Question

How many quarts of pure antifreeze must be added to 6 quarts of a 20% antifreeze solution to obtain a 60% antifreeze solution?

Answer 1

amount @ 1 = 6(.2-.6)/(1-.6) maybe ...

Answer 2

nope; gotta redo that one :)

Answer 3

60% of 6 + x is x + 20% of 6

as an equation this is
\[.6(x+6)=x+.2\times 6\]

Answer 4

right hand side of the equation says you have x amount of pure antifreeze plus the 20% of the 6 you already have. left hand side says you want the resulting amount to be 60% antifreeze. now you have to solve for x

Answer 5

i would multiply by 10 first but i am lazy.

Answer 6

lol work it out, x= 6
avg of 100 and 20 is 60 so it makes sense the num of quarts is equal

Answer 7

\[6(x+6)=10x+2\times 6\]
\[6x+36=10x+12\]
\[36-12=10x-6x\]
\[24=4x\]
\[6=x\]

Answer 8

lot of "6s" in this problem

Answer 9

i hate math

Answer 10

a+b = t

ax+by = tz

ax = (a+b)z - by

ax = az +bz -by

a(x-z) = b(z-y)

a = b(z-y)/(x-z) where b = known amount, and y = its %, z = final%; and x = unknown amounts %

a = 6(.6-.2)/(1-.6)

Answer 11

haha...

Answer 12

i like dumbcow's answer though very snappy!

Answer 13

so put 6?

Answer 14

6 seems to be the consensus

6 + 6(.2) = 12(.6)

6 + 1.2 = 7.2

7.2 = 7.2 its good

Answer 15

my first bout gave me -6 ... i wonder if that a side effect tho ...

Answer 16

it was 6.. thanks guys! this is awesome.