Pom9 Solutions Chapter 1
Pom9 Solutions Chapter 1
Pom9 Solutions Chapter 1
3 9 12 3 1 3 1× 2
a) + = b) + = +
10 10 10 8 4 8 4×2
6 3 2
= = +
5 8 8
1 5
=1 =
5 8
5 2 5×5 2 ×6 7 2 16 2
c) − = − d) 1 − = −
6 5 6×5 5×6 9 5 9 5
25 12 16 × 5 2 × 9
= − = −
30 30 9 ×5 5×9
13 80 18
= = −
30 45 45
62
=
45
17
=1
45
1 1
5 3 5 3/ ⎛3 2⎞ 3 3
a) × = × b) ⎜ ÷ ⎟ = ×
12 10 12 10 ⎝4 3⎠ 4 2
4 2
9
1 =
= 8
8
1
=1
8
7 1 23 13 2 2 2 16
c) 2 ×6 = × d) ÷2 = ÷
8 2 8 2 9 7 9 7
299 2 7
= = ×
16 9 16
11 1
= 18 2/ 7
16 = ×
9 16
8
7
=
72
The fraction of pizza left for Brad can be determined by subtracting the fractions that the others
have eaten from the 2 original pizzas.
⎛1 3 1 1⎞ ⎛ 1 × 8 3 × 3 1 × 6 1 × 12 ⎞
2−⎜ + + + ⎟ = 2−⎜ + + + ⎟
⎝3 8 4 2⎠ ⎝ 3 × 8 8 × 3 4 × 6 2 × 12 ⎠
⎛ 8 9 6 12 ⎞
= 2−⎜ + + + ⎟
⎝ 24 24 24 24 ⎠
35
= 2−
24
48 35
= −
24 24
13
=
24
13
Brad has of one pizza left.
24
a) 13 + ( −5) = 8 b) −7 + 2 = −5
7 − 11 = 7 + ( −11)
c) −8 + ( −15) = −23 d)
= −4
2 − 16 = 2 + ( −16 )
e) f) 8 − ( −7 ) = 8 + 7
= −14
= 15
g) −5 − ( −9 ) = −5 + 9 h) 100 × ( −4 ) = −400
=4
i) −7 × 7 = −49 j) −3 × ( −14 ) = 42
k) 42 ÷ ( −6 ) = −7 l) −28 ÷ 7 = −4
a) −3 ( 9 + 11) = −3 ( 20 ) b) 2 + 3 (10 − 4 ) = 2 + 3 ( 6 )
2 2
= −60 = 2 + 3 ( 36 )
= 2 + 108
= 110
2 2
e) ⎡ 2 − ( 6 + 3) 2 ⎤ = ⎡ 2 − ( 9 )2 ⎤ f) −15 + 8 × 7 − 32 ÷ 16 = −15 + 56 − 2
⎣ ⎦ ⎣ ⎦
= 39
= [ 2 − 81]
2
= ( −79 )
2
= 6241
b) 4, 17, 30, 43, 56, 69, 82… Add 13 to the previous term.
One nickel: 5¢
One dime: 10¢
One nickel, one dime: 15¢
One quarter: 25¢
One nickel, one quarter: 30¢
One dime, one quarter: 35¢
One nickel, one dime, one quarter: 40¢
Two quarters: 50¢
One nickel, two quarters: 55¢
One dime, two quarters: 60¢
One nickel, one dime, two quarters: 65¢
You can make 11 different sums of money with one nickel, one dime and two quarters.
a) 1 × 1 = 1
11 × 11 = 121
111 × 111 = 12 321
1111 × 1111 = 1 234 321
b) The digits increase from 1 up to the number of 1s in one of the factors, and then decrease to 1
so that the answer is symmetric.
c) There are nine 1s in the first factor. Following the pattern, the product is
12 345 678 987 654 321.
a) 11 × 37 = 407
22 × 37 = 814
33 × 37 = 1221
b) Add 407 to the previous term. The next three terms are 1628, 2035, and 2442.
c) 99 × 37 is the eighth term after 11 × 37 . Add 407 eight times to obtain 3663.
1 2 3
a) = 01
. = 0.2 = 0.3
9 9 9
If the numerator is less than 9, then the decimal will be that digit repeated infinitely after the
decimal place.
1 12 23
b) = 0.01 = 012
. = 0.23
99 99 99
If the numerator is less than 99, then the decimal will be that number written as a two-digit
number, repeated infinitely after the decimal place.
1 123 12345
c) = 0.00001 = 0.00123 = 012345
.
99 999 99 999 99 999
If the numerator is less than 99 999 then the decimal will be that number written as a five-digit
number, repeating infinitely after the decimal point.
a) Gina is 22 years, 3 months, and 17 days old on January 1, 2020. Sam is 25 years, 11 months,
and 9 days old on January 1, 2020.
b) To find the number of years, subtract the birth year from 2019, since 2020 has just started. To
find the number of months, subtract the month number from 12, for December of 2019. To find
the number of days, subtract the birthday number from 31, the number of days in December, and
add 1 for January 1st.
b) Since there are 3 players, there are 3 pairs: 1-2, 1-3, and 2-3. Three games are required.
c) With 4 players, there are 6 pairs: 1-2, 1-3, 1-4, 2-3, 2-4, and 3-4. Six games are required.
d) You could list all possible pairs, as in the first three parts of this question. Alternatively, you
can use reasoning and multiplication. With 10 players, each of the 10 can be matched with 9
others, resulting in 10 × 9 = 90 pairs. However, each pair has been counted twice. For example,
90
the pair 1-2 is the same as the pair 2-1. Hence there are = 45 pairs of players, requiring 45
2
games.
2× 2 = 4
3× 3 = 9
2 × 2 × 3 × 3 = 36
7 × 7 = 49
2 × 2 × 7 × 7 = 196
3 × 3 × 7 × 7 = 441
2 × 2 × 3 × 3 × 7 × 7 = 1764
All of these divide evenly into 8820. There are 7 perfect squares that divide evenly into 8820.
You are given that F is half of C. However, C can also only be one more than F, carrying the 1
from the previous column. Therefore, C must be 2 and F must be 1. Looking at the last column, T
must be a 0. E cannot be 3, 4 or 5, since that makes A a 0, 1 or 2, which have already been used.
This leaves 6, 8 or 9 for E. Try combinations, to find those that work.
630 860
1766 1788
2396 2648
Use the clues to place limits on what each letter may represent. For
example, H and D are perfect squares between 1 and 16, limiting their
values to 1, 4, 9 or 16. B, C, N, and R are greater than 12, limiting their
values to 13, 14, 15 or 16. Draw a large grid, and place the possible values
in each square. Then, eliminate those that lead to inconsistent sums. The
solution is shown.
a) Subtract 5 from the previous term. The next two terms are 0 and –5.
b) Subtract 4 from the previous term. The next two terms are –18 and –22.
1 1
c) Add to the previous term. The next two terms are 1 and 1 .
4 4
2 6 4
d) Subtract from the previous term. The next two terms are and .
5 5 5
e) Multiply the previous term by –2. The next two terms are 48 and –96.
f) Divide the previous term by 2. The next two terms are –12 and –6.
g) Subtract descending multiples of 5 from the previous term, starting with 20. The next two
terms are 50 and 50 (subtract 5, then 0).
h) Multiply the previous term by integers starting from 1, and increasing by 1 each time. The
next two terms are 360 (multiply by 5) and 2160 (multiply by 6).
1 25
π ( 5 ) = π . The
2
The area of the semicircle on the hypotenuse is
2 2
1 9
areas of the semicircles on the other two sides are π ( 3) = π and
2
2 2
1 16 9 16 25
π ( 4 ) = π . Note that π + π = π . The area of the semicircle
2
2 2 2 2 2
on the hypotenuse equals the sum of the areas of the semicircles on the other two sides.
Diagram B is correct. As the wheel moves forward, the height of the light will increase and
decrease smoothly.
a) The map is
divided into
sections where time
changes by one-
hour intervals.
Starting at the
original time zone,
count how many
time zones away the
other one is using
positive integers to
the right and
negative integers to the left. Add this integer value to the original time.
b) Halifax is +1 hour ahead of Toronto. If it is 3:00 P.M. in Toronto, it is 4:00 P.M. in Halifax.
1
a) Use the last two rows. In the fraction strip made of pieces,
7
1
shade three parts. In the fraction strip made of pieces, shade
8
4 3
four parts. Compare the shaded parts. > because four pieces
8 7
1 1
of are wider than three pieces of .
8 7
1 1 5
b) Place the piece and the piece side by side. They will have the same width as a
2 3 6
piece.
1 1
c) Twelve rows are needed to illustrate + . The lowest common denominator is 12.
3 4
d) The dark blue bars are getting smaller. When 1 is divided by larger and larger numbers, the
pieces become smaller and smaller. The number of pieces is the denominator of the fraction.
1 + 3 + 5 + 7 + 9 = 25
1 + 3 + 5 + 7 + 9 +11 = 36
c) There are 50 odd numbers from 1 to 99 inclusive. The sum is 502 = 2500.
d) There are 75 odd numbers from 1 to 150. There are 300 odd numbers from 1 to 600. The sum
of the odd numbers from 150 to 600 is 3002 – 752 = 84 375.
a)
Answers will vary, depending on the size of the classroom. One approach is to treat the puck as a
box of height 2.5 cm, and length and width both 7.4 cm. Find the volume of the puck, and divide
this into the volume of the classroom.
Sample answer:
2000
The round-trip distance on each ride is 16 km. Honi can make = 125 trips. Assuming 5
16
125
trips per week, the tires will last = 25 weeks, or about half a year.
5
You can use the guess and test strategy. Start by guessing that Joe ate 1 slice. Then, Emily ate 2
slices, Samir ate 4 slices, Kendra ate 1 slice and Fong ate 5 slices. Add the slices: 1 + 2 + 3 + 1 +
5 = 12. These add correctly.
2 1 3 1 1 1 5
Emily ate = , Samir ate = , Joe ate , Kendra ate , and Fong ate .
12 6 12 4 12 12 12
The snail must climb a total of 27 m. Each day, it climbs a net 1 m. It will take 23 days to reach a
point 4 m from the top of the pipe. On the 24th day, it will reach the top of the pipe.
A cat lives about 15 years, and has a heart rate of about 150 beats/min.
A cat's heart will beat about 1.18 billion times during its lifetime.
A polygon with 20 sides has 20 vertices. A diagonal is formed by connecting a vertex to another
vertex, excluding itself and the vertices on either side. Therefore, each of the 20 vertices can be
joined to 17 others. This implies that you have 20 × 17 = 340 diagonals. However, each diagonal
has been counted twice. Divide by 2 to obtain the correct answer of 170 diagonals.
For each part, the numerator and denominator of the fraction are two consecutive integers. The
numerator of the second fraction equals the denominator of the first. In each case the value of the
first fraction is less than that of the second.
1 2
a) <
2 3
2 3
b) <
3 4
3 4
c) <
4 5
4 5
d) <
5 6
b) If the driver cog makes half a turn, it turns through 15 teeth. The driving cog must make
15 3
, or turns.
20 4
c) If the driving cog turns through 5 turns, it turns through 100 teeth. The driver cog must make
100 1
, or 3 turns.
30 3
d) If the driver cog makes half a turn, it turns through 12 teeth. The driving cog must turn
12 3
through , or turns.
40 10
The same answers would result. The middle cog does not change the ratio of the driver cog to the
driving cog.
b)
c)
d) You cannot arrange the squares with a width of zero. The quotient 12 ÷ 0 is undefined.
2
a) Start at 0, and subtract three times to arrive at –2.
3
5
b) Start at 0 and subtract four times to arrive
4
at –5.
a) A calculator is appropriate for complicated math expressions involving square roots and
fractions.
395 = 90 224199
a) Subtract 6 to obtain the next term. The missing terms are –57 and –63.
b) Multiply by 3 to obtain the next term. The missing terms are 1215 and 3645.
c) Divide by –2 to obtain the next term. The missing terms are –4 and 2.
d) Subtract 3 to obtain the next term. The missing terms are –158 and –161.
e) Multiply by –2 to obtain the next term. The missing terms are –6144 and 12 288.
f) Add 24 to obtain the next term. The missing terms are 32 and 56.
1 1 1
a) Divide by increasing integers, starting with 2. The next three terms are , , and .
3 21 168
1 4 5
b) Subtract to obtain the next term. The next three terms are − , − , and − 2 .
3 3 3
1 1 1
c) Subtract to obtain the next term. The next three terms are 0 , − , and − .
4 4 2
1 1 1 1
d) Subtract to obtain the next term. The next three terms are , , and .
12 3 4 6
1 ⎛ 1⎞ 2
a) − + ⎜ − ⎟ = −
2 ⎝ 2⎠ 2
= −1
2 ⎛ 3⎞ 8 ⎛ 9⎞
b) − + ⎜ − ⎟ = − + ⎜ − ⎟
3 ⎝ 4⎠ 12 ⎝ 12 ⎠
17
=−
12
5
= −1
12
1 ⎛ 2 ⎞ 5 ⎛ 14 ⎞
c) + ⎜− ⎟ = + ⎜− ⎟
7 ⎝ 5 ⎠ 35 ⎝ 35 ⎠
9
=−
35
2 3 16 9
d) − + = − +
3 8 24 24
7
=−
24
3 5 3 ⎛ 5⎞
a) − = +⎜− ⎟
8 6 8 ⎝ 6⎠
9 ⎛ 20 ⎞
= + ⎜− ⎟
24 ⎝ 24 ⎠
11
=−
24
1 2 1 ⎛ 2⎞
b) − = +⎜− ⎟
2 3 2 ⎝ 3⎠
3 ⎛ 4⎞
= + ⎜− ⎟
6 ⎝ 6⎠
1
=−
6
⎛ 1⎞ 1 ⎛ 1⎞ ⎛ 1⎞
c) ⎜− ⎟ − = ⎜− ⎟ + ⎜− ⎟
⎝ 4⎠ 6 ⎝ 4⎠ ⎝ 6⎠
3 ⎛ 2⎞
= − + ⎜− ⎟
12 ⎝ 12 ⎠
5
=−
12
⎛ 4⎞ ⎛ 3 ⎞ ⎛ 4⎞ 3
d) ⎜ − ⎟ − ⎜ − ⎟ = ⎜ − ⎟ +
⎝ 5 ⎠ ⎝ 10 ⎠ ⎝ 5 ⎠ 10
8 3
=− +
10 10
5
=−
10
1
=−
2
Answers will vary. To multiply rational numbers in fraction form, multiply the numerators and
multiply the denominators. To divide rational numbers in fraction form, multiply the dividend by
the reciprocal of the divisor.
Examples:
2 3 6 2 3 2 4
× = ÷ = ×
3 4 12 3 4 3 3
1 8
= =
2 9
1 1
5 3 5 3
a) − × = − ×
6 10 6 10
2 2
1
=−
4
⎛ 1⎞ ⎛ 3⎞ 3
b) ⎜ − ⎟ × ⎜ − ⎟ =
⎝ 7 ⎠ ⎝ 5 ⎠ 35
⎛ ⎞ 3
⎛ 1⎞ 6 ⎜ 1 ⎟× 6
c) ⎜ 8 ⎟ 11 ⎜ − 8
− × =
⎟⎟ 11
⎝ ⎠ ⎜
⎝ 4 ⎠
3
=−
44
7 ⎛ 5⎞ 7 ⎛ 6⎞
÷ − = × −
8 ⎜⎝ 6 ⎟⎠ 8 ⎜⎝ 5 ⎟⎠
d)
⎛ 3 ⎞
7 ⎜ 6⎟
= × −
8 ⎜⎜ 5 ⎟⎟
4 ⎝ ⎠
21
=−
20
1
= −1
20
⎛ 5 ⎞ ⎛ 3⎞ ⎛ 5 ⎞ ⎛ 8⎞
e) ⎜− ⎟ ÷⎜− ⎟ = ⎜− ⎟×⎜− ⎟
⎝ 12 ⎠ ⎝ 8 ⎠ ⎝ 12 ⎠ ⎝ 3 ⎠
⎛ ⎞ ⎛ 2⎞
⎜ 5 ⎟ ⎜ 8⎟
= − × −
⎜⎜ 12 ⎟⎟ ⎜⎜ 3 ⎟⎟
⎝ 3 ⎠ ⎝ ⎠
10
=
9
1
=1
9
c) Use a calculator to multiply the thickness by 2 twenty times. The answer is 83 886.08 mm.
d) Answers will vary. As the number of folds increases, the thickness also increases, making it
more difficult to fold the paper.
Answers will vary. You can use fraction strips or a grid diagram.
1
a) A and B are each of the whole square.
4
1
C and E are each of the whole square.
16
1
D, F and G are each of the whole square.
8
1 1
b) i) A+B= +
4 4
1
=
2
1 1
ii) C + G = +
16 8
1 2
= +
16 16
3
=
16
1 1
iii) D + E = +
8 16
2 1
= +
16 16
3
=
16
1 1
iv) F − E = −
8 16
2 1
= −
16 16
1
=
16
1 1⎛1⎞
v) A= ⎜ ⎟
4 4⎝ 4⎠
1
=
16
1 1⎛1⎞ 1
vi) D − F= ⎜ ⎟−
2 2⎝8⎠ 8
1
=−
16
a) F = C + E
b) B = C + E + F
Subtract 4 to obtain the next term. To obtain the thousandth term, take 45, and subtract 4, 999
times to obtain 45 − 4 ( 999 ) = −3951 .
Subtract 7 to obtain the next term. From 100 to –600 is 700, or 100 subtractions of 7. Therefore,
–600 is the 101st term.
A cup is about 250 mL. A bathtub measures about 200 cm by 80 cm by 60 cm, for a volume of
960 000
200 × 80 × 60 = 960 000 mL. The number of cups required is = 3840 . About 4000 cups
250
are required to fill a bathtub.
Use a physical model. If you fold it once, and then cut it, you will have 3 pieces of string. If you
fold it twice, and cut it, you will have 5 pieces. If you fold it 3 times, and cut it, you will have 9
pieces. A general formula for the number of pieces after n folds is 2n + 1.
Let the first number be n. The next two numbers are n + 1 and n + 2. The sum of these numbers is
n + n + 1 + n + 2 = 3n + 3 . 3n is a multiple of 3, and must be divisible by 3. Adding 3 to 3n
produces another multiple of 3, and is divisible by 3.
Answers will vary. Example: Since a newspaper is made by folding a sheet in half, there will
always be an even number of pages because the number of pages equals two times the number of
sheets.
a) 5 × 2 + 8 − 3 = 15
b) 25 ÷ 5 + 11 = 25 − 9
1 1 11 1
c) + = −
2 3 12 2
2 ⎛ 1⎞ 1
d) ×⎜− ⎟ = −
3 ⎝ 8⎠ 12
3
c) is a fraction, but is greater than 1.
2
Select each square in turn, and move the knight. You can always find a
combination of moves that will land the knight on that square.
Answers will vary. Divide the surface area of each hallway by the surface area of one tile. A
sample answer is shown.
Fridgeway High School has two floors. Each has a hallway measuring 60 m long and 4 m wide.
Each tile measures 0.4 m by 0.4 m. The total area of the hallways is 2 × 60 × 4 = 480 m2 . The area
480
of a tile is 0.4 × 0.4 = 0.16 m2 . The number of tiles required is = 3000 .
0.16
The mass of a school bus is about 4800 kg. The average mass of a high school student is about 70
kg. The mass of a bus with 45 student passengers is 4800 + 45 ( 70 ) , or 7950 kg.
1 + 2 ( −2 ) + 3 ( 3) + 4 ( −4 ) + 5 ( 5 ) + 6 ( −6 ) + 7 ( 7 ) + 8 ( −8) + 9 ( 9 ) + 5 ( −10 ) = −5 .
1 + 2 ( −2 ) + 3 ( 3) + 4 ( −4 ) + 5 ( 5 ) + 6 ( −6 ) + 7 ( 7 ) + 8 ( −8 ) + 9 ( 9 ) + 10 ( −10 )
+11(11) + 12 ( −12 ) + 13 (13) + 9 ( −14 ) = −35.
You obtain two double Mobius strips that are looped together. The twist before taping results in
one strip inside the other.
Work backwards:
−392
Subtract −380 − 12 = −392 . Then, divide = 56 . The number is 56.
−7
Work backwards:
7 3 7 9
Add + = +
12 4 12 12
16
=
12
4
=
3
4 8
Then, multiply 2 × =
3 3
2
=2
3
2
The number is 2 .
3
Answers will vary. Working backward is very effective in these kinds of questions.
72 × 7 × 5 = 2520 .
a) 1 × 2 × 3 × 4 = 24
b) 52 − 1 = 24
c) 2 × 3 × 4 × 5 = 120
d) 112 − 1 = 120
e) 4 × 5 × 6 × 7 = 840
f) 292 − 1 = 840
g) When you multiply four consecutive natural numbers, the product equals one less than the
square of one more than the product of the first and last number.
Example: in parts a) and b) the product of the first and last number plus 1 is 1 × 4 + 1 = 5 . Square
the 5 and subtract 1 to get the same number as in a).
h) 5 × 6 × 7 × 8 = 1680
412 − 1 = 1680
10 × 11 × 12 × 13 = 17 160
1312 − 1 = 17 160
a) Karen used too much milk and sugar. She assumed that 500 mL is 5 L, rather than 0.5 L for
the milk. She also assumed that 125 g is 1.25 kg, rather than 0.125 kg.
The population of Ontario is about 12 000 000. Assume that four people share a pizza, and have
12 000 000
pizza once per month. This requires × 12 = 36 000 000 pizzas. Assume that an average
4
pizza measures about 0.6 m in diameter. The area is about 0.6 × 0.6 = 0.36 m2 . The total area of
the pizzas ordered in Ontario in a year is about 0.36 × 36 000 000 = 12 960 000 m 2 .
a) i)
ii)
b) Pair off the least and greatest numbers, moving toward the median. Then, put that single
number in the centre square and arrange the pairs around it.
a) Subtract 3 to obtain the next term. The next three terms are 0, –3, and –6.
b) Multiply by 2 to obtain the next term. The next three terms are 56, 112, and 224.
c) Add consecutive numbers starting from 1 to obtain the next term. The next three terms are 15,
20, and 26.
d) Subtract consecutive numbers starting from 4 to obtain the next term. The next three terms
are –19, –27, and –36.
340
The perimeter of the field is 2(100 + 70) = 340 m. The number of posts required is = 68 .
5
a) To find the number of different scores with three arrows, make a table and list all possible
patterns.
Calculate the time to travel 20 km, and then subtract this time from 7:30 P.M.
20
Time = Dave needs to catch the bus by 6:47 P.M.
28
0.714 h
43 min
New area = ( 2l )( 2 w )
= 4lw
Use a systematic trial on a calculator to help you find that 79 = 40 353 607 .
−402
Divide by 3: = −134 . The three consecutive integers must be –135, –134, and –133.
3
2 ⎛ 3 ⎞ 14 ⎛ 15 ⎞
+ − = + −
5 ⎜⎝ 7 ⎟⎠ 35 ⎜⎝ 35 ⎟⎠
a)
1
=−
35
2 ⎛ 1⎞ 2 1
b) − − ⎜ − ⎟ = − +
9 ⎝ 6⎠ 9 6
4 3
=− +
18 18
1
=−
18
1
2 1 2 1
c) − × =− ×
3 4 3 4
2
1
=−
6
7 ⎛ 3⎞ 7 ⎛ 7⎞
÷ −1 = ÷ −
12 ⎜⎝ 4 ⎟⎠ 12 ⎜⎝ 4 ⎟⎠
d)
7 ⎛ 4⎞
= ×⎜− ⎟
12 ⎝ 7 ⎠
1
⎛ 1 ⎞
7 ⎜ 4⎟
= × −
12 ⎜⎜ 7 ⎟⎟
3 ⎝ 1 ⎠
1
=−
3
11 5 ⎛ 1 ⎞ 7
− + +⎜− ⎟ = −
12 12 ⎝ 12 ⎠ 12
1 ⎛ 3⎞ ⎛ 1⎞ 7
+⎜− ⎟ +⎜− ⎟ = −
2 ⎝ 4⎠ ⎝ 3⎠ 12
7 ⎛ 2⎞ ⎛ 1⎞ 7
+⎜− ⎟ +⎜− ⎟ = −
12 ⎝ 3 ⎠ ⎝ 2 ⎠ 12