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 2a a c

2 2 2(120)a a

2 22 (120)a

2

2 (120)

2a

120

2a

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

ft121tiles

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.

2V 10500

Vp=N 42

ft

3Vp=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

sinopp

hyp

0sin 20500

x

ft

0sin 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

20tan 50

opp

adj

x

020 tan50 x

0

0 0

20 tan 50

tan 50 tan 50

x

0

20

tan 50x

16.78x ft the height of the part of the tree which remains standing

50o

y

x

20ft

19

sinopp

hyp

0 20sin 50

y

0sin 50 20y

0

0 0

sin 50 20

sin 50 sin 50

y

0

20

sin 50y → 26.11y ft the length of the broken part of the tree that forms 50

0.

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 line C

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

90tan

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

100tan 35

tan 35 100

opposite

adjective

ft

x

x ft

0

100

tan 35

142.81

ftx

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

tan 3065

opposite

adjective

x

ft

0tan 30 (65 )

37.53

x ft

x ft

The distance of the boat from the base of the light house

40ft 40ft

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:!

( , )( )!

nP 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

nn

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;

!( , ) ,

( )! !

nC 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

nnCr 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!

30,24050 10

120

50 10 252

nnCr

n r r

C

C

C

C ways

25