The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in

The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in Grade 7 or higher.

Problem C Retiring and Hiring A small technology company is beginning to expand. The company currently has 100 employees. At the end of each year for the next 4 years, 25 employees will retire and a new employee will be hired for each of the remaining employees. After this four year cycle of retiring and hiring is complete, determine the average number of employees that the company grows by each year. Extension: Using your results and assuming the pattern continues, predict the number of employees if this cycle of retiring and hiring were to continue for ten years.

Problem Problem C and Solution Retiring and Hiring A small technology company is beginning to expand. The company currently has 100 employees. At the end of each year for the next 4 years, 25 employees will retire and a new employee will be hired for each of the remaining employees. After this four year cycle of retiring and hiring is complete, determine the average number of employees that the company grows by each year. Solution The following chart will look at the process of retiring and hiring over the four year period. The number remaining will be 25 less than the number at the start of the year. The number hired in a year will be the same as the number remaining. The number of employees at the end of the year will be twice as many as the number of employees remaining. Year # of Employees # Retiring # Remaining # Hired # of Employees at Start of Year at End of Year 1 100 25 75 75 150 2 150 25 125 125 250 3 250 25 225 225 450 4 450 25 425 425 850 After four years of retiring and hiring the company employs 850 people. The number of employees increased by 850 100 = 750 people in four years. The average increase per year was 750 4 = 187.5 employees. Another way to look at this problem would be to take the average number of people hired per year and subtract the average number of people retiring in a year. The average number hired per year is (75 + 125 + 225 + 425) 4 = 850 4 = 212.5. The average retiring each year was 25. Therefore the average increase per year was 212.5 25 = 187.5, as above. Therefore the number of employees increased by approximately 188 per year. Solution to Extension Observe a couple of patterns in the column labelled # of Employees at End of Year. First, each number ends in 50. The leading digits are 1, 2, 4, and 8. These digits are powers of 2. In year 2, the leading digit is 2 1. In year 3, the leading digit is 2 2. In year 4, the leading digit is 2 3. The exponent appears to be one less than the year number. So in year 10, a good prediction for the leading digits would be 2 9 = 512. After 10 years of retiring and hiring, there would be 51 250 employees. (We can verify this prediction by continuing the table.)

Problem C Who Brought What When? Xavier, Yvonne and Zak arrived at school at three different times. They each brought one of their favourite snacks to share with the other two (one brought pretzels; one brought cookies; one brought licorice), and their favourite sports apparatus (one brought a baseball; one brought a soccer ball; one brought a football). We know a few other facts: 1. The first to arrive did not bring cookies. 2. Xavier arrived second and brought a football. 3. Yvonne arrived before Zak. 4. The person who brought cookies also brought a baseball. 5. The person who brought pretzels did not bring a soccer ball. Determine the order they arrived in, what they each brought for a snack, and which sports apparatus they each brought.

Problem Problem C and Solution Who Brought What When? Xavier, Yvonne and Zak arrived at school at three different times. They each brought one of their favourite snacks to share with the other two (one brought pretzels; one brought cookies; one brought licorice), and their favourite sports apparatus (one brought a baseball; one brought a soccer ball; one brought a football). We know a few other facts: 1. The first to arrive did not bring cookies. 2. Xavier arrived second and brought a football. 3. Yvonne arrived before Zak. 4. The person who brought cookies also brought a baseball. 5. The person who brought pretzels did not bring a soccer ball. Determine the order they arrived in, what they each brought for a snack, and which sports apparatus they each brought. Solution When solving logic problems, setting up a chart to fill in is generally a good way to start. Xavier Yvonne Zak Order of Arrival Snack Sports Apparatus Some of the given information is often more helpful than other information. For example, in the second statement we learn that Xavier arrived second and brought a football. Now the third statement gets us the fact that Yvonne arrived first and Zak arrived third. (This is true since Yvonne arrived before Zak and she could not arrive second leaving only the first and third spots left.) We can add this information to the chart. Order of Arrival Snack Sports Apparatus Xavier 2 nd football Yvonne 1 st Zak 3 rd We can combine the first statement and the fourth statement. Yvonne did not bring cookies. The person who brought cookies also brought a baseball. This cannot be Xavier since he brought a football. Therefore, Zak brought cookies and a baseball. Since there is only one piece of sports apparatus unaccounted for, Yvonne must have brought the soccer ball. We will add this new information to our chart.

Order of Arrival Snack Sports Apparatus Xavier 2 nd football Yvonne 1 st soccer ball Zak 3 rd cookies baseball We can now use the fifth statement to conclude that Xavier brought pretzels since the person bringing pretzels did not bring a soccer ball and Xavier is the only one without a snack accounted for other than Yvonne (who brought the soccer ball). Order of Arrival Snack Sports Apparatus Xavier 2 nd pretzels football Yvonne 1 st soccer ball Zak 3 rd cookies baseball The only snack unaccounted for is the licorice and Yvonne is the only one whose snack is unknown. Therefore, Yvonne brought licorice and our chart can be completed. Order of Arrival Snack Sports Apparatus Xavier 2 nd pretzels football Yvonne 1 st licorice soccer ball Zak 3 rd cookies baseball The information is summarized in the chart but will be stated below for completeness. Yvonne arrived first bringing licorice and a soccer ball. Xavier arrived second bringing pretzels and a football. Zak arrived third bringing cookies and a baseball.

Problem C Quick Change A die has the numbers 1, 2, 3, 4, 6, and 8 on its six faces. When this die is rolled, if an odd number appears on the top face, all of the odd numbers on the die magically double. For example, if the number appearing on the top face was a 1, the 1 and 3 on the original die would double and the other four numbers would remain the same. (This is illustrated below.) = However, if an even number appears on the top face as a result of the first roll, all of the even numbers on the die are halved. For example, if the number appearing on the top face was an 8, the 2, 4, 6, and 8 on the original die would change to half of their initial value and the other two numbers would remain the same. Suppose the die with 1, 2, 3, 4, 6, and 8 on its six faces is rolled once and changes as described above. The changed die is rolled again. No change occurs when the die is rolled this time. What is the probability that the number appearing on the top face after this roll is a 2?

Problem C and Solutions Quick Change Problem A die has the numbers 1, 2, 3, 4, 6, and 8 on its six faces. When this die is rolled, if an odd number appears on the top face, all of the odd numbers on the die magically double. For example, if the number appearing on the top face was a 1, the 1 and 3 on the original die would double and the other four numbers would remain the same. However, if an even number appears on the top face as a result of the first roll, all of the even numbers on the die are halved. For example, if the number appearing on the top face was an 8, the 2, 4, 6, and 8 on the original die would change to half of their initial value and the other two numbers would remain the same. Suppose the die with 1, 2, 3, 4, 6, and 8 on its six faces is rolled once and changes as described above. The changed die is rolled again. No change occurs when the die is rolled this time. What is the probability that the number appearing on the top face after this roll is a 2? Solution 1 In this solution, we will determine the possibilities for the first and second roll to count the total number of possible outcomes. We will then count the number of outcomes in which the second roll is a 2 and determine the probability. If the first roll is odd, the numbers on the die change from {1, 2, 3, 4, 6, 8} to {2, 2, 6, 4, 6, 8} as a result of doubling the odd numbers. If we write the possible first and second rolls as an ordered pair, then the following 12 combinations are possible: First roll 1: (1, 2), (1, 2), (1, 6), (1, 4), (1, 6), (1, 8) First roll 3: (3, 2), (3, 2), (3, 6), (3, 4), (3, 6), (3, 8) If the first roll is even, the numbers on the die change from {1, 2, 3, 4, 6, 8} to {1, 1, 3, 2, 3, 4} as a result of halving the even numbers. If we write the possible first and second rolls as an ordered pair, then the following 24 combinations are possible: First roll 2: (2, 1), (2, 1), (2, 3), (2, 2), (2, 3), (2, 4) First roll 4: (4, 1), (4, 1), (4, 3), (4, 2), (4, 3), (4, 4) First roll 6: (6, 1), (6, 1), (6, 3), (6, 2), (6, 3), (6, 4) First roll 8: (8, 1), (8, 1), (8, 3), (8, 2), (8, 3), (8, 4) There are 36 total possible outcomes. Of these outcomes, 8 have a second roll of 2. The probability of rolling a 2 on the second roll is 8 36 = 2 9.

Problem A die has the numbers 1, 2, 3, 4, 6, and 8 on its six faces. When this die is rolled, if an odd number appears on the top face, all of the odd numbers on the die magically double. For example, if the number appearing on the top face was a 1, the 1 and 3 on the original die would double and the other four numbers would remain the same. However, if an even number appears on the top face as a result of the first roll, all of the even numbers on the die are halved. For example, if the number appearing on the top face was an 8, the 2, 4, 6, and 8 on the original die would change to half of their initial value and the other two numbers would remain the same. Suppose the die with 1, 2, 3, 4, 6, and 8 on its six faces is rolled once and changes as described above. The changed die is rolled again. No change occurs when the die is rolled this time. What is the probability that the number appearing on the top face after this roll is a 2? Solution 2 In this solution, we solve the problem in a more theoretical manner using two known results from probability theory. 1. Two events are independent if the outcome of one event does not effect the outcome of the other. If two events are independent, then the probability of both events happening is the product of their two probabilities. 2. Two events are disjoint if they cannot happen at the same time. The probability of two disjoint events is the sum of their individual probabilities. There is a 2 6 = 1 chance of getting an odd number on the first roll. The die then changes from 3 {1, 2, 3, 4, 6, 8} to {2, 2, 6, 4, 6, 8} as a result of doubling the odd numbers. For each of the possible odd rolls, there is now a 2 6 = 1 chance of rolling a 2 on the second roll. Since the two 3 events are independent, we would multiply the two probabilities to obtain 1 3 1 3 = 1. This is 9 the probability of rolling a 2 on the second roll when the first roll is odd. There is a 4 6 = 2 chance of getting an even number on the first roll. The die then changes from 3 {1, 2, 3, 4, 6, 8} to {1, 1, 3, 2, 3, 4} as a result of halving the even numbers. For each of the possible even rolls, there is now a 1 chance of rolling a 2 on the second roll. Since the two 6 events are independent, we would multiply the two probabilities to obtain 2 3 1 6 = 2 18 = 1 9. This is the probability of rolling a 2 on the second roll when the first roll is even. Since the two cases are disjoint we add the probabilities to obtain 1 9 + 1 9 = 2. The probability 9 of rolling a 2 on the second roll is 2 9.

Problem C Circle-Go-Round A circle with centre O is drawn around OBD so that B and D lie on the circumference of the circle. BO is extended to A on the circle. Chord AC intersects OD and BD at F and E, respectively. If BAC = 19 and OF A = 99, determine the measure of BEC. ( BEC is marked x on the diagram.)

Problem C and Solutions Problem Circle-Go-Round A circle with centre O is drawn around OBD so that B and D lie on the circumference of the circle. BO is extended to A on the circle. Chord AC intersects OD and BD at F and E, respectively. If BAC = 19 and OF A = 99, determine the measure of BEC. ( BEC is marked x on the diagram.) Solution 1 In a triangle, the angles add to 180. So in OF A, F OA = 180 99 19 = 62 = a. BOA is a diameter and is therefore a straight line. Two angles on a straight line add to 180, so BOD = 180 a = 180 62 = 118 = b. O is the centre of the circle with B and D on the circumference. Therefore, OD and OB are radii of the circle and OD = OB. It follows that ODB is isosceles and ODB = OBD = c. Then in ODB, c + c + b = 180 2c + 118 = 180 2c = 62 c = 31 Opposite angles are equal so it follows that CEB = DEF = x and DF E = AF O = 99 = d. In F ED, BEC = 50. x + c + d = 180 x + 31 + 99 = 180 x + 130 = 180 x = 50

Problem A circle with centre O is drawn around OBD so that B and D lie on the circumference of the circle. BO is extended to A on the circle. Chord AC intersects OD and BD at F and E, respectively. If BAC = 19 and OF A = 99, determine the measure of BEC. ( BEC is marked x on the diagram.) Solution 2 In a triangle, the angle formed at a vertex between the extension of a side and an adjacent side is called an exterior angle. In the following diagram, XZW is exterior to XY Z. The exterior angle theorem states: the exterior angle of a triangle equals the sum of the two opposite interior angles. In the diagram, r = p + q. We will use this result and two of the pieces of information we found in Solution 1. In a triangle, the angles add to 180. So in OF A, F OA = 180 99 19 = 62 = a. O is the centre of the circle with B and D on the circumference. Therefore, OD and OB are radii of the circle and OD = OB. It follows that ODB is isosceles and ODB = OBD = c. F OA is exterior to ODB. F OA = ODB + OBD a = c + c 62 = 2c 31 = c CEB is exterior to EBA. CEB = EAB + EBA x = 19 + c BEC = 50. x = 19 + 31 x = 50

Problem C Square On ABCD is a square with area 64 m 2. E, F, G, and H are points on sides AB, BC, CD, and DA, respectively, such that AE = BF = CG = DH = 2 m. E, F, G, and H are connected to form square EF GH. Determine the area of EF GH.

Problem Problem C and Solution Square On ABCD is a square with area 64 m 2. E, F, G, and H are points on sides AB, BC, CD, and DA, respectively, such that AE = BF = CG = DH = 2 m. E, F, G, and H are connected to form square EF GH. Determine the area of EF GH. Solution The area of square ABCD is 64 m 2. Therefore the side lengths are 8 m since 8 8 = 64 and the area is calculated by multiplying the length and the width. Each of the smaller parts of the sides of square ABCD are 2 m so the longer parts of the sides are 8 2 = 6 m. Approach #1 to finding the area of square EF GH In right HAE, AE = 2 and AH = 6. We can use one side as the base and the other as the height in the calculation of the area of the triangle since they are perpendicular to each other. Therefore the area of HAE = AE AH = 2 6 = 6 m 2. Since each of the triangles has the same 2 2 base length and height, their areas are equal and the total area of the four triangles is 4 6 = 24 m 2. The area of square EF GH can be determined by subtracting the area of the four triangles from the area of square ABCD. Therefore the area of square EF GH = 64 24 = 40 m 2. Approach #2 to finding the area of square EF GH Some students may be familiar with the Pythagorean Theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. The longest side is located opposite the right angle. In right HAE, AE = 2, AH = 6 and HE is the hypotenuse. Therefore, HE 2 = AE 2 + AH 2 = 2 2 + 6 2 = 4 + 36 = 40 Taking the square root, HE = 40 m But EF GH is a square so all of its side lengths equal 40. The area is calculated by multiplying the length and the width. The area of EF GH = 40 40 = 40 m 2. Therefore the area of square EF GH is 40 m 2.

Problem C Triangle Triumph In the diagram, ABC is a right triangle with ABC = 90, BD = 6 m, AB = 8 m, and the area of ADC is 50% more than the area of ABD. Determine the perimeter of ADC. The P ythagorean T heorem states, In a right triangle, the square of the length of hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides. In the following right triangle, p 2 = r 2 + q 2.

Problem C and Solution Triangle Triumph Problem In the diagram, ABC is a right triangle with ABC = 90, BD = 6 m, AB = 8 m, and the area of ADC is 50% more than the area of ABD. Determine the perimeter of ADC. Solution Let a be the length of side DC, b be the length of side AC, and c be the length of side AD. Draw a line through A parallel to BC. The distance between this line and BC is 8 m and is the height of ABD and ADC. To find the area of a triangle, multiply the length of the base by the height and divide by 2. The area of ABD = AB BD 2 = 8 6 2 = 24 m 2. The area of ADC is 50% more than the area of ABD. Therefore, the area of ADC = area of ABD + 1 2 area of ABD = 24 + 12 = 36 m2. But the area of ADC = (AB)(DC) 2 = 8(DC) 2 = 4(DC). Therefore, 4(DC) = 36 and DC = 9 m. Then BC = BD + DC = 6 + 9 = 15 m. Since ABD has a right angle, AD 2 = AB 2 + BD 2 = 8 2 + 6 2 = 100. Then AD = 100 = 10 since AD > 0. Also, ABC has a right angle, so AC 2 = AB 2 + BC 2 = 8 2 + 15 2 = 64 + 225 = 289. Then AC = 289 = 17 since AC > 0. The perimeter of ADC = a + b + c = DC + AC + AD = 9 + 17 + 10 = 36 m The perimeter of ADC is 36 m.

Problem C The Great Tricycle Race Two paths are built from A to F as shown. The distance from A to F in a straight line is 100 m. Points B, C, D, and E lie along AF such that AB = BC = CD = DE = EF. The upper path, shown with a dashed line, is a semi-circle with diameter AF. The lower path, shown with a solid line, consists of five semi-circles with diameters AB, BC, CD, DE, and EF. Starting at the same time, Bev and Mike ride their tricycles along these paths from A to F. Bev rides along the upper path from A to F while Mike rides along the lower path from A to F. If they ride at the same speed, who will get to F first?

Problem Problem C and Solution The Great Tricycle Race Two paths are built from A to F as shown. The distance from A to F in a straight line is 100 m. Points B, C, D, and E lie along AF such that AB = BC = CD = DE = EF. The upper path, shown with a dashed line, is a semi-circle with diameter AF. The lower path, shown with a solid line, consists of five semi-circles with diameters AB, BC, CD, DE, and EF. Starting at the same time, Bev and Mike ride their tricycles along these paths from A to F. Bev rides along the upper path from A to F while Mike rides along the lower path from A to F. If they ride at the same speed, who will get to F first? Solution The circumference of a circle is found by multiplying its diameter by π. To find the circumference of a semi-circle, divide its circumference by 2. The length of the upper path is equal to half the circumference of a circle with diameter 100 m. The length of the upper path equals π 100 2 = 50π m. (This is approximately 157.1 m.) Each of the semi-circles along the lower path have the same diameter. The diameter of each of these semi-circles is 100 5 = 20 m. The length of the lower path is equal to half the circumference of five circles, each with diameter 20 m. The distance along the lower path equals 5 (π 20 2) = 5 (10π) = 50π m. Since both Bev and Mike ride at the same speed and both travel the same distance, they will arrive at point F at the same time. Neither wins the race since both arrive at the same time. The answer to the problem may surprise you. Most people, at first glance, would think that the upper path is longer. If you were to extend the problem so that Bev travels the same route but Mike travels along a lower path made up of 100 semi-circles of equal diameter from A to F, they would still both travel exactly the same distance, 50π m. Check it out!

Problem C One Step at a Time A circle with centre O has a point A on the circumference. Radius OA is rotated 20 clockwise about the centre, resulting in the image OB. Point A is then connected to point B. Radius OB is rotated 20 clockwise about the centre, resulting in the image OC. Point B is then connected to point C. The process of clockwise rotations continues until some radius rotates back onto OA. Every point on the circumference is connected to the points immediately adjacent to it as a result of the process. A polygon is created. n th point A A O 20 20 20 20 20 C B = B C F Construction E D F Resulting Polygon E D a) Determine the number of sides of the polygon. b) Determine the sum of the angles in the polygon. That is, determine the sum of the angles at each of the vertices of the polygon.

Problem Problem C and Solution One Step at a Time A circle with centre O has a point A on the circumference. Radius OA is rotated 20 clockwise about the centre, resulting in the image OB. Point A is then connected to point B. Radius OB is rotated 20 clockwise about the centre, resulting in the image OC. Point B is then connected to point C. The process of clockwise rotations continues until some radius rotates back onto OA. Every point on the circumference is connected to the points immediately adjacent to it as a result of the process. A polygon is created. a) Determine the number of sides of the polygon. b) Determine the sum of the angles in the polygon. That is, determine the sum of the angles at each of the vertices of the polygon. Solution Each time the process is repeated, another congruent triangle is created. Each of these triangles has a 20 angle at O, the centre of the circle. But a complete rotation at the centre is 360. Since each angle in the triangles at the centre of the circle is 20 and the total measure at the centre is 360, then there are 360 20 = 18 triangles formed. This means that there are 18 distinct points on the circumference of the circle and the polygon has 18 sides. An 18-sided polygon is called an octadecagon, from octa meaning 8 and deca meaning 10. The other two angles in the each of the congruent triangles are equal. (Two sides of the triangle are radii of the circle. The triangles are therefore isosceles.) The angles in a triangle sum to 180 so after the 20 angle is removed, there is 160 remaining for the other two angles. It follows that each of the other two angles in each triangle measures 160 2 = 80. The following diagram illustrates this information for the two adjacent triangles AOB and BOC. A O 20 20 Each angle in the polygon is formed by an 80 angle from one triangle and the adjacent 80 angle from the next triangle. For example, ABC = ABO + OBC = 80 + 80 = 160. There are 18 vertices in the octadecagon and the angle at each vertex is 160. Therefore the sum of the angles in the octadecagon is 18 160 = 2 880. Diagrams are provided on the next page to further support the solution. 80 80 80 80 C B

Angle at the Centre of a Circle 360 Completing the Construction L K M 80 80 80 J 80 80 80 80 80 N 80 80 80 80 20 20 20 20 20 20 20 20 20 20 20 20 80 80 80 80 20 20 20 20 20 20 80 80 80 80 80 80 80 80 80 80 80 80 I P H Q G R F S 80 80 80 E 80 80 80 80 80 A D B C The Octadecagon - 18 sided Polygon M N 160 160 P 160 Q 160 160 R 160 S 160 A L 160 160 B K 160 160 C J 160 I 160 160 H 160 G 160 F 160 160 E D Notice the vertices of the octadecagon are labelled A to S, but the letter O is missing since it was used in the original construction as the centre of the circle.

Problem C Overlapping Areas The area of ACD is twice the area of square BCDE. AC and AD meet BE at K and L respectively such that KL = 6 cm. If the side length of the square is 8 cm, determine the area of AKL.

Problem Problem C and Solutions Overlapping Areas The area of ACD is twice the area of square BCDE. AC and AD meet BE at K and L respectively such that KL = 6 cm. If the side length of the square is 8 cm, determine the area of AKL. Solution 1 In the first solution we will find the area of square BCDE, the area of ACD and the area of trapezoid KCDL. To find the area of a square, multiply the length times the width. To find the area of a trapezoid, multiply the sum of the lengths of the two parallel sides by the height and divide the product by 2. Area of square BCDE = 8 8 = 64 cm 2 Area ACD = 2 Area of Square BCDE = 2 64 = 128 cm 2 In trapezoid KCDL, the two parallel sides are KL and CD, and the height is the width of square BCDE, namely BC. Area of trapezoid KCDL = (KL + CD) BC 2 = (6 + 8) 8 2 = 14 8 2 = 56 cm 2 Area AKL = Area ACD Area of trapezoid KCDL = 128 56 = 72 cm 2 Therefore, the area of AKL is 72 cm 2.

Problem The area of ACD is twice the area of square BCDE. AC and AD meet BE at K and L respectively such that KL = 6 cm. If the side length of the square is 8 cm, determine the area of AKL. Solution 2 Construct the altitude of ACD intersecting BE at P and CD at Q. In this solution we will find the height of AKL and then use the formula for the area of a triangle to find the required area. To find the area of a square, multiply the length times the width. To find the area of a triangle, multiply the length of the base times the height and divide the product by 2. Area of square BCDE = 8 8 = 64 cm 2 Area ACD = 2 Area of Square BCDE = 2 64 = 128 cm 2 But Area ACD = CD AQ 2 128 = 8 AQ 2 128 = 4 AQ AQ = 32 cm We know that AQ = AP + P Q, AQ = 32 cm and P Q = 8 cm, the side length of the square. It follows that AP = AQ P Q = 32 8 = 24 cm. Area AKL = KL AP 2 Therefore, the area of AKL is 72 cm 2. = 6 24 2 = 72 cm 2

Problem C Floor Plan The rectangular floor plan of the first level of a house is shown in the following diagram. Both the laundry room and the dining room are square with areas of 4 m 2 and 25 m 2, respectively. The living room is rectangular with an area of 30 m 2. Determine the area of the kitchen.

Problem C and Solution Floor Plan Problem The rectangular floor plan of the first level of a house is shown in the following diagram. Both the laundry room and the dining room are square with areas of 4 m 2 and 25 m 2, respectively. The living room is rectangular with an area of 30 m 2. Determine the area of the kitchen. Solution Let the width of a room be the distance represented top to bottom on the diagram. Let the length of a room be the distance represented horizontally on the diagram. The dining room is a square and has an area of 25 m 2. Its length and width must both be 5 m since Area = 5 5 = 25 m 2. The width of the dining room and living room are the same. So the width of the living room is 5 m. But the area of the living room is 30 m 2 so the length of the living room is 6 m since Area = 5 6 = 30 m 2. The laundry room is a square and has an area of 4 m 2. Its length and width must both be 2 m since Area = 2 2 = 4 m 2. The width of the laundry room and kitchen are the same. So the width of the kitchen is 2 m. (Length of Laundry Room (Length of Living Room + Length of Kitchen) = + Length of Dining Room) 2 + Length of Kitchen = 6 + 5 Length of Kitchen = 9 m Since the width of the kitchen is 2 m and the length of the kitchen is 9 m, the area of the kitchen is 2 9 = 18 m 2.

Problem C Slightly Irregular In the following slightly irregular shape, AB = 50 cm, CD = 15 cm, EF = 30 cm; the area of the shaded triangle, DEF, is 210 cm 2 ; and the area of the entire figure, ABCDE, is 1000 cm 2. Determine the length of AE.

Problem Problem C and Solution Slightly Irregular In the following slightly irregular shape, AB = 50 cm, CD = 15 cm, EF = 30 cm, the area of the shaded triangle, DEF, is 210 cm 2 ; and the area of the entire figure, ABCDE, is 1000 cm 2. Determine the length of AE. Solution The first task is to mark the given information on the diagram. This has been completed on the diagram to the right. To find the area of a triangle, multiply the base length by the height and divide by 2. In DEF, the base, EF, has length 30 cm. The height of DEF is the perpendicular distance from EF (extended) to vertex D, namely GD. The area is given. So Area DEF = 30 GD 2 210 = 15 GD 14 = GD We know that EH = AB = 50, GH = DC = 15, and EH = EF + F G + GH. It follows that 50 = 30 + F G + 15 and F G = 5 cm. Now we can relate the total area to the areas contained inside. Area ABCDE = Area ABHE + Area CDGH + Area DF G + Area DEF 1000 = AB AE + DG DC + F G GD + 210 2 1000 = 50 AE + 14 15 + 5 14 + 210 2 1000 = 50 AE + 210 + 35 + 210 1000 = 50 AE + 455 1000 455 = 50 AE 545 = 50 AE 545 50 = AE AE = 10.9 cm.

Problem C A Little Tiling Tyler, the tiler, has an unlimited supply of square tiles. He has 1 cm by 1 cm tiles, 2 cm by 2 cm tiles, 3 cm by 3 cm tiles, and so on. Every tile has integer side lengths. A rectangular table top with an 84 cm by 112 cm surface is to be completely covered by identical square tiles, none of which can be cut. Tyler knows that he can cover the table top with 1 cm 1 cm tiles, 9 408 in total, since 84 112 = 9 408 cm 2. However, Tyler wants to use the minimum number of identical tiles to complete the job in order to reduce his overall material cost. Determine the minimum number of tiles required to completely cover the table top.

Problem Problem C and Solution A Little Tiling Tyler, the tiler, has an unlimited supply of square tiles. He has 1 cm by 1 cm tiles, 2 cm by 2 cm tiles, 3 cm by 3 cm tiles, and so on. Every tile has integer side lengths. A rectangular table top with an 84 cm by 112 cm surface is to be completely covered by identical square tiles, none of which can be cut. Tyler knows that he can cover the table top with 1 cm 1 cm tiles, 9 408 in total, since 84 112 = 9 408 cm 2. However, Tyler wants to use the minimum number of identical tiles to complete the job in order to reduce his overall material cost. Determine the minimum number of tiles required to completely cover the table top. Solution To use the smallest number of tiles we must use the largest tile possible. The square tile must have sides less than or equal to 84 cm. If it was greater than 84 cm, the tile would have to be cut to fit the width of the table. Since the tiles are square and must completely cover the top surface, the side length of the tile must be a number that is a factor of both 84 and 112. In fact, since we need the largest side length, we are looking for the greatest common factor of 84 and 112. The factors of 84 are The factors of 112 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. 1, 2, 4, 7, 8, 14, 16, 28, 56, and 112. The largest number common to both lists is 28. Therefore the greatest common factor of 84 and 112 is 28. The required tiles are 28 cm 28 cm. Since 84 28 = 3, the surface is 3 tiles wide. Since 112 28 = 4, the surface is 4 tiles long. The minimum number of tiles required is 3 4 = 12 tiles. The number of 28 cm 28 cm tiles required to cover the top of the table is 12. This is the minimum number of tiles required.

Problem C This is Sum Problem The number 90 can be expressed as the sum of 3 consecutive whole numbers. That is, 90 = 29 + 30 + 31. The number 90 can also be written as the sum of 4 consecutive whole numbers. That is, 90 = 21 + 22 + 23 + 24. Express the number 220 as the sum of 5 consecutive whole numbers and then as the sum of 8 consecutive whole numbers. This problem is extended this week in Problem D and Problem E of Problem of the Week.

Problem Problem C and Solution This is Sum Problem The number 90 can be expressed as the sum of 3 consecutive whole numbers. That is, 90 = 29 + 30 + 31. The number 90 can also be written as the sum of 4 consecutive whole numbers. That is, 90 = 21 + 22 + 23 + 24. Express the number 220 as the sum of 5 consecutive whole numbers and then as the sum of 8 consecutive whole numbers. Solution First, we want to express 220 as the sum of 5 consecutive whole numbers. When 90 is expressed as the sum of 3 consecutive whole numbers, the average is 90 3 = 30. Since 30 is a whole number and there is an odd number of consecutive whole numbers in the sum, 29, 30 and 31 will produce the correct sum. If we apply the same idea to 220, the average would be 220 5 = 44. Since 44 is a whole number and there is an odd number of consecutive whole numbers in the sum, 42, 43, 44, 45, and 46, should produce the correct sum. Checking, 42 + 43 + 44 + 45 + 46, we obtain 220 as required. Next, we want to express 220 as the sum of 8 consecutive whole numbers. When 90 is expressed as the sum of 4 consecutive whole numbers, the average is 90 4 = 22.5. We need consecutive whole numbers such that two are below and two are above 22.5. This gives the numbers 21, 22, 23, and 24, as in the example. This would only work for an even number of consecutive whole numbers if the average is half way between two consecutive whole numbers. Applying the same idea to 220, the average would be 220 8 = 27.5. This number is half way between 27 and 28. We would need four consecutive whole numbers below the average and four consecutive whole numbers above the average giving us the eight numbers 24, 25, 26, 27, 28, 29, 30, and 31. Checking, 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31, we obtain 220 as required. As a concluding note, it is not possible to express 220 as the sum of four consecutive whole numbers. 220 4 = 55. The number 55 is the average of the four numbers but 55 is a whole number. We could try 54 + 55 + 56 + 57 = 222 220 or 53 + 54 + 55 + 56 = 218 220. It is possible to express 220 as the sum of n consecutive whole numbers when n is odd provided that the average 220 n is a whole number. It is possible to express 220 as the sum of n consecutive whole numbers when n is even provided that the average 220 n is half way between two consecutive whole numbers.

Problem C Playing With Blocks Ten blocks are arranged as illustrated in the following diagram. Each letter shown on the front of a block represents a number. The sum of the numbers on any three consecutive blocks is 19. Determine the value of S.

Problem C and Solution Playing With Blocks Problem Ten blocks are arranged as illustrated in the following diagram. Each letter shown on the front of a block represents a number. The sum of the numbers on any three consecutive blocks is 19. Determine the value of S. Solution Since the sum of the numbers on any three consecutive blocks is the same, 4 + P + Q = P + Q + R 4+ P + Q = P + Q + R since P + Q is common to both sides R = 4 Again, since the sum of the numbers on any three consecutive blocks is the same, T + U + V = U + V + 8 T + U+ V = U+ V + 8 since U + V is common to both sides T = 8 Since the sum of any three consecutive numbers is 19: R + S + T = 19 4 + S + 8 = 19 substituting R = 4 and T = 8 S + 12 = 19 The value of S is 7. S = 7

Problem C Show Me The Money Over the past several years, Ima Saver collected nickels (5 cent coins) and dimes (10 cent coins) and put them in his pink piggy bank. Only nickels and dimes went in his bank. Finally, one day, the bank was full so Ima counted his money and discovered that he had exactly $10 in the bank. He also observed that he had 11 more nickels than dimes in his bank. How many coins were in Ima Saver s pink piggy bank?

Problem C and Solutions Show Me The Money Problem Over the past several years, Ima Saver collected nickels (5 cent coins) and dimes (10 cent coins) and put them in his pink piggy bank. Only nickels and dimes went in his bank. Finally, one day, the bank was full so Ima counted his money and discovered that he had exactly $10 in the bank. He also observed that he had 11 more nickels than dimes in his bank. How many coins were in Ima Saver s pink piggy bank? Solution 1 In this solution we will solve the problem without using equations. Ima had 11 more nickels than dimes. These 11 nickels are worth 11 5 = 55 or $0.55. The remaining $10.00 $0.55 = $9.45 would be made up using an equal number of nickels and dimes. Each nickel-dime combination is worth 15 or $0.15. By dividing $9.45 by $0.15 we determine the number of 15 cent combinations that are required to make the total. Since $9.45 $0.15 = 63 we need 63 nickel-dime pairs. That is, we need 63 nickels and 63 dimes to make $9.45. But there are 11 more nickels. Therefore, there is a total of 63 + 63 + 11 = 137 coins in his bank. Solution 2 In this solution we will solve the problem using an equation. Let n represent the number of nickels and (n 11) represent the number of dimes. Since each nickel is worth 5, the value of n nickels is (5n). Since each dime is worth 10, the value of n 11 dimes is 10(n 11). The bank contains a total value of $10 or 1 000. Therefore, Value of Nickels (in ) + Value of Dimes (in ) = Total Value (in ) 5n + 10(n 11) = 1 000 5n + 10n 110 = 1 000 15n = 1 000 + 110 15n = 1 110 n = 74 n 11 = 63 There are 74 nickels and 63 dimes for a total of 74 + 63 = 137 coins in his bank.

Problem C That s Odd 1 = 1 2 1 + 3 = 4 = 2 2 1 + 3 + 5 = 9 = 3 2 1 + 3 + 5 + 7 = 16 = 4 2 Did you know that the sum of the first n positive odd integers is n 2? The diagram above illustrates the first four possible sums. The sum of the first five positive odd integers would be 5 2 or 25. We can easily check to see that 1 + 3 + 5 + 7 + 9 = 25. When adding the first a positive odd integers to the first b positive odd integers, the sum is 180. If p is the largest odd number in the first set of numbers and q is the largest odd number in the second set of numbers, then determine the sum p + q.

Problem Problem C and Solution That s Odd Did you know that the sum of the first n positive odd integers is n 2? The sum of the first five positive odd integers would be 5 2 or 25. We can easily check to see that 1 + 3 + 5 + 7 + 9 = 25. When adding the first a positive odd integers to the first b positive odd integers, the sum is 180. If p is the largest odd number in the first set of numbers and q is the largest odd number in the second set of numbers, then determine the sum p + q. Solution Since there are a positive odd integers and the largest is p, then 1 + 3 + 5 + + p = a 2. Since there are b positive odd integers and the largest is q, then 1 + 3 + 5 + + q = b 2. We also know that when these two sets of odd numbers are added together, the sum is 180 so (1 + 3 + 5 + + p) + (1 + 3 + 5 + + q) = a 2 + b 2 = 180. One way to proceed is to pick values for a, determine a 2 and then determine if the remaining number required to sum to 180 is a perfect square. The results are summarized in the table below. a a 2 b 2 = 180 a 2 b (b > 0) Solution? 1 1 180-1=179 13.4 no 2 4 180-4=176 13.3 no 3 9 180-9=171 13.1 no 4 16 180-16=164 12.8 no 5 25 180-25=155 12.4 no 6 36 180-36=144 12 yes 7 49 180-49=131 11.4 no 8 64 180-64=116 10.8 no 9 81 180-81=99 9.9 no 10 100 180-100=80 8.9 no 11 121 180-121=59 7.7 no 12 144 180-144=36 6 yes 13 169 180-169=11 3.3 no If a = 14, then a 2 = 196. This produces a value greater than 180 and cannot be a possible solution. There appear to be two possible solutions. When a = 6 and b = 12, then a 2 + b 2 = 36 + 144 = 180. This means that adding the first 6 odd positive integers to the first 12 odd positive integers results in a sum of 180. So p is the sixth odd positive integer, namely 11, and q is the twelfth odd positive integer, namely 23. The sum, p + q, is 11 + 23 or 34. The second solution, a = 12 and b = 6, produces p = 23 and q = 11. The sum, p + q, is still 34.

Problem C Clock Talk My clock is a perfectly good clock. It keeps exact time. But it only has an hour hand. Today, in the afternoon, I looked at my clock and discovered that the hour hand was 7 8 of the distance between the 4 and the 5. Determine the exact time (hours, minutes and seconds).

Problem C and Solution Clock Talk Problem My clock is a perfectly good clock. It keeps exact time. But it only has an hour hand. Today, in the afternoon, I looked at my clock and discovered that the hour hand was 7 8 of the distance between the 4 and the 5. Determine the exact time (hours, minutes and seconds). Solution To solve this problem we note that in one hour, the hour hand travels 1 12 of a complete revolution while the minute hand travels a complete revolution or 60 minutes. Since the hour hand is 7 8 of the distance between the 4 and the 5, the minute hand will travel 7 8 of a complete revolution or 7 8 of 60 minutes which is 7 8 60 or 521 2 minutes. Since we want the time in hours, minutes and seconds, we need to convert 1 2 minute to seconds. The number of seconds may be obvious but the calculation, 1 2 minute 60 seconds 1 minute = 30 seconds, is provided for completeness. Therefore the precise time is 30 seconds after 4:52 p.m. This can be written 4:52:30 p.m. or 16:52:30 using the twenty-four hour clock.

Problem C Ride On, Ride On! A motorcycle and a delivery truck left a roadside diner at the same time. After travelling in the same direction for one and one-quarter hours, the motorcycle had travelled 25 km farther than the delivery truck. If the average speed of the motorcycle was 60 km/h, find the average speed of the delivery truck.

Problem Problem C and Solution Ride On, Ride On! A motorcycle and a delivery truck left a roadside diner at the same time. After travelling in the same direction for one and one-quarter hours, the motorcycle had travelled 25 km farther than the delivery truck. If the average speed of the motorcycle was 60 km/h, find the average speed of the delivery truck. Solution We can calculate distance by multiplying the average speed by the time. In one and one-quarter hours at 60 km/h, the motorcycle would travel 60 1 1 4 = 60 5 4 = 75 km. In the same time, the delivery truck travels 25 km less. The delivery truck has travelled 75 25 = 50 km. Since the distance travelled equals the average speed multiplied by the time, then the average speed will equal the distance travelled divided by the time travelled. Thus, the average speed of the delivery truck equals 50 1 1 4 = 50 5 4 = 50 4 5 = 40 km/h. Therefore the average speed of the delivery truck is 40 km/h. The calculations in this problem could be done using decimals by converting one and one-quarter hours to 1.25 hours.

Problem C Balloon Breaker A wall is covered with balloons. Each balloon has the number 5, 3 or 2 printed on it. You are given 7 darts to throw at the wall. If your dart breaks a balloon, then you earn the number of points printed on the balloon. If your dart does not break a balloon, then you are awarded 0 points for that shot. You win a prize if your total score is exactly 16 points on seven shots. If your total is over 16 or under 16, then you lose. Determine the number of different point combinations that can be used to win the game.

Problem Problem C and Solutions Balloon Breaker A wall is covered with balloons. Each balloon has the number 5, 3 or 2 printed on it. You are given 7 darts to throw at the wall. If your dart breaks a balloon, then you earn the number of points printed on the balloon. If your dart does not break a balloon, then you are awarded 0 points for that shot. You win a prize if your total score is exactly 16 points on seven shots. If your total is over 16 or under 16, then you lose. Determine the number of different point combinations that can be used to win the game. Solution 1 Let us consider cases. 1. You break three balloons with a 5 printed on them. You have a total of 3 5 = 15 points. There is no possible way to get 16 points since the other balloon values are 2 or 3. There is no way to win by breaking three (or more) balloons with a 5 printed on them. 2,. You break two balloons with a 5 printed on them. You have a total of 2 5 = 10 points. You need to get 16 10 = 6 points by breaking balloons with 2 or 3 printed on them. There are two ways to do this. Break three balloons with a 2 printed on them and miss on two shots or break two balloons with 3 printed on them and miss on three shots. There are 2 ways to win if you break two balloons with a 5 printed on them. 3. You break one balloon with a 5 printed on it. You have a total of 5 points. You need to get 16 5 = 11 points by breaking balloons with 2 or 3 printed on them. You cannot get 11 points breaking only balloons with a 2 printed on them and you cannot get 11 points breaking only balloons with a 3 printed on them. However, you can get 11 points by breaking one 3 and four 2 s or by breaking three 3 s and one 2. There are 2 ways to win if you break one balloon with a 5 printed on it. 4. You break no balloons with a 5 printed on it. You need to make 16 points by breaking only balloons with a 2 or 3 printed on them. You cannot get 16 points breaking only balloons with a 3 printed on them. You cannot get 16 points breaking only balloons with a 2 printed on them because you only have seven darts giving a maximum of 14 points. It is possible to get 16 points using combinations of 3 point and 2 point balloons. If you break two 3 point balloons and five 2 point balloons, then you win in seven shots. If you break four 3 point balloons and two 2 point balloons, then you have 16 points in six shots and would have to miss on one of your shots. There are 2 ways to win if you do not break any 5 point balloons. There are 0 + 2 + 2 + 2 = 6 combinations that allow you to win by getting 16 points in seven shots.

Problem A wall is covered with balloons. Each balloon has the number 5, 3 or 2 printed on it. You are given 7 darts to throw at the wall. If your dart breaks a balloon, then you earn the number of points printed on the balloon. If your dart does not break a balloon, then you are awarded 0 points for that shot. You win a prize if your total score is exactly 16 points on seven shots. If your total is over 16 or under 16, then you lose. Determine the number of different point combinations that can be used to win the game. Solution 2 In this solution we will complete a chart to determine the valid possibilities. Let the number of 5 point balloons be a, the number of 3 point balloons be b, and the number of 2 point balloons be c. We want 5a + 3b + 2c = 16 and a + b + c 7. Number Number Number Total Number WIN of 5 Point of 3 Point of 2 Point Points of Missed Shots or Balloons Balloons Balloons Scored Needed LOSE a b c 5a + 3b + 2c 16 a b c 3 0 0 15 LOSE 3 0 1 17 LOSE 3 1 0 18 LOSE 2 2 0 16 3 WIN 2 1 1 15 LOSE 2 1 2 17 LOSE 2 0 3 16 2 WIN 1 4 0 17 LOSE 1 3 0 14 LOSE 1 3 1 16 2 WIN 1 2 2 15 LOSE 1 2 3 17 LOSE 1 1 4 16 1 WIN 0 6 0 18 LOSE 0 5 0 15 LOSE 0 5 1 17 LOSE 0 4 2 16 1 WIN 0 3 3 15 LOSE 0 3 4 17 LOSE 0 2 5 16 0 WIN 0 1 6 15 LOSE 0 0 7 14 LOSE There are only 6 combinations that allow you to win by getting 16 points in seven shots. (In following a method like the above method, one must be careful to systematically examine all possible cases.)

Problem C Photo Fun Five of Santa s elves: Alpha, Beta, Delta, Epsilon and Gamma, are lined up in alphabetical order from left to right. They can each choose one of five festive hats to wear for a photo. The hats are identical except for colour. Three of the hats are red and two of the hats are green. How many different photos can be taken?

Problem Problem C and Solution Photo Fun Five of Santa s elves: Alpha, Beta, Delta, Epsilon and Gamma, are lined up in alphabetical order from left to right. They can each choose one of five festive hats to wear for a photo. The hats are identical except for colour. Three of the hats are red and two of the hats are green. How many different photos can be taken? Solution Since the elves are already organized in alphabetical order, we are looking for the number of different ways that we can distribute the hats among the elves. We will consider cases: 1. If the first elf gets a green hat, there are four ways to give out the second green hat. Once the green hats are distributed, the remaining three elves must each get a red hat. Therefore, there are 4 ways to distribute the hats so that the first elf receives a green hat. 2. If the first elf gets a red hat and the second elf gets a green hat, there are three ways to give out the second green hat. Once the green hats are distributed, the remaining two elves must each get a red hat. Therefore, there are 3 ways to distribute the hats so that the first elf receives a red hat and the second elf receives a green hat. 3. If the first two elves each get a red hat and the third elf gets a green hat, there are two ways to give out the second green hat. Once the green hats are distributed, the remaining elf must get a red hat. Therefore, there are 2 ways to distribute the hats so that the first two elves each receive a red hat and the third elf receives a green hat. 4. If the first three elves each get a red hat and the fourth elf gets a green hat, the fifth elf must get the second green hat. Therefore, there is only 1 way to distribute the hats so that the first three elves each receive a red hat and the fourth elf receives a green hat. There are no other cases to consider. The total number of ways to distribute the hats is the sum of the number of ways from each of the cases. Therefore, there are 4 + 3 + 2 + 1 = 10 ways to distribute the hats. There are 10 different photos that can be taken.

Problem C A Powerful Problem For The New Year 5 3 is a power with base 5 and exponent 3. 5 3 means 5 5 5 and equals 125 when expressed as an integer. When 5 2013 is expressed as an integer, what are the last three digits?

Problem C and Solution A Powerful Problem For The New Year Problem 5 3 is a power with base 5 and exponent 3. 5 3 means 5 5 5 and equals 125 when expressed as an integer. When 5 2013 is expressed as an integer, what are the last three digits? Solution Let s start by examining the last three digits of various powers of 5. 5 1 = 005 5 2 = 025 5 3 = 125 5 4 = 625 5 5 = 3 125 5 6 = 15 625 5 7 = 78 125 5 8 = 390 625 Notice that there is a pattern for the last three digits after the first two powers of 5. For every odd integral exponent greater than 2, the last three digits are 125. For every even integral exponent greater than 2, the last three digits are 625. The pattern continues so 5 9 will end 125 since the exponent 9 is odd and 5 10 will end 625 since the exponent 10 is even. This is easily verified since 5 9 = 1 953 125 and 5 10 = 9 765 625. For 5 2013, the exponent 2013 is greater than 2 and is an odd number. the last three digits of 5 2013 are 125.

Problem C Positively Perfect Prime Products A prime number is any number that has exactly two positive integer factors, 1 and the number itself. A composite number has more than two positive integer factors. The number 1 is neither prime nor composite. Four distinct prime numbers have a product of d10: a three-digit number with hundreds digit d. Determine all possible values of the sum of these four prime numbers.

Problem Problem C and Solution Positively Perfect Prime Products A prime number is any number that has exactly two positive integer factors, 1 and the number itself. A composite number has more than two positive integer factors. The number 1 is neither prime nor composite. Four distinct prime numbers have a product of d10: a three-digit number with hundreds digit d. Determine all possible values of the sum of these four prime numbers. Solution Since the product d10 ends in 0, it must be divisible by 10, which is the product of the two primes 2 and 5. When d10 is divided by 10, the quotient is d10 10 = d1. This two-digit number must be composite and must be the product of two distinct prime numbers, neither of which is 2 or 5. We can rule out any two-digit prime numbers for d1 since these numbers would only have one prime factor. Therefore, we can rule out 11, 31, 41, 61, and 71: the five prime numbers ending with a 1. Then, d cannot be 1, 3, 4, 6, or 7. Since d10 is a three-digit number, d 0 because d10 = 010 = 10 is not a three-digit number. The remaining possibilities for d are 2, 5, 8, and 9. If d = 2, then the two-digit number would be 21, which has prime factors 7 and 3. The four prime factors of d10 = 210 are 2, 3, 5, and 7, producing a sum of 2 + 3 + 5 + 7 = 17. If d = 5, then the two-digit number would be 51, which has prime factors 17 and 3. The four prime factors of d10 = 510 are 2, 3, 5, and 17, producing a sum of 2 + 3 + 5 + 17 = 27. If d = 8, then the two-digit number would be 81, which is the product 9 9. There are already four factors, two of which are composite so d10 = 810 cannot be expressed as the product of four distinct prime numbers. Therefore, 8 can be ruled out as a possible value for d. (Note that 810 = 2 3 3 3 3 5, which is the product of six prime numbers not all of which are distinct. If d = 9, then the two-digit number would be 91, which has prime factors 7 and 13. The four prime factors of d10 = 910 are 2, 5, 7, and 13, producing a sum of 2 + 5 + 7 + 13 = 27. However, we already have the sum 27. Since there are no other possible cases to consider, the only possible sums of the four distinct prime factors that multiply to d10 are 17 and 27.

Problem C Good to Plan and Prepare but... In preparing to write an examination, Stu Deus made the following observations: The exam had 20 questions. He estimated that he would spend 6 minutes per question. The exam would take him 2 hours to complete. However, during the actual examination, Stu discovered some difficult questions which each required 15 minutes to complete. He also discovered some questions which were much easier than he expected and only took him 2 minutes per question to complete. Seven of the questions, however, still required 6 minutes each to complete. Surprisingly, Stu was still able to complete the exam in 2 hours. How many of the 20 examination questions did Stu find difficult?

Problem Problem C and Solution Good to Plan and Prepare but... In preparing to write an examination, Stu Deus made the following observations: the exam had 20 questions, he would spend 6 minutes per question, and the exam would take him 2 hours to complete. However, during the actual examination, Stu discovered some difficult questions which each required 15 minutes to complete. He also discovered some questions which were much easier than he expected and only took him 2 minutes per question to complete. Seven of the questions, however, still required 6 minutes each to complete. Surprisingly, Stu was still able to complete the exam in 2 hours. How many of the 20 examination questions did Stu find difficult? Solution 1: Systematic Trial Since 7 of the questions required 6 minutes each to complete, it took Stu 7 6 = 42 minutes to complete these questions. The total exam took 2 hours or 120 minutes. He had 120 42 = 78 minutes to complete 20 7 = 13 questions. Let d represent the number of difficult questions and e represent the number of easier questions. We know that d + e = 13. Since each difficult question took 15 minutes, it took 15d minutes to complete all of the difficult questions. Since each easier question took 2 minutes, it took 2e minutes to complete all of the easier questions. Since Stu s total remaining time was 78 minutes, 15d + 2e = 78 minutes. At this point we can pick values for d and e that add to 13 and substitute into the equation 15d + 2e = 78 to find the combination that works. (We can observe that d < 6 since 15 6 = 90 > 78. If this were the case, then e would have to be a negative number.) If d = 3 then e = 13 3 = 10. The time to complete these would be 15 3 + 2 10 = 45 + 20 = 65 minutes and he would complete the exam in less than 2 hours. If d = 4 then e = 13 4 = 9. The time to complete these would be 15 4 + 2 9 = 60 + 18 = 78 minutes and he would complete the exam in exactly 2 hours. Therefore, Stu found 4 of the questions to be more difficult and time-consuming than he expected. Solution 2 involves algebra and equation solving.

Problem In preparing to write an examination, Stu Deus made the following observations: the exam had 20 questions, he would spend 6 minutes per question, and the exam would take him 2 hours to complete. However, during the actual examination, Stu discovered some difficult questions which each required 15 minutes to complete. He also discovered some questions which were much easier than he expected and only took him 2 minutes per question to complete. Seven of the questions, however, still required 6 minutes each to complete. Surprisingly, Stu was still able to complete the exam in 2 hours. How many of the 20 examination questions did Stu find difficult? Solution 2: Using Algebra and Equations Since 7 of the questions required 6 minutes each to complete, it took Stu 7 6 = 42 minutes to complete these questions. The total exam took 2 hours or 120 minutes. He had 120 42 = 78 minutes to complete 20 7 = 13 questions. Let d represent the number of difficult questions and (13 d) represent the number of easier questions. Since each difficult question took 15 minutes, it took 15d minutes to complete all of the difficult questions. Since each easier question took 2 minutes, it took 2(13 d) minutes to complete all of the easier questions. Since Stu s total remaining time was 78 minutes, 15d + 2(13 d) = 78 15d + 26 2d = 78 13d + 26 = 78 Subtracting 26 from both sides: 13d = 52 Dividing both sides by 13: d = 4 Therefore, Stu found 4 of the questions to be more difficult and time-consuming than he expected.

Problem C CUT! Kuri Uz is given some red licorice that is wrapped in a coil. Upon flattening the licorice, Kuri discovers that the total length is 60 cm. She then cuts the licorice into two pieces such that the ratio of the lengths of the two pieces is 7 : 3. Each piece is then bent to form a square. What is the total area of the two squares?

Problem Problem C and Solution CUT! Kuri Uz is given some red licorice that is wrapped in a coil. Upon flattening the licorice, Kuri discovers that the total length is 60 cm. She then cuts the licorice into two pieces such that the ratio of the lengths of the two pieces is 7 : 3. Each piece is then bent to form a square. What is the total area of the two squares? Solution Since the licorice is cut in the ratio 7 : 3, let the longer piece be 7x cm and the shorter piece be 3x cm. Then 7x + 3x = 60 or 10x = 60 and x = 60 10 = 6. Therefore, the longer piece is 7x = 7 6 = 42 cm and the shorter piece is 3x = 3 6 = 18 cm. Each of the two pieces is then bent to form a square. The perimeter of each square is the length of the licorice used to form it. The side length of the longer square is 42 4 = 10.5 cm and the side length of the shorter square is 18 4 = 4.5 cm. To find the area of each square, we multiply length by width. In effect, to find the area of the square, we square the side length. The area of the larger square is 10.5 10.5 = 10.5 2 = 110.25 cm 2 and the area of the smaller square is 4.5 4.5 = 4.5 2 = 20.25 cm 2. Therefore, if Kuri Uz cuts the licorice into pieces of length 42 cm and 18 cm, she can form two squares with area 110.25 cm 2 and 20.25 cm 2, respectively. The total area of the two squares is 110.25 + 20.25 = 130.5 cm 2. As a slight extension, the ratio of the area of the larger square to the area of the smaller square is 110.25 : 20.25 = 11025 : 2025 = 441 : 81 = 49 : 9 = 7 2 : 3 2. The ratio of the perimeters is 7 : 3 and the ratio of the areas is 7 2 : 3 2. In general, if the ratio of the perimeters of two squares is a : b, is it true that the ratio of the areas of the two squares is a 2 : b 2?

Problem C Trickling Down The number in an unshaded square in the grid below is obtained by adding the numbers connected to it from the row above. For example, the 5 in the second row is obtained by adding the numbers, 2 and 3, connected to it in the row above. The numbers trickle down until the final square containing 2x. Determine the value of x.

Problem Problem C and Solutions Trickling Down The number in an unshaded square in the grid is obtained by adding the numbers connected to it from the row above. For example, the 5 in the second row is obtained by adding the numbers, 2 and 3, connected to it in the row above. The numbers trickle down until the final square containing 2x. Determine the value of x. Solution 1 In the first solution a trial and error type solution will be presented. We will pick a value for x and then complete the table. Let x = 0. We would expect 2x = 2(0) = 0. However, after completing the table, 2x = 12 0. Therefore, x 0. Let x = 5. We would expect 2x = 2(5) = 10. After completing the table, 2x = 27 10. Therefore, x 5. For our next trial we should choose a number lower than x = 0. Let x = 5. We would expect 2x = 2( 5) = 10. However, after completing the table, 2x = 3 10. Therefore, x 5. Let x = 12. We would expect 2x = 2( 12) = 24. After completing the table, 2x = 24, the expected value. Therefore, x = 12. This solution is valid but not efficient. To reach the solution could take many, many trials. In the second solution, we will look at an algebraic solution.

Solution 2 For easy reference label all of the empty, unshaded squares in the diagram as shown. We can complete the second row using the addition rule for the table, a = 3 + x and b = x + 1. Moving to the third row, c = 5 + a = 5 + 3 + x = 8 + x and d = a + b = 3 + x + x + 1 = 4 + 2x. Finally, in the fourth row, 2x = c + d = 8 + x + 4 + 2x. We can now solve the equation. 2x = 8 + x + 4 + 2x 2x 2x = 8 + x + 4 + 2x 2x 0 = 12 + x Simplifying Subtracting 2x from both sides 0 12 = 12 + x 12 Subtracting 12 from both sides 12 = x Therefore, x = 12. An algebraic solution to this problem is much more efficient. Some students may not quite have the necessary background to complete this solution on their own at this point.

A list consists of the integers Problem C Can t Get There From Here 1, 2, 3, 5, 8, 13, 21, and 34. A second list of integers is created that contains all possible integers that can be formed by adding two different integers from the first list. The smallest integer in this list is 1 + 2 = 3 and the largest integer in this list is 21 + 34 = 55. A third list of integers contains all of the integers from 3 to 55 that are not contained in the second list. Determine the number of integers in the third list. For your information: The integers in the first list are numbers found in the F ibonacci Sequence. The rectangle shown above is often referred to as T he F ibonacci Rectangle. The numbers inside the squares correspond to the side length of the square that contains the number. The two smallest squares (the white one and the black one) in the diagram each have side length 1. The rectangle spirals out from these two smallest squares. You may wish to investigate this sequence further.

Problem Problem C and Solution Can t Get There From Here A list consists of the integers 1, 2, 3, 5, 8, 13, 21, and 34. A second list of integers is created that contains all possible integers that can be formed by adding two different integers from the first list. The smallest integer in this list is 1 + 2 = 3 and the largest integer in this list is 21 + 34 = 55. A third list of integers contains all of the integers from 3 to 55 that are not contained in the second list. Determine the number of integers in the third list. Solution Notice that in the first list, each integer after the first two integers is the sum of the previous two integers. It is easiest to determine the number of numbers in the second list. We are able to systematically do this. If 34 is added to each of the seven smaller integers in the first list, seven unique integers in the range 35 to 55 can be formed, namely 35, 36, 37, 39, 42, 47, and 55. If 21 is added to each of the six smaller integers in the first list, six unique integers in the range 22 to 34 can be formed, namely 22, 23, 24, 26, 29, and 34. If 13 is added to each of the five smaller integers in the first list, five unique integers in the range 14 to 21 can be formed, namely 14, 15, 16, 18, and 21. If 8 is added to each of the four smaller integers in the first list, four unique integers in the range 9 to 13 can be formed, namely 9, 10, 11, and 13. If 5 is added to each of the three smaller integers in the first list, three unique integers in the range 6 to 8 can be formed, namely 6, 7, and 8. If 3 is added to each of the two smaller integers in the first list, two unique integers in the range 4 to 5 can be formed, namely 4 and 5. If the first two integers in the first list are added, the number 3 is obtained. The second list contains 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 integers. The number of integers from 3 to 55 is the same as the number of integers from 1 to 53. There are 53 integers from 3 to 55. Therefore, the third list contains 53 28 = 25 integers.

Problem C E - I - E - I - O Old MacDonald had a farm, E-I-E-I-O, says the old children s song. But Old MacDonald did have a farm! And on that farm he had some horses, cows, pigs and 69 water troughs for the animals to drink from. Only horses drank from the horse troughs, exactly two horses for each trough. Only cows drank from cow troughs, exactly three cows per trough. And only pigs drank from pig troughs, exactly eight pigs per trough. Old MacDonald s farm has the same number of cows, horses and pigs. How many animals does Old MacDonald have on his farm?

Problem Problem C and Solutions E - I - E - I - O Old MacDonald had a farm, E-I-E-I-O, says the old children s song. But Old MacDonald did have a farm! And on that farm he had some horses, cows, pigs and 69 water troughs for the animals to drink from. Only horses drank from the horse troughs, exactly two horses for each trough. Only cows drank from cow troughs, exactly three cows per trough. And only pigs drank from pig troughs, exactly eight pigs per trough. Old MacDonald s farm has the same number of cows, horses and pigs. How many animals does Old MacDonald have on his farm? Solution 1 Let n represent the number of each type of animal. Since there are two horses for every horse trough, then n must be divisible by 2. Since there are three cows for every cow trough, then n must be divisible by 3. Since there are eight pigs for every pig trough, then n must be divisible by 8. Therefore, n must be divisible by 2, 3 and 8. The smallest number divisible by 2, 3, and 8 is 24. (This number is called the lowest common multiple or LCM, for short.) If there are 24 of each kind of animal, there would be 24 2 = 12 troughs for horses, 24 3 = 8 troughs for cows and 24 8 = 3 troughs for pigs. This would require a total of 12 + 8 + 3 = 23 troughs. Since there are 69 troughs and 69 23 = 3, we require 3 times more of each type of animal. That is, there would be 24 3 = 72 of each type of animal. The total number of animals is 72 + 72 + 72 or 216. We can check the correctness of this solution. Since there are two horses for every horse trough, there are 72 2 = 36 horse troughs. Since there are three cows for every cow trough, there are 72 3 = 24 cow troughs. Since there are eight pigs for every pig trough, there are 72 8 = 9 pig troughs. The total number of troughs is 36 + 24 + 9 = 69, as expected.

Solution 2 In this solution, algebra and equation solving will be used to solve the problem. Let n represent the number of each type of animal. Since there are n horses and there are two horses for every horse trough, then there would be n 2 troughs for horses. Since there are n cows and there are three cows for every cow trough, then there would be n troughs for cows. Since there are n pigs and there are eight pigs for every pig 3 trough, then there would be n troughs for pigs. Since there are 69 troughs in total, 8 n 2 + n 3 + n = 69 8 12n 24 + 8n 24 + 3n 24 23n 24 = 69 common denominator 24 = 69 simplify the fractions 23n = 24 69 multiply both sides by 24 23n = 1656 simplify n = 1656 divide both sides by 23 23 n = 72 There are 72 of each type of animal, a total of 216 animals. Solution 3 In this solution, ratios will be used to solve the problem. Let n represent the number of each type of animal. The ratio of the number of troughs required for the pigs to the number of troughs required for the cows is n 8 : n 3 = 3n 24 : 8n 24 = 3n : 8n = 3 : 8. Similarly, the ratio of the number of troughs required for the cows to the number of troughs required for the horses is 2 : 3 = 8 : 12. So the ratio of the number of troughs required for the pigs to the number required for the cows to the number required for the horses is 3 : 8 : 12. Let the number of troughs required for the pigs be 3k, for the cows be 8k and for the horses be 12k, for some positive integer value of k. Since the total number of troughs required is 69, then 3k + 8k + 12k = 69 23k = 69 k = 3 The number of troughs required for the pigs is 3k = 9. There are 8 pigs at each trough. There are a total of 9 8 = 72 pigs. Since there are the same number of each animal, there are also 72 cows and 72 horses. There are a total of 72 + 72 + 72 = 216 animals on Old MacDonald s farm.

Problem C The Bead Goes On Jill has been making two types of bracelets. The smaller bracelets contain four birthstone-coloured beads each and the larger bracelets contain seven birthstone-coloured beads each. So far Jill has used a total of 99 beads. How many of each type of bracelet has Jill made? (There may be more than one possible answer.)

Problem Problem C and Solution The Bead Goes On Jill has been making two types of bracelets. The smaller bracelets contain four birthstone-coloured beads each and the larger bracelets contain seven birthstone-coloured beads each. So far Jill has used a total of 99 beads. How many of each type of bracelet has Jill made? (There may be more than one possible answer.) Solution Let S represent the number of small bracelets and L represent the number of large bracelets. Since the small bracelets use 4 beads each, S bracelets would use 4 S or 4S beads in total. Since the large bracelets use 7 beads each, L bracelets would use 7 L or 7L beads in total. Since Jill has used a total of 99 beads, 4S + 7L = 99, with L and S integers greater than or equal to 0. We can also determine a maximum value for L. Each large bracelet uses 7 beads. We know that 99 7 =14.1 so L must be a positive integer less than or equal to 14. We could at this point check all of the possible integer values for L from 0 to 14. However, we can narrow down the possibilities even more. In the equation, 4S + 7L = 99, 4S will always be an even number since 4 times any number is always even. We then have an even number plus 7L equals the odd number 99. This means that 7L must be an odd number. (An even number plus an even number would be an even number, not an odd number.) For 7L to be an odd number, L must be odd. (If L is even, 7L would be even.) This observation reduces the possible values for L to the odd positive integers between 0 and 14, namely 1, 3, 5, 7, 9, 11, 13. Now we can test the possible values of L to see which ones, if any, produce a valid possibility for S. Number Number Number Number of Large of Beads of Beads of Small Valid or Invalid Bracelets Used Remaining Bracelets Possibility? L 7L 99 7L (99 7L) 4 1 7 1 = 7 99-7=92 92 4 = 23 valid, it is an integer 3 7 3 = 21 99-21=78 78 4 = 19.5 invalid, not an integer 5 7 5 = 35 99-35=64 64 4 = 16 valid, it is an integer 7 7 7 = 49 99-49=50 50 4 = 12.5 invalid, not an integer 9 7 9 = 63 99-63=36 36 4 = 9 valid, it is an integer 11 7 11 = 77 99-77=22 22 4 = 5.5 invalid, not an integer 13 7 13 = 91 99-91=8 8 4 = 2 valid, it is an integer There are four valid possibilities. Jill has either made 1 large bracelet and 23 small bracelets, or 5 large bracelets and 16 small bracelets, or 9 large bracelets and 9 small bracelets, or 13 large bracelets and 2 small bracelets.

Problem C Road Trip A cyclist rode 560 km in seven days. Each day she travelled 15 km more than the day before. Determine how far she rode on the seventh day.

Problem Problem C and Solutions Road Trip A cyclist rode 560 km in seven days. Each day she travelled 15 km more than the day before. Determine how far she rode on the seventh day. Solution 1 We will begin by representing the information on a diagram. The total distance is made up of seven trips with the same length as Day 1 plus 15+2(15)+3(15)+4(15)+5(15)+6(15) = 15+30+45+60+75+90 = 315 km. So seven days riding the same distance as Day 1 would total 560 315 = 245 km. On Day 1, the cyclist rode 245 7 = 35 km. On Day 7, the cyclist rode the Day 1 distance plus 6 15 = 90 km. The total distance travelled on the seventh day was 35 + 90 = 125 km. This first solution deliberately avoids algebra and equations. Many solvers would be able to reason the solution in a similar way. The second and third solutions will present more algebraic approaches.

Solution 2 Let x be the distance travelled on the first day. Then the cyclist rides x + 15, x + 30, x + 45, x + 60, x + 75, and x + 90 km on days two through seven, respectively. Then, x + (x + 15) + (x + 30) + (x + 45) + (x + 60) + (x + 75) + (x + 90) = 560 7x + 15 + 30 + 45 + 60 + 75 + 90 = 560 7x + 315 = 560 7x + 315 315 = 560 315 7x = 245 7x = 245 7 7 x = 35 The cyclist rode 35 km on the first day. On the seventh day, the cyclist rode x + 90 = 35 + 90 = 125 km. Solution 3 Let m be the distance travelled on the middle day, which is day four. On day five the cyclist would ride (m + 15) km; on day six the cyclist would ride m + 15 + 15 = (m + 30) km; and on day seven the cyclist would ride m + 30 + 15 = (m + 45) km. We can also work backwards by reducing the distance the cyclist rode by 15 km. On day three the cyclist would ride (m 15) km; on day two the cyclist would ride m 15 15 = (m 30) km; and on day one the cyclist would ride m 30 15 = (m 45) km. Then, m + (m + 15) + (m + 30) + (m + 45) + (m 15) + (m 30) + (m 45) = 560 7m + 15 + 30 + 45 15 30 45 = 560 7m + 15 15 + 30 30 + 45 45 = 560 7m = 560 7m = 560 7 7 m = 80 The cyclist rode 80 km on the fourth day. On the seventh day, the cyclist rode m + 45 = 80 + 45 = 125 km. This third solution always works well when there is an odd number of terms in a sequence that increases (or decreases) by a constant amount.

Problem C This One is For the Birds One hundred pigeons are to be housed in identical-sized cages under the following conditions: each cage must contain at least one pigeon; no two cages can contain the same number of pigeons; and no cages can go inside any other cage. Determine the maximum number of cages required to house the pigeons.

Problem Problem C and Solution This One is For the Birds One hundred pigeons are to be housed in identical-sized cages under the following conditions: each cage must contain at least one pigeon; no two cages can contain the same number of pigeons; and no cages can go inside any other cage. Determine the maximum number of cages required to house the pigeons. Solution In order to maximize the number of cages, each cage must contain the smallest number of birds possible. However, no two cages can contain the same number of pigeons. The simplest way to determine this number is to put one pigeon in the first cage and then let the number of pigeons in each cage after that be one more than the number of pigeons in the cage before it, until all 100 pigeons are housed. Put 1 pigeon in the first cage, 2 pigeons in the second cage, 3 pigeons in the third cage, and so on. After filling 12 cages in this manner, we have 1 + 2 + 3 + + 11 + 12 = 78 pigeons housed. After putting 13 pigeons in the thirteenth cage, 78 + 13 = 91 pigeons are housed. There are 9 pigeons left to house. But we already have a cage containing 9 pigeons. The remaining 9 pigeons must be distributed among the existing cages while maintaining the condition that no two cages contain the same number of pigeons. The most obvious way to do this is to put the 9 pigeons in the last cage which already contains 13 pigeons. This would mean that the final cage would contain 13 + 9 = 22 birds! A better solution might be to increase the number of birds in each of the final nine cages by one bird each. This solution is summarized below: Cage # 1 2 3 4 5 6 7 8 9 10 11 12 13 # of birds 1 2 3 4 6 7 8 9 10 11 12 13 14 The maximum number of cages required is 13. If you had 14 cages, with the first cage holding 1 pigeon and each cage after that holding one more pigeon than the cage before, you could house 105 pigeons, five more than the number of pigeons that we have.

Problem C Road Trip A cyclist rode 560 km in seven days. Each day she travelled 15 km more than the day before. Determine how far she rode on the seventh day.