Algebra Points: Some 40% acid solution is to be mixed with some 80% solution to get 200 L of a 50% solution.?

Question

How many liters of (((EACH))) should be used?

Best answer will be chosen tommorrow!

Answer 1

Mix 150L of the 40% solution with 50L of the 80% solution and you will get 200L of a 50% solution.

150*.4 + 50*.8 = 200*.5

Answer 2

Set up the equation as:
let x be the amount of 40% solution, so the amount of the 50% solution is 200-x (want 200L total)

.40x + .80(200-x) = .50 (200)

solve for x and get 150, so need 150 L of the 40% solution and 50 L of the 80 % solution.

Answer 3

Let there be X lt of 40% soln and Y lt of 80% soln. With the given information we can derive following equations:

1. X+Y = 200 (resulting volume of both soln = 200) and
2. 40*X/100 + 80*y/100 = 50*200/100 (ie X lt of 40% soln + Y lt of 80% soln results into 200 Lt of 50% soln)

Simplyfing the second equation we get X+2Y = 250 (just divide by 4)


Now solving these two equations simultaneously

X +2Y = 250
X+ Y = 200 (multiply this equation by -1 and add to first)

We get Y = 50, putting Y = 50 in X+Y = 200 we get X = 150

Hence 150 Lt of 40% soln and 50 Lt 0f 80% soln should be mixed.

Answer 4

You have two equations given.

Let x & y be the unkown no. of liters.

So from the first information

xl+yl = 200l

From the information given about the solution strengths

0.4*x + 0.8y = 0.5*200

0.4x + 0.8y = 100

4x+8y=1000

x+2y=250

x=250-2y

Then substitute x in your first equation
x + y = 200

250-2y+y=200

250 – y =200

y=50 litres

Then calculate x by substituting y

x+y=200
x+50 =200


x=150 litres

Answer 5

130L of the 40% acid
60L of the 80% solution