Lesson 18 RATIO AND PROPORTION 2This Lesson follows from Lesson 17. Here, we will see how to solve any proportion. In this Lesson, we will answer the following:
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"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. |
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(Euclid, VII. 13.)
(1 is half of 2; 5 is half of 10.) (1 is a fifth of 5; 2 is a fifth of 10.)
"12 is to 36 as 2 is to 6." (Why?) "12 is to 2 as 36 is to 6." (Why?) Example 3. Complete this proportion:
"5 is to 7 as 20 is to what number ?" 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: "4 is to 5 as what number is to 15?" 4 is to 5 as three 4's are to three 5's; as 12 is to 15. In fact, 4 is to 5 as any number of 4's are to an equal number of 5's. That is called the Theorem of the Same Multiple. It is a direct consequence of the Theorem of the Alternate Proportion. |
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(Euclid, VII. 17.) We have already seen that the ratio will also be preserved if we divide both terms by the same number. Example 5. Complete this proportion: Solution. 6 has been multiplied by 2 to give 12. Therefore the missing term must be two times 7. Example 6. Solve this proportion: Solution. 3 has been multiplied by 6. Therefore, 2 also must be multiplied by 6: In fact, consider these columns of the multiples of 2 and 3: Now, 2 is two thirds of 3. (Lesson 17.) 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, 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 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 ? Now, 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:
Solution. Here, we must look directly: 9 is a fifth of 45. And 2 is a fifth of 10.
Example 10. Common divisor. Complete this proportion:
Solution. Alternately, we see that 200 has been divided by 2. Therefore 12 also must be divided by 2:
Equivalently, to go from 200 to 100, we must take half. Therefore we must also take half of 12.
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? 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:
Since 35 = 5 × 7, the missing term is 5 × 3 = 15. There are 15 girls. Hence there are 20 boys. At this point, please "turn" the page and do some Problems. or |