Wednesday, May 10, 2017
(in North America and South America)
Thursday, May 11, 2017
(outside of North American and South America)
©2016 University of Waterloo
Evaluating,
Answer: (B)
Reading from the tallest bar on the graph, approximately 50 students play soccer.
Since this is larger than the number of students who play any of the other sports, then soccer is played by the most students.
Answer: (C)
Michael has $280 in $20 bills and so the number of $20 bills that he has is
Answer: (C)
There are only two different products of two positive integers whose result is 14.
These are
Since the two integers must be between 1 and 10, then the product must be
The sum of these two integers is
Answer: (D)
Written as a fraction, three thousandths is equal to
As a decimal, three thousandths is equal to
Answer: (E)
Solution 1:
Three vertices of the square are labelled
Since the given figure is a square, then
Since
The three angles in any triangle add to
Since
Solution 2:
The vertices of the square are labelled
Diagonal
Since these triangles are identical,
Since
That is,
Answer: (B)
Solution 1:
Written as a mixed fraction,
Since
The integer closest in value to
Solution 2:
Written as a decimal
Since 0.75 is closer to 1 than it is to 0, then
The integer closest in value to
Answer: (C)
When
(In fact, the value of
Answer: (D)
The mean (average) of three integers whose sum is 153 is
The mean of three consecutive integers equals the middle of the three integers.
That is, 51 is the middle integer of three consecutive integers and so the largest of these integers is 52.
(We may check that
Answer: (A)
Each of the 4 smaller triangles is equilateral and thus has sides of equal length.
Each of these smaller triangles has a perimeter of 9 cm and so has sides of length
In
Since
Therefore, the perimeter of
Answer: (E)
The denominators of the two fractions are 7 and 63.
Since
That is,
Therefore, the number that goes into the
Answer: (A)
If puzzles are bought individually for $10 each, then 6 puzzles will cost
Since the cost for a box of 6 puzzles is $50, it is less expensive to buy puzzles by the box than it is to buy them individually.
Buying 4 boxes of 6 puzzles gives the customer
Buying one additional puzzle for $10 gives the customer 25 puzzles at a minimum cost of
Answer: (A)
A translation moves (slides) an object some distance without altering it in any other way.
That is, the object is not rotated, reflected, and its exact size and shape are maintained.
Of the triangles given, the triangle labelled
Thus,
Answer: (D)
Since the time in Toronto, ON is 1:00 p.m. when the time in Gander, NL is 2:30 p.m., then the time in Gander is 1 hour and 30 minutes ahead of the time in Toronto.
A flight that departs from Toronto at 3:00 p.m. and takes 2 hours and 50 minutes will land in Gander at 5:50 p.m. Toronto time.
When the time in Toronto is 5:50 p.m., the time in Gander is 1 hour and 30 minutes ahead which is 7:20 p.m.
Answer: (A)
Henry was slower than Faiz and thus finished the race behind Faiz.
Ryan was faster than Henry and Faiz and thus finished the race ahead of both of them.
From fastest to slowest, these three runners finished in the order Ryan, Faiz and then Henry.
Toma was faster than Ryan but slower than Omar.
Therefore, from fastest to slowest, the runners finished in the order Omar, Toma, Ryan, Faiz, and Henry.
The student who finished fourth was Faiz.
Answer: (A)
The positive divisors of 20 are:
Of the 20 numbers on the spinner, 6 of the numbers are divisors of 20.
It is equally likely that the spinner lands on any of the 20 numbers.
Therefore, the probability that the spinner lands on a number that is a divisor of 20 is
Answer: (E)
Solution 1:
Since 78 is 2 less than 80 and 82 is 2 greater than 80, the mean of 78 and 82 is 80.
Since the mean of all four integers is 80, then the mean of 83 and
The integer 83 is 3 greater than 80, and so
That is,
(We may check that the mean of
Solution 2:
Since the mean of the four integers is 80, then the sum of the four integers is
Since the sum of 78, 83 and 82 is 243, then
Therefore,
Answer: (D)
A discount of 20% on a book priced at $100 is a
Thus for option (A), Sara’s discounted price is
A discount of 10% on a book priced at $100 is a
A second discount of 10% on the new price of $90 is a
Thus for option (B), Sara’s discounted price is
A discount of 15% on a book priced at $100 is a
A further discount of 5% on the new price of $85 is a
Thus for option (C), Sara’s discounted price is
A discount of 5% on a book priced at $100 is a
A further discount of 15% on the new price of $95 is a
Thus for option (D), Sara’s discounted price is
Therefore, the four options do not give the same price and option (A) gives Sara the best discounted price.
Answer: (A)
In the diagram, rectangles
The overlapping square
That is,
In rectangle
In rectangle
Since
Answer: (B)
Points
If Betty and Ann had met for the first time at point
When they meet, the time that Betty has been walking is equal to the time that Ann has been walking and so the ratio of Betty’s speed to Ann’s speed is equal to the ratio of the distance that Betty has walked to the distance that Ann has walked.
That is, if they had met for the first time at point
Similarly, if Betty and Ann had met for the first time at point
In this case, the ratio of their speed’s would be
When Betty and Ann actually meet for the first time, they are between
Thus Betty has walked less distance than she would have had they met at
That is, the ratio of Betty’s speed to Ann’s speed must be less than
We must determine which of the five given answers is a ratio that is less than
One way to do this is to convert each ratio into a mixed fraction.
That is, we must determine which of the five answers is less than
Converting the answers, we get
Of the five given answers, the only fraction that is less than
If Betty and Ann meet for the first time between
Answer: (B)
Solution 1:
The first and tenth rectangles each contribute an equal amount to the perimeter.
They each contribute two vertical sides (each of length 2), one full side of length 4 (the top side for the first rectangle and the bottom side for the tenth rectangle), and one half of the length of the opposite side.
That is, the first and tenth rectangles each contribute
Rectangles two through nine each contribute an equal amount to the perimeter.
They each contribute two vertical sides (each of length 2), one half of a side of length 4, and one half of the length of the opposite side (which also has length 4).
That is, rectangles two through nine each contribute
Therefore, the total perimeter of the given figure is
Solution 2:
One method for determining the perimeter of the given figure is to consider vertical lengths and horizontal lengths.
Each of the ten rectangles has two vertical sides (a left side and a right side) which contribute to the perimeter.
These 20 sides each have length 2, and thus contribute
Since these are the only vertical lengths contributing to the perimeter, we now determine the sum of the horizontal lengths.
There are two types of horizontal lengths which contribute to the perimeter: the bottom side of a rectangle, and the top side of a rectangle.
The bottom side of each of the first nine rectangles contributes one half of its length to the perimeter.
That is, the bottom sides of the first nine rectangles contribute
The entire bottom side of the tenth rectangle is included in the perimeter and thus contributes a length of 4.
Similarly, the top sides of the second rectangle through to the tenth rectangle contribute one half of their length to the perimeter.
That is, the top sides of rectangles two through ten contribute
The entire top side of the first rectangle is included in the perimeter and thus contributes a length of 4.
In total, the horizontal lengths included in the perimeter sum to
Since there are no additional lengths which contribute to the perimeter of the given figure, the total perimeter is
Solution 3:
Before they were positioned to form the given figure, each of the ten rectangles had a perimeter of
When the figure was formed, some length of each of the ten rectangles’ perimeter was “lost” (and thus is not included) in the perimeter of the given figure.
These lengths that were lost occur where the rectangles touch one another.
There are nine such locations where two rectangles touch one another (between the first and second rectangle, between the second and third rectangle, and so on).
In these locations, each of the two rectangles has one half of a side of length 4 which is not included in the perimeter of the given figure.
That is, the portion of the total perimeter of the ten rectangles that is not included in the perimeter of the figure is
Since the total perimeter of the ten rectangles before they were positioned into the given figure is
Answer: (D)
The units digit of the product
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Checking, we see that the product is correct and so
Answer: (B)
Given 8 dimes (10 coins) and 3 quarters (25 coins), we list the different amounts of money (in cents) that can be created in the table below.
When an amount of money already appears in the table, it has been stroked out.
FLAG: recreate complex table with strikethroughs, originally tikz drawing
We may ignore the first entry in the table, 0, since we are required to use at least one of the 11 coins.
We are left with 27 different amounts of money that can be created using one or more of the 8 dimes and 3 quarters.
Answer: (A)
We begin by constructing rectangle
The vertical sides
The horizontal sides
We determine the area of
To determine the horizontal side lengths of the right-angled triangles we count units along the
For example, since
Thus, the length of
Therefore the length of
Similarly, the length of
Since
Thus, the length of
To determine the vertical side lengths of the right-angled triangles we may count units along the
For example, since
Thus, the length of
Therefore the length of
Similarly, the length of
Since
Thus, the length of
The area of
The area of
The area of
The area of
Since
Finally, the area of
Answer: (B)
Solution 1:
The sum of the positive integers from 1 to
For example when
Using this expression, the sum of the positive integers from 1 to 2017, or
To determine the sum of the integers which Ashley has not underlined, we must subtract from 2 035 153 any of the 2017 integers which is a multiple of 2, or a multiple of 3, or a multiple of 5, while taking care not to subtract any number more than once.
First, we find the sum of all of the 2017 numbers which are a multiple of 2.
This sum contains 1008 integers and is equal to
Since each number in this sum is a multiple of 2, then this sum is equal to twice the sum
That is,
Using the formula above, the sum of the first 1008 positive integers is equal to
We may similarly determine the sum of all of the 2017 numbers which are a multiple of 3.
This sum is equal to
Since each of these numbers is a multiple of 3,
The sum of all of the 2017 numbers which are a multiple of 5 is equal to
We summarize this work in the table below.
Description | Sum | Result |
---|---|---|
All integers from 1 to 2017 | ||
Integers that are a multiple of 2 | ||
Integers that are a multiple of 3 | ||
Integers that are a multiple of 5 |
If we now subtract the sum of any of the 2017 integers which is a multiple of 2, or a multiple of 3, or a multiple of 5 from the sum of all 2017 integers, is the result our required sum?
The answer is no. Why?
There is overlap between the list of numbers that are a multiple of 2 and those that are a multiple of 3, and those that are a multiple of 5.
For example, any number that is a multiple of both 2 and 3 (and thus a multiple of 6) has been included in both lists and therefore has been counted twice in our work above.
We must add back into our sum those numbers that are a multiple of 6 (multiple of both 2 and 3), those that are a multiple of 10 (multiple of both 2 and 5), and those that are a multiple of 15 (multiple of both 3 and 5).
The sum of all of the 2017 numbers which are a multiple of 6 is equal to
The sum of all of the 2017 numbers which are a multiple of 10 is equal to
The sum of all of the 2017 numbers which are a multiple of 15 is equal to
We again summarize this work in the table below.
Description | Sum | Result |
---|---|---|
All integers from 1 to 2017 | ||
Integers that are a multiple of 2 | ||
Integers that are a multiple of 3 | ||
Integers that are a multiple of 5 | ||
Integers that are a multiple of 6 | ||
Integers that are a multiple of 10 | ||
Integers that are a multiple of 15 |
If we take the sum of all 2017 integers, subtract those that are a multiple of 2, and those that are a multiple of 3, and those that are a multiple of 5, and then add those numbers that were subtracted twice (the multiples of 6, the multiples of 10, and the multiples of 15), then we get:
The answer is still no, but we are close!
Consider any of the 2017 integers that is a multiple of 2, 3 and 5 (that is, a multiple of
Each number that is a multiple of 30 would have been underlined by Ashley, and therefore should not be included in our sum.
Each multiple of 30 was subtracted from the sum three times (once for each of the multiples of 2, 3 and 5), but then added back into our sum three times (once for each of the mutiples of 6, 10 and 15).
Thus, any of the 2017 integers that is a multiple of 30 must still be subtracted from 611 048 to achieve our required sum.
The sum of all of the 2017 numbers which are a multiple of 30 is equal to
Finally, the sum of the 2017 integers which Ashley has not underlined is
Solution 2:
We begin by considering the integers from 1 to 60.
When Ashley underlines the integers divisible by 2 and by 5, this will eliminate all of the integers ending in 0, 2, 4, 5, 6, and 8.
This leaves
Of these, the integers
Therefore, of the first 60 integers, only the integers
Among these 16 integers, we notice that the second set of 8 integers consists of the first 8 integers with 30 added to each.
This pattern continues, so that a corresponding set of 8 out of each block of 30 integers will not be underlined.
Noting that 2010 is the largest multiple of 30 less than 2017, this means that Ashley needs to add the integers
Before proceeding, we justify briefly why the pattern continues:
Every positive integer is a multiple of 30, or 1 more than a multiple of 30, or 2 more than a multiple of 30, and so on, up to 29 more than a multiple of 30. Algebraically, this is saying that every positive integer can be written in one of the forms
depending on its remainder when divided by 30.
Every integer with an even remainder when divided by 30 is even, since 30 is also even.
Similarly, every integer with a remainder divisible by 3 or 5 when divided by 30 is divisible by 3 or 5, respectively.
This leaves us with the formsNo integer having one of these forms will be underlined, since, for example, is one more than a multiple of 2 and 5 (namely, ) and is 2 more than a multiple of 3 (namely, ) so is not divisible by 2, 3 or 5.
The sum of the 8 integers in the first row of the table above is
Since each of the integers in the second row of the table is 30 greater than the corresponding integer in the first row, then the sum of the numbers in the second row of the table is
Similarly, the sum of the integers in the third row is
We note that
Therefore,
Here, we have used the fact that the integers from 1 to 66 can be grouped into 33 pairs each of which adds to 67, as shown here:
Answer: (A)
Michael has $280 in $20 bills and so the number of $20 bills that he has is
Answer: (C)
Evaluating, we get
Answer: (A)
Exactly 1 of the 5 equal sections contains the number 4.
Therefore, the probability that the spinner lands on 4 is
Answer: (E)
The number of grade 8 students on the chess team is
Answer: (B)
Since
Since
Therefore,
Answer: (B)
In terms of
Since the perimeter is 21, then
Answer: (B)
Reading from the graph, 20 students chose pink and 25 students chose blue.
The ratio of the number of students who chose pink to the number of students who chose blue is
After simplifying this ratio (dividing each number by 5),
Answer: (A)
Solution 1:
To get the original number, we reverse the steps.
That is, we add 6 and then divide by 3.
Therefore, the original number is
Solution 2:
If the original number is
When this result is decreased by 6, we get
Solving
Answer: (D)
Tian travels 500 m every 625 steps and so she travels
If Tian walks 10 000 steps at this same rate, she will walk a distance of
Since there are 1000 m in each kilometre, Tian will walk
Answer: (D)
Line segments
Since
The three interior angles of any triangle add to
Thus,
Answer: (B)
The volume of a rectangular prism is determined by multiplying the area of its base by its height.
The area of the base for the given prism is
Since the prism’s volume is 60 cm
Answer: (D)
Since
By the Pythagorean Theorem,
Since
Walking from
Thus, David walks
Answer: (D)
Each term of the sum
Since
Answer: (C)
If three of the students receive the smallest total number of pens possible, then the remaining student will receive the largest number of pens possible.
The smallest number of pens that a student can receive is 1, since each student receives at least 1 pen.
Since each student receives a different number of pens, the second smallest number of pens that a student can receive is 2 and the third smallest number of pens that a student can receive is 3.
The smallest total number of pens that three students can receive is
Therefore, the largest number of pens that a student can receive is
Answer: (C)
The even integers between 1 and 103 are
Since there are 51 even integers in the list
Next, we want to find a number
We notice that our lower bound, 4, is 3 greater than our original lower bound of 1.
By increasing each of the 51 even integers from above by 3, we create the first 51 odd integers which are greater than 4.
These odd integers are
Since there are 51 odd integers in the list
That is, the number of even integers between 1 and 103 is the same as the number of odd integers between 4 and 106.
Answer: (E)
Label points
Each shaded triangle is equilateral,
Therefore,
In quadrilateral
Therefore,
Join
Since
That is,
Similarly,
We may also join
Thus,
Since 3 of these 9 triangles are shaded, the fraction of the area of
Answer: (B)
The range of the players’ heights is equal to the difference between the height of the tallest player and the height of the shortest player.
Since the tallest player, Meghan, has a height of 188 cm, and the range of the players’ heights is 33 cm, then the shortest player, Avery, has a height of
Thus, answer (D) is a statement which provides enough information to determine Avery’s height, and so must be the only one of the five statements which is enough to determine Avery’s height.
(Can you give a reason why each of the other four answers does not provide enough information to determine Avery’s height?)
Answer: (D)
When Brodie and Ryan are driving directly towards each other at constant speeds of 50 km/h and 40 km/h respectively, then the distance between them is decreasing at a rate of
If Brodie and Ryan are 120 km apart and the distance between them is decreasing at 90 km/h, then they will meet after
Since
Answer: (E)
The mean age of three of the friends is 12 years and 3 months which is equal to
Since the mean equals the sum of the ages divided by 3, then the sum of the ages of these three friends is
The mean age of the remaining four friends is 13 years and 5 months or
Thus, the sum of the ages of these four friends is
The sum of the ages of all seven friends is
Answer: (E)
The units digit of the product
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Since
Thus, the units digit of
Therefore, the only possible value of
Substituting
Checking, we see that the product is correct and so
Answer: (B)
On the bottom die, the two visible faces are showing 2 dots and 4 dots.
Since the number of dots on opposite faces of this die add to 7, then there are 5 dots on the face opposite the face having 2 dots, and 3 dots on the face opposite the face having 4 dots.
Therefore, the top face of this bottom die (which is a face that is hidden between the dice) has either 1 dot on it or it has 6 dots on it.
On the second die from the bottom, the sum of the number of dots on the top and bottom faces (the faces hidden between the dice) is 7 since the number of dots on opposite faces add to 7. (We do not need to know which faces these are, though we could determine that they must have 1 and 6 dots on them.)
Similarly, on the third die from the bottom, the sum of the number of dots on the top and bottom faces (the faces hidden between the dice) is also 7.
Finally, the top face of the top die shows 3 dots, and so the bottom face of this die (which is a face that is hidden between the dice) contains 4 dots.
Therefore, the sum of the number of dots hidden between the dice is either
Of these two possible answers, 24 is the only answer which appears among the five given answers.
Answer: (C)
To give
To make
Thus, we must assign
Let
Since
Similarly, we assign
Since
Thus, the greatest possible value of
Answer: (D)
We introduce the letter M to represent a game that Mike has won, and the letter A to represent a game that Alain has won.
We construct a tree diagram to show all possible outcomes.
The M at the far left of the tree represents the fact that Mike won the first game.
The second “column” shows the two possible outcomes for the second game - Mike could win (M), or Alain could win (A).
The third, fourth and fifth columns show the possible outcomes of games 3, 4 and 5, respectively.
A branch of the tree is linked by arrows and each branch gives the outcomes of the games which lead to one of the players becoming the champion.
Once Mike has won 3 games, or Alain has won 3 games, the branch ends and the outcome of that final game is circled.
Since the first player to win 3 games becomes the champion, one way that Mike could become the champion is to win the second and third games (since he has already won the first game).
We represent this possibility as MMM, as shown along the top branch of the tree diagram.
In this MMM possibility, the final M is circled in the diagram, meaning that Mike has become the champion.
Since we are asked to determine the probability that Mike becomes the champion, we search for all branches through the tree diagram which contain 3 Ms (paths ending with a circled M).
The tree diagram shows six such branches: MMM, MMAM, MMAAM, MAMM, MAMAM, and MAAMM.
All other branches end with a circled A, meaning that Alain has won 3 games and becomes the champion.
Since each player is equally likely to win a game, then Mike wins a game with probability
Of the six ways that Mike can win (listed above), only one of these ends after 3 games (MMM).
The probability that Mike wins in exactly 3 games is equal to the probability that Mike wins the second game, which is
That is, the probability that Mike becomes the champion by winning the first 3 games is
Of the six ways that Mike can win, two of these end after 4 games (MMAM, MAMM).
The probability that Mike wins games two and four but Alain wins game three (MMAM) is equal to the probability that Mike wins the second game, which is
In this case, Mike becomes the champion with probability
Similarly, the probability that Mike becomes the champion by winning games three and four, but loses game two, is also
Finally, we determine the probability that Mike becomes the champion by winning in exactly 5 games (there are three possibilities: MMAAM, MAMAM and MAAMM).
Each of these three possibilities happens with probability
Therefore, if Mike has won the first game, then the probability that he becomes champion is
Answer: (C)
Let the area of the shaded region that lies outside of both semi-circles be
Let the area of the shaded region that lies inside of both semi-circles be
The sum of the areas of both semi-circles counts the shaded area
Therefore, if we subtract
That is, the area of quarter-circle
The area of quarter-circle
The area of the semi-circle drawn on
The area of the semi-circle drawn on
Thus,
We build square
Next, we will show that
Since
Beginning at point
After these two moves, we arrive at the point labelled
Since the diameter of the semi-circle drawn on
Therefore,
Beginning again at point
Since these are the same two moves we made previously (up 4 and right 4), then we must again arrive at
Since the diameter of the semi-circle drawn on
Therefore,
The two semi-circles intersect at exactly one point (other than point
Since we have shown that
Therefore,
Finally, we find the value of
First we construct
By symmetry,
Each of these equal areas,
That is,
The area of the shaded region is
Of the answers given,
Answer: (D)
Solution 1:
Let the number of black plates, gold plates, and red plates be
Brady is stacking 600 plates, and so
Rewrite this equation as
Since 600 is a multiple of 2, and
Since the right side of the equation is a multiple of 2, then the left side,
We are given that
Similarly, rewriting the equation as
Since the right side of the equation is a multiple of 6, then the left side,
That is, each of
So then the equation
Each solution to the equation
For example,
Since
If
If
Thus,
That is, once we assign values for
Thus, to determine the number of solutions to the equation
For example, above we showed that the three pairs
Continuing in this way, we determine all possible pairs
Value of |
Value of |
Number of plates: |
---|---|---|
100 | 0 | 600,0,0 |
99 | 0 | 594,0,6 |
99 | 1 | 594,6,0 |
98 | 0 | 588,0,12 |
98 | 1 | 588,6,6 |
98 | 2 | 588,12,0 |
97 | 0 | 582,0,18 |
97 | 1 | 582,6,12 |
97 | 2 | 582,12,6 |
97 | 3 | 582,18,0 |
⋮ | ⋮ | ⋮ |
0 | ||
1 | ||
2 | ||
⋮ | ⋮ | ⋮ |
We see from the table that if the value of
That is, when
Each additional decrease of 1 in
Therefore, the total number of solutions to
Using the fact that the sum of the first
Since each of these solutions corresponds to a way that Brady could stack the plates, there are 5151 ways that Brady could stack the plates under the given conditions.
Solution 2:
In a given way of stacking the plates, let
Then there are
Since the total number of plates in a stack is 600, then
We note that the numbers of black, gold and red plates completely determines the stack (we cannot rearrange the plates in any way), and so the number of ways of stacking the plates is the same as the number of ways of solving the equation
Since
When
Since
Since
Therefore, the possible values for
Since
When
When
Each time we increase
Thus, as we continue to increase
In other words, every even value for
Therefore, when
When
Again,
Therefore, the possible values of
Again, each value of
Therefore, when
Consider the case of an unknown value of
Again,
Also, the maximum possible value of
This means that there are
Again, each value of
Therefore, for a general
We make a table to summarize the possibilities:
# of solutions | |||
---|---|---|---|
Therefore, the total number of ways of stacking the plates is
Therefore, the number of ways that Brady could stack the plates is
Answer: (E)