Mathematics Support

Making Sense of Division

Earlier, students learned that a division such as 10÷2=? can be interpreted as “how many groups of 2 are in 10?” or “how much is in each group if there are 10 in 2 groups?” They also saw that the relationship between 10, 2 and the unknown number ("?") can also be expressed with multiplication:

This week, they use these ideas to divide fractions. For example, 6÷112=?can be thought of as “how many groups of 112 are in 6?” Expressing the question as a multiplication and drawing a diagram can help us find the answer.

From the diagram we can count that there are 4 groups of 112 in 6.

We can also think of 6÷112=? as “how much is in each group if there are 112equal groups in 6?” A diagram can also be useful here.

From the diagram we can see that if there are three 12-groups in 6. This means there is 2 in each 12 group, or 4 in 1 group.

In both cases 6÷112=4, but the 4 can mean different things depending on how the division is interpreted.

Here is a task to try with your student:

1. How many groups of 23 are in 5?
   1. Write a division equation to represent the question. Use a “?” to represent the unknown amount.
   2. Find the answer. Explain or show your reasoning.
    
2. A sack of flour weighs 4 pounds. A grocer is distributing the flour into equal-size bags.
   1. Write a question that 4÷25=? could represent in this situation.
   2. Find the answer. Explain or show your reasoning.
    

Solution:

1. 1. 5÷23=?
   2. 712. Sample reasoning: There are 3 thirds in 1, so there are 15 thirds in 5. That means there are half as many two-thirds, or 152 two-thirds, in 5.
    
2. 1. 4 pounds of flour are divided equally into bags of 25-pound each. How many bags will there be?
   2. 10 bags. Sample reasoning: Break every 1 pound into fifths and then count how many groups of 25 there are.
      
    

Many people have learned that to divide a fraction, we “invert and multiply.” This week, your student will learn why this works by studying a series of division statements and diagrams such as these:

• 2÷13=? can be viewed as “how many 13s are in 2?”
  

  Because there are 3 thirds in 1, there are (2⋅3) or 6 thirds in 2. So dividing 2 by 13 has the same outcome as multiplying 2 by 3.

• 2÷23=? can be viewed as “how many 23s are in 2?”
  

  We already know that there are (2⋅3) or 6 thirds in 2. To find how many 23s are in 2, we need to combine every 2 of the thirds into a group. Doing this results in half as many groups. So 2÷23=(2⋅3)÷2, which equals 3.

• 2÷43=? can be viewed as “how many 43s are in 2?”
  

  Again, we know that there are (2⋅3) thirds in 2. To find how many 43s are in 2, we need to combine every 4 of the thirds into a group. Doing this results in one fourth as many groups. So 2÷43=(2⋅3)÷4, which equals 112.

Notice that each division problem above can be answered by multiplying 2 by the denominator of the divisor and then dividing it by the numerator. So 2÷abcan be solved with 2⋅b÷a, which can also be written as 2⋅ba. In other words, dividing 2 by ab has the same outcome as multiplying 2 by ba. The fraction in the divisor is “inverted” and then multiplied.

Here is a task to try with your student:

1. Find each quotient. Show your reasoning.
   1. 3÷17
   2. 3÷37
   3. 3÷67
   4. 37÷67
    
2. Which has a greater value: 910÷9100 or 125÷625? Explain or show your reasoning.

Solution:

1. 1. 21. Sample reasoning: 3÷17=3⋅71=21
   2. 7. Sample reasoning: 3÷37=3⋅73=7
   3. 312. Sample reasoning: 3÷17=3⋅76=72. The fraction 67 is two times 37, so there are half as many 67s in 3 as there are 37s.
   4. 12. Sample reasoning: 37÷67=37⋅76=36
    
2. The have the same value. Both equal 10. 910÷9100=910⋅1009=10 and 125÷625=125⋅256=10.

Over the next few days, your student will be solving problems that require multiplying and dividing fractions. Some of these problems will be about comparison. For example:

• If Priya ran for 56 hour and Clare ran for 32 hours, what fraction of Clare’s running time was Priya’s running time?

  We can draw a diagram and write a multiplication equation to make sense of the situation.

  
  (fraction)⋅(Clare’s time)=(Priya’s time)
  ?⋅32=56
  We can find the unknown by dividing. 56÷32=56⋅23, which equals 1018. So Priya’s running time was 1018 or 59 of Clare’s.

Other problems your students will solve are related to geometry—lengths, areas, and volumes. For examples:

• What is the length of a rectangular room if its width is 212 meters and its area is 1114 square meters?
  

  We know that the area of a rectangle can be found by multiplying its length and width (?⋅212=1114), so dividing 1114÷212 (or 454÷52) will give us the length of the room. 454÷52=454⋅25=92. The room is 412meters long.

• What is the volume of a box (a rectangular prism) that is 312 feet by 10 feet by 14 foot?

  We can find the volume by multiplying the edge lengths. 312⋅10⋅14=72⋅10⋅14, which equals 708. So the volume is 708 or 868cubic feet.

Here is a task to try with your student:

1. In the first example about Priya and Clare’s running times, how many times as long as Priya’s running time was Clare’s running time? Show your reasoning.
2. The area of a rectangle is 203 square feet. What is its width if its length is 43feet? Show your reasoning.

Solution:

1. 95. Sample reasoning: We can write ?⋅56=32 to represent the question “how many times of Priya’s running time was Clare’s running time?” and then solve by dividing. 32÷56=32⋅65=1810. Clare’s running time was 1810or 95 as long as Priya’s.
2. 5 feet. Sample reasoning: 203÷43=203⋅34=204=5