Question

For the following problem, complete the table to the right Then solve the problem using a system of equations. Liters of solution Liters of pure dye A 20% dye solution is to be mixed with a 90% dye solution to get 140 L of a 50% solution. How many liters of the 20% and 90% solutions will be needed? Percent (as a decimal) 0.20 0.90 0.50 Y у 140 Complete the table below. Liters of solution Percent (as a decimal) Liters of pure dye 0.20 y 0.90 140 0.50 (Simplify your answers. Use integers or decimals for any numbers in the expression) To get 140 L of a 50% solution, liters of the 20% solution and liters of the 90% solution are needed.see attached picture.

Answer 1

20% is 20/100 = 0.2

90% is 90/100 = 0.9

Let,

x liters of 20% solution needed

y liters of 90% solution needed

Thus, we can say:

0.2 of x SUM with 0.9 y GIVES up 140 Liters of 50% solution

We can write:

$0.2x+0.9y=0.5(140)$

Also, we know we will have a total of 140 liters, so we can say:

x + y = 140

or

y = 140 - x

Now, substituting this into 1st and solving, we get:

$\begin{gathered} 0.2x+0.9y=0.5(140) \\ 0.2x+0.9(140-x)=0.5(140) \\ 0.2x+126-0.9x=70 \\ 126-70=0.9x-0.2x \\ 56=0.7x \\ x=\frac{56}{0.7} \\ x=80 \end{gathered}$

We know y = 140 - x, so

y = 140 - 80 = 60

Thus,

x = 80

y = 60

so,

we need 80 liters of 20% solution and 60 liters of 90% solution