# Word Problems

Math Tutor Joe is the Word Problem Whiz!

You learn strategies to analyze the language of the word problem, to simplify the problem into its basic parts, to restate the problem in the form of mathematic concepts, and to apply step-by-step problem solving techniques to arrive at the solution. Math Tutor Joe makes it easy. Here is a demonstration of how to turn a word problem into a solution.

You have a pile of 50 coins, which contains nickels, dimes, and quarters. When you count them, they add up to \$6.90. The number of dimes is one and a half times the number of quarters. How many nickels, dimes, and quarters do you have in the pile of coins?

How Math Tutor Joe Breaks it Down

Step One: What is the important information?

You are told in the word problem that the number of nickels, dimes, and quarters added together equals 50 coins, which you write as:
N + D + Q = 50

You are told in the word problem that the total value of the pile of coins is 690 cents, and you already know the monetary value of a nickel, a dime, and a quarter, which you write as:
5N + 10D + 25Q = 690 cents

Step Two: How can you restate this information so that everything in the problem is similar?

You are told in the word problem that the number of dimes is one and a half times the number of quarters, which you write as:
D = 1.5Q

You can restate the original equation for the number of coins, N + D + Q = 50, so that you solve for N, and replace D with 1.5Q, which you write as:
N = 50 – Q – 1.5Q or

N = 50 – 2.5Q

You can again restate the original equation for the number of coins, N + D + Q = 50, so that all the factors are in the form of quarters; you replace N with 50-2.5Q and D with 1.5Q, which you write as:
(50 – 2.5Q) + 1.5Q + Q
= 50 coins

Now, you can restate the equation for the original equation to solve for the value of each type of coin, which you write as:
5(50 – 2.5Q) + 10(1.5Q) + 25Q = 690

Step Three:

Now that everything in the equation is similar (in the form of quarters), you can solve for Q!

Using multiplication, addition, and subtraction, you can solve this equation for Q:
250 – 12.5Q + 15Q + 25Q
= 690

250 + 27.5Q = 690

27.5Q = 440

Q = 16

Step Four:

Now that you know what Q equals (16), you can plug it back into the equation and determine the values of D and N:

Use multiplication to solve for D and N, which you write as:
D = 1.5(Q)

D = 1.5(16)

D = 24

and

N = 50 – 2.5(Q)

N = 50 – 2.5(16)

N = 10

Step Five: Solve the Word Problem!

The number of nickels is 10, the number of dimes is 24, and the number of quarters is 16.

10 Nickels = 50 cents

24 Dimes = 240 cents

16 Quarters = 400 cents

The total value of the pile of coins is 690, or \$6.90