In a class of 100: 80 have at least one pen; 65 have at least one pen

Question

In a class of 100:

80 have at least one pen;
65 have at least one pencil;
70 have at least 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. 15
B. 20
C. 35
D. 50
E. 60

chetan2u · Accepted Answer

Max number is when all 65 who have pencil have other two items too => a=65

Least is when we fill up the 100s first before starting any overlap.

80 have pen, that is 20 don’t have pens.

65 have pencil. 
=> so fill up 20 who don’t have pen. This will leave 65-20 to have both. 
So 45 have both, and remaining 55 have only one of the twos. : 80-45=35 have pen only, and 20 have only pencil

70 have at least one slate. 
=> So fill up 55 who have just one item.
The remaining 70-55 have to be adjusted with those who have both.

Final status : 15 have all three, and remaining 85 have two items only. => b=15

a-b=65-15=50

D

sujoykrdatta · Answer

At least 1 pen = 80 
Means: pens only + only pen and pencil + only pen and slate + pen and pencil and slate = 80
Basically means: the circle for pens is 80
So no pen = 100-80=20

At least 1 pencil = 65
the circle for pencils is 65
So no pencil = 100-65=35

At least 1 slate = 70 
the circle for slates is 70
So no slate = 100-70=30

# Max of all 3 = min of 80,65,70 = 65

Min of all 3:

Let's think u have only pen and pencil 
Min overlap = x 
So, 80+65-x=100
x= 45
(We equate to 100 so that there is nothing that is neither pen nor pencil; this ensures that the overlap is also min) 
Now, think of overlap with this 45 and slate: 
Min overlap = y
So: 45+70-y = 100
y = 15

# So minimum overlap of all three = 15

Answer = 65-15 = 50
 Option D

Posted from my mobile device

Kinshook · Answer

In a class of 100:

80 have at least one pen;
65 have at least one pencil;
70 have at least 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 = 65; Since 65 have at least one pencil

For b:

---------------------100------------------
---------------80------------------
               ------------------70----------
-----30-----                -----35---------

b = 15

a - b = 65 - 15 = 50

IMO D

hr1212 · Answer

For similar questions, I was able to derive a neat formula for quick calculation =>

Max of all 3 = min(80, 65, 70) = 65
Min of all 3 = max(0, sum(80, 65, 70) - 2*Total) = 80 + 65 + 70 - 2*100 = 15

Max - Min = 65 - 15 = 50

Answer: D

Note - Use it cautiously after validating scenarios

For anyone interested in knowing how this formula works -

A  ∩ B + A ∩ B ∩ C = A + B - Total
B ∩ C + A ∩ B ∩ C = C + B - Total 
A ∩ C + A ∩ B ∩ C = A + C - Total

A ∩ B + B ∩ C + A ∩ C + A ∩ B ∩ C = Total ---- (only A = only B = only C = 0)

Combining these equations -

2A + 2B + 2C - 3*Total = A ∩ B + B ∩ C + A ∩ C + 3*A ∩ B ∩ C
 2A + 2B + 2C - 3*Total = Total - A ∩ B ∩ C + 3*A ∩ B ∩ C
 A ∩ B ∩ C = A + B + C - 2*Total

bumpbot · Answer

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In a class of 100: 80 have at least one pen; 65 have at least one pen

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