MODELING REAL WORLD SITUATIONS WITH LINEAR EQUATIONS

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 -----> number of adult tickets

y -----> number of kids tickets

Then, 

x + y  =  548 -----(1)

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

Solving (1) & (2),

x  =  202

y  =  346

The number of adults tickets sold is 202 and the number of kids tickets sold is 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, substitute x = 95 in (3).

(3)-----> y  =  150(95) + 10000

y  =  14250 + 10000

y  =  24250

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 -----> number of units sold at $6/unit

y -----> number of units sold at $8/unit

Then,

x + y  =  100 -----(1)

6x + 8y  =  660 -----> 3x + 4y  =  330 -----(2)

Solving (1) and (2),

x  =  70

y  =  30

The number of tickets sold at $6 per unit is 70 and the number of tickets sold at $8 per unit is 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. 

Then,

8x + 6y  =  33 -----(1)

4x - 5y  =  4.50 -----(2)

Solving (1) and (2),

x  =  3

y  =  1.5

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

Then,

y  =  120% of x

y  =  1.2x -----(2)

Substitute y  =  1.2x  in (1)

x + 1.2x  =  2640

2.2x  =  2640

x  =  1200

Substitute x  =  1200 in (2)

y  =  1.2(1200)

y  =  1440

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,

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),

x  =  66

y  =  84

The cost prices of two products are $66 and $84 

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