Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1207:30 AM PST
-08:30 AM PST
Join the conversation to learn more about INSEAD vs LBS
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
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. - Dec 16
08:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes! - Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
19 Aug 2022, 10:44
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
25% (05:33) correct
75%
(02:58)
wrong
based on 4
sessions
History
Date
Time
Result
Not Attempted Yet
The infinite sequence \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_{n} < a_{n-1}\), for all \(n > 1\). Does the sequence have more than 12 positive numbers ?
(1) \(\frac{a_{14}}{a_{13}}=\frac{\sqrt{10}}{\pi}\).
(2) The product of any two of the first 14 terms is positive.
(1) \(\frac{a_{14}}{a_{13}}=\frac{\sqrt{10}}{\pi}\).
(2) The product of any two of the first 14 terms is positive.
19 Aug 2022, 10:44
Kudos
Bookmarks
Official Solution:
The infinite sequence \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_{n} < a_{n-1}\), for all \(n > 1\). Does the sequence have more than 12 positive numbers ?
What does \(a_{n} < a_{n-1}\) tell us? Well, this means that every successive term (\(a_n\)) is less than the previous term (\(a_{n-1}\)). So, the sequence at hand is decreasing. (1) \(\frac{a_{14}}{a_{13}}=\frac{\sqrt{10}}{\pi}\).
What is all this about?
Firs of all, since the sequence is decreasing, then we know that \(a_{14} < a_{13}\). Now, IF \(a_{13}\) were positive, then after dividing \(a_{14} < a_{13}\) by \(a_{13}\) we'd have that \(\frac{a_{14}}{a_{13}} < 1\). But since \(\sqrt{10} > \pi\) (\(^*\) check below as to why?), then \(\frac{\sqrt{10}}{\pi} > 1\) and thus \(\frac{a_{14}}{a_{13}} >1\).
Therefore, our assumption that \(a_{13}\) is positive was wrong and in reality, \(a_{13} < 0\). But since \(a_{14} < a_{13}\), then \(a_{14} < a_{13} < 0\). The question asks whether the sequence has more than 12 positive numbers, and \(a_{14} < a_{13} < 0\) means that the sequence can have at most 12 positive numbers, NOT more than 12. We have a definite NO answer to the question.
Sufficient.
\(^*\) \(\pi \approx {\frac{22}{7}}\), and \(\pi^2 \approx {\frac{484}{49}}\), which is less than 10, so \(\sqrt{10} > \pi\).
(2) The product of any two of the first 14 terms is positive.
Well, the above would be true if ALL first 14 terms were positive (answer YES) as well as if ALL first 14 terms were negative (answer NO). So, this statement is not sufficient.
Answer: A
The infinite sequence \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_{n} < a_{n-1}\), for all \(n > 1\). Does the sequence have more than 12 positive numbers ?
What does \(a_{n} < a_{n-1}\) tell us? Well, this means that every successive term (\(a_n\)) is less than the previous term (\(a_{n-1}\)). So, the sequence at hand is decreasing. (1) \(\frac{a_{14}}{a_{13}}=\frac{\sqrt{10}}{\pi}\).
What is all this about?
Firs of all, since the sequence is decreasing, then we know that \(a_{14} < a_{13}\). Now, IF \(a_{13}\) were positive, then after dividing \(a_{14} < a_{13}\) by \(a_{13}\) we'd have that \(\frac{a_{14}}{a_{13}} < 1\). But since \(\sqrt{10} > \pi\) (\(^*\) check below as to why?), then \(\frac{\sqrt{10}}{\pi} > 1\) and thus \(\frac{a_{14}}{a_{13}} >1\).
Therefore, our assumption that \(a_{13}\) is positive was wrong and in reality, \(a_{13} < 0\). But since \(a_{14} < a_{13}\), then \(a_{14} < a_{13} < 0\). The question asks whether the sequence has more than 12 positive numbers, and \(a_{14} < a_{13} < 0\) means that the sequence can have at most 12 positive numbers, NOT more than 12. We have a definite NO answer to the question.
Sufficient.
\(^*\) \(\pi \approx {\frac{22}{7}}\), and \(\pi^2 \approx {\frac{484}{49}}\), which is less than 10, so \(\sqrt{10} > \pi\).
(2) The product of any two of the first 14 terms is positive.
Well, the above would be true if ALL first 14 terms were positive (answer YES) as well as if ALL first 14 terms were negative (answer NO). So, this statement is not sufficient.
Answer: A
