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05 Aug 2026, 23:05
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A home-goods warehouse inspected a shipment of ceramic mugs for glazing flaws. Of the mugs in the shipment, 70% were blue-glazed, and the rest were not blue-glazed. Of the mugs that were not blue-glazed, \(\frac{2}{5}\) had a glazing flaw. What percentage of the blue-glazed mugs did not have a glazing flaw?
(1) If one mug is selected at random from the shipment, the probability that it will have a glazing flaw is 0.32.
(2) If one mug with a glazing flaw is selected at random from the shipment, the probability that it will be blue-glazed is \(\frac{5}{8}\).
(1) If one mug is selected at random from the shipment, the probability that it will have a glazing flaw is 0.32.
(2) If one mug with a glazing flaw is selected at random from the shipment, the probability that it will be blue-glazed is \(\frac{5}{8}\).
05 Aug 2026, 23:05
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Official Solution:
A home-goods warehouse inspected a shipment of ceramic mugs for glazing flaws. Of the mugs in the shipment, 70% were blue-glazed, and the rest were not blue-glazed. Of the mugs that were not blue-glazed, \(\frac{2}{5}\) had a glazing flaw. What percentage of the blue-glazed mugs did not have a glazing flaw?
Use 100 mugs as a convenient total. Since \(70\%\) of the mugs were blue-glazed:
• Blue-glazed \(= 70\)
• Not blue-glazed \(= 30\)
Of the mugs that were not blue-glazed, \(\frac{2}{5}\) had a glazing flaw:
• So \(\frac{2}{5} * 30 = 12\) not-blue-glazed mugs had a glazing flaw.
The question asks what percentage of the 70 blue-glazed mugs did not have a glazing flaw.
(1) If one mug is selected at random from the shipment, the probability that it will have a glazing flaw is 0.32.
So \(32\%\) of the mugs had a glazing flaw. Since we are using 100 mugs, 32 mugs had a glazing flaw.
Of these 32 flawed mugs, 12 were not blue-glazed, so:
blue-glazed mugs with a flaw \(= 32 - 12 = 20\)
Therefore:
blue-glazed mugs without a flaw \(= 70 - 20 = 50\)
So the required percentage is:
\(\frac{50}{70} = \frac{5}{7}\)
Sufficient.
(2) If one mug with a glazing flaw is selected at random from the shipment, the probability that it will be blue-glazed is \(\frac{5}{8}\).
This means:
\(\frac{\text{blue-glazed mugs with a flaw}}{\text{all mugs with a flaw}} = \frac{5}{8}\)
We already know that 12 not-blue-glazed mugs had a glazing flaw.
Let \(B\) be the number of blue-glazed mugs with a glazing flaw. Then:
\(\frac{B}{B + 12} = \frac{5}{8}\)
\(B = 20\)
So 20 blue-glazed mugs had a glazing flaw.
Therefore:
blue-glazed mugs without a flaw \(= 70 - 20 = 50\)
So the required percentage is:
\(\frac{50}{70} = \frac{5}{7}\)
Sufficient.
Answer: D
A home-goods warehouse inspected a shipment of ceramic mugs for glazing flaws. Of the mugs in the shipment, 70% were blue-glazed, and the rest were not blue-glazed. Of the mugs that were not blue-glazed, \(\frac{2}{5}\) had a glazing flaw. What percentage of the blue-glazed mugs did not have a glazing flaw?
Use 100 mugs as a convenient total. Since \(70\%\) of the mugs were blue-glazed:
• Blue-glazed \(= 70\)
• Not blue-glazed \(= 30\)
Of the mugs that were not blue-glazed, \(\frac{2}{5}\) had a glazing flaw:
• So \(\frac{2}{5} * 30 = 12\) not-blue-glazed mugs had a glazing flaw.
The question asks what percentage of the 70 blue-glazed mugs did not have a glazing flaw.
(1) If one mug is selected at random from the shipment, the probability that it will have a glazing flaw is 0.32.
So \(32\%\) of the mugs had a glazing flaw. Since we are using 100 mugs, 32 mugs had a glazing flaw.
Of these 32 flawed mugs, 12 were not blue-glazed, so:
blue-glazed mugs with a flaw \(= 32 - 12 = 20\)
Therefore:
blue-glazed mugs without a flaw \(= 70 - 20 = 50\)
So the required percentage is:
\(\frac{50}{70} = \frac{5}{7}\)
Sufficient.
(2) If one mug with a glazing flaw is selected at random from the shipment, the probability that it will be blue-glazed is \(\frac{5}{8}\).
This means:
\(\frac{\text{blue-glazed mugs with a flaw}}{\text{all mugs with a flaw}} = \frac{5}{8}\)
We already know that 12 not-blue-glazed mugs had a glazing flaw.
Let \(B\) be the number of blue-glazed mugs with a glazing flaw. Then:
\(\frac{B}{B + 12} = \frac{5}{8}\)
\(B = 20\)
So 20 blue-glazed mugs had a glazing flaw.
Therefore:
blue-glazed mugs without a flaw \(= 70 - 20 = 50\)
So the required percentage is:
\(\frac{50}{70} = \frac{5}{7}\)
Sufficient.
Answer: D
