How to Solve Number Word Problems

The Solving Number Word Problems Series

Word Problems are difficult to many. The Solving Number Word Problems Series is the first series
of detailed tutorials on how to solve various number problems. Here are the posts.
(1) How to Solve Number Problems Mentally
This introduction discusses various strategies used to solve easy number word problems. Before you
solve a problem using paper and pencil, you should try to solve it first mentally.

(2) How to Solve Number Problems Part 1

This part solves the same numbers in (1) but using algebra. The objective of this part is to introduce how
to set up equations based on “word phrases.”
(3) How to Solve Number Problems Part 2
This part introduces more problems that are slightly more complicated than in (2). It also introduces
“number problems in disguise.”

(4) How to Solve Number Problems Part 3

This part of the series focus on how to solve consecutive numbers. Problem of consecutive numbers are
very common in math tests.

(5) How to Solve Number Problems Part 4

This post discusses more complicated problems and also introduces how to set up solutions to number
problems with fractions.

What’s more to come?

Maybe, I’ll have one more post for this series in the future. But for now, I will focus on the next topic which
is about age problems.

How To Solve Number Word Problems Mentally

Most students in high school would rush and get a pencil or pen and write the equation if they see word
problems such as Problem 1. I’m sure many of you will do the same. But you should really stop wasting
lead and ink because problems such as this can be solved mentally.
Many are poor in mental math because most of us did not develop the habit of solving the problem mentally
first, before getting a pencil and paper. Most of us, and our high school teachers too, are so obsessed and
always in a hurry to write “let x = something” when we see word problems. If you want to be a good problem
solver, the pencil and paper (and other tools) should be the last resort. Before getting a tool, try solving any
problem in your head first.

Before you get excited, take 3-5 minutes to solve Problem 1 in your head and see if you can get the right
answer before you continue reading.
Problem 1
One number is 3 more than the other. Their sum is 45. What are the numbers?
Now, let’s solve number word problems mentally!

Solution
First, we have two numbers. One is 3 more than the other. So, if the first number is, for example, 18, the
other number is 18+3 which is 21. This means, that if we subtract 3 from the larger number, then they will
be equal.

Analysis
 1. Facts: two numbers, sum = 45, one is 3 greater than the other.
2. If we subtract 3 from the greater number, the two numbers will be equal.
3. If we subtract 3 from the greater number, their sum will also decrease by 3. I’m sure you can do 45 – 3 in your
head. Now, we have sum = 42.
4. Now, that we have subtracted 3, the numbers are equal with a sum of 42. Well, we just divide 42 by 2 since the
two numbers are equal. 42/2 = 21.
So, the smaller number is 21, and the larger number is 21 + 3 = 24.

Check: Is one number 3 more than the other? Yes, 24 is 3 more than 21. Is the sum 45? Yes, 21 + 24 =
45.

Try solving mentally:

The sum of two numbers is 87. One is 5 more than the other. What are the numbers?

If you get this right, you should treat me for a cup of coffee. Starbucks?

Problem 2
The sum of two numbers is 53. One number is 7 less than the other. What are the numbers?
This problem is quite the same with Problem 1. The only difference is that the other number described is
less than the other number (not more than). Still, we solve mentally.

Analysis

1. Facts: two numbers, sum = 53, one number is 7 less than the other.
2. If we add 7 to the smaller number, the two numbers will be equal.
3. If we add 7 to the smaller number, the sum will also increase by 7. I’m su re 53 + 7 can be calculated mentally.
The new sum is now 60.
4. Now that we have added 7 to the smaller number, the two numbers are now equal. So, we divide the sum 60 by
2 which is equal to 30.
Therefore, the larger number is 60/2 = 30. We subtract 7 from 30 to get the smaller. Now, 30-7 = 23.

Check: Is 30 + 23 = 53? Oh yes. Is one number 7 less than the other? Yes, 23 is 7 less than 30.

Try solving mentally: The sum of two numbers is 71. One number is 9 less than the other. What are the
two numbers?

If you get this, that’s two coffee.

Problem 3
One number is twice the other number. Their sum is 45. What are the numbers?
To solve the problem above, consider the analogy. If a group of people can be represented by a circle
(Group 1), then a group which is twice its size can be represented by two circles (Group 2). This means
that the people can be divided equally into three groups.
So, we divide 45 by 3 which is equal to 15. The number twice it’s size is 30.

Check: Is 30 twice 15? Yes. Is the sum 45? Yes.

Try solving mentally: One number if thrice the other number. Their sum is 44. What are the numbers?

Tagalog Math Videos: How to Solve Number Problems Mentally

The three examples above show that many of the word problems that many of us usually fear are not that
difficult at all. I think a change of perspective and a lot of practice will help us solve these problems faster
and more accurately. Note that many problems are really number problems in disguise. For example, the
following problems can be solved mentally because they are the same as Problem 1.
 One log is cut into two. One is 3 meters longer than the other. Their total length is 45 meters. How long are the
logs?
 Susan is three years older then Mary. The sum of their ages is 45. What are their ages?
 Jack and Jill together have 45 dollars. Jack has 3 dollars more than Jill. How much does each of them have?
See, if you can solve the problems such as the three examples above, you can really solve a lot of
problems. My advice is practice, practice, and more practice.

How to Solve Number Word Problems Part 1

In the previous post, we have learned How to Solve Number Problems Mentally. In this post, we are going to
solve the same word problemsalgebraically. The objective of this post is for you to be able to learn how t o
set up equations based on given problems. Once you know how to set up equations for easy problems, it
will be easier for you to do so using harder problems which we will discuss in the latter parts of this series.
Note that before solving these problems, it is already assumed that you know how to solve equations.
Problem 1
One number is 3 more than the other. Their sum is 45. What are the numbers?
Scratch Work
The strategy in solving algebraic problems is to take a specific case. For instance, in the problem above, if
one number is say, , then the larger number is because it is greater than the first number. Since
we do not know the numbers yet, we can represent the smaller number by and the larger number
by .
The next clue is the word “sum,” their sum is . So, sum means you have to add the two numbers which
are and . In sentence form, the sum of and is or

in equation form. Now, we write the solution.

Solution
Let be the smaller number and be the larger number.
So, the smaller number is and the larger number is

Of course, after this, you can always do the checking by looking at the condit ions. Is the larger number
greater than ? Is the sum of two numbers ? Once the answer satisfies all the conditions in the given
problem, then it is correct.
Problem 2
The sum of two numbers is 53. One number is 7 less than the other. What are the numbers?
Scratch Work
In this example, if the larger number is , then the other number is less than the or . So,
if the number is , the smaller number is . The next sentence is their sum is , so we have to
add and forming the equation

We now write the solution.

Solution
Let be the larger number and be the smaller number.

Therefore, the larger number is and the smaller number is . Again, you can
check the answer by if it satisfies the conditions above.

Problem 3
One number is twice the other number. Their sum is 45. What are the numbers?
Scratch Work
If one number is , then the number twice it is , or . Therefore, if one number is , the number twice
it is or . Next, their sum is . It means that if you add and , the sum is . Or,
.

Solution
Let be the smaller number and be the larger number.

Dividing both sides by , we have

 .

Therefore, the smaller number is and the larger number is . Checking it, is indeed
twice , and yes, their sum is .

How to Solve Number Word Problems Part 2

This is the second part of the the Solving Number Word Problems Series. In this part, we will discuss how
to solve various number problems. Note that some of these problems are not really number problems per
se, but the strategy in solving them is technically the same. You could say that they are really “number
problems in disguise.”
We already had three problems in the first part of this series, so let’s solve the fourth problem.
Problem 4
If is subtracted from three times a number, then the result is . What is the number?
Scratch work
In the How to Solve Number Word Problems Part 1, I mentioned that sometimes, if it is hard to convert the
words in the problem to equations, it is helpful to think of a particular number. For example, in this problem,
the phrase is “three subtracted from three times a number.” So, if we choose a number, say for example,
, we want to subtract from three times . In numerical expression, that is . So, if a number is ,
the expression is . Now, this results to as stated above. Intuitively, it is saying
that is equal to . That’s our equation!
Solution
Let be the number.

is 8 less than 3 times the number.

Now, .

Check: The problem says that if is subtracted three times the number, the result is . Now, three
times is . Now, if we subtract from the result is and we are correct.

Problem 5
Separate into two parts such that the one exceeds the other by . What are the numbers?

Scratch Work
If we separate into two parts, and one part is, for example, , then the other part is which
is . This means that if one part of is , then the other part is .

Now it says that the larger number exceeds the smaller number by . This means that

larger number – smaller number = 24.

The only part left now is to choose which is larger, or . It won’t really matter.
Solution
Let be the larger and be the smaller number.

Then,
 .

Subtracting from both sides results to

Dividing both sides by , we get

So, the smaller number is and the larger number is .

Check: Is the sum of the two numbers is ? Does exceed by ? If both answers are yes, then
we are correct.
Problem 6
The sum of the ages of Abby, Bernice, and Cherry is . Bernice is twice as old as Abby, while Cherry is
4 years older than Abby. What are the ages of the three ladies?

Scratch Work
As I have mentioned above, some problems are really number problems in disguise. This problem is one
of them.

From the problem, it is easy to see that the youngest in the group is Abby. Let us say, Abby is . So,
Bernice is twice as old or years old. Then, Cherry is four years older than Abby
or .

So, from this analysis, if Abby is years old, then, Bernice is . Since Cherry is four years older than
Abby, then here age is .

In the first sentence, it says that the sum of the ages of the three ladies is . Therefore, we must add
their ages ( , and ) and equate it to . That is our equation.

Solution
Let be Abby’s age, be Bernice’s age and be Cherry’s age.

Dividing both sides by , we have .

So, Abby is , Bernice is and Cherry is years old.

Check: Is the sum of their ages ?

How to Solve Number Word Problems Part 3
This is the third part of The Number Word Problem Series. In this post, we will be solving number word
problems about consecutive numbers. In number word problem solving, consecutive numbers are numbers
that follow each other in order. Here are the examples of consecutive numbers (integers).
consecutive numbers: 4, 5, 6, 7, 8, …

consecutive odd numbers: -2, 0, 2, 4, …

consecutive odd numbers: 7, 9, 11, 13, 15, …

I am quite sure that you have solved consecutive numbers in your high school mathematics class.
In the previous post, we finished our 6th problem, so, we start with the seventh problem.
Problem 7
The sum of two consecutive numbers is . What are the two numbers?

Scratch Work
If is a number, then the next consecutive number is . Therefore, if two numbers are consecutive
and the smaller number is , then the next number is . Their sum is . This means that if we
add and , the result is equal to . From there, we can now form our equation.

Solution
Let be the smaller number and be the larger number.

Subtracting from both sides, we get

Dividing both sides by , we get

Therefore, the smaller number is and the larger number is

Check: .

Problem 8
The sum of three odd consecutive numbers is . What are the three numbers?

Scratch Work
Please take note that we are talking about odd numbers. So, if the smallest number is, say, , the next
odd number is . The next odd number is . So, if the smallest number is , we have

: next odd number

: largest odd number

Now, if you add these three numbers, the sum will be .

Solution
Let , , and be the consecutive odd numbers.

Subtracting from both sides we have

Dividing both sides by , we have

So, the numbers are , and .

Check: .

Problem 9
The sum of three even integers is . What are the three numbers?

Scratch Work
Observe from the introduction above, the consecutive even numbers also increase by 2 everytime. So, if
the smallest number is , then the other larger numbers are and .

Solution

Subtracting from both sides of the equation results to

Dividing both sides by , we have

Therefore, the three numbers are , , and .

Check: .

How to Solve Number Word Problems Part 4

This is the fourth part and the conclusion to the Number Word Problem Series. In the introduction to this
series, we have learned How to Solve Number Problems Mentally. In Part 1 and Part 2, we have discussed the
basic number word problems, and in Part 3, we have learned how to solve word problems about
consecutive numbers.
In this post, we discuss about more complicated problems especially problems that involve fractions. We
have already discussed 9 problems in the previous parts of this series, so, we now solve the 10th problem.

Problem 10
There are consecutive numbers. The sum of the second and the fourth number is . What is the largest
number?
Scratch Work
Since we have five consecutive numbers, if we let be the smallest, then the other numbers are
, , , and .

Now, the second number is and the fourth is . Their sum is . And that’s where we get our
equation.

Solution
Let , , , and be the five consecutive numbers.

Second Number:

Fourth Number:

Second Number + Fourth Number =

Subtracting 4 from both sides,

So, the five consecutive numbers are , , , and

The largest among them is .

Check: Yes, they are consecutive numbers and .

Problem 11
A number added to of itself is equal to . What is the number?

Scratch Work
If that number is equal to , then of that number is or . Therefore, if the number is , it’s one
fourth is . Now, if we add and , the sum is 80. That is where we get our equation.

Solution
Let be the number and a fourth of it as

To get rid of the fraction, we multiply everything by . This gives us

Check: of is . Now, . This means that we are correct.

Problem 12
One number exceeds the other by . One third the larger subtracted by one one half the smaller is equal
to . What are the numbers?

Scratch Work
One number exceeds the other by 22 means that the other number is 22 more than the smaller. So, if we
let be the smaller number, the larger number is .

Now, we will subtract one-third by one fourth of . The difference will be . In equation form we
have

Solution
Let be the smaller number and be the larger.

The least common multiple of the denominators of the fraction (3 and 2) is , so we multiply everything
with to eliminate the fractional parts.

By the distributive property, we have

Multiplying both sides by gives us which is the smaller number. The larger number
is .

Check: .

How to Solve Age Problems

The Age Problem Solving Series

One of the common types of word problems in mathematics and in many examinations is about Age
Problems. This series discusses various problem styles involving age problems and explains in details how
they are solved.
The Age Problem Solving Series
How to Solve Age Problems Part 1 discusses simple 2-person problems particularly present-past and
present-future age relationships.
How to Solve Age Problems Part 2 discusses a slightly more difficult 2-person problems particularly present-
past and present-future age relationships.
How to Solve Age Problems Part 3 discusses age problems that involves fractions.
I am planning to write a fourth part for this series in the near future, but for now, I will focus more on yet
uncovered topics.

How to Solve Age Problems Part 1

After a series of tutorials on word problems involving numbers, we now move to learning on how to
solve word problems involving age. Age problems are very similar to number problems, so if you have
finished reading The Number Word Problem Series, then it will be easier for you to solve the following age
problems.
Example 1
Benjie is thrice as old as his son Cedric. The sum of their ages is 64. How old are both of them?
Scratch Work
This is one of those age problems that are very similar to number problems. Let’s take a specific case.
If Cedric is say years old, then Benji is years old. This means that if Cedric is years old, then
Benjie is . If we add their ages, the result is .
Solution
Let be the age of Cedric and be the age of Benjie.

Cedric’s Age + Benjie’s Age = 64

Dividing both sides of the equation by gives us .

Therefore, Cedric is and Benjie is year sold.

Check
is thrice and . So, we are correct.

Example 2
Karen is years older than Nina. Five years from now, the sum of their ages will be . How old are both
of them?

Scratch Work
If Nina is years old, then Karen is years older, so her age will be . Five years from now, both of
their ages will increase by as shown on the table below.
Therefore, 5 years from now, the sum of their ages will be equal to

. Now this sum is equal to .

Solution
Let be Nina’s age and be Karen’s age. In 5 years, Nina will be years old and Karen will
be years old.

Now,

Subtracting from both sides of the equation, we have

Dividing both sides by we have

This means that Nina is and Karen is .

Check
is 6 more than and five years from now, . Therefore, we are correct.

Example 3
Sarah is twice as old as Jimmy. Three years ago, the sum of their ages is 39. How old are both of them
now?

Scratch Work
If Jimmy is years old, then Sarah’s age is twice his age, so Sarah is years old. Three years ago, both
are younger by years, so both their ages must be subtracted by .
Three years ago, the sum of their ages is . So, we add and and equate it to

Solution
Let be Jimmy’s age and be Sarah’s age.

Three years ago, Jimmy was years old and Sarah was years old.

Three years ago, the sum of Jimmy’s and Sarah’s age is

Adding to both sides of the equation results to

Dividing both sides by , we have

So, Jimmy is and Sarah is .

Check
Three years ago, Jimmy was years old and Sarah was years old. The sum
of their ages was .

How to Solve Age Problems Part 2

This is the second part of the Solving Age Problem Series. We will continue solving age problems that are
slightly more complicated that the first part. We have already discussed 3 problems in the first part of this
series, so we continue with the fourth problem.

Problem 4
Simon is four years older than Jim. The sum of their ages is 52. How old is Simon?

Scratch Work
This problem is a sort of review of first part of this series. Simon is older than Jim by years. So, if Jim
is years old, then Simon is years old. The sum of their ages is . This means that if
add and , then the sum is . That is the equation.
Solution
Let be Jim’s age and be Simon’s age.

Now,

Jim’s age + Simon’s age = 52 which means that

.
Simplifying we have

Subtracting from both sides, we have

So, Jim is years old. Now, the question asks for the age of Simon. Simon is years old.

Check
Simon is and Jim is , so he is indeed four years older. The sum of their ages
is which agrees with the given in the problem. Therefore, we are correct.

Problem 5
Allan is times as old as Leah. Five years from now, he will be times as old. How old is Allan?

Scratch Work
Now, if Leah is, for example, 7 years old, then Allan is years old. This means that if Leah is years
old, then Allan is years old. Five years from now, Leah will be years old and Allan will
be years old as shown on the table below.

Note that $latex $5 years from now, Allan will be three times as old as Leah. This means that if we
multiply Leah’s age by , then, their ages will be equal. That is, if we multiply by , it will be equal
to . In equation form,

which is the final equation.

Solution
Let be Leah’s age and be Allan’s age.

Five years from now, Leah will be years old and Allan will be years old.

Now, we multiply Leah’s age and equate it to that of Allan’s

By Distributive Property, we have

.
Putting all x’s to the right and all numbers to the left, we have

Dividing both sides by , we have

So, Leah is years old and Allan is

Check
Allan is and Leah is so he is indeed times as old. In years, Allan will be and Leah will be .
Thirty is indeed three times , so we are correct.

Problem 6
Philip is twice as old as Ben. If is subtracted from Philip’s age and is added to Ben’s age, then their
ages will be equal. How old are both of them?

Scratch Work
Ben is years old and Philip is . If we subtract from Philip’s age, it will become . If we
add to Ben’s age, it will be . Now, after the results to these operations, their ages will be equal
or

Solution
Let be Ben’s age and be Philip’s age.

So, Ben is and Philip is .

Check
Philip is and Ben is so, Philip is twice as old Ben. Subtracting from Philip’s age results to .
Adding to Ben’s age is . Well, equals , so we are correct.

How to Solve Age Problems Part 3

This is the third part of the Solving Age Problems Series. In this part, we will solve age problems with a
variety of formats and difficulty that are not discussed in the first two parts. We have already solved six
problems in the first and second part, so we start with the seventh problem.
Example 7
Bill is four times as old as Carol. One fifth of Bill’s age added to one half Carol’s age is equal to years.
How old are both of them?

Scratch work
Bill is older than Carol and he is four times older. This means that if Carol is years old, then Bill is years
old. Now, one fifth of Bill’s age is and one half of Carol’s age is . Add these together and you
get . Now, we have an equation.
Solution
Let be Carol’s age and be Bill’s age.

Simplifying, we have

Since we have a fraction, we can eliminate the denominator by multiplying everything with the least
common multiple of and which is . Multiplying both sides of the equation by , we have

This means that Carol is and Bill is .

Check
Bill is and Carol is . Yes, Bill is four times as old as Carol. One fifth of is . One half
of is and . So, we are correct.

Example 8
When a really smart math kid was asked about his age, he said:

“I am one fifth as old as my mother. In six years, I will be one-third as old.”

How old is the kid and his mother?

Scratch Work
The kids is one fifth as old as his mother. So, if the mother is years old, then the kid is Six years from
now, the ages of the mother and the kid respectively are and as shown in the table below.
As the kid said, in years, his age will be a third of his mother. This means that if we multiply his age by
$latex3$, then it will equal the age of his mother. In equation form, we have

Now, we write the solution.

Solution
Let be the mother’s age and be the kid’s age.

We simplify the right hand side by Distributive Property. This gives us

Now, to eliminate the fraction, we multiply both sides of the equation by .

Again, by distributive property, we have

Putting all the x’s on the left hand side and all the numbers on the right hand side, we have

So, the mother and and the kid is . A smart kid indeed, giving problems such as this at age
6.

Check
Left as an exercise.
Example 9
Donna is years older than Demi. One fifth of Donna’s age a year ago added to three fourth of Demi’s
age is equal to Demi’s age. How old is Donna?

Scratch Work

Demi is years old and Donna is . Now, Donna’s age a year a go is which is equal
to . How, one fifth of Donna’s age a year ago is and one fourth of Demi’s age is .

Now, these ages if added equal’s Donna’s age which is . Therefore, the equation is

Solution

Let be Demi’s age and be Donna’s age

Simplifying, the left hand side by distributive property, we have

Now, to eliminate the fractions, we multiply both sides of the equation by the least common multiple
of and which is . This will result to

Therefore, Demi is and Donna is .

How to Solve Consecutive Number Problems

How to Solve Consecutive Number Problems Part 1

This is the first of the Solving Consecutive Number Series, a series of post discussing word problems about
consecutive numbers.
Consecutive numbers are numbers that follow each other in order. In number problems in Algebra,
consecutive numbers usually have difference 1 or 2. Below are the types of consecutive numbers,

consecutive numbers – 5, 6, 7, 8, …

consecutive even numbers – 16, 18, 20, 22…

consecutive odd numbers – 3, 5, 7, 8, …

The symbol … means that the list may be continued.

Notice that consecutive numbers always increase by 1 in each term. If we make 5 as point of reference,
then, we can write the numbers above as

5, 5 + 1, 5 + 2, 5 + 3.

That means that if our first number is x, then the list above can be written as
x, (x + 1), (x + 2), (x + 3)
and so on.
As for the consecutive even and consecutive odd numbers above, with the smallest numbers as point of
reference, they can be written as

16, 16 + 2, 16 + 4, 16 + 6.
and

3, 3 + 2, 3 + 4, 3 + 6.
Notice that both consecutive odd and consecutive integers increase by 2 in each time. So, if we let x be
the first number, the terms can be written as
x, (x + 2), (x + 4), (x + 6)
and so on.
Now, that we know how to represent consecutive numbers, let us solve our first problem.

Example 1
The sum of two consecutive numbers is 81. What are the numbers?

Solution
Since there is no mention of odd or even, the terms only increase by 1. So, let

x = the first number

x + 1 = the second number.
The word sum means we have to add and the phrase “is 81” means that we have to equate the s um to
81. That is

first number + second number = 81.

Since the first number is x and the second number is (x + 1),

x + x + 1 = 81.
Solving the equation, we have

2x + 1 = 81.
Subtracting 1 from both sides, we have

2x = 80.
Dividing both sides by 2 results to

x = 40.
So, the smaller number is 40, and the larger number is 40 + 1 = 41. The consecutive numbers are 40, 41
and their sum is 81.

Example 2
The sum of three consecutive even numbers is 42. What are the numbers?

Solution
In this example, we have 3 consecutive even numbers. Recall that form above, consecutive even
numbers increase by 2 each time. So, let

x = first number
x + 2 = second number
x + 4 = second number.
Again, the problem mentioned the word sum, so we have to add. That is,

first number + second number + third number = 42.

Substituting the algebraic representation above, we have

x + (x + 2) + (x + 4) = 42
Solving the equation,

3x + 6 = 42.
Subtracting 6 from both sides, we have

3x = 36.
Dividing both sides of the equation by 3 results to

x = 12.
So, 12 is the smallest number, 12 + 2 = 14 is the second number and 12 + 4 = 16 is the largest number.
The consecutive numbers are 12, 14, and 16 and their sum is 42.

How to Solve Consecutive Number Problems Part 2

In the previous post, we have discussed the basics of consecutive number problems. We have learned that
in word problems in Algebra, consecutive numbers usually mean numbers increasing by 1. Consecutive
even numbers and consecutive odd numbers increase by 2. So, consecutive numbers whose smallest is x
are x, x + 1 and x + 2 and so on, while consecutive odd/even numbers whose smallest number is y are y,
y + 2, y + 4 and so on.
In this post, we begin with the third example in the series since we already had 2 examples in the
previous post.
Example 3
The sum of four consecutive numbers is 70. What are the numbers?

Solution
Let

= first number

= second number

= third number

= fourth number

Since we are talking about the sum of the four numbers, we add them. That is,

sum of four numbers = 70

Simplifying, we have

So, the smallest number is 16. Therefore, the four consecutive numbers are 16, 17, 18, and 19.

Check:

Example 4
The sum of 3 consecutive odd numbers is equal to 51. What are the numbers?

Solution
As we have discussed above, odd numbers increase by 2 each time (like 5, 7, 9, 11), so we let

= first number

= second number

= third number

Now, we add the numbers and equate to 51.

So the smallest odd number is 15. Therefore, the three consecutive odd numbers are 15, 17, and 19.

Check:

How to Solve Consecutive Number Problems Part 3

This is the third part of the Solving Consecutive Number Problems Series. In this post, we solve more
problems about consecutive numbers. We have already discussed four problems in the first
part and second part of this series, so we start with the fifth example.
Example 5
There are 3 consecutive odd numbers. Twice the smallest number is one more than the largest. What are
the numbers?

Solution
In the first post in this series, we have learned that odd numbers increase by 3 (e.g. 7, 9, and 11). So, let

= the smallest odd number

= the second odd number

= the largest odd number.

Now that we have represented the numbers, we now go to the second sentence. The second sentence
says that twice the smallest number is one more than the largest. The smallest number is , so twice the
smallest number is . Now, is one more than the largest number . This means that if we add 1
to the largest number, then they will be equal. That is,

Solving, we have
 .

So the consecutive numbers are 5, 7, and 9.

Check: Twice the smallest is 2(5) = 10 is one more than the largest which is 9. Therefore,
our answers are correct.
Example 6
There are three consecutive even integers. The sum of the first two integers is 16 more than the largest.
What are the numbers?
Solution
As we have discussed in the previous posts, the representations of consecutive odd numbers and
consecutive even numbers are the same. Consecutive even numbers such as 18, 20, 22 increase by 2
inch time.

= the smallest even number

= the second even number

= the largest even number

The second sentence states that the sum of the first two integers is 16 more than the largest. The sum of
the first two integers is and the largest integer is . Since the sum of the
first two integers is 16 more than the largest integer, if we add 16 to the largest integer, then they will be
equal. That is,

Simplifying,

Therefore, the consecutive integers are 18, 20, 22$.

Check: The sum of the first two integers 18 + 20 = 38 is 16 more than 22. Therefore, we are correct.

How to Solve Mixture Problems

How to Solve Mixture Problems Part 1

If you have followed this blog, then you would know that we have been tackling a lot of math word problems .
In this series, we will learn to solve another type of math word problem called mixture problems. Mixture
problems are easy if you know how to set up the equation.
Mixture problems can be classified into two, those which deals with percent and the other which deals
with price. In any case, the method of solving is almost the same. Of course, in working with percent, you
must be able to know the basics of percentage problems. Let’s have our first example.

Example 1
How many ml of alcohol does a 80 ml mixture if it contains 12% alcohol?

Solution and Explanation

In this example,the word mixture means that alcohol is mixed with another liquid (we don’t know what it is
and we don’t need to know). The total mixture contains 80 ml and we are looking for the pure alcohol
content which is 12% of the entire mixture. Therefore,
pure alcohol content = 12%× 80 ml = 0.12 x 80 ml = 9.6 ml

In the calculation above, we converted percent to decimal (12% to 0.12) and then multiply it with 80. This
means that in the 80 ml alcohol, 9.6 ml is pure alcohol.

Example 2
What is the total alcohol content of an 80 ml mixture containing 12% alcohol and a 110 ml mixture
containing 8% alcohol?

Solution and Explanation

In this problem, we need to “extract” the pure alcohol content of both mixtures and add them in order to
find the volume of the pure alcohol content of both mixtures.

In Example 1, we already know that it contains 9.6 ml of alcohol, therefore, we only need to solve for the
second mixture.

Pure alcohol content of Example 2 = 8% × 110 ml = 0.08 x 110 ml = 8.8 ml

In the calculation above, we converted 8% to decimal so it became 0.08. Multiplying 0.08 by 110 gives us
8.8 ml. Therefore,

total alcohol content = 9.6 ml + 8.8 ml = 18.4 ml.

That means that the two mixtures contain 18.4 ml of pure alcohol in total.
The second problem shows that if we have more than one mixture, and we want to find the total amount
of pure content, then we need to add the pure contents in the mixtures. Now, does the percentages add
up? Will the total mixture contain 20% alcohol? Let’s see.

The total amount of liquid = 110 ml + 80 ml = 190 ml

Total amount of alcohol = 18.4 ml

18.4 ml / 190 ml = 0.0968

As we can see, if we convert 0.0968 to percent, it becomes 9.68% and not 20%. In what case will they
add up?

How to Solve Mixture Problems Part 2

In the previous post, we have learned the basics of mixture problems. We have learned that if solutions
are added, then the pure content of the combined solution is equal to the sum of the all the amount of
pure content in the added solutions.
In this post, we are going to discuss two more mixture problems. We have already finished
two examples in the previous part, so we start with Example 3.
Example 3
How many liters of 80% alcohol solution must be added to 60 liters of 40% alcohol solution to produce a
50% alcohol solution.

Solution and Explanation

The first thing that you will notice is that we don’t know the amount of liquid with 80% alcohol solution. So,
if we let x = volume of the solution of liquid with 80% alcohol content. So, the amount of pure alcohol
content is 80% times x.

We also know that the amount of alcohol in the second solution is 60% times 40.

Now, if we let solution 1 be equal to the solution with 80% alcohol, solution2 with 40% alcohol,
and solution 3 be the combined solutions, we have

amt of alchohol in solution 1 = 0.8x

amt of alcohol in solution 2 = (0.4)(60)
amt of alchohol in solution 3 = 0.50(0.8x + 60)

Note that we have already converted the percentages to decimals in the calculation above: 80% = 0.8, 40%
= 0.4, and 50% = 0.5.
amt. of alcohol in solution 1 + amt. of alcohol in solution 2 = amt. of alcohol in solution 3

Solving, we have

0.8x + (0.4)(60) = 0.50(x + 60)

0.8x + 24 = 0.5x + 30

0.8x – 0.5x = 30 – 24

0.3x = 6

To get rid of the decimal, we multiply both sides by 10.

3x = 60

x = 60/3

x = 20.

This means that we need 20 liters of solution 1, the solution containing 80% alcohol. We need to combine
this to solution 2, to get solution 3 which has a 50% alcohol content.

In problems like this, you can check your answers by substituting the value of x to the original equation.
0.8x + (0.4)(60) = 0.50(x + 60)

0.8(20)+ (0.4)(60) = 0.50(20 + 60)

16 + 24 = 0.50(80)
40 = 40

Indeed, the amount of alcohol in the left hand side is the same as the amount of alcohol in the right hand
side.

How to Solve Mixture Problems Part 3

In the previous post, we have discussed three examples on how to solve mixture problems. In this post,
we are going to learn how to set up the givens in a mixture problem in a table, so it is easier to
solve. Let’s have the fourth example.
Example 4
How many liters of pure water must be added to 15 liters of a 20% salt solution to make a
5% salt solution?
Solution and Explanation
In the first column, we placed the liquid or solution. We have three kinds: water, the liquid with 20% salt,
and liquid with 5% salt. In the second column, we place the volumes of ths liquid. We do not know the
volume of water, so we represent it with x. Since we have to combine water with 20% salt solution, we
have to add the volumes of the two liquid. This makes sense since if we add more water, the amount of
salt is getting lesser in relation to the total volume of the liquid.

Again, the total amount of salt when water is combined with 20% salt solution should also be equal to the
total amount of salt in the 5% salt solution. That means that in the last column, we have to add the first
and the second row and then equated to the third row. That is,

3 = 0.05(x + 15)

3 = 0.05x + 0.75

2.25 = 0.05x

45 = x

That means that we need 45L of water to turn a 20% salt solution to a 5% solution.
Check:

3 = 0.05(x + 15)

3 = 0.05(45 + 15)

3 = 0.05(60)

3=3

How to Solve Mixture Problems Part 4

This is the fifth post of the Solving Mixture Problems Series on PH Civil Service Review. In this post, we
are going to solve a problem in which only the total amount of mixture is given.
Problem
A chemist creates a mixture with 5% boric acid and combined it with another mixture containing
40% boric acid to obtain a 800 ml of mixture with 12% boric acid. How much of each mixture did he use?
Solution and Explanation
Let

x = mixture with 5% boric acid

800 – x = mixture with 40% boric acid.

Note that if these two mixtures are combined, we will produce a mixture with 800 ml of solution with
12% boric acid. As we have learned in the previous tutorials, the amount of boric acid in the first mixture
added to the amount of boric acid in the second mixture is equal to the amount of boric acid in the
combined mixtures. That is,
(5% )(x) + (40% )(800-x) = (12% )(800).
Take note that x, 800 – x, and 800 are amount of the mixture and if these are multiplied by the percentage of boric
acid, then we will get the exact amount of pure boric acid.
Converting percent to decimals, we have

(0.05)(x) + (0.4)(800-x) = (0.12)(800)

0.05x + 320 – 0.4x = 96.
Simplifying, we have

-0.35x = -224
x= 640.
That means that we need 640 ml of mixture with 5% boric acid and 800 – 640 = 160 ml of mixture with
40% boric acid.

Check:
Amount of boric acid in mixture with 5% boric acid: (640 ml)(0.05) = 32ml
Amount of boric acid in mixture with 40% boric acid: (160 ml)(0.4) = 64ml
Amount of boric acid in mixture with 12% boric acid: (800 ml)(0.12) = 96ml

As we can see, 32 ml + 64 ml = 96 ml which means that we are correct.

How to Solve Mixture Problems Part 5

In the previous posts, we have learned how to solve mixture problems involving percentages and
liquid mixture problems. In this post, we are going to solve mixture problems involving prices. Although
these two types of problems are different, they are very similar when you set up the equation. Below is
our first problem.
Problem
A seller mixes 20 kilograms of candy worth 80 pesos per kilogram to candies worth 50 pesos per
kilogram. He sold at 60 pesos per kilogram.

After selling all the candies he discovered that he had no gain or loss. How much of the 50-pesos per
kilogram candies did he use?

Solution and Explanation

Let x = number of kilograms of candy worth 50-pesos per kilogram.

The total price of 20 kilograms of candy at 80 pesos per kilogram is (20 kg)(80 pesos/kg) = 1600 pesos.

The total price of x kilograms of candy at 50 pesos per kilogram is (x kg)(50 pesos/kg) pesos.

When we add these 2, the number of kilograms of candy is x + 20 (can you see why?) and it is sold at 60
pesos. So, its total price is (x + 20)(60 pesos/kg).

Using these facts, we have the following equation:

total price of 80pesos/kg candy + total price of 50pesos/kg candies = total pric e of 60pesos/kg candies.

Substituting the expressions above, we have

(20 kg)(80 pesos/kg) + (x kg)(50 pesos/kg) = (x + 20 kg)(60 pesos/kg)

1600 + 50x = 60(x + 20)

1600 + 50x = 60x + 1200
1600 – 1200 = 60x – 50x
400 = 10x
40 = x.

Therefore, he used 40 kilograms of candy worth 50 pesos per kilogram.

Check:

1600 + 50(40) = 60(40 + 20)

1600 + 2000 = 60(60)
3600 = 3600.
This means that we are correct.

How to Solve Mixture Problems Part 6

This is the 6th part and last part of the Solving Mixture Problems Series. In the previous 4 parts, we have
learned how to solve mixture problems involving percent and in part 5, we have learned how to solve
problems involving percents. In this post, we solve another problem involving percent.
A bakery owner wants to box assorted chocolates. Each box is to be made up of packs of black
chocolates worth $10 per pack and packs of ordinary chocolates worth $7 each pack. How many packs of
each kind should he use to make 15 packs which he can sell for $8 per pack?
Solution
Let x = number of $10 packs
15 – x = number of $7 packs

Multiplying the cost per pack and the number of packs we have

($10)(x) = total cost of $10 packs

($7)(15 – x) = total cost of the $7 packs
($8)(15) = total cost of all the chocolates

Now, we know that

total cost of $10 packs + total cost of the $7 packs = total cost of all the chocolates.

Substituting the values, we have

($10)(x) + ($7)(15 – x) = ($8)(15).

Eliminating the dollar sign and solving for x, we have

(10)(x) + (7)(15 – x) = (8)(15)

10x + 105 – 7x = 120
3x + 105 = 120
3x = 120 – 105
3x = 15
x = 5.
This means that we need 5 packs of $10 and 10 packs of $7 chocolates.

Check:
($10)(5) + ($7)(10) = ($8)(15)
$50 + $70 = $120
$120 = $120
Therefore, we are correct.

How to Solve Perimeter Problems

Introduction to the Concept of Perimeter

Aside from Algebra problems, the Civil Service Examination also contains geometric problems. In this post,
I am going to introduce you to the concept of perimeter.
The perimeter of a polygon is the total distance around it. If you have a rectangular pool, for example, and
you start walking from one corner along its edge (as near as you can), until you reach the place where you
started, then the distance you have walked is its perimeter.

Therefore, if you have a rectangular pool with length 20 meters and width 15 meters, then the perimeter
is meters. Remember, there are two long sides and two short sides, so you
have to add the numbers twice. Therefore, if a rectangle has a given length and width, it’s perimeter is
twice it’s length plus twice it’s width. Or, we can put the equation

Where is the perimeter of the rectangle and and are its length and width respectively.

Since the idea of perimeter is to add all the distance around the polygon, this means that a square with
side 5 has perimeter
.

This means that the perimeter of a square with side s is given by the formula

We don’t really need to derive all the formula once you already know the concept since all you have to do
is to add all the side lengths.

Perimeter of a Circle
The circle has also a perimeter, but we call it circumference. If you have a circular garden and you wall
around it one time, then the distance you have walked is its circumference.
The perimeter of a circle is where is it’s radius (distance from the center to the circle) and is
approximately equal to . If you want to ask where did the formula come from, it is beyond the scope
of this blog to derive it (I have written a derivation about it in Filipino).

How to Solve Perimeter of Rectangle Word Problems

The perimeter of a polygon is the sum of all the lengths of its sides. Since a rectangle has two pairs of
sides, if we call the longer side length and the shorter side width, then
Perimeter = length + length + width + width.
We can shorten this formula if we let be the perimeter, be the length and be the width:

You see, you don’t have to memorize the formula as long as you know the concept and you know the shape
of the polygon. In some of the examples, below, I did not show use the shorter formula since they can be
solved intuitively.
Problem 1
The length of a rectangle is centimeters and its width is centimeters. What is its width?

Solution
A rectangle has two pairs of sides, so, just add the length and width twice. That is

So, the correct answer is .

Problem 2
The perimeter of a rectangle is centimeters. It’s width is centimeters. What is its length?

Solution
As much as possible, train yourself to solve problems intuitively. This problem for example does not need
to use Algebra. Just draw the triangle and then label appropriately.
The two shorter sides are cm which sums up to centimeters. We subtract from and the
difference is . Now, the two sides will share centimeters equally, so each side is centimeters. The
longer side which is asked by the question is centimeters.

Problem 3
The with of a rectangular garden is meters less than its length. Its perimeter is meters. What are the
dimensions of the garden?

Solution
Let be the length and be the width of the rectangle.

Now, we know that the perimeter of a rectangle is described by the equation

Perimeter = length + length + width + width

Substituting, we have

Simplifying the right hand sides, we have

Adding to both sides gives us

Dividing both sides by gives us

So, length is cm and width is cm.

Problem 4
The perimeter of a rectangle is centimeters. It’s length is twice its width. What are the dimensions of the
rectangle?

Solution
Let be the width of the rectangle and be its length.

Perimeter = length + length + width + width

Adding the right hand side, we have

Diving both sides by results to

So, the width of the rectangle is and its length is centimeters. So the garden
is meter.

How to Solve Motion Problems

Introduction to Motion Problems

We have already finished learning how to solve number problems and age problems, so we continue with
learning motion problems. Motion problems deal with moving objects such as cars, planes, boats, etc and
their speed, distance traveled and time spent traveling. Walking and running are also usually asked in
motion problems. Motion problems also include headwind, tailwind, for planes and other flying objects and
upstream and downstream problems on boats traveling in rivers.
Average Speed and Actual Speed
In motion problems, you can usually read phrases like the “speed of a car is 60 kilometers per hour.” In
many of such problems, it appears that the cars are traveling at a constant speed. In reality, of course,
speed is not usually constant. In the course of the travel, a car will accelerate, decelerate, or even stop at
an intersection if the light is red. This means that when we talk about “60 kilometers per hour” in moition
problems, in reality, they mean average speed. That is, in one hour, a car usually travels 60 kilometers
given normal traffic conditions. Usually, in math problems, rate and speed means the same thing.

Distance = Rate Times Time

A car that travels 70 kilometers per hour can travel 140 kilometers in two hours. We can write this as an
equation shown below

140km = (70 km/hr)(2hrs).

Since 140 kilometers is the distance (d), 70 km/hr is the rate (r), and 2 hours is the time (t), it follows that
distance = rate time

or .
This is a basic formula and you don’t really need to memorize it if you understand the concept. Now, for
the examples, let’s solve two basic motion problems.
Problem 1
Aron goes to a convenient store every week using a bicycle. If his average speed using a bicycle is 30
km/hr, how far is the convenient store if it takes him 15 minutes to go there?

Solution 1
This problem can be solved by inspection, but we will setup the equation in solution 2. For inspection, 15
minutes is 1/4 of an hour, so if it takes Aron to travel 30 km per hour, then he will travel 1/4 of it given 1/4
of an hour. Therefore, 1/4 of 30 kilometers is 7.5 kilometers.
Solution 2
The second solution is needed so you would learn how to set up equations. Later problems will be a lot
harder than this, so you will need to set up the equations correctly.

distance = ?

rate = 30 km/h

time 1/4 hr

km.

Problem 2
It took 4.5 hours from Ninoy Aquino International Airport (Philippines) to Narita Airport (Japan). If the
route of the plane is around 2250 kilometers, what is its speed?

Solution

hours

kilometers

From the original equation, we have

To get , divide both sides of the equation by giving us

So, the airplane is traveling 500 kilometers per hour.

How to Solve Motion Problems Part 1

Now that I have introduced to you motion problems, let us solve the type of problems that usually appear in
textbooks as well as examinations. In this part of the series, we will learn how to analyze and solve a
problem involving objects moving in the same direction. This is our third problem in the How to Solve Moti on
Problems Series.
Problem 3
Car A left Math City going to English City at an average speed of 40 kilometers per hour. Two hours later,
Car B traveling 60 kilometers per hour leaves the same plac e for English City. In how many hours, will the
car B overtake Car A?
Solution 1
This problem can be solved manually by creating the table as shown in the next figure. In the first column,
we have the cars, and in the first row the number of hours. Car A traveled 40 kilometers and Car B travels
60 kilometers per hour. Since Car B only started traveling after two hours, so, after three hours, Car A has
traveled 120 kilometers and Car B just 60 kilometers. Now, since Car B is faster, it will eventually overt ak e
Car A which is after 6 hours shown below. Having the same distance traveled in this problem means that
Car B overtakes Car A since they are traveling the same route.

Now notice that the question in “how many hours” means that the counting starts when Car B left.
Therefore, the answer is 4 hours because in 4 hours, car B traveled 240 kilometers, the same distance
traveled by Car A.
Solution 2
In solving Motion problems, it is always helpful to create a table of the given distance, rate, and time. Th e
rate r of Car A is 40km/hr and Car B is 60 km/hr. Now, for the time, Car A started two hours earlier, so if,
for example, it has traveled 5 hours, Car B has only traveled 3 hours. It means that the time Car A traveled
is 2 hours more than that of Car B. This means that if Car B traveled hours, then car A
traveled hours. Now, as we have discussed in the previous post, , so for column in the
table below, we just multiply the rate and the time . Now what to do next?

At the exact time Car B overtakes Car A, which is what is asked above, their distance traveled will be
equal. This means that

distance traveled by Car A = distance traveled by Car B

or using the expressions above

Note that the in this equation is the time from the table above, so the answer will be in hours. We solve
the equation by simplifying the left hand side first by distributive property

Subtracting from both sides results to

 which simplifies to

This means that Car A overtakes car B in 4 hours after Car B left. This confirms the answer in Solution 1.

How to Solve Motion Problems Part 2

This is the second part of the How to Solve Motion Word Problem series. I suggest that you read first
the introduction to this series as well as the first part in order to understand better.
In the first part of this series, we discusses about a faster object overtaking a slower one who left first. In
this post, we continue to solve motion problems involving two objects traveling in the same direction and
one object faster than the other; this time, they left at the same time. We want to know that given a particular
time when they are a number of kilometers apart.
Problem 4
A freight train and a passenger train left the same station and traveled to the same direction. The passenger
train travels at an average speed of 80 kilometers per hour while the freight train travels at an average
speed of 65 kilometers per hour. In how many hours will they be 75 kilometers apart?
Solution 1
Like in the first part of this series, the problem can be solved using a table. The table below shows the
distance traveled by the train after each hour. In the 5th hour, the passenger train traveled 400 km while
the freight train traveled 325 kilometers. This means that after 5 hours, the distance between them is 75
kilometers.

Solution 2
An algebraic solution can also be done to solve the problem above. Let us create a table like what we have
done in the first part of this series. As shown in the table below, the rate of the passenger train is 80
kilometers per hour and the rate of the freight train is 65 kilometers per hour. The time traveled is the same
since they left the same location at the same time. The distance is the product of the rate and time so we
multiply them as shown below.

Now, that we have all the given in place, let us analyze the problem. We are asked for the number of hours
when the trains are 75 kilometers part. The phrase “75 kilometers apart” means the

d traveled by the passenger train – d traveled by the freight train = 75

or
Dividing both sides by 15, we have

Note that is the number of hours in the table, therefore, in 5 hours, the trains will be 75 kilometers apart.

How to Solve Motion Problems Part 3

This is the third part of the How to Solve Motion Problems Series, a part of the Word Problem Solving Series
of Ph Civil Service Reviewer. In Part 1 and Part 2 of this series, we discussed objects moving in the same
direction. In this part, we are going to discuss objects moving toward each other. We have already
discussed four problems in the previous parts of this series, so, we solve the fifth problem.
Problem 5
A car leaves City A and travels towards City B at an average speed of 60 kilometers per hour. At the same
time, another car leaves City B and travels towards city A at an average speed of 70 kilometers per hour.
If the two cars use the same route, and if the distance between two cities is 520 kilometers, how many
hours before they meet?
Solution 1
Again, this problem can be solved using a table. By now, you would have realized that most motion
problems can be solved by creating a table. You can use the table solution in case you cannot solve the
motion problem algebraically.

As we can see in the table above, for one hour, the two cars have traveled 130 kilometers toward each
other. Since the distance between the two cities is 520 kilometers, it will take them four hours.

Solution 2
In the algebraic solution, it is also important to create a table as shown below. Shown on the columns are
the rate, time, and distance. The car from City A travels at 60 kilometers per hour and the car from City B
70 kilometers per hour. The time they spend on the road is the same since they left the cities at the same
time. Then, the distance they traveled is the product of the rate and the time.

Note that at the exact time they meet, the would have traveled the total distance from A to B which is 520
kilometers. This means that if we add the distance they traveled, then the sum is 520 kilometers.
From the discussion above, the equation is

d traveled by car from City A + d traveled by car from City B = 520 km.

Substituting the expressions on the table, we have

Dividing both sides by 4, we have

Therefore, the two cars will meet in 4 hours after they have left the two cities.
Check
By the time they meet, the car from City A will have traveled 60(4) = 240 kilometers and car from City B will
have traveled 70(4) = 280 kilometers. If you add the distance traveled by both cards, the answer is 520
kilometers. Therefore, we are correct.

How to Solve Motion Problems Part 4

In Part 1, Part 2, and Part 3 of the How to Solve Motion Problems Series, we have learned how to solve
problems involving objects moving in the same direction as well as those which move toward each other.
In this post, we are going to learn about objects which move on opposite directions. The method in solving
this problem is very similar to the method used in Part 3 of this series.
We now solve the sixth problem in this series.
Problem 6
Two jet planes left Naria Airport at 9:00 am and travels in opposite directions. One jet travels at an average
speed of 450 kilometers per hour and the other jet travels at an average speed of 550 kilometers per hour.
By what time will the two jets be 2500 kilometers apart?

Solution 1
Just like in the previous parts, we can solve this problem using a table. As we can see, in the first hour, the
jets will be 1000 kilometers apart. After three hours they will be 3000 kilometers apart.
The question, however, is the time when they are 2500 kilometers. From the table above, since after each
hour, the distance traveled is 1000 kilometers, then 2500 kilometers will require 2 and a half hours. Now, 2
and a half hours after 9 am is 11:30 am.

Solution 2
We name the jet planes A and B as shown below. Plane A travels at 450 kilometers per hour and plane B
at 550 kilometers per hour. The time traveled by both planes, we let and since they both start and the
same time, they will have the same time (duration) traveled. The distance d is the product of the rate and
the time.
Note: The time of departure so (9:00 am) is irrelevant at first in solving the problem. It can only be used
after the answer is obtained.

Now that we have the table, let us examine the figure below. In the question, how many hours will the
two planes be 2500 kilometers apart. Since the planes are traveling in opposite directions, the
word “apart” in the problem means the distance traveled by the first plane (450x) and the distance
traveled by the second plane (550x). Therefore, we can form the equation
d traveled by Plane A + d traveled by plane B = 2500

If we substitute the values in the table in the preceding equation, then we have

hours

Meaning, if the planes are 2500 kilometers apart, two and a half hours would have past. Therefore, 2 and
a half hours from the time of the departure which is 9:00 am is 11:30 am.

Therefore the correct answer is 11:30 am. This confirms the answer in Solution 1.

How to Solve Coin Problems

Introduction to Coin Problems

Coin problems is one of the word problem types that may also appear in the next Civil Service Examination.
Coin problems may refer problems regarding actual coins or even problems involving bills. Althoug h
the Civil Service Examination is solely for Filipinos, nobody will prevent the creators of the exams using
American terms such as pennies, nickels, and dimes. In case you do not know, or you have forgotten,
a penny is equivalent to 1 cent, a nickel is equivalent to 5 cents, and a dime is equivalent to 10 cents.
Let us try to solve two problems as a teaser to this series.
Problem 1
Bingbong has 18 coins in his pocket. Three of them are nickels and five of them are pennies. If the remaining
coins are dimes, how much money does Bingbong in his pocket?

Solution
There are 3 nickels and a nickel is 5 cents, so the three nickels are worth 15 cents.

There are 5 pennies and each penny is 1 cent, so 5 pennies are worth 5 cents.

There are 10 coins left, each of which is a dime or 10 cents. Therefore, there is 1.00 peso.

So, Bingbong has 0.15 + 0.05 + 1.00 = P1.20.

Therefore, Bingbong has one peso and 20 cents.

Problem 2
Jamie has 18 bills in her wallet worth 20 pesos and 50 pesos. If the bills totaled to 660 pesos, how many
20-peso and 50-peso are there?

Solution
There are 18 bills and for example, there are five 20-peso bills, then we will be left with fifty-
peso bills. This means that if there are 20-peso bills, then there are 50-peso bills.

Now, if we multiply the amount and the number of coins, we have for the 20-peso coin
and for the 50-peso coin. If we add these total amounts, we have 660 pesos. Therefore, we
can form the equation

By distributive property, we have

Simplifying, we have

Subtracting 550 from both sides, we have

Dividing both sides by , we have .

Therefore there are twenty-peso coins and fifty-peso bills.

Check
The total amount for the 20-peso bill is 8(20) = 160.

The total amount for the 50-peso bill is 10(50) = 500.

How to Solve Coin Problems Part 2

I have already introduced how to solve coin problems, so in this post, we solve more problems about it. Coin
problems involve problems consisting coins, bills, and of course, any object of value. We have already
solved 2 problems in the previous post, so we continue with the third problem.
Problem 3
A box contains 32 bills consisting of Php20 and Php50. The total amount of money in the box is 1000
pesos. How many bills of each kind are there?

Note: For my readers from other countries, Php means Philippine pesos.
Solution
This problem is similar to problem 2, so we will not be solving it in details.

Let be the number of 50-peso bills. Since there are 32 bills, then the number of 20-peso bills
is Now, if we multiply the number of 50-peso bills by 50 pesos and multiply the number of 20-
peso bills by 20 pesos, we have

and

respectively.

If we add them, the total is 1000 pesos. That is,

Simplifying and solving for , we have

Subtracting both sides by 640, we have

Dividing both sides by 30, we have

Therefore, the number of 50-peso bills is 12. Now, since there are 32 bills, the number of 20-peso bills is
32-12 = 20.

Check
12(Php50) + 20(Php20) = Php600 + Php400 = Php1000

Problem 4
In a charity musical show, there are the same number of tickets sold worth $20, $50, and $100. The total
cost of the tickets is $5100.
Solution
Let be the number of tickets sold. Since there are the same number of tickets, if we multiply the
number of ticket to each of the price, we have

, and

which are the total cost of each kind. If we add them all together, then, it is the total cost of all
the tickets which is $5100. That is,

Simplifying, we have

Dividing both sides by , we have

This means that there are 30 tickets of each price that were sold.

Check
$20(30) + $50(30) + $100(30) = $600 + $1500 + $3000 = $5100.

How to Solve Simple Interest Problems

The Solving Simple Interest Problems Series

The Solving Simple Interest Problems is a series of tutorials on how to solve simple interest problems. In
solving interest problems you need to know the following terms:

the principal (P) is the money invested, the rate (R) of interest is the percentage of interest (that is the
number with percent sign), the time (T) and the interest (I) is the earnings or return of investment. The
interest is the product of the principal, the rate, and the time, or I = PRT is explained in the first part of the
series.
How to Solve Simple Interest Problems Part 1 discusses the basics of simple interest problems and the
terms used in such problems. Two examples worked examples are solved in which interests are both
unknowns.
How to Solve Simple Interest Problems Part 2 is a continuation of the simple interest discussion. One
problem is an example of a rate of interest which is not given yearly and the other one is an investment
made for more than a year.
How to Solve Simple Interest Problems Part 3 is a discussion of simple interest problems where the
unknowns are the principal and the rate.

How to Solve Simple Interest Problems Part 1

This is the first part of the Solving Simple Interest Problem Series for the Civil Service Examination.
Simple interest problems are usually included in many examinations such as the Civil Service Exams. It is
important that you practice solving these types of problems in order to increase your chance of passing
the exams.
Before solving simple interest problems, let us familiarize ourselves with the terms used in simple interest
problems. These are the money invested which is called the principal, the rate of interest which is the
percent and the interest which is the income or return of investment, and time. Time may vary depending
on the investment. It can range from months to years.

Example 1
Mr. Reyes invested Php50,000 at an interest rate of 3% per year.

a.) Identify the principal and rate of interest.

b.) Calculate the interest earned after 1 year.

Solution
For (a)
The money invested or principal is Php50,000, the interest rate is 3%, and the time is 1 year.

For (b)
We want to calculate 3% of Php50,000. To multiply, we must convert 3 percent to decimal which is equal
to 0.03.

interest = Php50,000 × 0.03

interest = Php1500

This means that for a year, the money earned Php1500.

Example 2
Ms. Gutierrez invested Php60,000 at a simple interest of 4% per year for 4 years.

a.) Identify the principal, rate of interest, and time.

b.) How much money will Ms. Gutierrez have after four years?

Solution
For (a),
The principal or money invested is Php60,000.
The rate of interest is 4%.
The time is 4 years.

For (b),
We need to calculate 4% of Php60,000. Just like above, we must first convert 4 percent to decimal which
is equal to 0.04.

Now,
interest (1 year) = P60,000 × 0.04 = 2,400
That is the interest for 1 year. To be able to calculate the interest for four years, we have

interest (4 years) = 2,400 × 4 = 9600.

So, the money Ms. Gutierrez will have by the end of four years is the Principal which is 60,000 and the
interest for 4 years which is 9600. So, in total, her money will be 69,600.

***

From the two problems above, we can see the interest (I) is the product of the principal (P), the rate (R),
and the time(T). Therefore, we can have the formula.

I=P×R×T

or simply
I = PRT.

How to Solve Simple Interest Problems Part 2

This is the second part of the Solving Simple Interest Problems Series. In the previous post, we have
discussed the basics of simple interest problems. We have learned that the simple interest (I) is equal to
the product of the amount of money invested or the principal (P), the percentage of interest or the rate (R)
and the time (T). Therefore, we can use the following formula:
I = PRT.

In this post, we are going to discuss more problems particularly interests that are not yearly and finding
unknowns other than interest.

Example 3
Dr. Lopez invested his Php120,000 in a bank that gives 2% interest every quarter. What is the interest of
his money if he is to invest it for 1 years?

Solution and Explanation

Notice that the interest is applied quarterly and not every year. Quarterly means every three months and
therefore it will be applied four times a year since there are four quarters every year. So in this case, the
time is 4. So,

P = Php120,000
R = 2%
T=4
I=?

Note that the rate percent must be converted to decimal by dividing it by 100. So 2% equals 0.02. Now,
using the formula, we have
I = PRT
I = (Php120,000)(.02)(4)
I = Php9600.00

So, the interest of the money for 1 year is Php9600.

Example 4
Danica invested here money amounting to Php150,000 in a bank that offers a 5% simple interest every
year. She went abroad and never made any deposit or withdrawal in her account. After coming back, she
immediately checked her account and found out that her money got an interest of Php37,500. How many
years was the money invested?
Solution and Explanation
In this problem, interest is given and time is unknown. Assigning the values we have

I = Php37,500
P = Php150,000
R = 5%
T = ?.

Using the formula, we have

I = PRT

Converting 5 percent to decimal and substituting, we have

37,500 = (150,000)(.05)(T)
37,500 = 7500(T).

Dividing both sides by 7500, we have

5 = T.

That means that the money was invested for 5 years.

How to Solve Simple Interest Problems Part 3

This is the third and last part of the Solving Simple Interest Problems Series. In the first and second part of
this series, we have learned how to solve simple interest problems. In this post, we continue with two
more problems, one with the rate missing and the other one with the principal missing.
Example 5
Mr. Wong deposited $15,000 in a bank for a certain rate of interest per year. After two years, the interest
was $900. What was the rate of interest in percent?

Solution and Explanation

We have the following given:

Principal (P) = $15,000

Interest (I) = 900
Time = 2 (years)
Rate (R) = ?

As we have learned in the previous posts, simple interest is the product of the principal, the rate, and time
or

I = PRT.

Substituting the given we have

900 = (15000)(R)(2)
900 = 30000R

Dividing both sides by 30000, we have

R = 900/30000
R = 0.03

The rate is 0.03 which is we need to convert to percent by multiplying it by 100. Therefore, the rate is 3%.

Example 6
Mrs. Lansangan invested a certain amount of money in a bank that gives 4% interest per year. She got
an interest of Php2400 after 3 years.
Solution and Explanation

Given
I = 2400
R = 4%
T = 3 years
Using the simple interest formula mentioned above, we have

I = PRT
2400 = (P)(4%)(3)

Converting 4 percent to decimal, we have

2400 = (P)(0.04)(3)
2400 = 0.12P

Dividing both sides by 0.12, we have

20000 = P

Therefore, Mrs. Lansangan invested Php20000.

How to Solve Investment Problems

How to Solve Investment Problems Part 1

In the Simple Interest Problems Series, we have learned how to calculatethe interest, rate, or time of the
principal invested. In that series, the principal is invested to a single bank or company. In this series, we
will learn how to calculate the interest of money invested at different companies (different rates). Before
starting with our first example, familiarize yourself with the following terms:
principal – the amount of money invested
rate of interest – the percent of interest yearly or any period of time (e.g. monthly, quarterly)
interest – income or return of investment

Hence, if Php10000 is invested at a bank with 3% interest is 300, Php10000 is the principal, 3% is the
rate of interest, and 300 is the interest.

Example 1
Mr. Molina invested Php100,000.00. A part of it was invested in a bank at 4% yearly interest and another
part of it at a credit cooperative at 7% yearly interest. How much investment he made in each if his yearly
income from the two investments is Php5950.00?

Scratch Work
If we let x be the money invested at a bank, then 100000 – x is the amount invested at the credit
cooperative. To calculate for the interest, we must apply the percent of each interest at each amount.
That is

(4%)(x) = yearly interest from at the bank

(7%)(100000 – x) = yearly interest from the credit cooperative

If we add the interest, it will amount to Php5950. So, here’s our solution.
Solution
Let x = amount of money invested at the bank
100000 – x = amount of money invested at the credit cooperative

Total Interest = Interest From the Bank + Interest from the Credit Cooperative
5950 = (4%)(x) + (7%)(100000 – x)
We need to convert percent to decimals in order to multiply. We do this by dividing the percentage by
100. So, 4% = 0.04 and 7% = 0.07. Substituting to the previous equation, we have

5950 = (0.04)(x) + (0.07)(100000 – x)

5950 = 0.04x + 7000 – 0.07x
5950 – 7000 = -0.03x
-1050 = -0.03x.

We can eliminate the decimals by multiplying by 100.

-105000 = -3x
Dividing both sides by -3, we have

35000 = x.

That means that Mr. Molina invested Php35000 in the bank and Php100000 – Php35000 = Php65000 in
the credit cooperative.

Check
(4%)(35000) + (7%)(65000) = (.04)(35000) + (.07)(65000)
= 1400 + 4550 = 5950

As we can see, our interest from the two investments is Php5950.00

How to Solve Investment Problems Part 2

This is the second part of the Solving Investment Problems Series. In thefirst part, we discussed in detail
the solution of a problem at two different rates of interest. In this post, we discuss another problem.
Problem
Mr. Reyes invested a part of Php70000 at 3% yearly interest and the remaining part at a 5% yearly
interest. The annual interest on the 3% investment is Php100 more than the annual interest on the 5%
investment. How much was invested at each rate?
Solution
If we let x be the amount invested at 3%, then, 70000 – x is the amount invested at 5%. The yearly
interest is the product of the rates and the amount invested so,
(3%)(x) = yearly interest of the amount invested at 3%
(5%)(70000 – x) = yearly interest of the amount invested at 5%

Now, the annual interest at 3% is 100 more than the annual interest at 5%. This means that if we add 100
to the yearly interest at 5%, the interests will be equal. That is,

(3%)(x) = (5%)(70000 – x) + 100.

Next, we convert percent to decimal by dividing the percentage by 100. So,

(0.03)(x) = (0.05)(70000 – x) + 100.

Simplifying, we have

0.03x = 3500 – 0.05x + 100

0.03x = 3600 – 0.05x
0.03x + 0.05x = 3600
0.08x = 3600.
Tip: You can calculate better by eliminating the decimal. You can do this by multiplying both sides by 100.

Dividing both sides by 0.08, we have

x = 45000.

This means that 45000 is invested at 3% yearly interest.

Now, the remaining amount is 70000 – 45000 = 25000.

This means that 25000 is invested at 5%.

Check:
Yearly interest of 45000 at 3% interest = (45000 x 0.03) = 1350

Yearly interest of 25000 at 5% interest = (25000 x 0.05) = 1250

As we can see, the interest at 45000 at 3% interest is 100 more than the interest of 25000 at 5% interest.

Therefore, we are correct.

How to Solve Investment Problems Part 3

This is the third part of the Solving Investment Problems Series. In this part, we discuss
solve invest problem which is very similar to the second part. We discuss an investment at two different
interests.
Problem
A government employee invested a part of Php60000 in bonds at 6% yearly interest and the remaining
part in stocks at a 5% yearly interest. The annual interest in stocks is Php850 less than the annual
interest in bonds. How much was invested at each rate?
Solution and Explanation
Let x = amount invested on bonds (6% yearly interest)
60,000 – x = amount invested on stocks (5% yearly interest).

The interest in bonds is 6% per year times the amount invested or

(0.06)(x)

when the percentage is converted to decimals.

In addition, the interest in stocks is 5% per year times the amount invested or

(0.05)(60000 – x)

when the percentage is converted to decimals.

Now, the interest in stocks is 850 less than the interest in bonds which means that if we subtract 850 from
the interest in bonds they will be equal. That is

interest in bonds – 850 = interest in stocks.

Substituting the expressions of each, we have

(0.06)(x) – 850 = (0.05)(60000 – x).

Simplifying, we have

0.06x – 850 = 3000 – 0.05x.

To eliminate the decimal numbers, we multiply everything by 100. The equation becomes

6x – 85000 = 300000 – 5x

We simplify by adding 85000 and 5x to both sides. The equation becomes.

11x = 385000.

Dividing both sides by 11, we have

x = 35000.

So, 35000 was invested in bonds and 60000-35000 = 25000 was invested in stocks.

Check:

35000 × 0.06 = 2100

25000 × 0.05 = 1250

Indeed, the amount invested in bonds is 2100 – 1200 = 850 less than the interest in stocks.

How to Solve Investment Problems Part 4

This is the fourth part of the Solving Investment Problems Series. In this part, we discuss a problem which
is very similar to the third part. We discuss an investment at two different interest rates.
Problem
Mr. Garett invested a part of $20 000 at a bank at 4% yearly interest. How much does he have to invest at
another bank at a 8% yearly interest so that the total interest of the money is 7%.
Solution and Explanation
Let x be the money invested at 8%

(1) We know that the interest of 20,000 invested at 4% yearly interest is

20,000(0.04)

(2) We also know that the interest of the money invested at 8% is

(0.08)(x)

(3) The interest of total amount of money invested is 7%. So,

(20,000 + x)(0.07)

Now, the interest in (1) added to the interest in (2) is equal to the interest in (3). Therefore,

20,000(0.04) + (0.08)(x) = (20,000 + x)(0.07)

Simplifying, we have
800 + 0.08x = 1400 + 0.07x

To eliminate the decimal point, we multiply both sides by 100. That is

80000 + 8x = 140000 + 7x
8x – 7x = 140000 – 80000
x = 60000

This means that he has to invest $60,000 at 8% interest in order for the total to be 7% of the entire
investment.

Check:
$20,000 x 0.04 = $800
$60,000 x 0.08 = 4800

Adding the two interest, we have $5600. We check if this is really 7% of the total investment.
Our total investment is $80,000.

Now, $80,000 x 0.07 = $5600.

How to Solve Work Problems

How to Solve Work Problems Part 1

This is the first part of the Solving Working Problems Series. In this post, we are going to discuss in details
the basics of work problems.
Work problems usually involve the time for two or more persons or machines to complete the same job
given the rate that they can work. For discussion purposes, let us have the following example.
Work problem:

Ariel can paint a house in 5 days and Ben can do the same job in 6 days. In how many days can they complete the
job if they work together?
Discussion and Scratch Work
If Ariel can finish the job in 5 days, then if he were to work one day, he would have completed 1/5 of the
job. If he works for two days, then he would have completed 2/5 of the job. Similarly, if Ben can finish the
same job in 6 days, if he were to work for one day, then he would have completed 1/6 of the job. If he works
for 2 days, he would have completed 2/6 of the job (or 1/3 of the job if reduced to lowest terms).
Suppose Ariel and Ben work together starting on a Monday. Then their progress can be described as shown
in the table below.

Now, since Ariel can finish the job in 5 days, with Ben working with him even at a slower rate, the job can
be finished less than 5 days (working together makes the completion shorter!). Looking at their progress in
the table below, it is quite clear that the job can be finished in less than 3 days (can you see why?). Examine
the table and see why this is so.

The Geometric Representation of Working Together

The rates of work of Ariel and Ben can be represented geometrically by comparing them to a length of a
rectangle. The length of the red rectangle (Ariel’s rate per day) is 1/6 the length the white rectangle, while
the length of the blue rectangle (Ben’s rate per day) is 1/5 the length of the white rectangle. Since we used
the white rectangle as our unit of measure, its length is equal to 1. In this problem, 1 represents the
completed job.

By the end of Monday (see next figure), the combined work done is shown as the length of the two
colored rectangles. So, we can say that the number these pairs that can fit horizontally into the white
rectangle is the number of days that the job will be completed.

By the end Tuesday, both of them have worked 1/5(2) + 1/6(2) as shown in the figure below, more
than half of the job.
In the next figure, notice that the job can be completed even if Ben work less than his usual rate. Of
course it is also possible to lessen Ariel’s work or lessen both of their work.

Since the pair of blue and red rectangle represents one day, this means that the job can be completed in
less than 3 days if they both work together.

The Tabular Representation of Working Together

In the first part of this post, I have mentioned that from the table, it can be seen that the job can be completed
in less than 3 days. Why? Because if you can see, 3/6 is already half of the job (50%), while 3/5 is more
than half (60%). This means that on the third day, they would have completed 110% of the job. Therefore,
the job will take less than 3 days.

The 110% is confirmed by adding the fractions. By the end of the Wednesday, they would have completed
33/30 which is equal to 110%.

Notice that both the tabular and geometric interpretation only gave us an approximation. This is why we
need Algebra to solve this type of problem.

So, how many days will it take to finish the job if they work together?

The Algebraic Representation of Working Together

We can form the equation using the table above. If they both work for one day, then they have
worked . If they both worked for two days, they have finished of the job. Since
we are looking for the number of days that they have worked together, which we represent with days, we
can form the expression . And since we are looking for the complete job, we will equate the
expression with 1. That is,

Finalizing the Solution

Now, we have an equation with fractions. As we have learned in solving equations with fractions, we need
to eliminate the denominators. To eliminate the denominators of the fractions, we multiply everything with
the least common denominator of 1/5 and 1/6 which is 30. That is,

Again this confirms the tabular and geometric representations that the job can be completed in less than
3 days.

How to Solve Work Problem Part 2

This is the second part of the Solving Work Problems Series. In the previous post, we have discussed in
detail the concept behind how to solve work problems. In this post, we are going to learn
more examplesand solve more complicated problems.
Problem 2
A hose can fill a pool in 3 hours, while a smaller hose can fill it in 5 hours. If the hoses are opened together
the same time, how many hours will they be able to fill the pool?

Solution
House A can fill the pool in 3 hours, so it can fill 1/3 of the pool in 1 hour.

House A can fill the pool in 5 hours, so it can fill 1/5 of the pool in 1 hour.

Together, they can fill 1/3 + 1/5 of the pool in 1 hour.

Let x be the number of hours to fill the pool. As we have done in the previous post, we set up the
following equation (read the previous post for details). That is,

Multiplying both sides by 15, the least common denominator of 1/5 and 1/3, we have
This means that the two hoses will fill the pool in 15/8 or 1 and 7/8 hours.

Problem 3
Chloe and Diane are gown designers in a prestigious company. Chloe and Diane can embellish a gown in
4 hours. Chloe alone can do the same task in 6 hours. How long will Diane be able to do the same task if
she were to work alone?

Solution
Chloe and Diane can finish the task in 4 hours, so they can finish 1/4 of the task in 1 hour.

Chloe alone can finish the task in 6 hours, so she can finish 1/6 of the task in 1 hour.

Diane can finish the task in x hours, so she can finish 1/x of the task in 1 hour.
Note that if we combine the work of Chloe (1/6) and Diane (1/x), their rate is 1/4 of the task. That is

We multiply the equation by 24x, the least common denominator of 6, 4, and x, we have

This means that Diane can finish the task alone in 12 hours.

How to Solve Work Problems Part 3

This is the third part of the Solving Work Problems Series. The first part of this series discussed in detail
the concept behind work problems and the second part discussed the basic work problems and their
solutions.
In this post, we discuss two more work problems. The first problem is about two persons who started to
work together and after a while, the other person stopped. The second problem is about filling a pool whose
outlet pipe is left open.

Sample Problem 4
Jack can dig a ditch alone in 5 days, while John alone can do it in 8 days. The two of them
started working together, but after two days, Jack left the job. How many more days do John need to work
to finish the job alone?
Solution
Jack can finish the job in 5 days, so he can finish 1/5 of the job in 1 day.

John can finish the job in 8 days, so he can finish 1/8 of the job in 1 day.

In 1 day, both of them can finish 1/5 + 1/8 of the job.

Since both of them worked for two days, they have worked 2(1/5) + 2(1/8) before Jack left.
Now, let x be the number of days John need to finish the job, so he needs to work 1/8(x) more days. The
whole job is equal to 1, so we form the following equation.

Multiplying both sides by 40, the least common denominator of the three fractions, we have

So, John still needs days or about .

Sample Problem 5
A swimming pool can be filled with water using an inlet pipe in 6 hours. It can be emptied using an outlet
pipe in 8 hours. One day, after emptying the pool, the owner opened the inlet pipe but forgot t o close the
outlet pipe. How many hours will it take to fill the pool with both pipes open?

Solution
The inlet pipe can fill the pool in 6 hours, so it can fill 1/6 of the pool in 1 hour.

The outlet pipe can empty the pool in 8 hours, so it can empty 1/8 of the pool in 1 hour.

If both the inlet and outlet pipes are opened, then in 1 hour the pool is 1/6 filled but emptied 1/8 of water.
So, the remaining water is 1/6 – 1/8.

Therefore, if we let x be the number of hours, then in x hours, the pool is filled with
.

So, to fill the entire pool, we equate the preceding equation with 1 (can you see why?).

Multiplying both sides of the equation by 24 which is the least common denominator of 1/6 and 1/8, we
have

.
Therefore, with both pipes open, the pool can be filled in 24 hours.

Discount Problems

How to Solve Discount Problems

The Civil Service Examinations offer various types of math problems which may change from one
examination to another. Two types of math problems that will likely appear are discount and interest
problems. In this post, we will tackle discount problems.

The tag price that you see on items are their marked price. The sale price is the price that you pay after the
discount has been made. If an item costs Php100 (marked price) and has a 10% discount, then you have
to subtract the 10% of 100 from 100. Therefore, that item will cost Php100 – Php10 = Php90. Php90 is the

sale price.
Note that in solving discount problems, you must know how to convert percent to decimals. You need to
convert percent to decimals (just divideby 100) in order to perform the calculation.
Sample Problem 1
A movie DVD which costs 600 is marked “25% off.” What is the discount? What is its sale price?

Solution
Discount = 25% of Php600 or (Php600 × 0.25) = Php150.00
Sale price = Php600 – Php150 = Php450.00

So, the discount is Php150 and the sale price is Php450.

Note that 0.25 is the decimal equivalent of 25%.

Sample Problem 2
Anna shops in an international store. A t-shirt with a tag price $42 is marked “save 20%.” How much will
Anna have to pay for the t-shirt if she were to buy it?

Solution
Discount = 20% of $42 or ($42 × 0.20) = $8.40
Sale Price = $42.00 – $8.40 = $33.60

Therefore, Anna will have to pay $33.60 if she wants to buy the t -shirt.

Although it seldom happens in the real world, a discounted price might also be given in a Civil
Service Exam problem. In this type of problem, the task is to find the original price (marked price) like the
problem below.
Sample Problem 3
After getting a 10% discount, Nina bought a sofa for only 7200. What was the original price of the sofa?

Solution
We can use this equation to solve the problem above.
 Marked Price – Discount Price = Sale Price

Now, if we let be the marked price of the sofa, then the discount price is 10% multiplied by or .

So, substituting to the equation above, we have

To eliminate the decimal point, multiply both sides by .

Dividing both sides by 9, we have .

Therefore, the marked price of the sofa is Php8000. In the next post, we will discuss about strategies and
short-cuts in solving discount problems.

Now that you have understood that concept of discount, you may want to read about the strategies and
shortcuts on how to solve discount problems or take a quiz on discount problems.

Strategies in Solving Discount Problems

In the previous post, we have learned how to solve discount problems. In this post, I am going to teach you
some strategies that will make solving faster. We know the Civil Service Examinations, as well as other
examinations, are always under time pressure. Being able to solve problems fast will be a great advantage.
We are going to solve the same problems, only this time, we are going to use strategies that would be able
to make solving discount problems faster. You are probably wondering why I didn’t teach this strategy the
first time. The answer is, you have to know the basics first, so if you forgot your strategy or shortcut, you
can always go back to the long method.
Sample Problem 1
A movie DVD which costs 600 is marked “25% off.” What is the discount? What is its sale price?
Discussion
In the previous post, we used decimals to solve this problem. However, percentages like 25%, 50%, 75%,
10%, 75% and the like are easy to convert to fractions. If you are familiar with their equivalent fractions, it is
easier to solve the problem above.
Solution
The equivalent of 25% is 1/4 and 1/4 is half of half. So, half of 600 is 300 and half of 300 is 150.
Therefore, the discount is P150 and the sale price is P600-150 = Php450.

You see, you can solve the problem above mentally.

Sample Problem 2
Anna shops in an international store. A t-shirt with a tag price $42 is marked “save 20%.” How much will
Anna have to pay for the t-shirt if she were to buy it?

Discussion
In the previous post, we multiplied $42 by 0.2 (or 20%), then subtract the result from 42. Note that if you
subtract the percentage first, the calculation will be easier. That is, if the discount is 20%, then, we only
have 20 pay 80%. Therefore, we just have to multiply 0.8 by 42.

Solution
The discount is 20% so we only need to pay 80% of the $42. So, (42)(0.8) = 33.60. This means that the
sale price is $33.60.

Sample Problem 3
After getting a 10% discount, Nina bought a sofa for only 7200. What was the original price of the sofa?

Discussion
Again, like in Problem 1, it is faster to convert 10% to fraction. The discount is 1/10, so this means that
Nina only paid 9/10 of the original price. So, we can set up the equation .

Solution
The discount of the sofa was 1/10, therefore, Nica paid 9/10 of the price which is 7200. Setting up the
equation, we have

So, the sofa cost Php8000.00

Discount Problem Quiz

Solve each problem as fast as you can. You can check your answer by clicking the red + button after each
question.

1. You were walking and saw the following sign in a shoe shop. How much is the discount and sale price
of a shoes worth Php1600.00?
Answer
2. A coffee shop label its 1 cup of frappuccino 10% discount. If 1 cup of frappucino costs Php150, how
much will the customer pay upon ordering?
Answer
3. After a 10% discount has been made, a necklace costs Php5400.00. What is the original price of the
necklace before the discount has been deducted?

Answer
4. A popular mall placed a big “up to 40% off” streamer in front of its entrance. You are planning to buy a
television worth Php21,550. What is the lowest possible amount that you will pay if you buy the
television?

Answer
5. You want to buy a mountain in a bicycle shop where you saw a “15% off flyer.” The price of the bike
was Php26350.00. You asked the owner what will be price after the discount and he replied that the tag
price was the discounted price. What was the original price of the bike?

Answer
6. What is the sale price of a laptop worth Php42000.00 which has a 5% discount?

Answer
7. A plate is worth Php120 each. The shop owner told you that he will give you a 10% discount of the total
price if you buy 1 dozen. How much will you have to pay if you buy one dozen of the plates?
Answer
8. After getting an 8% discount, a jacket costs 690. What is the original price?

Answer
9. You saw a cellular phone company placed a “5% off on all items”commercial on television. How much
will you have to pay for a cellular phone whose price is Php5250.00?
Answer
10. You saved 200 pesos for buying an external hard drive which is marked “5% discount.” How much
was the price of the hard without the discount?

Answer

Solving Quadratic Word Problems

How to Solve Quadratic Word Problems Part 1

In the previous posts, we have learned how to solve quadratic equations by getting the extracting the
square root, by factoring, and by quadratic formula. We continue this series by learning how to solve
math word problems using quadratic equations. Most of the time, we need to rewrite the equation to the
general form which is .
Problem 1
The product of two consecutive positive even numbers is 48. What are the numbers?

Solution and Explanation

This problem can be solved mentally and by simple guess and check; however, we will solve it
algebraically in order to illustrate the method of using quadratic equations.

Let
= smaller number
= larger number

From the given above, we can form the following equation.

Smaller number times larger number = 48

By the distributive property, this results to

Now, we need to make this equation in general form so we can factor easily. To do
this, we subtract 48 from both sides resulting to

By factoring, we need two numbers whose sum is 2 and product is -48 where the absolute value of the
larger number is greater than that of the smaller. With this restriction in mind, we have the following pairs
of factors whose product is -48.
{48, -1}, {24, -2}, {12, -4}, {8,-6}

From these pairs, 8 and -6 has a sum of 2. Therefore, the factors are
Equating to 0, we have

,
,

Since we are looking for positive integers, we will take and .

Therefore, the two consecutive numbers are 6 and 8.

Check: The numbers 6 and 8 are consecutive numbers and their product is 48. Therefore, we are c orrect.

How to Solve Quadratic Problems Part 2

In the previous post, we have used quadratic equations to solve a word problem involving consecutive
numbers. In this post, we discuss more quadratic problems. This is the second problem in the series.
Problem 2

Miel is 12 years older than Nina. The product of their ages is 540.
Solution
Let x = age of Nina
x + 12 = age of Miel

The product of their ages is 540, so we can multiply the expressions above and equate the product to
540. That is,
x(x + 12) = 540.

Multiplying the expressions, we have

Subtracting 540 from both sides, we obtain

We want to find two numbers whose product is -540 and whose sum is 12. Those numbers are -18 and
30.

This means that the factors are

(x – 18)(x + 30) = 0.

Equating each expression to 0, we have

x – 18 = 0, x = 18
x + 30 = 0, x = – 30.

Since we are talking about age, we take the positive answer x = 18.

This means that Nina is 18 years old. Therefore, Miel is 18 + 12 = 30 years old.
Solving Quadratic Word Problems in Algebra
Quadratic Equations are equations of the form ax^2 + bx + c = 0 where , and are real numbers
and . Depending on the form of the equation, you can solve for by extracting the quare
root, factoring, or using the quadratic formula.This type of equation appears in various problems that
involves multiplication and usually appears in the Civil Service Exams.
The following series details the method and strategies in solving problems involving quadratic equations.

How to Solve Quadratic Word Problems Part 1 is about solving problems involving consecutive integers. In
this problem, the product of consecutive numbers is given and factoring was used to solve the problem.
How to Solve Quadratic Word Problems Part 2 is about solving age problems. Just like in the first part, the
product of two ages are given and factoring was used to solve the problem.
How to Solve Quadratic Word Problems Part 3 is about equations to solve area problems. In this problem, a
walk is created around a rectangular garden. Both the areas of the garden and the walk is given and the
width of the walk was solved.

Venn Diagrams

Introduction to Venn Diagrams

According to some examinees, there are several items in the October 2014 Civil Service Exam that requires
the use of Venn diagrams, so I will discuss in details how to use Venn diagrams in a series of posts. In this
post, I am going to introduce its concept and its use as layman as possible. I will limit the discussion to the
preparation of solving word problems involving Venn diagrams.
Venn diagrams are used to represent logical relations between sets. In word problems involving diagrams,
elements of sets are usually people who choose or prefer a particular thing (e.g. color, food, hobbies). For
instance, we have two available desserts, biscuits and cookies, and then Abby ate a biscuit and Bella ate
a cake. Chubby, being extremely hungry, ate a biscuit and a cake. So, the situation can be represented as
follows using a Venn diagram.

Note that Abby is in the blue area which means that Abby chose biscuits only. Bella is in the red area which
means that she chose cookies only. Chubby is in the overlap of the two circles, which means that she chose
both. Now, if Debbie is on diet and she didn’t eat any dessert, then the Venn diagram would look like below.
The rectangle which includes all four friends, those who ate dessert and those who did not, is called
the universal set. Debbie’s name is outside the circles because she did not eat cookies and biscuits.
Now suppose there is a third dessert which is ice cream, and Emma ate all three, then the diagram would
look like the one in the next figure.

Suppose Fe, Girlie, and Haidee arrived. Fe ate ice cream, Girlie ate biscuits and ice cream, and Haidee
ate ice cream and cookies, the diagram would look like the one shown in the next figure.
How to Solve Venn Diagram Problems Part 1
According to examinees, some of the items that were included in t he recently concluded Civil
Service Exams are problems involving Venn Diagrams. So, in this series, we will discuss in detail how to
solve these types of problems. If you are not familiar with Venn diagrams, please read the Introduction to
Venn Diagrams.
Problems involving Venn Diagrams usually discusses choices of groups of people. Let’s have the first
example.

Problem 1
In a class of 40 students, 25 are taking English and 17 are taking Mathematics. If 7 are taking both
subjects,
(a) How many students are not taking English, and at the same time not taking Math?

(b) How many students are in English class only? in Math class only?
Solution
The most effective strategy in solving problems such as this is to create overlapping circles which is
usually known as a Venn Diagram. Venn diagrams are used to represent sets . In this problem, we are
actually talking about sets, their union, and their intersection. Below is the Venn diagram of the problem
above.

First, notice that the overlapped part belongs to both Math and English, so we can place the 7 students who
are taking both subjects in that overlap.
 Second, the problem says, there are 25 students taking
English. But from the diagram, 7 are taking both subjects. So, the only number of students who are taking
English and not taking Math is 25 – 7 = 18.

Third, there are 17 students who take Math, but 7 of those students are also taking English. Therefore,
there are only 10 students who take Math and not English.

So, all in all, there are 18 + 7 + 10 = 35 students who are either in English, Math or both. Since there are
40 students in all, 40 – 35 = 5 of them are not taking English and at the same time not taking Math.
This answers the first question.
If we want to represent the universal set which is all the 40 students, we can place the overlapping circles
inside a rectangle (or any other shape) and place 5 inside that rectangle but outside the circles as shown
below.
From above, it is also clear that 18 are only taking English and not Math, and 10 are taking Math and not
English. This answers the second question.

How to Solve Venn Diagram Problems Part 2

In the previous posts, the introduction and the second part of this series, we have learned the basics
of Venn Diagrams as well as solving the 2-circle Venn Diagram problem. In this post, we are going to solve
a more complicated problem which is composed of 3-circle Venn diagram problem.
Venn Diagram Problem
There are 100 students surveyed and asked which of the following subjects they take this
semester: Mathematics, English, or Biology. Below is the result of the survey.
 35 responded English
 50 responded Mathematics
 29 responded Biology
 12 responded Mathematics and English
 8 responded English and Biology
 11 responded Biology and Math
 5 responded all
Questions
1. How many students are not taking any of the three subjects?
2. How many students take Math, but not Biology or English?
3. How many students take Math and English, but not Biology?
Solution
As we have done in Solving Venn Diagram Problems Part 1, we started with the overlapping circles. In this
problem, we have four regions which overlap. The the easiest strategy is to start at the center, the part
where the three circles overlap. In short, we start from bottom to the top in the result above.
(1). Five students responded that they took all the subjects, so we put 5 at the center.
(2). Eleven responded Biology and Math. So, we should put it in the Biology -Math overlap. However, of
the 11 who takes Biology and Math, 5were also taking English as shown in (1). So, there are 11-5
= 6 students in the Math-Biology overlap.

(3). Eight responded Biology and English. But of those 8 taking Biology and English, 5 are also taking
Math. So, there are 8 – 5 = 3 students who are taking Biology and English.
Also, there are 12 students who are taking Math and English, and of those 12, 5 are also taking Biology,
so there are 12 – 5 = 7 who are taking Math and English. Now see the next figure to see how the Venn
diagram should look like after this step.

(4). Next, 29 students responded Biology. But notice that 6, 5, and 3 are already in the Biology circle. So,
we subtract those students from 29. That is, 29 – (6 + 5 + 3) = 29 – 14 = 15. So, there are 15 students
who take only Biology.
(5). Lastly, there are 50 students who are taking Math and 35 who are taking English. But in the Math circle
there are 6 + 5 + 7 students who are also taking the other subjects and in the English circle, there
are 3 + 5 + 7students who are also taking the other subjects. Therefore, we can have the following
calculations
Number of Students Who Take Only Math = 50 – (6 + 5 + 7) = 32
Number of Students Who Take Only English = 35 – (3 + 5 + 7) = 20
After answering the problems above, the Venn diagram should look like below.

Now, to answer the first question, how many students

did not take any of the three subjects, recall that there are 100 students who were surveyed. If we add all
the numbers in the diagram, 15 + 32 + 20 + 6 + 7 + 3 + 5, the sum is only 88. Therefore, 100 – 88 = 12
students did not take any of the three subjects.

For question 2, another way to rephrase it is how many students take only Math, so the answer is 32.
For question 3, the answer is 7 (can you see why?)
 Ratio Problems

How to Solve Word Problems Involving Ratio Part 1

In a dance school, 18 girls and 8 boys are enrolled. We can say t hat the ratio of girls to boys is 18:8 (read
as 18 is to 8). Ratio can also be expressed as fraction so we can say that the ratio is 18/8. Since we can
reduce fractions to lowest terms, we can also say that the ratio is 9/4 or 9:4. So, ratio can be a
relationship between two quantities. It can also be ratio between two numbers like 4:3 which is the ratio of
the width and height of a television screen.

Problem 1
The ratio of boys and girls in a dance club is 4:5. The total number of students is 63. How many girls and
boys are there in the club?

Solution and Explanation

The ratio of boys is 4:5 means that for every 4 boys, there are 5 girls. That means that if there are 2
groups of 4 boys, there are also 2 groups of 5 girls. So by calculating them and adding, we have

4+5=9
4(2) +5(2) =18
4(3) +5(3) =27
4(4) +5(4) = 36
4(5) +5(5) = 45
4(6) +5(6) =54
4(7) +5(7) =63

As we can see, we are looking for the number of groups of 4 and, and the answer is 7 groups of each. So
there are 4(7) = 28 boys and 5(7) = 35 girls.
As you can observe, the number of groups of 4 is the same as the number of groups of 5. Therefore, the
question above is equivalent to finding the number of groups (of 4 and 5), whose total number of persons
add up to 63.

Algebraically, if we let x be the number of groups of 4, then it is also the number of groups of 5. So, we
can make the following equation.

4 x number of groups + 5 x number of groups of 5 = 63

Or

4x + 5x = 63.

Simplifying, we have

9x = 63
x = 7.

So there are 4(7) = 28 boys and 5(7) = 35 girls. As we can see, we confirmed the answer above
using algebraic methods.
How to Solve Word Problems Involving Ratio Part 2
This is the second part of a series of post on Solving Ratio Problems. In the first part, we have learned
how to solve intuitively and algebraically problems involving ratio of two quantities. In this post, we are
going to learn how to solve a ratio problem involving 3 quantities.
Problem 2
The ratio of the red, green, and blue balls in a box is 2:3:1. If there are 36 balls in the box, how many
green balls are there?
Solution and Explanation
From the previous, post we have already learned the algebraic solutions of problems like the one shown
above. So, we can have the following:
Let be the number of grous of balls per color.

So, there are 6 groups. Now, since we are looking for the number of green balls, we multiply x by 3.

So, there are 6 groups (3 green balls per group) = 18 green balls.

Check:
From above, is the number of blue balls. The expression 2x represent the number of red balls,
so we have 2x = 2(6) = 12 balls. Therefore, we have 12 red balls, 18 green balls, and 6 blue balls.

We can check by adding them: 12 + 18 + 6 = 36.

This satisfies the condition above that there are 36 balls in all. Therefore, we are correct.

How to Solve Word Problems Involving Ratio Part 3

In the previous two posts, we have learned how to solve word problemsinvolving ratio
with two and three quantities. In posts, we are going to learn how to solve a slightly different problem
where both numbers are increased.
Problem
The ratio of two numbers is 3:5 and their sum is 48. What must be added to both numbers so that the
ratio becomes 3:4?

Solution and Explanation

First, let us solve the first sentence. We need to find the two numbers whose ratio is 3:5 and whose sum
is 48.

Now, let x be the number of sets of 3 and 5.

3x + 5x = 48
8x = 48
x=6

Now, this means that the numbers are 3(6) = 18 and 5(6) = 30.

Now if the same number is added to both numbers, then the ratio becomes 3:4.
Recall that in the previous posts, we have discussed that ratio can also be represented by fraction. So,
we can represent 18:30 as . Now, if we add the same number to both numbers (the numerator and the
denominator), we get . If we let that number y, then

Cross multiplying, we have

By the distributive property,

So, we add 18 to both the numerator and denominator of . That is,

Now, to check, is ? Yes, it is. Divide both the numerator and the denominator by 12 to reduce the
fraction to lowest terms.

How to Solve Word Problems Involving Ratio Part 4

This is the fourth and the last part of the solving problems involving ratio series. In this post, we are going
to solve another ratio word problem.

Problem

The ratio of two numbers 1:3. Their difference is 36. What is the larger number?
Solution and Explanation
Let x be the smaller number and 3x be the larger number.

3x – x = 36
2x = 36
x = 18

So, the smaller number is 18 and the larger number is 3(18) = 54.

Check:

The ratio of 18:54 is 1:3? Yes, 3 times 18 equals 54.

Is their difference 36? Yes, 54 – 18 = 36.
Therefore, we are correct.

Digit Problems

How to Solve Digit Problems Part I

Digit Problems is one of the word problems in Algebra. To be able to solve this problem, you must
understand how our number system works. Our number system is called the decimal number system
because the numbers in each place value is multiplied by powers of 10 (deci means 10). For instance, the
number 284 has digits 2, 8, and 4 but has a value of 200 + 80 + 4. That is,
.

As you can observe, when our number system is expanded, the hundreds digit is multiplied by 100, the
tens digit is multiplied by 10, and the units digit (or the ones digit) is multiplied by 1. Then, all those
numbers are added. The numbers 100, 10, and 1 are powers of 10: , , and
. So, numbers with , , and as hundreds, tens, units digits respectively has value

It is clear that this is also true for higher number of digits such as thousands, ten thousands, hundred
thousands, and so on.

Many of the given numbers in this type of problem have their digits reversed. As we can see, if 10t + u is
reversed, then it becomes . For instance, when reversed
is . Now, that we have already learned the basics, we proceed to our sample
problem.
Worked Example
The tens digit of a number is twice the units digit. If the digits are reversed, the new number is 18 less
than the original. What are the numbers?

Solution and Explanation

The tens digit of a number is twice the unit digit. This means that if we let the units digit be , then the
tens digit is . As we have mentioned above, we multiply the tens digit with 10 and the units digit with 1.
So, the number is

Now, when the digits are reversed, then x becomes the tens digit and becomes the ones digit. So, the
value of the number is

From the problem above, the number with reversed digit is 18 less than the original number. That means,
that if we subtract 18 from original number, it will equal the new number. That is,
So, the number is 42 and the reversed number is 24.

Check: 42 – 24 = 18.

How to Solve Digit Problems Part II

In the previous post, we have discussed the basics of digit problems. We have learned the
decimal number system or the number system that we use everyday. In this system, each digit is
multiplied by powers of 10. For instance, 871 means
.

Recall that .

In this post, we continue this series by providing another detailed example.

Problem
The sum of the digits of a 2-digit number is . If the digits are reversed, the new number is more than
the original number. What are the numbers?

Solution and Discussion

If the tens digit of the number is , then the ones digit is (can you see why?).

Since the tens digit is multiplied by , the original number can be represented as

Simplifying the previous expression, we have 10x – x + 9 = 9x + 9.

Now, if we reverse the number, then becomes the tens digit and the ones digit becomes . So,
multiplying the tens digit by 10, we have

Simplifying the expression we have 10 – 10x + x = 90 – 9x.

As shown in the problem, the new number (the reversed number) is more than the original number.
Therefore,

reversed number – original number = 45.

Substituting the expressions above, we have

90 – 9x – (9x + 9) = 45.
Simplifying, we have

Therefore, the tens digit of the original number is 2 and the ones digit is .

So, the original number is and the reversed number is .

Now, the problem says that the new number is more than the original number. And this is
correct since .