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

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Director
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In a class of 100 80 have atleast one pen 65 have [#permalink] New post 14 Nov 2007, 06:47
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A
B
C
D
E

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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,

a-b is:

A)35
B)20
C)15
D)50
E)60


Calculating a is straightforward. But I fail to understand the logic behind calculating b. Can someone please explain the logic behind calc. b? I know the answer. Thanks.
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Joined: 08 Nov 2007
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Re: PS: Pen, Pencil, Slate [#permalink] New post 14 Nov 2007, 07:44
GK_Gmat 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:

A)35
B)20
C)15
D)50
E)60


Calculating a is straightforward. But I fail to understand the logic behind calculating b. Can someone please explain the logic behind calc. b? I know the answer. Thanks.


I make it C?

The min number of people with both Pens and Pencils will be 50 - There are twenty people who don't have a pencil - if all of those have a pen, then there will be 50 left who have to have a pen.

Thus there will be 50 with both, and 50 with one or the other. As to the slate then, up to 50 people can have either the slate and the pen, or the slate and the pencil, but not all three. Thus there are 15 people at minimum who must fall into all three.

I think that makes sense!
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 [#permalink] New post 14 Nov 2007, 10:26
I get C

For max consider the max overlap = 80+70-100 = 50
for min consider the min overlap = 65+70-100 = 35

Therefore a-b is 50-35 = 15
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Re: PS: Pen, Pencil, Slate [#permalink] New post 15 Nov 2007, 11:08
GK_Gmat 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:

A)35
B)20
C)15
D)50
E)60


Calculating a is straightforward. But I fail to understand the logic behind calculating b. Can someone please explain the logic behind calc. b? I know the answer. Thanks.



I get D.

a=65 (the same as the number of people who have pencil)
b=15

to find b, it is a little complicated.
step 1: find the min no. of people who have both pen and pencil, let it be b1.

b1=100-(100-65)-(100-80)
=100-35-20
=45.

step 2: (similar concept as in step 1) use b1 to find the b.

b=100-(100-b1)-(100-70)
=100-55-30
=15.

therefore, a-b=50.

please post the OA, thanks.
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 [#permalink] New post 15 Nov 2007, 16:30
I got D ( I AM 100% CONFIDENT)
I agree with rrong19
a=65 (No issues)

80 Pen, 65 pencil, 70 Slate

Min Number of people who has pencil and pen
= (80+65 -100) = 45

Min number of people who has pencial, pen and slate
= (45+70)-100 = 15

b=15

a-b=50 (Option D)
  [#permalink] 15 Nov 2007, 16:30
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