Question

A candy store makes a 10-lb mixture of gummy worms, candy corn, and sourballs. The cost of gummy worms is $1.00 per pound, candy corn cost $3.00 per pound, and sourballs cost $1.50 per pound. The mixture calls for three times as many gummy worms as candy corn. The total cost of the mixture is $15.00. How much of each ingredient did the store use?

Answer 1

Answer:

6 grams of gummy worms, 2 grams of candy corn and 2 grams of sourballs.

Step-by-step explanation:

Let the number of pounds of each ingredient be as follows:

Gummy Worms = x pounds

Candy Corn = y pounds

Sourballs = z pounds

The store makes a mixtures of 10 pounds. This means the sum of x, y and z would be 10. Setting up the equation:

$x + y +z = 10$                                     (Equation 1)

The mixture calls for 3 times as many gummy worms as candy corn. This means amount of gummy worm will be 3 times the candy corn. Setting up the Equation:

$x=3y$                                                  (Equation 2)

Cost of gummy worms is $1.00 per pound, candy corn cost $3.00 per pound, and sourballs cost $1.50 per pound. So cost of x, y and z pounds would be:

1x , 3y and 1.5z, respectively. The total cost of mixture is $15. So we can set up the Equation as:

$x+3y+1.5z=15$                                (Equation 3)

Using the value of x from Equation 2, in Equations 1 and 3 give us following two equations:

$4y+z=10$                             By substitution in Equation 1. (Equation 4)$6y+1.5z=15$                        By substitution in Equation 3. (Equation 5)

Multiplying the Equation 4 by 1.5 and subtracting from Equation 5 gives us:

$6y +1.5z-1.5(4y+z)=15-1.5(10)\\\\ 6y+1.5z-6y-1.5z=15-15\\\\ 0=0$

When two sides of equations turn into something that is always positive, we conclude that there are infinite number of solutions. In such cases, we fix a variable and give different values to it, to find corresponding values of other variables. Lets re-write the solution in terms of z.

From Equation 4, we have:

$y=\frac{10-z}{4}$

From Equation 2, we have:

$x=3(\frac{10-z}{4} )$

Therefore, the solution set will be:

$(3(\frac{10-z}{4} ), \frac{10-z}{4} , z)$

Now in order to find any combination of ingredient, we give any value to z. Let, z is equal to 2 grams.

So,

x would be = 6 grams

y would be = 2 grams

So, one of the possible amount of ingredients that store can use is:

6 grams of gummy worms, 2 grams of candy corn and 2 grams of sourballs.