WORD PROBLEMS INVOLVING LINEAR EQUATIONS

About "Word Problems Involving Linear Equations"

Word Problems Involving Linear Equations :

Solving word problems on linear equations is sometimes a difficult job for some students.

Actually it is not. There is a simple trick behind it.

The picture given below tells us the trick.

Word Problems Involving Linear Equations - Examples

Example 1 :

Of two numbers, 1/5th of a the greater equal to 1/3rd of the smaller and their sum is 16. Find the numbers.  

Solution :

Let "x" and "y" be the two numbers such that

> y

Given : The sum of the two number is 16.

So, we have

x + y  =  16 ------(1)

Given : 1/5th of a the greater equal to 1/3rd of the smaller

So, we have

1/5 ⋅ x  =  1/3 ⋅ y

Simplify.

3x  =  5y ------(2)

Solving (1) and (2), we get

x  =  10 and y  =  6

Hence, the two numbers are 10 and 6.

Example 2 :

The wages of 8 men and 6 boys amount to $33. If 4 men earn $4.50 more than 5 boys, determine the wages of each man and boy. 

Solution :

Let "x" and "y" be the wages of each man and boy. 

Given : The wages of 8 men and 6 boys amount to $33.

So, we have

8x + 6y  =  33 ------(1)

Given : 4 men earn $4.50 more than 5 boys. 

So, we have

4x - 5y  =  4.50 ------(2)

Solving (1) and (2), we have

x  =  3 and y  =  1.5

Hence, the wages of each man and each boy are $3 and $1.50 respectively.

Example 3 :

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed. Find the number. 

Solution :

Because the number is between 10 and 100, it has to be a two digit number.

So, let "xy" be the number between 10 and 100. 

Given :  The number between 10 and 100 is five times the sum of its digits. 

xy  =  5(x + y) ------(1)

xy is a two digit number.

And x is in tens place and y is in ones place. 

So, we have

xy  =  10 ⋅ x + 1 ⋅ y

xy  =  10x + y

Then, we have

(1)------> 10x + y  =  5(x + y)

10x + y  =  5x + 5y

Simplify. 

5x - 4y  =  0 -------(2)

Given :  If 9 be added to it the digits are reversed.

xy + 9  =  yx

10 ⋅ x + 1 ⋅ y + 9  =  10 ⋅ y + 1 ⋅ x

10x + y + 9  =  10y + x

Simplify.

9x - 9y  =  - 9

Divide both sides by 9.

x - y  =  - 1 ------(3)

Solving (2) and (3), we get

x  =  4 and y  =  5

Hence, the required number is 45.

Example 4 :

The age of a man is three times  the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man.

Solution :

Let "x" be the present age of the man and "y" be the sum of the present ages of his two sons. 

Given : Present age of the man is 3 times the sum of the ages of 2 sons.

So, we have

x  =  3y ------(1) 

5 years hence, 

Age of the man  =  x + 5

Sum of the ages of his two sons  =  y + 5 + 5  =  y + 10

(There are two sons in y. So 5 is added twice)

Given : 5 years hence, age of the man will be double the sum of the ages of his two sons.

So, we have 

x + 5  =  2(y + 10)

Simplify. 

x + 5  =  2y + 20

x  =  2y + 15 ------(2) 

From (1), we can plug x  =  3y in (2). 

(2)------> 3y  =  2y + 15

Subtract 2y from both sides. 

y  =  15

Plug y  =  15 in (1). 

(1)------> x  =  3 ⋅ 15

x  =  45

Hence the present age of the man is 45 years.

Example 5 :

A trader has 100 units of a product. He sells some of the units at $6 per unit and the remaining units at $8 per units. He receives a total of $660 for all 100 units. Find the number units sold in each category. 

Solution :

Let x be the no. of units sold at $6/unit and y be the no. of units sold at $8/unit.

Given : The trader sells 100 units in all. 

So, we have 

x + y  =  100 ------(1)

Given : He receives a total of $660 for all 100 units.

So, we have 

6x + 8y  =  660

Divide both sides by 2.

3x + 4y  =  330 ------(2)

Solving (1) and (2), we get

x  =  70 and y  =  30

Hence, the no. of tickets sold at $6 per unit is 70 and the no. of tickets sold at $8 per unit is 30.

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