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A painter mixes two batches of paint to create a custom color. Batch A has a gloss finish concentration of G% by volume. Batch B has a gloss finish concentration of H% by volume. The painter combines M liters of Batch A with N liters of Batch B. What is the gloss finish percentage, by volume, of the resulting mixture?
(1) The ratio of G to H is 7 to 5.
(2) The ratio of M to N is 3 to 4.
(1) The ratio of G to H is 7 to 5.
(2) The ratio of M to N is 3 to 4.
13 Jan 2026, 09:28
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Official Solution:
A painter mixes two batches of paint to create a custom color. Batch A has a gloss finish concentration of G% by volume. Batch B has a gloss finish concentration of H% by volume. The painter combines M liters of Batch A with N liters of Batch B. What is the gloss finish percentage, by volume, of the resulting mixture?
(1) The ratio of G to H is 7 to 5.
We know \(G:H = 7:5\), so \(G = 7k\) and \(H = 5k\). The resulting mixture percentage still depends on the actual value of \(k\) and also on how much of each batch is used. Not sufficient.
(2) The ratio of M to N is 3 to 4.
We know \(M:N = 3:4\). The resulting mixture percentage would be \(\frac{3G + 4H}{7}\), which still depends on the actual values of G and H. Not sufficient.
(1)+(2) Substitute \(G = 7k\) and \(H = 5k\) into \(\frac{3G + 4H}{7}\). We get \(\frac{3 * 7k + 4 * 5k}{7} = \frac{41k}{7}\). Different values of \(k\) give different mixture percentages. Not sufficient.
Answer: E
A painter mixes two batches of paint to create a custom color. Batch A has a gloss finish concentration of G% by volume. Batch B has a gloss finish concentration of H% by volume. The painter combines M liters of Batch A with N liters of Batch B. What is the gloss finish percentage, by volume, of the resulting mixture?
(1) The ratio of G to H is 7 to 5.
We know \(G:H = 7:5\), so \(G = 7k\) and \(H = 5k\). The resulting mixture percentage still depends on the actual value of \(k\) and also on how much of each batch is used. Not sufficient.
(2) The ratio of M to N is 3 to 4.
We know \(M:N = 3:4\). The resulting mixture percentage would be \(\frac{3G + 4H}{7}\), which still depends on the actual values of G and H. Not sufficient.
(1)+(2) Substitute \(G = 7k\) and \(H = 5k\) into \(\frac{3G + 4H}{7}\). We get \(\frac{3 * 7k + 4 * 5k}{7} = \frac{41k}{7}\). Different values of \(k\) give different mixture percentages. Not sufficient.
Answer: E
