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is c even?

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by sanju09 » Thu Apr 09, 2009 4:03 am
If c and d are integers, is c even?

(1) c (d + 1) is even.

(2) (c + 2) (d + 4) is even.



OA C
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by bluementor » Thu Apr 09, 2009 4:24 am
If c and d are integers, is c even?

Statement 1: c (d + 1) is even.

if d is odd, c could even or odd. Insufficient.

Statement 2: (c + 2) (d + 4) is even.

if d is even, c could be even or odd. Insufficient.

Both statements together:

d = odd, c = even -> possible on both statements
d=even, c = even-> possible on both statements
d = odd, c = odd -> not possible in statement 2
d = even, c = odd -> not possible in statement 1

Hence, c is always even regardless of what d is. Sufficient.

Choose C.

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Re: is c even?

by iamcste » Thu Apr 09, 2009 4:24 am
sanju09 wrote:If c and d are integers, is c even?

(1) c (d + 1) is even.

(2) (c + 2) (d + 4) is even.



OA C

c (d + 1) is even.

c=1, d=1, c(d+1)=2 ( Even) C odd

c=2, d=1, c(d+1)=4 ( Even) , C is Even

Insufficient as C may be odd or even to get product even

(c + 2) (d + 4) is even.

c=1, d=2, (c + 2) (d + 4)=18 (Even) C is odd
c=2, d=1, (c + 2) (d + 4)=20 (Even) C is Even


Insufficient as C may be odd or even to get product even

Collectively,

Only when C is even, both c(d+1) and(c + 2) (d + 4) are even

c=1 (odd), d=1, c(d+1)=2 ( even) while (c + 2) (d + 4)= 15 (odd)

c=2, d=1, (c + 2) (d + 4)=20 (Even) while c(d+1) is 4 (Even)

so when C is even, both (c + 2) (d + 4) and c(d+1) are even

Sufficient

Choose C

Fastest way is to make a truth table of c and d with ( odd, even), ( even, odd), ( odd,odd) and ( even, even) combinations
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