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by hjafferi » Mon Jun 10, 2013 10:58 am
In a group of 30 students, 25 are taking mathematics, 22 English and 19 history. Every student is taking at least one of the courses. The greatest number of students who could be taking all three courses is x. The least number of students who could be taking all three courses is y. What is the value of x + y.

A. 17
B. 19
C. 22
D. 23
E. 24
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by GMATGuruNY » Mon Jun 10, 2013 2:25 pm
hjafferi wrote:In a group of 30 students, 25 are taking mathematics, 22 English and 19 history. Every student is taking at least one of the courses. The greatest number of students who could be taking all three courses is x. The least number of students who could be taking all three courses is y. What is the value of x + y.

A. 17
B. 19
C. 22
D. 23
E. 24
T = M + E + H - (ME + MH + EH) - 2(MEH).

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in M, everyone in E, and everyone in H:
Those in exactly 2 of the groups (ME + MH + EH) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (MEH) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
T = 30
M = 25
E = 22
H = 19.
Thus:
30 = 25 + 22 + 19 - (ME + MH + EH) - 2(MEH)
(ME + MH + EH) + 2(MEH) = 36.

MAXIMUM:
To maximize the value of MEH, we must MINIMIZE the value of ME + MH + EH.
If ME + MH + EH = 0, we get:
0 + 2(MEH) = 36
MEH = 18.

MINIMUM:
To MINIMIZE the value of MEH, we must MAXIMIZE the value of ME + MH + EH.
Since M=25, the maximum value of EH = 30-25 = 5.
Since E=22, the maximum value of MH = 30-22 = 8.
Since H=19, the maximum value of ME = 30-19 = 11.
Since the maximum value of ME + MH + EH = 11+8+5 = 24, we get:
24 + 2(MEH) = 36.
MEH = 6.

Thus:
x+y = 18+6 = 24.

The correct answer is E.

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by ygdrasil24 » Tue Jun 11, 2013 10:28 am
Can someone please share a graphic approach via venn diagrams?
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by tarunjohri » Thu Jun 13, 2013 1:57 am
GMATGuruNY wrote:
hjafferi wrote:In a group of 30 students, 25 are taking mathematics, 22 English and 19 history. Every student is taking at least one of the courses. The greatest number of students who could be taking all three courses is x. The least number of students who could be taking all three courses is y. What is the value of x + y.

A. 17
B. 19
C. 22
D. 23
E. 24
T = M + E + H - (ME + MH + EH) - 2(MEH).

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in M, everyone in E, and everyone in H:
Those in exactly 2 of the groups (ME + MH + EH) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (MEH) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
T = 30
M = 25
E = 22
H = 19.
Thus:
30 = 25 + 22 + 19 - (ME + MH + EH) - 2(MEH)
(ME + MH + EH) + 2(MEH) = 36.

MAXIMUM:
To maximize the value of MEH, we must MINIMIZE the value of ME + MH + EH.
If ME + MH + EH = 0, we get:
0 + 2(MEH) = 36
MEH = 18.

MINIMUM:
To MINIMIZE the value of MEH, we must MAXIMIZE the value of ME + MH + EH.
Since M=25, the maximum value of EH = 30-25 = 5.
Since E=22, the maximum value of MH = 30-22 = 8.
Since H=19, the maximum value of ME = 30-19 = 11.
Since the maximum value of ME + MH + EH = 11+8+5 = 24, we get:
24 + 2(MEH) = 36.
MEH = 6.

Thus:
x+y = 18+6 = 24.

The correct answer is E.

Check here for a similar problem:

https://www.beatthegmat.com/sets-t148362.html
Is the maximum position possible?
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