In a class of 100
80 have atleast one pen
65 have atleast one pencil
70 have atleast one slate
1) What is the min and max of people with two?
2) What is the min max of people with all 3?
Looking at 2: (What am I doing wrong?)
Max: 65. that's the most pencil
Min: Using formula: Total= 3 group +double 2Triple + Neither
100= (80+65+70) +D2Triple +0
115=D2T, rewrite T=(D+115)/2
To minimize T, you must min D. How do you minimize D?
Min/Max
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My nonformula approach:
Min Number of people who has pencil and pen
(80+65 100) = 45
Min number of people who has pencil, pen and slate
(45+70)100 = 15
Max is 65, no problem.
My question is: How do you do this formulaically?
If formula is Total = Group1 + Group 2 + Group3 Double 2Triple +Neither:
TotalGroup(1,2,3)=D2T
1) To MAX Double you MIN Triple and Min Neither: (per the formula: TotalGroup(1,2,3)+2TN=D)
2) Max double: 85,65,70. Pair 85 with 70; you get 70 pairs with 15 left over. Pair the remaining 15 with the 65; you get 15. Total Double= 85
115=852t
115=85+2t
T=15
Min Number of people who has pencil and pen
(80+65 100) = 45
Min number of people who has pencil, pen and slate
(45+70)100 = 15
Max is 65, no problem.
My question is: How do you do this formulaically?
If formula is Total = Group1 + Group 2 + Group3 Double 2Triple +Neither:
TotalGroup(1,2,3)=D2T
1) To MAX Double you MIN Triple and Min Neither: (per the formula: TotalGroup(1,2,3)+2TN=D)
2) Max double: 85,65,70. Pair 85 with 70; you get 70 pairs with 15 left over. Pair the remaining 15 with the 65; you get 15. Total Double= 85
115=852t
115=85+2t
T=15

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