beater wrote:a,b and c are integers. Is abc = 0?
1. a^2 - 2a = 0
a=2 or a=0
Insuf
2. b/c = (a+b)^2 / (a^2+2ab+b^2) - 1
b/c = (a+b)^2 / [(a+b)^2 -1]
b/(b-c) = (a+b)^2
This gives us that b/(b-c) > 0
b can c can be 0
0/-c = 0
b/b = 1
or they can be any number
a+b can be any nunber too.
SO IMO this is INSUF as well
together we know that
a is either 0 or 2
for a = 0 SUF
for a = 2
b/(b-c) = (a+b)^2
b/(b-c) = b^2
It still does not prove that either b or c is 0
I did not like this question, but I will go with E
abc
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logitech
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LGTCH
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cramya
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Stmt I
a2-2a = 0
a(a-2) = 0
a=0 a=2
No info on bc
INSUFF
Stmt II
b/c = (a+b)^2 / (a+b)^2-1
All we know is (a+b)^2-1 > 0
INSUFF
Stmt I and II
Many possibilities for a ,b,c (some can be 0) or non zero
INSUFF
Choose E)
a2-2a = 0
a(a-2) = 0
a=0 a=2
No info on bc
INSUFF
Stmt II
b/c = (a+b)^2 / (a+b)^2-1
All we know is (a+b)^2-1 > 0
INSUFF
Stmt I and II
Many possibilities for a ,b,c (some can be 0) or non zero
INSUFF
Choose E)
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Ian Stewart
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I'm pretty sure, just judging by how contrived the second statement is, that the bracketing on the right side is supposed to be different from how logitech and cramya have interpreted it above. I'd guess it's supposed to read:beater wrote:a,b and c are integers. Is abc = 0?
1. a^2 = 2a
2. b/c = (a+b)^2 / (a^2+2ab+b^2) - 1
b/c = [(a+b)^2 / (a^2+2ab+b^2)] - 1
which changes the answer.
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logitech
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Yeap, then second statement will give us b/c=o since C can not be 0, we know that b=o and abc=0 so answer would be BIan Stewart wrote:I'm pretty sure, just judging by how contrived the second statement is, that the bracketing on the right side is supposed to be different from how logitech and cramya have interpreted it above. I'd guess it's supposed to read:beater wrote:a,b and c are integers. Is abc = 0?
1. a^2 = 2a
2. b/c = (a+b)^2 / (a^2+2ab+b^2) - 1
b/c = [(a+b)^2 / (a^2+2ab+b^2)] - 1
which changes the answer.
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
