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Algebraic Expression Word Problems

In real life situations, we come across a lot of questions in verbatim. But mostly such questions can be answered by converting them into algebraic equations which can be solved by algebraic methods and translate the answer back to verbal form. An algebraic equation is a bridge that connects two algebraic expressions. The algebraic expression word problems are the models for such situations and are guide lines to solve practical problems in our daily life.

Algebraic Expression Examples

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Algebraic expression is an expression that contains variables and a finite number of algebraic operations. An algebraic expression word problem is a test for dealing a real life situation. You must be capable of translating the sentences into algebraic expressions and solve algebraically. Once you get an algebraic solution you must correctly transform that as a conclusion in verbal form. For a better understanding, let us study and solve a few algebraic expression word problem.

Solved Examples

Question 1: If 3 is subtracted from five times a number is equal to the sum of that number and 17. What is that number?

Solution:
Let us assume that the number be ‘x’.

Step 1:

Let us consider the first part of the word problem, 3 is subtracted from five times a number.

=> The algebraic expression for ‘five times the number’ is ‘5x’ and ‘3 is subtracted from five times a number’ implies an algebraic expression ‘5x – 3’.

Step 2:

Now let us attend to the other part of the word problem ‘the sum of that number and 17’. that is ‘x + 17’.

Step 3:

The next part is the ‘bridge’ or the conjunction between these two statements. Since it says as,' is equal to’ the algebraic operator for this ‘=’.

Thus the translated equation of the word problem is,

=> 5x – 3 = x + 17

Step 4:
 Solve for x,

=> 5x – 3 = x + 17

=> 5x - x = 17 + 3

=> 4x = 20

Divide each side by 4

=> $\frac{4x}{4} = \frac{20}{4}$

=> x = 5

Transforming back to verbatim we can say that the number is 5.

 

Question 2: 30 liters of 12% sugar solution is to be added with 30% sugar solution to produce a mixture which should have a 15% sugar. Find the quantity of 30% sugar solution that can be added.

Solution:
Given
In a mixture
30 liters of 12% sugar solution is to be added with 30% sugar solution

Let  'x'  liters of 30% sugar solution to be added.

Step 1:

The quantity of sugar solution after mixing will be (30 + x) liters.

The sugar content in 30 liters of 12% sugar solution + x liters of 30% sugar solution will be

=> 30 * 0.12 + x * 0.3) = 3.6 + 0.3x                    .......................................(1)

Step 2:
 
The sugar content in (30 + x) liters of 15% sugar solution will be

=> (30 + x) * 0.15 = 30 * 0.15 + 0.15x

= 4.5 + 0.15x

=> (30 + x) * 0.15 = 4.5 + 0.15x                          .........................................(2)

Step 3:
As per the problem statement,  (1) = (2) That is,

=> 3.6 + 0.3x =  4.5 + 0.15x
 
Solve for x,
                                                                 
Subtracting 0.15x from both sides,

=> 3.6 + 0.3x – 0.15x =  4.5 + 0.15x – 0.15x

=> 3.6 + 0.15x =  4.5
             
Subtracting 3.6 from both sides, we have

=> 3.6 + 0.15x – 3.6 =  4.5 – 3.6

=> 0.15x =  0.9                                    

Divide each side by 0.15

=> $\frac{0.15x}{0.15} =  \frac{0.9}{0.15}$

=>  x = 6

Answer:
The quantity of 30% sugar solution that can be added is 6 liters.