Question

how much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?

Answer 1

Let X be the total volume of the final mixture. Then you know that 60% of X is antifreeze. Hence
\[ 0.6X = 1 \ gallon \]
or
\[ X = \frac{1}{0.6} \ gallons \]
Now can you find the amount of water you need to add?

Answer 2

Notice that \( X = 1 \frac{2}{3} \) gallons. Hence how much water is added?

Answer 3

i'm sorry i'm totally clueless....i feel like x should be the amount of water?

Answer 4

X is the volume of the final mixture.

Answer 5

And you want 60% of X to be the anti-freeze.

Now 60% of X is
\[ \frac{60}{100} X \]
So far, so good?

Answer 6

so then x-1/.6 is the water?

Answer 7

No.

Answer 8

40 percent...?

Answer 9

X = (antifreeze) + (water)

Answer 10

And we want that
60% 40%
X = (anti-freeze) + (water)

Answer 11

But we also are given 1 gallon of anti-freeze. Therefore it must be that
60% of X = 1 gallon.

Yes?

Answer 12

right

Answer 13

Now 60% of X is
\[ \frac{60}{100}X \]
if that is equal to 1, then
\[ X = \frac{100}{60} = \frac{5}{3} \]
ok?

Answer 14

ok

Answer 15

Now because
X = (anti-freeze) + (water)
and X = 5/3
what must the volume of water be?

Answer 16

1-5/3?

Answer 17

X = (anti-freeze) + (water)

i.e.,
5/3 = 1 + (water)

i.e.,
(water) = 5/3 - 1 gallons
= 2/3 gallons

Answer 18

ohhh ok, man i am so bad at word problems

Answer 19

There's another way to solve this using proportions.

Answer 20

If 60% of the solution is antifreeze, then 40% is water and it must be that
\[ \frac{antifreeze}{water} = \frac{60}{40} \]
Now, as the volume of anti-freeze is 1 gallon, we have that
\[ \frac{1}{water} = \frac{60}{40} \]
Now solve for Water.

Answer 21

oh i think that one makes more sense to me

Answer 22

can i ask you another one or should i post it on the board again?

Answer 23

post a new question.