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07 Aug 2026, 10:13
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A cosmetics lab prepared a liquid formula that consisted of base liquid, fragrance oil, and gel stabilizer. The weight of the base liquid in the formula was increased by \(\frac{1}{3}\) by adding more base liquid, while the weights of the fragrance oil and gel stabilizer remained unchanged. Before this increase, what was the ratio of the weight of the base liquid to the total weight of the formula?
(1) After the addition of base liquid, the ratio of the weight of fragrance oil to the total weight of the formula decreased from \(\frac{1}{5}\) to \(\frac{1}{6}\).
(2) After the addition of base liquid, the ratio of the weight of base liquid to the total weight of the formula became \(\frac{2}{3}\).
(1) After the addition of base liquid, the ratio of the weight of fragrance oil to the total weight of the formula decreased from \(\frac{1}{5}\) to \(\frac{1}{6}\).
(2) After the addition of base liquid, the ratio of the weight of base liquid to the total weight of the formula became \(\frac{2}{3}\).
07 Aug 2026, 10:14
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Official Solution:
A cosmetics lab prepared a liquid formula that consisted of base liquid, fragrance oil, and gel stabilizer. The weight of the base liquid in the formula was increased by \(\frac{1}{3}\) by adding more base liquid, while the weights of the fragrance oil and gel stabilizer remained unchanged. Before this increase, what was the ratio of the weight of the base liquid to the total weight of the formula?
Let \(B\) be the weight of the base liquid before the increase, and let \(T\) be the total weight of the formula before the increase.
The question asks for \(\frac{B}{T}\).
Since the weight of the base liquid was increased by \(\frac{1}{3}\), the weight of base liquid added was \(\frac{B}{3}\).
So the new total weight of the formula was:
\(T + \frac{B}{3}\)
(1) After the addition of base liquid, the ratio of the weight of fragrance oil to the total weight of the formula decreased from \(\frac{1}{5}\) to \(\frac{1}{6}\).
Let \(F\) be the weight of the fragrance oil. Since only base liquid was added, \(F\) did not change.
Before the addition:
\(\frac{F}{T} = \frac{1}{5}\)
\(F = \frac{T}{5}\)
After the addition:
\(\frac{F}{T + \frac{B}{3}} = \frac{1}{6}\)
\(F = \frac{T + \frac{B}{3}}{6}\)
Therefore:
\(\frac{T}{5} = \frac{T + \frac{B}{3}}{6}\)
\(\frac{6T}{5} = T + \frac{B}{3}\)
\(\frac{T}{5} = \frac{B}{3}\)
\(\frac{B}{T} = \frac{3}{5}\)
Sufficient.
(2) After the addition of base liquid, the ratio of the weight of base liquid to the total weight of the formula became \(\frac{2}{3}\).
After the addition, the weight of the base liquid became:
\(B + \frac{B}{3} = \frac{4B}{3}\)
The new total weight of the formula was:
\(T + \frac{B}{3}\)
So statement (2) gives:
\(\frac{\frac{4B}{3}}{T + \frac{B}{3}} = \frac{2}{3}\)
\(\frac{4B}{3T + B} = \frac{2}{3}\)
\(12B = 6T + 2B\)
\(10B = 6T\)
\(\frac{B}{T} = \frac{3}{5}\)
Sufficient.
Answer: D
A cosmetics lab prepared a liquid formula that consisted of base liquid, fragrance oil, and gel stabilizer. The weight of the base liquid in the formula was increased by \(\frac{1}{3}\) by adding more base liquid, while the weights of the fragrance oil and gel stabilizer remained unchanged. Before this increase, what was the ratio of the weight of the base liquid to the total weight of the formula?
Let \(B\) be the weight of the base liquid before the increase, and let \(T\) be the total weight of the formula before the increase.
The question asks for \(\frac{B}{T}\).
Since the weight of the base liquid was increased by \(\frac{1}{3}\), the weight of base liquid added was \(\frac{B}{3}\).
So the new total weight of the formula was:
\(T + \frac{B}{3}\)
(1) After the addition of base liquid, the ratio of the weight of fragrance oil to the total weight of the formula decreased from \(\frac{1}{5}\) to \(\frac{1}{6}\).
Let \(F\) be the weight of the fragrance oil. Since only base liquid was added, \(F\) did not change.
Before the addition:
\(\frac{F}{T} = \frac{1}{5}\)
\(F = \frac{T}{5}\)
After the addition:
\(\frac{F}{T + \frac{B}{3}} = \frac{1}{6}\)
\(F = \frac{T + \frac{B}{3}}{6}\)
Therefore:
\(\frac{T}{5} = \frac{T + \frac{B}{3}}{6}\)
\(\frac{6T}{5} = T + \frac{B}{3}\)
\(\frac{T}{5} = \frac{B}{3}\)
\(\frac{B}{T} = \frac{3}{5}\)
Sufficient.
(2) After the addition of base liquid, the ratio of the weight of base liquid to the total weight of the formula became \(\frac{2}{3}\).
After the addition, the weight of the base liquid became:
\(B + \frac{B}{3} = \frac{4B}{3}\)
The new total weight of the formula was:
\(T + \frac{B}{3}\)
So statement (2) gives:
\(\frac{\frac{4B}{3}}{T + \frac{B}{3}} = \frac{2}{3}\)
\(\frac{4B}{3T + B} = \frac{2}{3}\)
\(12B = 6T + 2B\)
\(10B = 6T\)
\(\frac{B}{T} = \frac{3}{5}\)
Sufficient.
Answer: D
