Tuesday, February 28, 2017
(in North America and South America)
Wednesday, March 1, 2017
(outside of North American and South America)
©2016 University of Waterloo
Evaluating,
Alternatively, we note that
Answer: (C)
Evaluating,
Alternatively, we note that
Answer: (D)
The height of the snowman equals the sum of the lengths of the diameters of the three spheres.
Since the radii of the spheres are 10 cm, 20 cm and 30 cm, then the lengths of their diameters are 20 cm, 40 cm and 60 cm.
Thus, the height of the snowman is
Answer: (D)
We write each of the choices in lowest terms:
Next, we note that
The fraction with the greatest value will be the one that is equal to 1 minus the smallest fraction.
Since
Answer: (E)
Since 300 litres drains in 25 hours, then the rate at which water is leaving the tank equals
Answer: (A)
When Penelope folds the paper in half, the number of layers doubles.
Starting with 4 layers of paper, then after the next three folds, there are 8 and then 16 and then 32 layers of paper. Additional folds create more layers.
Of the given choices (which are all less than 32), only 16 is a possible number of layers.
Answer: (D)
By definition,
Answer: (B)
In option (A), the first and third cards have 0 numbers in common, so (A) is not correct.
In option (B), the second and third cards have 2 numbers in common, so (B) is not correct.
In option (C), the first and third cards have 2 numbers in common, so (C) is not correct.
In option (E), the first and third cards have 2 numbers in common, so (E) is not correct.
In option (D), the first and second cards share exactly 1 number (namely, 4), the first and third cards share exactly one number (namely, 7), and the second and third cards share exactly one number (namely, 2). Thus, (D) is correct.
Answer: (D)
Since the bill including a 13% tip was $226, then $226 is 113% of the bill before tax.
Thus, the bill before tax was
The amount of the tip is 15% of the bill before tax, or
Answer: (C)
Since
Therefore,
Therefore,
Since
Answer: (B)
Since 10 identical squares have a total area of
Since the area of each of the 10 identical squares is
The perimeter of the given figure equals 22 square side lengths (4 on the left side, 4 on the bottom, 4 on the right side, 2 separate ones on the top, and 8 in the “U” shape in the middle).
Therefore, the perimeter of the figure is
Answer: (C)
Since the mean of
Since the mean of
Therefore, the mean of the five integers is
Answer: (A)
First, we consider the column of units (ones) digits.
From this column, we see that the units digit of
By trying the possible digits from 0 to 9, we find that
When
From this column, we see that the units digit of
By trying the possible digits from 0 to 9, we find that
There is no carry created when
Looking at the remaining digits, we see that
Checking, if
Therefore,
Answer: (B)
Since Igor is shorter than Jie, then Igor cannot be the tallest.
Since Faye is taller than Goa, then Goa cannot be the tallest.
Since Jie is taller than Faye, then Faye cannot be the tallest.
Since Han is shorter than Goa, then Han cannot be the tallest.
The only person of the five who has not been eliminated is Jie, who must thus be the tallest.
Answer: (E)
Since there are 32 red marbles in the bag and the ratio of red marbles to blue marbles is
Since there are 56 blue marbles in the bag and the ratio of blue marbles to purple marbles is
Since the bag contains only red, blue and purple marbles, then there are
Answer: (E)
Since
Answer: (B)
First, we factor
Since 41 is a prime factor of 1230, then one of
Since all of the possibilities given for the minimum sum
Thus, we can let
Now we are looking for positive integers
The possible pairs of values for
The pair with the smallest sum is
Therefore, we set
Answer: (B)
Since
Since
Since the fractions are all positive and
Since
Answer: (C)
We start by drawing a graph that includes the point
We label the various points of intersection (
We want to determine the area of
Since
Since the slope of
Since the slope of
Since
We can view
Therefore, the area of this triangle is
Alternatively, we could note that the area of
Thus, the area of
Answer: (C)
Since there are 60 seconds in 1 minute, then
Since there are 60 minutes in 1 hour, then
Consider the distances that Car X and Car Y travel between the instant when the front of Car Y is lined up with the back of Car X and the instant when the back of Car Y is lined up with the front of Car X.
Since the length of Car X is 5 m and the length of Car Y is 6 m, then during this interval of time, Car Y travels
Since there are 1000 metres in 1 km, then 11 m is equivalent to
Since Car X travels at 90 km/h, then in
Since Car Y travels at 91 km/h, then in
Therefore,
Answer: (A)
We label the remaining unknown entries as
Now
In particular, we see first that
If
Thus,
In this case,
If
Thus,
In this case,
If
Thus,
In this case,
If
Thus,
If
If
If
Having considered all possible cases, we see that there is only one possible value for
Answer: (A)
Suppose that the height of the top left rectangle is
Since square
Since the width of the bottom rectangle is 42 (the side length of the square), then the perimeter of the bottom rectangle is
Suppose that the width of the top left rectangle is
Since each of the small rectangles has the same perimeter, then
Thus,
Since the side length of the square is 42, then the width of the top right rectangle is
Since the two rectangles on the top right have the same perimeter and the same width, then they must have the same height, which is equal to half of the height of the top left rectangle.
Thus, the height of each of the two rectangles on the right is
Finally, the perimeter of the top right rectangle must equal also equal
Thus,
This tells us that the shaded rectangle has dimensions
(We can check by substitution that the resulting rectangles in the original diagram are
Answer: (E)
To solve this problem, we need to use two facts about a triangle with positive area and side lengths
If the triangle is obtuse, then
Suppose that the unknown side length
Here, we must have
From the first inequality,
From the second inequality,
Suppose that the unknown side length
Here, we must have
From the first inequality,
From the second inequality,
Therefore, the possible values of
The sum of these possible values is
We still need to justify the second fact. We show that if the triangle is obtuse, then
Consider
with obtuse and with , and .
At, draw a perpendicular line segment with length . Draw and .
Since
is right-angled at , then , by the Pythagorean Theorem.
Nowis isosceles with .
Thus,.
But.
Since, then in , we have .
This means thatand so , as required.
Answer: (E)
We think of each allowable sequence of moves as a string of X’s, Y’s and Z’s. For example, the string ZZYXZ would represent moving Z one space to the right, then Z, then Y, then X, then Z.
For each triple
For example,
We want to find
Next, we note that if
We now make the key observation that if
This rule is true because:
Every allowable sequence counted by
If an allowable sequence counted by
Furthermore, if
The last two bullets together tell us that the number of sequences counted by
Similarly, the sequences counted by
Therefore,
Using this rule, we can create tables of values of
0 | 1 | 2 | 3 | |
---|---|---|---|---|
0 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 |
2 | 0 | 0 | 0 | 0 |
3 | 0 | 0 | 0 | 0 |
0 | 1 | 2 | 3 | |
---|---|---|---|---|
0 | 1 | 0 | 0 | 0 |
1 | 2 | 3 | 0 | 0 |
2 | 2 | 5 | 5 | 0 |
3 | 0 | 0 | 0 | 0 |
0 | 1 | 2 | 3 | |
---|---|---|---|---|
0 | 1 | 0 | 0 | 0 |
1 | 3 | 6 | 0 | 0 |
2 | 5 | 16 | 21 | 0 |
3 | 5 | 21 | 42 | 42 |
The non-zero entries are filled from top to bottom and left to right, noting that each entry is the sum of the entries (if applicable) to the left in the same table (this is
Also,
From these tables, we see that
Answer: (C)
We proceed using several steps.
Step 1: Least common multiples
For each positive integer
One way to calculate the least common multiple of a list of integers is to determine the prime factorization of each of the integers in the list and then to create the product of the largest of each of the prime powers that occurs among the integers in the list.
For example, if
The prime factorizations of the integers larger than 1 in this list are
We note that, for any list of integers, a common multiple,
Step 2: Connection between
We show that
Suppose that
Then
From Step 1,
Since
Since
Also, if
Putting this together,
Step 3: Re-statement of problem
Based on Step 2, we now want to find all positive integers
Step 4: If
Suppose that
Further, suppose that
Then
In this case,
This means that if
Similarly,
Step 5: Further analysis of
Suppose that
From Step 4, both
Since
In other words, one of
Furthermore, we know that
This means that the list
This means that
Since one of
Step 6: Second re-statement of problem
We want to find all positive integers
From Step 5, we know that either
This means that we want to find all positive integers
If
If
Step 7: Determining Nella numbers
We make two separate tables, one where
Range of |
Largest power of 2 less than |
Prime power? | Nella? | ||
---|---|---|---|---|---|
32 | 32 | 33 | No ( |
No | |
64 | 64 | 65 | No ( |
No | |
128 | 128 | 129 | No ( |
No | |
256 | 256 | 257 | Yes | See below | |
512 | 512 | 513 | No ( |
No | |
1024 | 1024 | 1025 | No ( |
No |
Note that
For each
For
Range of |
Largest power of 2 at most |
Prime power? | Nella? | ||
---|---|---|---|---|---|
32 | 32 | 31 | Yes | See below | |
64 | 64 | 63 | No ( |
No | |
128 | 128 | 127 | Yes | See below | |
256 | 256 | 255 | No ( |
No | |
512 | 512 | 511 | No ( |
No | |
1024 | 1024 | 1023 | No ( |
No |
Note that
For each
For
For each
For
Step 8: Final tally
From the work above, there are
Answer: (A)