SOLUTION: Hi, I have the following math problem and wondering if possible can get some help. Thank you in advance. A chemist has three different acid solutions. The first acid solution

Algebra.Com
Question 1175036: Hi, I have the following math problem and wondering if possible can get some help. Thank you in advance.
A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% and the third contains 75%. He wants to use all three solutions to obtain a mixture of 64 liters containing 50% acid, using 33 times as much of the 75% solution as the 35% solution. How many liters of each solution should be used?

Found 4 solutions by josgarithmetic, MathTherapy, greenestamps, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
SOLUTION CONC.       VOLUME(L)        SOLUTE
   25%                 x              0.25x
   35%                 y              0.35y
   75%                 33y            0.75*33y
                     64               


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Hi, I have the following math problem and wondering if possible can get some help. Thank you in advance.
A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% and the third contains 75%. He wants to use all three solutions to obtain a mixture of 64 liters containing 50% acid, using 33 times as much of the 75% solution as the 35% solution. How many liters of each solution should be used?
Another person did this before, used more than one variable, made a mistake that led to a WRONG answer. Now, another person has done it and still, 
you're given directions that'll lead to a WRONG answer!
Here are the CORRECT formulation and the CORRECT answers. So, throw the following in the GARBAGE:
Let the amount of 35% solution to mix, be T
Then the amount of 75% solution is, 33T
Also, amount of 25% solution to mix is, 64 - (T + 33T) = 64 - 34T
We then get: .35T + .75(33T + .25(64 - 34T) = .5(64)
.35T + 24.75T + 16 - 8.5T = 32
16.6T = 32 - 16
16.6T = 16
Amount of 35% solution to mix, or
Amount of 75% solution to mix =
Amount of 25% solution to mix =
Accept NO OTHER answers!
I leave the check for you to do!
Tutor @IKLEYN! He should be IGNORED. He's an IDIOT and a racist who NEVER corrects the other people who make a VAST number of errors. He's
behind the times and wants to tell everyone how to do mathematics as if he knows it. He's just an OLD RACIST GOAT who wants to NITPICK and
CRITICIZE other people whose RACE he doesn't like! Do you realize that she chooses to pick fights with you and I? He needs to be IGNORED
and I hope he would just get LOST or disappear one of these days!
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


For tutor @MathTherapy....

If you are going to have a contest with tutor @ikleyn to see which of you can post the most responses that say "accept no other answer", then you might take care to make sure that those answers are correct.

In your response to this question, you told the reader they could check your answer themselves. It might have been a good idea if you had checked it before you posted it. If you take a quick look now at what you posted, you will see a simple typographical error which leads to a wrong answer... and then you told the reader not to accept any different answer.

This is the second time in about a week that I have made a post asking you to stop the arrogance and simply supply the good responses you are capable of providing.

It's highly unpleasant to see frequent tutors on this forum proclaiming that their responses are the only ones that a reader should pay attention to. There are many different ways to solve most problems; and different students will find different solution methods more to their liking. You are doing a disservice to the reader when you tell them that they should ignore any responses to their questions than yours.


Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.

In his post, tutor @greenestamps mentioned my name disapprovingly, so I fill a necessity to answer.


Thanks to me and to tutor  @MathTherapy,  we reveal tens and tons errors and incorrect solutions in
the posts of other tutors.


So I think the visitors must be thankful to us for our checking and rechecking job.


Without this our job,  this forum would became a  GARBAGE  BOX  very soon and long time ago.


Not sure,  whether visitors would bring their problems to the forum then . . .


But with our efforts it still functions,  thanks to god . . .



            It is my comment to the post of @greenestamps . . .


@greenestamps,  it is possible that you forgot,  but I remember as I found excellent and brilliant solutions

to problems,  which you disclaimed as ill-suited.



Even me,  and even @MathTherapy,  and even YOU can make an error . . . ,
so do not gloat at each such occasion . . .




RELATED QUESTIONS

Greetings, I am an older adult who has always struggled with basic algebra. I am not... (answered by stanbon)
Hi my name is Baylee and I really need help in math and I was wondering if you can help... (answered by Alan3354)
Hi There I am new to math.... and need help please... to solve how to simplify the... (answered by math_helper)
Hello, I am stuck on a problem and I need some help please. The problem is the... (answered by Alan3354)
Hello. I would like some help solving this word problem. I'm really stuck. Can someone... (answered by josgarithmetic)
Hi, I was wondering if someone could help me with this problem. I am told to expressthe... (answered by mathie123)
Hi, I have a homework problem that states, " The cosine of some angle (theta), which is... (answered by stanbon)
Hi, I was wondering if you can help me with the following problem. I need to write the (answered by jim_thompson5910)
Hi there, I was just wondering if i could please have some help with answering a tafe... (answered by Boreal)