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In a class of 100 80 have atleast one pen 65 have atleast

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Manager
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In a class of 100 80 have atleast one pen 65 have atleast [#permalink]  30 Nov 2006, 19:59
In a class of 100
80 have atleast one pen
65 have atleast one pencil
70 have atleast one slate

If a is max no of classmates having all three
and b min no of classmates having all three,

a-b is:

A)35
B)20
C)15
D)50
E)60
Senior Manager
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Re: set theory problem [#permalink]  01 Dec 2006, 09:02
In a class of 100
80 have atleast one pen
65 have atleast one pencil
70 have atleast one slate

If a is max no of classmates having all three
and b min no of classmates having all three,

a-b is:

Max students having all three can not be greater than 65 (a)

For min strat with pen 80 have pen
For pencil assume 35 without pensil were the ones who had pen so now 80-35=45 have both

for slate assume 30 without slate were folks with pensil and pen
sp 45-30-15 (b)

a-b=65-15=50
Manager
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expln correct.

well done!
Senior Manager
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In a class of 100
80 have atleast one pen
65 have atleast one pencil
70 have atleast one slate

If a is max no of classmates having all three
and b min no of classmates having all three

For max:
65 have one pencil so 100 - 65 = 35 cannot have all three.
Therefore 65 = max = a

For min: If we look at the number of people in each group who dont have anything and assume that they are separate from one another then subtract them from the total, this will give us the min:

100 - [(100-80) + (100-65) + (100 - 70)] = 15 = min = b

a-b = 65 - 15 = 50
VP
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I think, to find the max is straightforward.

To summarize the formula of finding the minimum all 3 covered:

setA + setB + setC - 2*total_#_of_non_repeating_elements.
Manager
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Can anyone please give a simple explanation to this problem
Senior Manager
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Since 65 is the smallest no. among the three, there cannot be more than 65 who could have all the three items. So a=65

Since there are only 100 students in the class, And 80 have pen, there are 20 who do not have pen. Since 70 have slate there are 30 who do not have slate. Out of 65 who have pencil, there is a chance that these 20+30=50 students who do not have pen and slate. So there should be atleast 65-50=15 who have all the three. So b=15

hence a-b=65-15=50

hope it helps
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Sumithra wrote:
Since 65 is the smallest no. among the three, there cannot be more than 65 who could have all the three items. So a=65

Since there are only 100 students in the class, And 80 have pen, there are 20 who do not have pen. Since 70 have slate there are 30 who do not have slate. Out of 65 who have pencil, there is a chance that these 20+30=50 students who do not have pen and slate. So there should be atleast 65-50=15 who have all the three. So b=15

hence a-b=65-15=50

hope it helps

This is a cool way of doing the problem Thanks!
Manager
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tennis_ball wrote:
I think, to find the max is straightforward.

To summarize the formula of finding the minimum all 3 covered:

setA + setB + setC - 2*total_#_of_non_repeating_elements.

can you please explain what "totalnononrepeatingelements" means and how you came up with this formula ?
thnaks
Manager
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In a class of 100
80 have atleast one pen
65 have atleast one pencil
70 have atleast one slate

If a is max no of classmates having all three
and b min no of classmates having all three

For max:
65 have one pencil so 100 - 65 = 35 cannot have all three.
Therefore 65 = max = a

For min: If we look at the number of people in each group who dont have anything and assume that they are separate from one another then subtract them from the total, this will give us the min:

100 - [(100-80) + (100-65) + (100 - 70)] = 15 = min = b

a-b = 65 - 15 = 50

this seems to be the opposite of another explanation below, where the people who dont have anything from two sets are taken and put into the third, instead of separating.

the numbers however seem to match - is it a coincidence?
in any case, what is the basis for this formula?
what is the logic here that ensures that all three are minimum.
plz explain. thanks.
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trivikram wrote:
Sumithra wrote:
Since 65 is the smallest no. among the three, there cannot be more than 65 who could have all the three items. So a=65

Since there are only 100 students in the class, And 80 have pen, there are 20 who do not have pen. Since 70 have slate there are 30 who do not have slate. Out of 65 who have pencil, there is a chance that these 20+30=50 students who do not have pen and slate. So there should be atleast 65-50=15 who have all the three. So b=15

hence a-b=65-15=50

hope it helps

This is a cool way of doing the problem Thanks!

sumitra/trivikram,

What is the logic behind this solution?
In your approach, I understand you are distributing the 50 so as to minimize the overlap. But on what basis do you conclude that the remaining 15 (65-50) should be common to all three? What if some of them could be common to any two only(say 8 can have pencil and pen and 7 pencil+slate).

Please explain what is the underlying mathematical principle, relation or logic you are using here.
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Manager
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trahul4 wrote:

thanks, i've gone thru' it before - but unfortunately, the topic covered in that thread (Formula for "Only Both" (AUBUC = A+B+C - Only Both - 2 (All Three)) is unrelated to this problem.
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Re: set theory problem [#permalink]  07 Jul 2007, 14:55
Damager wrote:
In a class of 100
80 have atleast one pen
65 have atleast one pencil
70 have atleast one slate

If a is max no of classmates having all three
and b min no of classmates having all three,

a-b is:

Max students having all three can not be greater than 65 (a)

For min strat with pen 80 have pen
For pencil assume 35 without pensil were the ones who had pen so now 80-35=45 have both

for slate assume 30 without slate were folks with pensil and pen
sp 45-30-15 (b)

a-b=65-15=50

looked at all the solutions and tried to solve this problem again, but haven't figured it out.

for example, in the above case u are assuming 65 as max, but this has the consequence that not all 100 will have a pen, pencil or slate. (bcoz the balance 15 (in 80) and 5 (in 70) will not add up to 100. Of course, this is not a requirement mentioned in the question, but if you do away with this requirement, by the same token, you can distribute the excess that is not common outside of the three groups - in other words, you can have "minimum all three" = 0

what is the official answer? where on earth did this question come from?
sorry to beat up this half-dead horse, but its driving me nuts.
Re: set theory problem   [#permalink] 07 Jul 2007, 14:55
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