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25 May 2026, 07:50
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (03:18) correct
65%
(03:04)
wrong
based on 20
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Last year, \(\frac18\) of the employees of Benson’s Furniture were born outside of Thorndale. During the past year, for every three employees hired who were born outside of Thorndale, one employee born in Thorndale has left the company. Now \(\frac13\) of all employees of Benson’s were born outside of Thorndale. If the 75 employees hired during the last year were born outside of Thorndale, what is the smallest possible number of employees Benson’s could now have?
(A) 280
(B) 330
(C) 360
(D) 450
(E) 500
(A) 280
(B) 330
(C) 360
(D) 450
(E) 500
25 May 2026, 10:31
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This one's tagged Min-Max, and yeah, there is a min-max layer, but let me walk through the core of it first.
1. Let the original number of employees be N. Then N/8 were born outside Thorndale. For N/8 to be a whole number, N must be a multiple of 8.
2. 75 employees were hired outside Thorndale during the year. For every 3 hired outside, 1 Thorndale-born left. So the number who left = 75/3 = 25.
3. New total employees = N + 75 - 25 = N + 50.
4. New outside-born employees = N/8 + 75.
5. Now 1/3 of all employees are outside-born, so: (N/8 + 75) = (1/3)(N + 50). For 1/3 to yield a whole number, N + 50 must be divisible by 3.
6. Solve: Multiply through by 24. You get 3N + 1800 = 8N + 400, which gives 5N = 1400, so N = 280.
7. Check the constraints: N = 280 is divisible by 8. N + 50 = 330 is divisible by 3. Both hold.
8. Total now = 280 + 50 = 330. Outside now = 35 + 75 = 110, and 110/330 = 1/3. Confirmed.
Answer is (B) 330.
The "smallest possible" phrasing is a bit of a red herring here since the algebra gives you one unique N, not a range. Where people go wrong is forgetting the 1-to-3 ratio and subtracting 75 instead of 25, which throws off the new total entirely. I fell for that exact trap the first time I saw a problem set up this way.
1. Let the original number of employees be N. Then N/8 were born outside Thorndale. For N/8 to be a whole number, N must be a multiple of 8.
2. 75 employees were hired outside Thorndale during the year. For every 3 hired outside, 1 Thorndale-born left. So the number who left = 75/3 = 25.
3. New total employees = N + 75 - 25 = N + 50.
4. New outside-born employees = N/8 + 75.
5. Now 1/3 of all employees are outside-born, so: (N/8 + 75) = (1/3)(N + 50). For 1/3 to yield a whole number, N + 50 must be divisible by 3.
6. Solve: Multiply through by 24. You get 3N + 1800 = 8N + 400, which gives 5N = 1400, so N = 280.
7. Check the constraints: N = 280 is divisible by 8. N + 50 = 330 is divisible by 3. Both hold.
8. Total now = 280 + 50 = 330. Outside now = 35 + 75 = 110, and 110/330 = 1/3. Confirmed.
Answer is (B) 330.
The "smallest possible" phrasing is a bit of a red herring here since the algebra gives you one unique N, not a range. Where people go wrong is forgetting the 1-to-3 ratio and subtracting 75 instead of 25, which throws off the new total entirely. I fell for that exact trap the first time I saw a problem set up this way.
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