GMAT Changed on April 16th - Read about the latest changes here

It is currently 21 Apr 2018, 19:59

Close

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

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

S98-03

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 44599
S98-03 [#permalink]

Show Tags

New post 16 Sep 2014, 01:52
Expert's post
4
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

37% (01:32) correct 63% (01:24) wrong based on 30 sessions

HideShow timer Statistics

If \(x\), \(y\), and \(z\) are integers, with \(x \lt y \lt z\), what is the average (arithmetic mean) of \(x\), \(y\), and \(z\)?


(1) \((x + y)z = 5\)

(2) \(x + z \lt 3\)
[Reveal] Spoiler: OA

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 44599
Re S98-03 [#permalink]

Show Tags

New post 16 Sep 2014, 01:52
Official Solution:


Since the average of \(x\), \(y\), and \(z\) is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of \(x + y + z\)?" We also note the restrictions on the possible values of \(x\), \(y\), and \(z\) - the variables must be integers in ascending order from \(x\) to \(z\) (not necessarily consecutive). Moreover, they must be different integers, since the inequality \(x \lt y \lt z\) indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question.

Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set \(x = 0\), \(y = 1\), and \(z = 2\) meets all conditions \((x + z = 2 \lt 3\), all variables are integers and in ascending order), with \(x + y + z = 3\). Another set (\(x = -1\), \(y = 0\), and \(z = 1\)) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement.

Statement (1): INSUFFICIENT. The equation states that \(x + y\) (which must be an integer) multiplied by \(z\) (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and -1 and -5. (Don't forget about the negative possibilities). Keeping the conditions that \(x \lt y \lt z\), we can construct the only sets that work:

\(x + y = 1\) and \(z = 5\) (There's no way to assign \(z = 1\) and \(x + y = 5\) while preserving \(x \lt y \lt z\).)

\(x = 0\), \(y = 1\), \(z = 5\), \(sum = 6\)

\(x = -1\), \(y = 2\), \(z = 5\), \(sum = 6\)

\(x = -2\), \(y = 3\), \(z = 5\), \(sum = 6\)

\(x = -3\), \(y = 4\), \(z = 5\), \(sum = 6\)

\(x + y = -5\) and \(z = -1\)

\(x = -3\), \(y = -2\), \(z = -1\), \(sum = -6\)

Since there are 2 possible sums, this statement is insufficient.

Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of \(x + z\) for each case, keeping only the cases in which \(x + z\) is less than 3. Two cases remain.

Case 1: \(x = -3\), \(y = 4\), \(z = 5\), \(x + z = 2 \lt 3\), \(x + y + z = 6\)

Case 2: \(x = -3\), \(y = -2\), \(z = -1\), \(x + z = -4 \lt 3\), \(x + y + z = -6\)

Since the two cases yield different sums, we cannot determine a single value for that sum.


Answer: E
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
B
Joined: 26 Sep 2016
Posts: 7
Re: S98-03 [#permalink]

Show Tags

New post 15 Apr 2018, 12:24
Bunuel, hello!
If I am not mistaken we are looking for average. So if we have that sum everywhere is 6, then average is two, isnt it?



Bunuel wrote:
Official Solution:


Since the average of \(x\), \(y\), and \(z\) is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of \(x + y + z\)?" We also note the restrictions on the possible values of \(x\), \(y\), and \(z\) - the variables must be integers in ascending order from \(x\) to \(z\) (not necessarily consecutive). Moreover, they must be different integers, since the inequality \(x \lt y \lt z\) indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question.

Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set \(x = 0\), \(y = 1\), and \(z = 2\) meets all conditions \((x + z = 2 \lt 3\), all variables are integers and in ascending order), with \(x + y + z = 3\). Another set (\(x = -1\), \(y = 0\), and \(z = 1\)) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement.

Statement (1): INSUFFICIENT. The equation states that \(x + y\) (which must be an integer) multiplied by \(z\) (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and -1 and -5. (Don't forget about the negative possibilities). Keeping the conditions that \(x \lt y \lt z\), we can construct the only sets that work:

\(x + y = 1\) and \(z = 5\) (There's no way to assign \(z = 1\) and \(x + y = 5\) while preserving \(x \lt y \lt z\).)

\(x = 0\), \(y = 1\), \(z = 5\), \(sum = 6\)

\(x = -1\), \(y = 2\), \(z = 5\), \(sum = 6\)

\(x = -2\), \(y = 3\), \(z = 5\), \(sum = 6\)

\(x = -3\), \(y = 4\), \(z = 5\), \(sum = 6\)

\(x + y = -5\) and \(z = -1\)

\(x = -3\), \(y = -2\), \(z = -1\), \(sum = -6\)

Since there are 2 possible sums, this statement is insufficient.

Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of \(x + z\) for each case, keeping only the cases in which \(x + z\) is less than 3. Two cases remain.

Case 1: \(x = -3\), \(y = 4\), \(z = 5\), \(x + z = 2 \lt 3\), \(x + y + z = 6\)

Case 2: \(x = -3\), \(y = -2\), \(z = -1\), \(x + z = -4 \lt 3\), \(x + y + z = -6\)

Since the two cases yield different sums, we cannot determine a single value for that sum.


Answer: E
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 44599
Re: S98-03 [#permalink]

Show Tags

New post 16 Apr 2018, 00:50
Jahfors wrote:
Bunuel, hello!
If I am not mistaken we are looking for average. So if we have that sum everywhere is 6, then average is two, isnt it?



Bunuel wrote:
Official Solution:


Since the average of \(x\), \(y\), and \(z\) is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of \(x + y + z\)?" We also note the restrictions on the possible values of \(x\), \(y\), and \(z\) - the variables must be integers in ascending order from \(x\) to \(z\) (not necessarily consecutive). Moreover, they must be different integers, since the inequality \(x \lt y \lt z\) indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question.

Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set \(x = 0\), \(y = 1\), and \(z = 2\) meets all conditions \((x + z = 2 \lt 3\), all variables are integers and in ascending order), with \(x + y + z = 3\). Another set (\(x = -1\), \(y = 0\), and \(z = 1\)) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement.

Statement (1): INSUFFICIENT. The equation states that \(x + y\) (which must be an integer) multiplied by \(z\) (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and -1 and -5. (Don't forget about the negative possibilities). Keeping the conditions that \(x \lt y \lt z\), we can construct the only sets that work:

\(x + y = 1\) and \(z = 5\) (There's no way to assign \(z = 1\) and \(x + y = 5\) while preserving \(x \lt y \lt z\).)

\(x = 0\), \(y = 1\), \(z = 5\), \(sum = 6\)

\(x = -1\), \(y = 2\), \(z = 5\), \(sum = 6\)

\(x = -2\), \(y = 3\), \(z = 5\), \(sum = 6\)

\(x = -3\), \(y = 4\), \(z = 5\), \(sum = 6\)

\(x + y = -5\) and \(z = -1\)

\(x = -3\), \(y = -2\), \(z = -1\), \(sum = -6\)

Since there are 2 possible sums, this statement is insufficient.

Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of \(x + z\) for each case, keeping only the cases in which \(x + z\) is less than 3. Two cases remain.

Case 1: \(x = -3\), \(y = 4\), \(z = 5\), \(x + z = 2 \lt 3\), \(x + y + z = 6\)

Case 2: \(x = -3\), \(y = -2\), \(z = -1\), \(x + z = -4 \lt 3\), \(x + y + z = -6\)

Since the two cases yield different sums, we cannot determine a single value for that sum.


Answer: E


In once case the sum is 6 and in another case the sum is -6:

Case 1: \(x = -3\), \(y = 4\), \(z = 5\) --> \(x + y + z = 6\) --> the average = 6/3 = 2.

Case 2: \(x = -3\), \(y = -2\), \(z = -1\) --> \(x + y + z = -6\) --> the average = -6/3 = -2.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Re: S98-03   [#permalink] 16 Apr 2018, 00:50
Display posts from previous: Sort by

S98-03

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel



GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.