MODELING REAL WORLD SITUATIONS WITH LINEAR EQUATIONS

About "Modeling real world situations with linear equations"

Modeling real world situations with linear equations :

Linear equations can be used to solve many real world problems.

If we want to solve real world problems using linear equations, first the given real world situation has to be modeled with linear equations.

Let us see how the real world situations can be modeled with linear equations.

Modeling real world situations with linear equations - Examples

Example 1 :

A park charges \$10 for adults and \$5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of \$3750 ?

Solution :

x ------> no. of adult tickets,   y ------> no. of kids tickets

According to the question, we have x + y = 548 ---------(1)

And also, 10x + 5y = 3750 -------> 2x + y = 750 --------(2)

Solving (1) & (2), we get x = 202 and y = 346

Hence, the number of adults tickets sold  =  202

the number of kids tickets sold  =  346

Example 2 :

A manufacturer produces 80 units of a product at \$22000 and 125 units at a cost of \$28750. Assuming the cost curve to be linear, find the equation of the line and then use it to estimate the cost of 95 units.

Solution :

Since the cost curve is linear, its equation will be

y = Ax + B.

(Here y = Total cost, x = no. of units)

80 units at \$22000 --------> 22000 = 80A + B -------(1)

125 units at \$28750 --------> 28750 = 125A + B -------(2)

Solving (1) and (2), we get A  = 150 and B = 10000

So, the equation of the line is y = 150x + 10000 --------(3)

To find the cost of 95 units, plug x = 95 in (3).

(3) -------> y = 150(95) + 10000

y = 14250 + 10000

y = 24250

Hence, the cost of 95 units is \$24250

Example 3 :

A trader has 100 units of a product. A 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 :

"x" -------> no. of units sold at \$6/unit

"y" -------> no. of units sold at \$8/unit

According to the question, x + y = 100 ------------(1)

6x + 8y = 660 ------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  =  70

the no. of tickets sold at \$8 per unit  =  30

Example 4 :

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.

From the information given in the question, we have

8x + 6y = 33 ----------(1)

Wages of 4 men = 4x

Wages of 5 boys = 5y

According to the question, 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 5 :

Sum of incomes of A and B is \$2640. If B's income is 20% more than A, find the income of A and B.

Solution :

Let "x" and "y" be the incomes of A and B respectively.

Then,  x + y  =  2640  --------(1)

Given : B's income is 20% more than A

So, we have     y  =  120% of x

y  =  1.2x --------(2)

Plugging  y  =  1.2x  in (1) ------> x + 1.2x = 2640

2.2x  =  2640 --------> x  =  1200

Plugging   x  =  1200 in (2) ------> y = 1.2(1200)

y  =  1440

Hence, the incomes of A and B are \$1200 and \$1440

Example 6 :

A trader gains one third of the cost price as profit on a product and one fourth of the cost price as profit on other product. Total profit earned on these two products is \$43.  The sum of the cost prices of two products is \$150. Find the cost price of each product.

Solution :

Let "x" and "y" be the cost prices of the two products.

Then, we have   x + y  =  150 -------(1)

Given : One third of the cost price as profit on a product

So, profit on the first product = (1/3)x  =  x/3

Given : One fourth of the cost price as profit on the other product.

So, profit on the second product = (1/4)y  =  y/4

Total profit earned on these two products  =  \$43

x/3  +  y/4  =  43 -------> (4x + 3y) / 12  = 43

4x + 3y  =  516 --------(2)

Solving (1) and (2), we get x  =  66  and y  =  84

Hence, the cost prices of two products are \$66 and \$84

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