BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

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
Start Free 5-Day Trial Read 559 Reviews →

Data Sufficiency for 780+ Aspirants

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 amFollowed by: 9 members
Location: Delhi
Thanked: 84 times

by sureshbala » Sat Apr 11, 2009 3:24 am
Folks, sorry for the delay (I was on a vacation.)

hemantsood has given the correct answer to the previous question.

Here is the next one.

What is the sum of all the digits of the four-digit even number?

I. The sum of the first and the last digit is 6 more than the sum of the middle two digits.

II. The sum of the first three digits is 2 less than the last digit. The four digit number comprises 4 distinct digits and no zeros.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 232
Joined: Fri Jul 04, 2008 4:14 pmFollowed by: 1 member
Thanked: 14 times
GMAT Score:760

by rs2010 » Sat Apr 11, 2009 3:45 am
Let abcd be num
we need a+b+c+d
1. a+d=b+c+6 (not suff) 6000 or 8020
2. a+b+c=d-2 (not suff) as 1+2+3 then d=8 or 1+2+4 then d =9

Combining both number could be

2149 or 2438

Ans E
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 682
Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member
Thanked: 32 times

by Vemuri » Sat May 09, 2009 3:16 am
sureshbala wrote: What is the sum of all the digits of the four-digit even number?

I. The sum of the first and the last digit is 6 more than the sum of the middle two digits.

II. The sum of the first three digits is 2 less than the last digit. The four digit number comprises 4 distinct digits and no zeros.
IMO E.

Question is asking what is the value of a+b+c+d = ?

Stmt I : a+d=b+c+6 ==> a+a+d-6+d ==> 2(a+d)-6 (Insuff)

Stmt II: a+b+c = d-2 ==> d-2+d ==> 2d-2 (Insuff)

Combining both the statements, we still cannot determine the value of a+b+c+d
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members
Location: Toronto
Thanked: 1710 times
GMAT Score:800

by Stuart@KaplanGMAT » Sat May 09, 2009 8:29 am
hemantsood wrote:Let abcd be num
we need a+b+c+d
1. a+d=b+c+6 (not suff) 6000 or 8020
2. a+b+c=d-2 (not suff) as 1+2+3 then d=8 or 1+2+4 then d =9

Combining both number could be

2149 or 2438

Ans E
I haven't checked your math, but if you're correct about the 2 possible values, the answer is (C), since the question tells us that abcd is EVEN, which means that 2149 is NOT possible.

However, if you were just giving those as possible examples and there are others, the answer may in fact be (E).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Fri Feb 20, 2009 5:41 am

by tox4 » Thu May 14, 2009 6:35 am
So we have a number ABCD

I. NS (as in prev. posts)

II.
A+B+C=D-2

Now, since the number is even D can only be 0,2,4,6 or 8

0 - no zeros
2 - A+B+C=0, no zeros, not possible
4 - A+B+C=2, not possible, since # are distinct
6 - A+B+C=4, not possible, since # are distinct
8 - A+B+C=6, possible

So, SUFFICIENT

(B)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 752
Joined: Sun May 17, 2009 11:04 pm
Location: Tokyo
Thanked: 81 times
GMAT Score:680

by tohellandback » Tue Jun 09, 2009 12:46 am
IMO B,

1) example number 6000 or 1218 and many other numbers possible

2) numbers must contain 1,2,3 and 8 as last digit
1238,1328,2318...but sum of digits is same-15
so B is the answer

Thanks
The powers of two are bloody impolite!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Mon Jun 15, 2009 8:59 pm

by shashank.mehra » Wed Jun 17, 2009 5:40 pm
sureshbala wrote:Here is the next one...

The roots of a quadratic equation ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

1. Sum of the roots is positive.

2. Product of the roots is positive.
Alirhgt let R1 and R2 be the roots

Statement 1: R1 + R2 = +ve
Statement 2: R1* R2 = + ve. This statement is true when either both roots are negative or both are positive. Substitute both the cases in statement 1 (i.e. take either both positive or both negative), therefore, you see that both roots will have to be positive.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Wed Feb 14, 2007 1:34 am
Location: India, bangalore

by bizwizashish » Wed Jun 17, 2009 8:09 pm
B

Assume the number to be ABCD,,, where D can be anything from set of 2,4,6,8,0

1. A+D = B+C+6

numerous numbers available.. as B and C need not be different as well...

2. A+B+C = D-2

A != B != C ! = D ! = 0

Only possible value D = 8

Because the minimum sum of ABC has to be 6

hence B

what is OA ?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 160
Joined: Tue Jul 07, 2009 1:09 pmFollowed by: 1 member
Thanked: 1 times

by Sharma_Gaurav » Tue May 01, 2012 11:45 am
Answer is definitely B for this
1) many numbers are possible
2) only no possible are 1238, 2138, 3128 etc
and all digits add to 14 always . Hence option B is correct
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply