A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 80% . He wants to use all three solutions to obtain a mixture of 54 liters containing 45% acid, using 2 times as much of the 80% solution as the 35% solution. How many liters of each solution should be used? The chemist should use liters of 20% solution, liters of 35% solution, and liters of 80% solution.

Elisabeth Esparza

Elisabeth Esparza

Answered question

2022-07-29

A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 80% . He wants to use all three solutions to obtain a mixture of 54 liters containing 45% acid, using 2 times as much of the 80% solution as the 35% solution. How many liters of each solution should be used? The chemist should use liters of 20% solution, liters of 35% solution, and liters of 80% solution.

Answer & Explanation

umshikepl

umshikepl

Beginner2022-07-30Added 11 answers

Step 1
First acid = 20 %
Second acid = 35 %
Third acid = 80 %
x + y + z = 4 ( 1 )
and 20 x + 35 y + 80 z = 4 × 45 ( 2 )
z = 2 y ( 3 )
2 y z = 0
Step 2
solve 1 and 2 and 3 we get
x = 2 y   l i t ( 20 %   s o l u t i o n )
y = 10   l i t   ( 35 %   s o l u t i o n )
z = 20   l i t ( 80 %   s o l u t i o n )

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