18 = 2 × 8 + 6 – 4
28 = 4! + 8 + 2 – 6
9 = (8 + 6 + 4) ÷ 2
19 = 4! – 8 + 6 ÷ 2
29 = 8 × 4 – 6 ÷ 2
10 = 6 × 2 – 8 ÷ 4
20 = 8 + 6 + 4 + 2
30 = 4 × 6 + 8 – 2
etc.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 5 - CALCULATORS (INDIVIDUAL)
Marks: 2 to either or both competitors for the correct answer
Time: 90 seconds
You are reminded that the written questions are to be given simultaneously
to the respective pupils at the beginning of this section.
Year
7-9
1)
10-11 2)
Evaluate
when x = 3.456,
giving your answer correct to 4 significant figures.
Given that x is an acute angle whose sine is 0.2345, evaluate
giving your answer correct to 4 significant figures.
12
3)
The equation
between x = 2 and x = 2.5.
has a solution
Obtain this solution correct to 2 significant figures.
13
4)
In this question θ is measured in radians.
Given that θ1 = 1 and
,
find the limiting value of the sequence θ1, θ2, θ3, …
correct to 3 significant figures.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 6 - ALGEBRA AND CALCULUS (INDIVIDUAL)
Marks: 2 or 1 for opponent
Time: 60 seconds.
In Questions 1 to 6, [a, b] is defined to equal a2 – ab .
For example:
[4, 3] = 42 – 4 × 3 = 16 – 12 = 4.
Year
7-9
1)
Find the value of [-3, -2] .
2)
Find the value of [-2, -5] .
10-11 3)
4)
Find all the solutions of [2y, 3] = 0 .
Find all the solutions of [y, 3 – y] = 0 .
In Questions 5 and 6, simplify the expressions and give the answer in
fully factorized form, for example 3y(y + 2) rather than 3y2 + 6y .
12
5)
Simplify [1 – y, y + 1] .
6)
Simplify [2 + y, 2 – y] .
In Questions 7 and 8, you are reminded that |a| = a if a ≥ 0 and |a| = -a if a < 0.
13
)
8)
Find the coordinates of the lowest point of the graph of
y = |x + 4| – 2 .
Evaluate
.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
SECTION 7 - RACE (INDIVIDUAL)
Marks: 2 or 0
Time: 60 seconds.
Year
7-9
1)
10-11 2)
1444 is a square number. Find its positive square root.
1 kilogram is 2.2046 pounds. Find to the nearest integer how many
kilograms are equivalent to 55 pounds.
12
3)
Find what fraction of positive two-digit integers are square numbers,
giving your answer in lowest terms.
13
4)
The area bounded by the positive x-axis, the line y = x, and a line
with gradient -1 is 25 square units.
Find the equation of this third line.
7-9
5)
The sizes of the angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6.
Find the size of the largest angle.
10-11 6)
Divide the number of degrees in the internal angle at each vertex of
a regular octagon by the number of degrees in the largest angle of
a 3, 4, 5 triangle.
12
7)
Given that
, find the value of k .
13
8)
Find in degrees the acute angle for which 2sinθ = secθ .
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2011/2012
ROUND 2
ANSWERS (allow equivalent answers except where stated)
SECTION 1
SECTION 4
1.
Please see separate answer sheet
2.
3.
4.
5.
6.
7.
8.
30.5
11
6.50 a.m. [a.m. required]
or 06.50
(2½, -2¼)
(5, 9)
9, -8 [both required]
17
3.
4.
1.
2.
3.
4.
18
3
-6
0, 3/2 [both required]
0, 3/2 [both required]
2y(y – 1)
[must be completely factorized]
6.
(cm)
27.5 (cm)
1.192
-0.2782
2.2
1.11
SECTION 6
1.
2.
3.
4.
5.
SECTION 2
1.
2.
SECTION 5
2y(y + 2)
[must be completely factorized]
7.
8.
(-4, -2)
SECTION 3
SECTION 7
1.
2.
3.
4.
5.
6.
7.
8.
1.
2.
3.
4.
5.
6.
7.
8.
0.078183
357
1071
78183
1/18
1/36
1/18
1/9
38
25
1/15
y = 10 – x
120 (º)
1.5
7
45 (º)