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 →

beginners'bin#4

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members
Location: India
Thanked: 267 times
GMAT Score:760

beginners'bin#4

by sanju09 » Sat Mar 27, 2010 3:57 am
If x, y, and z are consecutive negative integers, and if x > y > z, which of the following must be a positive odd integer?
(A) x y z
(B) (x - y) (y - z)
(C) x - y z
(D) x (y + z)
(E) x + y + z
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Join the discussion
Log increate an account to reply.
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Mon Mar 15, 2010 6:15 amFollowed by: 1 member
Thanked: 32 times

by eaakbari » Sat Mar 27, 2010 4:02 am
IMO B
Last edited by eaakbari on Sat Mar 27, 2010 10:35 am, edited 1 time in total.
Top
Join the discussion
Log increate an account to reply.
Master | Next Rank: 500 Posts
Posts: 114
Joined: Mon Sep 22, 2008 3:51 am
Thanked: 8 times
GMAT Score:680

by Fiver » Sat Mar 27, 2010 9:18 am
Choose B
It will always yeild 1 as the result.
Top
Join the discussion
Log increate an account to reply.
Master | Next Rank: 500 Posts
Posts: 268
Joined: Wed Mar 17, 2010 2:32 am
Thanked: 17 times

by this_time_i_will » Sat Mar 27, 2010 9:25 am
Fastest way would be to take examples. But, we needd to ensure that we have considered all cases.

Take (-1, -2, -3) & (-2,-3,-4).

Only B fits the bill.
Top
Join the discussion
Log increate an account to reply.
Master | Next Rank: 500 Posts
Posts: 148
Joined: Tue Jun 09, 2009 11:06 pm
Thanked: 7 times

by pops » Sat Mar 27, 2010 7:42 pm
thats correct, whenever solving a question by keeping examples in place of variables (x,y,z) it is important that you take more than 1 example..
Here, for example, if you take -1,-2,-3 even D would give a positive, Odd Number... So, its better you take one more example -2,-3,-4 to arrive at conclusion !
Top
Join the discussion
Log increate an account to reply.
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Wed Mar 10, 2010 6:11 am
Thanked: 1 times

by suryapal » Sun Mar 28, 2010 5:25 am
yup its B
Top
Join the discussion
Log increate an account to reply.
Legendary Member
Posts: 809
Joined: Wed Mar 24, 2010 10:10 pmFollowed by: 4 members
Thanked: 50 times

by akhpad » Sun Mar 28, 2010 8:46 am
as x, y and z are consecutive -ve integer

x-y = 1 and y-z = 1

(x-y)(y-z) = 1

Ans B
Top
Join the discussion
Log increate an account to reply.
User avatar
Legendary Member
Posts: 537
Joined: Sun Jul 19, 2009 7:15 amFollowed by: 1 member
Location: Nagpur , India
Thanked: 41 times

by rockeyb » Sun Mar 28, 2010 10:22 am
An alternate approach :

We know x > y >z . And x , y and z are consecutive integers , we also know consecutive integers alternate between odd and even integers .

But we do not know which variable is odd / even .

So there can be only 2 cases .

Case 1 : y = even .
So x = odd and z = odd .

Lets look at the answer options .

(A) x y z -> Odd x Even x Odd = Even , eliminate

(B) (x - y) (y - z) -> Odd - Even = Odd , Even - Odd = Odd .
Odd x Odd = Odd .
(C) x - y z -> Odd -Even = Odd .

(D) x (y + z) -> Odd x Odd = Odd .

(E) x + y + z -> Odd + Even + Odd = Even.

So far we have B , C , D as contenders .

Lets look at Case 2 :

y = odd .

x= even and z = even .

Lets check answer options now.

(A) x y z -> even x odd x even = Even , eliminate

(B) (x - y) (y - z) -> (even-odd)(odd - even)
odd x odd = odd .
(C) x - y z -> even - Even = even .

(D) x (y + z) -> even x Odd =even.

(E) x + y + z -> Even + Odd + Even = Odd .


Only option B gives the Odd result in both the cases .

Plugging in will be much faster but some times you may forget some values .
"Know thyself" and "Nothing in excess"
Top
Join the discussion
Log increate an account to reply.
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members
Thanked: 173 times
GMAT Score:710

by gmatmachoman » Sun Mar 28, 2010 10:45 am
rockeyb wrote:An alternate approach :

We know x > y >z . And x , y and z are consecutive integers , we also know consecutive integers alternate between odd and even integers .

But we do not know which variable is odd / even .

So there can be only 2 cases .

Case 1 : y = even .
So x = odd and z = odd .

Lets look at the answer options .

(A) x y z -> Odd x Even x Odd = Even , eliminate

(B) (x - y) (y - z) -> Odd - Even = Odd , Even - Odd = Odd .
Odd x Odd = Odd .
(C) x - y z -> Odd -Even = Odd .

(D) x (y + z) -> Odd x Odd = Odd .

(E) x + y + z -> Odd + Even + Odd = Even.

So far we have B , C , D as contenders .

Lets look at Case 2 :

y = odd .

x= even and z = even .

Lets check answer options now.

(A) x y z -> even x odd x even = Even , eliminate

(B) (x - y) (y - z) -> (even-odd)(odd - even)
odd x odd = odd .
(C) x - y z -> even - Even = even .

(D) x (y + z) -> even x Odd =even.

(E) x + y + z -> Even + Odd + Even = Odd .


Only option B gives the Odd result in both the cases .

Plugging in will be much faster but some times you may forget some values .
@rockeyb

Rightly said!! I would say always try to plug in more distinct set of values to reach "conclusion".This hold true for DS also.
My inputs are 3 distinct sets.Even though it may take a extra 30-40 secs..u will certainly nail it!
Top
Join the discussion
Log increate an account to reply.
Master | Next Rank: 500 Posts
Posts: 182
Joined: Mon Apr 20, 2009 7:09 pmFollowed by: 1 member
Thanked: 1 times

by akahuja143 » Sun Mar 28, 2010 11:11 am
Plugging in numbers is easier and faster on this kind of probelms...

IMO B
Top
Join the discussion
Log increate an account to reply.
Post new topic Post Reply
• Page 1 of 1