Problem 51
Question
How many grams of a \(50 \%\) -silver alloy should be mixed with a \(75 \%\) -silver alloy to obtain 40 grams of a \(60 \%\) -silver alloy?
Step-by-Step Solution
Verified Answer
Answer: 24 grams of 50% silver alloy and 16 grams of 75% silver alloy should be mixed to obtain 40 grams of a 60% silver alloy.
1Step 1: Define the variables
Let \(x\) represent the amount of the \(50 \%\) -silver alloy (in grams), and \(y\) represent the amount of the \(75 \%\) -silver alloy (in grams).
2Step 2: Create an equation to represent the total weight of the mixture
Since the final mixture needs to be 40 grams, we can write the equation for the total weight as:
\(x + y = 40\)
3Step 3: Create an equation to represent the silver content of the mixture
The final mixture needs to be \(60 \%\) silver, so we can write an equation for the silver content as:
\(0.5x + 0.75y = 0.6(40)\)
This equation represents the total silver content in the final mixture (which is \(60 \%\) of 40 grams).
4Step 4: Simplify the equations
Simplify the second equation:
\(0.5x + 0.75y = 24\)
5Step 5: Solve the system of equations
We can solve this system of equations using the substitution or elimination method. Let's use the elimination method.
1. Multiply the first equation by \(0.5\) to make the coefficients of x in both equations equal:
\(0.5(x + y) = 0.5(40)\)
So, we get:
\(0.5x + 0.5y = 20\)
2. Subtract the first equation from the second equation:
\((0.5x + 0.75y) - (0.5x + 0.5y) = 24 - 20\)
After simplifying, we get:
\(0.25y = 4\)
3. Solve for y by dividing both sides by 0.25:
\(y = \frac{4}{0.25}\)
Hence, \(y = 16\).
4. Substitute the value of y back into the first equation (x + y = 40):
\(x + 16 = 40\)
5. Solve for x:
\(x = 40 - 16\)
Therefore, \(x = 24\).
6Step 6: Conclusion
24 grams of \(50 \%\) -silver alloy should be mixed with 16 grams of \(75 \%\) -silver alloy to obtain 40 grams of a \(60 \%\) -silver alloy.
Key Concepts
Percentage in Mixture ProblemsSystem of Equations in Mixture ProblemsAlgebraic Reasoning in Solving Mixture Problems
Percentage in Mixture Problems
Understanding percentage is crucial when dealing with mixture problems like the one described above. A percentage represents a way to express a number as a fraction of 100. In the scenario provided, we are dealing with different alloys of silver, each with varying percentages.
Here is how you can interpret the percentages in the problem:
Here is how you can interpret the percentages in the problem:
- For the 50% silver alloy, 50% of the weight is pure silver.
- For the 75% silver alloy, 75% of the weight is pure silver.
- The goal is to create a final mixture that is 60% silver, which means 60% of its total weight is pure silver.
System of Equations in Mixture Problems
A system of equations can be an effective tool for solving problems that involve multiple unknowns or variables, like the mixture problem in the exercise. Here, we used two variables, \(x\) and \(y\), for the different silver alloys, and expressed our conditions in terms of equations.
The following are the primary equations we used:
The following are the primary equations we used:
- Total weight equation: \(x + y = 40\). This equation conveys that the sum of the two different alloys must be 40 grams.
- Silver content equation: \(0.5x + 0.75y = 24\). This equation signifies that the total silver content in the resultant mixture needs to be 60% of 40 grams.
Algebraic Reasoning in Solving Mixture Problems
Algebraic reasoning allows you to strategically set up and solve equations derived from the problem statement. In this exercise, algebraic reasoning was primarily used for creating equations and solving them logically.
Here's how you can approach it:
When given a word problem, it's often useful to:
Here's how you can approach it:
When given a word problem, it's often useful to:
- Identify what you are solving for and let those be your variables. For this problem, we chose \(x\) and \(y\) for the different types of alloys.
- Create equations based on quantities and their relationships, such as the total mass and the percentage compositions.
- Choose a method to solve these equations. Here, the elimination method was used by aligning coefficients to subtract one equation from another, simplifying the process.
Other exercises in this chapter
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