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10

WORD PROBLEMS

Examples

Problems


WORD PROBLEMS fall into categories. Here are examples. The only difficulty will be translating verbal language into algebraic language. Practice.

Example 1.   Jane spent $42 for shoes.  This was $14 less than twice what she spent for a blouse.  How much was the blouse?

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

 Solution.   Always let x be the unknown number; that is, let x answer the question.  Let x, then, be how much she spent for the blouse.  

The problem states that "This" -- $42 -- was $14 less  than two times x.  Here is the equation:

2x − 14 = 42
 
2x = 42 + 14   (Lesson 9)
 
  = 56.
 
x = 56
 2
 
  = 28.

The blouse cost $28.

Example 2.   There are b boys in the class.  This is three more than four times the number of girls.  How many girls are in the class?

 Solution.   Let x answer the question; let x be the number of girls.

The problem states that b is three more than four times x:

4x + 3 = b
 
  Therefore,    
 
4x = b − 3
 
x = b − 3
   4

Here, the solution is not a number, because it will depend on the value of b.  This is a type of "literal" equation, which is very common in algebra.

Example 3.  The whole is equal to the sum of the parts.

The sum of two numbers is 84, and one of them is 12 more than the other.  What are the two numbers?

 Solution.   Again, we must let x answer the question.  But we are asked for two numbers.  Let x, then, be the first number.

We are told that the other number is 12 more, x + 12.

The problem states that their sum is 84:

  = 84

The line over x + 12 is a grouping symbol called a vinculum.  It saves us writing parentheses.

We have:

 
2x = 84 − 12
 
  = 72.
 
x = 72
 2
 
  = 36.

This is the first number.  Therefore the other number is

x + 12 = 36 + 12 = 48.

The sum of 36 + 48 is 84.

Example 4.   The sum of two consecutive numbers is 37.  What are they?

Solution.    Two consecutive numbers are like 8 and 9, or 51 and 52.

So, let x be the first number.  Then the next number is x + 1.

The problem states that their sum is 37:

  = 37

2x = 37 − 1
 
  = 36.
 
x = 36
 2
 
  = 18.

The two numbers are 18 and 19.

Example 5.   Divide $80 among three people so that the second will have twice as much as the first, and the third will have $5 less than the second.

Solution.   Let x be how much the first person gets.

Then the second gets twice as much, 2x.

And the third gets $5 less than that, 2x − 5.

Their sum is $80:

 

5x = 80 + 5
 
x = 85
 5
 
  = 17.

This is how much the first person gets.  Therefore the second gets

2x = 34.
 
        And the third gets
 
2x − 5 = 29.

The sum of 17, 34, and 29 is in fact 80.

Problems

Problem 1.   Julie has $50, which is eight dollars more than twice what John has.  How much has John?  (Compare Example 1.)

Let x answer the question -- how much John has.

Here is the equation.  (But first, make an effort yourself !)

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

2x + 8 = 50

Here is the solution:

x = $21

Problem 2.   Carlotta spent $35 at the market.  This was seven dollars less than three times what she spent at the bookstore; how much did she spend there?

Here is the equation.

3x − 7 = 35

Here is the solution:

x = $14

Problem 3.   There are b black marbles.  This is four more than twice the number of red marbles.  How many red marbles are there?  (Compare Example 2.)

Here is the equation.

2x + 4 = b

Here is the solution:

x =  b − 4
   2

Problem 4.    Janet spent $100 on books.  This was k dollars less than five times what she spent on lunch.  How much did she spend on lunch?

Here is the equation.

5x − k = 100

Here is the solution:

x =  100 + k
     5

Problem 5.  The whole is equal to the sum of the parts.

The sum of two numbers is 99, and one of them is 17 more than the other.  What are the two numbers?  (Compare Example 3.)

Here is the equation.

Here is the solution:

x  = 41
 
x + 17 = 58

Problem 6.   A class of 50 students is divided into two groups; one group has eight less than the other; how many are in each group?

Here is the equation.

Here is the solution:

x  = 29
 
x − 8 = 21

Problem 7.   The sum of two numbers is 72, and one of them is five times the other; what are the two numbers?

Here is the equation.

x + 5x = 72

Here is the solution:

x = 12.   5x = 60.

Problem 8.   The sum of three consecutive numbers is 87; what are they?  (Compare Example 4.)

Here is the equation.

Here is the solution:

28, 29, 30.

Problem 9.   A group of 266 persons consists of men, women, and children.  There are four times as many men as children, and twice as many women as children.  How many of each are there?

(What will you let x equal -- the number of men, women, or children?)

Let x = The number of children.  Then
 
4x = The number of men. And
 
2x = The number of women.
 
Here is the equation:

x + 4x + 2x = 266

Here is the solution:

x = 38.  4x = 152.  2x = 76.

Problem 10.   Divide $79 among three people so that the second will have three times more than the first, and the third will have two dollars more than the second.   (Compare Example 5.)

Here is the equation.

Here is the solution:

$11,  $33,  $35.

Problem 11.   Divide $15.20 among three people so that the second will have one dollar more than the first, and the third will have $2.70 more than the second.

Here is the equation.

Here is the solution:

$3.50,  $4.50,  $7.20.


Next Lesson:  Inequalities


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