Wednesday, May 15, 2019
(in North America and South America)
Thursday, May 16, 2019
(outside of North American and South America)
©2018 University of Waterloo
Erin receives $3 a day. To receive a total of $30, it will take Erin
Answer: (E)
Beginning at the origin
The point
Answer: (D)
One of four identical squares is shaded and so the fraction of square
Answer: (C)
Adding, we get
Answer: (D)
The mode is the amount of rainfall that occurs most frequently.
Reading from the graph, the daily rainfall amounts for Sunday through to Saturday are 6 mm, 15 mm, 3 mm, 6 mm, 3 mm, 3 mm, and 9 mm.
The mode for the amount of rainfall for the week is 3 mm.
Answer: (C)
When
Answer: (B)
When
Alternately, the result can be found by subtracting,
Answer: (A)
Solution 1
One third of 396 is
To finish the book, Joshua has
Solution 2
Joshua has read the first third of the book only, and so he has
Two thirds of 396 is
Joshua has 264 pages left to read.
Answer: (A)
One complete rotation equals
Therefore,
Answer: (D)
Solution 1
The mean of the numbers
Since each of the given answers has three numbers, then for the mean to equal 30, the sum of the three numbers must also equal
Of the given answers, only (D) has numbers whose sum is 90 (
Solution 2
Since 20 is 10 less than 30 and 40 is 10 more than 30, the mean of the numbers
In each of the given answers, 30 is the middle number in the list of three numbers.
Thus, for the mean of the three numbers to equal 30, the first and last numbers must be equal “distances” away from 30 (with one number being less than 30 and the other greater than 30).
Looking at answer (D), 23 is 7 less than 30 and 37 is 7 more than 30 and so the mean of these three numbers is 30. (We may check that this isn’t the case for each of the other four answers.)
Answer: (D)
Evaluating,
Answer: (B)
The width of rectangle
This distance is equal to the difference between their
Similarly, the height of rectangle
This distance is equal to the difference between their
The area of rectangle
Answer: (B)
The repeating pattern of ABCDEFG has 7 white keys.
Since the first white key is A, the pattern repeats after each number of keys that is a multiple of 7.
Since 28 is a multiple of 7, then the 28
Answer: (E)
On the spinner given, the prime numbers that are odd are
Since the spinner is divided into 8 equal sections, the probability that the arrow stops in a section containing a prime number that is odd is
Answer: (C)
Barry’s 12 coins include at least one of each of the 5 coins of different values.
The total value of these 5 coins is
Barry has the smallest total amount of money that he could have if each of his remaining
Thus, the smallest total amount of money that Barry could have is
Answer: (A)
Solution 1
The are 10 palindromes between 100 and 200:
The are 10 palindromes between 200 and 300:
Similary, there are 10 palindromes between each of 300 and 400, 400 and 500, 500 and 600, 600 and 700, 700 and 800, 800 and 900, 900 and 1000.
That is, there are 10 palindromes between each of the 9 pairs of consecutive multiples of 100 from 100 to 1000.
The number of palindromes between 100 and 1000 is
Solution 2
Each palindrome between 100 and 1000 is a 3-digit number of the form
Thus, there are 9 choices for the first digit
There are 10 choices for the second digit
Thus, there are 90 palindromes between 100 and 1000.
Answer: (B)
The first scale shows that has the same mass as
.
The second scale shows that has the same mass as
.
Therefore, the sum of the masses of and
is equal to the sum of the masses of
and
.
That is, the equal-arm scale shown below is balanced.
The left and right sides of this scale each contain a , and the scale remains balanced if one
is removed from each side of the scale (since they are equal in mass).
That is, the equal-arm scale shown below is balanced.
Therefore of the answers given, has the same mass as
.
Answer: (D)
The area of the rectangle with length
The area of the triangle with base 16 and height
The area of the rectangle is equal to the area of the triangle, or
Answer: (C)
Solution 1
Each of the four fractions is equal to the other three fractions, and each fraction has numerator 1, so then each denominator must be equal to the other three denominators.
That is,
If we let
Thus, the correct ordering is
Solution 2
Each of the four fractions is equal to the other three fractions, and each fraction has numerator 1, so then each denominator must be equal to the other three denominators.
That is,
Since
Since
Since
Thus, the correct ordering is
Answer: (E)
Each of 14 and 21 is a divisor of
Since
Since
So far, the positive divisors of
Since 2 and 3 are divisors of
Since
The positive divisors of
We are given that
The sum of these 8 positive divisors is
Answer: (D)
We begin by separating the given information, as follows:
From bullets 2 and 3, we can conclude that Alice owns more dogs than both Kathy and Bruce.
From bullet 4, we can conclude that answer (A) is not true.
From bullets 1 and 4, we can conclude that both Kathy and Bruce own more cats than Alice.
However, we cannot determine if Kathy owns more cats than Bruce, or vice versa.
Therefore, we cannot conclude that (B) or (C) must be true.
From bullet 2, we can conclude that (E) is not true.
Thus the statement which must be true is (D).
Answer: (D)
The single-digit divisors of 36 are:
The groups of 3 of these digits whose product is 36 are:
Next, we count the number of ways to arrange each of these 5 groups of digits.
The digits
The digits
The digits
The digits
The digits
The number of 3-digit positive integers having digits whose product is 36 is
Answer: (A)
Begin by constructing
The four segments
Label the intersections of the perpendicular pairs of these four segments as points
These diagonals divide square
Since 10 of these triangles are shaded, then
Answer: (A)
Solution 1
The ten moves have lengths 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
If the first move is vertical, then the five vertical moves have lengths 1, 3, 5, 7, 9 and the five horizontal moves have lengths 2, 4, 6, 8, 10.
If the first move is horizontal, then the five horizontal moves have lengths 1, 3, 5, 7, 9 and the five vertical moves have lengths 2, 4, 6, 8, 10.
If a horizontal move is to the right, then the length of the move is added to the
If a horizontal move is to the left, then the length of the move is subtracted from the
If a vertical move is up, then the length of the move is added to the
If a vertical move is down, then the length of the move is subtracted from the
Therefore, once the ten moves have been made, the change in one of the coordinates is a combination of adding and subtracting 1, 3, 5, 7, 9 and the change in the other coordinate is a combination of adding and subtracting 2, 4, 6, 8, 10.
For example, if the dot moves right 1, down 2, right 3, up 4, right 5, down 6, right 7, up 8, left 9, and up 10, then its final
We note that, in the direction with the moves of even length, the final change in coordinate will be even, since whenever we add and subtract even integers, we obtain an even integer.
In the other direction, the final change in coordinate will be odd, since adding or subtracting an odd number of odd integers results in an odd integer. (Since odd plus odd is even and odd minus odd is even, then after two moves of odd length, the change to date is even, and after four moves of odd length, the change to date is still even, which means that the final change after the fifth move of odd length is completed must be odd, since even plus or minus odd is odd.)
In the table below, these observations allow us to determine in which direction to put the moves of odd length and in which direction to put the moves of even length.
Choice | Change in |
Change in |
Horizontal moves | Vertical moves |
---|---|---|---|---|
(A) |
||||
(B) |
||||
(C) |
||||
(D) |
||||
(E) |
Since each of the locations (A), (C), (D), and (E) is possible, then the location that is not possible must be (B).
We note that in the case of (B), it is the change in the
In other words, we cannot obtain a total of 21 by adding and subtracting 1, 3, 5, 7, 9.
Can you see why?
Solution 2
Let
For example, if
Notice that if
First, assume the initial move is in the horizontal direction.
This means the second move will be in the vertical direction, the third will be horizontal, and so on.
In all, the first, third, fifth, seventh, and ninth moves will be horizontal and the others will be vertical.
Also, the first move is by one unit, the second is by two units, the third is by three units, and so on, so the horizontal moves are by
Each of these moves is either to the left or the right.
If all of the horizontal moves are to the right, then
If the point moves, left on the first move, right on the third, then left on the fifth, seventh, and ninth moves,
In this case, the final position is
In each of the five horizontal moves, the point either moves to the left or to the right.
This means there are two choices (left or right) for each of the five horizontal moves, so there are
Going through these carefully, the table below computes all possible values of
Some numbers appear more than once, but after inspecting the list, we see that when the first move is in the horizontal direction,
When the first move is horizontal, the second, fourth, sixth, eighth, and tenth moves will be vertical and have lengths
Using the same idea as the previous case, all
Again, some numbers appear more than once, but examination of the table shows that when the first move is horizontal, the possible values of
We now begin to inspect the possible answers. The point in (A) is
In this case, we have that
These values for
The point
The point
We have shown that the answer must be either (B) or (D).
Note that in both cases we have that
If the first move is vertical, then the third, fifth, seventh, and ninth moves are also vertical, and the other moves are horizontal.
Going through similar analysis as before, we will see that the restrictions on
That is,
The point
However, the point
Since
Answer: (B)
The rectangular prism has two faces whose area is
Therefore,
The prism is made up of
Each
Therefore,
Thus, we get
After dividing the numerator and denominator by
First, note that
This means
If
Since
What if
This can be rearranged to give
Again, this value of
Following this reasoning, if
If
Rearranging this gives
We continue in this way for all possible positive integer values of
The results are summarized in the table below.
According to the table,
We now consider what happens when
If
Since
Similarly, since
We conclude that the only possible positive integer values of
The sum of these numbers is
Answer: (B)
As a percentage, one half of a muffin is equivalent to
Answer: (E)
The sum of the angles in a triangle is
Thus
Answer: (B)
Place zero and the given answers in their correct locations on the number line.
The integer closest to 0 is
Answer: (A)
A number gives a remainder of 3 when divided by 5 if it is 3 more than a multiple of 5.
Since 88 is 3 more than 85, and 85 is a multiple of 5, then 88 gives a remainder of 3 when divided by 5.
We may check that this is the only answer which gives a remainder of 3 when divided by 5.
Answer: (D)
Recall that a prime number is an integer greater than 1 whose only divisors are 1 and itself.
The prime numbers between 10 and 20 are:
Thus, there are 4 integers between 10 and 20 which are prime numbers.
Answer: (E)
Reading from the graph, 15 vehicles had an average speed from 80 km/h to 89 km/h, 30 vehicles had an average speed from 90 km/h to 99 km/h, and 5 vehicles had an average speed from 100 km/h to 109 km/h.
The average speed of each of the remaining vehicles was less than 80 km/h.
The number of vehicles that had an average speed of at least 80 km/h was
Answer: (E)
Any positive integer that is divisible by both 3 and 7 is divisible by their product,
That is, we are being asked to count the number of positive integers less than 100 that are multiples of 21.
These integers are:
We note that the next multiple of 21 is
There are 4 positive integers less than 100 that are divisible by both 3 and 7.
Answer: (C)
The circumference,
If the circumference is 100, then
Answer: (C)
The area,
If the area of a triangle is 6, then substituting we get
Consider the base,
Since
The point which lies a perpendicular distance of 3 units above
Answer: (E)
Barry’s 12 coins include at least one of each of the 5 coins of different values.
The total value of these 5 coins is
Barry has the smallest total amount of money that he could have if each of his remaining
Thus, the smallest total amount of money that Barry could have is
Answer: (A)
An isosceles triangle has two sides of equal length.
Since we are given that two of the side lengths in an isosceles triangle are 6 and 8, then it is possible that the three side lengths are 6, 6 and 8.
In this case, the perimeter of the triangle is
The only other possibility is that the side lengths are
In this case, the perimeter is
Answer: (C)
The angles measuring
Thus,
The angles measuring
Since
Therefore,
Answer: (A)
For the five numbers
If
If
Therefore,
If
If
And finally we confirm that if
Answer: (C)
The first scale shows that has the same mass as
.
The second scale shows that has the same mass as
.
Therefore, the sum of the masses of and
is equal to the sum of the masses of
and
.
That is, the equal-arm scale shown below is balanced.
The left and right sides of this scale each contain a , and the scale remains balanced if one
is removed from each side of the scale (since they are equal in mass).
That is, the equal-arm scale shown below is balanced.
Therefore of the answers given, has the same mass as
.
Answer: (D)
Solution 1
After the arrow is spun twice, the possible outcomes are: red, red; red, blue; red, green; blue, red; blue, blue; blue, green, green, red; green, blue; green, green.
That is, there are 9 possible outcomes.
Of these 9 outcomes, 3 have the same colour appearing twice: red, red; blue, blue; green, green.
The probability that the arrow lands on the same colour twice is
Solution 2
First, we count the total number of possible outcomes given that the arrow is spun twice.
When the arrow is spun the first time, there are 3 possible outcomes (red, blue, green).
When the arrow is spun the second time, there are again 3 possible outcomes.
Thus when the arrow is spun twice, there are
Next, we count the number of outcomes in which the arrow lands on the same colour twice.
When the arrow is spun the first time, there are 3 possible outcomes (red, blue, green).
When the arrow is spun the second time, there is 1 possible outcome (since the colour must match the colour that the arrow landed on after the first spin).
Thus when the arrow is spun twice, there are
The probability that the arrow lands on the same colour twice is
Solution 3
The first spin will land on one of the three colours.
For the arrow to land on the same colour twice, the second spin must land on the colour matching the first spin.
Since exactly 1 of the 3 colours matches the first spin, the probability that the arrow lands on the same colour twice is
Answer: (D)
The lightbulb is used for exactly 2 hours every day and it will work for 24 999 hours.
Thus, the lightbulb will work for
How many weeks are there in 12 500 days?
Since
Specifically,
The lightbulb starts being used on a Monday and so the last day of 1785 complete weeks will be a Sunday.
The lightbulb will stop working 5 days past Sunday, which is a Friday.
Answer: (B)
Since
Both
That is,
In the final equation given,
Since
Answer: (B)
We begin by separating the given information, as follows:
From bullets 2 and 3, we can conclude that Alice owns more dogs than both Kathy and Bruce.
From bullet 4, we can conclude that answer (A) is not true.
From bullets 1 and 4, we can conclude that both Kathy and Bruce own more cats than Alice.
However, we cannot determine if Kathy owns more cats than Bruce, or vice versa.
Therefore, we cannot conclude that (B) or (C) must be true.
From bullet 2, we can conclude that (E) is not true.
Thus the statement which must be true is (D).
Answer: (D)
The horizontal line through
Joining
The
The
Using the Pythagorean Theorem,
(Alternatively, we could have drawn the vertical line through
Answer: (A)
Solution 1
Square
We require the total area of the shaded regions to equal the total area of the non-shaded regions, and so the total area of the shaded regions is to be half the area of the square, or 1800.
Since
To determine the area of each of the shaded regions, we begin by labelling each of the given lengths (as well as those that we can determine), and constructing diagonal
Quadrilateral
In
Thus, the area of
In
The area of
Therefore, the area of quadrilateral
Quadrilateral
The midpoint of
In
In
The area of
Therefore, the area of quadrilateral
Finally, adding the areas of the shaded regions, we get
Solution 2
As in Solution 1, we require the total area of the shaded regions to equal 1800.
We begin by labelling
Since
Since
We label each of the given lengths (as well as those that we can determine).
Quadrilateral
In
In trapezoid
The area of trapezoid
Therefore, the area of quadrilateral
Quadrilateral
The area of trapezoid
Adding the areas of the shaded regions, we get
Answer: (E)
Let the number of teams in Jen’s baseball league be
Each of these
Since there are 2 teams in each of these games, the total number of games played is
The total number of games played is 396, so
The numbers
Since
Answer: (A)
Let Rich’s 4-digit positive integer be
We begin by determining which of the 4 digits is erased.
If Rich erases the digit
The units digit of this sum is determined by adding the units digits of
In this case, the units digit of the sum is even since
The sum of these two integers is 6031 which has an odd units digit, and so the digit that is erased cannot be
Using a similar argument, the digit that is erased cannot be
Next, we determine the digits
The hundreds column of this sum gives that
If the sum is 0, then
If the sum is 20, then
If
Therefore,
The thousands column then gives
The hundreds column gives
If the carry from the tens column is 0, then
In the tens column, if
This tells us that the carry from the tens column cannot be 0.
If the carry from the tens column is 1, then
(Note that sum of the tens column has units digit 3 and so the carry cannot be 2.)
Answer: (B)
We begin by recognizing that each of the given answers has a common numerator of
Since
However, the product of the integers from 1 to 19 inclusive is equal to
That is, the common numerator
Next, we consider the result after dividing
Since
The product of a perfect square and some positive integer factor
An answer satisfying this condition is (E).
Why is
Rewriting,
Can you explain why each of the other four answers is not equal to a perfect square?
Answer: (E)
First, notice that the given list of 10 numbers has a sum of
If
This means
We also require that the number of groups be at least 2, so the number of groups is one of
If the number of groups is
Similarly, if the number of groups is
Therefore, there are either
Let’s first assume that there are five groups.
In this case, the total in each group must be
Since
Since the total in each group must be
The number 8 must be paired with 1 since adding any other positive integer to 8 will give a sum greater than 9.
That is, one of the groups has the numbers
The number 7 cannot be in a group with any number larger than 2, and 1 is already paired with 8, so another group must be
Continuing with this reasoning, we get that
Therefore, if there are 5 groups, they must be
As mentioned before,
There are 5 ways to do this:
We have shown that there are 5 ways to separate the numbers
Let’s now assume that there are three groups.
As we did when there were 5 groups, we ignore
Since there are three groups, the sum in each group is
We now find all groups which add up to
There are 17 of them, so we will label them using
Label | Group |
---|---|
A | {6, 9} |
B | {1, 5, 9} |
C | {2, 4, 9} |
D | {1, 2, 3, 9} |
E | {7, 8} |
F | {1, 6, 8} |
G | {2, 5, 8} |
H | {3, 4, 8} |
I | {1, 2, 4, 8} |
J | {2, 6, 7} |
K | {3, 5, 7} |
L | {1, 2, 5, 7} |
M | {1, 3, 4, 7} |
N | {4, 5, 6} |
O | {1, 3, 5, 6} |
P | {2, 3, 4, 6} |
Q | {1, 2, 3, 4, 5} |
Of these 17 groups, only
Therefore, any way of separating the integers into three groups must use exactly one of these four groups.
Assume
The groups which lack both
If
If
This forces the third group to be
Similarly, if
At this point, we can stop checking.
This is because, for example, if we assume
To summarize, if
If we assume
The groups satisfying this condition are
By the same reasoning as in the previous paragraph, the configurations including
The groups which have no members in common with
In total, we have found that there are
As before, placing
Adding this to the five from earlier, the number of ways to separate the list
Answer: (E)
We begin by labeling
We will now find the measures of all
Let
Also,
Rearranging this equation, we have
In
Since
Substituting this into the above equation, we have
Note that this means
Substituting
Since
Solving this, we get
Angles
Since
We now set
Since
Substituting this and
Finally, we have
In summary, the 12 angles in terms of
We are given that the measure of each angle in degrees is an integer.
In particular, since
We also know that
This means
Since
The integers
In this case, the twelve angles are
Therefore,
By factoring, we have that
Therefore, this number can only be prime when
We know that
Similarly,
When
When
In order to satisfy the condition that the measures of exactly
The prime numbers less than 30 are
We have already investigated what happens when
All 3 angles prime? | |||
---|---|---|---|
Of the angles that appear in the second two columns, the numbers which are prime are
Therefore, all three angles are prime when
Combining these with
Answer: (D)