The first thing I want to talk about is a problem-solving technique that motivated the parenthetical addendum to the title of this blog. During the years following my discovery of this technique, I heard about it from two other sources, but these two sources had not, to my knowledge, generalized the technique to situations other than the first example that I discuss below.
A traditional verbal problem like the following may be found in many textbooks. Farmer Jones has a number of chickens and sheep. The animals have a total of a hundred heads and 280 legs. How many of each animal does Farmer Jones have? The standard solution to this type of problem is algebraic, using either one variable of two. Using two variables, let c = # of chickens and let s = # of sheep, so c + s = 100 (because the # of animals = the # of heads) and 2c + 4s = 280 (because each chicken has two legs for a total of 2c legs and each sheep has 4 legs for a total of 4s legs). This system of equations may then be solved, yielding c = 60 and s = 40. Using one variable, let c = # of chickens, so 100 – c = # of sheep. Solving 2c + 4(100 – c) = 280 yields c = 60 and 100 – c = 40.
This kind of problem can be solved arithmetically, i.e., without any equations to solve at all. If 2 legs per animal were counted, there would be 200 legs counted. That leaves 280 – 200 = 80 additional legs that must belong to the sheep (because all the chicken legs were counted, but only 2 legs of each sheep were counted). At 2 legs per sheep that were not counted, there must be 80 ÷ 2 = 40 sheep, and therefore 60 chickens. Which way would you rather solve this problem?
Let’s see what other kinds of traditional verbal problems may be approached in this way. If apples cost 90 cents apiece and peaches cost 65 cents apiece and John paid $22.75 for 30 of these fruit, how many of each did he buy? Again, the traditional solution is algebraic with one variable in one equation or two variables in two equations. But you don’t have to know algebra to solve this problem. If all 30 of the fruit John bought were peaches, he would have paid 30 · $.65 = $19.50 . That leaves $22.75 – $19.50 = $3.25 more that must have been paid for the apples. At 90 – 65 = 25 cents more per apple than per peach, John must have paid $3.25 ÷ 25 cents = 13 apples, so he bought 30 – 13 = 17 peaches.
This method can be used with any problem in which there are 2 things that we can label A and B and we know the total # of A’s and B’s, and each A has one quantity of items associated with it and each B has another quantity of the same items associated with it, and we know the total # of all those items.
In the animals problem, the A’s are chickens, the B’s are sheep, each A is associated with the quantity of its legs (2) and each B is associated with the quantity of its legs (4). Since we know the total # of animals (100) and the total # of legs (280), we can solve the problem in the manner indicated.
In the fruit problem, the A’s are apples, the B’s are peaches, each A is associated with its price in cents (90) and each B with its price in cents (65). Since we know the total # of fruit (30) and the total price of the fruit in cents (2275), we can solve this problem in the same way.
Note: I’m not claiming that this arithmetic procedure is easier than the traditional algebraic one in every case where it can be applied. But if you know only one way of doing things, you have no choice but to use it even in those cases where another way might be easier. And the quickest way to develop any ease with a new method is to practice it as much as possible, even in those cases where the method you already know is easier.
Let’s try another problem. In a weight loss group of 30 men and women, each man lost 15 pounds and each woman lost 12 pounds. If the total weight lost by the whole group was 393 pounds, how many of each sex were in the group? If you want to try the arithmetic procedure yourself before I do, identify the A’s, B’s and associated quantities and then perform the appropriate operations before you read any further.
The A’s are men, the B’s are women, the quantities associated with each A and each B are 15 (pounds) and 12 (pounds), respectively. The total # of A’s and B’s is 30 and the total # of pounds associated with all the A’s and B’s is 393, so we can use our new method. If each person in the group lost 12 pounds, 360 pounds would be lost, so the additional 393 – 360 = 33 pounds must have been lost by the men. Since each man lost 3 pounds more than each woman, there must have been 33 ÷ 3 = 11 men in the group and 30 – 11 = 19 women.
Sometimes it’s not so obvious what the A’s, B’s and associated quantities are. Consider another weight-loss problem. The weight loss group consists of 12 men and 18 women and the weight lost by any intersex pair was 22 pounds. If the total weight lost by the whole group was 342 pounds, how much weight was lost by each man and how much by each woman?
Here, the A’s and B’s are the weights lost by each man and by each woman, respectively, the quantities associated with each A and B are 12 and 18, respectively and the total of all the weight lost is 342, so our method is still applicable. If there were only 12 women in the group, 12 · 22 = 264 pounds would be lost, so the additional 342 – 264 = 78 pounds must have been lost by the additional 18 – 12 = 6 women, so each woman lost 78 ÷ 6 = 13 pounds and each man lost 22 – 13 = 9 pounds.
If, instead of the intersex-pair weight-loss, we knew only that each woman lost 4 pounds more than each man, the A’s, B’s and associated quantities would still be the same, so we would reason that if each woman lost only as much as each man, then 18 · 4 = 72 fewer pounds would have been lost, so 342 – 72 = 270 pounds would have been lost. Therefore, each man lost 270 ÷ 30 = 9 pounds and each woman lost 9 + 4 = 13 pounds.
Practice Problems. Try these without using algebra. Determine four numbers: the total # of A’s and B’s, the total # of a specified item associated with the A’s and B’s, the # of the specified item associated with each A and the number of that item associated with each B. Then perform the appropriate operations with those numbers.
- If a cup of pudding has 135 calories and a cup of yogurt has 115 calories and 25 cups containing pudding or yogurt have 3,175 calories, how many of them contain pudding and how many contain yogurt?
- Bill went fishing and caught 14 trout and bass with a total weight of 82 pounds. If each trout weighs 5 pounds and each bass weighs 7 pounds, how many of each did Bill catch?
- A factory assembly line can be programmed to produce gadgets or gizmos. It takes 2 minutes and 24 seconds to produce a gadget and a minute and 20 seconds to produce a gizmo. If 194 items are produced in 6 hours, how many of each item are produced?
- On the first leg of Bob’s 9-day cross-country trip, he drove 425 miles each day, and on the second leg, he drove 375 miles each day. If he drove a total of 3,525 miles, of how many days did each leg of the trip consist?
- The first leg of Emily’s cross-country trip was 4 days and the second leg was 7 days. On each day of the first leg, she drove 40 miles more than she did on each day of the second leg. If she drove a total of 4230 miles, how many miles per day did she drive on each leg of her trip? (Hint: This is like the third weight-loss problem.)