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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
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Bunuel
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
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
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