A water purification plant processes raw water through three successive filtration stages:… — ACT

A water purification plant processes raw water through three successive filtration stages: Stage A, Stage B, and Stage C. Stage A removes $\frac{3}{8}$ of the impurities present in the raw water. Stage B removes $40\%$ of the remaining impurities. Stage C removes $0.75$ of the impurities that remain after Stage B. If $4.5$ kilograms of impurities are successfully removed in Stage C, how many kilograms of impurities were originally in the raw water?

Answer: 16 kg

Answer

The original amount of impurities in the raw water was

16

kilograms.

The correct answer is

16

kg. To find this, we express the remaining impurities at each stage as a fraction of the initial amount

x

. After Stage A,

\frac{5}{8}x

remains. In Stage B,

40\%

is removed, meaning

60\%

\frac{5}{8}x

, or

\frac{3}{8}x

, remains. In Stage C,

0.75

(or

\frac{3}{4}

) of this remainder is removed, which is

\frac{3}{4} \times \frac{3}{8}x = \frac{9}{32}x

. Setting

\frac{9}{32}x = 4.5

gives

x = 16

Step-by-Step Solution

Represent the initial mass of impurities in the raw water with a variable.

Let

x

be the initial kilograms of impurities.

Establishing a variable allows setting up an equation to track the changes through each stage.

Calculate the remaining impurities after Stage A.

Impurities remaining =

x - \frac{3}{8}x = \frac{5}{8}x

kg.

Stage A removes

\frac{3}{8}

of the initial impurities, so

1 - \frac{3}{8} = \frac{5}{8}

of the impurities remain.

Calculate the impurities removed and remaining after Stage B.

Impurities removed in Stage B =

0.40 \times \frac{5}{8}x = \frac{1}{4}x

kg. Impurities remaining after Stage B = \frac{5}{8}x - \frac{1}{4}x = \frac{3}{8}x$ kg.

Stage B removes

40\%

of the impurities that remained after Stage A. Subtracting the removed portion from the starting amount for this stage yields the remaining fraction.

Express the impurities removed in Stage C.

Impurities removed in Stage C =

0.75 \times \frac{3}{8}x = \frac{9}{32}x

kg.

Stage C removes

0.75

(or

\frac{3}{4}

) of the impurities remaining after Stage B.

Equate the Stage C expression to the given value and solve for

x

\frac{9}{32}x = 4.5 \implies x = 4.5 \times \frac{32}{9} = 16

kg.

The problem states that

4.5

kg of impurities are removed in Stage C, so solving this equation yields the initial value.

Key Concept

Solving multi-step word problems involving successive applications of fractions, decimals, and percentages.

Alternative Method

Working backwards from the final stage can simplify the calculations. Since Stage C removes

0.75

of the impurities remaining after Stage B, and this amount equals

4.5

kg, the amount remaining after Stage B is

\frac{4.5}{0.75} = 6

kg. Since Stage B removes

40\%

of the impurities remaining after Stage A, the

6

kg represents

100\% - 40\% = 60\%

of the impurities remaining after Stage A. Thus, the amount remaining after Stage A is

\frac{6}{0.60} = 10

kg. Finally, since Stage A removes

\frac{3}{8}

of the original impurities, the

10

kg represents

1 - \frac{3}{8} = \frac{5}{8}

of the original impurities. The original amount is therefore

10 \times \frac{8}{5} = 16

kg.

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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