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A R I T H M E T I C

Lesson 18

SOLVING PROPORTIONS

This Lesson follows from Lesson 17. Here, we will see how to solve any proportion.

 1. In a proportion, which are the corresponding terms? The 1st and 3rd terms, the 2nd and the 4th.

 12 = 510

"1 is to 2  as  5 is to 10."

1 is called the first term of the proportion, 2 is the second term, 5 is the third, and 10, the fourth.

We say that 5 corresponds to 1, and 10 corresponds to 2.

 2. What is the theorem of the alternate proportion? "If four numbers are proportional, then the  corresponding terms are also proportional." If then alternately,
 Example 1. 12 = 5 10

Directly:

"1 is to 2  as  5 is to 10."

(1 is half of 2;  5 is half of 10.)

Alternately:

"1 is to 5  as  2 is to 10."

(1 is a fifth of 5;  2 is a fifth of 10.)

 Example 2. 1236 = 2 6

Directly:

"12 is to 36  as  2 is to 6."   (Why?)

Alternately:

"12 is to 2  as  36 is to 6."   (Why?)

Example 3.   Complete this proportion:

 57 = 20 ?

"5 is to 7  as  20 is to  what number ?"

If we look directly at the ratio of 5 to 7, it is not obvious.  But if we look across, or alternately,

 57 = 2028

"5 is to 20  as  7 is to 28."

5 is a fourth of 20.  And 7 is a fourth of 28.

If we cannot solve a proportion directly, then we can solve it alternately.

Example 4.  The theorem of the same multiple.   Complete this proportion:

 45 = ? 15

"4 is to 5 as what number is to 15?"

Solution.  Alternately, we see that 5 is a third of 15 -- or we could say that 5 has been multiplied by 3.  Therefore, 4 also must be multiplied by 3 :

 45 = 12 3 × 4 15 3 × 5

 4 is to 5 as Three 4's are to three 5's as 12 is to 15.

In fact, as 4 is to 5, so any number of 4's are to an equal number of 5's.

That is called the theorem of the same multiple.  It follows directly from the theorem of the alternate proportion.

 3. What is the theorem of the same multiple? "If we multiply two numbers by the same number,  then the products will have the same ratio  as the numbers we multiplied."

(Euclid, VII. 17.)

We have already seen that a ratio will be preserved if we divide both terms by the same number.

Example 5.   Complete this proportion:

 67 = 12 ?

Solution.  6 has been multiplied by 2 to give 12.  Therefore the missing term must be two times 7.

 67 = 12 2 × 6  14 2 × 7

Example 6.   Solve this proportion:

 23 = ?18

Solution.  3 has been multiplied by 6.  Therefore, 2 also must be multiplied by 6:

 23 = 12 6 × 2  18 6 × 3

In fact, consider these columns of the multiples of 2 and 3:

2      3

4      6

6      9

8    12

10    15

12    18

14    21

And so on.

Now, 2 is two thirds of 3. (Lesson 17.)  And as 2 is to 3, so each multiple of 2 is to the same multiple of 3:

4 is two thirds of 6.

6 is two thirds of 9.

8 is two thirds of 12.

And so on.  In fact, those are the only natural numbers where the first will be two thirds of the second.

Example 7.   Name three pairs of numbers such that the first is three fifths of the second.

Solution.  The elementary pair are 3 and 5.  To generate others, take the same multiple of both:  6 and 10,  9 and 15,  12 and 20, and so on.

Example 8.   27 is three fourths of what number?

Solution.  Only a multiple of 3 can be three fourths of another number, which must be that same multiple of 4.

As 3 is to 4, so 27 is to ?

27 is 9 times 3; therefore the missing term is 9 times 4:

As 3 is to 4, so 27 is to 36.

27 is three fourths of 36.

Example 9.   Solve this proportion:

 9 45 = 2?

Solution.  Here, we must look directly:

9 is a fifth of 45.  And 2 is a fifth of 10.

 9 45 = 2 10

Example 10.  Common divisor.   Complete this proportion:

 12200 = ? 100

Solution.  Alternately, we see that 200 has been divided by 2.  Therefore 12 also must be divided by 2:

 12200 = 6  12 ÷ 2 100 200 ÷ 2

Equivalently, to go from 200 to 100, we must take half.  Therefore we must also take half of 12.

Summary
 To solve a proportion alternately, either divide both terms by the same number, as in the example above, or, as in Examples 5 - 8, multiply both terms by the same number.

As for the theorem of the common divisor, it is what we call the symmetrical version of the theorem of the same multiple. For, this proportion,

6 is to 100  as  12 is to 200,

in which the 3rd and 4th terms appear as doubles of the 1st and 2nd, is logically equivalent to this proportion,

12 is to 200  as  6 is to 100,

in which the 3rd and 4th terms appear as halves of the 1st and 2nd.

Example 11.   In a class, the ratio of girls to boys is 3 to 4.

There are 24 boys.  How many girls are there?

Solution.   Proportionally,

 GirlsBoys = 34 = ? 24

Note that 24 corresponds to the boys.

Now, 24 is 6 × 4.  Therefore, the number of girls is 6 × 3 = 18.

This is another way to approach Example 7 of the previous Lesson. And the following example is another way to approach Example 8 of that Lesson.

Example 12.  The whole is equal to the sum of the parts.   In a class, the number of girls is 75% of the number of boys.  There are 35 students. How many girls are there and how many boys?

Solution.   To say that girls are 75% of the boys, is to say that the ratio of

girls to boys is 3 to 4.  But that means that 3 out of every 7 students are girls (3 + 4 = 7), and 4 out of every 7 are boys.  Therefore form the proportion:

 _______ Girls _______Total number of students = 37 = ? 35

Since 35 = 5 × 7, the missing term is 5 × 3 = 15.

There are 15 girls.  And so there are 20 boys.

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