Question

Medal for who helps!! (please help, don't give the answer)

Answer 1

Show an equation and a solution. How much pure antifreeze must be added to 12 L of a 40% solution to obtain a 60% solution. (Remember that pure antifreeze is 100% = 1.)

Answer 2

Im guessing you do:
(12 \div 40) 60

Answer 3

let x = amount of pure antifreeze to be added to the mix

Answer 4

we start out with 12 L of a 40% solution of antifreeze

how much pure antifreeze do we have so far?

Answer 5

12L...?

Answer 6

wait nvm

Answer 7

that's how much of the solution (antifreeze mixed with other chemicals)

I want to know how much of just antifreeze alone we have

Answer 8

5 L right?

Answer 9

we have "12 L of a 40% solution "

so 40% of that 12 liters is

40% of 12 = 0.4 * 12 = 4.8

therefore we currently have 4.8 liters of pure antifreeze

Answer 10

My bad, my brains not working today >,<

Answer 11

no worries

Answer 12

x+12(.40)=(x+12)(.60)
x+4.8=.60x+7.2
x=.60x+2.4
.40x=2.4
x=6

Answer 13

so notice how if you were to divide the 4.8 liters (of the pure stuff) by 12 liters (of the mixed stuff) you'd get

4.8/12 = 0.4

which confirms that 4.8 is indeed 40% of 12

Answer 14

what we want to do is add x liters of the pure stuff to the original 4.8 liters

Answer 15

Yes i know, i just did my math wrong :p

Answer 16

in doing that, we also add x liters to the total 12 liters as well

Answer 17

so we go from 4.8 liters of pure stuff to 4.8+x

and for the mixed, we go from 12 to 12+x

-------------------------------------------

we will now have this fraction

\[\large \frac{4.8+x}{12+x}\]

and we set it equal to 0.6 because we want the percentage to be 60%

\[\large \frac{4.8+x}{12+x} = 0.6\]

from here you solve for x

Answer 18

uggh my answer doesn't look right, i keep getting 2.5 + x over x = .6

Answer 19

and if u take out the x's its 2.5 = .6

Answer 20

\[\large \frac{4.8+x}{12+x} = 0.6\]

\[\large 4.8+x = 0.6(12+x)\]

\[\large 4.8+x = 7.2+0.6x\]

Keep going