Finding the minimum cardinal value of intersection of 4 sets. In a battle, 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. Find the minimum value of x.
So, this is related to Classic Set Theory by George Cantor.
As far I've tried the minimum number is the intersection of two set and the maximum number is the set having the lowest cardinal number. But, in this case, there are four sets, so that doesn't seem to work.
To approach the number, I tried making Venn Diagrams, but that's way too complicated for it. 
Now, I'm clueless. I don't have any idea how to get this done, I doubt perhaps my approach was wrong.
Answer Given in Textbook is = 10.
Thanks in Advance
 A: For every 100 combactants,
Given,
$$ n(eye) = 70,$$
$$ n(ear) = 80,$$
So, as we know, the Minimum number of two non-empty sets can be Calculated using the Intersection Formula.
So,
$$ n(eye \cap ear) = n(eye) + n(ear) - n(U) $$
$$ = 70 + 80 - 100$$
(As we know $n(U)$ = 100, as calculations are being done for every 100 combactants, where 100 refers to Universal Set of all the 100 combactants)
$$ = 50 $$
Now, let Set $ eye \cap ear$ be set A.
So, the minimum number of Combatants who have got their eyes, ears and arm can be represented as $A \cap arm$
Thus,
$$ n(A \cap arm) = n(A) + n(arm) - n(U) $$
(Again, here U refers to Universal Set which contains all the hundred Combactants)
$$ = 50 + 75 - 100 $$
$$ = 25 $$
Thus, the minimum number of Combatants which have got their Eyes, Ears, and Arms Injured are 25.
Now, Let Set C be the set of the Minimum number of Combatants which have lost their Eyes, Ears, and Arms.
So, as we know, the minimum number of people who have lost all four will be $ C \cap leg $
Thus,
$$ n(C \cap leg) = n(C) + n(leg) - n(U) $$
$$= 25 + 85 - 100$$
$$= 110 - 100$$
$$ = 10 $$
Thus, The minimum possible Combatants which have lost all four are 10.
