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2021 Team Up Challenge
Solutions

June 2021

©2021 University of Waterloo


Team Paper Answer Key

Question Answer
1 7.5
2 3 cm
3 30
4 29
5 125
6 38 or 37.5%
7 7
8 6
9 circle
10 (17,4)
11 B, C, E
12 5
13 172
14 4
15 60

Crossnumber Puzzle Answer Key

5 7 B 3 1 B 6 2 1
1 B 1 B 6 5 4 B 5
7 3 1 B 4 B 6 2 B
B 6 B B 4 B B B 6
2 0 5 5 B 1 1 2 8
4 B B B 1 B B 5 B
B 4 3 B 0 B 1 8 1
3 B 8 5 2 B 9 B 4
2 0 4 B 9 3 B 4 7

Logic Puzzle Answer Key

Position in Photo
1 2 3 4
Location Park Blanche Dorothy Rose Sophia
Kitchen Blanche Dorothy Rose Sophia
Garden Rose Sophia Dorothy Blanche
Driveway Blanche Dorothy Sophia Rose
Basement Blanche Sophia Dorothy Rose

Relay Answer Key

Practice Relay
Player Player 1 Player 2 Player 3 Player 4
Correct Answer 5 3 17 24
Relay A
Player Player 1 Player 2 Player 3 Player 4
Correct Answer 110 165 101 808
Relay B
Player Player 1 Player 2 Player 3 Player 4
Correct Answer 9 6 50 138
Relay C
Player Player 1 Player 2 Player 3 Player 4
Correct Answer 5 13 57 228

Team Paper Solution

  1. From the graph, we can see that Ana spent 2+1+0.5+1+0+0.5+2.5=7.5 hours playing soccer.

    Answer: 7.5

  2. Since the four sides of a square are equal in length and the perimeter is 12 cm, then each side has length 124=3 cm.

    Answer: 3 cm

  3. If the symbol represents the number 5, then the expression becomes 5×5+5 which is equal to 30.

    Answer: 30

  4. The character starts by moving 5 steps.
    Inside the loop, the character moves 3 steps. Since the character repeats the loop 8 times, this is a total of 8×3=24 steps.
    In total, the character has moved 5+24=29 steps.

    Answer: 29

  5. Since 125×40=5000 (or 5000÷40=125), then Jeff would need to type for 125 minutes.

    Answer: 125

  6. Exactly 3 of the 8 equal sections are labelled with the letter B. Therefore, the probability that the spinner lands on B is 38 or 37.5%.

    Answer: 38 or 37.5%

  7. Recall that a prime number is an integer greater than 1 whose only divisors are 1 and itself.
    The prime numbers between 6 and 30 are: 7, 11, 13, 17, 19, 23, and 29.
    Thus, there are 7 integers between 6 and 30 which are prime numbers.

    Answer: 7

  8. Every triangular prism has exactly 6 vertices, as shown in the diagram.

    A prism with two identical triangles as its parallel end faces. Each of the three triangles has three vertices and these are joined in corresponding pairs to form the rectangular side faces of the prism.

    Answer: 6

  9. The pattern repeats after every 6 shapes, so we want to find the largest multiple of 6 that is less than 55. This is 54. The pattern will start again after the 54th shape. The 55th shape will then be the first shape in the pattern, which is .

    Answer:

  10. We start by reflecting the lizard over the dotted line as shown.

    The tail of the lizard is now at (10,4).

    We then translate the lizard 7 units to the right, as shown.

    The tail now has coordinates (17,4).

    Answer: (17,4)

  11. It is possible to trace over all the lines exactly once in Picture B, Picture C and Picture E.

    First we will look at Picture B.

    The middle dot in the leftmost column is highlighted and labelled Dot 1. The middle dot in the rightmost column is highlighted and labelled Dot 2.

    One way to do trace all the lines as described is to start from Dot 1 and trace over the line connecting Dot 1 to Dot 2 and then trace over all the remaining lines, finishing at Dot 2.

    There are other ways to do this, but the important thing is that we have to start at either Dot 1 or Dot 2 since these are the only dots with an odd number of lines coming out of them. Every time we pass through a dot (other than at the beginning or end), we use up an even number of lines. So, the only way to use up all the lines for a dot with an odd number of lines coming out of it is to make sure that dot is the first or last dot.

    Notice that Picture C and Picture E each have exactly two dots with an odd number of edges coming out of them. We can trace over all the lines exactly once in each of these pictures by starting at one of these dots and finishing at the other.

    In C, the following two dots are highlighted and labelled as odd: the dot at the midpoint of the horizontal base of the equilateral triangle, and the dot forming the top vertex of the triangle. In E, the following two dots are highlighted and labelled as odd: the extra dot on the top horizontal side of the base of the boat, and the dot joined to it on the triangle sail.

    Picture A and Picture D each have more than two dots with an odd number of lines coming out them, so there’s no way to trace over all the lines in these pictures exactly once.

    In A, the four dots on the circle are highlighted and labelled as odd. In D, the three dots forming the vertices of the triangle and the three dots at the midpoints of sides of the triangle are highlighted and marked as odd.

    Therefore, it is possible to trace over every line, as described, in Pictures B, C, and E.

    Answer: B, C, E

  12. Since the known numbers in the list are all different, then if the unknown number is different than each of these, there is no single mode. Therefore, the unknown number must be equal to one of the known numbers. It follows that the unknown number will also be the mode.
    If we arrange the known numbers in increasing order, we have: 2, 3, 5, 6, 9.
    If the unknown number is 2 or 3, then the median of the six numbers would be 3+52=4 which is not possible, since the median does not equal the mode.
    If the unknown number is 6 or 9, then the median of the six numbers would be 5+62=5.5 which is not possible, since the median does not equal the mode.
    If the unknown number is 5, then the median of the six numbers is 5 and the mode of the six numbers is also 5.
    Therefore, the unknown number is 5.

    Answer: 5

  13. We begin by considering the positive integers between 1 and 100. From 1 to 69, the digit 7 is written 7 times. From 70 to 79, the digit 7 is written 11 times. From 80 to 100, the digit 7 is written 2 times. Thus, from 1 to 100, the digit 7 is written 20 times.

    The next 10 times Jonah writes the digit 7 are as follows: 107,117,127,137,147,157,167,170,171,172 Therefore, Jonah’s favourite number is 172.

    Answer: 172

  14. Bea, Diane, and Gene have each met 7 people, therefore Hans has met Bea, Diane and Gene.
    Edgar and Foster have both met 3 people. Since they have met Bea, Diane, and Gene, then they have not met anyone else. So, Hans has not met Edgar or Foster.
    Amad has met 5 people. Since Amad cannot have met Edgar or Foster, then Amad has met Cho and Hans.
    Cho has met 4 people: Amad, Bea, Diane, and Gene.
    Therefore, Hans has met 4 people: Amad, Bea, Diane and Gene.

    Answer: 4

  15. Since the robot must travel though M we can count the number of paths from A to M and then count the number of paths from M to B.

    We begin by considering the following three paths from A to M.

    2 right, then 3 down. 1 right, then 3 down, then 1 right. 1 down, then 1 right, then 1 down, then 1 right, then 1 down.

    Notice that these paths involve two moves right and three moves down. In fact, every path from A to M will involve exactly two moves right and three moves down. What makes each path different is the order in which the robot makes these moves. For example, if R means move right and D means move down, then the sequence of moves R,R,D,D,D is different from the sequence of moves R,D,R,D,D.
    Counting the number of different possible sequences of moves, we get:

    There are a total of 10 ways the robot can get from A to M.

    Similarly, we can consider the following two paths from M to B.

    1 down, then 5 right. 2 right, then 1 down, then 3 right.

    Notice that these paths involve five moves right and one move down. In fact, every path from M to B will involve exactly five moves right and one move down. What makes each path different is the order in which the robot makes these moves. Counting the number of different orders that are possible, we get:

    There are a total of 6 ways the robot can get from M to B.

    We now consider three of the possible paths from A to B through M.

    Starting at A, move 2 right, then 3 down to M, then 2 right, then 1 down, then 3 right ending at B. Starting at A, move 2 right, then 4 down passing through M, then 5 right ending at B. Starting at A, move 2 right, then 3 down to M, then 4 right, then 1 down, then 1 right ending at B.

    Notice that, in each of these three paths, the robot takes the same path from A to M, but then takes a different path from M to B. What makes each path different is how the robot moves after he reaches M. Since each of the 10 paths from A to M can be paired with each of the six paths from M to B, there are a total of 10×6=60 paths the robot can take from A to B.

    Answer: 60

Crossnumber Puzzle Solution

Grid

5 7 B 3 1 B 6 2 1
1 B 1 B 6 5 4 B 5
7 3 1 B 4 B 6 2 B
B 6 B B 4 B B B 6
2 0 5 5 B 1 1 2 8
4 B B B 1 B B 5 B
B 4 3 B 0 B 1 8 1
3 B 8 5 2 B 9 B 4
2 0 4 B 9 3 B 4 7

Across Answers

  1. The range is 204147=57.

  2. 50% of 62=0.5×62=31.

  3. Since 2324=621648, the number is 621.

  4. From the grid, the hundreds digit of this number is 6. The only multiple of 109 with a hundreds digit of 6 is 654.

  5. The sum is 110+621=731.

  6. July and August each have 31 days. Thus the total number of days is 62.

  7. The result is 45×45+30=2055.

  8. The product is 12×94=1128.

  9. From the grid, the ones digit of this number is 3. Since the only factors of 731 are 1, 17, 43, and 731, the only two-digit factor with a ones digit of 3 is 43.

  10. From the grid, the ones digit of this number is 1. It follows that the hundreds digit is also 1. The tens digit will be determined from the clue to 14 DOWN.

  11. From the grid, the hundreds digit is 8 and the ones digit is 2. The median of these two numbers is 5. Thus, the number is 852.

  12. A dozen is a set of 12. So seventeen dozen is 17×12=204.

  13. The sum of the digits in this number must be 12, since the digit sum of 1128 is 12.
    From the grid, the tens digit of this number is 9. Thus, the number is 93.

  14. The smallest prime number greater than 43 is 47.

Down Answers

  1. The number is 16×32+5=517.

  2. The number is 45 of 2055=0.8×2055=1644.

  3. The digits are determined from the clues to 4 ACROSS, 7 ACROSS, and 10 ACROSS.

  4. An odd multiple of 5 has a ones digit of 5. The tens digit will be determined from the clue to 4 ACROSS.

  5. A rectangle with area 517 and width 47 has length 517÷47=11.

  6. In a square, each angle is equal to 90°. The sum of four angles, in degrees, is 90×4=360.

  7. The number is 62+6=68.

  8. From the grid, the tens digit of this number is 2. The only multiple of 8 with a tens digit of 2 is 24.

  9. The digits of 852 in reverse order are 258.

  10. Since 1000÷7=142.8, the largest 3-digit multiple of 7 is 142×7=994.
    Then 994+35=1029.

  11. The mean is 621+1472=384.

  12. The number is 4324=19.

  13. There are 7 days in a week, so 1029 days is 1029÷7=147 weeks.

  14. The value is 14+0.5×36=14+18=32.

Logic Puzzle Solution

Start by considering the second, eighth and first clues (in that order).

(2) Dorothy and Sophia are the only people to ever stand in Position 2.

Although we do not yet know which person occupies which Position 2 in each photo, we can make note that one of them does.

(8) The order for the garden photo is the reverse of the order for the driveway photo.

Since either Dorothy or Sophia occupy the second position in these photos, it follows that they must occupy the third position as well. As a result, Blanche and Rose must somehow occupy Positions 1 and 4 in these photos.

(1) Blanche is standing in Position 1 in four of the five photos.

Since the garden photo is the reverse of the driveway photo it is not possible for Blanche to be in Position 1 in both of these photos. Thus, in order to be standing in Position 1 in four photos, she must be standing in Position 1 in the park, kitchen and basement photos. Then, the fourth photo with Blanche in Position 1 will be either the garden or driveway photo.

The following partially-completed table contains what we have determined from these three clues.

Position in Photo
1 2 3 4
Location Park Blanche Dorothy/
Sophia
Kitchen Blanche Dorothy/
Sophia
Garden Blanche/
Rose
Dorothy/
Sophia
Dorothy/
Sophia
Blanche/
Rose
Driveway Blanche/
Rose
Dorothy/
Sophia
Dorothy/
Sophia
Blanche/
Rose
Basement Blanche Dorothy/
Sophia

Next, we consider the third clue.

(3) Rose is standing in the same position for the photos in the basement and the driveway, but she isn’t standing in that position in any other photos.

We previously determined that Rose is standing in either Position 1 or Position 4 for the driveway photo. Since she must be standing in the same position for the basement photo it follows that she must be standing in Position 1 or Position 4 in the basement photo. But, Blanche is standing in Position 1 in the basement photo, and so Rose must be standing in Position 4 for both the basement and the driveway photos.

Now that we know Rose is standing in Position 4 in the driveway photo, we also know that Blanche is in Position 1. Furthermore, since the garden photo is the reverse of the driveway photo, we know that in the garden photo Rose is in Position 1 and Blanche is in Position 4.

Furthermore, since Rose is not standing in the same position in the park and kitchen photos as she is in the driveway/basement photos, it follows that Rose is not standing in Position 4 in either the park or the kitchen photos. We already know that Blanche is in Position 1 in both photos, and that either Dorothy or Sophia are in Position 2 in both photos. Therefore, Rose must be standing in Position 3 for both the park and the kitchen photos.

Position in Photo
1 2 3 4
Location Park Blanche Dorothy/
Sophia
Rose
Kitchen Blanche Dorothy/
Sophia
Rose
Garden Rose Dorothy/
Sophia
Dorothy/
Sophia
Blanche
Driveway Blanche Dorothy/
Sophia
Dorothy/
Sophia
Rose
Basement Blanche Dorothy/
Sophia
Rose

Now, we consider the fourth clue.

(4) Dorothy is never standing in Position 1 or Position 4.

In both the park and the kitchen photo, the only possible positions left for Dorothy to stand in are Positions 2 and 4. Since Dorothy can never stand in Position 4, she must be standing in Position 2 for both of these photos. This leaves Sophia standing in Position 4 for both the park and the kitchen photos.

Position in Photo
1 2 3 4
Location Park Blanche Dorothy Rose Sophia
Kitchen Blanche Dorothy Rose Sophia
Garden Rose Dorothy/
Sophia
Dorothy/
Sophia
Blanche
Driveway Blanche Dorothy/
Sophia
Dorothy/
Sophia
Rose
Basement Blanche Dorothy/
Sophia
Rose

Notice that we have now determined the order in which the siblings are standing for the park and kitchen photos. That means, that the sixth clue is not actually needed. But, we can confirm that it is in fact satisfied, since the siblings are standing in the same order for the park and kitchen photos.

Now, we can use the fifth clue.

(5) Sophia is standing in Position 2 exactly twice.

Since the garden photo is the reverse of the driveway photo, Sophia is in Position 2 exactly once in these two photos. As a result, to be in Position 2 exactly twice, she must be in Position 2 in the basement photo. As a result, Dorothy must be standing in Position 3 in the basement photo.

Position in Photo
1 2 3 4
Location Park Blanche Dorothy Rose Sophia
Kitchen Blanche Dorothy Rose Sophia
Garden Rose Dorothy/
Sophia
Dorothy/
Sophia
Blanche
Driveway Blanche Dorothy/
Sophia
Dorothy/
Sophia
Rose
Basement Blanche Sophia Dorothy Rose

Finally, the seventh clue tells us where Dorothy and Sophia are standing in the garden and driveway photos.

(7) In three of the four photos where Blanche is standing in Position 1, Dorothy is standing next to her.

This clue is equivalent to saying that in three of the four photos where Blanche is standing in Position 1, Dorothy is standing in Position 2. Currently, we have determined that Dorothy is standing in Position 2 in exactly two photos where Blanche is standing in Position 1. Therefore, Dorothy must be standing in Position 2 in the driveway photo (where Blanche is standing in Position 1). As a result, Sophia will be standing in Position 3.

Since the garden photo is the reverse of the driveway photo, we know that in the garden photo Sophia is in Position 2 and Dorothy is in Position 3. This completes the logic puzzle.

Position in Photo
1 2 3 4
Location Park Blanche Dorothy Rose Sophia
Kitchen Blanche Dorothy Rose Sophia
Garden Rose Sophia Dorothy Blanche
Driveway Blanche Dorothy Sophia Rose
Basement Blanche Sophia Dorothy Rose

Relay Solution

(Note: Where possible, the solutions are written as if the value of N is not initially known, and then N is substituted at the end.)

Practice Relay

Answer: 5, 3, 17, 24

Relay A

Answer: 110, 165, 101, 808

Relay B

Answer: 9, 6, 50, 138

Relay C

Answer: 5, 13, 57, 228