
Checking Proportionality (page 3 of 7) Sections: Ratios, Proportions, Checking proportionality, Solving proportions There is some terminology related to proportions that you may need to know. In the proportion: ...the values in the "b" and "c" positions are called the "means" of the proportion, while the values in the "a" and "d" positions are called the "extremes" of the proportion. A basic defining property of a proportion is that the product of the means is equal to the product of the extremes. In other words, given the proportional statement: ...you can conclude that ad = bc. (This is, in effect, the crossmultiplication demonstrated on the previous page.) This relationship is occasionally turned into a homework problem, such as:
For these ratios to be proportional (that is, for them to be a true proportion when they are set equal to each other), I have to be able to show that the product of the means is equal to the product of the extremes. In other words, they are wanting me to find the product of 140 and 30 and the product of 24 and 176, and then see if these products are equal. So I'll check: 140 × 30
= 4200 While these values are close, they are not equal, so I know the original fractions cannot be proportional to each other. So the answer is that they are not proportional. The other technical exercise based on terminology is the finding of the "mean proportional" between two numbers. Mean proportionals are a special class of proportions, where the means of the proportion are equal to each other. An example would be: .^{1}/_{2} = ^{2}/_{4} Copyright © Elizabeth Stapel 20012011 All Rights Reserved ...because the means are both "2", while the extremes are 1 and 4. This tells you that 2 is the "mean proportional" between 1 and 4. You may be given two values and be asked to find the mean proportional between them.
I'll let "x" be the number that I'm looking for. Since x will also be both of the means, I'll set up my proportion with 3 and 12 as the extremes, and x as both means:
Now I'll solve for x: Since I am looking for
the mean proportional of 3 and 12,
you would figure that I would need to take the positive answer, so that
the mean proportional would be just the 6.
However, considering the fractions, either value would work: So, actually, there are two mean proportionals: –6 and 6 Your book (or instructor) may want you only to consider the positive mean proportional, since the positive value is between 3 and 12.
I'll set this up the same way as before, and solve: _{} (–3)(–12)
= x^{2} So there are again two mean proportionals: –6 and 6 Your book (or instructor) may only want "–6" as an answer.
Note the difference is signs; this problem is different from the ones that preceded it. But I can set up the proportion in the exact same way: Now I'll solve for x: (–3)(12)
= x^{2} Since I can't take the square root of a negative number, then there is no solution for the mean proportion of the two given values.
At first, you might think that this isn't possible, but it is. I'll just set up the proportion using fractions within fractions, and proceed normally: So the two mean proportionals are ^{–3}/_{4} and ^{3}/_{4}. Your book (or instructor) may only be looking for "^{ 3}/_{4} ". << Previous Top  1  2  3  4  5  6  7  Return to Index Next >>


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