5th Grade Math student-ebook-course-11

LESSON Reducing by Grouping
Factors Equal to 1
54 Dividing Fractions

Power Up Building Power

facts Power Up G
mental
a. Number Sense: 6 × 250 1 1 1 2  2  3  5
math b. Number Sense: 736 − 4080 8 2 223

1 c. Number Sense: 375 + 99
8
d. Money: $8.75 + $5.00

e. Fra18ctional Parts: 1 of 9 2235
2 223

f. Number Sense: 30 × 30

g. Geometry: Can you think of a time when a figure could have the same
value for both its perimeter and its area?

h. Calculation: 8 × 8, − 1, ÷ 9, × 7, + 1, ÷ 5, × 10

problem The PE class ran counterclockwise A
solving
around the school block, starting and 30 m
70 m
finishing at point A. Instead of running all 100 m 50 m
200 m 30 m
the way around the block, Nimah took

what she called her “shortcut,” shown

by the dotted line. How many meters did

Nimah save with her “shortcut?”

New Concepts Increasing Knowledge

reducing by The factors in the problem below are arranged in order from least to
grouping greatest. Notice that some factors appear in both the dividend and the
factors divisor.

equal to 1 1 1 2235
82 223
1
8

Since 2 ÷ 2 equals 1 and 3 ÷ 3 equals 1, we will mark the combinations of
factors equal to 1 in this problem.

2 ·2 ·3· 5
2 ·2 ·3

Looking at the factors this way, the problem becomes 1 ∙ 1 ∙ 1 ∙ 5, which is 5.
Verify Which property helps us know that 1 ∙ 5 = 5?

280 Saxon Math Course 1

Example 1

Reduce this fraction: 2225 1  1  1  2 2 3 1
2235 3 3 4 2

Solution

We will mark combinations of factors equal to 1.

1 · 3 2 · 5 a34  1 b 1 1
2 2
22 2 4·

2 ·2 ·3·5

By grouping factors equal to 1, the pro22blem22 22b32ec22o55m 32es 155  1 11 123, w1 hi2332ch is 23. 3
4

2  2  2  52  221 2323frd1ai555cv1itdioin32n11gs 123 When we divide 10 by 5, we are answering the question “How many 5s are 1
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2  
 2 221b12 1
2
2

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2
1 1
2

1 How m12 an34y 1 s are in 3 ? aQ34 43 1 Rb a 3 12 12ab34  1 b 1 1 1 1 1 1 1 1
2 2 4 2 4 2 2 2 2 2 2

Solution

Before we show the two-step process, we will solve the problem with our
fraction manipulatives. The question can be stated this way:

How many 1 s are needed to make ?1
2
4
11
44

225 2 We see that the answer is more than one but less than two. If we take one 1
235 2 cu34t int12o two equal par12ts ( ), then we43can fit t21he 31 and 2
3 32 parts23 ( ) together t34o make three fo12urths. 42
1  1  1  2 2 2 and another 1 first
2 3 o n55e 2 1
  2
  of t1he 1sm a1ller23 3
4

1

2

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4 2 2 2
4
11
44

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3  1 1 1 1 3 3  1 1 1  1  2 1
4 2 2 2 4 4 2 2 2 2

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