try setting up a 2 by 2 matrix, it gets much easier.



1) with the info given, we can find out the number of unbroken in box 2

2) solving for z which is 32, we can populate the matrix and find out the required info.

both statements are equally sufficient.

scoobydoo...fantastic

you made that look so easy...I feel like kicking my own ass for missing this

There is no need for a matrix to solve this question , its very basic logic

First statement is sufficient ( there is no problem with it)

Lets look at the second one

Before that what is given :

1.0 there are 55 bulbs
2.0 2 broken bulbs - First box
3.0 5 broken bulbs in the second
{ remember that the total (55) includes the broken ones and there distribution is given )

so from this we know that (2+good bulbs in A)+(5+good ones in B) =55

so B has three o extra bad bulbs

Then Statement II states that A has 12 more B , so we can find out how many good bulbs are there in B

Sufficient

So D