Download 2010 Knockout Semi-final

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Mathematics wikipedia , lookup

Addition wikipedia , lookup

Mathematics and architecture wikipedia , lookup

History of mathematics wikipedia , lookup

Arithmetic wikipedia , lookup

Secondary School Mathematics Curriculum Improvement Study wikipedia , lookup

Foundations of mathematics wikipedia , lookup

Ethnomathematics wikipedia , lookup

Elementary mathematics wikipedia , lookup

Transcript
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
SECTION 1 - STARTERS (INDIVIDUAL)
Marks: 2 marks to either or both competitors for the correct answer
Time: 30 seconds.
Year
7-9
1)
10-11 2)
Find the least possible number of coins in my change if I buy
three items costing £3.17 each with a £10 note.
Find the length of the hypotenuse of this triangle.
7 cm
24 cm
12
3)
Find the area under the curve
between x = 0, x = 1 and y = 0.
13
4)
Find the value of sin15ºcos15º.
7-9
5)
Find half of a half of a half of 100.
10-11 6)
A cylindrical plastic pipe has outer diameter 4 cm and inner
diameter 3 cm. Find the volume of plastic in 4 m of pipe.
12
7)
Find exact expressions for the solutions of the equation
x2 + 2x – 1 = 0 .
13
8)
Find the sum of the infinite geometric series
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
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)
12-13 4)
The figure shows a regular hexagon
with centre O.
Given that area A : area B = 1 : x ,
find x.
O
The figure shows a regular octagon
with centre O.
Given that area A : area B = 1 : x ,
find an exact expression for x.
A
B
O
The figure shows a circle with centre O
and one of its tangents.
Given that area A : area B = 1 : x ,
find an exact expression for x.
The figure shows a regular octagon
with centre O.
Given that area A : area B = 1 : x ,
find an exact expression for x.
[Hint: triangles OIK, JHO are
similar.]
A
B
A
B
45º
O
H
J
A
B
K
O
I
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
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)
A list starts with the year 2011 and then goes back 11 years at a
time, so that the list begins 2011, 2000, 1989, … .
Find the earliest year of the twelfth century to appear in the list.
2)
Find the least integer greater than 100 which is 5 more than a
multiple of 6 and 6 more than a multiple of 7.
10-11 3)
12
13
Find how many integers from 12 to 789 inclusive which
(like these two numbers) have digits which are consecutive
and increase from left to right.
4)
16 and 64 are two 2-digit squares where the units digit of one is the
tens digit of the other. Find how many other pairs like this there are.
5)
The numbers 1, 2, 3, 4 are arranged in random order.
Find the probability that 1 and 2 are next to each other.
6)
The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are arranged in random order.
Find the probability that the numbers 1, 2, 3 come in that order,
though not necessarily next to each other.
7)
Find how many distinct ordered pairs of integers (x, y) satisfy
the equation
[Reminder:
.]
8)
Find how many distinct ordered pairs of integers (x, y) satisfy
the equation
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
SECTION 4 - TEAM QUESTION
Time: 5 minutes.
Towers are made from unit cubes, called bricks.
The top layer has 1 brick, the second layer down is a 2 × 2 square
of 4 bricks, the third layer is a 3 × 3 square of 9 bricks, and so on.
The following table gives the total number of bricks in a tower with a given
number of layers, and also the number of bricks in a large cube of the same
height.
No. of layers
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
No. of bricks in tower
1
5
14
30
55
91
140
204
285
385
506
650
819
1015
1240
No. of bricks in large cube
1
8
27
64
125
216
343
512
729
1000
Your task is start with a tower of bricks and show how to make a set of smaller
towers and cubes that uses all of the bricks. There can be at most one tower or
cube of any particular size. For example
3 layer tower (14 bricks) = cube of side 2 (8 bricks) + 2 layer tower (5 bricks)
+ cube of side 1 (1 brick).
This solution is written as 8 + 5 + 1.
You are to do this for towers of from 5 to 15 layers, entering your solutions on
the sheet provided.
Marks: Give 1 point for each correct answer and -1 point for each wrong
answer; no penalty for omissions. The winning team scores 5, the other team
scores 5 – D, where D is the difference in points, with minimum score 0.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
SECTION 4 - TEAM QUESTION
Answer sheet
No. of layers
5
6
7
8
9
10
11
12
13
14
15
No. of bricks in tower
55
91
140
204
285
385
506
650
819
1015
1240
Solution
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
SECTION 4 - TEAM QUESTION
Solutions (not unique)
No. of layers
5
6
7
8
9
10
11
12
13
14
15
No. of bricks in tower
55
91
140
204
285
385
506
650
819
1015
1240
Solution
27 + 14 + 8 + 5 + 1
64 + 27
91 + 30 + 14 + 5
140 + 64
216 + 64 + 5
343 + 27 + 14 + 1
343 + 140 + 14 + 8 + 1
385 + 91 + 64 + 55 + 27 + 14 + 8 + 5 + 1
512 + 285 + 14 + 8
650 + 343 + 14 + 8
819 + 385 + 30 + 5 + 1
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
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)
Find the last four digits of the product
314 572 × 2 135 213 .
Find the exact value of the product
314 572 × 2 135 213
12
3)
Find the largest prime factor of 87 703 .
13
4)
A sequence x1, x2, x3, … is defined by
x1 = 12
for n ≥ 1 .
Find the value of x4 correct to the nearest integer.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
SECTION 6 - ALGEBRA AND CALCULUS (INDIVIDUAL)
Marks: 2 or 1 for opponent
Time: 60 seconds.
For Questions 1 – 4
If X is the point (a, b) and Y is the point (c, d) then the gradient of the line XY
is given by G(X, Y) =
d b
c a
Give all answers in the simplest form.
Year
7-9

1)
Find G(X, Y) when X is (3, 4) and Y is (7, 12) .
2)
Find G(X, Y) when X is ( 5 , 1) and Y is (7, 13) .
10-11 3)
4)
Find G(X, Y) when X is ( m , m(m  2) ) and Y is (3, 3) .
 X is ( a 2 , ab ) and Y is (1, a 2b) .
Find G(X, Y) when
 
12
5)
Find the gradient of the curve
1
2
y  6x 
x
at the point where x   12 .

6)  Find the gradient of the curve
 2
6
 x
y    2
2 x
at the point where x  2 .
In Questions 7 and 8 your answer should be an exact numerical expression.

13
7)
Find the gradient of the curve
y  sin 2 4x
at the point where
.
8)  Find the gradient of the curve
at the point where x  0

HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
SECTION 7 - RACE (INDIVIDUAL)
Marks: 2 or 0
Time: 60 seconds.
Year
7-9
1)
10-11 2)
Find three different positive whole numbers whose product is
three times their sum.
Write 7 4  4 4 as the product of prime factors.
12
3)
A piece of wire 120 cm long is cut into two parts. Each part is
then bent to form a square.
Find the maximum possible total area of these squares.
13
4)
A piece of wire 120 cm long is cut into two parts. Each part is
then bent to form an equilateral triangle.
Find the maximum possible total area of these triangles.
7-9
5)
Each internal angle of a regular polygon is 171º.
Find how many sides the polygon has.
10-11 6)
12
7)
A cone with base radius r has the same volume as a sphere
of radius r .
Find the height of the cone in terms of r .
Given that y 
10
0
13
8)
1
 1  x10 dx , find
dy
.
dx
The first ten terms of the arithmetic progression
199, 409, 619, … are all prime numbers.
Find a prime factor of the eleventh term.
HANS WOYDA MATHEMATICS QUIZ COMPETITION 2010/2011
SEMI-FINAL
ANSWERS (allow equivalent answers)
SECTION 1
SECTION 5
1.
2.
3.
4.
5.
6.
7.
8.
5
25
0.5
0.25
12½
700π (cm3)
-1 ± √2
2
1.
2.
3.
4.
SECTION 6
SECTION 2
1.
2.
3.
2
2 + 2√2
1.
2.
3.
2
1
4.

 5.
4.

SECTION 3
1.
2.
3.
4.
1109
125
15
3
5.
1/2
6.
7.
8.
1/6
8
12
SECTION 4
Please see solutions sheet
3 836
671 678 223 836
67
531

m 1
ab
a 1
10
2.5
6.
7.
4
8.

3 3
2
 SECTION 7
1.
2.
3.
4.
5.
6.
7.
8.
E.g. 2, 3, 5
3  5  11  13
450 (cm2)
200 3 (cm2)
40
4r
0
11 or 19