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21 Feb 2018, 02:10
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C
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The table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both chess and drama, 5 students are in both chess and math, and 6 students are in both drama and math. How many different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84
Attachment:
Capture.PNG [ 5.21 KiB | Viewed 43391 times ]
15 Mar 2019, 18:48
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broall
We can create the equation:
Total = #chess + #drama + #math - #chess and drama - #chess and math - #drama and math + #all three
Total = 40 + 30 + 25 - 10 - 5 - 6 + 0
Total = 74
Answer: C
21 Feb 2018, 20:20
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broall
Attachment:
vennclubs.png [ 11.74 KiB | Viewed 38634 times ]
In this case, I prefer a Venn diagram.
We have totals for each club, and the overlap numbers.
Find the students who are in ONLY one club with the Venn diagram.
Draw the diagram, and write in information.
Start with the most restrictive information: there are 0 in all three clubs
Next most restrictive: the number in the intersections (10, 5, 6)
Finally, derive the number of students who are in ONLY one club from CLUB TOTAL - TWO OVERLAPs, for each club
Add all the numbers inside the Venn diagram.
That is the number of different students in the three clubs.
GIVEN (write this info in the diagram)
Chess total = 40
Drama total = 30
Math total = 25
C + D = 10
C + M = 5
D + M = 6
All three = 0
DERIVE how many in ONLY ONE
Chess ONLY (purple shaded region):
Total must equal 40
(10 + 5 = 15) (gray regions) of those 40 are in other clubs.
Chess only: (40 - 15) = 25
Drama ONLY? (green shaded region)
Total must equal 30.
(10 + 6) = 16 (gray regions) of those 30 are in other clubs.
Drama only: (30 - 16) = 14
Math ONLY? (pink shaded region)
Total must equal 25
(5 + 6) = 11 (gray regions) are in other clubs.
Math only: (25 - 11) = 14
Add the numbers in each region:
25 + 14 + 14 + 10 + 6 + 5 = 74
Answer C
