Aptitude:Mixture and Alligation

(6) A jeweler has gold of 18 carats and another of 24 carats. How many grams of each type of gold must be mixed to create 50 grams of a blend that is 20 carats?

• (a) 20 grams of 18 carats gold and 30 grams of 24 carats gold
• (b) 25 grams of 18 carats gold and 25 grams of 24 carats gold
• (c) 30 grams of 18 carats gold and 20 grams of 24 carats gold
• (d) 35 grams of 18 carats gold and 15 grams of 24 carats gold

Answer

Correct Answer: (b) 25 grams of 18 carats gold and 25 grams of 24 carats gold
Let \( z \) grams be the amount of 18 carats gold and \( 50 – z \) grams be the amount of 24 carats gold. The equation based on the carat value is:
<br /> 18z + 24(50 - z) = 20 \times 50 \\<br /> 18z + 1200 - 24z = 1000 \\<br /> -6z = -200 \\<br /> z = 25<br />
Hence, 25 grams of 18 carats gold and 25 grams of 24 carats gold are needed.

(7) A fruit seller has apples of two varieties, one costing $0.60 each and the other costing $0.80 each. How many apples of each variety must be mixed to create a dozen apples costing $0.70 each?

• (a) 3 apples of $0.60 variety and 9 apples of $0.80 variety
• (b) 6 apples of $0.60 variety and 6 apples of $0.80 variety
• (c) 9 apples of $0.60 variety and 3 apples of $0.80 variety
• (d) 8 apples of $0.60 variety and 4 apples of $0.80 variety

Answer

Correct Answer: (b) 6 apples of $0.60 variety and 6 apples of $0.80 variety
Let \( w \) apples be the amount of $0.60 variety and \( 12 – w \) apples be the amount of $0.80 variety. The equation based on the cost is:
<br /> 0.60w + 0.80(12 - w) = 0.70 \times 12 \\<br /> 0.60w + 9.60 - 0.80w = 8.40 \\<br /> -0.20w = -1.20 \\<br /> w = 6<br />
Hence, 6 apples of $0.60 variety and 6 apples of $0.80 variety are needed.

(8) A motorist has two types of fuel, one with a mileage of 20 miles per gallon (mpg) and the other with a mileage of 30 mpg. How many gallons of each type of fuel must be mixed to drive 100 miles with an average mileage of 25 mpg?

• (a) 20 gallons of 20 mpg fuel and 5 gallons of 30 mpg fuel
• (b) 10 gallons of 20 mpg fuel and 10 gallons of 30 mpg fuel
• (c) 15 gallons of 20 mpg fuel and 5 gallons of 30 mpg fuel
• (d) 25 gallons of 20 mpg fuel and 2 gallons of 30 mpg fuel

Answer

Correct Answer: (b) 10 gallons of 20 mpg fuel and 10 gallons of 30 mpg fuel
Let \( u \) gallons be the amount of 20 mpg fuel and \( 100 – u \) gallons be the amount of 30 mpg fuel. The equation based on the mileage is:
<br /> 20u + 30(100 - u) = 25 \times 100 \\<br /> 20u + 3000 - 30u = 2500 \\<br /> -10u = -500 \\<br /> u = 50<br />
Hence, 10 gallons of 20 mpg fuel and 10 gallons of 30 mpg fuel are needed.

(9) A painter has two types of paint, one that covers an area of 12 square meters per liter and another that covers an area of 18 square meters per liter. How many liters of each type of paint must be mixed to paint a wall of 20 square meters with an average coverage of 15 square meters per liter?

• (a) 3 liters of 12 sq meters/liter paint and 2 liters of 18 sq meters/liter paint
• (b) 4 liters of 12 sq meters/liter paint and 1 liter of 18 sq meters/liter paint
• (c) 2 liters of 12 sq meters/liter paint and 3 liters of 18 sq meters/liter paint
• (d) 5 liters of 12 sq meters/liter paint and 1 liter of 18 sq meters/liter paint

Answer

Correct Answer: (b) 4 liters of 12 sq meters/liter paint and 1 liter of 18 sq meters/liter paint
Let \( v \) liters be the amount of 12 sq meters/liter paint and \( 20 – v \) liters be the amount of 18 sq meters/liter paint. The equation based on the coverage is:
<br /> 12v + 18(20 - v) = 15 \times 20 \\<br /> 12v + 360 - 18v = 300 \\<br /> -6v = -60 \\<br /> v = 10<br />
Hence, 4 liters of 12 sq meters/liter paint and 1 liter of 18 sq meters/liter paint are needed.

(10) A farmer has two types of fertilizer: one with 10% nitrogen and another with 20% nitrogen. How many kilograms of each type of fertilizer should be mixed to get 15 kilograms of fertilizer with 15% nitrogen content?

• (a) 7.5 kg of 10% and 7.5 kg of 20%
• (b) 5 kg of 10% and 10 kg of 20%
• (c) 6 kg of 10% and 9 kg of 20%
• (d) 8 kg of 10% and 7 kg of 20%

Answer

Correct Answer: (c) 6 kg of 10% and 9 kg of 20%
Let \( x \) kilograms be the amount of 10% fertilizer and \( 15 – x \) kilograms be the amount of 20% fertilizer. The equation based on nitrogen content is:
<br /> 0.10x + 0.20(15 - x) = 0.15 \times 15 \\<br /> 0.10x + 3 - 0.20x = 2.25 \\<br /> -0.10x = -0.75 \\<br /> x = 7.5<br />
Therefore, 6 kilograms of 10% fertilizer and 9 kilograms of 20% fertilizer are needed.