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Old 01-18-2013, 11:02 AM   #1
scottybuzz
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Maths question for you all...!

The avg. (arithmetic - mean) of a and b is ninety; the avg. (arithmetic mean) of a and c is 150; what is the total value of (b-c)/2?

Last edited by scottybuzz; 01-18-2013 at 11:16 AM..
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Old 01-18-2013, 11:03 AM   #2
Webmaster Advertising
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Three fiddy.
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Old 01-18-2013, 11:18 AM   #3
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quick bitches!
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Old 01-18-2013, 11:33 AM   #4
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-60





...........
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Old 01-18-2013, 11:54 AM   #5
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You can't solve something like this you have one equation with two unknowns. Because arithmetic average of a and b is ((a+b)/2) and you said it's 90, but you don't know from this, what is a or b
And the same is is valid for c and d, of course.
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Old 01-18-2013, 11:57 AM   #6
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there is no D.... unless he is seeking the sum of D as to which I thought would be X
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Old 01-18-2013, 11:57 AM   #7
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Macs suck!
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Old 01-18-2013, 12:00 PM   #8
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Math makes my brain hurt
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Old 01-18-2013, 12:06 PM   #9
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you just subtract 2nd equation from the first...

(a+b)/2=90
(a+c)/2=150
-------------------
((a+b)-(a+c))/2=90-150
(a+b-a-c))/2=-60
(b-c)/2=-60
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Last edited by woj; 01-18-2013 at 12:10 PM..
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Old 01-18-2013, 12:09 PM   #10
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Quote:
Originally Posted by woj View Post
you just subtract 2nd equation from the first...

(a+b)/2=90
(a+c)/2=150
-------------------
(b-c)/2=-60
You may be dead on, or dead wrong. I am too old to remember these equations. If I had to guess though, I would say you were RIGHT!
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Old 01-18-2013, 12:10 PM   #11
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Quote:
Originally Posted by woj View Post
you just subtract 2nd equation from the first...

(a+b)/2=90
(a+c)/2=150
-------------------
(a+b-(a+c))/2=90-150
(a+b-a-c))/2=-60
(a+b-a-c))/2=-60
(b-c)/2=-60
this is correct, and I said it already a few posts up
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Old 01-18-2013, 12:11 PM   #12
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there is no D.... unless he is seeking the sum of D as to which I thought would be X
Fuck, I didn't read it properly, in this case I agree with " d-null " it's -60
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Old 01-18-2013, 12:14 PM   #13
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Three fiddy.
Gets my vote...
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Old 01-18-2013, 12:15 PM   #14
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this is correct, and I said it already a few posts up
solution without proof is useless...
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Old 01-18-2013, 12:19 PM   #15
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it's whatever the rothchilds and rockefellars say it is!
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Old 01-18-2013, 12:19 PM   #16
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solution without proof is useless...
you are right, I thought it too simple to bother

a+b=180
a+c=300

a=180-b
a=300-c

180-b=300-c

-b=120-c

b=c-120

(b-c)/2


(c-120-c) /2

-60
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Old 01-18-2013, 12:47 PM   #17
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(a+b)/2=90
(a+c)/2=150

a+b=180
a+c=300

a=180-b
c+180-b=300

c-b=120
-120=b-c
-60=(b-c)/2
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Old 01-18-2013, 01:01 PM   #18
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Originally Posted by d-null View Post
you are right, I thought it too simple to bother

a+b=180
a+c=300

a=180-b
a=300-c

180-b=300-c

-b=120-c

b=c-120

(b-c)/2


(c-120-c) /2

-60
My math teachers would give us an F if we didn't "write it out".

And.... wait for it.... wait for it... most of us ARE simple. lmao
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Old 01-18-2013, 01:16 PM   #19
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Reading this thread I see why so many people ask to explain what is the ratio column in our affiliate report and how it is calculated.
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Old 01-18-2013, 07:21 PM   #20
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(a+b)/2=90; (a+c)/2=150
(a+b)/2=((a+c)/2)-60
a/2 + b/2=a/2 + c/2 -60
b/2=c/2 -60
b/2-c/2=c/2-c/2-60
(b-c)/2=-60
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Old 01-18-2013, 07:42 PM   #21
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tube sites
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Old 01-18-2013, 07:52 PM   #22
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at least 5 different methods
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Old 01-18-2013, 07:58 PM   #23
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Old 01-18-2013, 07:59 PM   #24
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Old 01-18-2013, 08:00 PM   #25
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where are the bones ?
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Old 01-18-2013, 08:01 PM   #26
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Old 01-18-2013, 08:09 PM   #27
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You may be dead on, or dead wrong.
Or in my case arithmetically challenged.

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Old 01-18-2013, 08:14 PM   #28
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Years and years ago there was a software you could feed the two starting equations to and it would do all the steps and writing it out for you. I think it ran on my Apple ][ (yes, showing my age) - anybody remember what it was called?
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