How many liters of a 10% saline solution must be added to to 4 liters of a 40% saline solution to obtain a 15% saline solution? a)20 l b)4 l c)2 l d)48 l plz explain the steps and how to get answer

It would take 20 liters of the 10% solution.

We will let x be the amount of 10% solution in our problem.  10% saline in x liters is 0.1x.  4 liters of a 40% solution would be 4*0.4 = 1.6.  We know that there are a total of x+4 liters in the final solution, and that it is 15% saline; that is 0.15(x+4).  Our equation is then:

0.1x + 0.4(4) = 0.15(x+4)
0.1x+1.6 = 0.15*x + 0.15*4
0.1x+1.6 = 0.15x+0.6

We cannot have a variable on both sides; we will subtract 0.1x from both sides:

0.1x + 1.6 - 0.1x = 0.15x + 0.6 - 0.1x
1.6 = 0.05x + 0.6

Subtract 0.6 from both sides:
1.6 - 0.6 = 0.05x + 0.6 - 0.6
1 = 0.05x

Divide both sides by 0.05:
1/0.05 = 0.05x/0.05
20 = x

x was the 10% solution, so there are 20 liters of it.

They lost 20 games hope this