2016 gauss contests - gauss contests (grades 7 and 8) wednesday, may 11, ... fiona dunbar mike eden...
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The CENTRE for EDUCATION
in MATHEMATICS and COMPUTINGcemc.uwaterloo.ca
2016 Gauss Contests(Grades 7 and 8)
Wednesday, May 11, 2016(in North America and South America)
Thursday, May 12, 2016(outside of North America and South America)
2015 University of Waterloo
2016 Gauss Contest Solutions Page 2
Centre for Education in Mathematics and Computing Faculty and Staff
Ed AndersonJeff AndersonTerry BaeShane BaumanSteve BrownCarmen BruniErsal CahitHeather CulhamSerge DAlessioJanine DietrichJennifer DoucetFiona DunbarMike EdenBarry FergusonJudy FoxSteve FurinoJohn GalbraithAlain Gamache
Robert GarbarySandy GrahamConrad HewittAngie HildebrandCarrie KnollJudith KoellerBev MarshmanMike MiniouBrian MoffatDean MurrayJen NelsonJ.P. PrettiKim SchnarrCarolyn SedoreIan VanderBurghTroy VasigaAshley WebsterTim Zhou
Gauss Contest Committee
Mark Bredin (Chair), St. Johns Ravenscourt School, Winnipeg, MBKevin Grady (Assoc. Chair), Cobden District P.S., Cobden, ONSarah Garrett, King George P.S., Guelph, ONJohn Grant McLoughlin, University of New Brunswick, Fredericton, NBJoAnne Halpern, Toronto, ONDavid Matthews, University of Waterloo, Waterloo, ONDavid Switzer, Sixteenth Ave. P.S., Richmond Hill, ONRachael Verbruggen, University of Waterloo, Waterloo, ONLaurissa Werhun, Parkdale C.I., Toronto, ONChris Wu, Zion Heights J.H.S., Toronto, ONLori Yee, William Dunbar P.S., Pickering, ON
2016 Gauss Contest Solutions Page 3
1. Evaluating, 333 + 33 + 3 = 366 + 3 = 369.
2. The day on which Tanner received the most text messages will be the day with the tallestcorresponding bar.Thus, Tanner received the most text messages on Friday.
3. Solution 1A number is a multiple of 7 if it is the result of multiplying 7 by an integer.Of the answers given, only 77 results from multiplying 7 by an integer, since 77 = 7 11.Solution 2A number is a multiple of 7 if the result after dividing it by 7 is an integer.Of the answers given, only 77 results in an integer after dividing by 7, since 77 7 = 11.
4. A positive fraction is larger than 12
if its denominator is less than two times its numerator.Of the answers given, 4
7is the only fraction in which the denominator, 7, is less than 2 times
its numerator, 4 (since 2 4 = 8).Therefore, 4
7is larger than 1
5. Rolling the cube does not change the size of the painted triangle.For this reason, we can eliminate answer (A).Rolling the cube does not change the number of painted triangles.For this reason, we can eliminate answers (D) and (E).Rolling the cube does not change the orientation of the painted triangle with respect to theface of the cube that it is painted on.For this reason, we can eliminate answer (C).Of the given answers, the cube shown in (B) is the only cube which could be the same as thecube that was rolled.
6. The measure of the three angles in any triangle add to 180.Since two of the angles measure 25 and 70, then the third angle in the triangle measures180 25 70 = 85.The measure of the third angle in the triangle is 85.
7. Each of the 30 pieces of fruit in the box is equally likely to be chosen. Since there are 10 orangesin the box, then the probability that the chosen fruit is an orange is 10
2016 Gauss Contest Solutions Page 4
8. Solution 1Since Alex pays $2.25 to take the bus, then 20 trips on the bus would cost Alex 20$2.25 = $45.Since Sam pays $3.00 to take the bus, then 20 trips on the bus would cost Sam 20$3.00 = $60.If they each take the bus 20 times, then in total Alex would pay $60 $45 = $15 less thanSam.
Solution 2Since Alex pays $2.25 to take the bus, and Sam pays $3.00 to take the bus, then Alex pays$3.00 $2.25 = $0.75 less than Sam each time they take the bus.If they each take the bus 20 times, then in total Alex would pay 20 $0.75 = $15 less thanSam.
9. Solution 1Travelling at a constant speed of 85 km/h, the entire 510 km trip would take Carrie510 85 = 6 hours.Since Carrie is halfway through the 510 km trip, then the remainder of the trip will take herhalf of the total trip time or 6 2 = 3 hours.Solution 2Carrie is halfway through a 510 km trip, and so she has half of the distance or 5102 = 255 kmleft to travel.Since Carrie travels at a constant speed of 85 km/h, then it will take her 255 85 = 3 hourslonger to complete the trip.
10. Since Q is halfway between P and R, then the distance between Pand Q is equal to the distance between Q and R.The distance between P and Q is 1 (6) = 1 + 6 = 5.Since P is 5 units to the left of Q, then R is 5 units to the right of Q.That is, R is located at 1 + 5 = 4 on the number line.
11. In the diagram, there are 4 rows of octagons and each row contains 5 octagons.Therefore, the total number of octagons in the diagram is 4 5 = 20.In the diagram, there are 3 rows of squares and each row contains 4 squares.Therefore, the total number of squares in the diagram is 3 4 = 12.The ratio of the number of octagons to the number of squares is 20 : 12 or 5 : 3.
12. The sum of the units column is Q+Q+Q = 3Q.Since Q is a single digit, and 3Q ends in a 6, then the only possibility is Q = 2.Then 3Q = 3 2 = 6, and thus there is no carry over to the tens column.The sum of the tens column becomes 2 + P + 2 = P + 4, since Q = 2.Since P is a single digit, and P + 4 ends in a 7, then the only possibility is P = 3.Then P + 4 = 3 + 4 = 7, and thus there is no carry over to thehundreds column.We may verify that the sum of the hundreds column is 3+3+2 = 8,since P = 3 and Q = 2.The value of P +Q is 3 + 2 = 5, and the final sum is shown.
2016 Gauss Contest Solutions Page 5
13. Since a cube is a rectangular prism, its volume is equal to the area of its base, lw, multipliedby its height, h.A cube has edges of equal length and so l = w = h.Thus, the volume of a cube is the product of three equal numbers.The volume of the larger cube is 64 cm3 and 64 = 4 4 4, so the length of each edge of thelarger cube is 4 cm.The smaller cube has edges that are half the length of the edges of the larger cube, or 2 cm.The volume of the smaller cube is 2 2 2 = 8 cm3.
14. Ahmed could choose from the following pairs of snacks: apple and orange, apple and banana,apple and granola bar, orange and banana, orange and granola bar, or banana and granola bar.Therefore, there are 6 different pairs of snacks that Ahmed may choose.
15. Sophia did push-ups for 7 days (an odd number of days), and on each day she did an equalnumber of push-ups more than the day before (5 more).Therefore, the number of push-ups that Sophia did on the middle day (day 4) is equal to theaverage number of push-ups that she completed each day.Sophia did 175 push-ups in total over the 7 days, and thus on average she did 175 7 = 25push-ups each day.Therefore, on day 4 Sophia did 25 push-ups, and so on day 5 she did 25 + 5 = 30 push-ups, onday 6 she did 30 + 5 = 35 push-ups and on the last day she did 35 + 5 = 40 push-ups.(Note: We can check that 10 + 15 + 20 + 25 + 30 + 35 + 40 = 175, as required.)
16. Since = 4+4+4, then by adding a to each side we get + = +4+4+4.Since + = + 4 + 4 + 4, then by adding a 4 to each side we get that+ +4 = +4+4+4+4.(Can you explain why each of the other answers is not equal to + +4?)
17. Each of the following four diagrams shows the image of triangle T after its reflection in thedotted line.
Thus, each of the triangles labelled A,B,D, and E is a single reflection of triangle T in someline.The triangle labelled C is the only triangle that cannot be a reflection of triangle T .
18. The mean (average) of the set of six numbers is 10, and so the sum of the set of six numbersis 6 10 = 60.If the number 25 is removed from the set, the sum of the set of the remaining five numbers is60 25 = 35.The mean (average) of the remaining set of five numbers is 35 5 = 7.
2016 Gauss Contest Solutions Page 6
19. The shaded and unshaded sections of the ribbon have equal length.Since there are 5 such sections, then each shaded and unshaded section has length equal to 1
of the length of the ribbon.All measurements which follow are made beginning from the left end of the ribbon.Point A is located 3 sections from the left end of the ribbon, or at a point 3 3
the length of the ribbon.Point D is located 4 sections from the left end of the ribbon, or at a point 4 3
the length of the ribbon.All points are equally spaced, and so points B and C divide the unshaded section betweenpoints A and D into 3 equal lengths.Since A is located 9
15of the ribbon length from the left end, and D is located 12
15of the ribbon
length from the left end, then B is located at 1015
of the ribbon length, and C is located at 1115
ofthe ribbon length.Thus, if Suzy makes a vertical cut at point C, the portion of the ribbon to the left of C will be1115
of the size of the original ribbon.We note that no point is located more than 2 sections from the right end of the ribbon.That is, no point is located more than 2 3
15along the length of the ribbon when measured
from the right end, and so measurements are taken from the left end of the ribbon.