Thursday, February 23, 2021
(in North America and South America)
Wednesday, February 24, 2021
(outside of North American and South America)
©2020 University of Waterloo
A rectangle with width 2 cm and length 3 cm has area
Answer: (E)
Calculating,
Answer: (A)
Expressed as a fraction, 25% is equivalent to
Answer: (B)
When
Answer: (B)
We note that
Answer: (C)
We draw an unshaded dot to represent the location of the dot when it is on the other side of the sheet of paper being shown. Therefore, the dot moves as follows:
It is worth noting that folding and unfolding the paper have no net effect on the figure. Thus, the resulting figure can be determined by rotating the original figure by
Answer: (E)
When
When
When
When
When
This means that
Answer: (D)
Suppose that the original integer has tens digit
This integer is equal to
When the digits are reversed, the tens digit of the new integer is
This new integer is equal to
Since the new two-digit integer minus the original integer is 54, then
Thus, the positive difference between the two digits of the original integer is 6. An example of a pair of such integers is 71 and 17.
Answer: (C)
The line with equation
The line with equation
This means that the new line has equation
To find its
Thus, the
Answer: (D)
Using exponent laws,
Since
Answer: (E)
Since the second number being added is greater than 300 and the sum has hundreds digit
From the ones column, we see that the ones digit of
This makes the following sum.
Since
Therefore,
Answer: (A)
A perfect square is divisible by 9 exactly when its square root is divisible by 3.
In other words,
In the list
Therefore, in the list
Answer: (E)
In an isosceles right-angled triangle, the ratio of the length of the hypotenuse to the length of each of the shorter sides is
Consider
Here,
Since
Consider
Here,
Since
Therefore, the perimeter of
Answer: (C)
Suppose that Natascha runs at
Since she cycles 3 times as fast as she runs, she cycles at
In 1 hour of running, Natascha runs
In 4 hours of cycling, Natascha cycles
Thus, the ratio of the distance that she cycles to the distance that she runs is equivalent to the ratio
Answer: (A)
Solution 1:
Since
Since
The largest multiple of 45 less than 2021 is
If
Here,
If
Therefore, if
If
Therefore, the minimum possible value of
Solution 2:
We re-write
Since
In particular, this tells us that
Since the sum of
Since
This means that
However,
The next multiple of 44 less than 1980 is
If
If
Since
Answer: (C)
The first few values of
This means that the only possible answer is choice (D), or 7.
To be complete, we explain why 7 cannot be the ones (units) digit of
For
The only odd factorial is
Since
For the ones (units) digit of
This is impossible as the first few factorials are shown above and every greater factorial has a ones (units) digit of 0, because it is a multiple of both 2 and 5.
Answer: (D)
Since the average of the two smallest integers in
Since the average of the two largest integers in
Suppose that the other five integers in the set
We would like this average to be as large as possible.
To make this average as large as possible, we want
What is the maximum possible value of
Let
Since
For
In this case, we can have
To make
If
This means that the maximum possible average of the integers in
Answer: (B)
Suppose that
The semi-circles with diameters
The area of each semi-circle with radius
Since the sum of the areas of the three semi-circles equals 5 times the area of the semi-circle with diameter
Since
In other words,
Therefore,
Answer: (B)
Since
Thus, the equation
Since the square of
If
In this case, since
If
In this case, since
We can check by direct substitution that
Since
Answer: (E)
Let
Since
Since
Thus,
Since
This means that
Consider
Since
Also,
Since
Since
Next, consider
Since
This means that
Since
Answer: (C)
Since
Since
While this answers the question, is there actually a function that satisfies the requirements? The answer is yes.
One function that satisfies the requirements of the problem is the function
Answer: (A)
The total surface area of the cone includes the circular base and the lateral surface.
For the given unpainted cone, the base has area
Thus, the total surface area of the unpainted cone is
Since the height and base are perpendicular, the lengths
For this cone, by the Pythagorean Theorem,
When the unpainted cone is placed in the container of paint so that the paint rises to a depth of 2 cm, the base of the cone (area
Also, the bottom portion of the lateral surface is covered in paint.
The unpainted portion of the cone is itself a cone with height 2.
When we take a vertical cross-section of the cone through its top vertex and a diameter of the base, the triangle formed above the paint is similar to the original triangle and has half its dimensions. The triangles are similar because both are right-angled and they share an equal angle at the top vertex. The ratio is
Therefore, the unpainted cone has radius 1.5 cm (half of the original radius of 3 cm) and slant height 2.5 cm (half of the original slant height of 5 cm).
Thus, the unpainted lateral surface area is the lateral surface area of a cone with radius 1.5 cm and slant height 2.5 cm, and so has area
This means that the painted lateral surface area is
Thus, the fraction of the total surface area of the cone that is painted is
Answer: (A)
Since each Figure is formed by placing two copies of the previous Figure side-by-side along the base and then adding other pieces above, the number of dots in the base of each Figure is two times as many as in the previous Figure.
Since each Figure is an equilateral triangle, then the number of dots in the Figure equals the sum of the positive integers from 1 to the number of dots in the base, inclusive. In other words, if the base of a Figure consists of
Since each Figure is formed by using three copies of the previous Figure and any new dots added are shaded dots, the number of unshaded dots in each Figure is exactly three times the number of unshaded dots in the previous Figure.
Since each dot is either shaded or unshaded, the number of shaded dots equals the total number of dots minus the number of unshaded dots.
Using these statements, we construct a table:
Figure | Dots in base | Dots in Figure | Unshaded dots | Shaded dots |
---|---|---|---|---|
1 | 2 | 3 | 3 | 0 |
2 | 4 | 10 | 9 | 1 |
3 | 8 | 36 | 27 | 9 |
4 | 16 | 136 | 81 | 55 |
5 | 32 | 528 | 243 | 285 |
6 | 64 | 2080 | 729 | 1351 |
7 | 128 | 8256 | 2187 | 6069 |
8 | 256 | 32896 | 6561 | 26335 |
9 | 512 | 131328 | 19683 | 111645 |
Therefore, the smallest value of
We note that since the number of dots in the base of Figure 1 is 2 and the number of dots in the base of each subsequent Figure is double the number of dots in the previous Figure, then the number of dots in the base of Figure
Since the number of unshaded dots in Figure 1 is 3 and the number of unshaded dots in each subsequent Figure is three times the number of unshaded dots in the previous Figure, then the number of unshaded dots in Figure
Therefore, a formula for the number of unshaded dots in Figure
Answer: (B)
The curves with equations
This equation has at least one real solution when the quadratic equation
(Note that when
This quadratic equation has at least one real solution when its discriminant is non-negative; that is, when
The equation
The inequality
The equation
The inequality
Re-phrasing the problem in an equivalent geometric way, we want to determine the probability that a point
Note that each of these two circles passes through the centre of the other circle.
Putting this another way, we want to determine the fraction of the area of the circle centred at the origin that is outside the circle centred at
Let
The area of each circle is
To determine the shaded area, we subtract the unshaded area from the area of the circle. By symmetry, the unshaded area below the
In the second diagram above, the origin is labelled
Since
Since
This means that
When
Since
Similarly, when
If
Thus, the total unshaded area inside the circle of area
Finally, we need to use the area of an equilateral triangle with side length
Therefore, the fraction of the circle that is shaded is
Of the given choices, this is closest to 61, or (E).
Answer: (E)
Suppose that
We determine the number of triples
Since the product
There are 6 ways in which this can happen: both factors in
Similarly, there are 6 ways of distributing each of the other squares of prime factors.
Since
Next, we include the condition no pair of
We count the number of triples with one pair equal, and subtract this number from
We do this by counting the number of these triples with
In order to have
Thus, for each of the 8 squared prime factors
Similarly, there will be
This means that there are
The original problem asked us to the find the number of triples
To convert triples
Therefore, the total number of triples
Answer: (C)