A.

Geometry Problems

1. Tata Jose plans to construct a bookcase with three shelves. The height of the case
is to be 60 cm more than the width. Only 6 meters of wood is available. What
should be the dimensions of the bookcase?
Solution:
The problem implies that 6 m (600 cm) plywood is to be used, no more, no less.
Defining the variable:
Let: w = width of the bookcase
w + 60 = height of the bookcase
Thought process:
(four times the width) + (two times the height) = 600
Equation:
4w + 2 (w + 60) = 600
Solving the Equation:
4w + 2 (w + 60) = 600
4w + 2w + 120 = 600
6w = 480
w = 80
w + 60 = 80 + 60 = 140
The width of the bookcase is 80 cm and the height is 140 cm.

Elementary Algebra
By: Catalina B. Manalo, page 278-279

Page 1
 2. The length of a rectangular lot is 8 meters more than its width. If the perimeter
is to be at least 76 meters and at most 124 meters, find the range of values that is
possible for the length and width of the rectangular lot.
Solution:
Step 1:
Let w = the width
w + 8 = the length
Step 2: 76 perimeter of the rectangular lot and at most 124
76 = 2w + 2 ( w + 8 ) = 124
Step 3:
76 = 2w + 2 ( w + 8 ) = 124
76 = 2w + 2w + 16 = 124
76 – 16 = 4w + 16 – 16 = 124 – 16
60 = 4w = 108
15 = w = 27
Then,
15 + 8 = w + 8 = 27 + 8
23 = w + 8 = 35

Source:
Elementary Algebra. Based on Basic Education Curriculum
By: Ligaya G. Irsigne, page 212-213

Page 2
 3. In a quadrilateral two angles are equal. The third angle is equal to the sum of
the two equal angles. The fourth angle is 60° less than twice the sum of the other
three angles. Find the measures of the angles in the quadrilateral.

Solution:

Step 1: Assign variables:

Let:

x = size of one of the two equal angles

Step 2: Write down the sum of angles in quadrilateral.

The sum of angles in a quadrilateral is 360°

Step 3: Plug in the values from the question.

360 = x + x + (x + x) + 2 (x + x + x + x) – 60

Combine like terms

360 = 4x + 2(4x) – 60
360 = 4x + 8x – 60
360 = 12x – 60

Isolate variable x
12x = 420
x = 35

The question requires the values of all the angles.

Substituting x for 35, you will get: 35, 35, 70, and 220

Answer: The values of the angles are 35°, 35°, 70° and 220°

Source:

http://www.onlinemathlearning.com/geometry-word-problems.html

Page 3
 4. The sum of the two non-right angles in a right triangle is of course 90 degrees. If
twice the first is 40 degrees more than three times the second. Find the
measurement of the angles of the right triangle.

Solution:

Let X = first angle

Let Y = second angle

Equation one: X + Y = 90

Equation two: 2X = 3Y + 40

From equation one we derive Y = 90 – X, then substitute this in equation two.

2X = 3(90-X) + 40

2X = 270 – 3X + 40

2X + 3X = 270 + 40

5X = 310

X=62

X = 62 first angle

Y = 90 – 62

= 28 second angle

Source:
http://hubpages.com/education/Solving-GeometryProblems-Involving-System-of-Equations-
In-Two-Variables

Page 4
 5. Two angles are supplementary. The bigger angle is thrice as large as the smaller
angle. Find the measurement of the angles.

Solution:

Let X = smaller angle

Let Y = bigger angle

The two angles are supplementary it means that their sum is equal to 180 degrees.

Equation one: X + Y = 180

Equation two: Y = 3X

Substitute equation two in equation one:

X +3X = 180

4X = 180

(1/4) 4X = 180 (1/4)

X = 45 smaller angle

Y = 3(45)

= 135 bigger angle

Source:

http://hubpages.com/education/Solving-Geometry-Problems-Involving-System-of-Equations-
In-Two-Variables

Page 5
B. TIME PROBLEM
1. An object starting from rest travels a distance of 75 meters in 5 seconds. If the
distance traveled varies directly as the square of time, how far will the object
travel in 15 seconds?
Solution:
Equation:

Time in seconds (t) 5 15

Distance in meters (d) 75 ?

Let d=distance
t= time
k= constant of variation
d = kt2
75 = k (25)
k=3
k=3
Then,
d = kt2
= (3) (152)
= 675 meters
In direct proportion
X1 = 5, y1 = 75, x2 = 15, y2 =?
25y2 = 225(75)
25y2 = 16875
y2 = 675 m.
So, the distance traveled in 15 seconds is 675 meters.

Source:
Algebra II- Functional approach
Julieta G. Bernabe, page 110

Page 6
 2. A boy traveled by train which moved at the speed of 30 mph. He then boarded a
bus which moved at the speed of 40 mph and reached his destination. The entire
distance covered was 100 miles and the entire duration of the journey was 3
hours. Find the distance he traveled by bus.

Solution:

Distance Speed Time

Train d 30 t
Bus 100-d 40 3-t

Let:

t = the time taken by the train

3 - t =the time taken by the bus

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

30t + 120- 40t=100

30t-40t=100-120

-10t=-20

t= 2

Substituting the value of t in 40(3-t), we get the distance traveled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d (100−d)
30
+ =3
40

Therefore: d = 60, which is the distance traveled by train.

100-60 = 40 miles is the distance traveled by bus.

Source:

http://www.mbacrystalball.com/blog/2015/08/14/time-distance-speed-problems/

Page 7
 3. Linda can finish cleaning her house in 45 minutes. If her brother will help, they
can finish it in 20 minutes. How long would it take her brother to clean the house
alone?
Representation:
Let h = the number of minutes spent by Linda’s brother in cleaning the house.
1
= part of the house cleaned by the brother in one minute.
h
1
= part of the house cleaned by Linda in one minute.
45
1
=part of the house cleaned by the two in one minute
20
1 1 1
Equation: + =
45 h h

180h ( 451 + 1h )=( 201 ) 180h

4h + 180 =9h
180=5h
36 = h
It should take Linda’s brother 36 minutes to clean the house alone.

Source:
High School Mathematics Concept and Operation, Pascua Landito, page 113

Page 8
 4. John took a drive to town at an average rate of 40 mph. In the evening, he drove
back at 30 mph. If he spent a total of 7 hours traveling, what is the distance
traveled by John?

Solution:
Step 1: Set up a rtd table.

r t d

Case 1

Case 2

Step 2: Fill in the table with information given in the question.

John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph.
If he spent a total of 7 hours traveling, what is the distance traveled by John?

Let: t = time to travel to town.

7 – t = time to return from town.

r t d

Case 1 40 t

Case 2 30 7–t

Step 3: Fill in the values for d using the formula d = rt

r t d

Case 1 40 t 40t
Case 2 30 7–t 30(7 – t)

Step 4: Since the distances traveled in both cases are the same, we get the equation:

40t = 30(7 – t)

Use distributive property

Page 9
 40t = 210 – 30t

Isolate variable t

40t + 30 t = 210

70t = 210

Step 5: The distance traveled by John to town is

40t = 120

The distance traveled by John to go back is also 120

So, the total distance traveled by John is 240

Answer: The distance traveled by John is 240 miles.

Source: http://www.onlinemathlearning.com/distance-problems.html

Page
10
 5. Two cars entered an Interstate highway at the same time and traveled toward
each other. The initial distance between cars was 390 miles.  First car was
running
at the speed 70 miles per hour, the second car was running at 60 miles per hour.
How long will it take for two cars to pass each other?
What distance each car will travel before passing?
Solution:
We will solve the problem by reducing it to the simple linear equation.
Let:
t =   the unknown time in hours for two cars to pass each other
Since the rate of the first car is 70 miles per hour, the distance it travels for time  
t = (70 ∙ t)
Since the rate of the second car is 60 miles per hour, the distance it travels for time
t = (60 ∙ t)
At the moment two cars pass each other, the sum of these distances is equal to the initial distance
between two cars, 390 miles
Therefore,
(70 ∙ t) + (60 ∙ t) = 390
(70 + 60) ∙ t = 390
130t = 390
130t 390
130
= 130

t=3
Check:
70 ∙ 3 = 210
60 ∙ 3 = 180
Answer.  It will take 3 hours for two cars to pass each other.
First car will travel 210 miles, and the second car will travel 180 miles before passing.

Page
11
Source: https://www.algebra.com/algebra/homework/word/travel/Travel-and-Distance-
problems.lesson
C. SPEED PROBLEMS
1. A train leaves from a station and moves at a certain speed. After 2 hours,
another train leaves from the same station and moves in the same direction at a
speed of 60 mph. If it catches up with the first train in 4 hours, what is the speed
of the first train?

Solution:
Let s = the speed of the first train

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4
hours,

Therefore, 6s = 60*4

Solving which gives s = 40.

So the slower train is moving at the rate of 40 mph.

Source: https://www.math10.com/en/algebra/word-problems.html
 

Page
12
 2. The distance between two towns is 380 km. At the same moment, a passenger car
and a truck start moving towards each other from different towns. They meet 4
hours later. If the car drives 5 km/hr faster than the truck, what are their
speeds?

Solution:

The main idea used in this kind of problems is that the distance equals speed multiplied by
time S = V ⋅ tS = V⋅t
V (km/hr) t (hr) S (km)
Car x+5 4 4(x +5)
Truck X 4 4x

4 (x + 5) + 4x = 3804(x + 5) + 4x = 380
4x + 4x = 380 − 204x + 4x = 380 − 20
8x = 3608x = 360
x = 3608x = 3608
x = 45x = 45

Therefore the truck's speed is 4545 km/hr, and the car's speed is 5050 km/hr.

Source:
https://www.math10.com/en/algebra/word-problems.html

Page
13
 3. A bus travels from town A to town B. If the bus's speed is 50 km/hr, it will arrive
in town B 42 min later than scheduled. If the bus increases its speed
by 509509 m/sec, it will arrive in town B 30 min earlier than scheduled. Find:
A) The distance between the two towns;
B) The bus's scheduled time of arrival in B;
C) The speed of the bus when it's on schedule.

Solution:

First we will determine the speed of the bus following its increase. The speed is increased
by 509509 m/sec =50⋅60⋅6091000=50⋅60⋅6091000 km/hr =20=20 km/hr.

Therefore,
The new speed is V=50+20=70V=50+20=70 km/hr.

If  x is the number of hours according to the schedule,

Then at the speed of 50 km/hr the bus travels from A to B within (x+4260) (x+4260) hr. When
the speed of the bus is V=70V=70 km/hr, the travel time is x−3060x−3060 hr.

Then,
50(x+4260)=70(x−3060)50(x+4260)=70(x−3060)
5(x+710)=7(x−12)5(x+710)=7(x−12)
72+72=7x−5x72+72=7x−5x
2x=72x=7
x=72x=72 hr.
So, the bus is scheduled to make the trip in 33 hr 3030 min.
The distance between the two towns is 70(72−12) =70⋅3=21070(72−12) =70⋅3=210 km and the
scheduled speed is 21072=6021072=60 km/hr.

Source:
https://www.math10.com/en/algebra/word-problems.html

Page
14
 4. A boat cruises downstream for 4 hours before heading back. After travelling
upstream for 5 hours, it is still 16 miles short of the starting point. If the speed of
the stream is 4 mph, find the speed of the boat in still water.
Let x= speed of boat in still water
Then x+4= speed of boat downstream
and x-4= speed of boat upstream

d r T
Downstream 4(x + 4) x+4 4
Upstream 5(x - 4) x-4 5

Distance traveled downstream=16+ distance traveled upstream

4(x + 4) = 16 + 5(x - 4)
4x + 16 = 16 + 5x - 20
20 = x
x = 20

The speed of the boat in still water is 20 mph.

Source:
Simplified College Algebra, Ferdinand P. Nocon, Page 99

Page
15
 5. Two airplanes left airports which are 600 miles apart and flew toward each
other. One plane flew 20 miles per hour faster than the other. If they passed each
other at the end of an hour and 12 minutes, what were their rates?

Solution:
Let x be the rate, in miles per hour of the slower plane
Then X+20 will be the rate of the faster plane. The distance that each flew
is found by multiplying its rate by time, 1 1/8 hours.
Therefore the total distance that they flew was 1 + 1 ( x+20), and this must be
600 miles. Thus,
x + x+ (20)= 600
x +24= 600
x = 600-24
x = 576
12x = 5(576)
12x = 2880
x = 2880/ 12
x = 240
x + 20= 260
That is, the rates of the two planes were 240 miles per hour and 260 miles per hour, respectively.

Source:
College Algebra, Paul R. Rider, Page 35

Page
16
 D. MISCELLANEOUS PROBLEMS

1. 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 x denotes 5% of x
x = $190(3)
x = $190.3
x = $600

Source:
http://www.nabla.hr/IA-LEquWordProblem3.htm

2. A mom bought for her son 3 pencils and 4 pens. Being arithmetically minded, she
quickly figured out that had she bought 4 pencils and 3 pens she would have spent
$1 less. If a pen is twice as expensive as a pencil, what is the price of each?

An implicit assumption that is seldom annunciated clearly is that all pencils cost the same as are
all pens.)

Let  x be the price of a pencil and y the price of a pen. We know that
y = 2x.

The cost of 3 pencils and 4 pens is $1 more than the he cost of 4 pencils and 3 pens translates
into
(3x + 4y) - (4x + 3y) = 1.

The equation is simplified to

y-x=1

Recollect that y = 2x, and replace y in the equation with 2x to obtain

2x - x = x = 1

So a pencil costs $1 and a pen $2 (which can be computed as either $1×2 or $1 + $1.)

Source:
http://www.cut-the-
knot.org/Outline/WordProblems/PensAndPencils.shtml#discussion

Page
17
 3. Tina is paid time-and-a-half for each hour worked over 40 hours in a week. Last
week she worked 45 hours and earned $380. What is her normal hourly rate?
Solution:
Let r represent Tina’s normal hourly rate. Then } 3 2 }r represents 1} 1 2 } times her normal
hourly rate (time-and-a-half). The following guideline can be used to help set up the equation.
Regular wages for first 40 hours + wages for 5 hours of overtime= Total wages
40r + 5(r)= $ 380

Solving this equation, we obtain

2(40r + 5(r) = 2(380)

2(40r) +2(5) = 760
80r + 15r = 760
95r = 760
r=8

Her normal hourly rate is $8 per hour.

Source:
http://math.sci.ccny.cuny.edu/document/show/2205

4. There are 51 students in a certain class. The number of females is 5 less than three
times the number of males. Find the number of females and the number of males in
the class.

Solution:

Let m = represent the number of males;

3m – 5 = represents the number of females.

The total number of students is 51, so the guideline is (number of males) PROBLEM 3
PROBLEM 4 8 Equations, Inequalities, and Problem Solving plus (number of females)
equals 51. Thus we can set up and solve the following equation,

m + (3m - 5) = 51
4m – 5 = 51
4m = 56
m =14

Therefore, there are 14 males and 3(14) 2 5 5 37 females.

Source:
http://math.sci.ccny.cuny.edu/document/show/2205

Page
18
 5. 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 22(=
X= = 2.5 kg

Therefore, there is 2.5 kg of dry grapes in 22 kg of fresh grapes.

Source:
http://www.nabla.hr/SP-SolvedProblemsU03C.htm

Page
19
 E. NUMBER PROBLEMS

1. Seven times a certain number is greater than 21. Find the number.
Solution:
Step 1: Let x = the number

Step 2: Seven times a number is greater then 21

7x > 21
Step 3:
7x > 21

7 x 21
>
7 7

x>3
Therefore, the solution set is: {x/x > 3}

Source:
Elementary Algebra: Based on Basic Education Curriculum, Ligaya G. Irsigne, page 210-211

Page
20
 2. The sum of three consecutive integers is 48. What are the integers?

Solution:
Let: x = the first integer
x + 1 = the second integer
x + 2 = the third integer

(x) + (x + 1) + (x + 2) = 48

first second third sum of the three

integer integer integer consecutive integers

x + x + 1 + x + 2 = 48
3x + 3 = 48
3x = 48 – 3
3x = 45
x=
x = 15
1st integer x = 15
2nd integer x + 1 = 15 + 1 = 16
3rd integer x + 2 = 15 + 2 = 17
Check:
15 + 16 + 17 = 48

Source:
Exploring, Valuing and Enjoying Yourself Through Mathematics,
Elisco E. Vigan, Conchita r. Ybañez, Dr. Gabriel G. Uriante, page 113.

Page
21
 3. The larger of two numbers is 5 less than twice the smaller, and their sum is 28.
What are the numbers?

Solution:

Defining the variable:

Let n = the smaller variable

2n – 5 = the larger number

Thought process:
(the smaller number) + (the larger number) = 28
Equation:
n + (2n – 5) = 28
Solving the equation:
n + (2n – 5) = 28
n + 2n – 5 = 28
3n = 33
n = 11
2n – 5 = 2 (11) – 5 = 17
Answering the question:
The numbers are 11 and 17.
Checking the answer:
The larger number is 17 is less than twice the smallest number 11, and the sum of 17 and
11 is 28.

Source:
Elementary Algebra, Catalina B. Maralo, page 276-277.

Page
22
4. Twice the larger of two numbers is three more than five times the smaller, and
the sum of four times the larger and three times the smaller is 71. What are the
numbers?

Solution: Let: the larger number:  x 

the smaller number:  y, twice the larger:  2x 

three more than five times the smaller:  5y + 3 

relationship between ("is"):  2x = 5y + 3

four times the larger:  4x 

three times the smaller:  3y 
relationship between ("sum of"):  4x + 3y = 71

Now I have two equations in two variables:

2x = 5y + 3 
4x + 3y = 71

I will solve, say, the first equation for x:

x = (5/2)y + (3/2)

Then I'll plug the right-hand side of this into the second equation in place of the "x":

4[ (5/2)y + (3/2) ] + 3y = 71 

10y + 6 + 3y = 71 
13y + 6 = 71 
13y = 65 
y = 65/13 = 5

Now that I have the value for y, I can solve for x:

x = (5/2)y + (3/2) 

x = (5/2)(5) + (3/2) 
x = (25/2) + (3/2) 
x = 28/2 = 14

As always, I need to remember to answer the question that was actually asked. The
solution here is not "x = 14", but is the following sentence:

The larger number is 14, and the smaller number is 5.

Source: http://purplemath.com/modules/numbprob.htm

Page
23
 5.  Two consecutive odd numbers are such that three times the first is 5 more than twice
the second.  What are those two odd numbers?
where we represent an odd number as 2n + 1.)
Solution:
Let the first odd number be 2n + 1.
Then the next one is 2n + 3 -- because it will be 2 more.
The problem states, that is, the equation is:
  3(2n + 1) = 2(2n + 3) + 5.
       That implies:
  6n + 3 = 4n + 6 + 5.
  2n = 8.
  n = 4.

Therefore the first odd number is 2 2· 4 + 1 = 9.· 4   The next one is 11.
And that is the true solution, because according to the problem:
3· 9 = 2· 11 + 5.

Source:
http://www.themathpage.com/alg/word-problems.htm

Page
24
 F. TIME PROBLEMS

1. A jeepney leaves Ronnie’s house at 60 cm per hour. Two hours later, a taxi
leaves the same place at 80 cm per hour. How long will it take the taxi to
overtake the jeepney?

Solution:
Distance is the product of the rate and time.
Let: t = the number of hours for the taxi to overtake the jeepney

Distance Rate Time

Jeepney 60 (t + 2) 60 t+2
Taxi 80t 80 t

In overtaking situations, distances traveled are equal. Hence, in this case,

(Distance the jeepney travels) = (Distance the taxi travels
60 (t + 2) = 80t
60t + 120 = 80t
120 = 20t
t=6
The taxi will take 6 hours to overtake the jeepney.

Source:
Elementary Algebra, Catalina B. Maralo, page 279-280.

Page
25
 2. One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes
are opened, they fill the pool in five hours. How long would it take to fill the pool
if only the slower pipe is used?

Convert to rates
hours to complete job:
  fast pipe: f
  slow pipe: 1.25f
  together: 5
completed per hour:
  fast pipe: 1/f
  slow pipe: 1/1.25f
  together: 1/5

adding their labor:

1
/f + 1/1.25f = 1/5

multiplying through by 5f:

5 + 5/1.25 = f 
5 + 4 = f = 9

Then 1.25f = 11.25, so the slower pipe takes 11.25 hour

Source:
http://purplemath.com/modules/workprob.htm

Page
26
 3. A farmer can plow his field in 5 days while his neighbor can do the same work in
4 days. How many days will be required if both men work together on the job?
Solution:
Let x= the number of days required if both men work together
= part of the work both can finish in one day
= part of the work farmer can do in one day
= part of the work his neighbor can do in one day
The equation will be: + =
4x+5x=20
9x=20
X= or 2
Therefore there 2 days required if both men work together on the job.
Source:
Algebra II Functional Approach, Julieta B. Bernabe, Page 98

4.  Two trains, traveling towards each other, left from two stations that are 900
miles apart, at 4 pm. If the rate of the first train is 72 mph and the rate of the
second train is 78 mph, at whatt time will they pass each other?

Solution:

 After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by 

D1 = 72 t and D2 = 78 t 

 After t hours total distance D traveled by the two trains is given by 

D = D1 + D2 = 72 t + 78 t = 150 t 

 When distance D is equal to 900 miles, the two trains pass each other. 

150 t = 900 

 Solve the above equation for t 

t = 6 hours.

Source:

http://www.analyzemath.com/math_problems/rate_time_dist_problems.html

Page
27
 5. Two cars left, at 8 am, from the same point, one traveling east at 50 mph and the
other travelling south at 60 mph. At what time will they be 300 miles apart?

A diagram is shown below to help you understand the problem. 

 The two cars are traveling in directions that are at right angle. Let x and y be the
distances traveled by the two cars in t hours. Hence 

x = 50 t and y = 60 t 

 Since the two directions are at right angle, Pythagora's theorem can used to find distance
D between the two cars as follows: 

D = sqrt ( x 2 + y 2 ) 

 We now find the time at which D = 300 miles by solving 

sqrt ( x 2 + y 2 ) = 300 

 Square both sides and substitute x and y by 50 t and 60 t respectively to obtain the
equation 

(50 t) 2 + (60 t) 2 = 300 2

 Solve the above equations to obtain 

t = 3.84 hours (rounded to two decimal places) or 3 hours and 51 minutes (to the nearest
minute) 

 The two cars will 300 miles apart at 

8 + 3 h 51' = 11:51 am.

Source: http://www.analyzemath.com/math_problems/rate_time_dist_problems.html

Page
28
 G. MONEY PROBLEM

1. Fifteen coins consisting of 25 centavos and 50 centavos worth P10. How many
are 25 centavo coins and 50 centavo coins?

Solution:

Let: x = the number of 25 centavo coins

.15 – x = the number of .50c coins
.25c (x) = amount in 25c coins
.50 (15 – x) = amount of 50c
.25x + .50 (.15 – x) = 5.00
.25x + 7.50 - .50x = 5.00
-.25x = -2.50
x = 10 (number of 25c)
15 – x = 5 (number of 50c)
x = 10
= P2.50
15 – x = 5 (50c) = P5.00
= P2.50
Check:
P0.25 (10) = P2.50
P0.25 (5) = P2.50
P5.00
Source:
Exploring, Valuing and Enjoying Yourself Through Mathematics
By: Elisco E. Vigan, Conchita r. Ybañez, Dr. Gabriel G. Uriante, page 114.

Page
29
 2. Mr. Ramos earns ₱300.00 per 2 hours for tutorial services. If she works 6
hours per week, how much is her income for one month?

Solution:
1st step: The earnings of Mrs. Tan for a month

Facts: ₱300 per 2 hours of work

6 hours per week of work

Find: One month earnings of Mrs. Tan

2nd step: Since there are 4 weeks in a month, multiply 6 hours by 4 to find out the total number of
hours Mrs. Tan works. Then, divide the number of hours by 2 and multiply it by ₱300.00

3rd step: a. 6*4= 24

b. 24/2= 12

c. ₱300.00*12= ₱3,600.00

4th step: Check the computation.

X = ₱3,600.00

The income of Mrs. Tan for one month is ₱3,600.00

Source:
Math Master Elementary Algebra, Aurora E. Tan, Page 18

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 3. Rheena bought a pair of shoes ₱15.00 less than the original price. She paid
₱9800.00 for the shoes. What was the original price of the pair of shoes?

Let x = the original price of the shoes

Equation: x-15=980
Solution:
x-15=980
x-15+15=980+15
x= 995

Check:
995- 15= 980

Answer:
The original price of the pair of shoes is ₱995.00

Solution:
Math Master Elementary Algebra, Aurora E. Tan, Page 18

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 4. You have 40 coins in nickels and dimes.  How many dimes do you have if you
have a total of $2.85?

Solution:

Our answer is 

        "We have ____ dimes"

Let 

        d = the number of dimes you have

then,

        40 - d  =  the number of nickels that you have.

The total money that you have is

        10d + 5(40 - d)  =  285            Value of dimes + Value of nickels = 285

        10d + 200 - 5d  =  285             Distributing the 5

        5d + 200  =  285                       10d - 5d = 5d

        5d  =  85                                    Subtracting 200 from both sides

        d = 17                                       Dividing by 5

We have 17 dimes.

Source:

https://www.ltcconline.net/greenl/courses/152a/lineareq/monmix.htm

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 5. Andre has more money than Bob.  If Andre gave Bob $20, they would have
the same amount.  While if Bob gave Andre $22, Andre would then have
twice as much as Bob.  How much does each one actually have?
Solution: 
Let x = be the amount of money that Andre has.  Let y be the amount that Bob has.
Always let x and y answer the question -- and be perfectly clear about what they
represent!
Now there are two unknowns.  Therefore there must be two equations.  (In general, the
number of equations must equal the number of unknowns.)  How can we get two equations
out of the given information?  We must translate each verbal sentence into the language of
algebra.
Here is the first sentence:
"If Andre gave Bob $20, they would have the same amount."
Algebraically:
1)   x − 20 = y + 20.
(Andre -- x -- has the same amount as Bob, after he gives him $20.)
Here is the second sentence:
"While, if Bob gave Andre $22, Andre would then have twice as much 
as Bob."

Algebraically:
(2)   x + 22 = 2(y − 22).

(Andre has twice as much as Bob -- after Bob gives him $22.)

To solve any system of two equations, we must reduce it to one equation in one of the
unknowns.  In this example, we can solve equation 1) for x --
x − 20 = y + 20
 
  implies   x = y + 40

-- and substitute it into equation 2):

y + 40 + 22 = 2(y − 22).

       That is,

y + 62 = 2y − 44,

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y − 2y = − 44 − 62,

 according to the techniques of Lesson 9,

−y = −106

y = 106.

Bob has $106.  Therefore, according to the exression for  x, Andre has

106 + 40 = $146.

Source:

http://www.themathpage.com/alg/word-problems3.htm

H. AGE PROBLEM

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 1. Miko is 5 years than Kyle. If both of their ages are whole numbers and the
sum is less than 37, what is the greatest possible age of each?
Solution:
Step 1: Let x = age of Miko
x + 5 = age of Kyle
Step 2: Miko’s age + Kyle’s age < 37
( x+ 5 ) + x < 37
Step 3: ( x + 5 ) + x < 37
2x + 5 < 37
2x + 5 – 5 < 37 – 5
2x < 32
<
x < 16
then, x + 5 = ( 16 ) + 5 = 21
therefore. The greatest possible age of Miko is 20 years old: for Kyle, 15 years old.

Source:
Elementary Algebra: Based on Basic Education Curriculum, Ligaya G. Irsigne, page 211

2. A father is four times as old as his daughter. If the sum of their ages is 35,
how old is the father? How old is the daughter?

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Solution:
Let x = the daughter’s age
4x = the father’s age
4x + x = 5x the sum of their ages
5x = 35
x=7 the daughter’s age
4 (x)
4 (7) = 28 the father’s age
4 (x) + x 4 (7) + 7
28 + 7 = 35 the sum of their ages

Source:
Exploring, Valuing and Enjoying Yourself Through Mathematics, By: Elisco E. Vigan,
Conchita r. Ybañez, Dr. Gabriel G. Uriante, page 115

3. Tony is six times as old as Imari now. Three years ago the sum of their ages
was 29. How old is each now?
Solution:

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Let x Imari’s age now
6x Tony’s age now
x–3 Imari’s age 3 years ago
6x – 3 Tony’s age 3 years ago
(x – 3) + (6x – 3) = 29 the sum of their ages 3 years ago
x – 3 + 6x – 3 = 29
7x = 29 + 6
7x = 35
x=
x=5 Imari’s age now
6 (x)
6 (5) = 30 Tony’s age now
x–3
5–3 = 2 Imari’s age 3 years ago
6 (x) – 3
30 – 3 = 27 Tony’s age 3 years ago

Source:
Exploring, Valuing and Enjoying Yourself Through Mathematics,By: Elisco E. Vigan,
Conchita r. Ybañez, Dr. Gabriel G. Uriante, page 115-116

4. Dandy was born 6 years before Nuel. In 2000, Dandy’s age was 2 years more
than four times the age of Nuel in 1995. How old will Nuel be in 2010?
Solution:

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 In some cases, a table of values helps in the solution of the problem.
Defining the variable:
Let: n = Nuel’s age in 1995

1995 2000 2010

Nuel n n+5 n + 15
Dandy n+6 4n + 2 n + 21

Thought process:
[(Dandy’s age in 1995 + 5)] = (Dandy’s age in 2000)
Equation:
n+6 +5 = 4n + 2
Solving the Equation:
n + 6 + 5 = 4n + 2
n + 11 = 4n + 2
11 – 2 = 4n – n
9 = 3n
n=3
n + 15 = 18
Answering the question:
Nuel will be 18 years old in 2010.
(Checking will be left as the student exercise)

Source:
Elementary Algebra, Catalina B. Maralo, page 277-278
5. In three more years, Miguel's grandfather will be six times as old as Miguel
was last year. When Miguel's present age is added to his grandfather's
present age, the total is 68. How old is each one now?

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This exercise refers not only to their present ages, but also to both their ages last year and
their ages in three years, so labeling will be very important. I will label Miguel's present
age as "m" and his grandfather's present age as "g". Then m + g = 68. Miguel's age "last
year" was m – 1. His grandfather's age "in three more years" will be g + 3. The
grandfather's "age three years from now" is six times Miguel's "age last year" or, in math:
g + 3 = 6(m – 1)
This gives me two equations with two variables:
m + g = 68 
g + 3 = 6(m – 1)
Solving the first equation, I get m = 68 – g. (Note: It's okay to solve for "g = 68 – m", too.
The problem will work out a bit differently in the middle, but the answer will be the same
at the end.) I'll plug "68 – g" into the second equation in place of "m":

g + 3 = 6m – 6 
g + 3 = 6(68 – g) – 6 
g + 3 = 408 – 6g – 6 
g + 3 = 402 – 6g 
g + 6g = 402 – 3 
7g = 399 
g = 57
Since "g" stands for the grandfather's current age, then the grandfather is 57 years old.
Since, m + g = 68, then m = 11, and Miguel is presently eleven years old.

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