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Source: — Data Sufficiency |
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by DanaJ » Fri Jan 30, 2009 1:56 pm
3a + 2b + c is an odd is equivalent to 3a + c is an odd number, since 2b is obviously an even number.
Now:
1. a is even and c is odd does not help us very much over here, since we cannot tell if b is odd or even.
2. if b is even, then we should use the fact that we've established that 3a + c is odd. These two pieces of information mean that 3a + c + b = 3a + b + c is odd (since 3a + c is odd and b is even). But we're actually looking for a + b + c, which is equal to 3a + b + c - 2a. Since 2a is always even and we know that 3a + b + c is odd, that means that a + b + c is always odd.

So I'd go with B.
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by ajmoney09 » Fri Jan 30, 2009 8:20 pm
I might be wrong but I think

the answer is A...

(e)(e) = Even
(O)(E) = Even
(O)(O) = Odd


(e)+(e) = Even
(O)+(E) = odd
(O)+(O) = even

Statement 1 says A is even and C is Odd

Which means

(Odd)(Even) + (even)(b) + odd = Odd

If we plug in an odd number for b then we get even as an answer making the statement untrue....

Although if we plug in and even number for b then we get an odd answer making the statement true....
OxE + ExE + O

E + E + O = Odd

Which then in turn gives us the answer for a+b+c = ?

thus the answer is A.
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by krisraam » Fri Jan 30, 2009 9:21 pm
IMO B.

3a + 2b + c is odd ==> 3a + c is odd(even + odd = odd)
==>a is even c is odd// a is odd c is even

so a is not sufficient.

b is even

a + b+c = even + (a+c) if a+c is odd a+b+c is odd.

3a + c is odd ==> a+c is always odd

So b alone is sufficient.

-raama
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by ontopofit » Sat Jan 31, 2009 1:01 am
gud one, B.
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by yalanand » Sat Jan 31, 2009 1:13 am
Let's try to get as much information as we can out of the question before we evaluate the statements. If 3a + 2b + c is an odd number, then if we subtract the even number 2b from it, we get 3a + c, which is an odd number. If we subtract the even number 2a from the odd number 3a + c, we get a + c, which is also an odd number. Thus, the question stem tells us that either a is odd and c is even, or a is even and c is odd.

To answer the question of whether a + b + c is an odd number, we need information about b.

Statement (1) is then insufficient. Statement (2) is sufficient, because if we know a + c is odd and b is even, then a + b + c is odd. So the answer is B.
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by Bidisha800 » Tue Feb 03, 2009 10:45 pm
3a+2b+c =odd
OR 2a+b+(a+b+c) =odd

2a is odd for all integer a

(a+b+c)= (odd-even) -b
=odd -b

if b is even then a+b+c is odd

if b is odd the a+b+c is even

(1) no information about b - insuff
(2) b is even -sufficient

(B)
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by gaggleofgirls » Wed Feb 04, 2009 8:30 pm
ajmoney09 wrote:
Statement 1 says A is even and C is Odd

Which means

(Odd)(Even) + (even)(b) + odd = Odd

If we plug in an odd number for b then we get even as an answer making the statement untrue....
Your flaw here is the even(odd) = even AND even(even) = even
SO, you can't tell if b is odd or even since either way, 2b is even

Therefore, we know a lot about a + b + c just from the stem.
We know it is either odd + ?? + even or even + ?? + odd. So the only thing we need to learn from the data is the sign of b.


1) Doesn't tell us any more than what the stem tells us.
2b has to be even. The expression 3a +2b +c has to be odd.
odd+even+odd = even so 3a and c must be odd/even or even/odd for the stem expression to be odd. For 3a to be odd, a must be odd (odd*odd=odd, odd*even=even). so, if a is odd, c must be even and if a is even, c must be odd.

In a DS question, if one of the choices tells us the same of less information than the stem, then it is useless meaning the answer won't be that choice alone (A or B, whichever it is), it won't be the combination (C) and it won't be either by itself (D), so it is either B or E at this point.

And, as we found working with the stem, we could have immediately looked at 1) and known it was insufficient, because all we need to know to answer the question is the sign of b and 1) doesn't give us any info about b.

2) knowing b's sign is what we need - sufficient. We don't need to solve any farther and actually figure out if a + b + c is odd or not.

Answer = B

-Carrie
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by mrsmarthi » Mon Feb 09, 2009 6:33 pm
here is my approach.

It is given 3a + 2b + c is odd. 2b will definitely be even irrespective of whether b is even or odd.

For the sum(3a + 2b + c) to be odd, either of 3a or c should be odd and the other even. If c is odd, then 3a should be even making a to be even. Otherwise, if c is even then 3a should be odd making the a to be odd.
So we know that, among a and c, one is definitely odd and one is definitely even. Now to find if a + b + c is even or odd, we need to know what b is since we already know that a and c are even odd combination.

From stmt 1, the value of b is not known. Hence insufficient.

From stmt 2, it is given that b is even. Hence it is sufficient to say whether a +b + c is even or odd.
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Re: integers

by x2suresh » Tue Feb 10, 2009 12:02 am
yalanand wrote:a, b, and c are integers and 3a + 2b + c is an odd number. Is a + b + c and odd number?

(1) a is even and c is odd
(2) b is even

B is the answer.

3a + 2b + c = a+b+c +(2a+b) = odd+(2a+b)

stat2:
b is even.. then expr= odd

sufficient
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