SOLVING LINEAR INEQUALITIES WORD PROBLEMS

We can follow the steps given below to solve word problems on linear inequalities.

Step 1 :

Read and understand the information carefully and translate the statements into linear inequalities. 

Step 2 :

Solve for the variable using basic operations like addition, subtraction, multiplication and division.

Step 3 :

Find the solution set and obtain some of the possible solutions.

Apart from the above steps, we have to make the following changes, when we multiply or divide each side of the inequality by a negative value. 

  •  If we have  <, then change it as  >
  •  If we have  >, then change it as  <
  •  If we have  , then change it as  
  •  If we have  , then change it as  

Problems

Problem 1 :

To secure A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If one scored 84, 87, 95, 91 in first four subjects, what is the minimum mark one scored in the fifth subject to get A grade in the course?

Solution :

Let "x" be the required mark in fifth subject

Four subjects marks are 84, 87, 95 and 91

The average of five subject marks would be grater than or equal to 90

(84 + 87 + 95 + 91 + x)/5   90

Multiply by 5 on both sides

357 + x    90(5)

357 + x    450

Subtract both sides by 357

x  ≥ 450 - 357

  93

So, the minimum mark is 93.

Problem 2 : 

A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?

Solution :

Quantity of solution at present  =  600 liters

Quantity of acid in liters in existing solution  =  12% of 600

  =  (12/100) 600

  =  72 liters

Let "x" be the required quantity of solution to be added.

Acid level in x liters of solution  =  30% of x

  =  (30/100)x

  =  0.3x

From this, we come to know that the resulting mixture will contain (600 + x) liters solution and (72 + 0.3x) liters acid.

Quantity of acid will be more than 15% in the new mixture

(72 + 0.3x) > 15% of (600 + x)

(72 + 0.3x) > 0.15 (600 + x)

72 + 0.3x > 90 + 0.15x

0.3x - 0.15x > 90 - 72

0.15x > 18

x > 18/0.15

x > 120 liters

Quantity of acid will be less than 18% in the new mixture

(72 + 0.3x) < 18% of (600 + x)

(72 + 0.3x) < 0.18(600 + x)

72 + 0.3x < 108 + 0.18x

0.3x - 0.18x < 108 - 72

0.12x < 36

x < 300

So, the acid level in the resulting mixture is between 120 L and 300 L.

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