Question

How much pure water(0%) must be mixed with 2 liters of a 40% solution of antifreeze to get a 25% antifreeze solution?

Answer 1

my equations for x+y=2 and .4y*2=.25

are these right?

Answer 2

Yes sir/mam your equations are right but make sure that your 4y doesnt have anything to do with the percentage or the liters this is strictly a math equation

Answer 3

so it would be .4y=.25*2?

Answer 4

but it says 2 liters of 40%

Answer 5

exactly so .4 * y = .25 * 2 so .25 * 2 = .5 ======= y= .5 or 50 %

Answer 6

but why is it .25*2?

Answer 7

.25 in 2 liters

Answer 8

ohh ok thanks

Answer 9

yw

Answer 10

What grade are you in??/

Answer 11

No spying or anything Im just saying I am learning this in 6th grade just wondering when I will have to learn it again

Answer 12

The way I would do this is:
we have 2 liters of 40% antifreeze and we add an unknown amount x liters of pure water:
x+2= y to end up with y liters total

the amount of antifreeze is 0*x (pure water) + 2*0.4 liters (40% of 2 liters)
we want this to equal 25% of y. Our second equation is
2*0.4 = 0.25 y
solve for y (the total amount of solution): y= 4*2*0.4= 3.2 liters
use the first equation: x+2= y, x+2= 3.2, x= 1.2 liters

Add 1.2 liters of pure water to the 2 liters of 40% solution to get 3.2 liters of 25% solution.

Answer 13

im in 7th

Answer 14

lol

Answer 15

The amount of antifreeze is the same in both cases, so

40(2)=25(2+x)

Answer 16

So.... 30=25x
6/5=x (amount added, in liters)

Answer 17

That is the same as the 1.2 liters calculated above, but I think this might be an easier calculation.