Posted 30 May 2007 - 01:56 PM

The number of soldiers who lost three body parts cannot be hundred. In fact, the maximum number of soldiers who lost three body parts can be only 80. Hence, as it has been reported in the first post, the statement that at a minimum, 100 soldiers lost 3 body parts is wrong.

According to me, the minimum number of people who lost four body parts will be 15.

Let w = number of people who lost exactly one body part

x = number of people who lost exactly two body parts

y = number of people who lost exactly three body parts

z = number of people who lost exactly four body parts

Since, the total number of soldiers is 100,

w + x + y + z = 100

And, when we add all the injuries (70 + 75 + 85 + 80), we get 310. In this 310, w is counted once, x is counted twice, y is counted thrice and z is counted four times.

Hence, w + 2x + 3y + 4z = 310 .... subtracting this equation from the first equation, we get...

x + 2y + 3z = 210

Now, in the above equation, we have to minimise z, which means we have to maximize y and x.

The set of injuries are 70, 75, 85, and 80. If we try to maximize y, then the maximum number of people with three injuries can be 75 (from the second, third and fouth group) and after this the standing will be 70, 0, 10 and 5 (subtracting 75 from second, third and fouth groups). Now, we can get another five people who have three injuries from the first, third and fourth groups which changes the group standings to 65, 0, 5 and 0. So, the maximum value of y can be 75 + 5 = 80. And with the new group standings the maximum value of x can be 5 (after which the group standings become 60, 0, 0 and 0). So, our last equation now is...

x + 2y + 3z = 210 where x = 5 and y = 80

3z = 45 -> z = 15