There are numerous types of word problems that can be modeled using linear equations. Some of these types are so popular, they are categorized under specific headings.We will take a look at a few examples of these more popular types of questions. Just as there are numerous types of questions, there are also numerous ways to set-up and solve each of these questions.
If your teacher asks for a particular format for solutions, be sure to prepare the question using that method.

Real-World Modeling Using Linear Equations
All algebraic solutions to real-world word problems have a few concepts in common.
Remember to:
• translate the words into algebraic symbols with variables (or variable expressions).
• tell everyone what your variables (or expressions) represent. (Let x = ..... , and so on)
• be aware of any units of measure that may be used in the problem.
• clearly state the linear equation you will be using.
• show your work.
• read carefully to be sure you are giving the desired answer.
• clearly state your answer.
• check your answer to be sure your results "make sense" and are mathematically correct.

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ex1 Type: NUMBER PROBLEM
The larger of two numbers is four times the smaller. If the larger number is increased by 12, the result is 5 times the smaller number. Find the numbers.

Notice that this question asks you to find BOTH numbers.

 

Check: 48 is four times 12.
48 + 12 = 5 (12)
60 = 60 True

Let x = the smaller number
4x = the larger number
4x + 12 = the larger number increased by 12
5x = five times the smaller number

Create the equation: 4x + 12 = 5x

Solve the equation:
wm1
The numbers are x = 12 and 4x = 48.

 

ex2 Type: PERCENT PROBLEM
A new design of sneakers is on sale for $58.80 after being marked down 40%. What was the original price of the sneakers?
You worked with percent word problems in middle school, so this should look familiar.

Remember that 40% is 0.40.

Check: Marked down 40% means cost is 60%.
60% of $98 = $58.80.

wsneakers
Let p = the original price
p - 0.40p = the sale price (price minus a 40% mark down)

Create the equation: p - 0.40p = 58.80

Solve the equation:
wm2
The original price was $98.00.

 

ex3 Type: AGE PROBLEM
Kyle is 6 years older than Melissa. Nine years ago he was twice Melissa's age. How old is Kyle now?


Notice that this question asks specifically for Kyle's age.
   Wsiblings
Let a = Melissa's age now (the smaller age)
a + 6 = Kyle's age now
a - 9 = Melissa's age nine years ago
(a + 6) - 9 = Kyle's age nine years ago

Create the equation: (a + 6) - 9 = 2(a - 9)

Solve the equation:
wm3
Kyle's age is a + 6 = 15 + 6 = 21.
Check: Now, Kyle is 21 and Melissa is 15. Nine years ago, Kyle was12 and Melissa was 6. Kyle was twice her age.

 

ex4 Type: DISTANCE PROBLEM
A snowshoe hiker, Aaron, and a cross country skier, Kevin, are racing to see who can arrive at a point 200 feet away. Aaron travels at a constant speed of 30 feet per minute for the entire distance. Confident Kevin travels at a constant speed of 40 feet per minute for the first 30 feet, stops to throw snowballs for 3 minutes, and then continues at 35 feet per minute for the remainder of the distance. Who won the race?

A picture is always a good idea with distance problems. Get a visual idea of what is happening.wskier2
Distance = Rate x Time
Think of "rate" as speed.
Let's investigate each boy's race.
Let t = Aaron's time in minutes.
Aaron's equation: 200 = 30t
wm4c
Aaron: 6.667 minutes
Kevin's time, m, occurs in sections.
First 30 feet: 30 = 40m
Last 170 feet: 170 = 35m
Don't forget he stops for 3 minutes. m = 3
wm4a          wm4b
Add up Kevin's times:
0.75 + 4.857 + 3 = 8.607 minutes
Aaron wins the race!

Check: 200 = 30(6.667) and 30=40(.75); 170=35(4.857).

 

ex5
 Type: CONSECUTIVE INTEGER PROBLEM
See more at Consecutive Integer Expressions.
Find three consecutive even integers whose sum is 18.

"Consecutive" means "one after the other". Consecutive integers (such as 1, 2, 3) are represented as x, x+1, x+2.

Consecutive even integers (such as 2, 4, 6) are represented as x, x+2, x+4.

Consecutive odd integers (such as 3, 5, 7) are represented as x, x+2, x+4.


Check: 4 + (4 + 2) + (4 + 4) = 18
4 + 6 + 8 = 18
18 = 18

Let n = the first even number
n + 2 = the second even number
n + 4 = the third even number

Create the equation:
n + (n + 2) + (n + 4) = 18

Solve the equation:
wm6
The numbers are 4, 6, and 8.

 

ex6 Type: MONEY PROBLEM
Mario has 20 coins in quarters and nickels. He has a total of $2.00. How many quarters and nickels does he have?

When working with money be careful to express the money all in cents (50¢) or all in dollars ($0.50). This example is expressed all in dollars.

wmoney

Check: 0.25q + 0.05(20 - q) = 2.00
0.25(5) + 0.05(20 - 5) = 2.00
1.25 + 1 - 0.25 = 2.00
2.00 = 2.00

Let q = number of quarters
20 - q = number of nickels
0.25q = money from quarters
0.05(20 - q) = money from nickels

Create the equation:
0.25q + 0.05(20 - q) = 2.00

Solve the equation:
wm7
There are 5 quarters and 15 nickels.

 

ex7 Type: MIXTURE PROBLEM (dry mixture)
Sunflower seeds sell for $0.50 a pound and cracked corn sells for $0.30 a pound. How many pounds of each will be needed to create a 40 pound mixture of birdseed selling at $16.40?
wbird Mixture problems are often solved in a chart (or table). It can, however, be difficult to remember how to set up the chart, and what to put where in the chart.
A simple diagram may help:
birdseed
Let s = pounds of sunflower seeds
0.50s = cost of sunflower seeds in mix
40 - s = pounds of corn
0.30(40 - s) = cost of corn in mix

Create the equation:
0.50s + 0.30(40 - s) = 16.40

Solve the equation:
wm8

The 40 pound mix is 22 pounds of sunflower seeds and 18 pounds of corn.

Check: .50(22) + .30(40-22) = 16.40

 

Type: MIXTURE PERCENT PROBLEM (liquid mixture)
Your chemistry lab requires a 20% saline solution. The only solutions available for your use are 10% saline and 40% saline solutions. You decide to mix the two solutions to create the 20% solution you need. If you need 8 liters of 20% saline solution, how many liters of the 10% saline and 40% saline solutions should you mix?

wchemistry

Notice how we used the physical amount of solutions we needed (in liters) and the amount of pure saline it would give to the mix (% times the liters). See diagram below.
       beaker3

Let x = liters of 10% solution needed
0.10x = amount of saline in mix (10%)
8 - x = liters of 40% solution needed
0.40(8 - x) = amount of saline in mix (40%)
0.20(8) = amount of saline we need
                  (8 liters of 20% solution)

Create the equation:
0.10x + 0.40(8 - x) = 0.20(8)

Solve the equation:
wmath9

ANSWER:
wm9a

Check: .10(5 1/3) + .40(2 2/3) = .2(8)



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