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20 May 2026, 03:37
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
81% (02:07) correct
19%
(02:38)
wrong
based on 42
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In a company, the average salary of all employees is $52,000 per year. The average salary of managers is $80,000, and the average salary of non-managers is $45,000. If there are 140 employees in total, how many managers are there?
A. 24
B. 28
C. 32
D. 36
E. 40
A. 24
B. 28
C. 32
D. 36
E. 40
20 May 2026, 06:42
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This is a classic weighted average problem.
Recall the weighted average ratio formula:
\(\frac{Managers}{Non-managers} = \frac{45-52}{52-80} = \frac{-7}{-28} = \frac{1}{4}\).
Now, put everything in a ratio table:
Hence, (B) 28.
Recall the weighted average ratio formula:
\(\frac{Managers}{Non-managers} = \frac{45-52}{52-80} = \frac{-7}{-28} = \frac{1}{4}\).
Now, put everything in a ratio table:
| Managers | Non-managers | Total employees | |
| Ratio values | \(1\) | \(4\) | \(5\) |
| Actual values | \(1\times 28 = 28\) | \(140\, (= 5\times 28)\) |
Hence, (B) 28.
20 May 2026, 10:59
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Answer: B (28 managers)
This is a weighted average / mixture problem. I'd go straight algebra here, cleaner to see what's happening.
1. Let M = number of managers. Then non-managers = 140 - M.
2. Set up the weighted average equation:
80,000(M) + 45,000(140 - M) = 52,000(140)
3. Simplify:
80,000M + 6,300,000 - 45,000M = 7,280,000
35,000M = 980,000
M = 28
The trap on this one is C (32). People eyeball the averages and think "52K is closer to 45K than to 80K, so there should be more non-managers" and then just pick something in the middle-ish range. The algebra takes 30 seconds and saves you from that.
The quicker way if you're comfortable with weighted averages: the overall average of 52K is 7K above the non-manager average (45K) and 28K below the manager average (80K). So the ratio of managers to non-managers is 7:28 = 1:4. With 140 total employees, that's 28 managers and 112 non-managers. Same answer either way.
When you see two groups with known averages and a combined average, just pick whichever method clicks faster for you under time pressure.
This is a weighted average / mixture problem. I'd go straight algebra here, cleaner to see what's happening.
1. Let M = number of managers. Then non-managers = 140 - M.
2. Set up the weighted average equation:
80,000(M) + 45,000(140 - M) = 52,000(140)
3. Simplify:
80,000M + 6,300,000 - 45,000M = 7,280,000
35,000M = 980,000
M = 28
The trap on this one is C (32). People eyeball the averages and think "52K is closer to 45K than to 80K, so there should be more non-managers" and then just pick something in the middle-ish range. The algebra takes 30 seconds and saves you from that.
The quicker way if you're comfortable with weighted averages: the overall average of 52K is 7K above the non-manager average (45K) and 28K below the manager average (80K). So the ratio of managers to non-managers is 7:28 = 1:4. With 140 total employees, that's 28 managers and 112 non-managers. Same answer either way.
When you see two groups with known averages and a combined average, just pick whichever method clicks faster for you under time pressure.
