Thursday, April 16, 2015
(in North America and South America)
Friday, April 17, 2015
(outside of North American and South America)
©2015 University of Waterloo
The
Substituting
Line 1 has
Solution 1
The equation of line 2 with slope
Line 2 passes through
Substituting, we get
The equation of line 2 is
Solution 2
A line with slope
Since line 2 has slope
We find the coordinates of point
This gives
That is, line 2 has
If we let the base of
Since
The
Therefore,
Finally, the area of
The total number of students at School A is the sum of the number of students who received a ride and the number of students who did not, or
Since 330 of 750 students received a ride, the percentage of students at School A who received a ride is
Solution 1
At school B, 30% of 240 students or
If 50% or
Therefore,
Solution 2
As a percent, the difference between 50% of students receiving a ride and the 30% of students who did receive a ride is 20%.
Therefore, 20% of 240 students or
Solution 1
At school C, 45% of 200 students or
At school D,
The total number of students at School C and School D is
The total number of students receiving a ride at School C and School D is
Since 57.6% of the combined group of students from the two schools received a ride, then
Multiplying both sides of this equation by 500, we get
Solution 2
At school C, 45% of 200 students or
The total number of students at School C and School D is
Since 57.6% of the combined group of students from the two schools received a ride, then
Out of the 288 students who received a ride, 90 students were from School C and so the remaining
Since there are 300 students at School D, the percentage of students receiving a ride is
Therefore, the value of
At school E,
At school F,
The total number of students at School E and School F is
The total number of students receiving a ride at School E and School F is
Between 55% and 60% of the 450 students from the two schools received a ride.
Since 55% of 450 is 247.5 and 60% of 450 is 270, then
Solving
Since
Since
(If
We first note that the product of an even integer and any other integers, even or odd, is always an even integer.
Let
If
The only remaining possibility is that both
If
In this case,
Therefore, for any integers
Since
Written as ordered pairs
Therefore, the factor pairs whose sum is odd are:
Since both
Let
We must first determine all factor pairs
We begin by considering the parity (whether each is even or odd) of the factors
Since 2 is even, then
If
However if
That is, if
We say that the factors
If
That is, if
Now we are searching for all factor pairs
Written as a product of its prime factors,
Since exactly one of
That is, either
In both cases,
We now determine all factor pairs
These are
Therefore
Since
This leaves 11 factor pairs
Each of these 11 factor pairs
To see this, let
For example when
That is, the factor pair
Each of the 11 pairs
We determine the corresponding ordered pair
8 | 1125 | ||
24 | 375 | ||
40 | 225 | ||
72 | 125 | ||
75 | 120 | ||
45 | 200 | ||
3 | 3000 | ||
5 | 1800 | ||
9 | 1000 | ||
15 | 600 | ||
25 | 360 |
There are 11 ordered pairs
Since
Since
Thus,
We make a table to track the position of the square as it rotates about the triangle:
After Move # |
Coinciding Vertices |
Second Vertex of on Side of Square |
Centre of Rotation of Next Move |
Angle of Rotation of Next Move |
---|---|---|---|---|
Initial | ||||
1 | ||||
2 | ||||
3 | ||||
4 | ||||
5 | ||||
6 | ||||
7 | ||||
8 |
The information in this table comes from the diagrams here:
Each angle of rotation is either
Therefore,
(We continued the table through the 8th move to be more useful in part (c).)
After 0 moves, the square has
After 8 moves, the square has
Therefore, through 8 moves, the net change of position of the square was
moving clockwise around 2 sides of the triangle. After these 8 moves,
the square is in a similar position, relative to the triangle, to its
initial position:
Starting from after the 8th move, the square starts in this similar
position, and so 8 more moves will take the square again to a similar
position. The net change will again be that the square has moved
clockwise around 2 sides of the triangle. (We note from part (b) that
after 8 moves is the first time (after the original position) that
vertex
Thus, after 16 moves, the square will have
Using similar reasoning, after 24 moves, the square will have
Therefore, we need to determine the total distance travelled by
This total distance equals 3 times the distance travelled by
This is because the relative position of the square with respect to the
triangle (
Since
Also, the distance from
Since
Also, the distance from
We make a chart of the rotation undergone by
Move # |
Centre of Rotation |
Distance of Centre from |
Angle of Rotation of Square |
---|---|---|---|
1 | 1 | ||
2 | 0 | ||
3 | 1 | ||
4 | 2 | ||
5 | |||
6 | |||
7 | |||
8 | 2 |
During each rotation, each point on the square (except for the point of the square in contact with the centre of rotation) is rotated about the centre of rotation through the angle of rotation of the square.
Thus, during each rotation, the distance travelled by
Therefore, the distance travelled by