Memorandum
Fractions
Ordering, compering and equivalence
Exercise 1
1.
a) One sixth, which we can write as:
1
6
b) One seventh, which we can write as:
1
7
c) You should realise that when a whole is divided into several pieces, the name
of the piece (Fraction) is determined by the number of pieces the wholes has
been divided into. For example, five equal pieces and each piece is a fifth, six
equal pieces and each piece is a sixth, seven equal pieces and each piece is
a seventh. It is very important that you as speak about the “1sixth” and not
“one over six” while this may seem trivial, “one over six” does not convey the
meaning of the fraction in the same way that 1-sixth does.
1
.It follows that 1-sixth is larger than 1- seventh since breaking a whole into 7
6
pieces will result in smaller pieces than breaking it into 6 pieces.
2.
3.
1
14
4. a) 1-fourth; 1-fourth is bigger than 1-firth because when you divide a whole into 4
pieces the pieces are bigger than when you divide a whole into 5 pieces.
b) 2-sixths; 2-sixth is bigger than 1-sevenths because 1-sixth is bigger than 1seventh, so double a smaller piece will be smaller than double a bigger piece.
3
, Since 1-sixth is bigger than 1-seventh, it follows that 3-sixth will be bigger than
6
3-sevenths.
c)
d)
8
9
6
1
8
1
7
7
9
9
1
1
7
9
, Because + = 1 whole and + = 1 whole. It follows that because > a
6
8
7
9
bigger piece has been taken away from 1 to get , therefore
is bigger. Although,
this may not appear to be the most efficient method, it provides evidence of
conceptual understanding and the ability to reason. At this stage it is more important
that you
e)
5
9
3
4
5
4
7
7
9
9
4
4
7
9
, Because + = 1 whole and + = 1 whole. It follows that because >
a bigger piece has been taken away from 1 to leave
3
7
, therefore
5
9
is bigger.
Alternatively, 3 is less than one-half of 7 and 5 is greater than one – half of 9 so
less than one-half and
5
9
is greater than one-half and it follows that
5.
0
1
2
3
4
4
4
0
6. a)
b)
1
2
3
3
1
1
1
3
3
4
7.
0
0
0
8. a)
1
4
b)
c)
d)
1
5
3
5
3
6
1
2
3
4
4
4
1
1
2
3
4
5
5
5
5
1
1
2
3
4
5
6
6
6
6
6
1
5
9
3
> .
7
3
7
is
e)
5
6
4
f)
6
Equivalence
Exercise 2
1.
a) Paired discussion
b) The rectangles can be divided into fifths, tenths or fifteenths
c) Possible answers
1
2
=
5
10
2. a)
b)
c)
1
2
1
4
1
6
3
=
15
=
4
=
20
5
25
2
3
4
5
4
6
8
10
2
3
8
12
= = = =
= =
=
=
2
12
=
3
18
4
=
16
=
4
24
=
6
30
=
7
35
6
12
5
20
=
=
5
30
=
6
24
=
6
36
3. a) Class discussion
b) Class discussion
c) Class discussion
In this question, you should notice that equivalent fractions are useful when
comparing the size of fractions and that a convenient denomination of the equivalent
fraction is the lowest common multiple of all denominations of fractions being
compared.
Calculating a fraction of a quantity of a quantity
Exercise 3
1.
Fractions
of the box
Number of
smarties
1
12
1
6
1
4
1
3
1
2
1
2
3
4
6
For example,
1
12
means how many smarties there will be in each group when the
box of smarties is shared or grouped into 12 equal groups.
1
means how many
6
smarties there will be in each group when the box of smarties is shared or grouped
into 6 equal groups etc. You should notice that to calculate the number of smarties,
you should share (or divide) the total by the denomination of the fraction, i.e. sixths
and the whole must be shared / divided into six equal parts, fourths and the whole
must be shared/divided into four equal parts etc.
b)
Fractions of
the box
Number of
smarties
Fractions of
the box
Number of
smarties
Fractions of
the box
Number of
smarties
2
12
3
12
4
12
5
12
6
12
2
3
4
5
6
7
12
8
12
9
12
2
6
3
6
7
8
9
4
6
4
6
5
6
2
4
3
4
2
3
8
10
6
9
8
4 2
6 3 2 8 4 2 9 3
c) = =; = = ; = = ; =
12 6
12 6 3 12 6 3 12 4
1
12
of a box of smarties is the number of smarties in each group if the smarties in the
box are divided into twelve equal groups.
2
of a box of smarties is the number of
12
smarties in two of these groups. You should notice that determining n-twelfths of a
box of smarties involves first determining the number of smarties in each twelfth and
then the number of smarties in n groups.
2. a) 10km
b)30km
c) 5km
d)
5
=.
60
1
12
For question 2 can be answered in two ways. You can calculate the distance of the
hike completed each day in order to calculate the distance left on the last day and
hence, the fraction of the distance left on the last day. The wording of this question
encourages you to do this. Another way is to calculate the fraction left after each day
and use this to determine the distance covered.
3. 10524
4. 445
5. a) 80
b) 96
c) 120
d) 160
Fractions in calculations
Exercise 4
1.a) 5-sevenths
b) 4 – firths
7
c)
8
8
d)
e)
11
43
100
2. You should notice that when you add fractions of the same type, for example
tenths and tenths or fifths and fifths, the answer will be the same type, i.e a – tenths
+ b-tenths = (a + b) – tenths.
3.a) 1
b) 1
1
10
c) 1
d) 1
1
5
e) 1
f) 1
2
6
8-tenths + 2-tenths is the same as 8 + 2 and it is completing the ten, which in this
8 3 8 2 1
1
case will make it a whole. + = + + =1
10 10 10 10 10
10
This is bringing the tens but will vary according to the fraction:
1
3-fifths + 3-fifths = 5-fifths + 1-fifths =1
5
You need to learn to complete or ‘fill’ a fraction as many parts of the fraction are
needed. For example, 5 fifths are needed to make a whole.
Exercise 5
1.
a)
Fractions of the
box
Number of
smarties
Fractions of the
box
Number of
smarties
1
2
1
3
1
4
1
5
1
6
1
8
60
40
30
24
20
15
2
3
2
5
3
5
4
5
5
6
3
8
80
48
72
96
100
45
b) 60
c) 40
d) 100
e)
5
6
f)20
g) 60
h)
i)
1
2
8
15
j) At this stage, you are expected that when the denominations of the fractions being
added are different, the denomination of the answer fraction is also different.
2. Class discussion
3. a)
b)
7
12
11
30
9
c) 10
d)
e)
19
24
8
15
Exercise 6
1. a) Class discussion
b) Class discussion
In this exercise you should notice the denominator of the answer is the product of the
denominators of the fractions being added.
It then guides you into realising that determining the lowest common denominator of
the fractions is more useful that simply multiplying the denominators together.
Because 7 is not a factor of 120, it is no longer useful to divide a whole into 120
parts. More useful ways of dividing a whole, i.e. into multiples of both denominators.
2.
a)
31
36
23
b) 56
c)
33
35
55
d) 63
e)
82
99
In this question, the denominator of the answer is the product of the denominator in
the question. You can determine the number of smarties to use by simply multiplying
the denominators in the fraction.
Learners referring to the lowest common multiple at this stage need not to be
discouraged: hoping for that.
3.Class discussion
4.a) Class discussion
b) Class discussion
5. a)
e)
f)
g)
h)
i)
j)
8
8
3
4
8
11
c)
9
or
5
b)
d)
6
16
=1 18
15
16
11
3
8
8
=1
21
5
16
16
=1
16
=2
8
1
8
7
16
In this exercise, the denomination of the fraction is a multiple of the denomination of
the other fraction. Decimal fractions are included to encourage you to reflect on
1
3
previous knowledge. It is expected that you know 0,5 = and 0,75 =
2
4
6.a) Class discussion
b) class discussion
The methods described by Sean and Collin encourage conceptual understanding
about adding mixed numbers. These methods may look long on paper, but are
efficient in mental calculations.
1
1
It is important that you can break up mixed numbers, i.e recognise that 3 2is 3 and 2,
in the same way as you recognise, for example 25 = 20 + 5. With fractions we ‘fill up
the whole numbers’ in the same way as ‘filling up tens’ with units.
3
7. a) 4
8
g) 6
3
16
1
b) 6
8
1
c) 4
f) 3
1
6
i) 5
16
1
d) 10
e) 7
h) 6
j) 8
8
1
5
16
1
9
3
l) 6
6
16
3
k) 6
3
15
4
Exercise 7
3
1.a)
b)
c)
d)
e)
f)
8
13
1
12
g) 0
24
9
h)
16
3
3
12
=
1
4
j) 0
10
3
8
The calculations in this question require you to use equivalent fractions in the same
way as you did when adding. Decimal fractions are included to encourage you to
1
3
reflect on previous knowledge. It is expected that you know 0,5 = and 0,75 =
2
4
7
7
2.Jessica’s method encourages you to recognise that 6 = 6+ .
8
8
For this calculation and the next question, you do not need to break up a whole
number into fractions in order to subtract from the whole.
3.a) 3
b) 3
c) 4
2
5
2
8
4
=3
1
4
2
=4
10
5
2
d) 4
e) 7
f) 4
=4
12
1
6
1
4
7
10
4. a) Class discussion
b) Class discussion
This calculation and those in the next question are different because you are
3
to 1
4
4
1
1
because it is easier to subtract a whole number. But he also has to add to 3 .
4
2
required to take a bigger fraction away from a smaller fraction. Travis added
3
2
1
Samir broke 1 into 1 and
4
4
4
1
the remaining .
4
2
5. a) 2
4
b) 4
1
=2
2
1
=
5
6
7
d) 1
8
from 3
4
1
e) 2
2
f)
8
10
c) 2
12
2
. He subtracted the 1
6 3
=
8 4
g) 5
3
4
3
h) 1
5
Exercise 8
5
1.a) 1
12
5
b) 1
42
c) 1
d)1
1
15
1
28
g)
5
36
h)
11
24
i) 6
j) 1
7
e) 1
30
k) 5
3
f) 1 20
l)
5
6
1
4
9
20
7
12
2
4
1
and then takes away
2. a) 10 km
b)
c)
1
2
1
12
d) 5km
1
e)
12
This question is like that done in question 2 of exercise 3, except that it asks you to
calculate fractions left of the hike instead of distance left of the hike. These answers
can be compared. For example, in exercise 3, you will have calculated that there are
30km left after day 3. In this exercise, you should calculate that half of the hike is left
after day 3. These answers agree because half of 60km is 30km.