A friend asked me for help with this math problem. I immediately got the answer by figuring it in my head, but that does not help my friend. I need to be able to show the equation one would set up to solve this problem. Again, I am not looking for the answer but HOW to get the answer. The word problem: How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution?

the answer is 40 liters right? if so, this is what i did, 50%+20%=70% 70%/2=35% it requires equal 50% and 20% to make 35% /forums/images/graemlins/dunno.gif not sure if this is right but thats how i did it

Lucky for you my wife and I are smart. Hopefully this is right: 1. Take the original mixture of 20/80 and times it by 40(litres) which equals 800/3200. 2. Through process of elimination, we figured that 40(litres)of the 50/50 solution(= 2000/2000) added to our original 40 litres of the 20/80 solution(= 800/3200) divided by total amount of litres (80 litres) equals 35/65. I have a headache now. Thanks very much.

[ QUOTE ] How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution? [/ QUOTE ] x=amount of 50% solution y=amout of 35% solution x+40=y (.5)x+(.2)40=(.35)y In the second equation we know y=x+40 from the first equation. Substitute it into the second equation. (.5)x+(.2)40=(.35)(x+40) Now solve for x and you need to add 40L like already said you just don't have to use trial & error.

[ QUOTE ] [ QUOTE ] How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution? [/ QUOTE ] x=amount of 50% solution y=amout of 35% solution x+40=y (.5)x+(.2)40=(.35)y In the second equation we know y=x+40 from the first equation. Substitute it into the second equation. (.5)x+(.2)40=(.35)(x+40) Now solve for x and you need to add 40L like already said you just don't have to use trial & error. [/ QUOTE ] This is correct.

[ QUOTE ] [ QUOTE ] How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution? [/ QUOTE ] x=amount of 50% solution y=amout of 35% solution x+40=y (.5)x+(.2)40=(.35)y In the second equation we know y=x+40 from the first equation. Substitute it into the second equation. (.5)x+(.2)40=(.35)(x+40) Now solve for x and you need to add 40L like already said you just don't have to use trial & error. [/ QUOTE ] I agree as to this being the correct way to go about this. Kinda BS that they start with 40 and you find 40 is the number to add. It rarely works like that. I got all pissed when I double checked the numbers. Then again I am a real dork for math (I can't resist a problem).

I HATE word problems !!!!! /forums/images/graemlins/angryfire.gif /forums/images/graemlins/yikes.gif Give me an equation problem any day!!

[ QUOTE ] Real life math problems are in the form of word problems. [/ QUOTE ] I know - but I'm still allowed to HATE them !! /forums/images/graemlins/histerical.gif