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RB Astillero                                       College Algebra                           Word Problems


WORD PROBLEMS INVOLVING LINEAR EQUATIONS IN ONE UNKNOWN

A word problem describes a situation involving both known and unknown quantities. Many real-
life problems can be solved by expressing them as word problems that often lead to equations. The
equations that represent a word problem are called mathematical models. The success in working
out word problems depends largely on one’s ability to translate it into a mathematical model.

TRANSLATING FROM WORDS TO MATHEMATICAL EQUATIONS
Usually there are key words and phrases in a verbal problem that translate into mathematical
expressions involving addition, subtraction, multiplication, and division. Translations of some
commonly used expressions follow.




Examples:
Translate each verbal sentence into an equation.
    1. Twice a number, decreased by 3, is 42.                          Answer: 2x - 3 = 42.
    2. The product of a number and 12, decreased by 7, is 105.       Answer: 12x - 7 = 105.
                                                                                 ௫
    3. The quotient of a number and the number plus 4 is 28.           Answer:       = 28.
                                                                                 ௫
                                                                                 ௫
    4. The quotient of a number and 4, plus the number, is 10.         Answer: + x = 10.
                                                                                 ସ

                                                                                                         1
RB Astillero                               College Algebra                              Word Problems


PROBLEM-SOLVING TECHNIQUES
Probably the most famous study of problem-solving techniques was developed by George Polya
(1888-1985). Among his many publications was the modern classic How to Solve It. In this book, Polya
proposed a four-step process for problem solving.

Polya’s Four-Step Process for Problem Solving
       1. Understand the problem. You must first decide what you are to find.
       2. Devise a plan. Here are some strategies that may prove useful.
                   If a formula applies, use it.
                   Write an equation and solve it. 
                  Draw a sketch.
                   Make a table or a chart.
                   Look for a pattern.
                   Use trial and error.
                   Workbackward.
       3. Carry out the plan. This is where the algebraic techniques you learned can be helpful.
       4. Look back and check. Is your answer reasonable? Does it answer the question that was
              asked?

STEPS IN SOLVING WORDED PROBLEMS
Although no standard technique for solving a word problem is prescribed, the following six steps
are recommended. These steps specifically apply Polya’s techniques to word problems in linear
equations.

Step 1 Read the problem carefully until you understand what is given and what is to be found.
Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed.
       Write down what the variable represents. If necessary, express any other unknown values in
       terms of the variables.
Step 3 Write an equation using the variable expression(s). Step
4 Solve the equation.
Step 5 State the answer to the problem. Does it seem reasonable?
Step 6 Check the answer in the words of the original problem.

TYPES OF WORD PROBLEMS
There are many types of word problems, but in this module you will only learn those types that are very
common in engineering applications. The following examples will illustrate how a word problem
is solved using the suggested six steps.

Example 1. Geometry Problem
The length of a rectangle is 1 cm more than twice the width. The perimeter of the rectangle is 110 cm.
Find the length and the width of the rectangle.
Solution:
Step 1 Read the problem. What must be found? The length and width of the rectangle.
       What is given? The length is 1 cm more than twice the width; the perimeter is 110 cm.



                                                                                                     2
RB Astillero                                    College Algebra                                Word Problems


Step 2Assign a variable. Choose a variable: let W = the width; then 1 + 2W = the length.


                         W


                                  1 + 2W

Step 3Write an equation. The perimeter of a rectangle is given by the formula P = 2L + 2W. Substituting
        P = 110, L = 1 + 2W, the required equation is
                                 110 = 2(1 + 2W) + 2W
Step 4 Solve the equation obtained in Step 3.
                               110 = 2 + 4W + 2W
                               110 = 2 + 6W
                               110 - 2 = 6W
                               108 = 6W
                                 ଵ଴
                                       =W
                                   ଴
                              18 = W
                              W = 18 [by symmetric equality axiom]
Step 5 State the answer. The width of the rectangle is 18 cm and the length is 1 + 2(18) = 37 cm.
Step 6 Check the answer by substituting these dimensions into the words of the original problem.

Example 2: Motion Problem
Two cars leave the same place at the same time, one going east and the other west. The eastbound car
averages 40 mph, while the westbound car averages 50 mph. In how many hours will they be 300 mi
apart?
Solution:
Step 1 Read the problem. We are looking for the time it takes for the two cars to be 300 mi apart.
Step 2 Assign a variable. A sketch shows what is happening in the problem: The cars are going in opposite
       directions.




        Let x = the time (in hours) the cars will be 300 mi apart. When the expressions for rate and time are
        entered, fill in each distance by multiplying rate by time using the formula d = rt.




                                                                                                           3
RB Astillero                                     College Algebra                            Word Problems


Step 3 Write an equation. From the sketch in step 2, the sum of the two distances is 300. Thus the
      equation is 40x + 50x = 300.

Step 4Solve the equation.
                         40x + 50x = 300
                               90x = 300
                                 x = ଷ଴଴
                                      ଽ଴


                                  x= ଵ = 3ଵ
                                     ଷ
                                      ଴
                                          ଷ

Step 5 State the answer. The cars travel 3ଵ hr or 3 hr 20 min.
                                                ଷ
Step 6 Check. The eastbound car traveled 40(ଵ ) = ସ଴ mi and the westbound car traveled 50(ଵ ) = ହ଴ mi,
                                             ଴      ଴                                      ଴      ଴
                                                ଷ      ଷ                                    ଷ     ଷ
       for a total of ସ଴ + ହ଴ = ଽ ଴ = 300 mi, as required.
                        ଴    ଴     ଴
                     ଷ      ଷ     ଷ

Example 3. Mixture Problem
A chemist must mix 8 L of a 40% acid solution with some 70% solution to get a 50% solution. How much of the
70% solution should be used?
Solution:
Step 1 Read the problem. The problem asks for the amount of 70% solution to be used.
Step 2 Assign a variable. Let x = the number of liters of 70% solution to be used. The information in the
       problem is illustrated in the figure below.




       Use the given information to complete a table as shown below.




       The numbers in the last column were found by multiplying the strengths and the numbers of
       liters. The number of liters of pure acid in the 40% solution plus the number of liters in the 70%
       solution must equal the number of liters in the 50% solution.

Step 3 Write an equation.
                     3.2 + 0.70x = 0.50(8 + x)
Step 4 Solve.        3.2 + 0.70x = 4 + 0.50x
                             0.20x = 0.8

                                                                                                         4
RB Astillero                                  College Algebra                               Word Problems


                                  x=4
Step 5 State the answer. The chemist should use 4 L of the 70% solution.
Step 6 Check. 8 L of 40% solution plus 4 L of 70% solution is 8(0.40) + 4(0.70) = 6 L of acid.
        Similarly, 8 + 4 or 12 L of 50% solution has 12(0.50) = 6L of acid in the mixture. The total
        amount of pure acid is 6 L both before and after mixing, so the answer checks.

Example 4. Geometry Problem
Find the value of x, and determine the measure of each angle in the triangle.
Figure:




Solution:
Step 1 Read the problem. We are asked to find the measure of each angle. Step
2 Assign a variable. Let x represent the measure of one angle.
Step 3 Write an equation. The sum of the three measures shown in the figure must be 180°.
                                  x + (x+20) + (210 -3x) =180
Step 4 Solve.                     2x + 230 - 3x = 180
                                        - x = 180 - 230
                                        - x = - 50
                                          x = 50
Step 5 State the answer. One angle measures 50°, another measures x + 20 = 50 + 20 = 70 , and the third
measures 210 - 3(50) = 60.
Step 6 Check. Since 50° + 70 + 60 = 180, the answers are correct.

Example 5. Number Relation Problem
Find three consecutive integers such that the sum of the first and third, increased by 3, is 50 more than
the second.
Solution:
Let x represent the first of the unknown integers. Then x + 1 will be the second, and x + 2 will be the
      third. The equation we need can be found by going back to the words of the original problem.




Solving the equation:          x + x + 2 + 3 = x + 1 + 50
                               2x + 5 = x + 51
                               2x - x = 51 - 5
                                    x = 46
Therefore, the three consecutive integers are 46, 47, and 48. (ans.)

Check: 46 + 48 + 3 = 47 + 50
                   97 = 97

                                                                                                        5
RB Astillero                                                  College Algebra              Word Problems


Example 6. Number Relation Problem
In a certain class, 12 more than ଵ of all the students are taking a science course and ଵ of those taking
                                              ଷ                                        ସ
a science course are also taking mathematics. If ଽ    of all the students in the class are taking
                                                   ଵ଴
mathematics, how many students are in the class?
Solution:
Let x = the number of students in the class.
     ଵ
         x + 12 = number of students who are taking science course.
     ଷ
     ଽ
         x = number of students who are taking mathematics.
     ଵ଴
     ଵଵ
      ( x + 12) = ଵ x + 3 = number of students who are taking both science and mathematics.
     ସ ଷ            ଵଶ

     By Venn Diagram:




                                  ଵ                   ଵ         ଽ       ଵ
                              ( x + 12)                 x+3         x - ( x + 3
                                  ଷ                   ଵଶ       ଵ଴       ଵଶ
                                      ଵ
                                  -(   ଵx
                                        ଶ   + 3)



                                       science                mathematics



                         ଵ
                         x + 12+ ଽ x - ( ଵ x + 3 x
                                                 =
                     ଷ                   ଵ଴        ଵଶ
                     ଵ        ଽ          ଵ
                         x+        x-        x - x +9 = 0
                     ଷ        ଵ଴         ଵଶ
                     ଵ ଴      ଴
                                             x = -9
                                   ଽ    x=-9
                                        x = 48

Therefore, there are 48 students in the class.
Check: number of students taking science = ଵ   (48) + 12 = 28
                                                               ଷ
           Number of students taking math                  = ଽ (48) = 27
                                                               ଵ଴
           Number of students both taking math & science = ଵ 28= 7
                                                            
                                                                                   ସ
           Number of students taking science only = 28 - 7 = 21
           Number of students taking math only = 27 - 7 = 20
           Total number of students in the class = 7 + 21 + 20 = 48 [check]



                                                                                                       6
RB Astillero                                       College Algebra                     Word Problems


Example 7: Age Problem
A 30-year old father has 4-year old son. In how many years will the father be three times as old as
his son?
Solution:
Let x = number of years, from now, by which the father’s age will be thrice that of his son.
                                  Present age Future age
                 Father                30          30 + x
                 Son                    4           4+x

                       30 + x = 3(4 + x) [why?]
                       30 + x = 12 + 3x
                             x=9
Therefore, 9 years from now the father will be thrice as old as his son.
Check: In 9 years the father will be (30 + 9 =) 39 yrs. of age while his son will be (4 + 9 =) 13 yrs.
       old. 39 is 3 times as much as 13, i.e. 3(13) = 39 as required.

Example 8: Age Problem
The sum of Kim’s and Kevin’s ages is 18. In 3 years, Kim will be twice as old as Kevin. What are their
ages now?
Solution:
Let x = present age of Kim.
    18 - x = present age of Kevin.
                                   Present age Future age
                         Kim                   x             x+3
                                                          (18 - x) + 3
                        Kevin              18 - x
                                                            = 21 - x

                               2(21 - x) = x + 3
                               42 - 2x = x + 3
                                    3x = 39
                                     x = 13
                                 18 - x = 5
Therefore, Kim is 13 yrs. old and Kevin is 5 yrs. old.
Check. In 3 years Kim will be 16 yrs. old while Kevin will be 8 yrs. old; 16 = 2(8).

Example 9. Rate of Work Problem
Lito can do a piece of work in 8 days, while Noli can do the same work in 10 days. In how many days
can they finish the job working together?
Solution:
Let x = the number of days it will take them to finish the job together.
    ଵ
        = the part of the job accomplished by Lito in 1 day.
    ଵ
         = the part of the job accomplished by Noli in 1 day.
    ଵ଴
                                 ଵ
                                     + ଵ =ଵ
                                      ଵ଴   ௫

                                                                                                      7
RB Astillero                                       College Algebra                                  Word Problems


                                             ଵ
                                         =
                                    ସ଴       ௫
                                     ଽ       ଵ
                                         =
                                        ସ଴ ସ௫
                                        x=  ଴
                                             ଽ   =4ସ
                                                   ଽ

Therefore, they will finish the job working together in 4 ସ days.
                                                                  ଽ
Check. The part of the job they will be accomplish in 1 day if working together is ଽ               which is equal
               ଵ   ଵ
                                                                                              ସ଴
           to +
                   ଵ଴

Example 10. Rate of Work Problem
Two pipes A and B operate independently at their respective constant rates. Pipe A alone takes 5
hours to fill the tank. When pipes A and B are used simultaneously, it takes them 2 hours to fill the tank.
How long will it take pipe B alone to fill the tank?
Solution:
Let x = number of hours it would take pipe B working alone to fill the tank.
     ଵ
         = part of the tank filled in by B in 1 hour.
     ௫
     ଵ
         = part of the tank filled in by A in 1 hour.
     ହ
     ଵ
         = part of the tank filled in by A and B in 1 hour.
     ଶ
                           ଵ
                           ௫   + ଵ= ଵ
                                 ହ  ଶ

                          LCD = 10x
                         10 + 2x = 5x
                            10 = 3x
                            3x = 10
                             x=ଵ  ଴
                                    ଷ
Therefore, it will take pipe B alone to fill the tank in ଵ or 3ଵ
                                                          ଴     hours.
                                                              ଷ       ଷ
Check. Pipes A and B running simultaneously will fill the tank in భଵ            =
                                                                                    బశభఱ   = మఱ ହ଴ 2 hours.
                                                                                             ଵ= =
                                                                              భయబ
                                                                          ఱ         ఱబ       ఱబ    ଶହ
                                                                          భ
Example 11. Digit Problem
The tens digit of a number is 3 less than the units digit. If the number is divided by the sum of the
digits, the quotient is four and the remainder is 3. What is the original number?
Solution:
Let x = the units digit.
     x - 3 = the tens digit.
    2x - 3 = sum of digits.
   10(x-3) + x = 10x - 30 + x = 11x - 30 = the number.

                           ଵ௫
                           ଵ              ଷ
                                    = 4 + ଶ௫
                           ଴
                               ଶ௫
                           LCD = 2x - 3
                           11x - 30 = 4(2x - 3) + 3
                           11x - 30 = 8x - 12 + 3
                           11x - 8x = -9 + 30

                                                                                                                    8
RB Astillero                                     College Algebra                                        Word Problems


                      3x = 21
                      x=7
                    x-3=4
Therefore, the number is 47.
                  ଷ
Check: 47  11 = 4
                             ଵଵ

Example 12. Clock Problem
At what time between 4 and 5 o’clock will the hands of a clock be (a) opposite each other? (b)
coincident?
Solution:
Let x = the number of minute spaces that the minute hand travels.
    ௫
         = the number of minute spaces that the hour hand travels.
    ଵଶ




                                                20                                                            20x




                                           ௫                                                                  ௫
                                          ଵଶ                                                                 ଵଶ
                   30
               x
                                                                                        Figure b
               Figure a
    (a)From figure a:                                               (b) From figure b:
        x = 20 + ௫ + 30                                                 x = ௫ + 20
                        ଵଶ                                                     ଵଶ
          (x = 50 + ௫ 12
                     )                                                    ( x = ௫ + 20)12
                         ଵଶ                                                        ଵଶ
          12x = 600 + x                                                  12x = x + 240
          11x = 600                                                       11x = 240
            x = ଴଴
                 ଴                                                          x = ଶସ଴
                    ଵଵ                                                             ଵଵ
               x = 54଴minute spaces                                         x = 21ଽ minute spaces
                         ଵଵ                                                             ଵଵ
Therefore, the time at which the hands will be opposite each other is 4: 54଴                      o’clock and the time
                                                                                             ଵଵ
               the hands will be together (coincident) is 4: 21ଽ        o’clock.
                                                                   ଵଵ
Check. Use real watch and check to see if the hands of the clock are opposite or together at the
          corresponding time thus obtained.



                                                                                                                     9
RB Astillero                                 College Algebra                               Word Problems




                6.1
Solve the following problems using the steps suggested in this module.
1. Find three consecutive integers whose sum is 378.
2. One-fifth of a certain number is two more than one-sixth of the number. Find the number.
3. Find two consecutive even integers such that twice the larger is forty less than three times the
    smaller.
4. Find two consecutive odd integers such that the difference of their squares is 160.
5. Find three consecutive integers so that twice the first added to five times the second exceeds
    three times the third by 51.
6. The sum of two numbers is 9 and their difference is 6. What are the numbers?
7. Find two numbers whose sum is 7, given that one is 3 times the other.
8. The smaller of two numbers is 9 less than the larger, and their sum is 37. Find the numbers.
9. A bird perching on top of a pole watched as a flock of birds passed by in front of him. After a
    while another flock was passing by. Out of curiosity he asked the leader of the second group
    how many were them. The leader replied, “My group is half as much as the first group, but if
    you join us we’ll be hundred in all including the first group.” How many birds were in the first
    and second group?
10.A sum of P2.45 is composed of 5-centavo and 25-centavo coins, there being twice as many 5-
    centavo coins as 25-centavo coins. How many 25-centavo coins are there?
11.The attendance in a certain rock concert was 737 persons. If there were 289 more males than
    females, how many females were there?
12.The length of a rectangle is 13 inches greater than its width. Its perimeter is 8 feet. Find its
    dimensions.
13.A vertical pole was broken by the wind. The upper part, still attached, reached a point on the
    level ground 15 feet from the base. If the upper part is 9 feet longer than the lower part, how tall
    was the pole?
14.If the width of a rectangle is 2 cm more than one-half its length and its perimeter is 40 cm, what
    are the dimensions?
15.The longest side of a triangle is twice as long as the shortest side and 2 cm longer than the third
    side. If the perimeter of the triangle is 33 cm, what is the length of each side?
16.The width and length of a rectangle are respectively two less and three more than the length of
    the side of a square. If the area of the rectangle is 4 more than the area of the square, find the
    dimensions of the rectangle and the area of the square.
17.The length of the rectangle is one more than the width. If the dimensions are both decreased by
    2 units, the area of the new rectangle is 30 sq. units less than the area of the original rectangle. Find
    the area of the original rectangle.
18.Each equal side of an isosceles triangle is 3ଵ inches longer than the base. The perimeter of the
                                                   ଶ
   triangle is 2 feet. Find the length of the base.
19.A man left one-fourth of his state to his wife, one-fifth to each of his two sons, one-eighth to his
   daughter and the remainder, an amount of P32, 000 to charity. Find the amount of the estate.
20.A worker is paid P180.00 for each day of work but is fined P15.00 if he does not report work.
   How many days did he work if his net pay in 16 days is P2, 100?


                                                                                                        10
RB Astillero                                  College Algebra                             Word Problems


21.Yen is twice as old as Ben. Four times Yen’s age four years ago is one year less than three
    times Ben’s age five years from now. How old is each now?
22.Jay is twelve years older than Mario. If Mario’s age two years from now is half Jay’s age six
    years from now, how old will each be ten years from now?
23.When Fidel and Janette were married, his age was ଷ of her age. If on their 50th wedding
                                                                ଶ
    anniversary, Fidel’s age will be       of Janette’s age, how old will each of them be on their 50th
                                       ଴
   wedding anniversary?
24.A man is three times as old as his son. Four years ago he was four times as old as his son was at
   that time. How old is the son?
25.The sum of the digits of a two-digit number is 15. If the digits are reversed the new number is
   nine less than the original number. Find the original number.
26.The units digit of a two-digit number is 5 more than the tens digit. If the digits are reversed and
   the new number is divided by the original number the quotient is 2 and the remainder is 7. What
   is the original number?
27.The tens digit of a two-digit number is two-thirds that of the units digit. Reversing the digits
   give rise to a new number that is 27 more than the original. Find the original number.
28.A storekeeper has candies that sell for P600 and P900 per kilogram. How many kilograms of
   each must be mixed together to make 100 kilograms of candy that can be sold for P720per
   kilogram?
29.How many liters of pure alcohol must be added to 15 liters of 20 % solution to obtain a mixture
   which is 30 % alcohol?
30.How many cubic meters of butterfat must be removed from 500 cubic meters of milk
   containing 10 % butterfat to reduce its content to 4% butterfat?
31.How much water must be added to 200 gallons of mixture which is 80 % alcohol to reduce it to
   a 75 % mixture?
32.How many ounces of pure nickel must be added to 150 ounces of alloy 70 % pure to make an
   alloy which is 85 % pure?
33.A math teacher prepared a test consisting of 50 problems worth 2 points, 5 points, and 10 points
   each. If the number of 2-point problems is thrice the number of 10-point problems, and the
   number of 5-point problems is 10 less than twice the number of 2-point problems, how many
   points is the entire test?
34.Two motorists, one travelling 5 kilometers per hour (kph) faster than the other, leave the same
   place at the same time and travel in opposite directions. Find the rate of each if they are 195 km
   apart after 3 hours?
35.Mark leaves the house for school walking at the rate of 4 kph. Two hours later Joyce follows
   Mark at 4.5 kph. How long will it take Joyce to overtake Mark?
36.A train travelling at 60 kph covers a distance in 3 hours. By how many kilometers per hour
   must it rate be increased so that it will travel the distance in one-half hour less time?
37.Two buses which are 80 kilometers apart start to travel at the same time toward each other. If
   the first bus travels at the rate of 55 kph and the second one at 45 kph, after how many hours
   will they meet?
38.A car starts at 7 A.M. and travels at 80 kph. Another car starts at 10 A.M. from the same place
   and goes in the same direction. How fast will the second car have to go to catch up with the first
   car by 4 P.M. of the same day?



                                                                                                      11
RB Astillero                               College Algebra                              Word Problems


39.A man walks a certain distance at the rate of 5 miles an hour and returns at the rate of 4 miles
   an hour. If the total time that it takes him is 3 hours and 9 minutes, what is the total distance
   that he walks?
40.Two automobiles set out simultaneously from places 100 miles apart. They traveled toward
   each other and passed at the end of 1 hour and 40 minutes. The speed of one automobile was 5
   miles per hour greater than of the other. Find the speed of each.
41.A car 15 feet long overtakes a truck 30 feet long which is traveling at the rate of 45 miles per
   hour. How fast must the car travel to pass the truck in 3 seconds?
42.A bullet is fired at target and the sound of its impact is heard 6 seconds later. If it travels at the
   rate of 2200 feet per second and sound travels at the rate of 1100 feet per second, how far away
   is the target?
43.A and B start at the same time from two places 136 km apart and travel toward each other. A
   travels 10 kph and B 8 kph. If B rest 1 hour on the way, in how many hours will they meet?
44.A company has three machines A, B, C used in the production of bolts. A can produce 1000
   bolts in 2 hours, B in 2ଵ hours and C in 6 hours. If A and B worked for 1 hour and then B and C
                           ଶ
    finished the job, how long did it take to produce 1000 bolts?
45.Twenty laborers could finish a certain job in 35 days. After 11 days, five laborers quit the work
    and are not replaced until 4 days. How many more laborers should be added to complete the job
    on time?
46.A, B, and C can do a piece of work in 10 days. A and B can do it in 12 days, A and C in 20 days.
    How many days would it take each to do the work alone?
47.A tank can be filled by two pipes in 4 and 6 hours, respectively. It can be emptied by a third
    pipe in 5 hours. In what time can an empty tank be filled if the three pipes are open?
48.At what time between 2 and 3 o’clock will the hands of the clock be at right angle with each
    other?
49.It is between 3 and 4 o’clock, and in 20 minutes the minute hand will be as much after the hour
    hand as it is now behind it. What is the time?
50.How soon after 12:00 P.M. will the hands of the clock extend in opposite directions?




                                                                                                    12

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Word+problems+le1

  • 1. RB Astillero College Algebra Word Problems WORD PROBLEMS INVOLVING LINEAR EQUATIONS IN ONE UNKNOWN A word problem describes a situation involving both known and unknown quantities. Many real- life problems can be solved by expressing them as word problems that often lead to equations. The equations that represent a word problem are called mathematical models. The success in working out word problems depends largely on one’s ability to translate it into a mathematical model. TRANSLATING FROM WORDS TO MATHEMATICAL EQUATIONS Usually there are key words and phrases in a verbal problem that translate into mathematical expressions involving addition, subtraction, multiplication, and division. Translations of some commonly used expressions follow. Examples: Translate each verbal sentence into an equation. 1. Twice a number, decreased by 3, is 42. Answer: 2x - 3 = 42. 2. The product of a number and 12, decreased by 7, is 105. Answer: 12x - 7 = 105. ௫ 3. The quotient of a number and the number plus 4 is 28. Answer: = 28. ௫ ௫ 4. The quotient of a number and 4, plus the number, is 10. Answer: + x = 10. ସ 1
  • 2. RB Astillero College Algebra Word Problems PROBLEM-SOLVING TECHNIQUES Probably the most famous study of problem-solving techniques was developed by George Polya (1888-1985). Among his many publications was the modern classic How to Solve It. In this book, Polya proposed a four-step process for problem solving. Polya’s Four-Step Process for Problem Solving 1. Understand the problem. You must first decide what you are to find. 2. Devise a plan. Here are some strategies that may prove useful.  If a formula applies, use it.  Write an equation and solve it.  Draw a sketch.  Make a table or a chart.  Look for a pattern.  Use trial and error.  Workbackward. 3. Carry out the plan. This is where the algebraic techniques you learned can be helpful. 4. Look back and check. Is your answer reasonable? Does it answer the question that was asked? STEPS IN SOLVING WORDED PROBLEMS Although no standard technique for solving a word problem is prescribed, the following six steps are recommended. These steps specifically apply Polya’s techniques to word problems in linear equations. Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. If necessary, express any other unknown values in terms of the variables. Step 3 Write an equation using the variable expression(s). Step 4 Solve the equation. Step 5 State the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem. TYPES OF WORD PROBLEMS There are many types of word problems, but in this module you will only learn those types that are very common in engineering applications. The following examples will illustrate how a word problem is solved using the suggested six steps. Example 1. Geometry Problem The length of a rectangle is 1 cm more than twice the width. The perimeter of the rectangle is 110 cm. Find the length and the width of the rectangle. Solution: Step 1 Read the problem. What must be found? The length and width of the rectangle. What is given? The length is 1 cm more than twice the width; the perimeter is 110 cm. 2
  • 3. RB Astillero College Algebra Word Problems Step 2Assign a variable. Choose a variable: let W = the width; then 1 + 2W = the length. W 1 + 2W Step 3Write an equation. The perimeter of a rectangle is given by the formula P = 2L + 2W. Substituting P = 110, L = 1 + 2W, the required equation is 110 = 2(1 + 2W) + 2W Step 4 Solve the equation obtained in Step 3. 110 = 2 + 4W + 2W 110 = 2 + 6W 110 - 2 = 6W 108 = 6W ଵ଴ =W ଴ 18 = W W = 18 [by symmetric equality axiom] Step 5 State the answer. The width of the rectangle is 18 cm and the length is 1 + 2(18) = 37 cm. Step 6 Check the answer by substituting these dimensions into the words of the original problem. Example 2: Motion Problem Two cars leave the same place at the same time, one going east and the other west. The eastbound car averages 40 mph, while the westbound car averages 50 mph. In how many hours will they be 300 mi apart? Solution: Step 1 Read the problem. We are looking for the time it takes for the two cars to be 300 mi apart. Step 2 Assign a variable. A sketch shows what is happening in the problem: The cars are going in opposite directions. Let x = the time (in hours) the cars will be 300 mi apart. When the expressions for rate and time are entered, fill in each distance by multiplying rate by time using the formula d = rt. 3
  • 4. RB Astillero College Algebra Word Problems Step 3 Write an equation. From the sketch in step 2, the sum of the two distances is 300. Thus the equation is 40x + 50x = 300. Step 4Solve the equation. 40x + 50x = 300 90x = 300 x = ଷ଴଴ ଽ଴ x= ଵ = 3ଵ ଷ ଴ ଷ Step 5 State the answer. The cars travel 3ଵ hr or 3 hr 20 min. ଷ Step 6 Check. The eastbound car traveled 40(ଵ ) = ସ଴ mi and the westbound car traveled 50(ଵ ) = ହ଴ mi, ଴ ଴ ଴ ଴ ଷ ଷ ଷ ଷ for a total of ସ଴ + ହ଴ = ଽ ଴ = 300 mi, as required. ଴ ଴ ଴ ଷ ଷ ଷ Example 3. Mixture Problem A chemist must mix 8 L of a 40% acid solution with some 70% solution to get a 50% solution. How much of the 70% solution should be used? Solution: Step 1 Read the problem. The problem asks for the amount of 70% solution to be used. Step 2 Assign a variable. Let x = the number of liters of 70% solution to be used. The information in the problem is illustrated in the figure below. Use the given information to complete a table as shown below. The numbers in the last column were found by multiplying the strengths and the numbers of liters. The number of liters of pure acid in the 40% solution plus the number of liters in the 70% solution must equal the number of liters in the 50% solution. Step 3 Write an equation. 3.2 + 0.70x = 0.50(8 + x) Step 4 Solve. 3.2 + 0.70x = 4 + 0.50x 0.20x = 0.8 4
  • 5. RB Astillero College Algebra Word Problems x=4 Step 5 State the answer. The chemist should use 4 L of the 70% solution. Step 6 Check. 8 L of 40% solution plus 4 L of 70% solution is 8(0.40) + 4(0.70) = 6 L of acid. Similarly, 8 + 4 or 12 L of 50% solution has 12(0.50) = 6L of acid in the mixture. The total amount of pure acid is 6 L both before and after mixing, so the answer checks. Example 4. Geometry Problem Find the value of x, and determine the measure of each angle in the triangle. Figure: Solution: Step 1 Read the problem. We are asked to find the measure of each angle. Step 2 Assign a variable. Let x represent the measure of one angle. Step 3 Write an equation. The sum of the three measures shown in the figure must be 180°. x + (x+20) + (210 -3x) =180 Step 4 Solve. 2x + 230 - 3x = 180 - x = 180 - 230 - x = - 50 x = 50 Step 5 State the answer. One angle measures 50°, another measures x + 20 = 50 + 20 = 70 , and the third measures 210 - 3(50) = 60. Step 6 Check. Since 50° + 70 + 60 = 180, the answers are correct. Example 5. Number Relation Problem Find three consecutive integers such that the sum of the first and third, increased by 3, is 50 more than the second. Solution: Let x represent the first of the unknown integers. Then x + 1 will be the second, and x + 2 will be the third. The equation we need can be found by going back to the words of the original problem. Solving the equation: x + x + 2 + 3 = x + 1 + 50 2x + 5 = x + 51 2x - x = 51 - 5 x = 46 Therefore, the three consecutive integers are 46, 47, and 48. (ans.) Check: 46 + 48 + 3 = 47 + 50 97 = 97 5
  • 6. RB Astillero College Algebra Word Problems Example 6. Number Relation Problem In a certain class, 12 more than ଵ of all the students are taking a science course and ଵ of those taking ଷ ସ a science course are also taking mathematics. If ଽ of all the students in the class are taking ଵ଴ mathematics, how many students are in the class? Solution: Let x = the number of students in the class. ଵ x + 12 = number of students who are taking science course. ଷ ଽ x = number of students who are taking mathematics. ଵ଴ ଵଵ ( x + 12) = ଵ x + 3 = number of students who are taking both science and mathematics. ସ ଷ ଵଶ By Venn Diagram: ଵ ଵ ଽ ଵ ( x + 12) x+3 x - ( x + 3 ଷ ଵଶ ଵ଴ ଵଶ ଵ -( ଵx ଶ + 3) science mathematics ଵ x + 12+ ଽ x - ( ଵ x + 3 x = ଷ ଵ଴ ଵଶ ଵ ଽ ଵ x+ x- x - x +9 = 0 ଷ ଵ଴ ଵଶ ଵ ଴ ଴ x = -9 ଽ x=-9 x = 48 Therefore, there are 48 students in the class. Check: number of students taking science = ଵ (48) + 12 = 28 ଷ Number of students taking math = ଽ (48) = 27 ଵ଴ Number of students both taking math & science = ଵ 28= 7  ସ Number of students taking science only = 28 - 7 = 21 Number of students taking math only = 27 - 7 = 20 Total number of students in the class = 7 + 21 + 20 = 48 [check] 6
  • 7. RB Astillero College Algebra Word Problems Example 7: Age Problem A 30-year old father has 4-year old son. In how many years will the father be three times as old as his son? Solution: Let x = number of years, from now, by which the father’s age will be thrice that of his son. Present age Future age Father 30 30 + x Son 4 4+x 30 + x = 3(4 + x) [why?] 30 + x = 12 + 3x x=9 Therefore, 9 years from now the father will be thrice as old as his son. Check: In 9 years the father will be (30 + 9 =) 39 yrs. of age while his son will be (4 + 9 =) 13 yrs. old. 39 is 3 times as much as 13, i.e. 3(13) = 39 as required. Example 8: Age Problem The sum of Kim’s and Kevin’s ages is 18. In 3 years, Kim will be twice as old as Kevin. What are their ages now? Solution: Let x = present age of Kim. 18 - x = present age of Kevin. Present age Future age Kim x x+3 (18 - x) + 3 Kevin 18 - x = 21 - x 2(21 - x) = x + 3 42 - 2x = x + 3 3x = 39 x = 13 18 - x = 5 Therefore, Kim is 13 yrs. old and Kevin is 5 yrs. old. Check. In 3 years Kim will be 16 yrs. old while Kevin will be 8 yrs. old; 16 = 2(8). Example 9. Rate of Work Problem Lito can do a piece of work in 8 days, while Noli can do the same work in 10 days. In how many days can they finish the job working together? Solution: Let x = the number of days it will take them to finish the job together. ଵ = the part of the job accomplished by Lito in 1 day. ଵ = the part of the job accomplished by Noli in 1 day. ଵ଴ ଵ + ଵ =ଵ ଵ଴ ௫ 7
  • 8. RB Astillero College Algebra Word Problems ଵ = ସ଴ ௫ ଽ ଵ = ସ଴ ସ௫ x= ଴ ଽ =4ସ ଽ Therefore, they will finish the job working together in 4 ସ days. ଽ Check. The part of the job they will be accomplish in 1 day if working together is ଽ which is equal ଵ ଵ ସ଴ to + ଵ଴ Example 10. Rate of Work Problem Two pipes A and B operate independently at their respective constant rates. Pipe A alone takes 5 hours to fill the tank. When pipes A and B are used simultaneously, it takes them 2 hours to fill the tank. How long will it take pipe B alone to fill the tank? Solution: Let x = number of hours it would take pipe B working alone to fill the tank. ଵ = part of the tank filled in by B in 1 hour. ௫ ଵ = part of the tank filled in by A in 1 hour. ହ ଵ = part of the tank filled in by A and B in 1 hour. ଶ ଵ ௫ + ଵ= ଵ ହ ଶ LCD = 10x 10 + 2x = 5x 10 = 3x 3x = 10 x=ଵ ଴ ଷ Therefore, it will take pipe B alone to fill the tank in ଵ or 3ଵ ଴ hours. ଷ ଷ Check. Pipes A and B running simultaneously will fill the tank in భଵ = బశభఱ = మఱ ହ଴ 2 hours. ଵ= = భయబ ఱ ఱబ ఱబ ଶହ భ Example 11. Digit Problem The tens digit of a number is 3 less than the units digit. If the number is divided by the sum of the digits, the quotient is four and the remainder is 3. What is the original number? Solution: Let x = the units digit. x - 3 = the tens digit. 2x - 3 = sum of digits. 10(x-3) + x = 10x - 30 + x = 11x - 30 = the number. ଵ௫ ଵ ଷ = 4 + ଶ௫ ଴ ଶ௫ LCD = 2x - 3 11x - 30 = 4(2x - 3) + 3 11x - 30 = 8x - 12 + 3 11x - 8x = -9 + 30 8
  • 9. RB Astillero College Algebra Word Problems 3x = 21 x=7 x-3=4 Therefore, the number is 47. ଷ Check: 47  11 = 4 ଵଵ Example 12. Clock Problem At what time between 4 and 5 o’clock will the hands of a clock be (a) opposite each other? (b) coincident? Solution: Let x = the number of minute spaces that the minute hand travels. ௫ = the number of minute spaces that the hour hand travels. ଵଶ 20 20x ௫ ௫ ଵଶ ଵଶ 30 x Figure b Figure a (a)From figure a: (b) From figure b: x = 20 + ௫ + 30 x = ௫ + 20 ଵଶ ଵଶ (x = 50 + ௫ 12 ) ( x = ௫ + 20)12 ଵଶ ଵଶ 12x = 600 + x 12x = x + 240 11x = 600 11x = 240 x = ଴଴ ଴ x = ଶସ଴ ଵଵ ଵଵ x = 54଴minute spaces x = 21ଽ minute spaces ଵଵ ଵଵ Therefore, the time at which the hands will be opposite each other is 4: 54଴ o’clock and the time ଵଵ the hands will be together (coincident) is 4: 21ଽ o’clock. ଵଵ Check. Use real watch and check to see if the hands of the clock are opposite or together at the corresponding time thus obtained. 9
  • 10. RB Astillero College Algebra Word Problems 6.1 Solve the following problems using the steps suggested in this module. 1. Find three consecutive integers whose sum is 378. 2. One-fifth of a certain number is two more than one-sixth of the number. Find the number. 3. Find two consecutive even integers such that twice the larger is forty less than three times the smaller. 4. Find two consecutive odd integers such that the difference of their squares is 160. 5. Find three consecutive integers so that twice the first added to five times the second exceeds three times the third by 51. 6. The sum of two numbers is 9 and their difference is 6. What are the numbers? 7. Find two numbers whose sum is 7, given that one is 3 times the other. 8. The smaller of two numbers is 9 less than the larger, and their sum is 37. Find the numbers. 9. A bird perching on top of a pole watched as a flock of birds passed by in front of him. After a while another flock was passing by. Out of curiosity he asked the leader of the second group how many were them. The leader replied, “My group is half as much as the first group, but if you join us we’ll be hundred in all including the first group.” How many birds were in the first and second group? 10.A sum of P2.45 is composed of 5-centavo and 25-centavo coins, there being twice as many 5- centavo coins as 25-centavo coins. How many 25-centavo coins are there? 11.The attendance in a certain rock concert was 737 persons. If there were 289 more males than females, how many females were there? 12.The length of a rectangle is 13 inches greater than its width. Its perimeter is 8 feet. Find its dimensions. 13.A vertical pole was broken by the wind. The upper part, still attached, reached a point on the level ground 15 feet from the base. If the upper part is 9 feet longer than the lower part, how tall was the pole? 14.If the width of a rectangle is 2 cm more than one-half its length and its perimeter is 40 cm, what are the dimensions? 15.The longest side of a triangle is twice as long as the shortest side and 2 cm longer than the third side. If the perimeter of the triangle is 33 cm, what is the length of each side? 16.The width and length of a rectangle are respectively two less and three more than the length of the side of a square. If the area of the rectangle is 4 more than the area of the square, find the dimensions of the rectangle and the area of the square. 17.The length of the rectangle is one more than the width. If the dimensions are both decreased by 2 units, the area of the new rectangle is 30 sq. units less than the area of the original rectangle. Find the area of the original rectangle. 18.Each equal side of an isosceles triangle is 3ଵ inches longer than the base. The perimeter of the ଶ triangle is 2 feet. Find the length of the base. 19.A man left one-fourth of his state to his wife, one-fifth to each of his two sons, one-eighth to his daughter and the remainder, an amount of P32, 000 to charity. Find the amount of the estate. 20.A worker is paid P180.00 for each day of work but is fined P15.00 if he does not report work. How many days did he work if his net pay in 16 days is P2, 100? 10
  • 11. RB Astillero College Algebra Word Problems 21.Yen is twice as old as Ben. Four times Yen’s age four years ago is one year less than three times Ben’s age five years from now. How old is each now? 22.Jay is twelve years older than Mario. If Mario’s age two years from now is half Jay’s age six years from now, how old will each be ten years from now? 23.When Fidel and Janette were married, his age was ଷ of her age. If on their 50th wedding ଶ anniversary, Fidel’s age will be of Janette’s age, how old will each of them be on their 50th ଴ wedding anniversary? 24.A man is three times as old as his son. Four years ago he was four times as old as his son was at that time. How old is the son? 25.The sum of the digits of a two-digit number is 15. If the digits are reversed the new number is nine less than the original number. Find the original number. 26.The units digit of a two-digit number is 5 more than the tens digit. If the digits are reversed and the new number is divided by the original number the quotient is 2 and the remainder is 7. What is the original number? 27.The tens digit of a two-digit number is two-thirds that of the units digit. Reversing the digits give rise to a new number that is 27 more than the original. Find the original number. 28.A storekeeper has candies that sell for P600 and P900 per kilogram. How many kilograms of each must be mixed together to make 100 kilograms of candy that can be sold for P720per kilogram? 29.How many liters of pure alcohol must be added to 15 liters of 20 % solution to obtain a mixture which is 30 % alcohol? 30.How many cubic meters of butterfat must be removed from 500 cubic meters of milk containing 10 % butterfat to reduce its content to 4% butterfat? 31.How much water must be added to 200 gallons of mixture which is 80 % alcohol to reduce it to a 75 % mixture? 32.How many ounces of pure nickel must be added to 150 ounces of alloy 70 % pure to make an alloy which is 85 % pure? 33.A math teacher prepared a test consisting of 50 problems worth 2 points, 5 points, and 10 points each. If the number of 2-point problems is thrice the number of 10-point problems, and the number of 5-point problems is 10 less than twice the number of 2-point problems, how many points is the entire test? 34.Two motorists, one travelling 5 kilometers per hour (kph) faster than the other, leave the same place at the same time and travel in opposite directions. Find the rate of each if they are 195 km apart after 3 hours? 35.Mark leaves the house for school walking at the rate of 4 kph. Two hours later Joyce follows Mark at 4.5 kph. How long will it take Joyce to overtake Mark? 36.A train travelling at 60 kph covers a distance in 3 hours. By how many kilometers per hour must it rate be increased so that it will travel the distance in one-half hour less time? 37.Two buses which are 80 kilometers apart start to travel at the same time toward each other. If the first bus travels at the rate of 55 kph and the second one at 45 kph, after how many hours will they meet? 38.A car starts at 7 A.M. and travels at 80 kph. Another car starts at 10 A.M. from the same place and goes in the same direction. How fast will the second car have to go to catch up with the first car by 4 P.M. of the same day? 11
  • 12. RB Astillero College Algebra Word Problems 39.A man walks a certain distance at the rate of 5 miles an hour and returns at the rate of 4 miles an hour. If the total time that it takes him is 3 hours and 9 minutes, what is the total distance that he walks? 40.Two automobiles set out simultaneously from places 100 miles apart. They traveled toward each other and passed at the end of 1 hour and 40 minutes. The speed of one automobile was 5 miles per hour greater than of the other. Find the speed of each. 41.A car 15 feet long overtakes a truck 30 feet long which is traveling at the rate of 45 miles per hour. How fast must the car travel to pass the truck in 3 seconds? 42.A bullet is fired at target and the sound of its impact is heard 6 seconds later. If it travels at the rate of 2200 feet per second and sound travels at the rate of 1100 feet per second, how far away is the target? 43.A and B start at the same time from two places 136 km apart and travel toward each other. A travels 10 kph and B 8 kph. If B rest 1 hour on the way, in how many hours will they meet? 44.A company has three machines A, B, C used in the production of bolts. A can produce 1000 bolts in 2 hours, B in 2ଵ hours and C in 6 hours. If A and B worked for 1 hour and then B and C ଶ finished the job, how long did it take to produce 1000 bolts? 45.Twenty laborers could finish a certain job in 35 days. After 11 days, five laborers quit the work and are not replaced until 4 days. How many more laborers should be added to complete the job on time? 46.A, B, and C can do a piece of work in 10 days. A and B can do it in 12 days, A and C in 20 days. How many days would it take each to do the work alone? 47.A tank can be filled by two pipes in 4 and 6 hours, respectively. It can be emptied by a third pipe in 5 hours. In what time can an empty tank be filled if the three pipes are open? 48.At what time between 2 and 3 o’clock will the hands of the clock be at right angle with each other? 49.It is between 3 and 4 o’clock, and in 20 minutes the minute hand will be as much after the hour hand as it is now behind it. What is the time? 50.How soon after 12:00 P.M. will the hands of the clock extend in opposite directions? 12