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PS · Problem Solving
705-805
percent
In a club, the number of man engineers to woman who are both engineer and doctor is equal and the ratio of man engineers to woman who are both engineer and doctor is the same to the ratio of man doctors to the woman doctors. If members in the club are man among the total number of 1000 members, what is the total number of engineers in the club?
Answer Choices
Correct answer marked belowA
500
600
Correct
C
400
D
200
E
100
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Official Explanation
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Let:
- = man engineers
- = woman engineers
- = man doctors
- = woman doctors
- = women who are both engineer and doctor
Given:The number of man engineers to woman who are both engineer and doctor is equal: .
The ratio is the same as the ratio . Since , the ratio is , so , implying .
of members are men. Total members = 1000, so men = , women = 400.
Since all men are either engineers or doctors (or both), but the problem implies non-overlapping categories for men (as it separates ME and MD without mention of overlap), assume men are only engineers or only doctors: .
Similarly, women are in categories: WE, WD, and WED. But note: WED are women who are both, so they are counted in both WE and WD. However, total women = WE + WD - WED = 400.
From above: and .
Let , then .
From men total: .
From women total: WE + WD - WED = 400. But WE includes women who are only engineers and those who are both? The problem does not explicitly give WE separately. However, note: WED are women who are both, so they are part of WD as well. The total women equation using given variables: WE is not directly given, but we have WD = y and WED = x. Also, all women doctors are WD, which includes WED. So WD = y = (women only doctors) + WED. Similarly, WE = (women only engineers) + WED. But we lack info to split.
Consider the total engineers: Engineers = ME + WE. But WE = (women engineers) = ?
We know WED = x, and let women only engineers = a, so WE = a + x.
Similarly, let women only doctors = b, so WD = b + x = y.
Total women: a + b + x = 400 (since a, b, x are disjoint).
Also, y = b + x.
From men: x + y = 600 -> x + (b + x) = 600 -> 2x + b = 600.
From women: a + b + x = 400.
We need total engineers = ME + WE = x + (a + x) = a + 2x.
From 2x + b = 600 -> b = 600 - 2x.
From a + b + x = 400 -> a + (600 - 2x) + x = 400 -> a + 600 - x = 400 -> a = x - 200.
Then total engineers = a + 2x = (x - 200) + 2x = 3x - 200.
We need another equation. Notice the ratio condition is already used. Possibly all members are either engineers or doctors (or both), meaning total engineers + total doctors - both = 1000. But both are only WED? Actually, both are women who are both engineer and doctor, so both = x.
Total engineers = E, total doctors = D. Then E + D - x = 1000.
But D = MD + WD = y + y = 2y = 2(b + x) = 2b + 2x.
And E = ME + WE = x + (a + x) = a + 2x.
So (a + 2x) + (2b + 2x) - x = 1000 -> a + 2b + 3x = 1000.
Substitute a = x - 200 and b = 600 - 2x:
(x - 200) + 2(600 - 2x) + 3x = 1000 -> x - 200 + 1200 - 4x + 3x = 1000 -> (x - 4x + 3x) + (1200 - 200) = 1000 -> 0x + 1000 = 1000 -> 0 = 0, which is always true. So x can vary.
We need an integer solution with positive a, b, x. a = x - 200 >= 0 -> x >= 200. b = 600 - 2x >= 0 -> x <= 300. Also, total engineers = 3x - 200. For x between 200 and 300, engineers ranges from 400 to 700. But only one option fits: 600.
If engineers = 600, then 3x - 200 = 600 -> 3x = 800 -> x = 800/3, not integer.
Wait, check options: 500, 600, 400, 200, 100. Possibly x must be integer.
Try each:
If engineers = 600, then 3x - 200 = 600 -> x = 800/3 ≈ 266.67, not integer.
If engineers = 500, then 3x - 200 = 500 -> 3x = 700 -> x = 700/3 ≈ 233.33, not integer.
If engineers = 400, then 3x - 200 = 400 -> 3x = 600 -> x = 200, integer. Then a = x - 200 = 0, b = 600 - 2*200 = 200. Valid.
If engineers = 200, then 3x - 200 = 200 -> 3x = 400 -> x = 400/3, not integer.
If engineers = 100, then 3x - 200 = 100 -> 3x = 300 -> x = 100, then a = -100, invalid.…
- = man engineers
- = woman engineers
- = man doctors
- = woman doctors
- = women who are both engineer and doctor
Given:
Since all men are either engineers or doctors (or both), but the problem implies non-overlapping categories for men (as it separates ME and MD without mention of overlap), assume men are only engineers or only doctors: .
Similarly, women are in categories: WE, WD, and WED. But note: WED are women who are both, so they are counted in both WE and WD. However, total women = WE + WD - WED = 400.
From above: and .
Let , then .
From men total: .
From women total: WE + WD - WED = 400. But WE includes women who are only engineers and those who are both? The problem does not explicitly give WE separately. However, note: WED are women who are both, so they are part of WD as well. The total women equation using given variables: WE is not directly given, but we have WD = y and WED = x. Also, all women doctors are WD, which includes WED. So WD = y = (women only doctors) + WED. Similarly, WE = (women only engineers) + WED. But we lack info to split.
Consider the total engineers: Engineers = ME + WE. But WE = (women engineers) = ?
We know WED = x, and let women only engineers = a, so WE = a + x.
Similarly, let women only doctors = b, so WD = b + x = y.
Total women: a + b + x = 400 (since a, b, x are disjoint).
Also, y = b + x.
From men: x + y = 600 -> x + (b + x) = 600 -> 2x + b = 600.
From women: a + b + x = 400.
We need total engineers = ME + WE = x + (a + x) = a + 2x.
From 2x + b = 600 -> b = 600 - 2x.
From a + b + x = 400 -> a + (600 - 2x) + x = 400 -> a + 600 - x = 400 -> a = x - 200.
Then total engineers = a + 2x = (x - 200) + 2x = 3x - 200.
We need another equation. Notice the ratio condition is already used. Possibly all members are either engineers or doctors (or both), meaning total engineers + total doctors - both = 1000. But both are only WED? Actually, both are women who are both engineer and doctor, so both = x.
Total engineers = E, total doctors = D. Then E + D - x = 1000.
But D = MD + WD = y + y = 2y = 2(b + x) = 2b + 2x.
And E = ME + WE = x + (a + x) = a + 2x.
So (a + 2x) + (2b + 2x) - x = 1000 -> a + 2b + 3x = 1000.
Substitute a = x - 200 and b = 600 - 2x:
(x - 200) + 2(600 - 2x) + 3x = 1000 -> x - 200 + 1200 - 4x + 3x = 1000 -> (x - 4x + 3x) + (1200 - 200) = 1000 -> 0x + 1000 = 1000 -> 0 = 0, which is always true. So x can vary.
We need an integer solution with positive a, b, x. a = x - 200 >= 0 -> x >= 200. b = 600 - 2x >= 0 -> x <= 300. Also, total engineers = 3x - 200. For x between 200 and 300, engineers ranges from 400 to 700. But only one option fits: 600.
If engineers = 600, then 3x - 200 = 600 -> 3x = 800 -> x = 800/3, not integer.
Wait, check options: 500, 600, 400, 200, 100. Possibly x must be integer.
Try each:
If engineers = 600, then 3x - 200 = 600 -> x = 800/3 ≈ 266.67, not integer.
If engineers = 500, then 3x - 200 = 500 -> 3x = 700 -> x = 700/3 ≈ 233.33, not integer.
If engineers = 400, then 3x - 200 = 400 -> 3x = 600 -> x = 200, integer. Then a = x - 200 = 0, b = 600 - 2*200 = 200. Valid.
If engineers = 200, then 3x - 200 = 200 -> 3x = 400 -> x = 400/3, not integer.
If engineers = 100, then 3x - 200 = 100 -> 3x = 300 -> x = 100, then a = -100, invalid.…
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Your ability to apply GMAT math concepts to solve a multi-step problem quickly and accurately. At the 705-805 level, accuracy and speed both matter.
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Overcomplicating the setup — most GMAT problems reward a direct, organized approach over brute calculation.
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705-805
Type
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