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This topic has expert replies
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Min/Max

by yellowho » Mon Mar 14, 2011 5:42 am
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) +D-2Triple +0
-115=D-2T, re-write T=(D+115)/2

To minimize T, you must min D. How do you minimize D?

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by srcc25anu » Mon Mar 14, 2011 12:56 pm
All 3 maximum can be 65
All 3 minimum can be 15

2 items maximum = when all 3 minimum therefore 2 items maximum = 100-15 = 85
2 items minimum = 0

Whats the OA?

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by yellowho » Mon Mar 14, 2011 7:33 pm
My non-formula 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:

Total-Group(1,2,3)=-D-2T

1) To MAX Double you MIN Triple and Min Neither: (per the formula: Total-Group(1,2,3)+2T-N=-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=-85-2t
115=85+2t
T=15

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by DarkKnight » Tue Mar 15, 2011 10:12 am
Whats the OA?

I m getting:

For 2 only - MIN=1, MAX=115
For 3 only - MIN=0, MAX=57

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by yellowho » Sat Mar 19, 2011 11:00 pm
Sorry I did this all wrong. The problem said [u]at least[/u] 65. So max all 3 is actually 100%. Disregard everything I wrote.

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