Solutions
(Grade 7 and 8)
Wednesday, May 13, 2015 (in North America and South America)
Thursday, May 14, 2015 (outside of North American and South America)
©2014 University of Waterloo
The circle is divided into 4 equal regions. Since 1 of these 4 regions is shaded, then the fraction of the circle that is shaded is
Answer: (C)
Evaluating,
Answer: (D)
Reading from the graph, Phil ran 4 km, Tom ran 6 km, Pete ran 2 km, Amal ran 8 km, and Sanjay ran 7 km. Therefore, Pete ran the least distance.
Answer: (C)
The equal-arm balance shows that 2 rectangles have the same mass as 6 circles. If we organize these shapes into two equal piles on both sides of the balance, then we see that 1 rectangle has the same mass as 3 circles.
Answer: (B)
Of the possible answers, the length of your thumb is closest to 5 cm.
Answer: (E)
There are 100 centimetres in 1 metre. Therefore, there are
Answer: (A)
The length of the side not labelled is equal to the sum of the two horizontal lengths that are labelled, or
Answer: (D)
The average (mean) number of points scored per game multiplied by the number of games played is equal to the total number of points scored during the season.
Therefore, the number of games that Hannah played is equal to the total number of points she scored during the season divided by her average (mean) number of points scored per game, or
Answer: (A)
The positive divisors of 20 are:
Answer: (B)
Using the digits
Answer: (A)
Solution 1
At Gaussville School,
Solution 2
At Gaussville School,
Answer: (B)
The first fold creates 2 layers of paper. The second fold places 2 sets of 2 layers together, for a total of 4 layers of paper. Similarly, the third fold places 2 sets of 4 layers of paper together, for a total of 8 layers of paper.
That is, each new fold places 2 sets of the previous number of layers together, thereby doubling the previous number of layers. The results of the first five folds are summarized in the table below.
Number of folds | 0 | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|---|
Number of layers | 1 | 2 | 4 | 8 | 16 | 32 |
After the sheet has been folded in half five times, the number of layers in the folded sheet is 32.
Answer: (B)
Solution 1
The multiples of 5 between 1 and 99 are:
Solution 2
To create an even multiple of 5, we must multiply 5 by an even whole number (since 5 is odd, multiplying 5 by an odd whole number creates an odd result).
The smallest positive even multiple of 5 is
The largest even multiple of 5 less than 99 is
That is, multiplying 5 by each of the even numbers from 2 to 18 results in the only even multiples of 5 between 1 and 99. Since there are 9 even numbers from 2 to 18 (inclusive), then there are 9 even whole numbers between 1 and 99 that are multiples of 5.
Answer: (C)
Consider the value of
Since a 3 already occurs in the second row, then
Since a 1 already occurs in the third column, then
Since
Therefore, a 2 and a 3 already occur in the second row and so
Answer: (E)
The rectangle has area
Each of the two congruent unshaded triangles has area
Answer: (E)
The total value of one quarter, one dime and one nickel is
Since you have equal numbers of quarters, dimes and nickels, you can separate your coins into piles, each containing exactly 1 quarter, 1 dime, and 1 nickel.
Each pile has a value of 40¢, and since
Therefore, you have 11 dimes.
Note: You can check that
Answer: (B)
The original cube (before the corner was cut off) had 12 edges.
Cutting off the corner does not eliminate any of the 12 edges of the original cube.
However, cutting off the corner does add 3 edges that were not present originally, the 3 edges of the new triangular face. Since no edges of the original cube were lost, but 3 new edges were created, then the new solid has
Answer: (D)
To find the image of
Since
That is, point
Similarly, after a reflection across the
That is, point
Answer: (B)
Since the number of digits that repeat is 6, then the digits
Answer: (C)
Since the sum of the measures of the three angles in any triangle is
The measures of the two unknown angles are in the ratio
That is, the larger of the two unknown angles measures
We may check that
Answer: (C)
We begin by choosing the largest number in each row,
However, the numbers in
Thus, the largest of the five answers given, 75, is not possible.
Note: In assuring that we take one number from each row, this choice of numbers,
Of the five answers given, the next largest answer is 73.
Since
For example, the list
That is, to obtain a sum of 73 while choosing exactly one number from each row, we must choose at least three of the numbers from
However, since two numbers in
Any other replacement would give a sum less than 73, which would require the replacement of a number with a larger number in another row to compensate. This is impossible since each row is represented in
Of the five answers given, the next largest answer is 71.
By choosing the numbers,
Thus, 71 is the largest possible sum that satisfies the given conditions. Note: There are other choices of five numbers which also give a sum of 71 and satisfy the given conditions.
Answer: (C)
Since the perimeter of the square is
The width of the rectangle was doubled to produced the side of the square with length
Therefore, the width of the rectangle is half of the side length of the square, or
The length of the rectangle was halved to produce the side of the square with length
Answer: (D)
Solution 1
Every 4-digit palindrome is of the form
Every 5-digit palindrome is of the form
That is, for every 4-digit palindrome
For example if
Solution 2
Every 4-digit palindrome is of the form
There are 9 choices for the first digit
Once the first two digits of the 4-digit palindrome are chosen, then the third and fourth digits are also determined (since the third digit must equal the second and the fourth must equal the first).
That is, there are 90 4-digit palindromes.
Every 5-digit palindrome is of the form
There are 9 choices for the first digit
Once the first three digits of the 5-digit palindrome are chosen, then the fourth and fifth digits are also determined (since the fourth digit must equal the second and the fifth must equal the first).
That is, there are 900 5-digit palindromes.
Thus, the ratio of the number of 4-digit palindromes to the number of 5-digit palindromes is
Answer: (E)
We can determine which triangle has the greatest area by using a fixed side length of 4 for each of the identical squares and using this to calculate the unknown areas.
We begin by constructing
The area of
Since
Since
Since
Since
Next, we construct
The area of
Since
Since
Since
Since
Construct
The area of
As we previously determined, the area of square
Since
Since
Construct
Since
Construct
Since
The areas of the 5 triangles are
Answer: (A)
All 2-digit prime numbers are odd numbers, so to create a reversal pair, both digits of each prime must be odd (so that both the original number and its reversal are odd numbers).
We also note that the digit 5 cannot appear in either prime number of the reversal pair since any 2-digit number ending in 5 is not prime.
Combining these two facts together leaves only the following list of prime numbers from which to search for reversal pairs:
This allows us to determine that the only reversal pairs are: 13 and 31, 17 and 71, 37 and 73, and 79 and 97.
(Note that the reversal of 11 does not produce a different prime number and the reversal of 19 is 91, which is not prime since
Given a reversal pair, we must determine the prime numbers (different than each prime of the reversal pair) whose product with the reversal pair is a positive integer less than 10 000.
The product of the reversal pair 79 and 97 is
We continue in this way, analyzing the other 3 reversal pairs, and summarize our results in the table below.
Prime Number | Product of the Prime Number with the Reversal Pair | |||
---|---|---|---|---|
13 and 31 | 17 and 71 | 37 and 73 | 79 and 97 | |
2 | greater than 10 000 | |||
3 | ||||
5 | greater than 10 000 | |||
7 | ||||
11 | greater than 10 000 | |||
13 | can’t use 13 twice | |||
17 | ||||
19 | ||||
23 | ||||
29 | greater than 10 000 | |||
Total | 8 | 4 | 2 | 0 |
In any column, once we obtain a product that is greater than 10 000, we may stop evaluating subsequent products since they use a larger prime number and thus will exceed the previous product.
In total, there are
Answer: (B)
Evaluating,
Answer: (A)
Since there are 60 minutes in an hour, then 40 minutes after 10:20 it is 11:00. Therefore, 45 minutes after 10:20 it is 11:05.
Answer: (E)
Of the possible answers, the length of your thumb is closest to 5 cm.
Answer: (E)
Reading from the graph, Phil ran 4 km, Tom ran 6 km, Pete ran 2 km, Amal ran 8 km, and Sanjay ran 7 km.
Ordering these distances from least to greatest, we get Pete ran 2 km, Phil ran 4 km, Tom ran 6 km, Sanjay ran 7 km, and Amal ran 8 km.
In this ordered list of 5 distances, the median distance is in the middle, the third greatest. Therefore, Tom ran the median distance.
Answer: (B)
Solution 1
Since
Solution 2
When multiplying
Since
Answer: (E)
The two equal widths, each of length 4, contribute
The two lengths contribute the remaining
Since the two lengths are equal, they each contribute
Answer: (B)
To begin, there are 4 circles and 2 rectangles on the left arm, balanced by 10 circles on the right arm.
If we remove 4 circles from each side of the equal-arm scale, the scale will remain balanced (since we are removing the same mass from each side).
That is, the 2 rectangles that will remain on the left arm are equal in mass to the 6 circles that will remain on the right arm. Since 2 rectangles are equal in mass to 6 circles, then 1 rectangle has the same mass as 3 circles.
Answer: (B)
Solution 1
A 5% increase in 160 is equal to
Therefore, Aidan’s height increased by 8 cm over the summer.
His height at the end of the summer was
Solution 2
Since Aidan’s 160 cm height increased by 5%, then his height at the end of the summer was
Answer: (A)
When
Answer: (E)
Solution 1
Evaluating using a denominator of 12,
Solution 2
Since
Answer: (C)
Consider the following diagram.
Straight angles measure
Therefore,
The three interior angles of any triangle add to
Thus,
Opposite angles have equal measures.
Since the angle measuring
Answer: (C)
Since Zara’s bicycle tire has a circumference of 1.5 m, then each full rotation of the tire moves the bike 1.5 m forward.
If Zara travels 900 m on her bike, then her tire will make
Answer: (C)
To find the image of
Since
That is, point
Similarly, after a reflection across the
That is, point
Answer: (B)
In the table below, we determine the total value of the three bills that remain in Carolyn’s wallet when each of the four bills is removed.
Bill Removed | Sum of the Bills Remaining |
---|---|
$5 | $10+$20+$50=$80 |
$10 | $5+$20+$50=$75 |
$20 | $5+$10+$50=$65 |
$50 | $5+$10+$20=$35 |
It is equally likely that any one of the four bills is removed from the wallet and therefore any of the four sums of the bills remaining in the wallet is equally likely.
Of the four possible sums, $80, $75, $65, and $35, two are greater than $70.
Therefore, the probability that the total value of the three bills left in Carolyn’s wallet is greater than $70, is
Answer: (A)
In the table below, we list the mass of each dog at the end of each month.
Month | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Walter’s mass (in kg) | 12 | 14 | 16 | 18 | 20 | 22 | 24 | 26 | 28 | 30 | 32 | 34 | 36 |
Stanley’s mass (in kg) | 6 | 8.5 | 11 | 13.5 | 16 | 18.5 | 21 | 23.5 | 26 | 28.5 | 31 | 33.5 | 36 |
After 12 months have passed, Stanley’s mass is 36 kg and is equal to Walter’s mass. (Note that since Stanley’s mass is increasing at a greater rate than Walter’s each month, this is the only time that the two dogs will have the same mass.)
Answer: (D)
First, we must determine the perimeter of the given triangle.
Let the unknown side length measure
Since the triangle is a right-angled triangle, then by the Pythagorean Theorem we get
Therefore the perimeter of the triangle is
Since the 4 sides of the square are equal in length, then each measures
Answer: (D)
Since the number of digits that repeat is 6, then the digits
Answer: (C)
Using the definition of
Answer: (C)
Solution 1
Originally there are 3 times as many boys as girls, so then for every 3 boys there is 1 girl and
That is, the number of boys in the room is
Next we consider each of the 5 possible answers, in turn, to determine which represents the total number of children in the room originally.
If the original number of children in the room is 15 (as in answer (A)), the number of boys is
Since it is not possible to have 11.25 boys in the room, then we know that 15 is not the correct answer.
If the original number of children in the room is 20 (as in answer (B)), the number of boys is
If the number of boys in the room was originally 15, then the number of girls was
Next we must check if there will be 5 times as many boys as girls in the room once 4 boys and 4 girls leave the room.
If 4 boys leave the room, there are 11 boys remaining. If 4 girls leave the room, there is 1 girl remaining and since there are not 5 times as many boys as girls, then 20 is not the correct answer.
If the original number of children in the room is 24 (as in answer (C)), the number of boys is
If the number of boys in the room was originally 18, then the number of girls was
If 4 boys leave the room, there are 14 boys left and if 4 girls leave the room, then there are 2 girls left.
Since there are not 5 times as many boys as girls, then 24 is not the correct answer.
If the original number of children in the room is 32 (as in answer (D)), the number of boys is
If the number of boys in the room was originally 24, then the number of girls was
If 4 boys leave the room, there are 20 left and if 4 girls leave the room, then there are 4 left.
Since there are 5 times as many boys as girls, then we know that the original number of children is 32. (Note: We may check that the final answer, 40, gives 30 boys and 10 girls originally and when 4 boys and 4 girls leave the room there are 26 boys and 6 girls which again does not represent 5 times as many boys as girls.)
Solution 2
Originally there are 3 times as many boys as girls, so if there are
If 4 boys leave the room, there are
If 4 girls leave the room, there are
At this point, there are 5 times as many boys as girls in the room.
That is,
Therefore, the original number of girls in the room is 8 and the original number of boys is
Answer: (D)
Solution 1
Call the given vertex of the rectangle
Point
Point
Point
Point
The only point remaining is
Point
How might we show that no two vertices of a 3 by 4 rectangle are 7 units apart? (See Solution 2).
Solution 2
The distance between any two adjacent vertices of a 3 by 4 rectangle
The distance between any two opposite vertices of a rectangle (such as
In the right-angled triangle
That is, the greatest distance between any two vertices of a 3 by 4 rectangle is 5 units.
As shown and explained in Solution 1, the distance between
Answer: (B)
In square
The area of
Since this area equals 18, then
The side of the square,
The area of
Square
Answer: (E)
Let the number of adult tickets sold be
Since the price for each adult ticket is $12, then the revenue from all adult tickets sold (in dollars) is
Since the number of child tickets sold is equal to the number of adult tickets sold, we can let the number of child tickets sold be
In dollars, the combined revenue of all adult tickets and all child tickets is
Since the total number of tickets sold is 120, and
Since the price for each senior ticket is $10, then the revenue from all senior tickets sold (in dollars) is
Thus the combined revenue from all ticket sales is
The total revenue from the ticket sales was $1100 and so
Solving this equation, we get
Therefore, the number of senior tickets sold for the concert was
We may check that the number of tickets sold to each of the three groups gives the correct total revenue.
Since the number of adult tickets sold was equal to the number of child tickets sold which was equal to
The revenue from 50 adult tickets is
The revenue from 50 child tickets is
The revenue from 20 senior tickets is
Answer: (B)
The list of integers
The sum of the 5 integers is
Since this average is a whole number, then
How small and how large can the sum
We know that
Since
Using the fact that
Since
The multiples of 5 between 29 and 47 are
When
The only ordered pair
Value of |
Value of |
Ordered Pairs |
---|---|---|
30 | ||
35 | ||
40 | ||
45 |
The number of ordered pairs
Answer: (E)
The two joggers meet every 36 seconds.
Therefore, the combined distance that the two joggers run every 36 seconds is equal to the total distance around one lap of the oval track, which is constant.
Thus the greater the first jogger’s constant speed, the greater the distance that they run every 36 seconds, meaning the second jogger runs less distance in the same time (their combined distance is constant) and hence the smaller the second jogger’s constant speed. Conversely, the slower the first jogger’s constant speed, the less distance that they run every 36 seconds, meaning the second jogger must run a greater distance in this same time and hence the greater the second jogger’s constant speed.
This tells us that if the first jogger completes one lap of the track as fast as possible, which is in 80 seconds, then the second jogger’s time to complete one lap of the track is as slow as possible.
We will call this time
Similarly, if the first jogger completes one lap of the track as slowly as possible, which is in 100 seconds, then the second jogger’s time to complete one lap of the track is as fast as possible.
We will call this time
Finding the value of
Recall that
If the first jogger can complete one lap of the track in 80 seconds, then in 36 seconds of running, the first jogger will complete
In this same 36 seconds, the two joggers combined distance running is 1 lap, and so the second jogger runs
Finding the value of
Recall that
If the first jogger can complete one lap of the track in 100 seconds, then in 36 seconds of running, the first jogger will complete
In this same 36 seconds, the two joggers combined distance running is 1 lap, and so the second jogger runs
Determining the product of the smallest and largest integer values of
Since the second jogger completes 1 lap of the track in at most
The second jogger completes 1 lap of the track in at least
Answer: (A)
Let the alternating sum of the digits be
If the 7-digit integer is
This sum can be grouped into the digits which contribute positively to the sum, and those which contribute negatively to the sum.
Rewriting the sum in this way, we get
Taking the 4 digits which contribute positively to
Similarly, taking the 3 digits which contribute negatively to
We determine the largest possible value of
That is, the largest possible alternating sum is
We determine the smallest possible value of
That is, the smallest possible alternating sum is
Since
Case 1: The alternating sum of the digits is 11, or
If
That is, the difference between two integers is odd only if one of the integers is even and the other is odd (we say that
However, if one of
But we know that
Therefore, it is not possible that
Case 2: The alternating sum of the digits is 0, or
If
Since
We find all groups of 3 digits, chosen from the digits 1 to 7, such that their sum
There are exactly 4 possibilities:
In each of these 4 cases, the digits from 1 to 7 that were not chosen,
4 digits whose sum is |
3 digits whose sum is |
2 examples of 7-digit integers created from these |
---|---|---|
Consider the first row of numbers in this table above.
Each arrangement of the 4 digits
Two such arrangements are shown (you may check that
Since there are
Each of these 144 arrangements is different from the others, and since
Similarly, there are also 144 arrangements that can be formed with each of the other 3 groups of integers that are shown in the final 3 rows of the table.
That is, there are a total of
The total number of 7-digit integers that can be formed from the integers 1 through 7 is equal to the total number of arrangements of the integers 1 through 7, or
Answer: (E)