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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
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.
26 Jun 2018, 19:16
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
80% (01:07) correct
20%
(01:15)
wrong
based on 61
sessions
History
Date
Time
Result
Not Attempted Yet
If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?
(1) The average (arithmetic mean) of John's highest and lowest scores was 118.
(2) John's highest score was 154.
(1) The average (arithmetic mean) of John's highest and lowest scores was 118.
(2) John's highest score was 154.
Show SpoilerDoubt
This question is #90 from Jeff Sackmann's Statistics-Sets 100 Challenge Q's.
I'm a bit stumped by OA on this question. I don't understand why it's not C. My thought process:
a) Say we have X, Y, Z. If X is 140, Z could be 96 to average 118, and then Y would be 100, making Y the median because X>Y>Z. If X is 120, and Z is 116, they average to 118, Y would still be 100 and Z would be the median because X>Z>Y. Not Sufficient
b) Clearly not sufficient alone.
Combined - If you know what the highest (X) is, and you know what the average of X and Z is and the average of X, Y, Z, then one could determine the median. Am I missing something?
I'm a bit stumped by OA on this question. I don't understand why it's not C. My thought process:
a) Say we have X, Y, Z. If X is 140, Z could be 96 to average 118, and then Y would be 100, making Y the median because X>Y>Z. If X is 120, and Z is 116, they average to 118, Y would still be 100 and Z would be the median because X>Z>Y. Not Sufficient
b) Clearly not sufficient alone.
Combined - If you know what the highest (X) is, and you know what the average of X and Z is and the average of X, Y, Z, then one could determine the median. Am I missing something?
26 Jun 2018, 20:15
Kudos
Bookmarks
TippingPoint93
If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?
The median of a set with odd number of terms is the middle term, when arranged in ascending/descending order.
Say, the three scores are x, y, and z, where \(x \leq y \leq z\). We are told that x + y + z = 3*112 and asked to find the value of y.
(1) The average (arithmetic mean) of John's highest and lowest scores was 118 --> x + z = 2*118. Substitute this in the above equation: 2*118 + y = 3*112 --> y = 100. Sufficient.
(2) John's highest score was 154 --> z = 154. Not sufficient.
Answer: A.
26 Jun 2018, 21:22
Kudos
Bookmarks
Bunuel
Hi Bunuel,
Thanks for responding. If the average of the lowest and highest is 118, then the third term will always be 100. [I understand this part]
But couldn't the largest term be 140 and the smallest be 96? making 100 the median? And couldn't the largest term also be 120 with other term 116, still averaging to 118, thus, making 100 the smallest term and hence not the median?
