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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

The minimum of the integers x, y, and z is 10 and their aver

Expert replies
by Max@Math Revolution » Thu Apr 26, 2018 12:12 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

[GMAT math practice question]

The minimum of the integers x, y, and z is 10 and their average is 11. What is the greatest possible value of their maximum?

A. 10
B. 11
C. 12
D. 13
E. 14
Join the discussion
Source: — Problem Solving |

by Keith@ThePrincetonReview » Thu Apr 26, 2018 10:21 am
Max@Math Revolution wrote:[GMAT math practice question]

The minimum of the integers x, y, and z is 10 and their average is 11. What is the greatest possible value of their maximum?

A. 10
B. 11
C. 12
D. 13
E. 14
Hi Max,

The least of three integers is 10. Let's let x = 10.

The question stem provides the average of the three integers, so we can write the equation (10 + y + z)/3 = 11.
Simplify the equation, so that 10 + y + z = 33, and y + z = 23.

The question asks for the greatest possible value of the greatest integer.
To maximize the value of one of the remaining integers, minimize the value of the other.
Since the least of the three integers is 10, neither y nor z can have a value less than 10.
If y = 10, and y + z = 23, then z = 13.

The correct answer is choice D.
Join the discussion

by Scott@TargetTestPrep » Fri Apr 27, 2018 9:30 am
Max@Math Revolution wrote:[GMAT math practice question]

The minimum of the integers x, y, and z is 10 and their average is 11. What is the greatest possible value of their maximum?

A. 10
B. 11
C. 12
D. 13
E. 14
Since the average is 11, the sum of the three integers is 11 x 3 = 33.

Since the smallest possible integer in the set is 10, we can let two of the integers equal 10, so the maximum integer is 33 - (2 x 10) = 13.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Join the discussion

by Max@Math Revolution » Sun Apr 29, 2018 5:23 pm
=>

Assume x ≤ y ≤ z.
( x + y + z ) / 3 = 11 and x = 10
We have 10 + y + z = 33 or y + z = 23.
In order to have the greatest maximum number, y must be the minimum which is 10.
10 + z = 23.
z = 13.

Therefore, D is the answer.

Answer : D
Join the discussion

by swerve » Mon Apr 30, 2018 9:58 am
Given minimum integer is = 10

Average of 3 integers x, y, and z =11

Therefore Total =11∗3=33

ie; (x + y + z = 33)

The greatest value is possible if the other two values are minimum.

Let xx and y= 10

Therefore the greatest possible value of their maximum;

x + y + z = 33

10 + 10 + z = 33

20 + z = 33

z = 33 - 20 = 13. Option D.

Regards!
Join the discussion

by Brent@GMATPrepNow » Mon Apr 30, 2018 10:21 am
Max@Math Revolution wrote:[GMAT math practice question]

The minimum of the integers x, y, and z is 10 and their average is 11. What is the greatest possible value of their maximum?

A. 10
B. 11
C. 12
D. 13
E. 14
Key concept: If we know the sum of a set of numbers, and we want to MAXIMIZE the biggest number in the set, we must MINIMIZE all of the other numbers.

GIVEN: Average of x, y, and z is 11
So, (x + y + z)/3 = 33
This means x + y + z = 33
Great! We know the sum of the values.

In order to MAXIMIZE the biggest number in the set, we must MINIMIZE all of the other numbers.
We're told that 10 is the MINIMUM value in the set.
So, let's let TWO of the values equal 10
Say x = 10 and y = 10
We have now MINIMIZED two of the three values.

Since we know that x + y + z = 33, we can now write 10 + 10 + z = 33
Solve to get: z = 13
So, the MAXIMUM value is 13.

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion