Word Problems relate mathematical concepts to real world situations. Algebra word problems make use of many an algebraic models to answer questions rising in real life situations.

Starting from elementary word problems, which essentially involve solving simple equations, students solve various applications of algebra during their study of the subject.

Let us look into few example problems which are solved at various levels of study of Algebra.

## Elementary Algebra Word Problems

Elementary algebra word problems can be solved using one variable and forming simple equations which can be solved in one or two steps.

### Solved Examples

Question 1: 5 added to 4 times a number is 29. Find the number.
Solution:

Let the number be assumed as n.
5 added to 5 times n can be written as 4n + 5
So the equation is formed as 4n + 5 = 29
The equation is solved in two steps as shown below.

Hence the number = 6.

Question 2: The entry tickets at a Carnival cost 5 dollar for a child. Mr.Owens bought for his family 2 adult tickets and three child tickets and paid 35 dollars. How much does one adult ticket cost?
Solution:

Let us assume the cost of one adult ticket as x dollars.
So it costs Mr.Owens, 2x + 3(5)  = 2x + 15 for two adult tickets and 3 child tickets. As he pays 35 dollars for this, we form the equation as
2x + 15 = 35.  The equation is solved in two steps.

Hence the cost of an adult's ticket = 10 dollars.

## Pre Algebra Word Problems

In Pre Algebra, students learn the concept of measurements in two dimensional and three dimensional space. Let us solve one problem on measurements and one age related problem here.

### Solved Examples

Question 1: The length of the legs of an isosceles triangle is 4 meters more than its base. If the Perimeter of the triangle is 44 meters, find the lengths of the sides of the triangle.
Solution:

Let us assume the base measures x meter. Hence each of the legs measure (x + 4) meters. The Perimeter of a triangle is sum of the three sides. The equation is formed and solved as follows:

The length of the base is solved as 12 meters. Hence each of the two legs measure 16 meters.

Question 2: Ann's mother Sara is 7 years older than thrice Ann's age. If the sum of their ages is 39, find the age of both Sara and Ann.
Solution:

Suppose Ann is x years old.
Then Sara's age = 3x + 7.
Equation is formed using the sum of their ages
x + 3x + 7 = 39

The equation is solved and Ann's age is found to be 8
Hence Sara's age = 3(8) + 7 = 24 + 7 = 31.

## Coin Word Problems Algebra

Coin word problems involve objects of value like coins, stamps etc.. These problems are solved either using a linear equation in one variable or using a system of linear equations.

### Solved Example

Question: Andrew has 150 coins in his Piggy Bank, all consisting of dimes and quarters. If the total worth of the coins is 30 dollars, how many dimes and quarters does Andrew have?
Solution:

Let d and q be correspondingly the number of dimes and quarters in the Piggy Bank.
We can form two equations, one on the total number of coins and another on the value of the coins.
x + y = 150            ----------------------------------(1)
0.10x + 0.25y = 30  ---------------------------------(2)
The system can be solved using substitution as follows:

Number of Dimes = 50
Number of Quarters = 100

## Algebra Distance Word Problems

Solving word problems involving time, speed distance equation is a common task in algebra. The equation relating distance speed and time is distance = speed x time.

A distance, time speed table is useful in solving this type of problems. Let us solve a distance problem, using a rational equation.

### Solved Example

Question: An airplane flew a distance of 1500 Km against a wind speed of 100 km/hr. On the return flight the airplane flew with the wind of same speed and took 1 hour and 15 minutes less.  Find the speed of the airplane, assuming the speed remained the same during both the trips.

Solution:

The value required as answer is generally assumed as a variable in word problems.
So, let the speed of the airplane = x Km/hr.
Let us form a distance, speed, time table for the situation required.

 Direction of  the wind Distance Speed of the     Flight Time taken  for the flight With wind 1500 Km x + 100 $\frac{1500}{x+100}$ Against wind 1500 Km x - 100 $\frac{1500}{x-100}$

The equation is formed, using the difference in the times.

$\frac{1500}{x-100}$ - $\frac{1500}{x+100}$ = $1\frac{1}{4}$  (The time difference is written in hours)

$4(x - 100)(x + 100)$ . ($\frac{1500}{x-100}$ - $\frac{1500}{x+100}$) = $\frac{5}{4}$ $\times$ $4(x - 100)(x + 100)$ (Multiplied by LCD)

6000(x + 100) - 6000(x -100) = 5x2 - 50000   (Distributive Property)

6000x + 600000 - 6000x + 600000 = 5x2 - 50000

5x2 = 1250000   (Equation simplified)

x2 = 250000      ⇒      x = 500

Hence the speed of the airplane = 500 Km/hr

## Integrated Algebra Word Problems

Let us solve a word problem using a linear model.

### Solved Example

Question: The linear model P(d) = 62.5d + 2117 is used to find the pressure (lb/ft2) at d feet below the surface of the water.
(a) What does the constant 2117 represent?
(b) What information do you get from the number 62.5?
(b) What is the pressure 200 ft below water surface?
Solution:

The model is given in slope intercept form y = mx + b , where m is the slope and b the y intercept.
The constant 2117, which can be viewed as the y intercept is the function value when d = 0.
This means the pressure on the surface of water = 2117 lb/ft2.

The number 62.5 can be related to 'm' the slope in a linear equation. It is the rate at which the pressure is increasing for every ft below the water surface.

P(d) = 62.5d + 2117
P(200) = 62.5(200) + 2117 = 12,500 + 2,117 = 14,617 lb/ft2.

## Intermediate Algebra Word Problems

Word problems based on Quadratic models are solved in Intermediate Algebra, which is also known as Algebra 2.

### Solved Example

Question: A tour organizer serves 450 customers in a day, if they charge 10 dollars per person. The company estimates that it will lose 10 customers for every increase of 0.50 dollars in the fare. Find the fare that would yield the maximum income to the company.
Solution:

Let us assume the number of 0.50 dollar increase as x.
Then the rate charged per passenger = 10 + 0.5x
Number of passengers at this rate = 450 - 10x
Hence the income of the company on a day  P(x) = (450 - 10x)(10 + 0.5x)
= -5x2 + 125x + 4500
P(x) is a quadratic function with negative leading term. Hence the maximum function value is reached at the vertex.

The x coordinate of the vertex  = $\frac{-125}{-10}$ = 12.5   (Using the formula $\frac{-b}{2a}$ to find the x coordinate of the vertex.)

Thus the company makes maximum profit with 12.5 times of 0.5 dollar increase.
Increase in fare from 10 dollars = 12.5 x 0.5 = 6.25
Hence the maximum income comes when the fare is fixed at 10 + 6.25 = 16.25 dollars.

## Linear Algebra Word Problems

### Solved Example

Question: Four points on a cubic polynomial are given as (-1, -10), (1, -4) (2, 2) and (0,-4). Find the equation to the polynomial.
Solution:

The cubic equation can be assumed as f(x) =  ax3 + bx2 + cx + d
Substituting the four points in the equation, we get a system of linear equations in four variables as
-a + b - c + d = -10
a + b + c + d = -4
8a + 4b + 2c + d = 2
d = -4

Substituting d = -4 in the first three equations the above system can be reduced to a system of  three variables,
-a + b - c = -6         ----------(1)
a + b + c = 0          ----------(2)
8a + 4b + 2c = 6     ----------(3)

The above system can be solved using the row operations on augmented matrix

Augmented Matrix for the system

$\begin{bmatrix} -1 & 1 & -1 & & -6\\ 1 & 1 & 1 & & 0\\ 8 & 4 & 2 & & 6 \end{bmatrix}$
The above matrix can be reduced to
$\begin{bmatrix} 1 & 0& 0 & & 2\\ 0& 1& 0& & -3\\ 0& 0 & 1 & & 1 \end{bmatrix}$

Hence a = 2, b = -3 and c = 1.

The required polynomial equation is

f(x) = 2x3 - 3x2 + x - 4. Answer