80% and 90% pure acid solutions are mixed to obtain 20 litres of 87% pure acid solution. Find the quantity (in litres) of 80% pure acid solution taken to form the mixture.

Question

Prepp Team · Accepted Answer

Calculating Quantity of Acid Solution in a Mixture

This problem involves mixing two acid solutions of different concentrations to obtain a final solution of a specific concentration and volume. We need to determine the quantity of the less concentrated acid solution used in the mixture.

Let's break down the problem and set up equations based on the given information about the acid solutions and their mixture.

Setting up Variables and Equations for Acid Mixture

We are mixing an 80% pure acid solution and a 90% pure acid solution to get 20 litres of an 87% pure acid solution.

Let \(x\) be the quantity (in litres) of the 80% pure acid solution.
Let \(y\) be the quantity (in litres) of the 90% pure acid solution.

The total volume of the mixture is 20 litres. So, the first equation is based on the total quantity:

\[x + y = 20 \quad (Equation \, 1)\]

The amount of pure acid in each solution is the percentage purity multiplied by the quantity. The total amount of pure acid in the mixture must equal the total volume multiplied by the final percentage purity.

Amount of acid in \(x\) litres of 80% solution = \(0.80x\) litres.
Amount of acid in \(y\) litres of 90% solution = \(0.90y\) litres.
Total amount of acid in 20 litres of 87% solution = \(0.87 \times 20\) litres.

So, the second equation is based on the total amount of pure acid:

\[0.80x + 0.90y = 0.87 \times 20\]

\[0.80x + 0.90y = 17.4 \quad (Equation \, 2)\]

Solving the System of Equations

Now we have a system of two linear equations with two variables:

\(x + y = 20\)
\(0.80x + 0.90y = 17.4\)

We want to find the value of \(x\) (the quantity of 80% pure acid solution). We can solve this system using substitution or elimination. Let's use substitution.

From Equation 1, we can express \(y\) in terms of \(x\):

\[y = 20 - x\]

Substitute this expression for \(y\) into Equation 2:

\[0.80x + 0.90(20 - x) = 17.4\]

Distribute 0.90 into the parenthetical term:

\[0.80x + (0.90 \times 20) - (0.90 \times x) = 17.4\]

\[0.80x + 18 - 0.90x = 17.4\]

Combine the terms with \(x\):

\[(0.80 - 0.90)x + 18 = 17.4\]

\[-0.10x + 18 = 17.4\]

Now, isolate the term with \(x\) by subtracting 18 from both sides:

\[-0.10x = 17.4 - 18\]

\[-0.10x = -0.6\]

Finally, solve for \(x\) by dividing both sides by -0.10:

\[x = \frac{-0.6}{-0.10}\]

\[x = 6\]

So, the quantity of the 80% pure acid solution taken is 6 litres.

We can also find the quantity of the 90% solution, \(y\), using \(y = 20 - x\):

\[y = 20 - 6 = 14\]

This means 14 litres of the 90% pure acid solution were used.

Verification of the Acid Mixture

Let's verify if mixing 6 litres of 80% solution and 14 litres of 90% solution gives 20 litres of 87% solution:

Total volume = 6 litres + 14 litres = 20 litres (Correct)
Amount of acid from 80% solution = \(0.80 \times 6 = 4.8\) litres.
Amount of acid from 90% solution = \(0.90 \times 14 = 12.6\) litres.
Total amount of acid in mixture = \(4.8 + 12.6 = 17.4\) litres.
Percentage purity of mixture = \(\frac{\text{Total acid}}{\text{Total volume}} \times 100 = \frac{17.4}{20} \times 100\).
\(\frac{17.4}{20} \times 100 = 0.87 \times 100 = 87\%\) (Correct)

The calculations are verified, and the quantity of the 80% pure acid solution is indeed 6 litres.

Solution Type	Quantity (Litres)	Percentage Purity	Amount of Acid (Litres)
80% pure acid	\(x = 6\)	80% (0.80)	\(0.80 \times 6 = 4.8\)
90% pure acid	\(y = 14\)	90% (0.90)	\(0.90 \times 14 = 12.6\)
Mixture	\(x + y = 20\)	87% (0.87)	\(4.8 + 12.6 = 17.4\) (\(0.87 \times 20\))

Revision Table: Acid Solution Mixture Concepts

Understanding how to solve mixture problems involving percentages or concentrations is important for quantitative aptitude.

Concept	Explanation	Application in Problem
Percentage Purity / Concentration	Represents the proportion of the pure substance (acid) in the total solution volume, usually expressed as a percentage.	Used to calculate the amount of pure acid in a given quantity of solution (e.g., 80% purity means 0.80 litres of acid per litre of solution).
Total Volume Equation	The sum of the volumes of the individual components equals the total volume of the mixture.	\(x + y = 20\), where \(x\) and \(y\) are volumes of the two solutions and 20 is the total mixture volume.
Total Amount of Pure Substance Equation	The sum of the amounts of the pure substance from each component equals the total amount of the pure substance in the mixture.	\(0.80x + 0.90y = 0.87 \times 20\), equating the total acid from initial solutions to the total acid in the final mixture.
Solving Systems of Equations	Methods like substitution or elimination are used to find unknown variables when multiple equations relate them.	We used substitution to find \(x\) (quantity of 80% solution) after setting up the two equations.

Additional Information on Mixture Problems

Mixture problems often involve combining substances with different concentrations, prices, or properties to achieve a desired outcome for the mixture. They can be solved using algebraic equations as demonstrated above, or sometimes using alternative methods like the alligation method (for two components being mixed).

Key steps usually involve:

Identifying the quantities and concentrations/properties of the components being mixed.
Identifying the total quantity and desired concentration/property of the final mixture.
Setting up equations based on total quantity/volume and total amount of the key substance/value.
Solving the system of equations to find the unknown quantities.

Always check your answer by plugging the calculated quantities back into the original problem conditions to ensure they satisfy all requirements.

Prepp Team · Answer

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Prepp Team · Answer

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