Question

Sarah would like to make a 6 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of a mixture that is 80% peanuts and 20% almonds and several pounds of a mixture that is 50% peanuts and 50% almonds.
a. What is the system that models this situation?
b. What is the solution to the system? How many pounds of the 80/20 mixture? How many pounds of the 50/50 mixture?

Answer 1

Answer:  

a.  The system of equations that models the situation is....

    $0.80x+0.50y=3.60\\\\ 0.20x+0.50y=2.40$

b.  The solution to the system:  x = 2 and y = 4

The amount of 80/20 mixture is 2 pounds and the amount of 50/50              mixture is 4 pounds.

Step-by-step explanation:

Suppose, the amount of 80/20 mixture is  $x$ pounds and the amount of 50/50 mixture is  $y$ pounds.

So, the amount of peanuts in 80/20 mixture $= 0.80x$ pound and the amount of almonds in 80/20 mixture $=0.20x$ pound.

And the amount of peanuts in 50/50 mixture $=0.50y$ pound and the amount of almonds in 50/50 mixture $=0.50y$ pound.

Now, Sarah would like to make a 6 pounds nut mixture that is 60% peanuts and 40% almonds.

So, the amount of peanuts in that mixture $=(6\times 0.60)=3.60$ pounds

and the amount of almonds in that mixture $=(6 \times 0.40)= 2.40$ pounds.

So, the system of equations will be.........

$0.80x+0.50y=3.60 ...................(1)\\\\ 0.20x+0.50y=2.40...................(2)$

Subtracting equation (2) from equation (1), we will get.....

$(0.80x+0.50y)-(0.20x+0.50y)=3.60-2.40\\ \\ 0.60x=1.20\\ \\ x= \frac{1.20}{0.60}=2$

Now, plugging this $x=2$ into equation (1), we will get......

$0.80(2)+0.50y=3.60\\ \\ 1.60+0.50y=3.60\\ \\ 0.50y=3.60-1.60=2\\ \\ y=\frac{2}{0.50}=4$

So, the amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.