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Sarah would like to make a 6lb 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 50/50 mixture? SHOW YOUR WORK!

Respuesta :

tramserran

Answer:    [tex]\left \{ {{A+B\ =\ 6}\atop{0.8A+0.5B\ =\ 0.6(6)}} \right.[/tex]

                 2 lbs of 80/20 and 4 lbs of 50/50 mixtures

Step-by-step explanation:

Let A represent the quantity of the 80/20 mixture.  Let B represent the quantity of the 50/50 mixture.

Let's solve based on peanuts:

  • A has 80% peanuts = 0.8
  • B has 50% peanuts = 0.5
  • Mixture has 60% peanuts = 0.6

First, set up the equations:

  • Quantity: A + B = 6
  • Peanut % x Quantity: 0.8A + 0.5B = 0.6(6)

Now, solve the system above using substitution: A + B = 6  ⇒  B = 6 - A

0.8A + 0.5(6 - A) = 0.6(6)

0.8A + 3 - 0.5A = 3.6

           3 + 0.3A = 3.6

                 0.3A = 0.6

                       A = 2

Lastly, plug in A (above) into the Quantity equation to solve for B:

A + B = 6

(2) + B = 6

       B = 4


SIDE NOTE: You could have solved based on almonds by setting up the system as:

  • Quantity: A + B = 6
  • Almond % x Quantity: 0.2A + 0.5B = 0.4(6)

which would give the same answer (A = 2, B = 4)


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