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26 Apr 2026, 13:46
Kudos
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A factory makes identical parts in two production runs.
In Run 1, the factory produced \(n\) parts, where \(n > 0\), and \(p\) percent of them failed inspection.
In Run 2, the factory produced 24,000 more parts than in Run 1, and \(q\) percent of them failed inspection.
In which run did more parts fail inspection?
(1) \(np < 360,000\)
(2) \(q = 15\)
In Run 1, the factory produced \(n\) parts, where \(n > 0\), and \(p\) percent of them failed inspection.
In Run 2, the factory produced 24,000 more parts than in Run 1, and \(q\) percent of them failed inspection.
In which run did more parts fail inspection?
(1) \(np < 360,000\)
(2) \(q = 15\)
26 Apr 2026, 13:46
Kudos
Bookmarks
Official Solution:
A factory makes identical parts in two production runs.
In Run 1, the factory produced \(n\) parts, where \(n > 0\), and \(p\) percent of them failed inspection.
In Run 2, the factory produced 24,000 more parts than in Run 1, and \(q\) percent of them failed inspection.
In which run did more parts fail inspection?
Failed parts in Run 1 = \(\frac{np}{100}\)
Failed parts in Run 2 = \(\frac{(n + 24,000)q}{100}\)
We need to determine which is larger.
(1) \(np < 360,000\)
This implies Run 1 failed parts = \(\frac{np}{100} < 3,600\). But \(q\) is unknown, so Run 2 failures could be less than 3,600 or greater than 3,600. Not sufficient.
(2) \(q = 15\)
This implies Run 2 failures = \(\frac{(n + 24,000) * 15}{100}\). But \(n\) and \(p\) are unknown, so we cannot compare Run 1 and Run 2. Not sufficient.
(1)+(2):
From (1), Run 1 failures < 3,600.
From (2), Run 2 failures = \(\frac{(n + 24,000) * 15}{100} = \frac{15n}{100} + 3,600\). Since \(n > 0\), \(\frac{15n}{100} > 0\), so Run 2 failures > 3,600.
Therefore Run 2 had more failed parts.
Answer: C
A factory makes identical parts in two production runs.
In Run 1, the factory produced \(n\) parts, where \(n > 0\), and \(p\) percent of them failed inspection.
In Run 2, the factory produced 24,000 more parts than in Run 1, and \(q\) percent of them failed inspection.
In which run did more parts fail inspection?
Failed parts in Run 1 = \(\frac{np}{100}\)
Failed parts in Run 2 = \(\frac{(n + 24,000)q}{100}\)
We need to determine which is larger.
(1) \(np < 360,000\)
This implies Run 1 failed parts = \(\frac{np}{100} < 3,600\). But \(q\) is unknown, so Run 2 failures could be less than 3,600 or greater than 3,600. Not sufficient.
(2) \(q = 15\)
This implies Run 2 failures = \(\frac{(n + 24,000) * 15}{100}\). But \(n\) and \(p\) are unknown, so we cannot compare Run 1 and Run 2. Not sufficient.
(1)+(2):
From (1), Run 1 failures < 3,600.
From (2), Run 2 failures = \(\frac{(n + 24,000) * 15}{100} = \frac{15n}{100} + 3,600\). Since \(n > 0\), \(\frac{15n}{100} > 0\), so Run 2 failures > 3,600.
Therefore Run 2 had more failed parts.
Answer: C
