Wednesday, February 24, 2016
(in North America and South America)
Thursday, February 25, 2016
(outside of North American and South America)
©2015 University of Waterloo
Evaluating,
Answer: (E)
According to the graph, 7 of the 20 teachers picked “Square” as their favourite shape.
Thus,
(We can also note that the numbers of teachers who chose “Triangle”, “Circle” and “Hexagon” were 3, 4 and 6. The sum of these totals is indeed
Answer: (E)
Evaluating,
Answer: (C)
Since each of Bill’s steps is
To walk 12 m, Bill thus takes
Answer: (D)
Solution 1
Since
Thus,
Solution 2
Since
Therefore,
Since the sum of the angles in
Answer: (A)
Since the slope of the line through points
From this,
Answer: (C)
When Team A played Team B, if Team B won, then Team B scored more goals than Team A, and if the game ended in a tie, then Team A and Team B scored the same number of goals.
Therefore, if a team has 0 wins, 1 loss, and 2 ties, then it scored fewer goals than its opponent once (the 1 loss) and the same number of goals as its oppponent twice (the 2 ties).
Combining this information, we see that the team must have scored fewer goals than were scored against them.
In other words, it is not possible for a team to have 0 wins, 1 loss, and 2 ties, and to have scored more goals than were scored against them.
We can also examine choices (A), (B), (D), (E) to see that, in each case, it is possible that the team scored more goals than it allowed.
This will eliminate each of these choices, and allow us to conclude that (C) must be correct.
(A): If the team won 2-0 and 3-0 and tied 1-1, then it scored 6 goals and allowed 1 goal.
(B): If the team won 4-0 and lost 1-2 and 2-3, then it scored 7 goals and allowed 5 goals.
(D): If the team won 4-0, lost 1-2, and tied 1-1, then it scored 6 goals and allowed 3 goals.
(E): If the team won 2-0, and tied 1-1 and 2-2, then it scored 5 goals and allowed 3 goals.
Therefore, it is only the case of 0 wins, 1 loss, and 2 ties where it is not possible for the team to score more goals than it allows.
Answer: (C)
Solution 1
We calculate the value of each of the five words as follows:
The value of
The value of
The value of
The value of
The value of
Of these, the word with the largest value is
Solution 2
We determine the word with the largest value by comparing the given words.
Since
Similarly, the value of
Therefore, the two possibilities for the word with the largest value are
The value of
Thus, the word with the largest value is
(Alternatively, we could note that 2
Answer: (D)
Solution 1
We write out the numbers in the sequence until we obtain a negative number:
Thus, Grace writes 11 positive numbers in the sequence.
Solution 2
The
Therefore, the
The first positive integer
Therefore, Grace writes 11 positive numbers in the sequence.
Answer: (A)
Solution 1
We label the players as A, B, C, D, and E.
The total number of matches played will be equal to the number of pairs of players that can be formed times the number of matches that each pair plays.
The possible pairs of players are AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE. There are 10 such pairs.
Thus, the total number of matches played is
Solution 2
Each student plays each of the other 4 students 3 times and so plays in
Since there are 5 students, then the students play in a total of
But each match includes 2 students, so each match is counted twice in the number of matches in which students play (60). Thus, the total number of matches player is
Answer: (C)
Extend
Since
Thus,
By the Pythagorean Theorem,
Now
Also,
Therefore,
Since
Answer: (E)
Since Alejandro selects one ball out of a box that contains 30 balls, then the possibility that is the most likely is the one that is satisfied by the largest number of balls in the box.
There are 3 balls in the box whose number is a multiple of 10. (These are 10, 20, 30.)
There are 15 balls in the box whose number is odd. (These are the numbers whose ones digits are 1, 3, 5, 7, or 9.)
There are 4 balls in the box whose number includes the digit 3. (These are 3, 13, 23, and 30.) There are 6 balls in the box whose number is a multiple of 5. (These are 5, 10, 15, 20, 25, 30.)
There are 12 balls in the box whose number includes the digit 2. (These are 2, 12, and the then integers from 20 to 29, inclusive.)
The most likely of these outcomes is that he selects a ball whose number is odd.
Answer: (B)
We note that
If we compare two fractions with equal positive numerators, the fraction with the smaller positive denominator will be largest of the two fractions.
Therefore,
(Alternatively, we could note that since
Answer: (C)
Since
Therefore,
When written out, this integer consists of a 1 followed by 120 zeros.
Answer: (A)
We note that
For an integer to be divisible by each of
The smallest such positive integer is
Answer: (E)
The three diagrams below show first the original position of the triangle, the position after it is reflected in the
The final position is seen in (D).
Answer: (D)
Since the perimeter of square
Therefore,
Since the perimeter of
Since
Therefore, the perimeter of pentagon
Answer: (C)
Solution 1
Suppose that the three integers are
From the first two equations,
Since
Since
Since
The three original numbers are 407, 591 and 643.
The difference between the largest and smallest of these integers is
Solution 2
Suppose that the three numbers are
Since
Since
Therefore,
This tells us that
Since
But
Answer: (E)
The number of points on the circle equals the number of spaces between the points around the circle.
Moving from the point labelled 7 to the point labelled 35 requires moving
Since the points labelled 7 and 35 are diametrically opposite, then moving along the circle from 7 to 35 results in travelling halfway around the circle.
Since 28 spaces makes half of the circle, then
Thus, there are 56 points on the circle, and so
Answer: (C)
Solution 1
Suppose that, when the
Since the students are being put in groups of 2, an incomplete group must have exactly 1 student in it.
Therefore,
Since the number of complete groups of 2 is 5 more than the number of complete groups of 3, then there were
Since there was still an incomplete group, this incomplete group must have had exactly 1 or 2 students in it.
Therefore,
If
In this case,
If
In this case,
If
If
Since the difference between the number of complete groups of 3 and the number of complete groups of 4 is given to be 3, then it must be the case that
In this case,
Solution 2
Since the
The first few integers larger than 1 that are not divisible by 2, 3 or 4 are 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, and 35.
In each case, we determine the number of complete groups of each size:
5 | 7 | 11 | 13 | 17 | 19 | 23 | 25 | 29 | 31 | 35 | |
# of complete groups of 2 | 2 | 3 | 5 | 6 | 8 | 9 | 11 | 12 | 14 | 15 | 17 |
# of complete groups of 3 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
# of complete groups of 4 | 1 | 1 | 2 | 3 | 4 | 4 | 5 | 6 | 7 | 7 | 8 |
Since the number of complete groups of 2 is 5 more than the number of complete groups of 3 which is 3 more than the number of complete groups of 4, then of these possibilities,
In this case,
(Since the problem is a multiple choice problem and we have found a value of
Answer: (B)
Suppose that Jackie had played
Since she scored an average of 20 points per game over these
In her last game, she scored 36 points and so she has now scored
But, after her last game, she has now played
Therefore, we can also say that her total number of points scored is
Thus,
This tells us that after 16 games, Jackie has scored
For her average to be 22 points per game after 17 games, she must have scored a total of
This would mean that she must score
Answer: (A)
Since the track is circular with radius 25 km, then its circumference is
In the 15 minutes that Alain drives at 80 km/h, he drives a distance of
When Louise starts driving, she drives in the opposite direction to Alain.
Suppose that Alain and Louise meet for the first time after Louise has been driving for
During this time, Louise drives at 100 km/h, and so drives
During this time, Alain drives at 80 km/h, and so drives
Since they start
Therefore,
Suppose that Alain and Louise meet for the next time after an additional
During this time, Louise drives
In this case, the sum of the distances that they drive is the complete circumference of the track, or
The length of time between the first and second meetings will be the same as the amount of time between the second and third, and between the third and fourth meetings.
Therefore, the total time that Louise has been driving when she and Alain meet for the fourth time will be
Answer: (C)
If the four sides that are chosen are adjacent, then when these four sides are extended, they will not form a quadrilateral that encloses the octagon. (See Figure 1.)
Figure 1
If the four sides are chosen so that there are exactly three adjacent sides that are not chosen and one other side not chosen, then when these four sides are extended, they will not form a quadrilateral that encloses the octagon. (See Figures 2 and 3.)
Figure 2
Figure 3
Any other set of four sides that are chosen will form a quadrilateral that encloses the octagon. This is true based on the following argument:
Suppose that side
These two sides will either be adjacent (Figure 4) or have one unchosen side between them (Figure 5) or will have two unchosen sides between them (Figure 6). (They cannot have three or four adjacent unchosen sides between them from the previous argument. They cannot have more than four adjacent unchosen sides between them since four sides have to be chosen.)
Figure 4
Figure 5
Figure 6
In each of these cases, sides
Continue the process to choose
Since each consecutive pair of chosen sides meets between the two sides and on or outside the octagon, then these four meeting points (between
Figure 7
We are told that there is a total of 70 ways in which four sides can be chosen.
We will count the number of ways in which four sides can be chosen that do not result in the desired quadrilateral, and subtract this from 70 to determine the number of ways in which the desired quadrilateral does result.
There are 8 ways in which to choose four adjacent sides: choose one side to start (there are 8 ways to choose one side) and then choose the next three adjacent sides in clockwise order (there is 1 way to do this).
There are 8 ways to choose adjacent sides to be not chosen (after picking one such set, there are 7 additional positions into which these sides can be rotated). For each of these choices, there are 3 possible choices for the remaining unchosen sides. (Figures 2 and 3 give two of these choices; the third choice is the reflection of Figure 2 through a vertical line.) Therefore, there are
Therefore, of the 70 ways of choosing four sides, exactly
Thus, the probability that four sides are chosen so that the desired quadrilateral is formed is
Answer: (B)
Suppose that a number
Suppose that these 19 integers are
Then
This tells us that
Also, since this is exactly the list of integers that is included strictly between
In other words, neither
This tell us that
Therefore, we have that
Next, we use
To have
But if
Furthermore,
Next, note that
When
When
Since we need
Therefore,
So we limit our search for
When
When
When
When
When
Therefore, the numbers
Finally, we must determine the sum of all such
The integers
The
We continue by including those satisfying
Each of these 31 numbers equals 23 plus a fraction between 0 and 1.
With the exception of the one number with denominator 2, each of the fractions can be paired with another fraction with the same denominator to obtain a sum of 1. (For example,
Therefore, the sum of all of these
Finally, the sum of all
Answer: (C)
Using the given rules, the words that are 1 letter long are A, B, C, D, E.
Using the given rules, the words that are 2 letters long are AB, AC, AD, BA, BC, BD, BE, CA, CB, CD, CE, DA, DB, DC, DE, EB, EC, ED.
Let
Let
Suppose
Consider a word of length
Since two vowels cannot occur in a row, the second letter of this word must be B, C or D.
This means that every word of length
Also, each word of length
Therefore,
Consider a word of length
Since the same letter cannot occur twice in a row, then the second letter of this word is either a vowel or a different consonant than the first letter of the word.
Each word of length
Each word of length
Therefore,
We note from above that
The equations
Since
We want to determine
We continue these calculations and make a table:
Therefore, there are 199 776 words of length 10 that begin with a vowel.
Answer: (E)