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operation@

by CITI29 » Tue Jul 29, 2008 2:49 pm
for all the integers x and y , the operation @ is defined by
x@y=(x+2)^2 + (y+3)^2 . What is the value of integer t?
1)t@2 =74
2)2@t =80

what am I missing in this question? I thot the ans is 'e'...but its 'c'.Pls explain.
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Re: operation@

by rolodex » Tue Jul 29, 2008 10:28 pm
CITI29 wrote:for all the integers x and y , the operation @ is defined by
x@y=(x+2)^2 + (y+3)^2 . What is the value of integer t?
1)t@2 =74
2)2@t =80

what am I missing in this question? I thot the ans is 'e'...but its 'c'.Pls explain.
From S1: (t+2)^2 + (2+3)^2 = 74
t^2 + 4t -45 = 0;
t = -9 and t = 5 SO MAYBE CASE AD are out

Similarily solve for statemetn 2 and get the values for t. if there is a common value in S1 and S2 then C. IF not then E.
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by sibbineni » Wed Jul 30, 2008 3:58 pm
x@y=(x+2)^2 + (y+3)^2
S1: here
t@2 =74
x=t and y=2

(t+2)^2 + (2+3)^2 = 74
t^2 + 4t -45 = 0;
t = -9 and t = 5 (insuff)

S2: here
2@t =80

x=2 and y=t

t^2+6t-55=0

t=-11 or t=5 (Insuff)


Together

t=5 SO C should be the answer
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Re: operation@

by lunarpower » Wed Jul 30, 2008 11:57 pm
the math on the preceding posts appears to be quite correct, but, watch it with the following claim:
rolodex wrote:Similarily solve for statemetn 2 and get the values for t. if there is a common value in S1 and S2 then C. IF not then E.
wrong.

this is not how it works.

the two statements (1) and (2) will NEVER be inconsistent with each other. in other words, there will ALWAYS be at least one value of 't' that is common to the two statements.
if you don't get any common values - say, t = 1 or 2 for the first statement, and t = 3 or 4 for the second statement - then you have solved at least one of the statements incorrectly, and should redo the problem if you have the time.

the answer will be (e) if there is STILL MORE THAN ONE SOLUTION to the combination of the two statements (1) and (2). that's what "insufficient" means: there's still more than one possibility for the solution, despite all the information that's available.

summa veritas: (assuming that both individual statements have proved insufficient)
* combined statements yield more than 1 common value --> (e)
* combined statements yield exactly 1 common value --> (c)
* combined statements yield no common values --> you have egg on your face
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