Mixture problems are a common question type on the GMAT exam, particularly within the Quantitative section. These questions involve combining different substances or items with varying properties (such as concentration, cost, or quality) to achieve a desired composition or value in the resulting mixture. Mixture problems test a candidate’s ability to understand and manipulate weighted averages and ratios.
These problems often require setting up and solving algebraic equations or employing logical reasoning to deduce the correct answer. They are crucial because they simulate real-world scenarios in business and economics, such as blending different types of coffee beans to achieve a specific flavor profile or cost, or mixing paints to get a particular shade.
Example 1: Paint Mixture Problem
Question: A painter needs to create a specific shade of purple by mixing red and blue paint. The painter mixes part of a 24-liter red paint container with part of a 36-liter blue paint container to make 48 liters of purple paint. The resulting mixture is 25% red paint. How many liters of blue paint did the painter use?
Solution:
Let ( x ) be the liters of red paint used, and ( y ) be the liters of blue paint used. We know:
[ x + y = 48 ]
[ \frac{x}{48} = 0.25 ] (since 25% of the mixture is red paint)
From the second equation, ( x = 12 ) liters (25% of 48 liters). Substituting ( x = 12 ) in the first equation:
[ 12 + y = 48 ]
[ y = 36 ]
Answer: The painter used 36 liters of blue paint.
Example 2: Coffee Blend Problem
Question: A coffee shop owner blends two types of coffee beans, Bean X costing $30 per kilogram and Bean Y costing $45 per kilogram, aiming to produce a 50-kilogram blend that costs $38 per kilogram. How many kilograms of each type of bean must the owner use?
Solution:
Let ( x ) be the kilograms of Bean X and ( y ) be the kilograms of Bean Y. We know:
[ x + y = 50 ]
[ 30x + 45y = 38 \times 50 ]
[ 30x + 45y = 1900 ]
Solving the equations simultaneously, we simplify the cost equation:
[ 6x + 9y = 380 ]
[ 2x + 3y = 126.67 ] (dividing all terms by 3 for simplicity)
From ( x + y = 50 ), express ( y ) in terms of ( x ):
[ y = 50 – x ]
Substitute in the simplified cost equation:
[ 2x + 3(50 – x) = 126.67 ]
[ 2x + 150 – 3x = 126.67 ]
[ -x = -23.33 ]
[ x = 23.33 ] (approximately)
[ y = 50 – 23.33 = 26.67 ] (approximately)
Answer: The owner must use approximately 23.33 kilograms of Bean X and 26.67 kilograms of Bean Y.
These examples illustrate the analytical thinking and algebraic manipulation required to solve mixture problems on the GMAT. Mastery of these problems involves practice and a solid understanding of basic algebraic concepts and the principles of mixing ratios and concentrations.
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