Solving Word Problems- compilation

1
Gov. Alfonso D. Tan College
Maloro, Tangub City
In Partial Fulfilment
Of the Subject Requirement in
Integrated Mathematics
(Math 128)
COMPILATION OF WORD PROBLEMS
Submitted to
Mr. Alemar C. Mayordo
Instructor
Submitted by
Elton John B. Embodo

2
Introduction
Learning Mathematics takes a step by step process. You cannot teach a child with some
complex lessons when he is not yet taught with the basic ones. He must learn and practice first
the basics before learning the complex. Teaching Mathematics should not only be embedded
inside the classroom setting, instead it should also be extended to the real world. As a knowledge
transporter in the field of Mathematics, one should have enough knowledge to be able to
integrate a certain lesson to the other fields of science, most importantly to the real life situations.
One way to easily integrate Mathematics to the real word is dealing with some Word Problems.
This compilation is designed to show how to integrate Mathematics to the real world
situations. Specifically, inside this compilation are the following areas of Mathematics: Algebra,
Geometry, Statistics and Trigonometry. In each area, two topics are being discussed with five
word problems. Each word problem is the application of each topic that is being discussed in
each area with the step by step process of solution.
To the students, God Bless!
The Author

3
Acknowledgement
Special Thanks
To Mr. Vincent S. Montebon
For sharing his thoughtful ideas
On how to solve some of the word problems!
Most importantly to our Almighty Father Jesus Christ
For giving me more strength, knowledge
And wisdom to make this
Compilation beautiful.

4
Dedication
I would like to dedicate this compilation to
Those people who inspire me to
Become enthusiastic, dynamic,
Alive, alert and
A Better
Person.
To my instructor, Alemar C. Mayordo for requiring
Me to create this Compilation. I’ve
Learned a lot
From this.
And to myself for doing
Such a great job!
Keep it up!

5
Table of Contents
Algebra
Equation in One Variable......................................................................................... 1
Word Problem 1........................................................................................... 1
Word Problem 2........................................................................................... 2
Equation in Two Variables...................................................................................... 3
Word Problem 3........................................................................................... 3
Word Problem 4........................................................................................... 4
Word Problem 5........................................................................................... 5
Geometry
Plane Figures............................................................................................................ 7
Word Problem 1........................................................................................... 7
Word Problem 2........................................................................................... 8
Solid Figures............................................................................................................ 9
Word Problem 3........................................................................................... 9
Word Problem 4........................................................................................... 10
Word Problem 5........................................................................................... 10
Trigonometry
Right Triangle.......................................................................................................... 11
Word Problem 1.......................................................................................... 11
Word Problem 2.......................................................................................... 12
Angle of Elevation and Angle of Depression.......................................................... 13
Word Problem 3........................................................................................... 14
Word Problem 4.......................................................................................... 14

6
Word Problem 5.......................................................................................... 15
Statistics
Permutation.............................................................................................................. 16
Word Problem 1........................................................................................... 16
Word Problem 2........................................................................................... 16
Combination
Word Problem 3........................................................................................... 17
Word Problem 4........................................................................................... 17
Word Problem 5........................................................................................... 18

7
Chapter 1
ALGEBRA
Equations in One Variable
Equation in One Variable is an algebraic expression in a form of ax + b = 0, where a and
b are constants and x is the variable. In this equation, we usually deal with finding the value of x
to solve a certain problem.
Word Problems involving Equation in One Variable
1. Phoebe spends 2 hours training for an upcoming race. She runs full speed at 8 miles per
hour for the race distance; then she walks back to her starting point at 2 miles per hour.
How long does she spend walking? How long does she spend running?
Let x be the time she spent running. Since she spent 2 hours all together, she must
have spent 2 – x hours walking.
Time  Speed = distance
running X  8 = 8x
Walking 2 – x  2 = 2(2 – x)
Since she ran out, then turned around and walked back, her running and walking distances must
be equal.
running:
x hours
8 miles per hour
8x miles
walking:
2 – x hours
2 miles per hour
2(2 – x) miles
Distances are equal

8
Set the distances equal and solve for x:
8 2(2 )x x 
8 4 2x x 
10 4x 
0.4x 
She spent 0.4 hours running and 2 – 0.4 = 1.6 hours walking.
2. Two planes, which are 2400 miles apart, fly toward each other. Their speeds differ by 60
miles per hour. They pass each other after 5 hours. Find their speeds.
Time  Speed = distance
First plane 5  x = 5x
Second plane 5  x + 60 = 5(x + 60)
2400
Since the planes started 2400 miles apart, when they pass each other they must have combined to
cover the 2400 miles.
So the sum of their faces is equal to 240:
First plane Second plane
240 miles

9
5 5( 60) 2400x x  
5 5 300 2400x x  
10 300 2400x 
10 2400 300x  
10 2100x 
210x 
One plane’s speed is 210 miles per hour. The other plane’s speed is 210 + 60 = 270 miles per
hour.
Equations in Two Variables
Equation in Two Variables is an algebraic expression in a form of ax + by = 0, where a
and b are constants and x and y are the variables. In this equation, we usually deal with finding
the value of x and y to solvea certain problem.
Word Problems involving Equation in Two Variables
3. How many litres of 20% alcohol solution should be added to 40 litres of a 50% alcohol
solution to make a 30% solution?
Let x be the quantity of the 20% alcohol solution to be added to the 40 litres of a
50% alcohol. Let y be the quantity of the final 30% solution. Hence
x + 40 = y

10
We shall now express mathematically that the quantity of alcohol in x litres plus
the quantity of alcohol in the 40 litres is equal to the quantity of alcohol in y litres. But remember
the alcohol is measured in percentage.
x (20%) + 40(50%) = y (30%)
Substitute y by x + 40 in the last equation to obtain.
x (20%) + 40(50%) = y + 40 (30%)
Change the percentage into decimal number.
x (.20) + 40 (.50) = x + 40 (.30)
Simplify
.20x +20 = .30 x + 12
.20x - .30x = 12 – 20
-.10x = -8
x = 80 litres
Therefore, 80 litres of 20% alcohol is to be added to 40 litres of a 50% alcohol solution
to make a 30% solution.
4. Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed to
a 90% Silver alloy to obtain a 500g of a 91% silver alloy?
Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to
make the 500 grams at 91%. Hence
x + y = 500

11
The number of grams of pure silver in x plus the number of grams of pure silver
in y is equal to the number of grams of pure silver in the 500 grams. The pure silver is given in
percentage forms. Hence
x (92.5%)+ y(90%) = 500 (91%)
Substitute y by 500 – x in the last equation to write
x (92.5%)+ 500 – x (90%) = 500 (91%)
Simplify and solve
92.5x + 45000 – 90x = 45500
2.5x = 500
x = 200 grams
Therefore, 200 grams of Sterling Silver is needed to make the 91% alloy.
5. John wants to make a 100ml of 55 alcohol solution mixing a quantity of a 2% alcohol
solution with a 7% alcohol solution. What are the quantities of each of the two solutions
(2% and 7%) he has to use?
Let x and y be the quantities of the 2% and 7% alcohol solutions to be used to
make 100 ml. Hence
x + y = 100
We now write mathematically that the quantity of alcohol in x ml plus the
quantity of alcohol in y ml is equal to the quantity of alcohol in 100ml.
x (2%) + y (7%) = 100 (5%)
The first equation gives y = 100 – x. Substitute in the last equation to obtain
x (2%) + 100 – x (7%) = 100 (5%)

12
Change the percentage to decimal and then simplify.
.2x + 70 – .7x = 50
-.5x = - 20
x = 40ml
Substitute x by 40 in the first equation to find y;
y = 100 – x
y = 100 – 40
y = 60ml
Therefore the quantities of each of the two solutions (2% and 7%) he has to use are 40ml
and 60ml respectively.

13
Chapter 2
GEOMETRY
Plane Figures
These are geometric figures with two dimensions: length and width. Usually, they refer
to polygons and circles.
Word Problems involving Plane Figures
1. The official distance between homeplate the second based in the baseball diamond is
120ft. Fin the area of the official ball diamond and the distances between the bases. (The
official ball diamond is in the form of square).
2 2 2
a a c 
2 2 2
(120)a a 
2 2
2 (120)a 
2
2 (120)
2
a 
120
2
a 
60 2a 
120ft
By Pythagorean Theorem
To find the area of the official ball diamond
, where a is
Distance between the bases =

14
2. Mr. Montebon wants to beautify his classroom grade 7 sampaguita. He wants to tile the
floor of the classroom. He later measures the floor’s area and he found out that it’s equal
to 484 square feet. Each tile that he had just bought has a length equal to 2 feet.
a. How many tiles needed to tile the entire floor area?
b. If each tile costs 196 Php, how much he has spent?
Let’s find first the area of each tile, since the tile is always square, then:
2
2
2
2
4
A s
A
A ft



a. Entire floor area
Area of each tile
2
2
484
4
ft
ft
121tiles
b. Amount that he has spent = 196 Php * 121 tiles = 23, 716 Php
floor
A = 484 square
feet
tile
2feet
number of tiles need to tile
the entire floor
=
Area of each tile

15
Solid Figures
These are geometric figures with three dimensions: length, width and height.
Word Problems involving Solid Figures
3. A classroom is 30ft long, 24ft wide and 14ft high. If the 42 pupils are assigned to the
room, how many cubic feet of air space does this room allow for each pupil?
Given: l = 30ft, w = 25ft and h = 14ft, N = 42 pupils
Find: the volume air allowed for each pupil.
Step1: find the volume of the room (V), since the room is rectangular, then:
V = l * w * h
3
V=30ft(25ft)(14ft)
V=10500ft
Step 2: Calculate for the volume of air allowed for each pupil.
2
V 10500
Vp=
N 42
ft

3
Vp=250ft Per pupil
Volume of air allowed for each
pupil
=
Volume of the room
Number of Pupils

16
4. Ms. Abiso plans to give her co-teacher Mr. Elton a gift during this Christmas Season. She
puts the gift inside the square box which highs 10 inches. Then, she wants it covered with
a gift wrapper. What will be the amount of gift wrapper she needed to use to wrap the
box?
Since the figure being referred above is a cube, then we will be dealing with
finding its surface area.
2
2
SA=6s
6(10 )
600
SA inches
SA squareinches


Therefore, 600 square inches the surface area of the box and at the same time the amount
of gift wrapper needed to wrap the box
5. Elton John was being told by his mother to fetch water. He has to fill up the circular
cylinder bucket with water in its full amount using gallons. If the full amount of each
gallon is equal to1.125πcubic feet. How many gallons of water need to fill-up the bucket
until it becomes full given its height equal to 5 feet and radius equal to 1.5feet?
Solution:
2
2
(1.5) (5)
(225)(5)
11.24
V r h
V
V
V cubicfeet








11.25
1.125
Vb cubicfeet
Vg cubifeet



= 10 gallons of water needed to fill-up the bucket in its full amount
Volume of the bucket
Volume of each gallon
= Number of gallons

17
Chapter 3
TRIGONOMETRY
Right Triangle
Right Triangle is a triangle having one right angle which measures exactly 90o
or
2

.
It is convenient to denote the vertices as A, B, C; the angles  and  ,  were  and  are
acute angles and  is the right angle; and denote the sides opposite as a, b, c respectively.
Word Problems involving Right Triangle
1. A man drives 500ft along the road which is inclined 20o
with the horizontal. How high
above his point?
500ft
x
C A
B
a
b
c

18
sin
opp
hyp
 
0
sin 20
500
x
ft

0
sin 20 (500 )x ft
171x ft The height above his starting point
2. A tress was broken over by storms form a right triangle with the ground. If the broken
part makes an angle 50o
with the ground and if the top of the tree is now 20ft from its
base, how tall was the trees?
0
tan
20
tan50
opp
adj
x
 

0
20 tan50 x
0
0 0
20 tan50
tan50 tan50
x

0
20
tan50
x 
16.78x ft the height of the part of the tree which remains standing
50o
y
x
20ft

19
sin
opp
hyp

0 20
sin50
y

0
sin50 20y 
0
0 0
sin50 20
sin50 sin50
y

0
20
sin50
y  → 26.11y ft the length of the broken part of the tree that forms 500
.
Angle of Elevation
It is the angle  that a line of sight makes above the horizontal line.
Angle of Depression
It is the angle  that a line of sight makes below the horizontal line.
Height of the tree = x +y
= 16.78ft + 26.11ft
= 42.89ft
Of Elevation
Of Depression
Line of sight
Line of sight
Horizontal line
Horizontal lineC
A B
A
C
B

20
3. A tree 90ft tall casts a shadow 125ft long. Find the angle of elevation of the sum.
1
0
tan
90
tan
125
tan 0.72
0.72 tan
35.75
opposite
adjacent
ft
ft
ft
ft










4. From the top of a light house, 100ft above the sea, the range of depression of a boat is
350
. How far is the boat from the light house?
90ft
125ft
350
350
100ft
x

21
Find the distance of the boat to the base of the light house.
0
0
tan
100
tan35
tan35 100
opposite
adjective
ft
x
x ft
 


0
100
tan35
142.81
ft
x
x ft


5. Two building are 65ft apart. From the roof of the shorter building, 40ft in height, the
angle of elevation to the roof of the taller building is 300
. How high is the taller building?
Solve for x
0
tan
tan30
65
opposite
adjective
x
ft
 

0
tan30 (65 )
37.53
x ft
x ft


The distance of the boat from the base of the light house
40ft40ft
300
x
65ft
The distance from the top of the
smaller building to the top of the
taller building
Height of the taller = x + 40ft
= x + 40ft
= 37.53ft + 40ft
= 77.53ft

22
Chapter 3
STATISTICS
Permutation
It refers to the different possible arrangements of a set of objects. The number of
permutations of n objects taken r at a time is:
!
( , )
( )!
n
P n r
n r


.
1. Ten runners join a race, in how many possible ways, can they be arranged as first,
second, and third placers?
!
Pr
( )!
10!
Pr
(10 3)!
10 9 8 7!
Pr
7!
Pr 720
n
n
n r
n
n
n ways




  


2. In how many ways can 5 people arrange themselves in a row for a picture taking?
5! 120ways

23
Combinations
It is the number of ways of selecting from a set when the order is not important. The
number of combinations of n objects then r at a time is given by;
!
( , ) ,
( )! !
n
C n r n r
n r r
 

3. How many different sets of 5 cards each can be formed from a standard deck of 52
cards?
!
,
( )! !
52!
(52 5)!5!
52 51 50 49 48 47!
47!5!
311,875, 200
120
2,598,960
n
nCr n r
n r r
nCr
nCr
nCr
nCr ways
 



    



4. From a population of 50 households, in how many ways can a researcher select a
sample with a size of 10?
10
50!
(50 10)!10!
50!
50 10
40!10!
50 49 48 47 46 45 44 43 42 41 40!
50 10
40!10!
50 10 1.027 10
nCr
C
C
C



         

 

24
5. In a 10-item Mathematics problem solving test, in how many ways can you select 5
problems to solve?
!
( )! !
10!
10 5
(50 5)!5!
10 9 8 7 6 5 4 3 2 1!
50 10
(5!)5 4 3 2 1!
30,240
50 10
120
50 10 252
n
nCr
n r r
C
C
C
C ways




        

   



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