Poll of the Day > Solve this math problem please.

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HornedLion
10/11/20 8:00:33 PM
#1:


30 + B + C + D = 872

B is greater than C, D, & 30

C is greater D & 30 but less than B.

D is greater than 30 but less than B and C.

Normally I would be all over this type of problem but its late and Im run down. Can anyone figure it out?

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Far-Queue
10/11/20 8:01:40 PM
#2:


42

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Ogurisama
10/11/20 8:03:20 PM
#3:


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JOExHIGASHI
10/11/20 8:06:18 PM
#4:


Done

B=557
C=63
D=169

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HornedLion
10/11/20 9:08:12 PM
#5:


JOExHIGASHI posted...
Done

B=557
C=63
D=169

D is greater than 30 but less than B and C.

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Greenfox111
10/11/20 9:16:06 PM
#6:


i don't know how math is supposed to work but

B= 372
C = 290
D = 180

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streamofthesky
10/11/20 9:20:13 PM
#8:


Wouldn't there be a shit ton of possible answers?

B > C > D > 30

B + C + D = 872 - 30
B + C + D = 842

There's plenty of options for B, C, and D that meet those requirements. Example:
B = 285
C = 284
D = 273

If you wanted only one solution for each letter, you'd need more constraints than that or more info in general...
EDIT: I will at least mention that B has to be greater than or equal to 282 (assuming they have to be whole numbers and fractions/decimal values aren't allowed) in order for a solution to work. That's one parameter you can determine for sure, at least. (anything lower and C and D of any values lower than B won't add up to enough)
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Greenfox111
10/11/20 9:20:36 PM
#9:


yeah what that guy said ^

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DrPrimemaster
10/11/20 9:24:25 PM
#10:


It seems like there are a lot of things that make this work are you missing something? Am I missing something?
30
D = 31
C = 32
B = 779

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dainkinkaide
10/11/20 9:42:33 PM
#11:


30 + B + C + D = 872
B > C > D > 30

D = 279; C = 281; B = 282 OR D = 31; C = 32; B = 779 OR D = 31; C = 405; B = 406

282 B 779; 32 C 405; 31 D 279

This is, of course, only solving for integer solutions.

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