A factory makes identical parts in two production runs.

Question

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

Bunuel · Accepted Answer

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 = np/100
Failed parts in Run 2 = (n + 24,000)q/100

We need to determine which is larger.

(1) np < 360,000

This implies Run 1 failed parts = 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 = (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 = (n + 24,000) * 15/100 = 15n/100 + 3,600. Since n > 0, 15n/100 > 0, so Run 2 failures > 3,600.

Therefore Run 2 had more failed parts.

Answer: C.

Adit_ · Answer

Two equations: np/100 240q+nq Statement 1: np<360,000 It means, np/100<3600 --> Failed INSUFFICIENT Statement 2: q=15 240q+nq = 3600+15n INSUFFICIENT Combining the two: 3600+15n>3600 as n is positive SUFFICIENT Answer: Option C

Edskore · Answer

Great question — this one tests a classic Data Sufficiency concept: comparing two quantities when information is split between the two statements.

Key concept being tested: Algebraic inequalities in a two-part comparison (which value is larger?), and recognizing how Statement 1 and Statement 2 each carry a different piece of the puzzle.

Setup:
Let the number of parts that failed in each run = the quantity we're comparing.
- Run 1 failures = n × (p/100) = np/100
- Run 2 failures = (n + 24,000) × (q/100)

Step 1 — Evaluate Statement (1): np < 360,000
This tells us Run 1 failures = np/100 < 3,600. We know Run 1 produced fewer than 3,600 defective parts. But we have zero information about q or Run 2's failure count. Not sufficient on its own.

Step 2 — Evaluate Statement (2): q = 15
Now we can express Run 2 failures = (n + 24,000) × 15/100 = (15n + 360,000)/100.
But we don't know n or p, so we can't compare this with np/100. Not sufficient on its own.

Step 3 — Combine (1) + (2):
- Run 1 failures = np/100
- From Statement 1: np < 360,000, so Run 1 failures < 3,600
- Run 2 failures = (15n + 360,000)/100 = 15n/100 + 3,600

Since n > 0 (given), we know 15n/100 > 0, which means Run 2 failures > 3,600.
And Run 1 failures < 3,600.
Therefore, Run 2 had strictly more failed parts. Sufficient!

Answer: C

Common trap: Most students see Statement 2 (q = 15) and think "now I know the percentage, that must be enough." Wrong — you still need to know n and p to compute Run 1's failures. The trap here is confusing knowing a rate with knowing an absolute count.

Takeaway: In DS questions comparing absolute values, a percentage alone (Statement 2) is almost never sufficient — you need to combine it with information about the actual quantities involved (Statement 1).

A factory makes identical parts in two production runs.

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GMAT Data Sufficiency (DS) Questions

A factory makes identical parts in two production runs.

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