Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average [#permalink]
Show Tags
18 May 2017, 00:28
1
1
00:00
A
B
C
D
E
Difficulty:
75% (hard)
Question Stats:
60% (02:22) correct 40% (03:42) wrong based on 108 sessions
HideShow timer Statistics
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average [#permalink]
Show Tags
18 May 2017, 01:12
To maximize z; from the given condition we shall take z=y Mean (z,y,x) = 40; i.e Sum = 120 Median= y= x-13 From A; If z=35, y=35 and x= 48 (35+13); sum= 70+48 = 118 From B; If z=36, y=36 and x=49 (36+13); sum=72+49 = 121 (which is more than sum) For remaining options sum will be more than 120 Hence Answer A.
x, y, and z are positive integers such that x ≥ y ≥ z. If the average [#permalink]
Show Tags
18 May 2017, 07:43
Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?
A. 35 B. 36 C. 37 D. 39 E. 40
average (arithmetic mean) of x,y, and z is 40 i.e. x+y+z = 40*3 = 120
median is (x–13) but since x ≥ y ≥ z so median of (x, y, z) = y = (x-13)
i.e. x+(x-13)+z = 120 i.e. 2x+z = 133
for z to be greatest, x must be smallest and z will be greatest when it is equal to y i.e. (x-13)
2x+(x-13) = 133 i.e. 3x = 146 i.e. x = 48.66
i.e. x min = 49 y and z max = ((120-49)/2 = 35.5 i.e. y = 36 and x max = 35
Answer: option A
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average [#permalink]
Show Tags
18 May 2017, 07:49
1
Top Contributor
3
Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?
A. 35 B. 36 C. 37 D. 39 E. 40
Since y is the middlemost value among the 3 numbers, we know that y is the median. Since we're told that the median = x-13, we can conclude that y = x-13
The average (arithmetic mean) of x,y, and z is 40 So, (x+y+z)/3 = 40 Multiply both sides by 3 to get: x + y + z = 120 Replace y with x-13 to get: x + x-13 + z = 120 Simplify: 2x - 13 + z = 120 Add 13 to both sides: 2x + z = 133 Solve for z to get: z = 133 - 2x
We're told that y ≥ z So, we can conclude that x-13 ≥ 133 - 2x Add 2x to both sides: 3x - 13 ≥ 133 Add 13 to both sides: 3x ≥ 146 Divide both sides by 3 to get: x ≥ 48.666...
Since x is an INTEGER, the SMALLEST possible value of x is 49 We already concluded that z = 133 - 2x We can see that we can MAXIMIZE the value of z by MINIMIZING the value of x The SMALLEST possible value of x is 49 So, plug in x = 49 to get: z = 133 - 2(49) = 35
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average [#permalink]
Show Tags
19 May 2017, 03:18
1
Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?
A. 35 B. 36 C. 37 D. 39 E. 40
we have to maximize z; to do that take y=z Mean (x,y,z) = 40; (x+y+z)/3=40 x+y+z = 120 Median= y= x-13 given; 2x-13+z=120 2x+z=133 as we took y=z=x-13 we get 3x=146 but by this x cant be integer instead of taking z= x-13 if we take x-14 we will have 3x= 147 which gives us x=49,y=36 and z=35. Hence Answer A.
_________________
_________________ Rules for posting in verbal Gmat forum, read it before posting anything in verbal forum Giving me + 1 kudos if my post is valuable with you
The more you like my post, the more you share to other's need.
“The great secret of true success, of true happiness, is this: the man or woman who asks for no return, the perfectly unselfish person, is the most successful.” -Swami Vivekananda
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average [#permalink]
Show Tags
22 May 2017, 18:34
1
Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?
A. 35 B. 36 C. 37 D. 39 E. 40
We are given that the average of x, y, and z is 40; thus:
(x + y + z)/3 = 40
x + y + z = 120
We know that x ≥ y ≥ z and the median is x - 13. The median of three numbers is the second largest number, so y is the median. That is, y = x - 13. However, if we want to determine the greatest possible value of z (i.e., the smallest number), we want z to be as close to y (i.e., the median) as possible. Thus, we can let z be x - 13 also and we have:
x + x - 13 + x - 13 = 120
3x = 146
x = 146/3 = 48.66
Since x must be an integer, the smallest value of x is 49, y = x - 13 = 36, and thus the greatest value of z is 120 - (49 + 36) = 35.
Answer: A
_________________
Scott Woodbury-Stewart Founder and CEO
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions