] 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
integers
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Source: Beat The GMAT — Data Sufficiency |
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ronniecoleman
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iamcste
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As per your conclusions,ronniecoleman wrote:]
together
a is even so b has to be ODD
SO
IMO C
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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ronniecoleman
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iamcste wrote:As per your conclusions,ronniecoleman wrote:]
together
a is even so b has to be ODD
SO
IMO C
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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iamcste
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ronniecoleman wrote:iamcste wrote:As per your conclusions,ronniecoleman wrote:]
together
a is even so b has to be ODD
SO
IMO C
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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cramya
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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)
Whats the OA?
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)
Whats the OA?
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vittalgmat
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theprophet
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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.
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.
