Answer:
210L volume of Mixture A with 50% acid part
70L volume of Mixture B with 10% acid part
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