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.
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
lightbulbs
Source: Beat The GMAT — Data Sufficiency |
scoobydoo...fantastic 
you made that look so easy...I feel like kicking my own ass for missing this
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
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
















