Question

a drugist has one solution that is 10% iodine and another that is 50% iodine. How much of each should he use to get 100mL of a mixture that is 20% iodine?

Answer 1

let x = amount of 10% iodine and y = amount of 50% iodine

Answer 2

since you make a final mix of 100 mL, we know that

x+y = 100

and you can solve for y to get y = 100 - x

Answer 3

you want to make a final mix of 20% iodine, so you want 100*0.2 = 20 mL of pure iodine

there are 0.1x mL of pure iodine from the first solution and 0.5y mL of pure iodine from the second

Answer 4

so 0.1x + 0.5y = 20

Answer 5

is tht the answer

Answer 6

no, it helps you find the answer

Answer 7

you need to solve for x and y

Answer 8

can u go futher for me

Answer 9

0.1x + 0.5y = 20

0.1x + 0.5(100 - x) = 20

now solve for x

Answer 10

distribute?

Answer 11

yes

Answer 12

0.3?

Answer 13

0.1x + 0.5y = 20

0.1x + 0.5(100 - x) = 20

0.1x + 50 - 0.5x = 20

-0.4x + 50 = 20

Did you get this far?

Answer 14

0.1+0.5(100-x)=20
0.6x(100-x)=20
60x-x=20
+x +x
------------
60x=20
--- ---
60 60

x=0.3

Answer 15

0.1x + 0.5y = 20

0.1x + 0.5(100 - x) = 20

0.1x + 50 - 0.5x = 20

-0.4x + 50 = 20

-0.4x = 20-50

-0.4x = -30

x = -30/(-0.4)

x = 75

Answer 16

nice work jim

Answer 17

thanks