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Learning Math Home
Session 9: Solutions
Session 9 Part A Part B Part C Homework
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Session 9 Materials:

A B C 


Solutions for Session 9, Part A

See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6| A7

Problem A1


The area is 3/12 (or 1/4) of a square mile.


The area is 6/15 (or 2/5) of a square mile.


The area is 8/36 (or 2/9) of a square mile.

<< back to Problem A1


Problem A2

If the factors are each less than 1, the product cannot be larger than its factors. The part that overlaps the horizontally shaded area and the vertically shaded area is the area of the product, and each of the original areas will be larger than the area shaded by both.

Note that this only works with "proper" fractions, where the numerator is less than the denominator (and therefore the fraction is less than 1).

<< back to Problem A2


Problem A3

In essence, you are looking to find the length of a rectangle whose one dimension is 3/4 and whose area is equal to 2/3 -- in other words, 3/4 • x = 2/3, or 2/3 3/4 = x.

You can start by marking 2/3 vertically. Next, mark 3/4 horizontally.

Then subdivide everything into ninths.

Then rearrange the top (purple) pieces from the original 2/3 area to fit the 3/4 height. The result will be the purple rectangle, whose length is 8/9:

2/3 3/4 = 8/9

<< back to Problem A3


Problem A4

Since we model division as the reverse of multiplication, the first fraction is represented as an overlap of two areas. It will be smaller than the quotient for the same reason that the multiplication area is smaller than either of the fractions multiplied. In some cases, the first fraction will be smaller than both the quotient and second fraction. (The illustration in the solution to Problem A3 is an example of such a case.)

If the first fraction happens to be larger than the second one in the division problem, the result will be an improper fraction (where the numerator is larger than the denominator).

<< back to Problem A4


Problem A5

All the answers are the same, 2, because the units are the same for each division problem.

<< back to Problem A5


Problem A6

The original question is "How many 3/4s are there in 3/5?" The common denominator is 20: 3/4 = 15/20, and 3/5 = 12/20. The question is now "How many 15/20s are there in 12/20?," which is easier to answer because the units are the same: 12/20 15/20 = 12 15 = 12/15 = 4/5. The answer is 4/5.

<< back to Problem A6


Problem A7

Both 0.6 and 0.2 can be expressed in units of tenths: "How many 2/10s are there in 6/10?" The answer to this question must be the same as "How many 2s are there in 6?" The answer to both questions will always be 3, regardless of the particular units involved.

<< back to Problem A7


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