Question

How much candy at $1.16 a pound should be mixed with candy worth 86 cent a pound in order to obtain a mixture of 60 pounds of candy worth a dollar a pound?

Answer 1

Let the weights of the two candies be repres. by x and y.

Then x + y = 60, or x = 60 - y

($1.16 / lb) x + ($0.86 / lb) y = ($1.00 / lb) (60 lb) = $60

Then 1.16(60-y) + 0.86y = 60
   69.6 - 1.16y + 0.86y = 60                                9.6
     9.6 = 0.3y                         Solving for y, y = ------- = 32 lb
                                                                            0.3

                                               Then x = (60-32) lb = 28 lb

Answer 2

Answer:

28 pounds of the candy costing $1.16 and 32 pounds of the candy costing $0.86.

Step-by-step explanation:

To Find: How much candy at $1.16 a pound should be mixed with candy worth 86 cent a pound in order to obtain a mixture of 60 pounds of candy worth a dollar a pound?

Let  x  be the candy costing $1.16 a pound.

Let  y  be the candy costing $0.86 per pound. (We'll work in dollars for the problem.)

So we know that:

Equation (1)$x+y=60$

Now we want the average cost to be $1 per pound.

So,To get the average cost we need to know the total cost and divide by the total pounds.

Total cost: 1.16x + 0.86y

Since Total pounds: 60

Average cost : 1

So,  $\frac{1.6x+0.86y}{60} =1$

Equation (2):  $1.16x+0.86y=60$

multiply equation (1) by  −0.86  to get equation (3):

Equation (3):  $-0.86x-0.86y=-51.6$

Add equation (2) and equation (3)

$0.3x =8.4\\\\x=28$

Substitute into equation (1):  

$28+y=60$

$y=32$

Thus 28 pounds of the candy costing $1.16 and 32 pounds of the candy costing $0.86.