POTWC 11 Combined7 8

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.

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Problem of the Week

Problem C

What is the Deal Anyway?

Three playing cards are placed in a row, from left to right. One card is a club(♣), one card is a diamond (♦), and one card is a heart (♥). The number oneach card is different. One card is a four, one card is a five and one card is aneight.

Using the following clues, determine the exact order of the cards, from left toright, including the suit and number.

1. The club is somewhere to the right of the heart.

2. The 5 is somewhere to the left of the heart.

3. The 8 is somewhere to the right of the 4.


Problem of the Week

Problem C and Solution

What is the Deal Anyway?

Problem

Three playing cards are placed in a row, from left to right. One card is a club(♣), one card is a diamond (♦), and one card is a heart (♥). The number oneach card is different. One card is a four, one card is a five and one card is aneight. Using the following clues, determine the exact order of the cards, fromleft to right, including the suit and number.

1. The club is somewhere to the right of the heart.2. The 5 is somewhere to the left of the heart.

3. The 8 is somewhere to the right of the 4.

Solution

There are six ways to order the suits:

(♣,♦,♥), (♣,♥,♦), (♦,♣,♥), (♦,♥,♣), (♥,♣,♦), and (♥,♦,♣).

The first clue tells us that the club is somewhere to the right of the heart. Wecan eliminate the first three from the above list since the club is to the left ofthe heart in each case. There are now only three ways to order the suits:

(♦,♥,♣), (♥,♣,♦), and (♥,♦,♣).

The second clue says that the 5 is somewhere to the left of the heart. Thismeans that the heart cannot be the leftmost card. This eliminates the last twopossibilities from the above list. The only possibility is (♦,♥,♣) and the 5must be the five of diamonds giving us (5♦,♥,♣).

The third clue tells us that the 8 is somewhere to the right of the 4. With onlytwo spots to decide, we can conclude that the 8 must be in the rightmost(third) spot and the last two cards are the 4 of hearts and the 8 of clubs. Thisgives us (5♦, 4♥, 8♣).

∴ the cards are dealt in the following order: 5 of diamonds, 4 of hearts, and8 of clubs.


Problem of the Week

Problem C

Fractions to the Max

In the expressiona

b+

c

d+

e

f, each letter is replaced by a different digit from

1,2,3,4,5,6. Each digit can be used exactly once.

What is the largest possible value of this expression?

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Problem of the Week

Problem C and Solutions


Problem

In the expressiona

b+

c

d+

e

f, each letter is replaced by a different digit from 1,2,3,4,5,6. Each

digit can be used exactly once. What is the largest possible value of this expression?

Solution 1

The largest fractions will be created by putting the three smallest numbers, 1,2, and 3, in the denominators and then placing the numbers 4, 5, and 6 in thenumerators. We will do this in every possible way, determine the sums andchoose the largest.

There are six different possible sums in which 1, 2, and 3 are the denominatorsand 4, 5, 6 are the numerators.

(1)4

1+

5

2+

6

3=

24

6+

15

6+

12

6=

51

6

(2)4

1+

6

2+

5

3=

24

6+

18

6+

10

6=

52

6

(3)5

1+

4

2+

6

3=

30

6+

12

6+

12

6=

54

6

(4)5

1+

6

2+

4

3=

30

6+

18

6+

8

6=

56

6

(5)6

1+

4

2+

5

3=

36

6+

12

6+

10

6=

58

6

(6)6

1+

5

2+

4

3=

36

6+

15

6+

8

6=

59

6

Therefore the largest possible value of the expression is59

6. It should be noted

that this approach would not be practical if more numbers were involved. Besure to look at solution 2 for a more logical approach.


Solution 2

We can start by observing that to get a fraction with the highest value we needa 6 in the numerator. The choice of denominators is possibly obvious as well.61 = 6, 6

2 = 3, 63 = 2, 6

4 = 1.5, and 65 = 1.2. 6

1 is the largest fraction and anynumerator other than 6 will produce a lower value.

Now we have four numbers left to place: {2,3,4,5}.

Of these remaining numbers, since 5 is the largest it should go in thenumerator. Then 5

2 = 2.5, 53

.= 1.7, and 5

4 = 1.25. 52 is the largest fraction and

any numerator other than 5 will produce a lower value.

Now we have two numbers left to place: {3,4}.

Our only two choices for the third fraction are 43

.= 1.3, and 3

4 = 0.75. So 43 is

the third fraction.

We can now determine the largest possible sum.

Largest Possible Sum =6

1+

5

2+

4

3

=36

6+

15

6+

8

6

=59

6or 9

5

6

∴ the largest possible sum that can be made from the numbers 1, 2, 3, 4, 5,

and 6 in the expressiona

b+

c

d+

e

fis 59

6 or 956 (approximately 9.83).


Problem of the Week

Problem C

Can You DIGIT?

The three digit number 5A4 is divisible by 4 and the three digit number 37B isdivisible by 3.

Determine the largest positive difference between 5A4 and 37B.


Problem of the Week


Can You DIGIT?

Problem

The three digit number 5A4 is divisible by 4 and the three digit number 37B isdivisible by 3. Determine the largest positive difference between 5A4 and 37B.

Solution

The largest positive difference will occur when 5A4 is a large as possible and37B is as small as possible. We are therefore looking for the largest possiblevalue of A and the smallest possible value of B.

A number is divisible by 4 if the last two digits of the number are divisibleby 4. For 5A4 to be divisible by 4, the only possible values of A are 0, 2, 4, 6,and 8 since 04, 24, 44, 64, and 84, respectively, are the only two digit numbersthat end in 4 and are divisible by 4. (04 is technically not a two-digit numberbut a larger number could end in these two digits.) Since we want 5A4 to be aslarge as possible, A = 8 and 5A4 = 584.

A number is divisible by 3 if the sum of its digits is divisible by 3. For 37B tobe divisible by 3, the only possible values of B are 2, 5, and 8.

• When B = 2 the number is 372. It is divisible by 3 since 3 + 7 + 2 = 12 isdivisible by 3.• When B = 5 the number is 375. It is divisible by 3 since 3 + 7 + 5 = 15 isdivisible by 3.• When B = 8 the number is 378. It is divisible by 3 since 3 + 7 + 8 = 18 isdivisible by 3.

No other three digit numbers of the form 37B will be divisible by 3 since noother values of B, other than 2, 5, and 8, give a digit sum that is divisible by 3.Since we want 37B to be as small as possible, B = 2 and 37B = 372.

The largest positive difference occurs when A = 8 and B = 2 so that5A4 − 37B = 584 − 372 = 212.

∴ the largest positive difference is 212.


Problem of the Week

Problem C

Pressure’s On

Willy Makit is participating at the local level of a competition in which hecompetes in 10 events. In each event the maximum possible score is 10 points.To advance to the regional level a competitor must earn a minimum total scoreof 75 points.

On the first five events, Willy had an average score of 6.8. On the next threeevents his average score was 7.2.

What must Willy average on his final two events in order to have a total scoreof exactly 75 points?


Problem of the Week


Pressure’s On

Problem

Willy Makit is participating at the local level of a competition in which hecompetes in 10 events. In each event the maximum possible score is 10 points.To advance to the regional level a competitor must earn a minimum total scoreof 75 points. On the first five events, Willy had an average score of 6.8. On thenext three events his average score was 7.2. What must Willy average on hisfinal two events in order to have a total score of exactly 75 points?

Solution

To determine an average, we add the numbers together and divide the total bythe number of numbers.

Average =Total

Number of Numbers

So to determine the total we would multiply the average by the number ofnumbers.

Total = Average × Number of Numbers

Total Score for First 5 Events = 6.8 × 5 = 34.0

Total Score for Next 3 Events = 7.2 × 3 = 21.6

Total Score to Move On = 75.0

Total Score Needed for Last Two Events = 75.0 − 34.0 − 21.6 = 19.4

Average Score for Final 2 Events = 19.4 ÷ 2 = 9.7

Willy needs to average 9.7, near perfect scores, on his last two events to have atotal score of exactly 75 points in order to advance to the regional round of thecompetition. The pressure is definitely on!


Problem of the Week

Problem C

Fair Game

Seven identical balls are numbered 1 to 7 and then placed in a bag. The bag isshaken so the balls are mixed up. A player draws four of the balls from the bagand then determines the sum of the numbers on the balls. The player wins thegame if the sum of the numbers on the balls is odd.

Determine the probability that a player wins the game. Express your answer infraction form.


Problem of the Week


Fair Game

Problem

Seven identical balls are numbered 1 to 7 and then placed in a bag. The bag is shaken so theballs are mixed up. A player draws four of the balls from the bag and then determines the sumof the numbers on the balls. The player wins the game if the sum of the numbers on the ballsis odd. Determine the probability that a player wins the game. Express your answer in fractionform.

Solution

In order to determine the probability, we must determine the number of ways to obtain a sum that isodd and divide it by the total number of possible selections of four balls from the bag.

We must systematically count all of the possibilities. One must be careful not to miss any of thepossibilities. We will systematically list the possible selections.

Select 1 and 2 and two higher numbers: 1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267.Select 1 and 3 and two higher numbers: 1345, 1346, 1347, 1356, 1357, 1367.Select 1 and 4 and two higher numbers: 1456, 1457, 1467.Select 1 and 5 and two higher numbers: 1567.Select 2 and 3 and two higher numbers: 2345, 2346, 2347, 2356, 2357, 2367.Select 2 and 4 and two higher numbers: 2456, 2457, 2467.Select 2 and 5 and two higher numbers: 2567.Select 3 and 4 and two higher numbers: 3456, 3457, 3467.Select 3 and 5 and two higher numbers: 3567.Select 4 and 5 and two higher numbers: 4567.

By counting the outcomes from each case, there are 10 + 6 + 3 + 1 + 6 + 3 + 1 + 3 + 1 + 1 = 35possible selections of four balls from the bag. We must now determine how many of these selectionshave an odd sum. We could take each of the possibilities, determine the sum and then count thenumber which produce an odd sum. However, it may be slightly easier to look at how a sum is oddwhen adding four numbers. The sum of four numbers is odd in two instances: there are three evennumbers and one odd number or there are three odd numbers and one even number.

Selections with three even numbers and one odd number:

1246, 2346, 2456, 2467

Selections with three odd numbers and one even number:

1235, 1237, 1257, 1345, 1347, 1356, 1367, 1457, 1567, 2357, 3457, 3567

The total number of selections where the sum is odd is 4 + 12 = 16. Therefore the probability of

winning, by selecting four balls with an odd sum, is16

35. A game is considered fair if the probability

of winning is 50%. In this game the winning probability, as a percentage, is approximately 46%.


Problem of the Week

Problem C

Pane - Filled Problem

A contractor is building new houses. Each house has a large window withwidth 2 feet and height 6 feet to be made up of six identical 1 foot by 2 feetglass panes.

The contractor wants each house to look slightly different. The window size isfixed, but he plans on arranging the glass panes in the windows so that no twowindows have the same configuration. Two possible arrangements of the glasspanes are given below.

Assuming that the glass panes cannot be cut, how many differentarrangements can be made?


Problem of the Week



Problem

A contractor is building new houses. Each house has a large window with width 2 feet andheight 6 feet to be made up of six identical 1 foot by 2 feet glass panes. The contractor wantseach house to look slightly different. The window size is fixed, but he plans on arranging theglass panes in the windows so that no two windows have the same configuration. Assumingthat the glass panes cannot be cut, how many different arrangements can be made?

Solution

Let’s consider the ways that the glass panes can be arranged. First, notice that there mustalways be an even number of window panes that have a vertical orientation (standing on end).

• All glass panes are horizontal (and none are vertical)

This can only be done one way.

• Four glass panes are horizontal and two are vertical

There could be 0, 1, 2, 3 or 4 horizontal panes below the two vertical panes. So there are5 ways that four glass panes are horizontal and two are vertical.

• Two glass panes are horizontal and four are vertical

We need to consider sub cases:• Case 1: There are no horizontal panes between the vertical panes.

There could be 0, 1 or 2 horizontal panes below the four vertical panes. So there are3 ways that two glass panes are horizontal, four are vertical and there are nohorizontal panes between the vertical panes.

• Case 2: There is one horizontal pane between the vertical panes.There could be 0 or 1 horizontal panes below the bottom two vertical panes. Sothere are 2 ways that two glass panes are horizontal, four are vertical and there isone horizontal pane between the vertical panes.

• Case 3: There are two horizontal panes between the vertical panes.There cannot be any horizontal panes below the bottom two vertical panes. Sothere is 1 way that two glass panes are horizontal, four are vertical and there aretwo horizontal panes between the vertical panes.

• All glass panes are vertical


Therefore, the total number of different configurations of the glass panes is1 + 5 + (3 + 2 + 1) + 1 = 13.

So the contractor can build 13 houses before he has to start duplicating window patterns.


Problem of the Week

Problem C

What’s Your Angle Anyway I?

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In 4ABD, C is on AB such that AC = CB = CD and ∠BCD = 70◦.

Determine the measure of ∠ADB.

If you want to solve a more general version of this problem, consider solvingthis week’s Grade 9 and 10 Problem of the Week “What’s Your Angle AnywayII?”.

If you want to further extend this problem, consider solving this week’s Grade11 and 12 Problem of the Week “What’s Your Angle Anyway III?”.


Problem of the Week



Problem

In 4ABD, C is on AB such that AC = CB = CD and∠BCD = 70◦. Determine the measure of ∠ADB.

Solution

Since ACB is a straight line, ∠ACD + ∠DCB = 180◦

but ∠DCB = 70◦ so ∠ACD = 110◦.

In 4ACD, since AC = CD, 4ACD is isosceles and∠CAD = ∠CDA = x◦.

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The angles in a triangle sum to 180◦ so in 4ACD

∠CAD + ∠CDA + ∠ACD = 180◦

x◦ + x◦ + 110◦ = 180◦

2x = 70

x = 35

Similarly, in 4BCD, since CB = CD, 4BCD is isosceles and∠CBD = ∠CDB = y◦.

The angles in a triangle sum to 180◦ so in 4CBD

∠CBD + ∠CDB + ∠BCD = 180◦

y◦ + y◦ + 70◦ = 180◦

2y = 110

y = 55

Then ∠ADB = ∠CDA + ∠CDB = x◦ + y◦ = 35◦ + 55◦ = 90◦.

∴ the measure of ∠ADB is 90◦.

See the next page for an interesting idea.


It turns out that it is not necessary to find the angles in the problem.

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Here is a second solution to the problem.

In 4CAD, since CA = CD, 4CAD is isosceles and ∠CAD = ∠CDA = x◦.

In 4CBD, since CB = CD, 4CBD is isosceles and ∠CBD = ∠CDB = y◦.

The angles in a triangle sum to 180◦ so in 4ABD

∠BAD + ∠ADB + ∠ABD = 180◦

x◦ + (x◦ + y◦) + y◦ = 180◦

2x◦ + 2y◦ = 180◦

x◦ + y◦ = 90◦

But ∠ADB = ∠ADC + ∠CDB = x◦ + y◦ = 90◦.



Problem of the Week

Problem C

A Griddy Performance

A (0,3)

B (4,0)

C (7,4)

x

y

Three of the vertices of square ABCD are located at A(0,3), B(4,0), andC(7,4).

(a) Determine the coordinates of the fourth vertex, D.

(b) Determine the area of square ABCD.


Problem of the Week



Problem

Three of the vertices of square ABCD are located at A(0,3),B(4,0), and C(7,4).



A (0,3)

B (4,0)

C (7,4)

D (3,7)

x

y

Solution

To determine the coordinates of D, observe that to get from A to B, you would go down 3units and right 4 units. To get from B to C, you move 3 units to the right and then 4 units up.Continuing the pattern, go up 3 units and left 4 units you get to D(3,7). Continuing, as acheck, go left 3 units and down 4 units, and you arrive back at A.

B (4,0)

C (7,4)

D (3,7)

x

y

A (0,3)

Determining the area of ABCD without using the Pythagorean Theorem

Draw a box with horizontal and vertical sides so that each vertex of the square ABCD is onone of the sides of the box. This creates a large square with sides of length 7 containing fouridentical triangles and square ABCD. Each of the triangles has a base 4 units long and height3 units long.

Area ABCD = Area of Large Square − 4 × Area of One Triangle

= Length × Width − 4 × (Base × Height ÷ 2)

= 7 × 7 − 4 × (4 × 3 ÷ 2)

= 49 − 4 × 6

= 49 − 24

= 25 units2

D is located at (3,7) and the area of the square is 25 units2. (See the next page for asolution to the area problem using the Pythagorean Theorem.)


B (4,0)

C (7,4)

D (3,7)

x

y

A (0,3)

O (0,0)

Determining the area of ABCD using the Pythagorean Theorem

Since ABCD is a square, it is only necessary to find the length of one side. We can determinethe area by squaring the length of the side.

Let the origin be O(0,0). Then OAB forms a right triangle. OA, the distance from the originto point A on the y-axis, is 3 units. OB, the distance from the origin to point B on the x-axis,is 4 units.

Using the Pythagorean Theorem, we can find AB2 which is AB × AB, the area of the square.

AB2 = OA2 + OB2

= 32 + 42

= 9 + 16

= 25

∴ the area of the square is 25 units2.


Problem of the Week

Problem C

Stack ’Em Up

Three cubes with side lengths 1 m, 2 m and 3 m are stacked on top of eachother as shown.

Determine the total surface area of the stack, including the bottom.


Problem of the Week


Stack ’Em Up

Problem

Three cubes with side lengths 1 m, 2 m and 3 m are stacked on top of eachother as shown. Determine the total surface area of the stack, including thebottom.

Solution

To determine the areas we will primarily use Area = length× width.

Each cube has four exposed square sides so the total area of all the sides is4×(1×1)+4×(2×2)+4×(3×3) = 4×(1)+4×(4)+4×(9) = 4+16+36 = 56 m2.

To determine the exposed top area of each of the cubes look down on the towerand see a cross-section like the one below.

This exposed area is exactly the same as the side area of one face of the largestcube. Therefore, the top exposed area is 3× 3 = 9 m2. The top area and thebottom area are the same. Therefore, the bottom area is 9 m2.

The total surface area is 56 + 9 + 9 = 74 m2.

Extension: Three cubes with side lengths x, y and z are stacked on top ofeach other in a similar manner to the original problem such that0 < x < y < z. Show that the total surface area of the stack, including thebottom, is 6z2 + 4y2 + 4x2.


Problem of the Week

Problem C

Irregular Area

x

y

The dots on the diagram are one unit apart, horizontally and vertically.

Determine the area of the figure.


Problem of the Week


Irregular Area

Problem

The dots on the diagram are one unit apart, horizontally and vertically. Determine the area ofthe figure.

Solution

On the diagram, draw two horizontal lines and two vertical lines toform a rectangle such that the vertices of the irregular area are onthe sides of the rectangle. Label the four vertices of the rectangleand the four vertices of the irregular area as shown on the diagram.

To find the area of the irregular shape, determine the area of therectangle and subtract the area of the four triangles that are notpart of the area of the irregular shape.

Since the dots are one unit apart horizontally and vertically we candetermine the various lengths:

x

y

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AB = 5, BC = 4, AC = 9 CD = 4, DE = 3, CE = 7

EF = 7, FG = 2, EG = 9 GH = 2, HA = 5, GA = 7

To find the area of the rectangle, multiply the length AC by the width CE obtaining9× 7 = 63 units2.

Each of the corners of the rectangles has a right angle. So each of the four triangles is rightangled and we can use the lengths of the two sides that meet at the right angle in thecalculation of the area of the triangle. Using Area = base× height÷ 2, we calculate the areas:

Area 4ABH =AB ×HA

2=

5× 5

2=

25

2= 12.5 units2

Area 4BCD =BC × CD

2=

4× 4

2=

16

2= 8 units2

Area 4DEF =DE × EF

2=

3× 7

2=

21

2= 10.5 units2

Area 4FGH =FG×GH

2=

2× 2

2=

4

2= 2 units2

Therefore the area of the irregular shape is 63− 12.5− 8− 10.5− 2 = 63− 33 = 30 units2.


Problem of the Week

Problem C

Be Mine Valentine

A valentine is constructed by pasting two white semi-circles, each withradius 3 cm, and a white triangle onto a 12 cm square sheet of red paper asshown below. (The dashed line and the right angle symbols will not actually beon the finished card.)

You are going to write your valentine a message in red ink on the white regionof the card.

Determine the total amount of area available for your special valentine greeting.


Problem of the Week


Be Mine Valentine

Problem

A valentine is constructed by pasting two white semi-circles, each withradius 3 cm, and a white triangle onto a 12 cm square sheet of red paper. Youare going to write your valentine a message in red ink on the white region ofthe card. Determine the total amount of area available for your specialvalentine greeting.

Solution

Place the given information on the diagram.

12 cm

3 cm

12 cm

The total area for writing the message is the area of the two semi-circles plusthe area of the white triangle.

Since there are two semi-circles of radius 3 cm, the total area is the same as thearea of a full circle of radius 3 cm. The area of the two semi-circles isπr2 = π(3)2 = 9π cm2.

The height of the triangle is the length of the square minus the radius of thesemi-circle. Therefore the height of the triangle is 12− 3 = 9 cm. The base ofthe triangle is 12 cm, the width of the square. The area of the triangle is12base× height = 1

2(12)(9) = 54 cm2.

The total area for writing the message is (9π + 54) cm2. This area isapproximately 82.3 cm2. Happy Valentine’s Day.


Problem of the Week

Problem C

Check the Perimeter

A median is a line segment drawn from the vertex of a triangle to themidpoint of the opposite side.

In 4ABC, a median is drawn from A meeting BC at M . The perimeter of4ABC is 24. The perimeter of 4ABM is 18. The perimeter of 4ACM is 16.

Determine the length of the median AM .

A

B CM


Problem of the Week


Check the Perimeter

Problem

A median is a line segment drawn from the vertexof a triangle to the midpoint of the opposite side.In 4ABC, a median is drawn from A meetingBC at M . The perimeter of 4ABC is 24. Theperimeter of 4ABM is 18. The perimeter of4ACM is 16. Determine the length of themedian AM .

A

B CM

t r

p q

m

Solution 1

Let AB = t, BM = p, MC = q, CA = r, and AM = m.

The perimeter of 4ABM = t + p + m = 18.

The perimeter of 4ACM = q + r + m = 16.

The perimeter of 4ABC = t + p + q + r = 24.

Consider the perimeter of 4ABM plus the perimeter of 4ACM . This is equalto (t + p + m) + (q + r + m) = t + p + m + q + r + m = (t + p + q + r) + 2×m.

So, if we add the perimeter of 4ABM to the perimeter of 4ACM , we willobtain the perimeter of 4ABC plus two times the length of the median, m.

In other words, since the perimeter of 4ABM is 18, the perimeter of 4ACMis 16 and the perimeter of 4ABC is 24, we find that 18 + 16 = 24 + 2×m. Itfollows that 34 = 24 + 2×m and 2×m must be 10. Therefore m, the length ofthe median, is 5.

In solution 2, we take a more algebraic approach to solving the problem. Itinvolves more formal equation solving.


Problem


A

B CM

t r

p q

m

Solution 2


The perimeter of 4ABM = t + p + m = 18 or t + p = 18−m. (1)

The perimeter of 4ACM = q + r + m = 16 or q + r = 16−m. (2)

The perimeter of 4ABC = t + p + q + r = 24. (3)

We will now combine the results.

t + p = 18−m (1)

q + r = 16−m (2)

So t + p + q + r = 18−m + 16−m, by combining the two results.

But the perimeter is t + p + q + r = 24.

∴ 24 = 18−m + 16−m

24 = 34− 2m

24− 34 = 34− 34− 2m

−10 = −2m

−10

−2=−2m

−2

5 = m

The length of the median AM = m = 5.


Problem of the Week

Problem C

Face It

A regular die has faces numbered 1, 2, 3, 4, 5, and 6. The numbers on the facesare arranged so that opposite faces total seven. For example, the facecontaining 2 is opposite the face containing five.

The four dice shown have been placed so that the two numbers on the facestouching each other always total nine. The face labelled P is the front of onedie as shown. What is the number on the face labelled P?

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Problem of the Week


Face It

Problem

A regular die has faces numbered 1, 2, 3, 4, 5, and 6. The numbers on the faces are arranged sothat opposite faces total seven. For example, the face containing 2 is opposite the facecontaining five. The four dice shown have been placed so that the two numbers on the facestouching each other always total nine. The face labelled P is the front of one die as shown.What is the number on the face labelled P?

Solution

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On Dice 1 , since 5 is on the front, there is a 2 on the back. Since 4 is on the top, there is a 3on the bottom. That leaves a 6 and 1 for the sides. Since the sides facing each other add to 9,the right side of dice 1 must be a 6. If it were a 1, the left face of dice 2 would have to be 8and that is not possible. Therefore the right side of dice 1 must be a 6.

That means that the left side of dice 2 must be a 3 since the sides facing each other total 9. Ifthe left side of dice 2 is 3, then the right side of dice 2 must be a 4 since opposite sides addto 7.

Then the left side of dice 3 must be a 5. If 5 is on the left side, 2 is on the right side. Since 4is on the top of dice 3 , there is a 3 on the bottom. That leaves 1 and 6 for the front and backof dice 3 . The front must be 6 in order for the numbers on the front of dice 3 and the backof dice 4 to total 9.

Since the front of dice 3 is a 6, the back of dice 4 must be a 3. If the back of dice 4 is a 3,then the front of dice 4 must be a four. But the front of dice 4 is P . Therefore P = 4.


Problem of the Week

Problem C

All Things Being Equal

!

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Square BCDE and 4ACD have equal areas. Square BCDE has sides oflength 12 cm. AD intersects BE at F .

Determine the area of quadrilateral BCDF .


Problem of the Week



Problem

Square BCDE and 4ACD have equal areas. SquareBCDE has sides of length 12 cm. AD intersects BE at F .Determine the area of quadrilateral BCDF .

!

"

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!"#$%Solution

The area of square BCDE = 12× 12 = 144 cm2. The area of 4ACD equals the area of squareBCDE. Therefore area 4ACD = 144 cm2.

The area of a triangle is calculated using the formula base× height÷ 2. It follows that:

Area 4ACD = (CD)× (AC)÷ 2

144 = 12× AC ÷ 2

144 = 6× AC

24 cm = AC

But AC = AB +BC so 24 = AB + 12 and it follows that AB = 12 cm.

Area 4ACD = Area Square BCDE

Area 4ABF +Area Quad. BCDF = Area 4DEF +Area Quad. BCDF

Since the area of quadrilateral BCDF is common to both sides of the equation, it follows thatthe area of 4ABF equals the area of 4DEF .

Area 4ABF = Area 4DEF

(AB)× (BF )÷ 2 = (DE)× (EF )÷ 2

12×BF ÷ 2 = 12× FE ÷ 2

6×BF = 6× FE

∴ BF = FE

But BF + FE = BE = CD = 12 so BF = FE = 6 cm. Then the area of4DEF = DE × FE ÷ 2 = 12× 6÷ 2 = 36 cm2.

The area of quadrilateral BCDF = area of square BCDE − area 4DEF

= 144− 36

= 108 cm2

Therefore the area of quadrilateral BCDF is 108 cm2.


Problem of the Week

Problem C


A (0,3)

B (4,0)

C (7,4)

x

y





Problem of the Week



Problem




A (0,3)

B (4,0)

C (7,4)

D (3,7)

x

y

Solution


B (4,0)

C (7,4)

D (3,7)

x

y

A (0,3)





= 7 × 7 − 4 × (4 × 3 ÷ 2)

= 49 − 4 × 6

= 49 − 24

= 25 units2



B (4,0)

C (7,4)

D (3,7)

x

y

A (0,3)

O (0,0)





AB2 = OA2 + OB2

= 32 + 42

= 9 + 16

= 25



Problem of the Week

Problem C

Stack ’Em Up




Problem of the Week


Stack ’Em Up

Problem


Solution








Problem of the Week

Problem C

Always Have A Good Altitude

! "

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$

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An altitude is a line segment drawn from a vertex of a triangle to the oppositeside such that the line segment is perpendicular to the opposite side.

In 4ABC, CD and BE are altitudes. AB = 16 cm, AC = 12 cm andCD = 6 cm.

Determine the length of BE.


Problem of the Week



Problem

An altitude is a line segment drawn from a vertex ofa triangle to the opposite side such that the linesegment is perpendicular to the opposite side.In 4ABC, CD and BE are altitudes. AB = 16 cm,AC = 12 cm and CD = 6 cm. Determine the lengthof BE. ! "

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$

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Solution

The area of a triangle is determined using the formula base× height÷ 2. Theheight of the triangle is the length of an altitude and the base of the triangle isthe length of the side to which a particular altitude is drawn.

Area 4ABC =(CD)× (AB)

2

=6× 16

2

= 48 cm2

But, Area 4ABC =(BE)× (AC)

2

48 =(BE)× 12

2

48 = 6×BE

8 cm = BE

Therefore, the length of altitude BE is 8 cm.


Problem of the Week

Problem C

Irregular Area

x

y

The dots on the diagram are one unit apart, horizontally and vertically.

Determine the area of the figure.


Problem of the Week


Irregular Area

Problem

The dots on the diagram are one unit apart, horizontally and vertically. Determine the area ofthe figure.

Solution

On the diagram, draw two horizontal lines and two vertical lines toform a rectangle such that the vertices of the irregular area are onthe sides of the rectangle. Label the four vertices of the rectangleand the four vertices of the irregular area as shown on the diagram.

To find the area of the irregular shape, determine the area of therectangle and subtract the area of the four triangles that are notpart of the area of the irregular shape.

Since the dots are one unit apart horizontally and vertically we candetermine the various lengths:

x

y

!

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&

'(

AB = 5, BC = 4, AC = 9 CD = 4, DE = 3, CE = 7

EF = 7, FG = 2, EG = 9 GH = 2, HA = 5, GA = 7

To find the area of the rectangle, multiply the length AC by the width CE obtaining9× 7 = 63 units2.

Each of the corners of the rectangles has a right angle. So each of the four triangles is rightangled and we can use the lengths of the two sides that meet at the right angle in thecalculation of the area of the triangle. Using Area = base× height÷ 2, we calculate the areas:

Area 4ABH =AB ×HA

2=

5× 5

2=

25

2= 12.5 units2

Area 4BCD =BC × CD

2=

4× 4

2=

16

2= 8 units2

Area 4DEF =DE × EF

2=

3× 7

2=

21

2= 10.5 units2

Area 4FGH =FG×GH

2=

2× 2

2=

4

2= 2 units2

Therefore the area of the irregular shape is 63− 12.5− 8− 10.5− 2 = 63− 33 = 30 units2.


Problem of the Week

Problem C

Be Mine Valentine

A valentine is constructed by pasting two white semi-circles, each withradius 3 cm, and a white triangle onto a 12 cm square sheet of red paper asshown below. (The dashed line and the right angle symbols will not actually beon the finished card.)

You are going to write your valentine a message in red ink on the white regionof the card.

Determine the total amount of area available for your special valentine greeting.


Problem of the Week


Be Mine Valentine

Problem

A valentine is constructed by pasting two white semi-circles, each withradius 3 cm, and a white triangle onto a 12 cm square sheet of red paper. Youare going to write your valentine a message in red ink on the white region ofthe card. Determine the total amount of area available for your specialvalentine greeting.

Solution

Place the given information on the diagram.

12 cm

3 cm

12 cm

The total area for writing the message is the area of the two semi-circles plusthe area of the white triangle.

Since there are two semi-circles of radius 3 cm, the total area is the same as thearea of a full circle of radius 3 cm. The area of the two semi-circles isπr2 = π(3)2 = 9π cm2.

The height of the triangle is the length of the square minus the radius of thesemi-circle. Therefore the height of the triangle is 12− 3 = 9 cm. The base ofthe triangle is 12 cm, the width of the square. The area of the triangle is12base× height = 1

2(12)(9) = 54 cm2.

The total area for writing the message is (9π + 54) cm2. This area isapproximately 82.3 cm2. Happy Valentine’s Day.


Problem of the Week

Problem C

Fund Raising - One Step at A Time

A temporary set of stairs has been constructed to get dignitaries on and off astage at a fundraising event. The steps are of equal depth and equal height.The entire structure is 120 cm high, 120 cm from front to back and 100 cmwide.

The stairs are to be carpeted and the two sides are to be painted. One of thetwo sides that is to be painted is shaded in the diagram. The back and bottomof the structure will not be seen and will not be carpeted or painted.

120 cm

120 cm

100 cm

Determine the total area to be carpeted and the total area to be painted.


Problem of the Week


Fund Raising - One Step at A Time

Problem

A temporary set of stairs has been constructed to get dignitaries on and off a stage at afundraising event. The steps are of equal depth and equal height. The entire structure is120 cm high, 120 cm from front to back and 100 cm wide. The stairs are to be carpeted andthe two sides are to be painted. One of the two sides that is to be painted is shaded in thediagram. The back and bottom of the structure will not be seen and will not be carpeted orpainted. Determine the total area to be carpeted and the total area to be painted.

Solution

In order to solve both parts of the problem, the depth and height of each individual step mustbe calculated. The entire structure is 120 cm from front to back and four steps cover the entiredepth. Therefore, each step is 120 cm ÷ 4 = 30 cm wide and high.

The area to carpet is made up of 8 identical rectangles, each 30 cm wide and 100 cm long.Using the formula Area = Length×Width, the area of one rectangle is 30 × 100 = 3 000 cm2.The area of all surfaces to be carpeted is 8 × 3 000 = 24 000 cm2.

There are many different ways to find the area of the side pieces. Onesolution would be to break the figure into four equal width rectangles,one four steps high, one three steps high, one two steps high and thefinal rectangle one step high.

The area of one side would then be30 × (4 × 30) + 30 × (3 × 30) + 30 × (2 × 30) + 30 × (1 × 30)= 30× 120 + 30× 90 + 30× 60 + 30× 30 = 3 600 + 2 700 + 1 800 + 900= 9 000 cm2.

The total area of the two sides to be painted is 2×9 000 = 18 000 cm2.120 cm

120 cm

100 cm

A second method to calculate the area of one side involves breakingthe figure into triangles. Draw a diagonal from the top left cornerto the bottom right corner. This diagonal line would hit the bottomcorner of each step as shown in the diagram. The larger triangle hasa base and height of 120 cm. Each of the four smaller triangles has abase and a height of 30 cm, the width and height of each step.

Using the formula Area = Base×Height÷ 2, the area= 120 × 120 ÷ 2 + 4 × 30 × 30 ÷ 2 = 7 200 + 1 800 = 9 000 cm2.The total area of the two sides to be painted is then2 × 9 000 = 18 000 cm2.

120 cm

120 cm

100 cm

∴ the total area to carpet is 24 000 cm2 and the total area to paint is 18 000 cm2.


Problem of the Week

Problem C

Wired

A piece of wire 60 cm in length is to be cut into two parts in the ratio 3 : 2.Each part is bent to form a square.

Determine the ratio of the area of the larger square to the smaller square.


Problem of the Week


Wired

Problem

A piece of wire 60 cm in length is to be cut into two parts in the ratio 3 : 2. Each part is bentto form a square. Determine the ratio of the area of the larger square to the smaller square.

Solution

Let the length of the longer piece of wire be 3x cm and the length of the shorter piece of wirebe 2x cm. Then 3x + 2x = 60 or 5x = 60 and x = 12 follows.

Then the longer piece of wire is 3x = 3(12) = 36 cm and the smaller piece of wire is2x = 2(12) = 24 cm. These two lengths correspond to the perimeters of the respective squares.

Each of the wires is bent to form a square. The length of each side of the square is theperimeter of the square divided by 4. Therefore the side length of the larger square is36 ÷ 4 = 9 cm and the side length of the smaller square is 24 ÷ 4 = 6 cm.

The area of a square is calculated by squaring the side length. The area of the larger square is92 = 81 cm2 and the area of the smaller square is 62 = 36 cm2.

The ratio of the area of the larger square to the area of the smaller square is 81 : 36. This ratiocan be simplified by dividing each term by 9. The ratio in simplified form can then be writtenas 9 : 4.

Therefore the ratio of the area of the larger square to the area of the smaller square is 9 : 4.

An observation:

The ratio of the area of the larger square to the area of the smaller square is 9 : 4 = 32 : 22. Isit a coincidence that the ratio of the area of the larger square to the area of the smaller squareis equal to the squares of each term in the given ratio?

Also notice that the ratio of the area of the larger square to the area of the smaller square isequal to the ratio of the square of the perimeter of the larger square to the square of theperimeter of the smaller square. In this case, the perimeter of the larger square is 36 cm andthe perimeter of the smaller square is 24 cm. Then 362 : 242 = 1296 : 576 = 1296

144: 576144

= 9 : 4.

It is left to the solver to see if these two results are true in general.


Problem of the Week

Problem C

Tunnel Vision

A train 1000 metres long travels through a 3000 metre tunnel. Thirty secondspass from the time the last car has just completely entered the tunnel until thetime when the front of the engine emerges from the other end.

Determine the speed of the train, in kilometres per hour.


Problem of the Week


Tunnel Vision

Problem

A train 1000 metres long travels through a 3000 metre tunnel. Thirty secondspass from the time the last car has just completely entered the tunnel until thetime when the front of the engine emerges from the other end. Determine thespeed of the train, in kilometres per hour.

Solution

A diagram to represent the problem will make it very easy to visualize.

3000 m tunnel

1000 m train

At the time the entire train is just inside the tunnel, there is3000 − 1000 = 2000 metres left to travel. The engine has to travel 2000 metresin 30 seconds. We can calculate the speed of the train by dividing the distancetravelled by the time required to travel the distance.

The speed of the train is 2000 m ÷ 30 seconds =200

3m/s.

Now our task is to convert from m/s to km/h. We will do this in two steps:first convert metres to kilometres and then convert seconds to hours.

(1)200 m

3 s=

200 m

3 s× 1 km

1000 m=

200 km

3000 s=

1 km

15 s

(2)1 km

15 s=

1 km

15 s× 60 s

1 min× 60 min

1 h=

3600 km

15 h=

240 km

1 h

The train is travelling at a speed of 240 km/h.


Problem of the Week

Problem C


A contractor is building new houses. Each house has a large window withwidth 2 feet and height 6 feet to be made up of six identical 1 foot by 2 feetglass panes.

The contractor wants each house to look slightly different. The window size isfixed, but he plans on arranging the glass panes in the windows so that no twowindows have the same configuration. Two possible arrangements of the glasspanes are given below.

Assuming that the glass panes cannot be cut, how many differentarrangements can be made?


Problem of the Week



Problem

A contractor is building new houses. Each house has a large window with width 2 feet andheight 6 feet to be made up of six identical 1 foot by 2 feet glass panes. The contractor wantseach house to look slightly different. The window size is fixed, but he plans on arranging theglass panes in the windows so that no two windows have the same configuration. Assumingthat the glass panes cannot be cut, how many different arrangements can be made?

Solution

Let’s consider the ways that the glass panes can be arranged. First, notice that there mustalways be an even number of window panes that have a vertical orientation (standing on end).

• All glass panes are horizontal (and none are vertical)


• Four glass panes are horizontal and two are vertical

There could be 0, 1, 2, 3 or 4 horizontal panes below the two vertical panes. So there are5 ways that four glass panes are horizontal and two are vertical.

• Two glass panes are horizontal and four are vertical

We need to consider sub cases:• Case 1: There are no horizontal panes between the vertical panes.

There could be 0, 1 or 2 horizontal panes below the four vertical panes. So there are3 ways that two glass panes are horizontal, four are vertical and there are nohorizontal panes between the vertical panes.

• Case 2: There is one horizontal pane between the vertical panes.There could be 0 or 1 horizontal panes below the bottom two vertical panes. Sothere are 2 ways that two glass panes are horizontal, four are vertical and there isone horizontal pane between the vertical panes.

• Case 3: There are two horizontal panes between the vertical panes.There cannot be any horizontal panes below the bottom two vertical panes. Sothere is 1 way that two glass panes are horizontal, four are vertical and there aretwo horizontal panes between the vertical panes.

• All glass panes are vertical


Therefore, the total number of different configurations of the glass panes is1 + 5 + (3 + 2 + 1) + 1 = 13.

So the contractor can build 13 houses before he has to start duplicating window patterns.


Problem of the Week

Problem C

Pushy Pushy

Manny Kerr runs a company that specializes in grooming and caring for lawns.Manny has a push mower and powerful riding mower.

At one location it takes him five hours to cut the entire lawn with the pushmower but only 70 minutes with the powerful riding mower. After 90% of thelawn was cut using the powerful riding mower, the remainder was cut using thepush mower.

How many minutes did it take Manny to cut the entire lawn?


Problem of the Week


Pushy Pushy

Problem

Manny Kerr runs a company that specializes in grooming and caring for lawns.Manny has a push mower and powerful riding mower. At one location it takeshim five hours to cut the entire lawn with the push mower but only 70 minuteswith the powerful riding mower. After 90% of the lawn was cut using thepowerful riding mower, the remainder was cut using the push mower. Howmany minutes did it take Manny to cut the entire lawn.

Solution

It takes Manny 70 minutes to cut 100% of the lawn with the powerful ridingmower. It would take him 90% of 70 minutes or 0.90 × 70 = 63 minutes to cut90% of the lawn with the powerful riding mower.

Since he cuts 90% with the powerful riding mower, he cuts 100% − 90% = 10%using the push mower. It takes him 5 hours or 5 × 60 = 300 minutes to cut100% of the lawn with the push mower. It would then take him0.10 × 300 = 30 minutes to cut 10% of the lawn with the push mower.

It would then take him a total of 63 + 30 = 93 minutes to cut the entire lawnusing the powerful riding mower for 90% of the job and the push mower for10% of the job.


Problem of the Week

Problem C


!

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Problem of the Week



Problem


!

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!"#$%Solution




144 = 12× AC ÷ 2

144 = 6× AC

24 cm = AC






(AB)× (BF )÷ 2 = (DE)× (EF )÷ 2

12×BF ÷ 2 = 12× FE ÷ 2

6×BF = 6× FE

∴ BF = FE



= 144− 36

= 108 cm2



Problem of the Week

Problem C

Wine Wine Wine

Vino is a winemaker. One day he filled a 45 litre container with wine. Heremoved 9 litres of wine and replaced it with 9 litres of water. Next heremoved 9 litres of the mixture and replaced it with 9 litres of water.

Determine the ratio of wine to water in Vino’s final mixture.


Problem of the Week


Wine Wine Wine

Problem

Vino is a winemaker. One day he filled a 45 litre container with wine. Heremoved 9 litres of wine and replaced it with 9 litres of water. Next heremoved 9 litres of the mixture and replaced it with 9 litres of water.Determine the ratio of wine to water in Vino’s final mixture.

Solution

We need to determine the amount of wine and the amount of water in the finalmixture.

Vino starts with 45 litres of wine and no water. After removing 9 litres of wineand adding 9 litres of water, he has 45 − 9 = 36 litres of wine and 9 litres of

water. So36

45=

4

5of the new mixture is wine and

9

45=

1

5of the new mixture is

water.

He then removes 9 litres of the new mixture,4

5of which is wine and

1

5of which

is water. So Vino removes4

5× 9 =

36

5or 7

1

5litres of wine and

1

5× 9 =

9

5or 1

4

5litres of water.

Before adding another 9 litres of water he has 36 − 71

5=

180

5− 36

5=

144

5

or 284

5litres of wine and 9 − 1

4

5=

45

5− 9

5=

36

5or 7

1

5litres of water.

After adding the additional water he has 9 + 71

5= 9 +

36

5=

45

5+

36

5=

81

5or

161

5litres of water.

The final ratio of wine to water is 284

5: 16

1

5=

144

5:

81

5= 144 : 81 = 16 : 9.

∴ the final ratio of wine to water is 16:9.


Problem of the Week

Problem C


In the expressiona

b+

c

d+

e

f, each letter is replaced by a different digit from

1,2,3,4,5,6. Each digit can be used exactly once.

What is the largest possible value of this expression?

! !!

"

#

$

%

&


Problem of the Week



Problem

In the expressiona

b+

c

d+

e

f, each letter is replaced by a different digit from 1,2,3,4,5,6. Each

digit can be used exactly once. What is the largest possible value of this expression?

Solution 1

The largest fractions will be created by putting the three smallest numbers, 1,2, and 3, in the denominators and then placing the numbers 4, 5, and 6 in thenumerators. We will do this in every possible way, determine the sums andchoose the largest.

There are six different possible sums in which 1, 2, and 3 are the denominatorsand 4, 5, 6 are the numerators.

(1)4

1+

5

2+

6

3=

24

6+

15

6+

12

6=

51

6

(2)4

1+

6

2+

5

3=

24

6+

18

6+

10

6=

52

6

(3)5

1+

4

2+

6

3=

30

6+

12

6+

12

6=

54

6

(4)5

1+

6

2+

4

3=

30

6+

18

6+

8

6=

56

6

(5)6

1+

4

2+

5

3=

36

6+

12

6+

10

6=

58

6

(6)6

1+

5

2+

4

3=

36

6+

15

6+

8

6=

59

6

Therefore the largest possible value of the expression is59

6. It should be noted

that this approach would not be practical if more numbers were involved. Besure to look at solution 2 for a more logical approach.


Solution 2

We can start by observing that to get a fraction with the highest value we needa 6 in the numerator. The choice of denominators is possibly obvious as well.61 = 6, 6

2 = 3, 63 = 2, 6

4 = 1.5, and 65 = 1.2. 6

1 is the largest fraction and anynumerator other than 6 will produce a lower value.

Now we have four numbers left to place: {2,3,4,5}.

Of these remaining numbers, since 5 is the largest it should go in thenumerator. Then 5

2 = 2.5, 53

.= 1.7, and 5

4 = 1.25. 52 is the largest fraction and

any numerator other than 5 will produce a lower value.

Now we have two numbers left to place: {3,4}.

Our only two choices for the third fraction are 43

.= 1.3, and 3

4 = 0.75. So 43 is

the third fraction.

We can now determine the largest possible sum.

Largest Possible Sum =6

1+

5

2+

4

3

=36

6+

15

6+

8

6

=59

6or 9

5

6

∴ the largest possible sum that can be made from the numbers 1, 2, 3, 4, 5,

and 6 in the expressiona

b+

c

d+

e

fis 59

6 or 956 (approximately 9.83).


Problem of the Week

Problem C


A (0,3)

B (4,0)

C (7,4)

x

y





Problem of the Week



Problem




A (0,3)

B (4,0)

C (7,4)

D (3,7)

x

y

Solution


B (4,0)

C (7,4)

D (3,7)

x

y

A (0,3)





= 7 × 7 − 4 × (4 × 3 ÷ 2)

= 49 − 4 × 6

= 49 − 24

= 25 units2



B (4,0)

C (7,4)

D (3,7)

x

y

A (0,3)

O (0,0)





AB2 = OA2 + OB2

= 32 + 42

= 9 + 16

= 25



Problem of the Week

Problem C

Stack ’Em Up




Problem of the Week


Stack ’Em Up

Problem


Solution








Problem of the Week

Problem C


! "

#

$

%





Problem of the Week



Problem


#

$

%

Solution



2

=6× 16

2

= 48 cm2


2

48 =(BE)× 12

2

48 = 6×BE

8 cm = BE



Problem of the Week

Problem C

Pressure’s On

Willy Makit is participating at the local level of a competition in which hecompetes in 10 events. In each event the maximum possible score is 10 points.To advance to the regional level a competitor must earn a minimum total scoreof 75 points.

On the first five events, Willy had an average score of 6.8. On the next threeevents his average score was 7.2.

What must Willy average on his final two events in order to have a total scoreof exactly 75 points?


Problem of the Week


Pressure’s On

Problem

Willy Makit is participating at the local level of a competition in which hecompetes in 10 events. In each event the maximum possible score is 10 points.To advance to the regional level a competitor must earn a minimum total scoreof 75 points. On the first five events, Willy had an average score of 6.8. On thenext three events his average score was 7.2. What must Willy average on hisfinal two events in order to have a total score of exactly 75 points?

Solution

To determine an average, we add the numbers together and divide the total bythe number of numbers.

Average =Total

Number of Numbers

So to determine the total we would multiply the average by the number ofnumbers.

Total = Average × Number of Numbers

Total Score for First 5 Events = 6.8 × 5 = 34.0

Total Score for Next 3 Events = 7.2 × 3 = 21.6

Total Score to Move On = 75.0

Total Score Needed for Last Two Events = 75.0 − 34.0 − 21.6 = 19.4

Average Score for Final 2 Events = 19.4 ÷ 2 = 9.7

Willy needs to average 9.7, near perfect scores, on his last two events to have atotal score of exactly 75 points in order to advance to the regional round of thecompetition. The pressure is definitely on!


Problem of the Week

Problem C

Prime Picking

A natural number greater than 1 is said to be prime if its only factors are 1 anditself. For example, the number 7 is prime since its only factors are 1 and 7.

A perfect square is an integer created by multiplying an integer by itself. Thenumber 25 is a perfect square since it is 5 × 5 or 52.

Determine the smallest perfect square that has three different prime numbersas factors.


Problem of the Week


Prime Picking

Problem

A natural number greater than 1 is said to be prime if its only factors are 1and itself. For example, the number 7 is prime since its only factors are 1 and7. A perfect square is an integer created by multiplying an integer by itself.The number 25 is a perfect square since it is 5 × 5 or 52. Determine thesmallest perfect square that has three different prime numbers as factors.

Solution

The problem itself is not very difficult once you determine what it is asking. Sobefore looking at the solution, let us examine some perfect squares.

The numbers 4 and 9 are both perfect squares that have only one primenumber as a factor, 4 = 22 and 9 = 32.

The number 36 is a perfect square since 36 = 62. However, the number6 = 2× 3 so 36 = (2× 3)2 = 2× 3× 2× 3 = 22 × 32. So 36 is the product of thesquare of each of two different prime numbers. In fact, since 2 and 3 are thesmallest prime numbers, 36 is the smallest perfect square that has two differentprime factors. In order to create a perfect square we must find the product ofthe squares of prime numbers.

Using this idea, we can create the smallest perfect square with three differentprime factors by squaring each of the three smallest primes, 2, 3, and 5, andthen multiplying these squares together. So the smallest perfect square thatuses three different prime factors would be 22 × 32 × 52 = 4 × 9 × 25 = 900. Itshould be noted that 900 = 302 = (2 × 3 × 5)2.

∴ the smallest perfect square with three different prime factors is 900.


Problem of the Week

Problem C

Powerful Factorials At Work!

The product of the positive integers 1 to 3 is 3× 2× 1 = 6 and can be writtenin an abbreviated form as 3!. We say 3 factorial. So 3! = 6.

The product of the positive integers 1 to 17 is 17× 16× 15× · · · × 3× 2× 1and can be written in an abbreviated form as 17!. We say 17 factorial.The · · · represents the product of all the missing integers between 15 and 3.

In general, the product of the positive integers 1 to n is n!. Note that 1! = 1.

Determine the units digit in the sum 1! + 2! + 3! + · · · + 18! + 19! + 20!.


Problem of the Week



Problem

The product of the positive integers 1 to 3 is 3 × 2 × 1 = 6 and can be writtenin an abbreviated form as 3!. We say 3 factorial. So 3! = 6. The product ofthe positive integers 1 to 17 is 17× 16× 15× · · · × 3× 2× 1 and can be writtenin an abbreviated form as 17!. We say 17 factorial. The · · · represents theproduct of all the missing integers between 15 and 3. In general, the product ofthe positive integers 1 to n is n!. Note that 1! = 1. Determine the units digit inthe sum 1! + 2! + 3! + · · · + 18! + 19! + 20!.

Solution

At first glance it may look as if there is a great deal of work to do. However byexamining several factorials we will discover otherwise.

1! = 1

2! = 2 × 1 = 2

3! = 3 × 2 × 1 = 6

4! = 4 × 3 × 2 × 1 = 24

5! = 5 × 4 × 3 × 2 × 1 = 120

Now 6! = 6 × (5 × 4 × 3 × 2 × 1) = 6 × 5! = 6(120) = 720.And 7! = 7 × (6 × 5 × 4 × 3 × 2 × 1) = 7 × 6! = 7(720) = 5040.An interesting observation surfaces, 8! = 8 × 7!, 9! = 9 × 8!, and so on.

Furthermore, 5! has a units digit of 0. Every factorial above 5! will also have aunits digit of 0 since multiplying a number by another number whose units digitis zero produces a zero in the units digit of the product. So no factorials above4! will have a units digit other than zero. Therefore, the units digit of the sum1! + 2! + 3! + 4! · · · + 18! + 19! + 20! will come from the units digit of 1!, 2!, 3!and 4!. Calculating the sum 1! + 2! + 3! + 4! we obtain 1 + 2 + 6 + 24 = 33.The units digit of the required sum is the units digit of 33, namely 3.

∴ the units digit of the sum 1! + 2! + 3! · · · + 18! + 19! + 20! is 3.


Problem of the Week

Problem C

Puzzling Product

In the product shown below, the letters F and L represent different digits from1 to 9. Determine the value of F and L.

F 8× 3 L

2 7 3 0


Problem of the Week


Puzzling Product

Problem

In the product shown below, the letters F and L represent different digits from1 to 9. Determine the value of F and L.

F 8× 3 L

2 7 3 0

Solution

In a multiplication question there are three parts: the multiplier, multiplicand

and product. In our problem, F8 is the multiplier, 3L is the multiplicand, and2730 is the product.

The units digit of the product 2730 is 0. The units digit of a product is equalto the units digit of the result obtained by multiplying the units digit of themultiplier and multiplicand.

So 8× L must equal a number with units digit 0. The only choices for L are 0and 5 since no other single digit times 8 produces a number ending in zero.

However, if L = 0, the units digit of the product is 0 and the remaining threedigits of the product are produced by multiplying 3×F8. But 3×F8 producesa number ending in 4, not 3 as required. Therefore L 6= 0 and L must equal 5.

We are then multiplying F8 by 35 to create the product 2730. To determine Fwe can work backwards. When we divide 2730 by 35, the quotient is 78. But78 and F8 are the same number so F = 7.

∴ F = 7 and L = 5.


Problem of the Week

Problem C

Tunnel Vision

A train 1000 metres long travels through a 3000 metre tunnel. Thirty secondspass from the time the last car has just completely entered the tunnel until thetime when the front of the engine emerges from the other end.

Determine the speed of the train, in kilometres per hour.


Problem of the Week


Tunnel Vision

Problem

A train 1000 metres long travels through a 3000 metre tunnel. Thirty secondspass from the time the last car has just completely entered the tunnel until thetime when the front of the engine emerges from the other end. Determine thespeed of the train, in kilometres per hour.

Solution

A diagram to represent the problem will make it very easy to visualize.

3000 m tunnel

1000 m train

At the time the entire train is just inside the tunnel, there is3000 − 1000 = 2000 metres left to travel. The engine has to travel 2000 metresin 30 seconds. We can calculate the speed of the train by dividing the distancetravelled by the time required to travel the distance.

The speed of the train is 2000 m ÷ 30 seconds =200

3m/s.

Now our task is to convert from m/s to km/h. We will do this in two steps:first convert metres to kilometres and then convert seconds to hours.

(1)200 m

3 s=

200 m

3 s× 1 km

1000 m=

200 km

3000 s=

1 km

15 s

(2)1 km

15 s=

1 km

15 s× 60 s

1 min× 60 min

1 h=

3600 km

15 h=

240 km

1 h

The train is travelling at a speed of 240 km/h.


Problem of the Week

Problem C

Coin Collection

Mai (pronounced “my”) Zur has been saving quarters, loonies and toonies forseveral years in her piggy bank. No other types of coins are in her bank. Aloonie is a Canadian one dollar coin and the toonie is a Canadian two dollarcoin.

One third of the coins in the bank are quarters and one fifth of the coins areloonies. There are 14 toonies in the bank.

Determine how much money Mai Zur has saved in her piggy bank.


Problem of the Week


Coin Collection

Problem

Mai (pronounced “my”) Zur has been saving quarters, loonies and toonies for several years inher piggy bank. No other types of coins are in her bank. A loonie is a Canadian one dollar coinand the toonie is a Canadian two dollar coin. One third of the coins in the bank are quartersand one fifth of the coins are loonies. There are 14 toonies in the bank. Determine how muchmoney Mai Zur has saved in her piggy bank.

Solution

One of the key sentences in the problem is “No other types of coins are in her bank.” Using thefractions given it will be possible to determine what fraction of the whole is made up bytoonies.

The fraction of toonies = 1 − 1

3− 1

5=

15

15− 5

15− 3

15=

7

15.

We can now determine the total number of coins in the bank. Since 14 toonies are in the bankand 7

15of the coins are toonies,

7

15of the coins = 14 coins.

Dividing by 7,1

15of the coins = 14 ÷ 7 = 2 coins.

Multiplying by 15,15

15of the coins = 2 × 15 = 30 coins.

There are 30 coins in the bank. We can now determine the number of quarters and loonies.

The number of quarters =1

3× 30 = 10 and the number of loonies =

1

5× 30 = 6.

To determine the amount of money in the bank we multiply the value of a particular coin bythe quantity of that coin and add the three values together.

Amount in the Bank = Value of Quarters + Value of Loonies + Value of Toonies

= $0.25 × 10 + $1.00 × 6 + $2 × 14

= $2.50 + $6.00 + $28.00

= $36.50

∴ Mai Zur has a total of $36.50 in her bank.


Problem of the Week

Problem C

More for Less

Harry Wirks is currently paid $567 for working a 45 hour week. His weeklysalary is to be increased by 10% but his hours are to be reduced by 10%.

Calculate the change from his old hourly rate of pay to his new hourly rate ofpay.

! !


Problem of the Week


More for Less

Problem

Harry Wirks is currently paid $567 for working a 45 hour week. His weekly salary is to beincreased by 10% but his hours are to be reduced by 10%. Calculate the change from his oldhourly rate of pay to his new hourly rate of pay.

Solution 1

To calculate the hourly rate of pay divide the weekly salary by the number of hours worked.

Harry’s old hourly rate of pay is $567÷ 45 h = $12.60/h.

New Weekly Salary = Old Weekly Salary + 10% of Old Weekly Salary

= $567 + 0.1× $567

= $567 + $56.70

= $623.70

New Number of Hours Worked = Old Hours Worked− 10% of Old Hours Worked

= 45 h− 0.1× 45 h

= 45 h− 4.5 h

= 40.5 h

Harry’s new hourly rate of pay is $623.70÷ 40.5 h = $15.40/h.

The change in his hourly rate of pay is $15.40/h − $12.60/h = $2.80/h.

∴ Harry’s hourly rate increased $2.80/h.

Solution 2

In the second solution we will use a more concise calculation. Harry’s weekly salary is 10%more than his old weekly salary. So Harry earns 110% of his old weekly salary. Harry’s hoursare reduced by 10% of his old hours so he now works 90% of his old hours. To calculate hischange in hourly rate we can take his new hourly rate and subtract his old hourly rate.

Change in Hourly Rate = New Hourly Rate − Old Hourly Rate

= New Salary÷ Hours Worked−Old Salary÷Old Hours Worked

= ($567× 1.10)÷ (45× 0.9)− $567÷ 45

= $623.70÷ 40.5− $567÷ 45

= $15.40/h− $12.60/h

= $2.80/h



Problem of the Week

Problem C

Creating Order

Each of the digits 1, 2, 3, 4 are used once to create a four-digit number.Twenty-four different four-digit numbers can be formed this way. If thesetwenty-four numbers are then listed from the smallest to the largest, in whatposition is the number 3 142?

◦◦◦

◦◦◦


Problem of the Week


Creating Order

Problem


Solution

We could list all twenty-four numbers and then arrange them in order to findthe position of 3 142, but there is an easier way to solve this problem.

The number 3 142 will be in the list after all the numbers that begin with a 1and after all the numbers that begin with a 2.

How many of the numbers in the list begin with a 1?

There are 24 four-digit numbers that can be made from the digits 1, 2, 3 and 4.There are 4 possibilities for the first digit so there are 24 ÷ 4 = 6 numbers thatbegin with a 1. The numbers are 1 234, 1 243, 1 324, 1 342, 1 423 and 1 432.

Similarly, there are 6 numbers in the list that begin with a 2, 6 that begin witha 3, and 6 that begin with a 4. We could list all 6 of the numbers in the listthat begin with a 2, but this is not necessary.

So the first 6 numbers in the ordered list begin with a 1, the next 6 numbersbegin with a 2, and the following 6 numbers begin with a 3. There are 12numbers in the ordered list before the first number beginning with a 3.

In order, the numbers that begin with a 3 are 3 124, 3 142, 3 214, 3 241, 3 412,3 421. We see that 3 142 is the second number in the ordered list of numbersbeginning with a 3.

Therefore, the number 3 142 is in the fourteenth position in the ordered list.


Problem of the Week

Problem C

Pushy Pushy

Manny Kerr runs a company that specializes in grooming and caring for lawns.Manny has a push mower and powerful riding mower.

At one location it takes him five hours to cut the entire lawn with the pushmower but only 70 minutes with the powerful riding mower. After 90% of thelawn was cut using the powerful riding mower, the remainder was cut using thepush mower.

How many minutes did it take Manny to cut the entire lawn?


Problem of the Week


Pushy Pushy

Problem

Manny Kerr runs a company that specializes in grooming and caring for lawns.Manny has a push mower and powerful riding mower. At one location it takeshim five hours to cut the entire lawn with the push mower but only 70 minuteswith the powerful riding mower. After 90% of the lawn was cut using thepowerful riding mower, the remainder was cut using the push mower. Howmany minutes did it take Manny to cut the entire lawn.

Solution

It takes Manny 70 minutes to cut 100% of the lawn with the powerful ridingmower. It would take him 90% of 70 minutes or 0.90 × 70 = 63 minutes to cut90% of the lawn with the powerful riding mower.

Since he cuts 90% with the powerful riding mower, he cuts 100% − 90% = 10%using the push mower. It takes him 5 hours or 5 × 60 = 300 minutes to cut100% of the lawn with the push mower. It would then take him0.10 × 300 = 30 minutes to cut 10% of the lawn with the push mower.

It would then take him a total of 63 + 30 = 93 minutes to cut the entire lawnusing the powerful riding mower for 90% of the job and the push mower for10% of the job.


Problem of the Week

Problem C


!

"

# $

%&




Problem of the Week



Problem


!

"

# $

%&

!"#$%Solution




144 = 12× AC ÷ 2

144 = 6× AC

24 cm = AC






(AB)× (BF )÷ 2 = (DE)× (EF )÷ 2

12×BF ÷ 2 = 12× FE ÷ 2

6×BF = 6× FE

∴ BF = FE



= 144− 36

= 108 cm2



Problem of the Week

Problem C


!"

!

"

# $

In 4ABD, C is on AB such that AC = CB = CD and ∠BCD = 70◦.

Determine the measure of ∠ADB.

If you want to solve a more general version of this problem, consider solvingthis week’s Grade 9 and 10 Problem of the Week “What’s Your Angle AnywayII?”.

If you want to further extend this problem, consider solving this week’s Grade11 and 12 Problem of the Week “What’s Your Angle Anyway III?”.


Problem of the Week



Problem

In 4ABD, C is on AB such that AC = CB = CD and∠BCD = 70◦. Determine the measure of ∠ADB.

Solution

Since ACB is a straight line, ∠ACD + ∠DCB = 180◦

but ∠DCB = 70◦ so ∠ACD = 110◦.

In 4ACD, since AC = CD, 4ACD is isosceles and∠CAD = ∠CDA = x◦.

!"

!

"

# $

#

#

$

$

The angles in a triangle sum to 180◦ so in 4ACD

∠CAD + ∠CDA + ∠ACD = 180◦

x◦ + x◦ + 110◦ = 180◦

2x = 70

x = 35

Similarly, in 4BCD, since CB = CD, 4BCD is isosceles and∠CBD = ∠CDB = y◦.

The angles in a triangle sum to 180◦ so in 4CBD

∠CBD + ∠CDB + ∠BCD = 180◦

y◦ + y◦ + 70◦ = 180◦

2y = 110

y = 55

Then ∠ADB = ∠CDA + ∠CDB = x◦ + y◦ = 35◦ + 55◦ = 90◦.


See the next page for an interesting idea.


It turns out that it is not necessary to find the angles in the problem.

!"

!

"

# $

#

#

$

$

Here is a second solution to the problem.

In 4CAD, since CA = CD, 4CAD is isosceles and ∠CAD = ∠CDA = x◦.

In 4CBD, since CB = CD, 4CBD is isosceles and ∠CBD = ∠CDB = y◦.

The angles in a triangle sum to 180◦ so in 4ABD

∠BAD + ∠ADB + ∠ABD = 180◦

x◦ + (x◦ + y◦) + y◦ = 180◦

2x◦ + 2y◦ = 180◦

x◦ + y◦ = 90◦

But ∠ADB = ∠ADC + ∠CDB = x◦ + y◦ = 90◦.



Problem of the Week

Problem C

Can You DIGIT?

The three digit number 5A4 is divisible by 4 and the three digit number 37B isdivisible by 3.

Determine the largest positive difference between 5A4 and 37B.


Problem of the Week


Can You DIGIT?

Problem

The three digit number 5A4 is divisible by 4 and the three digit number 37B isdivisible by 3. Determine the largest positive difference between 5A4 and 37B.

Solution

The largest positive difference will occur when 5A4 is a large as possible and37B is as small as possible. We are therefore looking for the largest possiblevalue of A and the smallest possible value of B.

A number is divisible by 4 if the last two digits of the number are divisibleby 4. For 5A4 to be divisible by 4, the only possible values of A are 0, 2, 4, 6,and 8 since 04, 24, 44, 64, and 84, respectively, are the only two digit numbersthat end in 4 and are divisible by 4. (04 is technically not a two-digit numberbut a larger number could end in these two digits.) Since we want 5A4 to be aslarge as possible, A = 8 and 5A4 = 584.

A number is divisible by 3 if the sum of its digits is divisible by 3. For 37B tobe divisible by 3, the only possible values of B are 2, 5, and 8.

• When B = 2 the number is 372. It is divisible by 3 since 3 + 7 + 2 = 12 isdivisible by 3.• When B = 5 the number is 375. It is divisible by 3 since 3 + 7 + 5 = 15 isdivisible by 3.• When B = 8 the number is 378. It is divisible by 3 since 3 + 7 + 8 = 18 isdivisible by 3.

No other three digit numbers of the form 37B will be divisible by 3 since noother values of B, other than 2, 5, and 8, give a digit sum that is divisible by 3.Since we want 37B to be as small as possible, B = 2 and 37B = 372.

The largest positive difference occurs when A = 8 and B = 2 so that5A4 − 37B = 584 − 372 = 212.

∴ the largest positive difference is 212.


Problem of the Week

Problem C


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#

$

%





Problem of the Week



Problem


#

$

%

Solution



2

=6× 16

2

= 48 cm2


2

48 =(BE)× 12

2

48 = 6×BE

8 cm = BE



Problem of the Week

Problem C

Prime Picking

A natural number greater than 1 is said to be prime if its only factors are 1 anditself. For example, the number 7 is prime since its only factors are 1 and 7.

A perfect square is an integer created by multiplying an integer by itself. Thenumber 25 is a perfect square since it is 5 × 5 or 52.

Determine the smallest perfect square that has three different prime numbersas factors.


Problem of the Week


Prime Picking

Problem

A natural number greater than 1 is said to be prime if its only factors are 1and itself. For example, the number 7 is prime since its only factors are 1 and7. A perfect square is an integer created by multiplying an integer by itself.The number 25 is a perfect square since it is 5 × 5 or 52. Determine thesmallest perfect square that has three different prime numbers as factors.

Solution

The problem itself is not very difficult once you determine what it is asking. Sobefore looking at the solution, let us examine some perfect squares.

The numbers 4 and 9 are both perfect squares that have only one primenumber as a factor, 4 = 22 and 9 = 32.

The number 36 is a perfect square since 36 = 62. However, the number6 = 2× 3 so 36 = (2× 3)2 = 2× 3× 2× 3 = 22 × 32. So 36 is the product of thesquare of each of two different prime numbers. In fact, since 2 and 3 are thesmallest prime numbers, 36 is the smallest perfect square that has two differentprime factors. In order to create a perfect square we must find the product ofthe squares of prime numbers.

Using this idea, we can create the smallest perfect square with three differentprime factors by squaring each of the three smallest primes, 2, 3, and 5, andthen multiplying these squares together. So the smallest perfect square thatuses three different prime factors would be 22 × 32 × 52 = 4 × 9 × 25 = 900. Itshould be noted that 900 = 302 = (2 × 3 × 5)2.

∴ the smallest perfect square with three different prime factors is 900.


Problem of the Week

Problem C


The product of the positive integers 1 to 3 is 3× 2× 1 = 6 and can be writtenin an abbreviated form as 3!. We say 3 factorial. So 3! = 6.

The product of the positive integers 1 to 17 is 17× 16× 15× · · · × 3× 2× 1and can be written in an abbreviated form as 17!. We say 17 factorial.The · · · represents the product of all the missing integers between 15 and 3.

In general, the product of the positive integers 1 to n is n!. Note that 1! = 1.

Determine the units digit in the sum 1! + 2! + 3! + · · · + 18! + 19! + 20!.


Problem of the Week



Problem

The product of the positive integers 1 to 3 is 3 × 2 × 1 = 6 and can be writtenin an abbreviated form as 3!. We say 3 factorial. So 3! = 6. The product ofthe positive integers 1 to 17 is 17× 16× 15× · · · × 3× 2× 1 and can be writtenin an abbreviated form as 17!. We say 17 factorial. The · · · represents theproduct of all the missing integers between 15 and 3. In general, the product ofthe positive integers 1 to n is n!. Note that 1! = 1. Determine the units digit inthe sum 1! + 2! + 3! + · · · + 18! + 19! + 20!.

Solution

At first glance it may look as if there is a great deal of work to do. However byexamining several factorials we will discover otherwise.

1! = 1

2! = 2 × 1 = 2

3! = 3 × 2 × 1 = 6

4! = 4 × 3 × 2 × 1 = 24

5! = 5 × 4 × 3 × 2 × 1 = 120

Now 6! = 6 × (5 × 4 × 3 × 2 × 1) = 6 × 5! = 6(120) = 720.And 7! = 7 × (6 × 5 × 4 × 3 × 2 × 1) = 7 × 6! = 7(720) = 5040.An interesting observation surfaces, 8! = 8 × 7!, 9! = 9 × 8!, and so on.

Furthermore, 5! has a units digit of 0. Every factorial above 5! will also have aunits digit of 0 since multiplying a number by another number whose units digitis zero produces a zero in the units digit of the product. So no factorials above4! will have a units digit other than zero. Therefore, the units digit of the sum1! + 2! + 3! + 4! · · · + 18! + 19! + 20! will come from the units digit of 1!, 2!, 3!and 4!. Calculating the sum 1! + 2! + 3! + 4! we obtain 1 + 2 + 6 + 24 = 33.The units digit of the required sum is the units digit of 33, namely 3.

∴ the units digit of the sum 1! + 2! + 3! · · · + 18! + 19! + 20! is 3.


Problem of the Week

Problem C

A Crafty Little Problem

A knot is tied near one end of a piece of string and beads are strung to form anecklace. The beads are placed on the string in the following sequence: 1 red, 1green, 1 blue, 2 red, 2 green, 2 blue, 3 red, 3 green, 3 blue, with the number ofeach colour increasing by one every time a new group of beads is placed. Thediagram illustrates how the first 18 beads are strung.

How many of the first 160 beads are blue?

R

RGGG

R

RR R

G

G G

B

BB

B

BB


Problem of the Week


A Crafty Little Problem

Problem

A knot is tied near one end of a piece of string and beads are strung to form anecklace. The beads are placed on the string in the following sequence: 1 red, 1green, 1 blue, 2 red, 2 green, 2 blue, 3 red, 3 green, 3 blue, with the number ofeach colour increasing by one every time a new group of beads is placed. Howmany of the first 160 beads are blue?

Solution

An equal number of red, green and blue beads occur after

3(1) = 3 beads are placed,

3(1) + 3(2) = 3 + 6 = 9 beads are placed,

3(1) + 3(2) + 3(3) = 3 + 6 + 9 = 18 beads are placed, and so on.

The greatest total that can be placed with equal numbers of red, green andblue beads is 3(1) + 3(2) + 3(3) + · · · + 3(9) = 3 + 6 + 9 + · · · + 27 = 135. Atthis point there are 135 ÷ 3 = 45 beads of each colour. We are at a point wherewe have strung 9 red beads, 9 green beads and 9 blue beads in the last threegroups.

The next groups will have 10 in them if there are enough beads. There are160− 135 = 25 beads left to place. We are able to place 10 red beads leaving 15beads left to place. At this point there are 55 red beads. We can place 10 greenbeads leaving 5 beads left to place. At this point there are 55 green beads. Thefive remaining beads must be blue making the total number of blue beads 50.

∴ there are 50 blue beads on the necklace.


Problem of the Week

Problem C

Wired

A piece of wire 60 cm in length is to be cut into two parts in the ratio 3 : 2.Each part is bent to form a square.

Determine the ratio of the area of the larger square to the smaller square.


Problem of the Week


Wired

Problem

A piece of wire 60 cm in length is to be cut into two parts in the ratio 3 : 2. Each part is bentto form a square. Determine the ratio of the area of the larger square to the smaller square.

Solution

Let the length of the longer piece of wire be 3x cm and the length of the shorter piece of wirebe 2x cm. Then 3x + 2x = 60 or 5x = 60 and x = 12 follows.

Then the longer piece of wire is 3x = 3(12) = 36 cm and the smaller piece of wire is2x = 2(12) = 24 cm. These two lengths correspond to the perimeters of the respective squares.

Each of the wires is bent to form a square. The length of each side of the square is theperimeter of the square divided by 4. Therefore the side length of the larger square is36 ÷ 4 = 9 cm and the side length of the smaller square is 24 ÷ 4 = 6 cm.

The area of a square is calculated by squaring the side length. The area of the larger square is92 = 81 cm2 and the area of the smaller square is 62 = 36 cm2.

The ratio of the area of the larger square to the area of the smaller square is 81 : 36. This ratiocan be simplified by dividing each term by 9. The ratio in simplified form can then be writtenas 9 : 4.

Therefore the ratio of the area of the larger square to the area of the smaller square is 9 : 4.

An observation:

The ratio of the area of the larger square to the area of the smaller square is 9 : 4 = 32 : 22. Isit a coincidence that the ratio of the area of the larger square to the area of the smaller squareis equal to the squares of each term in the given ratio?

Also notice that the ratio of the area of the larger square to the area of the smaller square isequal to the ratio of the square of the perimeter of the larger square to the square of theperimeter of the smaller square. In this case, the perimeter of the larger square is 36 cm andthe perimeter of the smaller square is 24 cm. Then 362 : 242 = 1296 : 576 = 1296

144: 576144

= 9 : 4.

It is left to the solver to see if these two results are true in general.


Problem of the Week

Problem C

At A Prime Age

A mother has three children. Each of their ages is a different prime number.The sum of their ages is 41 and the difference between two of their ages is 16.

Determine the ages of the three children.


Problem of the Week


At A Prime Age

Problem

A mother has three children. Each of their ages is a different prime number.The sum of their ages is 41 and the difference between two of their ages is 16.Determine the ages of the three children.

Solution

A prime number is any number with exactly two different positive factors, 1and the number itself. A composite number has at least three different positivefactors. The number one is neither prime nor composite.

We can start by listing all of the prime numbers less than 41. The possibleprimes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and 37. We could actuallyeliminate some of the higher primes from this list since there are three differentprimes in the sum.

Now we will look for all prime pairs from this list that differ by 16. The pairsinclude 3 and 19, 7 and 23, and 13 and 29. We will look at each of these pairsand determine the third number so that the sum of the three ages is 41.

For the pair 3 and 19, the sum is 22. The third age would be 41 − 22 = 19.The ages of the three children would be 3, 19, and 19 but all three childrenhave a different age. Therefore this solution is not acceptable.

For the pair, 7 and 23, the sum is 30. The third age would be 41 − 30 = 11.The ages of the three children would be 7, 11, and 23. Each of these numbersis a different positive prime. Therefore this is a possible solution. But are thereother solutions?

For the pair, 17 and 33, the sum is 50. The sum is already over 41 so this isnot a possible solution.

Therefore the only possible ages for the children are 7, 11 and 23.

For further thought: How would the problem change if the word prime wasremoved from the problem and replaced with “a positive integer”?


Problem of the Week

Problem C

Check the Perimeter

A median is a line segment drawn from the vertex of a triangle to themidpoint of the opposite side.

In 4ABC, a median is drawn from A meeting BC at M . The perimeter of4ABC is 24. The perimeter of 4ABM is 18. The perimeter of 4ACM is 16.

Determine the length of the median AM .

A

B CM


Problem of the Week


Check the Perimeter

Problem


A

B CM

t r

p q

m

Solution 1


The perimeter of 4ABM = t + p + m = 18.

The perimeter of 4ACM = q + r + m = 16.

The perimeter of 4ABC = t + p + q + r = 24.

Consider the perimeter of 4ABM plus the perimeter of 4ACM . This is equalto (t + p + m) + (q + r + m) = t + p + m + q + r + m = (t + p + q + r) + 2×m.

So, if we add the perimeter of 4ABM to the perimeter of 4ACM , we willobtain the perimeter of 4ABC plus two times the length of the median, m.

In other words, since the perimeter of 4ABM is 18, the perimeter of 4ACMis 16 and the perimeter of 4ABC is 24, we find that 18 + 16 = 24 + 2×m. Itfollows that 34 = 24 + 2×m and 2×m must be 10. Therefore m, the length ofthe median, is 5.

In solution 2, we take a more algebraic approach to solving the problem. Itinvolves more formal equation solving.


Problem


A

B CM

t r

p q

m

Solution 2


The perimeter of 4ABM = t + p + m = 18 or t + p = 18−m. (1)

The perimeter of 4ACM = q + r + m = 16 or q + r = 16−m. (2)

The perimeter of 4ABC = t + p + q + r = 24. (3)

We will now combine the results.

t + p = 18−m (1)

q + r = 16−m (2)

So t + p + q + r = 18−m + 16−m, by combining the two results.

But the perimeter is t + p + q + r = 24.

∴ 24 = 18−m + 16−m

24 = 34− 2m

24− 34 = 34− 34− 2m

−10 = −2m

−10

−2=−2m

−2

5 = m

The length of the median AM = m = 5.


Problem of the Week

Problem C

More for Less

Harry Wirks is currently paid $567 for working a 45 hour week. His weeklysalary is to be increased by 10% but his hours are to be reduced by 10%.

Calculate the change from his old hourly rate of pay to his new hourly rate ofpay.

! !


Problem of the Week


More for Less

Problem

Harry Wirks is currently paid $567 for working a 45 hour week. His weekly salary is to beincreased by 10% but his hours are to be reduced by 10%. Calculate the change from his oldhourly rate of pay to his new hourly rate of pay.

Solution 1

To calculate the hourly rate of pay divide the weekly salary by the number of hours worked.

Harry’s old hourly rate of pay is $567÷ 45 h = $12.60/h.

New Weekly Salary = Old Weekly Salary + 10% of Old Weekly Salary

= $567 + 0.1× $567

= $567 + $56.70

= $623.70

New Number of Hours Worked = Old Hours Worked− 10% of Old Hours Worked

= 45 h− 0.1× 45 h

= 45 h− 4.5 h

= 40.5 h

Harry’s new hourly rate of pay is $623.70÷ 40.5 h = $15.40/h.

The change in his hourly rate of pay is $15.40/h − $12.60/h = $2.80/h.


Solution 2

In the second solution we will use a more concise calculation. Harry’s weekly salary is 10%more than his old weekly salary. So Harry earns 110% of his old weekly salary. Harry’s hoursare reduced by 10% of his old hours so he now works 90% of his old hours. To calculate hischange in hourly rate we can take his new hourly rate and subtract his old hourly rate.

Change in Hourly Rate = New Hourly Rate − Old Hourly Rate

= New Salary÷ Hours Worked−Old Salary÷Old Hours Worked

= ($567× 1.10)÷ (45× 0.9)− $567÷ 45

= $623.70÷ 40.5− $567÷ 45

= $15.40/h− $12.60/h

= $2.80/h



Problem of the Week

Problem C

Creating Order


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Problem of the Week


Creating Order

Problem


Solution

We could list all twenty-four numbers and then arrange them in order to findthe position of 3 142, but there is an easier way to solve this problem.

The number 3 142 will be in the list after all the numbers that begin with a 1and after all the numbers that begin with a 2.

How many of the numbers in the list begin with a 1?

There are 24 four-digit numbers that can be made from the digits 1, 2, 3 and 4.There are 4 possibilities for the first digit so there are 24 ÷ 4 = 6 numbers thatbegin with a 1. The numbers are 1 234, 1 243, 1 324, 1 342, 1 423 and 1 432.

Similarly, there are 6 numbers in the list that begin with a 2, 6 that begin witha 3, and 6 that begin with a 4. We could list all 6 of the numbers in the listthat begin with a 2, but this is not necessary.

So the first 6 numbers in the ordered list begin with a 1, the next 6 numbersbegin with a 2, and the following 6 numbers begin with a 3. There are 12numbers in the ordered list before the first number beginning with a 3.

In order, the numbers that begin with a 3 are 3 124, 3 142, 3 214, 3 241, 3 412,3 421. We see that 3 142 is the second number in the ordered list of numbersbeginning with a 3.

Therefore, the number 3 142 is in the fourteenth position in the ordered list.

POTWC 11 Combined7 8

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