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Mixture problems 2 ways!

Mixture problems 2 ways!.

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Mixture problems 2 ways!

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  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. 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? • Mixture problems are CA CA! • CA+CA=CA or, more specifically, C1A1+C2A2=CTAT • C= Concentration or Cost depending on the problem • A=Amount • The subscripts are the 1st item, the 2nd item, and the total (or mixture) • The following graphic simply gets all of the numbers organized in order to put them into the formula. 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 C1A1+C2A2=CTAT 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. 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? • Same problem, different way: This graphic also organizes the information, but goes one step further and takes care of some of the arithmetic too. 1 mix 2 15-10 = 5 30-15= 15 10 15 30 C x 10-x A 10 L 10-x 10-7.5 2.5 Of the 30% solution 5x = 15(10-x) 5x = 150 – 15x +15x +15x 20x = 150 20 20 x = 7.5 L Of the 10% solution

  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? 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 C1A1+C2A2=CTAT $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? 1 mix 2 $1.20-1.00= .20 $1.00-.90= .10 $1.20 $1.00 $0.90 C x 10 A 10 + x .20x = .10 • 10 .20x = 1 .20 .20 x = 5 Of the $1.20 chocolate

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