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.
Jun 0808:00 PM EDT
-10:00 PM EDT
A powerful GMAT course taught live online + 6 months of access to TTP OnDemand GMAT Masterclass included! Mon/Wed June 8, 2026 →August 12, 2026 8:00pm-10:00pm EST
Jun 1006:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
Jun 1111:00 AM EDT
-01:00 PM EDT
TTP GMAT OnDemand gives serious students 400+ hours of expert video instruction, the full TTP course, AI support, weekly office hours, and a 715+ score guarantee—all built for elite GMAT score improvement.
06 Aug 2025, 04:13
Kudos
Bookmarks
Ann and Bob each randomly select an integer from the integers \(x\) to \(y\), inclusive. What is the probability that the integer Ann selects is greater than the one Bob selects?
(1) \(y - x = 9\)
(2) \(y = -20\)
(1) \(y - x = 9\)
(2) \(y = -20\)
06 Aug 2025, 04:13
Kudos
Bookmarks
Official Solution:
Ann and Bob each randomly select an integer from the integers \(x\) to \(y\), inclusive. What is the probability that the integer Ann selects is greater than the one Bob selects?
(1) \(y - x = 9\)
This implies they are choosing a number from a set of 10 integers, for example, from 1 to 10, inclusive.
Then:
• Total outcomes = 10 * 10 = 100
• Equal cases (both choose the same number) = 10
• Remaining cases = 90
• Out of those 90, half will have Ann’s number < Bob’s, and half will have Ann’s number > Bob’s.
• So, required probability = 45/100 = 9/20
As you can see, the only thing we needed to know is the sample size. Sufficient.
(2) \(y = -20\)
This is clearly insufficient. We have no idea how many integers they’re choosing from.
Answer: A
Ann and Bob each randomly select an integer from the integers \(x\) to \(y\), inclusive. What is the probability that the integer Ann selects is greater than the one Bob selects?
(1) \(y - x = 9\)
This implies they are choosing a number from a set of 10 integers, for example, from 1 to 10, inclusive.
Then:
• Total outcomes = 10 * 10 = 100
• Equal cases (both choose the same number) = 10
• Remaining cases = 90
• Out of those 90, half will have Ann’s number < Bob’s, and half will have Ann’s number > Bob’s.
• So, required probability = 45/100 = 9/20
As you can see, the only thing we needed to know is the sample size. Sufficient.
(2) \(y = -20\)
This is clearly insufficient. We have no idea how many integers they’re choosing from.
Answer: A
