Just as there is more than one way to skin a cat, there is more than one way to solve coin problems. After I had worked on the following problem, the book did not explain the answer. That's o.k.; I discovered a method that worked anyway. If you are studying Math A (or helping someone who is), take a few minutes with the following problem, which I quote:
In Jill's purse the ratio of the number of quarters to the number of dimes to the number of nickels is 3 to 4 to 2. If the value of these coins is $8.75, how many of each coin does Jill have?
My thinking went something like this: For the first three quarters that are in the purse, there will be four dimes and two nickels. Now I have one group of coins, consisting of nickels, dimes, and quarters. That group is worth $1.25 (3 quarters = $0.75, 4 dimes = $0.40, two nickels = $0.10). How many of these coin groups, worth $1.25 each, are in the purse? To find the answer, we divide $8.75 (total value) by $1.25 (value of each group). That makes 7 groups of coins.
3 quarters per group x 7 groups = 21 quarters = $5.25
4 dimes per group x 7 groups = 28 dimes = $2.80
2 nickels per group x 7 groups = 14 nickels = $0.70
Total: $8.75
One of the presents that I received this Christmas (2003) was a book called Building Mental Muscle (David Gamon, Ph.D., and Allen D. Bragdon. It is a good companion to Brain Bombardment, my first book. One chapter, devoted to math, offers a word problem in algebra. It does not require a knowledge of any math beyond high school algebra, but it requires insight. I am quoting it from page 236 of the revised edition:
A little girl and her brother are hunting Easter eggs. The boy says to his sister: "Give me seven of your eggs and I'll have twice as many as you!" To which the little girl replies: "Give me seven of yours and we'll have the same amount." How many does each have?
A hint is in upside-down lettering at the bottom of the page. I am happy to say that I did not have to consult it. Before you read further, you may want to try solving the problem.
How did you do? I found the solution, but with more trouble than I had expected. Since the book does not explain a method for solving the problem (only the answer and a hint), I'll explain it here.
Like many of these word problems in algebra, this problem may have more than one way to arrive at the correct answer. My own way involved what the algebra books refer to as linear equations in two variables. A linear equation, as it name implies, appears as a straight line when graphed.
Before we set up any equations, we must not overlook one point: the boy has 14 more eggs (not 7) than the girl has. To be even, the girl will gain seven eggs; but the boy will lose 7 at the same time if he grants the girl's request.
We have two unknowns. Let x = the no. of eggs that the girl has now. Let y = the no. of eggs that the boy has now. Then y = x + 14. We are off to a good start, but it is not enough. y could equal 114, and x could equal 100; or y could equal 500, and x could equal 486. We need a second equation to help us.
Let y + 7 = the no. of eggs that the boy will have if he gets his way. Let x - 7 = the no. of eggs that the girl will have if the boy gets his way. Then the boy will have twice the no. of eggs that the girl has:
y + 7 = 2(x - 7)
Now we can use our two equations together. When solving similar problems, I put the second equation under the first to aid visualization:
y = x + 14
y + 7 = 2(x-7)
Here comes the tricky part. Since y = x + 14, get rid of the y in the lower equation and replace it with the x + 14. Now that we have an equation with just one variable, it's a piece of cake:
x + 14 + 7 = 2(x - 7)
x + 21 = 2x - 14
Switcheroo. Put the -14 on the left side of the equation, changing signs along the way. Put the x on the right side of the equation, changing signs along the way. Now 35 = x, the number of eggs that the girl has not; 49 = the no. of eggs that the boy has now. If the boy gives the girl seven of his eggs, they will each have 42. If the girl gives the boy 7 of her eggs, she will have 28, and he will have 56 (i.e., twice as many).
Despite some annoying misprints and errors, which I have begun to catalog, the Math A blue book (Barron's) is teaching me a lot. At first I was disappointed that the answer key lacked an explanation to the word problems. Then I decided to write my own explanation. Here is the Barron's word problem that I have in mind (from Section 2.3, Solving Equations with Like Terms):
Concert tickets purchased in advance cost $4.50 each, and tickets purchased ... on the day of the concert cost $8.00 each. The total ... collected in ticket sales was the same as if every ticket purchased had cost $6.00. If 180 tickets were purchased in advance, what was the total number of tickets purchased for the concert?
The problem above is a disguised form of the mixture problems that I studied in high school algebra during the 1960's. (Usually the problems contained peanuts or coffee beans.) Try to work it out before reading any further....
Setting up the problem is often harder than doing the actual computation. I began by letting x represent the total number of tickets sold. Then x - 180 equals the number of tickets sold on the day of the concert. Now I have 180 tickets sold at $4.50 each, plus the number of tickets (x - 180) sold at $6.00 each. I can now use the following expression for the amount of money collected from ticket sales: $4.50(180) + $8.00(x - 180).
But what does that expression equal in terms of x? It equals 6x because we already know that $6.00 was the average price of each ticket. Now we can write the actual equation:
x = 4.5(180) + 8(x - 180)Start of Math Class
6x = 810 + 8x - 1440
-2x = -630 (Remember to change signs when moving terms to the other side of the = sign.)
x = 313
There were 315 tickets (answer).