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plz post your reasoning can we use decimal numbers to [#permalink]
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08 Aug 2006, 15:48
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This topic is locked. If you want to discuss this question please repost it in the respective forum. plz post your reasoning
can we use decimal numbers to disprove statement
Is A+B+C EVEN ?
ST1 A C  B IS EVEN
ST2 ( A C) B IS ODD
thanks
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e+e+e=e
o+o+o=o
e+o+o=e
o+e+e=o
(1) acb=even, so either all even or there are 2 odds and in either case the subtraction is same as addition, so addition must be even
SUFF
(2) (ac)b=odd, for odd result in multiplication, both factors must be odd..., so here (ac)=odd and b must be odd, because there is no evens in multiplication with odd oucome... however, (ac)...either a or c could be odd, so we will get o+o+e, which makes outcome even...
SUFF
so I say D... let's see if I screwed up



Manager
Joined: 26 Jun 2006
Posts: 152

mandy,
If it does not say numbers must be integers, then yes, you can try using decimals or fractions.
Anyway,
I am getting (C)
ST1. ACB is even, or ACB=2k
If we choose A,B,C all integers and even, then A+B+C is even
However, if we use fractions, it does A+B+C does not have to be even
Let's see: A=2.2, B=0.1, C=0.1 and ABC=2 even. But A+B+C is not
ST2. (AC)*B is odd
It follows that either
(1) AC is even and B is odd
or (2) AC is odd and B is even
or (3) AC is a fraction and B is integer such that (AC)*B=odd
or (4) AC is integer and B is a fraction such that (AC)*B=odd
Not sufficient.
Now, combine ST1 and ST2.
Neither AC nor B can be fractions, otherwise ST1 is not valid.
So, all numbers  A, B, and C  must be integers.
Moreover, either 1 or all 3 must be odd (otherwise ST2 is violated.)
Therefore, A+B+C is even
(C) is sufficient



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Re: DS even [#permalink]
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08 Aug 2006, 18:35
mandy wrote: can we use decimal numbers to disprove statement? Is A+B+C EVEN ?
ST1. A C  B IS EVEN ST2. ( A C) B IS ODD
E. good question. if A, B and C are not whole numbers/integers or are fractions, we cannot say.
initially i also thought B but after going through your question, should be E.
this is indeed another good question............................



Manager
Joined: 26 Jun 2006
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Prof,
Can you give me two examples where both ST1 and ST2 are satisfied and we get 2 different answers for the main question?
I am still convinced that it should be (C)...



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v1rok wrote: Prof, Can you give me two examples where both ST1 and ST2 are satisfied and we get 2 different answers for the main question?
I am still convinced that it should be (C)... if a = 4.5, b = 1 and c = 1.5 st 1. a  c  b = 4.5  1.5  1 = 2.........even st 2. (a  c) b = (4.5  1.5) 1 = 3........odd a + b + c = 4.5 + 1.5 + 1 = 7 ...odd. if a = 4, b = 1 and c = 1 st 1. a  c  b = 4  1  1 = 2.. even st 2. (a  c) b = (4  1) 1 = 3 .. odd a + b + c = 4 + 1 + 1 = 6 ... even. Quote: Is A+B+C EVEN ?
ST1. A C  B IS EVEN ST2. ( A C) B IS ODD



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This is E.
NEVER assume integers until explicitly stated.
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Manager
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THANKS FOR yours answers but i thought that even number have to be integer
becaause even is an integer when divided by 2
oa will come
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ps_dahiya wrote: This is E.
NEVER assume integers until explicitly stated.
Good ANSWER oa is E
plz can you explain me why
thanks
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Manager
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consider AD/BCE,
A
let A be 5.1,C be 2.1,A be 1
ABC=5.112.1=2(even)
but A+B+C=8.2 not even...
but if A is 6,B is 1 C is 3 both stmts are even hence A and D goes out
now consider B,
noe (5.12.1)1=3,odd
consider the same above combinations above,we will get yes for one
combinaton and no for another combination.hence B goes out
similarly for c we will get yes for 6,1,3 while no for 5.1,2.1,1
hence C also goes out
and giving us the answer E



CEO
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mandy wrote: ps_dahiya wrote: This is E.
NEVER assume integers until explicitly stated. Good ANSWER oa is E plz can you explain me why thanks
A+B+C is even does not mean that A,B and C are also integers. A = 1.5 B = 1.75 and C = 0.75 yields A+B+C = 4 which is even. These are simpler values. more complex would be A = 1.564325465879256 and B and C also like that........ this way the answer will be certainly E.
Moral of the Story: ALL VARIABLES ARE REAL NUMBERS UNLESS EXPLICITLY STATED
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Manager
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ps_dahiya wrote: mandy wrote: ps_dahiya wrote: This is E.
NEVER assume integers until explicitly stated. Good ANSWER oa is E plz can you explain me why thanks A+B+C is even does not mean that A,B and C are also integers. A = 1.5 B = 1.75 and C = 0.75 yields A+B+C = 4 which is even. These are simpler values. more complex would be A = 1.564325465879256 and B and C also like that........ this way the answer will be certainly E. Moral of the Story: ALL VARIABLES ARE REAL NUMBERS UNLESS EXPLICITLY STATED
so I understood well the whole sum is an even integer it does not mean
that each part each  has to be an integer
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