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# Cost of War

### #31

Posted 09 December 2007 - 05:47 AM

Because out of 100 people if 70 people lost an

no more or no less than 70.

It cant be more than 70 because no more then 70 lost an eye.

Get it?

Because its straining my brain to think how to explain it.

---------------------------------------------------------------------------------------------

Nm I total misunderstood the question!

### #32

Posted 24 January 2008 - 06:13 PM

**THE ANSWER IS 10!!!!!**

**DEAL WITH IT!!!!!!!**

sorry about all the mads, but the answer really is 10 and there isn't nothing you can do to change it. sorry.

### #33

Posted 29 January 2008 - 12:42 AM

This yielded 25 who did NOT lose an EAR and 30 who did NOT lose an EYE.

Add the uninjured and subtract from 100. (100-(25+30))

Since 55 were free of one injury or the other, 45 lost BOTH and EYE and an EAR.

Now taking this group as a new group that I called EYE&EAR, I again subtracted the injured from the total

number of soldiers and compared this group with ONE of the remaining injury groups.

I already know that 55 did NOT lose EYE&EAR and since 85 lost a LEG, 15 did not.

Add the uninjured and subtract from 100. (100-(55+15))

Since 70 were free of injury to EYE & EAR & LEG, 30 lost EYE & EAR & LEG.

Now taking this group as a new group that I called EYE & EAR & LEG, I again subtracted the injured from the

total number soldiers and compared this group with the remaining group.

I already know that 70 did NOT lose EYE & EAR & LEG and since 80 lost an ARM, 20 did not.

Add the uninjured and subtract from 100. (100-(70+20))

Since 90 were free of injury to EYE & EAR & LEG & ARM, 10 suffered injuries to all four areas.

### #34

Posted 11 February 2008 - 09:05 PM

Anyway, I used a 10x10 grid in Excel and simply filled the cells using I,E,L,A for eye, ear, leg and arm. And it becomes quickly apparent that indeed, there is a minimum of 10 people with all four injuries and all the others are suffering with three.

### #35

Posted 13 February 2008 - 09:10 PM

**The answer is 45**. Here's the logic:

Within the 100 number line, you are overlapping 4 lines of different lengths and are trying to get the least possible amount of simultaneous intersections.

|------------------------------------100-------------------------------|

xxxxxxxxxx|----------------------------85--------------------------|

|------------------------75----------------------|xxxxxx

**25**xxxxxx

|--------------------------80------------------------|xxxxxxxxxxxxx

xxxxxxxx

**30**xxxxxxxx|---------------------70--------------------|

To get the least number of intersections, each of the shortest lines (70 and 75) has to start at an opposite end of the number line. Regardless of where you position the longer lines, the minimum length of intersection between all four lines is the intersection of the shortest 2 lines starting from opposite ends of the number line. This length is 45 (75-30 or 70-25).

Think about it. Try any other positioning and you will always get a number larger than 45.

### #36

Posted 13 February 2008 - 10:26 PM

**the answer is 10**... as written several times before ... eg. check oranfry's post for lines

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### #37

Posted 14 February 2008 - 03:56 PM

the answer is 10... as written several times before ... eg. check oranfry's post for lines

Thanks for your response.

**You're right. My logic was flawed**. I based it on the assumption that the lines have to be contiguous. If I break the lines to avoid intersections in a more efficient way, the

**common intersection reduces to 10**.

It sounds more reasonable too. 45 seemed like an awefully high figure

Cheers.

### #38

Posted 26 February 2008 - 08:02 PM

I'd broken down the injuries as follows:

Let eye=Y

Let ear=E

Let leg=L

Let arm=A

Y=70/100, or 14/20

E=1/4, or 5/20

L=17/20

A=4/5, or 16/20

Just by looking at the Y and E variables, 1/20 of the soldiers are unaccounted for, meaning 5 of them would have only one or the other, but not both. Because those five would not have both injuries, they can't have all four injuries.

Can someone please show me where I made my error or where the logic is faulty? I'm not picking it up at all, though I'm sure it's something really obvious.

**Edited by els-chan, 26 February 2008 - 08:07 PM.**

### #39

Posted 26 February 2008 - 08:14 PM

I'm not doubting that the correct answer is 10, however I definitely came up with a different answer (0) and I'm hoping someone can show me the flaw in my logic (without pointing me back to the answer given - I get the adding up thing, thanks.)

I'd broken down the injuries as follows:

Let eye=Y

Let ear=E

Let leg=L

Let arm=A

Y=70/100, or 14/20

E=1/4, or 5/20

L=17/20

A=4/5, or 16/20

Just by looking at the Y and E variables, 1/20 of the soldiers are unaccounted for, meaning 5 of them would have only one or the other, but not both. Because those five would not have both injuries, they can't have all four injuries.

Can someone please show me where I made my error or where the logic is faulty? I'm not picking it up at all, though I'm sure it's something really obvious.

Nevermind. I'm retarded. I'm calculating 75% as 25% and I don't know why. PLEASE DISREGARD.

I'd edit it but the edit feature isn't working properly for some reason.

### #40

Posted 26 February 2008 - 08:25 PM

the edit feature is working properly - exactly as I set it up - you are allowed to edit your own posts within 10 minutes after submitting themNevermind. I'm retarded. I'm calculating 75% as 25% and I don't know why. PLEASE DISREGARD.

I'd edit it but the edit feature isn't working properly for some reason.

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