GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Feb 2019, 13:10

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 in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
  • Free GMAT Prep Hour

     February 20, 2019

     February 20, 2019

     08:00 PM EST

     09:00 PM EST

    Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST
  • Online GMAT boot camp for FREE

     February 21, 2019

     February 21, 2019

     10:00 PM PST

     11:00 PM PST

    Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

If p, q, and r are different positive integers such that p + q + r = 6

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

Hide Tags

 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52971
If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

Show Tags

New post 04 May 2015, 04:55
2
15
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

40% (02:24) correct 60% (02:47) wrong based on 243 sessions

HideShow timer Statistics

If p, q, and r are different positive integers such that p + q + r = 6, what is the value of x ?

(1) The average of x^p and x^q is x^r.

(2) The average of x^p and x^r is not x^q.

_________________

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

Senior Manager
Senior Manager
avatar
Joined: 02 Mar 2012
Posts: 297
Schools: Schulich '16
Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

Show Tags

New post 04 May 2015, 06:14
2
E

p,q and r can be any of the numbers 1,2 or 3

lets take p=1,q=2,r=3

1. x+x^2=2x^3

insuff as x can be 0 or 1

2 x+x^3=2x^2

insuf as x can be 0 or 1

taking togerther.

x can be 0 or 1

SO E
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52971
Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

Show Tags

New post 11 May 2015, 05:10
1
3
Bunuel wrote:
If p, q, and r are different positive integers such that p + q + r = 6, what is the value of x ?

(1) The average of x^p and x^q is x^r.

(2) The average of x^p and x^r is not x^q.


MANHATTAN GMAT OFFICIAL SOLUTION:

Before beginning work on the statements, note that p, q, and r are different positive integers that add up to 6. Only one set of three numbers satisfies that condition: 1, 2, and 3. Therefore, in some order, p, q, and r are 1, 2, and 3.

(1) INSUFFICIENT: Three cases are possible:

The average of x and x^2 is x^3;
The average of x and x^3 is x^2;
The average of x^2 and x^3 is x.
The first case gives the equation {x+x^2}/2=x^3 , or x+x^2=2x^3. This is a factorable quadratic equation:

2x^3-x^2-x=0
x(2x^2-x-1)=0
x(2x+1)(x-1)=0

This equation gives three solutions for x: x = 0, x = –1/2, and x = 1. Because there are multiple possible values for x, the statement is not sufficient. (There’s no need to investigate the other possible cases at this point, since you’ve just determined that this statement is insufficient.)

(2) INSUFFICIENT: Just about any value for x will make this statement true (that the average of x^pxp and x^rxr is not x^qxq). For instance, if p = 1, q = 2, and r = 3, then the only requirement is that the average of x and x^3x3 not be equal to x^2x2. Try random values of x: If x = 2, then the average of x and x^3x3 is (2 + 8)/2 = 5, which is not equal to x^2=4x2 = 4. If x = 3, then the average of x and x^3x3 is (3 + 27)/2 = 15, which is not equal to x^2=9x2 = 9. Because there are multiple possible values for x, the statement is not sufficient.

(1) AND (2) INSUFFICIENT: In order to determine the values of x that satisfy both statements, revisit the cases derived in statement 1, this time solving for all of them.

Case #1: If the average of x and x^2 is x^3, then, as shown above, x can be any of 0, –1/2, 1.

Case #2: If the average of x and x^3 is x^2, then {x+x^3}/2=x^2, or x+x^3=2x^2. This is also a factorable equation: x^3-2x^2+x=0 ––> x(x^2-2x+1)=0 ––> x(x-1)^2=0. Therefore, x can be either 0 or 1.
Case #3: If the average of x^2 and x^3 is x, then {x^2+x^3}/2=x, or x^2+x^3=2x. This too is a factorable equation: x^3+x^2-2x=0 ––> x(x^2+x-2)=0 ––> x(x+2)(x-1)=0. Therefore, x can be any of 0, –2, 1.

There are thus four possible values of x according to statement 1: x = –2, –1/2, 0, 1.

For the values x = 0 or 1, the average of two terms is equal to the third term in all 3 cases, so these two values can’t satisfy statement 2 (where the average of the first two terms cannot equal the third term). Discard the values 0 and 1.

On the other hand, if x = –2, p and q are 2 and 3 (in either order), and r = 1, then both statements are satisfied. The average of x^p and x^q is x^r: (4 + -8)/2 = -2, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (4 + -2)/2 does not equal -8, so statement 2 is true. Therefore, x could equal –2.

What about the final case? If x = –1/2, p and q are 1 and 2 (in either order), and r = 3, then both statements are also satisfied. The average of x^p and x^q is x^r: (–1/2 + 1/4)/2 = –1/8, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (–1/2 + –1/8)/2 does not equal 1/4, so statement 2 is true. Therefore, x can still be either –2 or –1/2. Both statements together are still not sufficient to answer the question.

The correct answer is (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

Current Student
avatar
B
Joined: 23 May 2013
Posts: 186
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5
GMAT ToolKit User
Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

Show Tags

New post 07 Jul 2016, 09:16
Bunuel wrote:
Bunuel wrote:
If p, q, and r are different positive integers such that p + q + r = 6, what is the value of x ?

(1) The average of x^p and x^q is x^r.

(2) The average of x^p and x^r is not x^q.


MANHATTAN GMAT OFFICIAL SOLUTION:

Before beginning work on the statements, note that p, q, and r are different positive integers that add up to 6. Only one set of three numbers satisfies that condition: 1, 2, and 3. Therefore, in some order, p, q, and r are 1, 2, and 3.

(1) INSUFFICIENT: Three cases are possible:

The average of x and x^2 is x^3;
The average of x and x^3 is x^2;
The average of x^2 and x^3 is x.
The first case gives the equation {x+x^2}/2=x^3 , or x+x^2=2x^3. This is a factorable quadratic equation:

2x^3-x^2-x=0
x(2x^2-x-1)=0
x(2x+1)(x-1)=0

This equation gives three solutions for x: x = 0, x = –1/2, and x = 1. Because there are multiple possible values for x, the statement is not sufficient. (There’s no need to investigate the other possible cases at this point, since you’ve just determined that this statement is insufficient.)

(2) INSUFFICIENT: Just about any value for x will make this statement true (that the average of x^pxp and x^rxr is not x^qxq). For instance, if p = 1, q = 2, and r = 3, then the only requirement is that the average of x and x^3x3 not be equal to x^2x2. Try random values of x: If x = 2, then the average of x and x^3x3 is (2 + 8)/2 = 5, which is not equal to x^2=4x2 = 4. If x = 3, then the average of x and x^3x3 is (3 + 27)/2 = 15, which is not equal to x^2=9x2 = 9. Because there are multiple possible values for x, the statement is not sufficient.

(1) AND (2) INSUFFICIENT: In order to determine the values of x that satisfy both statements, revisit the cases derived in statement 1, this time solving for all of them.

Case #1: If the average of x and x^2 is x^3, then, as shown above, x can be any of 0, –1/2, 1.

Case #2: If the average of x and x^3 is x^2, then {x+x^3}/2=x^2, or x+x^3=2x^2. This is also a factorable equation: x^3-2x^2+x=0 ––> x(x^2-2x+1)=0 ––> x(x-1)^2=0. Therefore, x can be either 0 or 1.
Case #3: If the average of x^2 and x^3 is x, then {x^2+x^3}/2=x, or x^2+x^3=2x. This too is a factorable equation: x^3+x^2-2x=0 ––> x(x^2+x-2)=0 ––> x(x+2)(x-1)=0. Therefore, x can be any of 0, –2, 1.

There are thus four possible values of x according to statement 1: x = –2, –1/2, 0, 1.

For the values x = 0 or 1, the average of two terms is equal to the third term in all 3 cases, so these two values can’t satisfy statement 2 (where the average of the first two terms cannot equal the third term). Discard the values 0 and 1.

On the other hand, if x = –2, p and q are 2 and 3 (in either order), and r = 1, then both statements are satisfied. The average of x^p and x^q is x^r: (4 + -8)/2 = -2, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (4 + -2)/2 does not equal -8, so statement 2 is true. Therefore, x could equal –2.

What about the final case? If x = –1/2, p and q are 1 and 2 (in either order), and r = 3, then both statements are also satisfied. The average of x^p and x^q is x^r: (–1/2 + 1/4)/2 = –1/8, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (–1/2 + –1/8)/2 does not equal 1/4, so statement 2 is true. Therefore, x can still be either –2 or –1/2. Both statements together are still not sufficient to answer the question.

The correct answer is (E).



I actually think the logic from Manhattan is backwards. If we would have arrived at the conclusion that one or both of the statements were sufficient on their own after choosing values for p,q, and r, then we would have to keep checking the other arrangement of values of p,q, and r to make sure that we weren't missing any cases. If we pick one arrangement, however, and determine that x could still have multiple values with both statements taken into account, then we can safely say that we need more information (as in hsbinfy's approach).
Math Revolution GMAT Instructor
User avatar
V
Joined: 16 Aug 2015
Posts: 6966
GMAT 1: 760 Q51 V42
GPA: 3.82
Premium Member
Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

Show Tags

New post 08 Jul 2016, 21:01
There are 4 variables (p, q, r and x) and 1 equation (p+q+r=6) in the original condition. In order to match the number of variables and the number of equations, we need 3 equations. Since con 1) and con 2) each has 1 equation, there is a high chance that E is the correct answer. Using 1) & 2), from x+x^3=2x or x^3+x^2=2x, we get x=-1/2, -2. The answer is not unique and the condition is not sufficient. Hence, the correct answer is E.


- For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Intern
Intern
avatar
B
Joined: 14 Aug 2018
Posts: 5
Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

Show Tags

New post 23 Oct 2018, 07:11
how could be solved in 2 mins? It is tough
GMAT Club Bot
Re: If p, q, and r are different positive integers such that p + q + r = 6   [#permalink] 23 Oct 2018, 07:11
Display posts from previous: Sort by

If p, q, and r are different positive integers such that p + q + r = 6

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


cron
Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| 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®.