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Mixture problems 2 ways!. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than.

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Presentation on theme: "Mixture problems 2 ways!. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than."— Presentation transcript:

1 Mixture problems 2 ways!

2 You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30%solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

3 Mixture problems are CA CA! CA+CA=CA or, more specifically, C 1 A 1 +C 2 A 2 =C T A T C= Concentration or Cost depending on the problem A=Amount The subscripts are the 1 st item, the 2 nd item, and the total (or mixture) The following graphic simply gets all of the numbers organized in order to put them into the formula. CA 110x 23010-x T1510 The amounts of 1 & 2 must add to make the total amount! In this problem, we know the total (10 liters) So, we make #1 = x, then #2 must be 10-x so that 1 & 2 together add to 10. Now, we put the numbers into the formula C 1 A 1 +C 2 A 2 =C T A T 10x + 30(10-x) = 15 10 10x + 300-30x = 150 -20x + 300 = 150 -300 -300 -20x = -150 -20 -20 X= 7.5 (this is the 10% solution) 10-x 10-7.5 2.5 (this is the 30% solution)

4 Same problem, different way: You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use? 12mix 1030 C A x10-x 10 L 15 15-10 = 5 30-15= 15 5x = 15(10-x) 5x = 150 – 15x +15x 20x = 150 20 x = 7.5 L Of the 10% solution 10-x 10-7.5 2.5 Of the 30% solution This graphic also organizes the information, but goes one step further and takes care of some of the arithmetic too.

5 How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound? CA 1$1.20x 2$0.9010 T$1.00x + 10 The amounts of 1 & 2 must add to make the total amount! In this problem, we know the amount of #2 is 10 pounds, So, we make #1 = x, then the total must be x +10. Now, we put the numbers into the formula C 1 A 1 +C 2 A 2 =C T A T $1.20x +.90 10 = $1.00(x+10) 1.20x + 9 = 1.00x + 10 -1.00x -1.00x.20x +9 = 10 -9 -9.20x = 1.20.20 X= 5 pounds of the $1.20 chocolate

6 How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound? 12mix $1.20$0.90 C A x10 10 + x $1.00 $1.20-1.00=.20 $1.00-.90=.10.20x =.10 10.20x = 1.20 Of the $1.20 chocolate x = 5


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