A chemist has three different acid solutions, The first solution contains 20 % acid, the second contains 35 % acid, and the third contains 55%
acid.
She created 108 liters of a 40 % acid mixture, using all three solutions. The number of liters of
55 % solution used is 3 times the number of liters of 35 % solution used.
How many liters of each solution was used?
The chemist used
liters of 20% solution,
liters of 35 % solution, and
liters of 55 % solution.
Question Help: Message
instructor
Calculator

9514 1404 393

Step-by-step explanation:

Let a, b, c represent the quantities of 20%, 35%, and 55% solutions, respectively. Then for the given mix, we have ...

  a + b + c = 108 . . . . . total number of liters

  .20a +.35b +.55c = 0.40·108 . . . . . liters of acid

  0a - 3b + c = 0 . . . . . 55% solution was 3 times 35% solution

__

Solving these three equations by your favorite method gives ...

  (a, b, c) = (36, 18, 54)

The chemist used 36 liters of 20%, 18 liters of 35%, and 54 liters of 55%.

Comment on alternate approach

Since the ratio of 55% to 35% is 3 : 1, that mix will have an acid content of ...

  (3(.55) +1(.35))/4 = 0.50 = 50%

So, the final mix is equivalent to some quantity of 50% mix being added to some quantity of 20% mix. The fraction that is 50% mix* will be (40-20)/(50-20) = 2/3 of the total, or (2/3)(108 L) = 72 L.

Now we know that (1/3)(108 L) = 36 L of 20% solution is needed and (1/4)(72 L) = 18 L of 35% solution is needed. 3 times that is 54 L of 55% solution.

__

* After you work a few mixture problems, you can see the pattern. If we start with an equation for x = fraction of 50% to use with 20% to make 40%, we get ...

  x(50%) +(1 -x)(20%) = 40%

  x(50% -20%) = 40% -20% . . . . . subtract 1(20%) and factor out x

  x = (40% -20%)/(50% -20%)

The key here is to see where the numbers 20%, 40%, and 50% show up in the fraction.

`1234567890-=/*-qwertyuiop[]asdfghjkl; '\\zxcvbnm,./+ /

you forgot an image

step-by-step explanation: