How do you find the equation to this mixture problem?

Question

A 16% acid solution is to be mixed with a 21% acid solution. How many liters of

each solution should be used to make 80 liters of a 20% acid solution?

what is the equation to this problem and may you please show me the steps to how you got it. thank you!!

Answer 1

It's easy.

You can use a system of equations.

First step: The total amount of acid solutions has to be equal to 80 liters.

It gives us the first equation: x + y = 80,

where x  is the needed quantity of the 16% solution and y that for the 21% one.

Second step: The final solution has to have 20% acid and the total amount of acid  in this solution has to equal the amount in x plus y.

-> x*0.16 + y*0.21 = 80*0.2

Third step: You can introduce the first equation in the second by replacing y by 80-x: x + y = 80 -> y = 80 -x

By solving the obtained equation you find that you need: x = 16 liters and y = 80 -16= 64 liters.

Answer 2

Your goal is to solve for two unknowns ( the volume of the first solutoin and the volume of the second solution)

Create equations.

x is the volume of solution 1 in liters, y is the volume of solution 2 in liters

x + y = 80

16x+21y=20(80)=1600

Somehow, you need to use these two equations to get an equation with only one variable.

subtract 16 times the first equation from the second equation.

    16x + 21y = 1600

    16x + 16y = 1280

---------------------------------

      0x + 5y = 320

y = 64

Now, plug that into one of the original equations.

x + 64 = 80

x = 16

there is your answer.

you need 16 liters of the first solution and 64 liters of the second solution.

Answer 3

16% acid solution means that for every 100L of water, there's 16L acid.

21% acid solution means that for every 100L of water, there's 21L acid.

We want a 20% acid solution, which means for ever 100L water there's 20L acid. But we only want 80L.

So let's call the L of the 16% solution "S" and the L of the 21% solution "T". We know that:

16%*S+21%*T = 20%*80=16

And:

S+T=80

So now we just solve the system:

S+T=80

S=80-T

16%*(80-T)+21%*T=16

It's easier to convert the %'s to decimals first (divide by 100), so we get:

0.16*(80-T) + 0.21*T=16

12.8 - 0.16T + 0.21T=16

12.8 + 0.05T = 16

0.05T = 3.2

T = 64

S+T=80

S+64=80

S=16

So we need 16L of the 16% solution and 64L of the 21% solution.

-IMP ;) :)  

Answer 4

Let x = liters of 16% solution, y = liters of 21% solution

1) x + y = 80 {total solution)

y = 80 - x

Add the solutions

2) .16x + .21y = .20(80)

.16x + .21(80 - x) = 16 {substitute for y from equation 1}

.16x + 16.8 - .21x = 16

-.05x = -.8

x = 16

y = 80 - 16 = 64

check

.16(16) + .21(64) = .20(80)

2.56 + 13.44 = 16

16 = 16

Answer 5

No. of liters of 16% acid solution (x):

0.16x + 0.21(80 - x) = 0.2(80)

16x + 21(80 - x) = 20(80)

16x + 1,680 - 21x = 1,600

21x - 16x = 1,680 - 1,600

5x = 80

x = 16

No. of liters of 21% acid solution:

= 80 - 16

= 64

Answer: 16% acid solution, 16 liters; 21% acid solution, 64 liters

Proof (new mixture is 20% acid solution, 80 liters):

= 100%([0.16{16 liters} + 0.21{64 liters}]/80 liters)

= 100%([2.56 liters + 13.44 liters]/80 liters)

= 100%(16 liters/80 liters)

= (100%/5)([16/16]/[80/{5 * 16}])

= 20%(1/[80/80])

= 20%(1/1)

= 20%