Linear Equations and Word Problems
    Word problems that lead to simple linear equations
      Work problems
      Time and travel (distance) problems
      Miscellaneous word problems
Word problems that lead to simple linear equations
The general procedure to solve a word problem is:
1. Set the unknown.
2. Write equation from the text of the problem.
3. Solve the equation for the unknown.
Work problems
Example:  One of two workers can finish a job for 15 days, and other for 10 days, how long will it take to
finish the job working together.

Solution:   The first worker does 1/15 of the job per day, and the second worker does 1/10 of the job per day.

If working together they complete the job in x days, then
                                                              10x  + 15x = 150
                                                                         25x = 150
                                                                             x = 6 days.
Example:  Suppose a person A can finish a job in 12 days. He worked three days when a person B joins him
to help. Suppose the person B can finish the same job in 15 days. How long will they take to finish the job?

Solution:   If working together they complete the job in x days, then

Example:  A tank can be filled with three pipes:
                 - first pipe (alone) takes 10 hours (a = 10) to fill the tank,
                 - second pipe takes 12 hours (b = 12),
                 - third pipe takes 15 hours (c = 15)  to fill the tank.
How long would it take to fill the tank with water,
               a)  if all pipes are opened at the same time,
               b)  if second and third pipes fill the tank while at the same time water leaking from the tank
through the first pipe?

Solution:  a) first pipe fills 1/a part of the tank per hour, second pipe fills 1/b and third pipe fills 1/c part of the

tank per hour. It will take x hours to fill the thank, thus
Time and travel problems - Distance, rate (or speed) and time relations
Use formulas:
distance = rate time,  
Example:  A rider has to catch a pedestrian up who is already 7 hours on his route. How long it will take if
the rider travels at the rate of 12 kilometers per hour and the pedestrian travels at 5 kilometers per hour?

Solution:  The same distance rider travels x hours, the pedestrian travels (x + 7) hours,

since,  distance = rate time,  then     12 x = 5 (x + 7)

                                                             12x = 5x + 35
                                                               7x = 35
                                                                 x = 5 hours.
Example:  To travel the distance between two stations a passenger train, traveling at rate of 12 meters per second, takes 16 minutes and 40 seconds less than a freight train moving at speed of 8 meters per second.
What is the distance?

Solution:  We equate the times denoting the distance by x,

  where, 16 minutes and 40 seconds = 1000 seconds
                                  2x + 2400 = 3x
                                                x = 24000 meters = 24 kilometers
Example:  A walker walking at the rate of 1 kilometer in 12 minutes travels a distance from A to B and spend
the same time as a cyclist who travels 10 kilometers longer distance riding at speed of 1 km in four and the
half minutes. What is the distance from A to B?

Solution:    Walker took a way of x kilometers at the rate of 1/12 kilometers per minute through the time

of x/ (1/12) minutes.
The cyclist travels (x + 10) kilometers at the rate of 1/(4 and 1/2) kilometers per minute through the time
   of    (x + 10) / [1/(4 and 1/2)] minutes.  
They traveled the same period of time, thus
Miscellaneous word problems
Example:  If fresh grapes contain 90% water and dried 12%, how much dry grapes we get from 22 kg of fresh grapes?
Solution:  Fresh grapes contain 90% water and 10% dry substance. 
                Dry grapes contain 12% water and 88% dry substance. 
    22 kg of fresh grapes = x kg of dry grapes, so 
Example:  An amount decreased 20% and then increased 50%, what is the total increase in relation to initial value.
Solution:  An initial amount x decreased by 20%,  
The obtained amount increased by 50%,  
The difference in relation to the initial value x,  
         shows increase by 20%.
Example:  From a total deducted is 5% for expenses and the remainder is equally divided to three persons.
What was the total if each person gets $190?
Solution:  If x denotes the total then  
Example:  Into 10 liters of the liquid A poured is 4 liters of the liquid B and 6 liters of the liquid C.
From obtained mixture D poured is out 3 liters, how many liters of the liquid C remains in the mixture D?
Solution:    A + B + C = D      =>     10 l + 4 l + 6 l  = 20 l
Intermediate algebra contents
Copyright 2004 - 2020, Nabla Ltd.  All rights reserved.