Download 2010 Round 3 - NLCS Maths Department

HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
SECTION 1 - STARTERS (INDIVIDUAL)
Marks: 2 marks to either or both competitors for the correct answer
Time: 30 seconds.
Year
13
1)
Given that secθ + tanθ = 4, find the value of secθ – tanθ .
12
2)
Given that a + b = 11.2 and a – b = 7,
find the value of a2 – b2 .
10-11 3)
The diagonals of a rhombus are 12 cm and 16 cm long.
Find the length of each side of the rhombus.
7-9
4)
Multiply the mean of the numbers 2, 3, 5, 7, 11 by their median.
13
5)
Find all the solutions of the equation y3 – y2 – 42y = 0.
12
6)
Given that cosθ =
10-11 7)
7-9
8)
and θ is acute, find the value of tanθ .
Solve the equation
Calculate
, simplifying your answer.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
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)
Each diagonal of a square is x units long.
Find the area of the square in terms of x .
Each longer side of a kite is twice as long as each shorter side.
The shorter sides meet at right angles.
Find the cosine of the angle between the two longer sides.
PA and PB are tangents to the
circle whose centre is C.
CP cuts the circle at Q.
Angle APC is 25º.
Find the size of angle AQB .
A
C
Q
P
B
12-13 4)
The circles (x – 1)2 + (y – 3)2 = 100
and
(x – 12)2 + (y – 6)2 = k
intersect at two points. At each of these points the tangents to the
two circles are perpendicular.
Calculate the value of k .
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
SECTION 3 - MENTAL ARITHMETIC AND PROBABILITY
(INDIVIDUAL)
Marks: 2 or 1 to opponent
Time: 60 seconds
All questions are to be done mentally
Year
7-9
1)
Calculate 12 + 32 + 42 + 72 + 82 .
2)
Calculate 22 + 32 + 52 + 62 + 92 .
10-11 3)
4)
The cube root of 6859 is an integer. Find it.
The cube root of 12167 is an integer. Find it.
Give the answers to Questions 5 – 8 as fractions in simplest form.
In Questions 5 and 6, a bag contains five balls numbered 2, 3, 5, 7 and 9.
Two balls are drawn at random from the bag and not replaced - so the two
numbers are different.
12
5)
Find the probability that the product of the two numbers is even.
6)
Find the probability that the sum of the two numbers is even.
In Questions 7 and 8, the same bag contains the numbers 2, 3, 5, 7 and 9.
A ball is drawn at random from the bag, the number is noted, the ball is
replaced, and another ball is drawn – so the same number could be drawn twice.
13
7)
Find the probability that the sum of the two numbers is even.
8)
Find the probability that the product of the two numbers is even.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
SECTION 4 - TEAM QUESTION
Time: 5 minutes.
You are to make the number 36 from each given set of four integers on the sheet
provided, using +, –, ×, ÷ or brackets. The four integers do not have to be used
in order, but must all be used, and repeated integers used as often as they are
given. Integers must not be used as parts of a larger number (so for example 1
and 2 cannot be used to make 12).
Examples: {1, 2, 3, 15} gives 36 = 2(15 + 3) × 1
{2, 2, 3, 3} gives 36 = 2 × 2 × 3 × 3
Marks:
Give 1 point for each correct answer, with no penalty for incorrect answers or
omissions. Winning team scores 5 marks, losing team scores 5 – D marks, where
D is the difference between the scores of the two teams. Minimum mark 0.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
SECTION 4 - TEAM QUESTION
Answer Sheet
{1, 2, 3, 4}
{1, 2, 3, 5}
{1, 2, 3, 6}
{1, 2, 3, 7}
{1, 2, 4, 8}
{1, 2, 3, 9}
{1, 2, 3, 11}
{1, 2, 3, 13}
{2, 5, 7, 9}
{2, 5, 6, 7}
{2, 5, 8, 8}
{2, 6, 7, 8}
{1, 2, 5, 5}
{2, 2, 6, 6}
{2, 3, 7, 7}
{2, 2, 8, 8}
{3, 3, 3, 3}
{3, 3, 3, 4}
{3, 3, 5, 6}
{3, 3, 6, 6}
{3, 3, 6, 7}
{3, 5, 8, 9}
{3, 3, 9, 9}
{4, 4, 4, 4}
{5, 5, 5, 6}
{6, 6, 6, 6}
{3, 4, 5, 6}
{4, 5, 6, 7}
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
36 =
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
SECTION 4 - TEAM QUESTION
Solutions (not unique)
{1, 2, 3, 4}
{1, 2, 3, 5}
{1, 2, 3, 6}
{1, 2, 3, 7}
{1, 2, 4, 8}
{1, 2, 3, 9}
{1, 2, 3, 11}
{1, 2, 3, 13}
{2, 5, 7, 9}
{2, 5, 6, 7}
{2, 5, 8, 8}
{2, 6, 7, 8}
{1, 2, 5, 5}
{2, 2, 6, 6}
{2, 3, 7, 7}
{2, 2, 8, 8}
{3, 3, 3, 3}
{3, 3, 3, 4}
{3, 3, 5, 6}
{3, 3, 6, 6}
{3, 3, 6, 7}
{3, 5, 8, 9}
{3, 3, 9, 9}
{4, 4, 4, 4}
{5, 5, 5, 6}
{6, 6, 6, 6}
{3, 4, 5, 6}
{4, 5, 6, 7}
36 = (1 + 2) × 3 × 4
36 = (5 + 1) × 3 × 2
36 = 1 × 2 × 3 × 6
36 = (7 – 1) × 2 × 3
36 = 4(8 + 2 – 1)
36 = (1 + 2)(3 + 9)
36 = (11 – 2)(3 + 1)
36 = 3 × 13 – 2 – 1
36 = 9 × 2(7 – 5)
36 = (7 + 5) × 6 ÷ 2
36 = 8 × 5 – 8 ÷ 2
36 = 6 × 7 – 8 + 2
36 = 5(5 + 2) + 1
36 = 6 × 6 + 2 – 2
36 = (7 – 3)(7 + 2)
36 = (8 – 2)(8 – 2)
36 = (3 + 3)(3 + 3)
36 = 4(3 + 3 + 3)
36 = 3 × 6(5 – 3)
36 = 6 × 6 + 3 – 3
36 = 7 × 6 – 3 – 3
36 = 9 × 8 ÷ (5 – 3)
36 = (9 – 3)(9 – 3)
36 = 4(4 + 4) + 4
36 = 5 × 5 + 5 + 6
36 = 6 × 6 + 6 – 6
36 = 6(5 + 4 – 3)
36 = 6(7 – 5 + 4)
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
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)
Given that
12 + 32 + 42 + 62 + 72 + 82 = 28 – mn ,
where m and n are positive integers,
find two possible pairs of values of m and n.
Write 13 – 22 + 33 – 42 + 53 – 62 + 73 – 82 + 93
as the product of prime factors.
12
3)
Evaluate
, giving your answer as a fraction.
13
4)
Evaluate
,
giving your answer as a fraction in lowest terms.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
SECTION 6 - ALGEBRA AND CALCULUS (INDIVIDUAL)
Marks: 2 or 1 for opponent
Time: 60 seconds.
Throughout this round [x] is defined to be the largest integer less than or equal
to x.
For example, [6.2] = 6, [-7.3] = -8 .
Year
7-9
1)
Evaluate [-1 ] + [1 ] .
2)
Evaluate [-2 ] × [3 ] .
10-11 3)
12
13
Evaluate [7.5 ÷ 2.5] – [7.5] ÷ [2.5] .
4)
Evaluate [2.52] – [2.5]2 .
5)
Given that [y] = y – 0.5, solve the equation
2.5[y] – [2.5]y = 3.5 .
6)
Given that [y] = y – 0.5, solve the equation
3.5[y] + [3.5]y = 14.5 .
7)
Sketch the graph of y = [x]2 for 0 ≤ x ≤ 4 .
8)
Evaluate
.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
SECTION 7 - RACE (INDIVIDUAL)
Marks: 2 or 0
Time: 60 seconds.
Year
7-9
1)
Find, correct to the nearest integer, the square root of
the number of hours in a week.
10-11 2)
Find, correct to the nearest integer, the square root of
the number of inches in a metre.
12
3)
Assume that 1 mile = 1.6 km and that 1 gallon = 4.5 litres.
A car travels 45 miles per gallon of petrol.
Find how many kilometres per litre this is.
13
4)
Given that x percent of single-digit prime numbers are even,
find the square root of x .
7-9
5)
The sizes of the angles of a triangle are in the ratio 1 : 3 : 6 .
Find the size of the largest angle.
10-11 6)
The area of a circle is 250 cm2.
Find, to the nearest centimetre, the length of the diameter.
, and k ≠ 1, find the value of k .
12
7)
Given that
13
8)
Find the coordinates of the minimum point on the graph of
y = 3x2 – 12x + 5 .
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
ROUND 3
ANSWERS (allow equivalent answers)
SECTION 1
SECTION 5
1.
2.
3.
4.
5.
6.
7.
8.
1.
1/4
78.4
10 (cm)
28
0, 7, -6 (all required)
5/12
3.75
3
2.
3.
Two of m = 3, n = 4
or m = 9, n = 2 or m = 81, n = 1
5 × 13 × 17
4.
SECTION 6
SECTION 2
1.
2.
3.
4.
2
x /2
3/4
115 (º)
30
SECTION 3
1.
2.
3.
4.
5.
6.
7.
8.
139
155
19
23
2/5
3/5
17/25
9/25
SECTION 4
Please see question sheet
1.
2.
3.
4.
5.
6.
7.
-1
-9
-0.5
2
9.5
2.5
Essential points:
Step function
with steps of
increasing size.
Be generous!
y
9
4
1
0
8.
14
SECTION 7
1.
2.
3.
4.
5.
6.
7.
8.
13
6
16
5
108 (º)
18 (cm)
9
(2, -7)
1
2
3
4
x