Overview
Some very common word problems are problems involving digits and problems involving coins. Like motion problems, these word problems can also be solved using equations in two variables.
Setting up a Digit Problem
Any two-digit number can be written as 10x +y where x is the number in the tens column and y is the number in the ones column. Think of expanded notation where a number like 34 can be expressed as 3 times 10 plus 4. The reverse, 43, can be expressed as 4 times 10 plus 3. Common types of digit problems have the sum of the digits equal to a certain number. For example, both 3 + 4 and 4 +3 equal 7. However, if the number 34 is reversed to 43, it is 9 more.
Solving a Digit Problem
Suppose the sum of the digits of a two-digit number is 5. If the digits are reversed, the new number is 27 more than the original number. To set up the problem, find the first equation, x +y =5. If the digits are reversed, 10y +x = 27 +10x +y. Simplify the second equation, so that all the x’s and y’s are on one side of the equation, as 9x-9y =-27. Then there are 2 equations, x +y = 5 and 9x-9y =-27. 9(x +y) = 5(9), or 9x +9y = 45. They can then be solved using the addition method, so that 9x +9y =45 + 9x-9y =-27. 9x +9x = 18x, 9y-9y cancels out as zero, and 45-27 is 18. 18x = 18 means that x equals 1. If x equals 1, then 1 +y = 5, or y equals 4. The original number is 14, and 14 +27 is 41, so the problem checks.
Setting up a Coin Problem
Coin problems are similar to digit problems. There are different types of coins and a total, and that relationship can be expressed with one equation. There are also a number more for one type of coins than another type, and that relationship can be expressed with a second equation. The system of equations can then be solved.
Solving a Coin Problem
Suppose there are 20 coins, some quarters and some dimes. The value of all 20 coins is $3.05. How many quarters and how many dimes are there? Let q equal the number of quarters and d equal the number of dimes. The first equation is q +d = 20. The second equation is 25q +10d = 305. Each quarter is 25 cents and each dime is 10 cents. The number of cents total in $3.05 is 305. The number of dimes can also be expressed in a slightly different form, as d = 20-q. Using the substitution method, 25q +10(20-q) = 305, which can be solved as 25q +200-10q =305, rearranged as (25q-10q) or 15q = 305-200 or 105. If 15q =105, then q equals 7. The number of dimes is 20-7 or 13. To check, 25(7) or 175 plus 10(13) or 130, equals 305.
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