mixtures- math bible sum

[vkb16]

A certain solution is 60% methanol. How many ounces of methanol must be added to 80 ounces of the original solution so that the resulting solution is 75% methanol?

OA is 48

Simple, but I didnt get why the ans is what it is, and not 80.

thanks

[ssmiles08]

Total original solution: 80
60% of that solution is Methanol: .6(80) = 48

x ounces are added to total solution; so new solution is now: 80 + x

75% of the new solution is Methanol: .75(80 + x)

set up equation; 48 + x = .75(80 + x)

.25x = 12; x = 48 ounces.

[shahdevine]

A certain solution is 60% methanol. How many ounces of methanol must be added to 80 ounces of the original solution so that the resulting solution is 75% methanol?

OA is 48

Simple, but I didnt get why the ans is what it is, and not 80.

thanks

different way to think about it is in terms of part vs. whole.

amount of methanol/total solution=percent of methanol
and change in percent methanol is attributed to the x you are adding

so amount of methanol=.6(80)+x and total solution = 80+x

therefore (.6(80)+x)/(80+x)=3/4
now simplify
4(.6(80)+x)=3(80+x)
4(48+x)=240+3x
192+4x=240+3x
x=48

you got this man!

[ssuarezo]

A certain solution is 60% methanol. How many ounces of methanol must be added to 80 ounces of the original solution so that the resulting solution is 75% methanol?

OA is 48

Simple, but I didnt get why the ans is what it is, and not 80.

thanks

Again, rule of three:

60% of 80 = 48, so 32 is another component.

I want 32 to be 25%, so, the other 75% will be methanol:

32 -- 25%
x -- 75%
x = 96, but, 48 are already there, so 96-48 = 48 .

my procedure is too easy to be truth .. I'll keep verifying if it always work.