Answer this Algebra problem please!?

Question

a chemist wants to make 2 L of an 8% acid solution by mixing a 10% acid solution and a 5 % acid solution. how many liters of the solution should the chemist use?

Answer 1

First you set up an equation showing the mixture using the percents as whole numbers (you can only use this by multiplying both sides by 100 to get rid of the decimals. I'm leaving the L only in the first step)

x L *10 + (2-x) L *5=2 L * 8

Then simplify:
10x + 10-5x=16
5x+10=16
5x=6
x=1 1/5 L of the 10% solution.

Then subtract from the 2 L to find the 5% solution.
you will need:

1 1/5 L or 1.2 L of the 10% solution

and

4/5 L or 0.8L of the 5% solution

Answer 2

The average strength of the solution to be used to create a new 8% strength solution is = (10% + 5%)/2 = 7.5%

The desired strength of the proposed new solution is 8%. Its relative strength against the component solution is = 8/7.5 = 1.0666666666 (using 10 decimal places).

The new proposed solution's volume is liters. Dividing this by 1.0666666666 gives us .0.9375 liters which is the portion of
the volume of the new solution that will contribute a 7.5% strength to the new solution. The other portion that will contribute an 8% strength is therefore = 2 L - 0.9375 L = 1.0625 L

Answer to Problem: 0.9375 L of the solution (with strength of 7.5%) should therefore be used to produce a 2L 8% solution.

Proof:

New Strength = [7.5%(.9375 L) + 8% (1.0625 L)]/2

= (7.03% + 8.5%)/2 = 7.765 (or approximately 8% strength).



The

Answer 3

I would multiply both sides by X. This would give me x^2 - 8 = 2x. I would then set everything equal to zero. x^2 -2x -8 = 0. You can then factor, graph, or use the quadratic equation to solve. I would probably factor or graph because those are the easiest for me. All will work just fine though. Since everyone else has already given an answer, I will say that there are two answers: x = 4 and -2.

Answer 4

We need 2 liters of 8% solution. If X is the amount of 5% acid solution, and Y is the amount of 10% acid solution, then:

x + y = 2 liters
y = 2 - x liters [1]

Now, let focus on the acid of each amount of solution:

5% solution --> x*0.05 [2]
10% solution --> y*0.1 = (2-x)*0.1 [3]

We need 2 liters of 8% solution, so the amount of acid is

8%solution --> 2 l * 0.08 = 0.16 l [4]

if we add the acid in both amount of solution, we get the 0,16 l of acid on the 2 liters of solution, so [2] + [3] = [4]:

0.05 x + [0.1 (2-x)] = 0.16
0.05 x + [0.2 - 0.1x] = 0.16
-0.05 x + 0.2 = 0.16
-0.05 x = 0.16 - 0.2
-0.05 x = -0.04
x = 0.04/0.05
x = 0.8 l of 5% acid solution

[1] y = 2 - x --> y = 2 - 0.8
y = 1.2 l of 10% acid solution

Answer 5

Let x be the number of litres of 10% acid solution
Let y be the number of litres of 5% acid solution
Now firstly x+y=2
And of course 0.1x+0.05y=0.08(2)
Rearrange the first equation to get x=2-y
Now substitute
0.1(2-y)+0.05y=0.16
0.2-0.1y+0.05y=0.16
-0.05y=-0.04
y=0.8
x=1.2
Answer 1.2 litres of 10% solution and 0.8 litres of the 5% solution.

Answer 6

x= amount of 10% acid
2- x = amount of 5% acid
.1x+.05(2-x) = .08*2
.1x +.1 -.05x = .16
.05 x = .06
x = 1.2 L of 10% acid
2-x =.8L of 5% acid