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A marketing class of a college has a total strength of 30. The class f
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05 Jan 2024, 01:04
05 Jan 2024, 01:04
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A marketing class of a college has a total strength of 30. The class formed three groups: G1, G2, and G3, which have 10, 10, and 6 students, respectively. If no student of G1 is in either of the other two groups, what is the greatest possible number of students who are in none of the groups?
(A) 6
(B) 7
(C) 8
(D) 10
(E) 14
(A) 6
(B) 7
(C) 8
(D) 10
(E) 14
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Joined: 11 Nov 2023
Posts: 207
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WE:Business Development (Advertising and PR)
Re: A marketing class of a college has a total strength of 30. The class f
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19 Jan 2024, 14:17
19 Jan 2024, 14:17
1
Rather than the standard Venn Diagram, I thought it might be interesting to visualize the overlapping groups using blocks.
We're given that G1 = 10, G2 = 10, G3 = 6. Let x = the # of students who are in none of the groups.
Say a = only in G1, b = in only G1+G2, c = only in G2, d = only in G1+G3, e = students in all three, f = only in G2+G3, g = only in G3.
G1 = a + b + d + e
G2 = b + c + e + f
G3 = d + e + f + g
x = 30 - (G1+G2+G3-b-d-2e-f)
Since "no student of G1 is in either of the other two groups", this means b, d, and e = 0.
Thus, a = 10 and x = 30 - (G1+G2+G3-f)
Now, let's consider what happens when the overlap between G2+G3, or f, varies, and how it affects x.
In the above, the blocks are arranged such that G1 does not overlap with the other two, and G2+G3 overlap by 2. Thus, f = 2.
If f = 2, then x = 30 - (10+10+6-2) = 6
However, we can see that in order to maximize x, we can fully overlap G3 over G2, such that f = 6.
If f = 6, then x = 30 - (10+4+6) = 10
Attachment:
File comment: G1, G2, and G3 represented as blocks
overlapping_blocks.png [ 20.18 KiB | Viewed 490 times ]
overlapping_blocks.png [ 20.18 KiB | Viewed 490 times ]
We're given that G1 = 10, G2 = 10, G3 = 6. Let x = the # of students who are in none of the groups.
Say a = only in G1, b = in only G1+G2, c = only in G2, d = only in G1+G3, e = students in all three, f = only in G2+G3, g = only in G3.
G1 = a + b + d + e
G2 = b + c + e + f
G3 = d + e + f + g
x = 30 - (G1+G2+G3-b-d-2e-f)
Since "no student of G1 is in either of the other two groups", this means b, d, and e = 0.
Thus, a = 10 and x = 30 - (G1+G2+G3-f)
Now, let's consider what happens when the overlap between G2+G3, or f, varies, and how it affects x.
Attachment:
File comment: Blocks overlapped so that x = 6
overlap_x6.png [ 21.72 KiB | Viewed 493 times ]
overlap_x6.png [ 21.72 KiB | Viewed 493 times ]
In the above, the blocks are arranged such that G1 does not overlap with the other two, and G2+G3 overlap by 2. Thus, f = 2.
If f = 2, then x = 30 - (10+10+6-2) = 6
Attachment:
File comment: Blocks overlapped so that x = 10
overlapping_x10.png [ 22.54 KiB | Viewed 497 times ]
overlapping_x10.png [ 22.54 KiB | Viewed 497 times ]
However, we can see that in order to maximize x, we can fully overlap G3 over G2, such that f = 6.
If f = 6, then x = 30 - (10+4+6) = 10
gmatclubot
Re: A marketing class of a college has a total strength of 30. The class f [#permalink]
19 Jan 2024, 14:17
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