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Question

Difficulty: MediumSolving Linear Equations

In a chemistry laboratory, a beaker contains a mixture of water and acid. The volume of water in the beaker is 33 liters more than twice the volume of acid. After 55 liters of water are added to the beaker, the ratio of the volume of water to the volume of acid is 55 to 22. If no acid was added or removed, what was the initial volume of water, in liters, in the beaker?

  1. A
    5
  2. B
    16
  3. C
    19
  4. 35Answer
  5. E
    50

Answer

35
The correct answer of 35 liters is found by setting up the linear equation where the initial volume of acid is aa and the initial volume of water is 2a+32a + 3. Adding 5 liters of water gives 2a+82a + 8 liters of water. Setting the ratio of water to acid to 52\frac{5}{2} and solving gives a=16a = 16. Finally, substituting a=16a = 16 into the expression for the initial volume of water (2a+32a + 3) yields 35 liters.

Step-by-Step Solution

1
Define variables for the initial volumes of acid and water based on the given relationships.
Let aa be the initial volume of acid in liters. The initial volume of water is 2a+32a + 3 liters.
This translates the statement 'the volume of water is 3 liters more than twice the volume of acid' into algebraic expressions.
2
Set up a linear equation representing the ratio after adding 5 liters of water.
The new volume of water is (2a+3)+5=2a+8(2a + 3) + 5 = 2a + 8 liters. The ratio of water to acid is 2a+8a=52\frac{2a + 8}{a} = \frac{5}{2}.
This uses the new state of the mixture to form an equation that can be solved for aa.
3
Solve the linear equation for aa by cross-multiplying and isolating the variable.
2(2a+8)=5a  ⟹  4a+16=5a  ⟹  a=162(2a + 8) = 5a \implies 4a + 16 = 5a \implies a = 16.
Cross-multiplication removes the fractions and allows for standard term isolation.
4
Calculate the initial volume of water.
Initial water volume = 2a+3=2(16)+3=352a + 3 = 2(16) + 3 = 35 liters.
The question asks for the initial volume of water, which is represented by 2a+32a + 3, not the volume of acid aa.

Key Concept

Solving linear equations in one variable derived from word problems.

Alternative Method

Instead of solving algebraically, one could test the answer choices. For example, testing 35 liters of water means the initial acid is (35−3)/2=16(35 - 3) / 2 = 16 liters. Adding 5 liters of water gives 40 liters of water. The ratio of water to acid is 40:1640 : 16, which simplifies to 5:25 : 2. This confirms 35 is correct.
Estimated Time:1m 30s
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Topics

Solving Linear Equations

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