A Rich Man once decided to hold a party at his place. He invited 100 guests and arranged for 100 plates of food. The guests were of three social groups: A, B and C.
The members of group A said that they will eat 2 plates each,
The members of group B said that they will eat 4 plates each.
Seeing that the rich man was getting worried, the members of Group C volunteered to eat 4 persons to a plate, thereby solving the man's problem.
Now, no plates were left and all the guests had eaten food.
How many members of each group were there?
Enter the answer as
$0$
if you think no answer is possible.
Enter the answer as
$a*b*c$
if there is a unique answer.
Enter the answer as the difference between the highest and lowest values of
$a*b*c$
if there are more than
$1$
answers.
Details and assumptions :
There is at least 1 member of each group.
Each plate in case of group C must be used by exactly 4 members i.e. you cannot have only 2 members eating from the last plate.
$a,b,c$ refer to number of members of A,B and C present at the party.
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First of all, Note that the number of plates left for group C is an even number and hence can be representated as $2k$ for some 'k'.
Now, $\frac{1}{4}c = 2k$ ;
==> $c = 8k$ .
So, number of people of group C is a multiple of 8.
Now, note that for $c\geq 80$ , we have average number of plates left $\geq 4$ that is either all remaining members are of group B or else, its impossible. We dont want either case since, we must have at least 1 member of each group.
Now, for $c\leq 56$ , we have average number of plates left $<2$ and hence the required number is
$>56 ; <80$ and a multiple of 8.
There remain just 2 options:
CASE 1:
$c = 64$
We have;
$a+b = 36$
$2a + 4b = 84$
Solving we get; $a=30; b=6$
CASE 2:
$c = 72$
We have;
$a+b = 28$
$2a + 4b = 82$
Solving, we get; $a=15 ; b=13$
Hence the answer is :
$(72*15*13)-(64*30*6) = 14040 - 11520 = \boxed{2520}$