(Euclid, VII. 17.)
4 is four fifths of 5. But each 4 is four fifths of each 5. Two 4's, then, upon adding them, will be four fifths of two 5's:
Four fifths of 5 + Four fifths of 5 = Four fifths of (5 + 5).
This is the essence of the proof of the alternate proportion.
Example 5. Complete this proportion:
Solution. Look at it alternately. 12 is two times 6. Therefore the missing term must be two times 7.
6
7 |
= |
12
2 × 6
14
2 × 7
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Example 6
. Complete this proportion:
Solution. 3 has been multiplied by 6. Therefore, 2 also must be multiplied by 6:
2
3 |
= |
12
6 × 2
18
6 × 3
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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 16.) And as 2 is to 3, so each multiple of 2 is to that 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, these 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 such 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 the same multiple of 4.
As 3 is to 4, so 27 is to ?
Since 27 is 9 × 3, therefore the missing term is 9 × 4:
As 3 is to 4, so 27 is to 36.
27 is three fourths of 36.
Example 9. Complete this proportion:
Solution. Here, we must look directly:
9 is a fifth of 45. And 2 is a fifth of 10.
Example 10. Complete this proportion:
Solution. Looking alternately, we see that 200 has been divided by 2. Therefore 12 also must be divided by 2:
12
200 |
= |
6
12 ÷ 2
100
200 ÷ 2
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Summary
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To solve a proportion alternately, either both terms must be divided by the same number, as in the example above, or, as in Examples 5 - 8, both terms must be multiplied by the same number.
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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,
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 here 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:
There are 15 girls. Hence there are 20 boys.
