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 →

The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible val

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 1223
Joined: Sat Feb 15, 2020 2:23 pmFollowed by: 1 member
by BTGModeratorVI » Wed Dec 16, 2020 12:37 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

Answer: B
Source: official guide
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

Re: The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible

by Brent@GMATPrepNow » Mon Dec 28, 2020 8:17 am
BTGModeratorVI wrote: ↑
Wed Dec 16, 2020 12:37 pm
 The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

Answer: B
Source: official guide
The average (arithmetic mean) of the positive integers x, y, and z is 3
We can write: (x+y+z)/3 = 3

Multiply both sides of the equation by [m]3[/m] to get: x + y + z = 9

At this point we need to find the greatest possible value of z
Important: Keep in mind that the three numbers are DIFFERENT POSITIVE INTEGERS

Since we know that x+y+z=9, we can MAXIMIZE the value of z by MINIMIZING the values of x and y
We know that x is the smallest value.
Since x must be a positive integer, the smallest possible value of x is 1
Since y must be different from x, the smallest possible value of y is 2
At this point we have maximized the value of z

If x=1 and y=2, then z=6

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2545
Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

Re: The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible

by swerve » Mon Dec 28, 2020 4:11 pm
BTGModeratorVI wrote: ↑
Wed Dec 16, 2020 12:37 pm
 The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

Answer: B
Source: official guide
The smallest positive integer is 1 and the 2nd smallest 2.

So, \(3\cdot 3 = 3+x \Longrightarrow x= 6\).

Therefore, B is the correct answer.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1