In each test each student was given a nonnegative integer score with a maximum possible score of 10.. Adrian noticed that in each test, only one student scored higher than he did and nob[r]
(1)April 23, 2008
NAME: SOLUTIONS GENDER:
PLEASE PRINT (First name Last name) M F
SCHOOL: GRADE:
(7,8,9)
• You have 90 minutes for the examination The test has two parts: PART A –short answer; and PART B – long answer The exam has pages including this one
• Each correct answer to PART A will score points You must put the answer in the space provided No part marks are given
• Each problem in PART B carries points You should show all your work Some credit for each problem is based on the clarity and completeness of your answer You should make it clear why the answer is correct PART A has a total possible score of 45 points PART B has a total possible score of 54 points
• You are permitted the use of rough paper Geome-try instruments are not necessary References includ-ing mathematical tables and formula sheets are not
permitted Simple calculators without programming or graphic capabilities are allowed Diagrams are not drawn to scale They are intended as visual hints only
• When the teacher tells you to start work you should read all the problems and select those you have the best chance to first You should answer as many problems as possible, but you may not have time to answer all the problems
MARKERS’ USE ONLY
PART A ×5 B1 B2 B3 B4 B5 B6 TOTAL (max: 99)
BE SURE TO MARK YOUR NAME AND SCHOOL AT THE TOP OF THIS PAGE
THE EXAM HAS PAGES INCLUDING THIS COVER PAGE
(2)PART A: SHORT ANSWER QUESTIONS
A1 The sum of two prime numbers is9 What is their product?
14
A2 Find the largest positive integerX such that2/7 is smaller than7/X
24
A3 In a class, students like hockey and 11 students like soccer Of these, students
25 like both hockey and soccer On the other hand, 10 students in the class like neither
hockey nor soccer How many students are in the class?
A4 Notice that 2008 has the property that it is divisible by its first and last digits but
2032 not by its second or third digits (because 2008/2 = 1004 and 2008/8 = 251 are both
whole numbers, but 2008/0 is not a whole number) What is the smallest number greater than2008 with this same property?
A5 Two squares have totalarea 85 cm2
and total perimeter 52 cm What is the area in
49 cm2
(3)A6 Shannon receives a bouquet of flowers containing two kinds of flower One kind of
102 flower has13petals and9leaves The other kind has8petals and11leaves Altogether
in her bouquet there are100petals How many leaves are there in the bouquet?
A7 The area of triangleABE is 56 m2
8 A
B
C D E
PointsCand Dlie on sideBE such thatCDis twice as long asBC andDE is twice as long asCD What is the area in m2
of triangle ABC?
A8 Twelve friends were at a restaurant Each person ordered four items from the menu
6 No two people ordered exactly the same set of four items At least how many different
items are on the restaurant’s menu?
A9 A circle is inscribed in a right-angled isosceles triangle
6 perimeter is6 + 4√2
(4)PART B: LONG ANSWER QUESTIONS
B1 Richard needs to go from his house to the park by taking a taxi There are two taxi companies available The first taxi company charges an initial cost of $10.00, plus
$0.50for each kilometre travelled The second taxi company charges an initial cost of $4.00, plus$0.80for each kilometre travelled Richard realizes that the cost to go to the park is the same regardless of which taxi company he chooses What is the distance in km from his house to the park?
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
The initial charge differs by $6 but the charge per km differs by $0.30 For the two costs to be the same, the park is 6/0.3 = 20 km away from his house
(5)B2 A radio station runs a contest in which each winner will get to attend two Flames playoff games and to take one guest to each game The winner does not have to take the same guest to the two games Luckily,five school friends Alice, Bob, Carol, David and Eva are all winners of this contest Show how each winner can choose two others from this group to be his or her guests, so that each pair of thefive friends gets to go to at least one playoff game together
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
I will replace names with theirfirst initial Here is one solution:
A chooses B and then C.
B chooses C and then D.
C chooses D and then E.
D chooses E and then A.
(6)B3 A class was given two tests In each test each student was given a nonnegative integer score with a maximum possible score of 10 Adrian noticed that in each test, only one student scored higher than he did and nobody got the same score as he did But then the teacher posted the averages of the two scores for each student, and now there was more than one student with an average score higher than Adrian
(a) [4 points] Give an example (using exact scores) to show that this could happen
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
Adrian could get scores of and Student A could get 10 and 6, while Student B could get and 10 All other students could get scores less than in both tests Then only Student A did better than Adrian in the first test, and only Student B did better than Adrian in the second test
But the averages are:
7 for Adrian and
8 for both Students A and B,
so both A and B did better than Adrian on average.
(b) [5 points] What is the largest possible number of students whose average score could be higher than Adrian’s average score? Explain clearly why your answer is correct
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
At mosttwo studentscan have a higher average score than Adrian This is because at most two students scored higher than Adrian on a test Every-one else scored lower than Adrian on both tests,
(7)B4 A rectangle with dimensions 6cm by8cm is drawn A circle is drawn circumscribing this rectangle A square is drawn circumscribing this circle A second circle is drawn that circumscribes this square
6
8 10
6
What is the area in cm2
of the bigger circle?
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
The first circle has diameter √62 + 82
= 10 cm and thus radius5 cm Therefore the square has side length10cm and thus diagonal10√2cm Therefore the second circle has radius5√2 cm
The answer isπ5√22= 50π cm2
(8)B5 Ais a two-digit whole number that does not contain zero as a digit Bis a three-digit whole number, andA%of B is400 Find all possible values ofA andB
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
We haveAB/100 = 400 meaning AB = 40000 = 26 54
A is a two-digit factor that does not contain a zero, so A cannot have both and as a factor Also A > 40, otherwise B≥1000 will have more than three digits The only possibility isA = 26
ThenB= 54
The only solution is A = 26= 64
and B = 54 = 625
(9)B6 (a) [2 points] Find a rectangle with the following two properties: (i) its perimeter is an odd integer; and (ii) none of its sides is an integer
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
We needx, ynot integers such that2(x+y)is an odd integer For example, the dimensions can bex=
4 by y=
4, and then its perimeter is4
1
=
(b) [2 points] Find a rectangle with the following two properties: (i) its area is an even integer; and (ii) none of its sides is an integer
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
We need x, y not integers such that xy is an even integer For example, the dimensions can be x= 32 by y = 43, and then its area is
2 · =
(c) [5 points] Find a quadrilateral (not necessarily a rectangle) with the following three properties: (i) its perimeter is a positive integer; (ii) its area is a positive integer; and (iii) none of its sides is an integer
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
SOLUTION:
Here is a rectangle solution (taken from a student’s solution): a2−√2by2+√2
rectangle has perimeter8 and area2−√2 +√2= 4−2 = For a non-rectangleexample, consider a rhombus
made up of four right triangles with sides
2
and and hypotenuse32
+ 22 =
2
Then the sides of the rhombus are all 2,
the perimeter is 45
= 10, and the area is41
2
2 =
2
2
2
A solution can be shown to exist a bit more simply by considering a square with side 5/4 say The perimeter of the square is and the area of the square is 25/16 which is greater than So if we squeeze two opposite corners of the square together, we get a rhombus whose perimeter remains equal to but whose area can be reduced to arbitrarily close to zero Thus at some point we can make the area exactly equal to