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06 Aug 2026, 10:58
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Alloy A is 30 percent copper by weight, and Alloy B is 12 percent copper by weight. If Alloy A and Alloy B are combined to produce an alloy that is 18 percent copper by weight, how many kilograms of Alloy B are used?
(1) To produce the final alloy, 40 kilograms of Alloy A are used.
(2) The final alloy weighs 120 kilograms.
(1) To produce the final alloy, 40 kilograms of Alloy A are used.
(2) The final alloy weighs 120 kilograms.
06 Aug 2026, 10:58
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Official Solution:
Alloy A is 30 percent copper by weight, and Alloy B is 12 percent copper by weight. If Alloy A and Alloy B are combined to produce an alloy that is 18 percent copper by weight, how many kilograms of Alloy B are used?
(1) To produce the final alloy, 40 kilograms of Alloy A are used.
Let \(b\) be the number of kilograms of Alloy B used.
Copper from Alloy A \(= 30\%\) of 40 \(= 12\) kilograms.
Copper from Alloy B \(= 12\%\) of \(b = 0.12b\).
Since the final alloy is \(18\%\) copper:
\(12 + 0.12b = 0.18(40 + b)\)
\(12 + 0.12b = 7.2 + 0.18b\)
\(4.8 = 0.06b\)
\(b = 80\)
Sufficient.
(2) The final alloy weighs 120 kilograms.
Let \(a\) be the number of kilograms of Alloy A used, and let \(b\) be the number of kilograms of Alloy B used.
Since the final alloy is \(18\%\) copper:
\(0.30a + 0.12b = 0.18(a + b)\)
\(0.30a + 0.12b = 0.18a + 0.18b\)
\(0.12a = 0.06b\)
\(b = 2a\)
So Alloy B must be twice Alloy A. Since the final alloy weighs 120 kilograms, Alloy B must be \(\frac{2}{3}\) of the final alloy:
\(\frac{2}{3} * 120 = 80\) kilograms
Sufficient.
Answer: D
Alloy A is 30 percent copper by weight, and Alloy B is 12 percent copper by weight. If Alloy A and Alloy B are combined to produce an alloy that is 18 percent copper by weight, how many kilograms of Alloy B are used?
(1) To produce the final alloy, 40 kilograms of Alloy A are used.
Let \(b\) be the number of kilograms of Alloy B used.
Copper from Alloy A \(= 30\%\) of 40 \(= 12\) kilograms.
Copper from Alloy B \(= 12\%\) of \(b = 0.12b\).
Since the final alloy is \(18\%\) copper:
\(12 + 0.12b = 0.18(40 + b)\)
\(12 + 0.12b = 7.2 + 0.18b\)
\(4.8 = 0.06b\)
\(b = 80\)
Sufficient.
(2) The final alloy weighs 120 kilograms.
Let \(a\) be the number of kilograms of Alloy A used, and let \(b\) be the number of kilograms of Alloy B used.
Since the final alloy is \(18\%\) copper:
\(0.30a + 0.12b = 0.18(a + b)\)
\(0.30a + 0.12b = 0.18a + 0.18b\)
\(0.12a = 0.06b\)
\(b = 2a\)
So Alloy B must be twice Alloy A. Since the final alloy weighs 120 kilograms, Alloy B must be \(\frac{2}{3}\) of the final alloy:
\(\frac{2}{3} * 120 = 80\) kilograms
Sufficient.
Answer: D
