How many liters of a 50% acid solution must be mixed with a 10% acid solution to get 280 L of a 40% acid solution

210L volume of Mixture A with 50% acid part

70L volume of Mixture B with 10% acid part

Step-by-step explanation:

Let,

x be volume of Mixture A with 50% acid part

In mixture A, 0.5x is volume of acid part

y be volume of Mixture B with 10% acid part

In mixture B, 0.1x is volume of acid part

Now, Mixture A is mixed with Mixture B and Mixture C is created.

The total volume of Mixture C is

x+y=280

x=280-y (Equation 1)

In mixture C, 40% is acid part

Part of acid in Mixture C is 40% of 280L

=0.4X280

=112L

We know that,

Acid part of Mixture A + Acid part of Mixture B = Acid part of Mixture C

0.5x+0.1y=112

5x+y=1120 (Equation 2)

From equation 1 and equation 2,

5x+y=1120

5 (280-y) + y=1120

1400-5y+y=1120

280=4y

y=70L

Replacing value of y in any equation,

5x+y=1120

5x+70=1120

x=210L

Thus, 210L of a 50% acid solution must be mixed with 70L of 10% acid solution to get 280 L of a 40% acid solution