paired weighings, different approaches

Paired Weighings, Different ApproachesAuthor(s): Dick Stanley, Uri Treisman and Harris S. ShultzSource: Mathematics in School, Vol. 20, No. 5 (Nov., 1991), pp. 12-14Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30216548 .

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IRED

WEICHINGS

different approaches by Dick Stanley, University of California Berkeley, Uri Treisman, University of California Berkeley

and Harris S. Shultz, California State University, Fullerton

The Paired Weighings Problem Three men, Tom, Dick, and Harry, step on scales so that each has one foot on the same scale as each of the other two. It is assumed that the weight of each is equally divided between the two scales he shares. Thus, each number in Figure 1 represents the average of the weights of the two men who occupy the scale. For example, the average of Tom's weight and Dick's weight is 167. The problem is to determine the weight of each man.

First we approach the problem algebraically. If T, D, and H represent their respective weights, then

T+D - 167

2

T+H = 174 2

D+H - 192

2

This is a system of three linear equation in three unknowns, T+ D = 334

T + H= 348

D + H= 384,

which can be solved by the standard methods.

Dieting With some insight, this problem can be handled without algebra. Suppose Harry were to go on a diet. Since his weight is divided evenly between each of his two scales,

Tom

167

Dick

192 174 Harry

Fig. 1

the readings on each will be diminished by 1 pound for each 2 pounds he loses. Thus, if Harry should lose 14 pounds, the paired weighings will be as shown in Figure 2. The loss of 14 pounds was deliberately chosen so that Tom's two scales show the same weight, 167. Since, when paired with Tom, Dick and "skinny" Harry are now identical, Dick and "skinny" Harry must now weigh the same. But, since the average of their weights is 185, each

167 Dick Tom

167 185

/Skinny Harry\

Fig. 2

must weight 185. Thus, Dick weighs 185 pounds and the undiminished Harry must have weighed 14 pounds more than Dick, that is, 199 pounds. Tom's weight can be

Mathematics in School, November 1991 12



determined in a similar fashion. If the original (Figure 1) Harry were to lose 50 pounds, then the paired weighings will be as shown in Figure 3. Since Tom and Harry now weigh the same, Tom's weight must be 149 pounds. This

167 Tom

149

Dick

167

More Skinny Harry

Fig. 3

discussion demonstrates that, although algebra is a power- ful and versatile tool, there are situations where less "ad- vanced" approaches yield simpler and more interesting solutions.

Guess and Check The problem also can be approached by a "guess and check" method. Looking again at Figure 1, let us guess that Tom's weight is 160. Since the average of Tom's weight and that of Harry is 174, Harry would weigh 188. From this we would infer that Dick's weight would be 196 (192 is the average of 188 and 196). This, in turn, would tell us that the average of the weights of Tom and Dick would be 178. However, 178 is 11 larger than their actual average of 167. So, we lower our guess of Tom's weight by 11. That is, we guess that Tom's weight is 160-11= 149. Since the average of Tom's weight and that of Harry is 174, Harry would weigh 199. From this we would infer that Dick's weight would be 185, which is correct since the average of the weights of Tom and Dick is, in fact, 167.

More than luck is involved here. By reducing the guess of Tom's weight by 11 we increase the guess of Harry's by 11 and, therefore, reduce the guess of Dick's by 11. It is easy to see that any change in a guess of Tom's weight results in an equal change in the average of the weights of Tom and Dick. This method of "guess and check" can be extended to cases in which four or more people and scales are involved. Suppose Angie, Janet, Michelle, and Tanya step on scales as indicated in Figure 4. If we guess that Angie's weight is, say, 110, then Janet's weight would be 98, Michelle's 134, and Tanya's 118. This would give Angie and Tanya an average weight of 114, which is 8 less than their actual average of 122. So, we increase by 8 our guess of Angie's weight. If Angie's weight is 118, then Janet's weight would be 90, Michelle's 142, and Tanya's 110. Again, the average of Angie's and Tanya's weights would be 114, that is, 8 less than their actual average of 122. Indeed, any guess x of Angie's weight will lead to an average of 114 for Angie and Tanya since Janet would weigh 2(104) - x, Michelle would weigh 2(116 - 104) + x, and Tanya would weigh

2(126 - 116 + 104)- x= 2(114)- x.

Thus, the situation indicated by Figure 4 is impossible. On the other hand, if the paired weighings are as

indicated in Figure 5, then any guess of Angie's weight would lead to a solution. That is, there are an infinite number of solutions. A necessary and sufficient condition

122 Angle 104

Janet Tanya

126 Michelle

116

Fig. 4

for this to happen is that the sum of the weights on two non-adjacent scales equal the sum of the weights on the other two scales (in our example, 114+ 116= 104+ 126).

114 Angie

104

Tanya Janet

126 116 Michelle

Fig. 5

In the case of five scales, there will always be a unique solution. For example, looking at Figure 6, let us guess that Art weighs 160. Then Bob would weigh 200, Cal would weigh 168, Dan would weigh 172, and Ed would weigh 128, so that the average of the weights of Art and Ed would be 144. Since 144 is 18 less than their actual average, we revise our guess of Art's weight to be 160 + 18 = 178. Bob's weight would then be 182, Cal's 186, Dan's 154, and Ed's 146. The average of the weights of Ed and Art would then be the given amount of 162.

Ed

150

Dan

162

Art

180

Bob

170 184 Cal

Fig. 6

Mathematics in School, November 1991 13



Using Variables but Still No Formal Algebra So far we have used approaches that apply good thinking strategies to solving a type of problem without the need of any formal algebra. Such work is important for students to see whether they are before, in, or after algebra. It counterbalances an approach of many textbooks that applies mechanical algebraic techniques to solving a type of problem without the need of any thinking.

As an extension of our approach we proceed to show how the same problem situation can be very effectively analyzed using variables and not numbers. There is still no formal algebra, but students will be able to see how natural and useful it is as an aid to thinking about a problem to let certain quantities be represented by letters.

V

Tom

U

Dick

W Harry

Fig. 7

Consider again the situation of three men Tom, Dick, and Harry, only this time with scale readings U, V, and W, as indicated in Figure 7. Suppose we start with all three men off the scales, so we have the scale readings U= V= W= 0. If Tom gets on first, his weight is just the sum of the scale readings U and V. Suppose next Dick gets on the scales. This creates new scale readings V and W but doesn't affect U. How can Tom's weight be determined from the scale readings now? Clearly U+ V gives too high a reading for Tom's weight, since Dick has added some- thing to reading V. How much has Dick added to V? The same amount he has added to W! Since no one else is on scale designated to be shared by Dick and Harry, we see that Dick has added exactly W to each of his scales. So V-W is what the scale shared by Tom and Dick read before Dick got on and we can conclude that

Tom's weight = U + V- W.

What happens now when Harry gets on? He will add equal amounts to readings U and W. But since U and W have coefficients + 1 and - 1, respectively, in the above formula for Tom's weight, it is still valid! The same argument holds for Dick's weight and Harry's weight. So we have proved quite generally that each person's weight is found by adding the two adjacent scale readings and subtracting the scale reading across the way.

The important point to be made is that we have found a general result and have used symbols to express it, but have used only common sense, and no formal algebra. This is a kind of reasoning that students should be encouraged to use along with the algebra they are learning. It will mean that when they come to solve this problem using simultaneous equations they will have much better insight into the overall situation. They will not just be applying a memorized technique blindly.

Consider again the situation of four women, Angie, Janet, Michelle, and Tanya, stepping on scales as indicated in Figure 8. Again, we want to see if we can determine each person's weight from the readings U, V, W, and X. As before, let's start with everyone off the scales. If Tanya steps on, her weight will be given by T= U+ V. When Angie steps on too, we can reason just as before that

Tanya's weight will now be T= U + V- W. When Michelle steps on, she adds to the reading U by just the reading X,

V Angie

W

Tanya Janet

U X Michelle

Fig. 8

so that we now have T= U+ V-W- X. So far so good. But what happens when Janet steps on? She will increase the readings W and X, making the value for T too small. But, since W and X both have coefficient - 1 in the formula T= U+ V- W- X, there is no canceling out as before. We get instead

T= U+ V-W-X+J,

where J denotes Janet's weight. This is not good, since Tanya's weight is not being determined from the readings of the scales U, V, W, and X alone; we need to know Janet's weight also.

Is there some other formula for T in terms only of U, V, W, and X that will work? It turns out that the answer is no. In fact, there is really an essential difference in the four scale situation from the three scale situation. In the four scale case, different values of the women's weights are compatible with the same scale readings U, V, W, and X. This surely renders fruitless any attempt to determine T solely from U, V, W, and X. To give an example of this kind of situation, suppose all four women are on the scales, and that Tanya and Janet suddenly lose 10 pounds each, while Angie and Michelle suddenly gain 10 pounds each. (Perhaps they transfer some objects they are holding.) Thinking about the situation we see readily that this shift in weights has no effect on any of the scale readings. So surely the scale readings can't be used to find the weights.

As an extension students can show that the deterministic analysis for three scales works for any odd number of scales, and that in general all the scale readings are necessary to determine any given person's weight. Similarly, the non-deterministic situation holds for any even number of scales.

If students carry out this kind of reasoning for them- selves, it will have real benefits for their ability to reason symbolically, and hence a real benefit as they learn algebra. Lest anyone think of this as a contrived exercise unrelated to practical situations, we should point out that in various applications it actually happens that meter readings rep- resenting averages of real phenomena must be used to reconstruct the individual values that gave rise to the averages. See the discussion on page 8 by Ivars Peterson.

Reference Peterson, Ivars (1988). The Mathematical Tourist. W. H. Freeman and

Company, New York.

14 Mathematics in School, November 1991



paired weighings, different approaches

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