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# Cost of War

## 62 posts in this topic

Posted · Report post

But no-one has mentioned the over time factor. Thats when an amputation leads to another loss of limb, which happened alot in pre-mid 20th century wars. They all died, except for, maybe 1 or 2.

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Posted · Report post

But no-one has mentioned the over time factor. Thats when an amputation leads to another loss of limb, which happened alot in pre-mid 20th century wars. They all died, except for, maybe 1 or 2.

well, the question was not asking who died ... it was as follows:

What is the minimum number of soldiers who must have lost all 4?

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Posted · Report post

Indeed, however, gangreen could claim more body parts.

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Posted · Report post

Indeed, however, gangreen could claim more body parts.

which has no immediate impact on the minimum number of soldiers who must have lost all 4 body parts, does it?

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Posted · Report post

You know, my favorite part about questions like this is that all the correct solutions people come up with are exactly the same mathematically. Adding up the uninjured and finding the intersection are the same mathematically.

Note: I shortened the mathematics heavily because I don't want to take a page to explain.

100 - 70 = 30

100 - 75 = 25

100 - 80 = 20

100 - 85 = 15

30 + 25 + 20 + 15 = 90

100 - 90 = 10

Lumping all these calculations together and simplifying:

100 - (100 - 70) - (100 - 75) - (100 - 80) - (100 - 85) = 10

100 - 100 + 70 -100 + 75 - 100 + 80 - 100 + 85 = 10

70 + 75 + 80 + 85 - 100 - 100 - 100 = 10

Intersection (shortened heavily):

70 + 75 - 100 = 45

45 + 80 - 100 = 25

25 + 85 - 100 = 10

Lumping all these equations together and simplifying:

70 + 75 - 100 = 45

(70 + 75 - 100) + 80 - 100 = 25

70 + 75 + 80 - 100 - 100 = 25

(70 + 75 + 80 - 100 - 100) + 85 - 100 = 10

70 + 75 + 80 + 85 - 100 - 100 -100 = 10

As you see, the last line is the same in both methods. The only difference between the two methods is the order of operations. This also leads to the solution I got, add them all together (70 + 75 + 80 + 85) and subtract 300.

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Posted · Report post

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!

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Posted · Report post

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.

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Posted · Report post

I began by subtracting the numbers who lost an eye or an ear from the total number of soldiers.

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.

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Posted · Report post

I guess I'm still tired for the weekend, but I couldn't rap my head around the math, and I couldn't work it out mentally so that it wasn't coming out as 0 (which I think is typically the answer to this question when you don't really work at it - logically it would appear to be the minimum is zero.)

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.

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Posted · Report post

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.

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Posted · Report post

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

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Posted · Report post

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.

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Posted (edited) · Report post

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.

Edited by els-chan
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Posted · Report post

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.

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Posted · Report post

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.

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 them

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Posted · Report post

100-70=30

100-75=25

100-85=15

100-80=20 100-90=10 (100-30,25,15,20= 10)

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Posted · Report post

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.

its okay i still believe your a smart person

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Posted · Report post

Wouldn't the minimum number of soldiers that lost all 4 limbs be 1?

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Posted · Report post

To me an interesting problem is to create a generic equation to solve any problem like this.

I think this is the answer but would be interested for someone to confirm.

Let:

T = total number of soldiers

N = number of different types of injuries

X = number of soldiers injured for each injury (index i from 1 to N)

Z = minimum number of soldiers with all N injuries

Z = (T - (N)(T)) + summation i=1 to N (X)

for this problem

Z = (100 - (4)(100)) + 70 + 75 + 80 + 85 = 10

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Posted · Report post

Indeed, however, gangreen could claim more body parts.

this is only a puzzle, just stick with what it gave you, and dont think about all the other possibilities of gangreen etc

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Posted · Report post

If you want to get into semantics,

I think that the answer is 70 because the question is not well stated enough for the answer to be logically 10.

Maybe if it was 100 'out of' the answer would be 10.

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Posted · Report post

its 10 no duh!!

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Posted · Report post

if you name all 100 solders first one being named 1 and so on this is easy

1-70 lost an eye

1-75 lost an ear

1-85 lost a leg

20-100 lost an arm

so there are 20 who are missing an eye an ear and a leg â€¦but not an arm

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Posted · Report post

This is super easy. If the least common injury totals 70, it cant exceed that number because if only 70 had injury A and B,C, and D are more than that, the average of those becomes 77.5 but bearing in mind only 70 had A and 75 had B, it can only be the least.

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Posted · Report post

This is super easy. If the least common injury totals 70, it cant exceed that number because if only 70 had injury A and B,C, and D are more than that, the average of those becomes 77.5 but bearing in mind only 70 had A and 75 had B, it can only be the least.

??????????????

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