How To Solve Word Problems in Algebra

Whether you’re taking the UPCAT, Civil Service Exam, NMAT, or other written examinations, word problems will always be part of the Mathematics subtest. They’re among the most frequently asked questions and also one of the most challenging.

In this section, we will be providing some tips on how to approach these word problems using algebra. In particular, we are going to solve number problems, age problems, work problems, rate problems, and so on.

Hopefully, you can answer word problems with ease after reading this chapter.

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General Tips in Solving Word Problems Using Algebra

Here are some quick tips to help you improve your problem-solving skills in mathematics:

  1. Read the problem carefully and determine what is being asked. It is crucial that you know what you need to solve because it will lead you to what process you have to perform in order to answer the problem.
  2. Identify the given and define them. Identify all quantities that are provided in the problem. Make sure that you know what these quantities are and how they relate to the problem.
  3. Set up a mathematical expression that represents the problem. Your mathematical expression can be an equation or inequality. It is essential that you make the correct expression so that you can obtain the precise answer.
  4. Solve the mathematical expression you have made. Ensure that you know the techniques on how to solve equations and inequalities. If you are not confident enough with your arithmetic and algebra of equations and inequalities, it is highly suggested to practice them first before attempting to answer these word problems. 

Keep the above-mentioned things in mind while we are discussing how to solve different word problems. 

 

Number Problems

Number problems are a type of word problem that asks you to determine a certain number or quantity using the descriptions given in the problem. 

To solve number problems, you should be familiar with some keywords used in algebra. These keywords such as sum, increased by, decreased by, of, and ratio of allow you to determine what mathematical operation/s is involved in the problem.

For instance, a number problem that contains the statement “the sum of a number and 5” gives a clue that the problem involves the addition of numbers.

Let us try to solve some number problems:

Sample Problem 1

Twice a number increased by 5 is 125. What is the number?

Solution:

The problem above is asking for an unknown number.

Let x be that unknown number.

Recall that the keyword “twice” means “double” or a certain number is multiplied by 2. Thus, twice the unknown number is 2x.

Twice that unknown number increased by 5 can be represented by 2x + 5.

Therefore, our equation would be 2x + 5 = 125. If we solve for x in this equation, we will be able to determine the unknown number in the problem.

Solving the equation:

2x + 5 = 125

2x  = -5 + 125 Transposition method

2x = 120

(2x)/2 = 120/2 Dividing both sides of the equation by 2

x = 60

Since x represents the unknown number in the problem, and we have computed that x = 60, then the unknown number is 60.

Sample Problem 2

When a number is increased by its reciprocal, the result is 2. Determine the number.

Solution:

The problem is asking for an unknown number.

Let x be that unknown number.

As stated in the problem, if the number is increased by its reciprocal, we will obtain 2. The reciprocal of x is simply 1/x. Thus, our equation would be:

x + 1/x = 2

To solve this equation, we have to get rid of the denominator first. This can be achieved by multiplying both sides of the equation by the Least Common Denominator (LCD) which is x:

x(x + 1/x) = x(2)

x2 + 1  = 2x

Notice that after we get rid of the denominator, the resulting equation is a quadratic equation. We can write it in standard form as follows:

x2 + 1 = 2x

x2 – 2x + 1 = 0 Transposition method

Now, let us solve for x in x2 – 2x + 1 = 0 to determine the unknown number:

x2 – 2x + 1 = 0 

(x – 1)(x – 1) = 0 By factoring

x – 1 = 0 x – 1 = 0 Set each factor to 0

x = 1 x = 1

Based on our solution, the unknown number in the problem is 1

You can also verify that the sum of 1 and its reciprocal is 2. The reciprocal of 1 is just 1. Hence, 1 + 1 = 2. 

Sample Problem 3

The sum of four consecutive positive whole numbers is 78. Determine the largest number among the consecutive numbers.

Solution.

Again, this problem is asking for an unknown number. 

Consecutive numbers pertain to a set of numbers that follow each other continuously in order from smallest to largest. For example, 8, 9, 10, and 11 are consecutive numbers since these numbers follow each other in an increasing manner.

Now, the problem is asking us to find four consecutive positive whole numbers whose sum is 78. 

Let x represent the smallest or the first number of these four consecutive numbers.

x – smallest/first number

Since the numbers are consecutive, the second number of these four consecutive numbers must be x + 1 since you have to add 1 to the first number (which is x) to obtain the next number.

x + 1 – second number

Now, we represent the third number as x + 2 since we have to add 2 to the first number (which is x) to obtain the third number

x + 2 – third number

Lastly, we represent the fourth number or the largest number as x + 3 since we have to add 3 to the first number (which is x) to obtain the fourth or the largest number.

x + 3 – fourth number

So far, we have the following to represent the four consecutive numbers:

x – first number

x + 1 – second number

x + 2 – third number

x + 3 – fourth number

Recall that the sum of these four consecutive numbers is 78. Thus, we have this equation:

first number + second number + third number + fourth number = 78

x + (x + 1) + (x + 2) + (x + 3) = 78

We can simplify the equation above as:

4x + 6 = 78

Let us solve for the value of x in the equation above:

4x + 6 = 78

4x = -6 + 78 Transposition method

4x = 72

(4x)/4 = 72/4 Dividing both sides of the equation by 4

x = 18

Since x represents the smallest or the first number and we have obtained x = 18, then the first number should be 18. It follows that the second number is 19, the third number is 20, and the fourth or the largest number is 21. 

Therefore, the answer to the problem is 21.

 

Age Problems

One of the most commonly asked questions in mathematics proficiency exams is those that involve age problems. These problems require you to determine the certain age of a person using details given in the problem.

Age problems are usually solvable using your intuition alone and without the need to apply algebraic concepts. However, relying on your intuition is not a reliable way to solve age problems, not to mention it can be time-consuming. Furthermore, solutions obtained through this process tend to be unorganized, making it difficult to answer the given problem.

Using algebra, you can provide a more systematic approach to solving age problems in less amount of time.

Sample Problem 1

Letty is three times as old as Bert. The sum of Bert’s and Letty’s ages is 48. How old is Letty?

Solution:

The problem is asking for the age of Letty given that she is three times older than Bert.

Let x be Bert’s age.

Since Letty is three times as old as Bert, we can write Letty’s age as 3x.

The sum of Bert’s and Letty’s ages is 48. Thus, our equation would be:

Bert’s age + Letty’s age = 48

x + 3x = 48

4x = 48

Using the equation, we have to solve for x to determine Bert’s age:

4x = 48

(4x)/4 = 48/4 Dividing both sides of the equation by 4

x = 12

Since x represents Bert’s age, then Bert is 12 years old. Now, we know that Letty is three times older than Bert. Therefore, Letty’s age is 3(12) = 36.

Letty is 36 years old. 

Sample Problem 2

This year, Albert is 11 years older than Isaac. Three years ago, the sum of their ages was 67. Determine how old Isaac and Albert are this year.

Solution:

The problem is asking for Albert’s and Isaac’s ages in the current year.

Let x be Isaac’s age in the present year.

The problem states that this year, Albert is 11 years older than Isaac. Hence, we can represent Albert’s age this year as x + 11.

This means that three years ago, Isaac’s age can be represented by x – 3. On the other hand, Albert’s age can be represented by x + 11 – 3 = x + 8.

To summarize:

PresentPast (3 years ago)
Albertxx – 3
Isaacx + 11x + 8
Sum67

The sum of the ages of Albert and Isaac in the past is 67. Hence, we can set up our equation as follows:

Albert’s age (past) + Isaac’s age (past) = 67

(x – 3) + (x + 8) = 67

We can simplify the equation above as:

2x + 5 = 67

Solving for x in the equation above:

2x + 5 = 67

2x = -5 + 67 Transposition method

2x = 62

2x/2 = 62/2 Dividing both sides of the equation by 2

x = 31

Since x represents Isaac’s age in the present, then Isaac is 31 years old in the present. Meanwhile, since Albert is 11 years older than Isaac in the present, then Albert is 31 + 11 = 42 years old in the present.

Hence, Isaac is 31 years old while Albert is 42 years old in the present.

Sample Problem 3

Five years from now, the sum of Richard’s age and Mike’s age will be 21. In the present year, Richard’s age is five years higher than twice Mike’s age. Determine Mike’s age in the present.

Solution:

The problem is asking for Mike’s age in the present.

It implies that Mike is younger than Richard. Hence, we let x be Mike’s age in the present.

Richard’s age is 5 years higher than twice Mike’s age. Twice of Mike’s age can be represented by 2x. Thus, Richard’s age in the present can be represented by 2x + 5.

Five years from now, Mike’s age will be x + 5 while Richard’s age will be 2x + 10.

To summarize:

PresentFuture (five years from now)
Mike’s agexx + 5
Richard’s age2x + 52x + 10
Sum21

The problem states that in the future (5 years from now), Mike’s age and Richard’s age will sum up to 21.

Thus, we have this equation:

Mike’s age (future) + Richard’s age (future) = 21

(x + 5) + (2x + 10) = 21

Combining like terms in the equation above:

3x + 15 = 21

Solving for x in the equation above:

3x + 15 = 21

3x = -15 + 21 Transposition method

3x = 6

(3x)/3 = 6/3 Dividing both sides of the equation by 3

x = 2

Looking back at our table of Richard’s and Mike’s ages, x represents Mike’s age in the present. Therefore, Mike is 2 years old in the present.

Sample Problem 4

This year, Ravel is 12 years younger than Claude. 12 years ago, Claude’s age was five times Ravel’s age. Determine how old Claude is in the present.

Solution:

The problem is asking for Claude’s age in the present.

It seems that Ravel is younger than Claude, so we let x be Ravel’s age in the present.

Ravel is 12 years younger than Claude. This also means that Claude is 12 years older than Ravel. Hence, let x + 12 represent Claude’s age.

Ravel’s age 12 years ago can be represented by x – 12. On the other hand, Claude’s age can be represented by x + 12 – 12 = x.

To summarize:

PresentPast (12 years ago)
Ravel’s agexx – 12
Claude’s agex + 12x

The problem states that 12 years ago, Claude’s age was five times Ravel’s age. 

Claude’s age = 5(Ravel’s age)

x = 5(x – 12) 

We can simplify the equation above as follows:

x = 5(x – 12) 

x = 5x – 60 Distributive Property

Let us solve the equation above:

x = 5x – 60

x – 5x = -60 Transposition method

-4x = -60

(-4x)/-4 = (-60)/(-4) Dividing both sides by -4

x = 15

Looking back at the table, x represents Ravel’s present age while x + 12 represents Claude’s present age. Using the value of x we have obtained from the solution (which is x = 15):

x + 12

(15) + 12 = 27

Thus, Claude is 27 years old today.

 

Work Problems

Solving a work problem involves figuring out how long it will take a person or a group of people to finish a certain task (or work) using the details provided in the problem.

There is a guiding formula or equation that we can use when solving work problems. This equation tells us that the sum of the reciprocal of the time it takes for A and B to finish a task is equal to the reciprocal of the time it takes for A and B to work together to finish the same task.

algebra word problems 1

Let us try to solve some work problems to see how this equation/formula works:

Sample Problem 1

Jacques can finish cleaning a garage in 2 hours. On the other hand, Kath can finish cleaning the same garage in 5 hours. Suppose that Jacques and Kath work together to clean the same garage, how long would it take for them to accomplish it?

Solution:

The problem is asking for the time it would take for Jacques and Kath to finish cleaning the garage if they work together.

Jacques can finish cleaning the garage in 2 hours.

Meanwhile, Kath can finish cleaning the garage in 5 hours.

We let t represent the time it would take for Jacques and Kath to finish cleaning the garage if they work together.

Recall that in a work problem, we can obtain the time it would take for two persons to finish a certain task by solving this equation:

algebra word problems 2

Therefore, to solve the problem, we can use the same equation/formula:

algebra word problems 3

Using the values given in the problem:

algebra word problems 4

Solving for the value of t:

½ + ⅕ = 1/t 

To make the equation above much easier to solve, we multiply both sides of the inequality by the LCD (the LCD is 10t):

10t(½ + ⅕) = 10t(1/t) 

5t + 2t = 10

Now, let us solve the resulting linear equation above to determine the value of t:

5t + 2t = 10

7t = 10

(7t)/7 = 10/7 Dividing both sides of the equation by 7

t = 10/7 

Based on our computation above, it would take 10/7 or 1 3/7 hours to completely clean the garage if Jacques and Kath decide to work together.

Sample Problem 2

There are two pipes in a large water tank: Pipe C and Pipe D. Pipe C puts water into the tank while Pipe D spills water from the tank. If only Pipe C is opened, it will fill the tank in 2 hours. Meanwhile,  if only Pipe D is opened, it will empty the tank in 3 hours. Suppose that for a certain purpose, both Pipe C and Pipe D are opened at the same time, how many hours will it take for the tank to be full?

Solution:

The problem is asking how long it would take for Pipe C to fill the tank given that Pipe D is also open.

algebra word problems 5

We know that Pipe C puts water into the tank while Pipe D spills water from the tank. This means that we should modify our formula for the work problem as follows:

algebra word problems 6

The reason why we made the operation in the formula a subtraction sign is that Pipe D is spilling water so it would take Pipe C longer to fill the tank.

Let t be the time it would take for the tank to be full given that Pipe C and Pipe D are opened at the same time.

It takes 2 hours for Pipe C to fill the tank while it takes 3 hours for Pipe D to empty the tank. Inputting these values to the equation:

½  – ⅓ = 1/t 

Now, let us solve for the value of t:

½ – ⅓ = 1/t

We multiply both sides of the equation by 6t to remove the denominators in the equation:

6t(½ – ⅓) = 6t(1/t)

3t – 2t = 6

t = 6

Based on our computation above, it would take 6 hours for the water tank to be full if both Pipe C and Pipe D are opened at the same time.

 

Rate Problems

Rate problems refer to word problems that involve moving objects. Most of the rate problems will ask you about the distance a certain moving object covers (distance), how long it would take for a moving object to reach a certain point (time), or how fast the moving object is (rate or speed). 

Since we are dealing with the distance, rate, and time of a moving object, it is essential that we are familiar with the distance formula for moving objects.

Distance Formula

Distance = Rate x Time

The distance formula tells us that a moving object (e.g., car, bus, airplane, cyclist, etc.) will cover a distance that is equivalent to the product of the rate or the speed of the object and the time or the duration of the movement of the object.

For example, if a car travels at a rate of 60 kph for 2 hours, the distance that the car will cover is:

Distance = Rate x Time

Distance = 60 x 2

Distance = 120 km

Keep the distance formula in mind because we will be using it a lot to solve rate problems.

Sample Problem 1

Two cars, Car A and B,  leave two different cities that are 400 km apart. Car A and Car B are moving toward each other at a speed of 60 kph and 50 kph, respectively. How many hours will it take for these two cars to meet or pass each other?

Solution:

The problem is asking how many hours it will take for Cars A and B to meet given their respective rates.

It is helpful to create an illustration for this problem:

algebra word problems 7

Based on our illustration above, the total distance covered by the cars is equal to 400 km. 

Let t be the number of hours it will take for the cars to meet or pass each other.

We can create a table so that we can summarize the details of the given situation.

Rate (in kph)TimeDistance (Rate x Time)
Car A60t60t
Car B50t50t
Total400

Based on our table above, the respective distances covered by both cars are 60t and 50t.

Again, as shown in the illustration, the total distance they will have covered (the sum of their respective distances at time t) by the time the two cars meet is 400 km.

Thus, our equation would be:

60t + 50t = 400

110t = 400

(110t)/110 = 400/110

t = 40/11 or 3.64 

Hence, the cars will meet after about 3.64 hours.

Sample Problem 2

Fred and Ludwig love cycling. One day they decided to test how long it would take for them to be 35 km apart if they start at a certain point and move away from each other. Fred moves at a rate of 15 kph while Ludwig moves at a rate of 20 kph. Compute how long it would take for them to be 35 km apart.

Solution:

The problem is asking for the number of hours it will take for Fred and Ludwig to be 35 km apart given their respective rates.

If we try to illustrate the problem above, it will look like this:

algebra word problems 8

It is clearly seen that the total distance covered by Fred and Ludwig is equal to 35 km if they are already 35 km apart from each other.

Let t represent the time it takes for Ludwig and Fred to be 35 km apart given their respective rates.

To summarize:

RateTimeDistance (Rate x Time)
Fred15t15t
Ludwig20t20t
Total35

Based on our table above, the respective distances covered by Fred and Ludwig in order to be 35 km apart is 15t and 20t.

The sum of the distances covered by Fred and Ludwig when they are 35 km apart is also equal to 35 km. Thus, we have this equation:

15t + 20t = 35

35t = 35

t = 1

It means that it will only take 1 hour for Fred and Ludwig to be 35 km apart given their respective rates.

 

Other Word Problems

To enrich your arsenal of problem-solving techniques, let us try to answer some word problems that don’t belong to any of the categories we have discussed so far.

Problem 1: A bag in a department store has been tagged with a 20% discount. If you buy the bag with the said discount, you only have to pay PHP 960. What is the original price of the bag?

Solution:

We are tasked to find the original price of the bag.

Let x be the original price of the bag (i.e., the price of the bag without the discount).

The discount amount (i.e., the amount to be deducted from the original price) can be represented by 0.20x.

The problem states that if the bag was bought with a discount, you only have to pay PHP 960. Hence, PHP 960 is the discounted price of the bag.

Note that

Original Price – Discount Amount = Discounted Price

Hence, our equation would be:

x – 0.20x = 960

0.80x = 960

Dividing both sides of the equation by 0.80:

(0.80x)/0.80 = 960/0.80

x = 1200

Therefore, the original price of the bag (price without the discount) is PHP 1200.

Problem 2: You are planning to buy flower pots from an online shopping site as a gift to your mother. Each flower pot costs PHP 350 and you have to pay a fixed amount of PHP 100 as a delivery fee. What is the maximum number of flower pots you can buy given that you only have a PHP 1000 budget?

Solution:

The problem is asking for the largest number of flower pots your budget of PHP 1000 can buy.

Let x be the number of flower pots.

Then, let 350x represent the total cost of buying x flower pots.

Since there’s a fixed service fee of PHP 100, the total amount you have to pay to the online shopping site would be 350x + 100.

Now, 350x + 100 must be less than or equal to 1000 since our budget is only PHP 1000. In other words, the total amount we need to pay to the online shopping site should not exceed 1000.

Hence, we have this inequality:

350x + 100 ≤ 1000

Let us solve the inequality above:

350x ≤ -100 + 1000

350x ≤ 900

(350x)/350 ≤ 900/350

x ≤ 900/350

x ≤ 2.57

We write our answer in decimal form to make our interpretation much easier to understand.

The solution set of the inequality is x ≤ 2.57 or the set of all numbers less than or equal to 2.57. This means that the number of flower pots that you can buy with your PHP 1000 budget can be any number less than 2.57. 

However, we are looking for the maximum number of flower pots we can buy. Take note that the number of flower pots should be a whole number (a fractional or decimal number of flower pots does not make sense). Since our solution set is x ≤ 2.57 or the set of all numbers less than or equal to 2.57 and the largest whole number in this set is 2, then the maximum number of flower pots we can buy is 2.

The answer to this problem is 2 flower pots

Problem 3: Vince bought a shirt from a supermarket in his town. He also bought a headset that costs PHP 700 lower than twice the price of the shirt. The total amount that Vince paid for the shirt and the headset is PHP 1100. How much is the shirt that Vince bought?

Solution:

We are tasked to find the price of the shirt that Vince bought.

Let x be the price of the shirt.

Since the price of the headset is PHP 700 lower than twice the price of the shirt, then we can represent the price of the headset as 2x – 700.

The total amount that Vince paid for the shirt and headset was PHP 1100:

Price of the shirt + Price of the headset = 1100

x + (2x – 700) = 1100

Let us solve the equation above:

3x – 700 = 1100

3x = 700 + 1100 Transposition method

3x = 1800

(3x)/3 = 1800/3 Dividing both sides of the equation by 3

x = 600

Hence, the price of the shirt is PHP 600.

 

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Jewel Kyle Fabula

Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music.

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