Download Division of Mixed Numbers

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Arithmetic Revisited
Lesson 4:
Part 4: Dividing Mixed Numbers
1. Converting the Numbers Into Improper Fractions
A nice way to solve a new problem is to convert it into an equivalent
problem that we already know how to solve. In this respect, since we
already know how to divide fractions, it would seem logical to divide mixed
numbers by first converting them into improper fractions. For example,
suppose we want to divide 4 12 by 2 12 , We could begin by rewriting 4 12 ƒ 2 12
as 92 ƒ 52 . We would then use the “invert and multiply” rule to obtain
9
2
9
4
2 ‚ 5 . or 5 . Finally we divide 9 by 5 and obtain 1 5 as the answer.
Notes:
• The definition of division hasn't changed. 4 12 ƒ 2 12 means the
number we must multiply by 2 12 to obtain 4 12 as the product. A
quick check shows that:
2 12 ‚ 1 45 œ 52 ‚ 95 œ 92 œ 4 12 .
• It is often helpful to estimate an answer before proceeding with
the actual computation. Noticing that 4 12 is “around” 4 and 2 12 is
“around” 2, we can estimate that the answer should be “around”
4 ƒ 2 or 2. Actually since 5 ƒ 2 12 is exactly 2, 4 12 ƒ 2 12 must be a
“little less” than 2. Thus, 1 45 is a plausible answer.
• What the quotient tells us is that not only is 4 12 almost twice as
much as 2 12 but it's exactly 1 45 times as much.
• As a practical application, suppose we can buy 2 12 pounds of
cheese for $4 12 . Then the price per pound (that is, “dollars per
pound”) is obtained by dividing $4 12 by 2 12 pounds. The quotient
tells us that the cost of the cheese is $1 45 or $1.80 per pound.
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© Herbert and Ken Gross 1999
Lesson 4 Part 4
Page 2
• As a check notice that $1.80 per pound, 2 pounds would cost
$3.60 and 12 pound would cost $0.90; and we then see that the total
cost is $3.60 $0.90 or $4.50
Lesson 4 Part 4
Page 3
Practice Problem #1
Use the above method to express 6 13 ƒ 1 34 as a mixed number.
Answer: 3 13
21
Solution:
6 13 ƒ 1 34
œ
19
3
ƒ 47
œ
19
3
‚ 74
œ
76
21
œ
76 ƒ 21
œ
3 13
21
Notes:
• To check that our answer is at least reasonable, we observe that
rounded off to the nearest whole number 6 13 œ 6 and 1 34 œ 2. Hence
our answer should be “reasonably close to” 6 ƒ 2 or 3.
• Once we obtain our answer, we can check to see if it's correct by
13
3
1
remembering that 6 13 ƒ 1 34 œ 31 13
21 means that 31 21 ‚ 1 4 œ 6 3 .
3
Thus to check our answer we compute the value of 31 13
21 ‚ 1 4 and
verify that it is equal to 6 13
3
31 13
21 ‚ 1 4
œ
76
21
‚ 74
œ
76 ‚ 7
21 ‚ 4
œ
2 ‚2 ‚19 ‚ 7
3 ‚ 7 ‚2 ‚ 2
œ
2 ‚
2 ‚19 ‚ 7
3 ‚
7 ‚
2 ‚
2
œ
19
3
œ
6 13
Lesson 4 Part 4
Page 4
2. Using the Mixed Numbers as Adjectives Modifying the Same Noun.
To divide 4 12 by 2 12 , we may think of both mixed numbers as modifying,
say, “a carton of books” where each carton contains 2 books1. In this
case, 4 12 cartons is another name for 9 books and 2 12 cartons is
another name for 5 books. Hence we may visualize the problem in
terms of the following steps:
4 12 ƒ 2 12
œ
œ
œ
4 12 cartons ƒ 2 12 cartons
9 books ƒ 5 books
9ƒ5
œ
1 45
While converting mixed numbers to improper fractions gives us the desired
result when we divide two mixed numbers, from a mathematical
perspective, it would be nice to know that the arithmetic of mixed numbers
is self-contained (at least in the sense that we aren't forced to rewrite
mixed numbers as improper fraction in order to do arithmetic). From a
teaching point of view, we might want to illustrate how we can divide mixed
numbers in ways that may be more intuitive to students. In this context:
Multiplying Both Numbers in a Ratio by the Same (Non Zero) Number:
Students might find it interesting to see that we can divide mixed
numbers by “translating” the problem into an equivalent whole number
problem without having to refer to such things as books and cartons.
The key is that it is still a fact that we do not change a quotient when
we multiply both the dividend and the divisor by the same (non zero)
whole number. To this end, if we are given the division problem
4 12 ƒ 2 12 we simply multiply both numbers by # to obtain:
4 12 ƒ 2 12
œ
œ
Ð4 12 ‚ 2Ñ ƒ Ð2 12 ‚ 2Ñ
*ƒ&
œ
1 45
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1More generally we would use the (least) common multiple of both denominators. For example, had the
problem been % "$ ƒ # &( , we would have assumed that each carton contained 21 books and then multiplied
both the dividend and the divisor by 21.
Lesson 4 Part 4
Page 5
Practice Problem #2
Use the above method to write 6 13 ƒ 2 35 as a mixed number.
Answer: 2 17
39
Solution:
Since the denominators are 3 and 5, we can eliminate them by
multiplying both numbers by 15.
15 ‚ 6 13
"& ‚ 2 35
œ
15 ‚ Ð6 13 Ñ
œ
œ
œ
Ð15 ‚ 6Ñ Ð15 ‚ 13 Ñ
90 5
95
œ
15 ‚ Ð2 35 Ñ
œ
œ
œ
Ð15 ‚ 2Ñ Ð15 ‚ 35 Ñ
30 9
39
Hence 6 13 ƒ 2 35 œ 95 ƒ 39 œ 2 17
39
As a check we see that
3
2 17
39 ‚ 2 5
œ
95
39
‚ 13
5
œ
95 ‚ 13
39 ‚ 5
œ
19‚5 ‚ 13
3 ‚ 13 ‚ 5
œ
19‚
5 ‚ 13
3 ‚
13 ‚ 5
œ
19
3
œ
6 13 2
________________________________________________________________________________
2In terms of relative size, the above result tells us that 6 1 is approximately 2 1 times as great as 2 3 (It is
3
2
5
actually 2 17
times as great)
39
Lesson 4 Part 4
Page 6
An Interesting Aside:
We don't usually think about using common denominators when we
want to divide two fractions but we can. That is, the denominator of
a fraction is the noun and when we divide two numbers that modify
the same noun, the nouns “cancel”. Thus if we decide to rewrite the
mixed numbers as improper fractions we obtain:
6 13 ƒ 2 35
œ
19
3
ƒ 13
5
œ
19‚5
3‚ 5
œ
95
15
œ
95 fifteenths ƒ 39 fifteenths
œ
95 ƒ 39
œ
2 17
39
‚3
ƒ 13
5‚3
ƒ 39
15
Enrichment Note
While the topic might be beyond the scope of the elementary school
student, it is interesting to note that the long division algorithm for whole
numbers, as a form of repeated subtraction, also applies to the division of
mixed number. For example,
6 13 ƒ 2 35 œ Ð Ñ
means the same thing as
2 35 ‚ Ð Ñ œ 6 13
Using trial-and error to ”fill in the blank” we see that
Lesson 4 Part 4
Page 7
2 35 ‚ 1 œ 2 35
2 35 ‚ 2 œ 4 65 œ 5 15
Ã
less than 6 13
2 35 ‚ 3 œ 6 95 œ 7 43
5
Ã
greater than 6 13
Ú
Ý In other words, 6 13 is more
Ã Û than twice as big as 2 3
5
Ý
Ü but less than three times as big.
5
3
2
The difference between 6 13 and 5 15 is 6 15
5 15
œ 1 15
. Hence if we use the
long division algorithm we see that:
2
R œ 1 15
2
2 35
6 13
5 15
2
1 15
and if we now write the remainder over the divisor we see that
2
1 15
1
3
63 ƒ 25 œ 2 3
25
We can simplify the complex fraction
2
1 15
2 35
by multiplying numerator and
denominator by 15. That is
2
2
15 ‚ Ð1 15
Ñ œ 15 ‚ Ð1 15
Ñ œ 15 2 œ 17
and
15 ‚ Ð2 35 Ñ œ 15 ‚ Ð2 35 Ñ œ 30 9 œ 39
Hence
6 13
ƒ 2 35
œ2
2
1 15
2 35
17
17
œ # 39
œ 2 39