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by vscid » Sun Dec 21, 2008 7:34 am
If A and B are integers, is B even?

1] 3a + 4b is even
2] 3a + 5b is even
The GMAT is indeed adaptable. Whenever I answer RC, it proficiently 'adapts' itself to mark my 'right' answer 'wrong'.
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by ronniecoleman » Sun Dec 21, 2008 7:42 am
] 3a + 4b is even
b can be anything odd or even...

Insuff


2] 3a + 5b is eve

a is even so b has to be even
a is odd so b has to be odd

Hence insuff

together

a is even so b has to be ODD

SO

IMO C
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by iamcste » Sun Dec 21, 2008 8:34 am
ronniecoleman wrote:]
together

a is even so b has to be ODD

SO

IMO C
As per your conclusions,
a=Even =assume 2 , b=Odd=1

3a+4b=10
3a+5b=11

Contradicts the facts given

so, even though Ans is C, Conclusions are incorrect IMO

Combining equations "3a+4b" and "3a+5b" are even only when both a and b are even and not when B is odd as you mentioned.

Pls let me know if I have missed anything
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by ronniecoleman » Sun Dec 21, 2008 9:10 am
iamcste wrote:
ronniecoleman wrote:]
together

a is even so b has to be ODD

SO

IMO C
As per your conclusions,
a=Even =assume 2 , b=Odd=1

3a+4b=10
3a+5b=11

Contradicts the facts given

so, even though Ans is C, Conclusions are incorrect IMO

Combining equations "3a+4b" and "3a+5b" are even only when both a and b are even and not when B is odd as you mentioned.

Pls let me know if I have missed anything

when combining both the statments i need to take the value of A that satisfies both the equation...

3a+ 4b
now 4b is always even
3a has to be even for its sum to be even
so a has to be even...
because ODD*ODD = ODD

3a+ 5b

now it can odd + odd
even + even

3a = a can be odd or even but common solution suggest a has to be even

now 3( even) + 5 ( even) for its sum to be even
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by iamcste » Sun Dec 21, 2008 9:20 am
ronniecoleman wrote:
iamcste wrote:
ronniecoleman wrote:]
together

a is even so b has to be ODD

SO

IMO C
As per your conclusions,
a=Even =assume 2 , b=Odd=1

3a+4b=10
3a+5b=11

Contradicts the facts given

so, even though Ans is C, Conclusions are incorrect IMO

Combining equations "3a+4b" and "3a+5b" are even only when both a and b are even and not when B is odd as you mentioned.

Pls let me know if I have missed anything

when combining both the statments i need to take the value of A that satisfies both the equation...

3a+ 4b
now 4b is always even
3a has to be even for its sum to be even
so a has to be even...
because ODD*ODD = ODD

3a+ 5b

now it can odd + odd
even + even

3a = a can be odd or even but common solution suggest a has to be even

now 3( even) + 5 ( even) for its sum to be even


Do you mean "3a = a can be odd or even but common solution suggest b has to be even

If yes, thats what I meant "b" has to be even unlike said in your earlier post "B has to be odd"
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by cramya » Sun Dec 21, 2008 12:40 pm
Stmt I

3a + 4b -> even (odd+odd->even or even+even->even)

3a-> odd 4b->odd or 3a->even 4b->even

4b can never be odd so 4b has to be even but b could be odd or even

For 3a to be even a has to be even

From stmt I all we know for sure is a is even . b can be odd or even

INSUFF

Stmt II

3a + 5b is even

3a-> odd 4b->odd or 3a->even 5b->even

a and b are bpth odd or both even

INSUFF

Stmt I and II

a is even

3a+4b->even

4b = even-3a

even - 3a is always evene so for 4 b to be even b could be odd or even

3a+5b->even

5b = even-3a

even-3a is always even so 5b has to be even and therefore b has to be even since 5* b(odd) will be odd

B IS EVEN

Choose C)
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by vittalgmat » Mon Dec 22, 2008 2:55 am
Excellent explanation Cramya,
I cant add anything to this..
Got C and used the same logic.
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by theprophet » Mon Aug 23, 2010 12:03 am
If you consider both together, the simplest way to do it is
3a + 5b is even 3a + 4b is even, if you subtract even from even you get even
which 3a + 5b - 3a - 4b = b which is even.
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