Bunuel wrote:
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. What is the minimum value of x?
A. 10
B. 12
C. 15
D. 20
E. 25
Are You Up For the Challenge: 700 Level QuestionsI believe the four limbs here mean the four organs given - eye, ear, leg and arm.
For ease of understanding, let us take eye, ear, arm and leg as a, b, c and d respectively.
Now if we want to minimise the value of x, which is people with all the four injuries, then distribute the injuries one after another in a way that it fills up 100% and then only moves to a person having second disease.
So 70% lost an eye. Next 80% lost an ear.
So the 100-70 or 30 are filled up for ear alone and then the remaining 80-30 of ear injuries are given to those who had already lost an eye.
At the end of this distribution, we have 50% with both injuries, 20% with just eye and 30% with just ear.
Next 75% lost an arm.
So we will fill up those with just one injury first. => So 20% now have injury in eye and arm, while 30% have in eye and arm. But we still have 75-20-30 or 25 arm injuries to fill up, who will go to those with eye and ear.
In the end of this distribution we have 25% with 3 injuries and remaining 75% with 2 injuries.
Next 85% lost a leg.
First we fill up those with two injuries, that is 75%, so 85-75 or 10% are still left and we have everyone with 3 injuries. Thus, the renaming 10% will have to be given to those people who already have 3 injuries.
So x has to be 10.
If you understand the concept of distribution above, the method is straightforward Add all of them => 70+80+75+85=310%
So 100% have three injuries, leading to 300%. So remaining 10% will make 10% of people to have all 4 injuries.
If you were looking for maximum value of x, we first give a set of people all the injuries, so the minimum of 4 will become the maximum value of x. Here 70 is the least of 70, 80, 75 and 85. So Max value of x is 70.