Question

one canned juice drink is 20% orange juice, another is 10% O.J. How many liters of each should be mixed together to get 10L that is 11% O.J....please help!!!!!

Answer 1

let

x = amount of 20% OJ
y = amount of 10% OJ

Answer 2

you need 10 L total, so

x+y = 10

solve for y to get

y = 10 - x

Answer 3

if you have x liters of 20% OJ, you will have 0.2x liters of pure OJ (and no water)

if you have y liters of 10% OJ, you will have 0.1y liters of pure OJ (and no water)

Answer 4

combine the two and you'll have 0.2x+0.1y liters of pure OJ

you want a mix that's 11% OJ and you want 10 L of this mix, so you want 0.11*10 = 1.1 liters of pure OJ

Answer 5

This means that your second equation is

0.2x + 0.1y = 1.1

Answer 6

x+y=10
0.20x+0.10y=(10)(0.11)
20x+10x=110

Answer 7

if you multiply everything in 0.2x + 0.1y = 1.1 by 10, you will get

0.2x + 0.1y = 1.1

10*0.2x + 10*0.1y = 10*1.1

2x + y = 11

Answer 8

Now you plug in y = 10 - x

2x + y = 11

2x + 10-x = 11

and you would solve for x

Answer 9

x+y=11
-20(x+y)=-20(11)
-20x-20y=-220

Answer 10

so what do you get when you solve

2x + 10-x = 11

Answer 11

-8=11so -8/11?

Answer 12

no

Answer 13

2x + 10-x = 11

x + 10 = 11

x = ??

Answer 14

1

Answer 15

x+y = 10

1+y = 10

y = 10-1

y = 9

Answer 16

So you need

1 liter of the 20% OJ
9 liters of the 10% OJ