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Source: — Data Sufficiency |

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by scoobydooby » Sat Jun 20, 2009 9:45 am
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.
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by hetavdave » Sat Jun 20, 2009 9:48 am
scoobydoo...fantastic :)

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

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by ghacker » Sun Jun 21, 2009 10:05 am
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