Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
11 Oct 2018, 03:13
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
46% (01:36) correct
54%
(01:17)
wrong
based on 185
sessions
History
Date
Time
Result
Not Attempted Yet
If the average (arithmetic mean) of the three different positive integers x,y and z is 3, what is the median of the three integers?
(1) x=5
(2) y=3
(1) x=5
(2) y=3
11 Oct 2018, 03:56
Kudos
Bookmarks
T1101
Answer:
Given that the average of three numbers is 3.
1) since x= 5, this means other two numbers must be less than 5 (because mean of xyz is 3, hence there sum must be 9, so if x =5, =>. y+z = 4, so neither y nor z can be greater than 5, also 4)
Lets try to find out what will be our second largest number (lets say y).
Can Y be 4?, No since y+z must be equal to 4 and both y and z must be positive
Can Y be 3? Yes, since for y=3, we can have z=1 and y+z will be equal to 4. ALSO WITH THIS WE HAVE OUR MEDIAN AS 3.
Can Y be 2? No, because then for having y+z = 4, z must be 2 which can not be true as all three numbers must be different.
so we can only have cases with either Y=3 and Z=1 (as shown) or vice-versa. In any of the two case, MEDIAN WILL BE 3. So option 1 is sufficient.
2) For y=3 and mean of X,Y,Z = 3, Y must be at the center. Why?
because if X and Z will be greater than Y, the 3 CANT BE THE MEAN ( as y=3 will be smallest). Sim, will be the case wit X,Z lower than Y.
Hence Y must be in the middle . This is also sufficient.
so both conditions are sufficient. Hence choice D is right
15 Oct 2018, 01:16
Kudos
Bookmarks
T1101
(1) x=5
x+y+z=9
x=5
y+z=4
x,y,z can be arranged in
x,y,z
or
x,z,y ( as x is the greatest)
the y,z pairs can be (3,1)
median when (3,1) = 3
in both the cases we can determine the median
sufficient
(2) y=3
x+z=6
the x,z pairs can be (2,4)(3,3)(5,1)
median when (2,4) = 3
median when (5,1) = 3
in each of the case , we can determine the median
sufficient
D
