Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Status: The last round
Joined: 18 Jun 2009
Posts: 1207
Concentration: Strategy, General Management

If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
Updated on: 14 Aug 2012, 00:31
Question Stats:
63% (02:37) correct 37% (02:55) wrong based on 589 sessions
HideShow timer Statistics
If x + x + y = 7 and x + y  y =6 , then x + y = A. 3 B. 4 C. 5 D. 6 E. 9
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Originally posted by Hussain15 on 13 Jun 2010, 05:41.
Last edited by Bunuel on 14 Aug 2012, 00:31, edited 1 time in total.
Edited the question.




Math Expert
Joined: 02 Sep 2009
Posts: 56266

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
13 Jun 2010, 07:01
Hussain15 wrote: If x + x + y = 7 and x + y  y =6 , then x + y =
A. 3 B. 4 C. 5 D. 6 E. 9 If \(x\leq{0}\), then \(x + x + y = 7\) becomes: \(xx+y=7\) > \(y=7>0\), but then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) > hence \(x>0\). Similarly if \(y\geq{0}\), then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), but then \(x + x + y = 7\) becomes: \(x+x+y=12+y=7\) > \(y=5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) > hence \(y<0\). So \(x>0\) and \(y<0\): \(x+x+y=7\) becomes: \(x+x+y=7\) > \(2x+y=7\); \(x+yy=6\) becomes: \(xyy=6\) > \(x2y=6\). Solving: \(x=4\) and \(y=1\) > \(x+y=3\). Answer: A. I feel there is an easier way, but world cup makes it harder to concentrate.
_________________




Manager
Joined: 04 May 2010
Posts: 81
WE 1: 2 yrs  Oilfield Service

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
13 Jun 2010, 06:42
I did it by brute force, considering all possibilities. But I'm sure someone can come up with a way to quickly identify the signs of x and y.
x + x + y = 7 ....1
x + y  y = 6 .....2
Consider x < 0 => x = x Substitute in 1
x  x + y = 7 y = 7
Substitute in 2
x + 7  7 = 6 x = 6
But this violates x < 0 So our assumption was incorrect.
Consider x > 0 and y > 0 => x = x and y = y
Substitute in 2
x + y  y = 6 x = 6
Substitute in 1
6 + 6 + y = 7 y = 5
This violates y > 0 so our assumption was incorrect.
Consider x > 0 and y < 0 => x = x and y = y
Substituting in 1 and 2:
2x + y = 7 .....3 x  2y = 6 .....4
Solving we get x = 4 and y = 1 x + y = 3
Pick A.




Intern
Affiliations: NYSSA
Joined: 07 Jun 2010
Posts: 32
Location: New York City
Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program)
WE 1: Senior Associate  Thomson Reuters
WE 2: Analyst  TIAA CREF

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
18 Jun 2010, 07:23
Bunuel wrote: Hussain15 wrote: If x + x + y = 7 and x + y  y =6 , then x + y =
A. 3 B. 4 C. 5 D. 6 E. 9 If \(x\leq{0}\), then \(x + x + y = 7\) becomes: \(xx+y=7\) > \(y=7>0\), but then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) > hence \(x>0\). Similarly if \(y\geq{0}\), then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), but then \(x + x + y = 7\) becomes: \(x+x+y=12+y=7\) > \(y=5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) > hence \(y<0\). So \(x>0\) and \(y<0\): \(x+x+y=7\) becomes: \(xx+y=7\) > \(2x+y=7\); \(x+yy=6\) becomes: \(xyy=6\) > \(x2y=6\). Solving: \(x=4\) and \(y=1\) > \(x+y=3\). Answer: A. I feel there is an easier way, but world cup makes it harder to concentrate. Why when y<0 do we get 2y?



Math Expert
Joined: 02 Sep 2009
Posts: 56266

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
18 Jun 2010, 07:28
GMATBLACKBELT720 wrote: Bunuel wrote: Hussain15 wrote: If x + x + y = 7 and x + y  y =6 , then x + y =
A. 3 B. 4 C. 5 D. 6 E. 9 If \(x\leq{0}\), then \(x + x + y = 7\) becomes: \(xx+y=7\) > \(y=7>0\), but then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) > hence \(x>0\). Similarly if \(y\geq{0}\), then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), but then \(x + x + y = 7\) becomes: \(x+x+y=12+y=7\) > \(y=5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) > hence \(y<0\). So \(x>0\) and \(y<0\): \(x+x+y=7\) becomes: \(xx+y=7\) > \(2x+y=7\); \(x+yy=6\) becomes: \(xyy=6\) > \(x2y=6\). Solving: \(x=4\) and \(y=1\) > \(x+y=3\). Answer: A. I feel there is an easier way, but world cup makes it harder to concentrate. Why when y<0 do we get 2y? When \(y<0\), then \(y=y\) and \(x+yy=6\) becomes: \(xyy=6\) > \(x2y=6\).
_________________



Intern
Affiliations: NYSSA
Joined: 07 Jun 2010
Posts: 32
Location: New York City
Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program)
WE 1: Senior Associate  Thomson Reuters
WE 2: Analyst  TIAA CREF

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
18 Jun 2010, 08:17
^ Wow... confusing as heck. Essentially saying that (hypothetical number here) 3 = (3), thats fine. But I kept thinking you would apply this to y and essentially make it +y since and make them both +y...



Director
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 950

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
14 Jun 2011, 03:25
for x>0 and x < 0 2x+ y = 7 and y = 7
for y>0 and y<0 x=6 and x2y=6
for x,y<0 solution exists. solving x=4 and y = 1
3



Intern
Joined: 09 Nov 2011
Posts: 1

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
02 Jul 2013, 06:38
If x + x + y = 7 and x + y  y =6 , then x + y =
can be done in 2.3 mins :
there are 4 cases to be tested : 1) x is ve and y is ve substituting in the equation , we get xx+y=7 and xyy=6 solve for x and y we get x=20 and y=7 , so x+y=27 REJECT
2)x is +ve and y is +ve
substitute in the equation, we ger x+x+y=7 and x+yy=6 solve for x and y we get x=6 and y=5 ,therefore x+y=1 not on list so REJECT
3) x is ve and y is +ve
substitute , we get xx=y=7 and x+yy=6 solve fo x and y we get x=6 and y=7, x+y=13 not on list so REJECT
4) x is +ve and y is ve
substitute , we get x+x=y=7 and xyy=6 solve for x and y , we get x=4 and y= 1 ,x+y=3 , ANSWER CHOICE



Manager
Joined: 14 Jun 2011
Posts: 67

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
02 Jul 2013, 11:22
I solved in 1.36 mins
_________________
Kudos always encourages me



Intern
Joined: 04 May 2013
Posts: 44

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
05 Jul 2013, 11:19
Bunuel wrote: GMATBLACKBELT720 wrote: Bunuel wrote: If \(x\leq{0}\), then \(x + x + y = 7\) becomes: \(xx+y=7\) > \(y=7>0\), but then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) > hence \(x>0\).
Similarly if \(y\geq{0}\), then \(x + y  y =6\) becomes: \(x+yy=6\) > \(x=6>0\), but then \(x + x + y = 7\) becomes: \(x+x+y=12+y=7\) > \(y=5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) > hence \(y<0\).
So \(x>0\) and \(y<0\): \(x+x+y=7\) becomes: \(xx+y=7\) > \(2x+y=7\); \(x+yy=6\) becomes: \(xyy=6\) > \(x2y=6\).
Solving: \(x=4\) and \(y=1\) > \(x+y=3\).
Answer: A.
I feel there is an easier way, but world cup makes it harder to concentrate.
Why when y<0 do we get 2y? When \(y<0\), then \(y=y\) and \(x+yy=6\) becomes: \(xyy=6\) > \(x2y=6\). Sorry, still confusing me. I understand the first y i.e. y = y But y < 0, so wouldn't x + y  y = x  y  (y), which would make it just x?



Math Expert
Joined: 02 Sep 2009
Posts: 56266

Re: Interesting Absolute value Problem
[#permalink]
Show Tags
05 Jul 2013, 11:30
jjack0310 wrote: Bunuel wrote: GMATBLACKBELT720 wrote: Why when y<0 do we get 2y?
When \(y<0\), then \(y=y\) and \(x+yy=6\) becomes: \(xyy=6\) > \(x2y=6\). Sorry, still confusing me. I understand the first y i.e. y = y But y < 0, so wouldn't x + y  y = x  y  (y), which would make it just x? When y<0, then y=y: correct. But y must stay as it is. Consider this, suppose we have only xy=6, and I tell you that y is negative would you rewrite the equation as x(y)=6? Hope it's clear.
_________________



Senior Manager
Joined: 13 May 2013
Posts: 414

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
09 Jul 2013, 16:24
If x + x + y = 7 and x + y  y =6 , then x + y =
x>0, y>0
x + (x) + y = 7 2x+y=7
x + y  y = 6 x + y  y = 6 x=6
2x+y=7 2(6)+y=7 y=5
INVALID as 6>0 but 5 is not > 0
x>0, y<0
x + (x) + y = 7 2x+y=7
x + y  y = 6 x + (y)  y = 6 x  2y = 6 x = 6+2y
2(6+2y) + y = 7 12+4y+y=7 12+5y=7 5y+5 y=1
2x+y=7 2x + (1) = 7 2x  1 = 7 2x=8 x=4
VALID as 4>0 and 1<0
therefore;
x+y = (4)+(1) = 3
(A)



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9437
Location: Pune, India

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
28 Sep 2015, 00:10
Hussain15 wrote: If x + x + y = 7 and x + y  y =6 , then x + y =
A. 3 B. 4 C. 5 D. 6 E. 9 In such a question, you can use some brute force and get to the answer too. How long it takes depends on how quickly you observe the little things. x + x + y = 7 x + y  y =6 Both equations yield about the same result though in one y is positive and in the other it is negative. x and y are positive and assuming x is positive, a negative y would pull down the first equation and pump up the second one to give almost equal values. The difference is very small also signifies that the negative variable might have a very small value. Since the options give the value of x + y as 3/4/5... etc, it is likely that we are dealing with small number pairs such as (4, 2), (3, 2), (4, 1) etc. Since the first equation has 7 as the result, both variables will not be even. A couple of quick iterations brought me to (4, 1). So x + y = 3
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 08 Sep 2012
Posts: 59
Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
24 Oct 2015, 07:05
VeritasPrepKarishma wrote: Hussain15 wrote: If x + x + y = 7 and x + y  y =6 , then x + y =
A. 3 B. 4 C. 5 D. 6 E. 9 In such a question, you can use some brute force and get to the answer too. How long it takes depends on how quickly you observe the little things. x + x + y = 7 x + y  y =6 Both equations yield about the same result though in one y is positive and in the other it is negative. x and y are positive and assuming x is positive, a negative y would pull down the first equation and pump up the second one to give almost equal values. The difference is very small also signifies that the negative variable might have a very small value. Since the options give the value of x + y as 3/4/5... etc, it is likely that we are dealing with small number pairs such as (4, 2), (3, 2), (4, 1) etc. Since the first equation has 7 as the result, both variables will not be even. A couple of quick iterations brought me to (4, 1). So x + y = 3 Hi Karishma, I tried another approach, but got stuck  can you hep me solve by this method? x + x + y = 7 => x + y = 7  x => Whatever be the value of x, x will be always nonnegative => x + y has to be less than 7 => This eliminates option E x + y  y = 6 => Now, if y is positive, then this equation becomes x = 6 => on substituting in above eqn, we get y = 5 => x+y = 1 => Not present in any of the options => y is negative => x  y  y = 6 => x  2y = 6 => x = 6 + 2y Now x + y = 7  6 + 2y If we take y = 1, we get x + y = 3 = Option A If we take y = 2, we get x + y = 5 = Option C I am getting both options here  where am I going wrong here? Or this approach incorrect?
_________________
+1 Kudos if you liked my post! Thank you!



CEO
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
24 Oct 2015, 07:55
sagar2911 wrote:
Hi Karishma,
I tried another approach, but got stuck  can you hep me solve by this method? x + x + y = 7 => x + y = 7  x => Whatever be the value of x, x will be always nonnegative => x + y has to be less than 7 => This eliminates option E x + y  y = 6 => Now, if y is positive, then this equation becomes x = 6 => on substituting in above eqn, we get y = 5 => x+y = 1 => Not present in any of the options => y is negative => x  y  y = 6 => x  2y = 6 => x = 6 + 2y Now x + y = 7  6 + 2y If we take y = 1, we get x + y = 3 = Option A If we take y = 2, we get x + y = 5 = Option C I am getting both options here  where am I going wrong here? Or this approach incorrect? Let me try to answer. You are making a mistake with using 6+2y After you get x=6+2y and substitute in equation (1), you get 6+2y+6+2y+y=7 >6+3y+6+2y = 7 > 3y+6+2y = 1. Now the point to note is that the 'nature' of 6+2y changes at y=3 and thus you need to evaluate 6+2y for values smaller than 3 and for values greater than 3. You have already assumed that y<0 in order to get x=6+2y, so your ranges to consider become 3 \(\le\)q y < 0 and y < 3 Case 1: 3 \(\le\) y < 0, giving you 6+2y \(\geq\) 0 > 3y+6+2y = 1 > 3y+6+2y = 1 > 5y=5 > y=1. Acceptable value giving you x=6+2y = 4 > x+y = 3 Case 2: y<3 > 6+2y = (6+2y) > 3y+6+2y = 1 > 3y62y = 1 > y=7 this contradicts the assumption that y<3 , making this out of scope. Thus the only value of x+y = 3.



Manager
Joined: 08 Sep 2012
Posts: 59
Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
24 Oct 2015, 09:52
Engr2012 wrote: sagar2911 wrote:
Hi Karishma,
I tried another approach, but got stuck  can you hep me solve by this method? x + x + y = 7 => x + y = 7  x => Whatever be the value of x, x will be always nonnegative => x + y has to be less than 7 => This eliminates option E x + y  y = 6 => Now, if y is positive, then this equation becomes x = 6 => on substituting in above eqn, we get y = 5 => x+y = 1 => Not present in any of the options => y is negative => x  y  y = 6 => x  2y = 6 => x = 6 + 2y Now x + y = 7  6 + 2y If we take y = 1, we get x + y = 3 = Option A If we take y = 2, we get x + y = 5 = Option C I am getting both options here  where am I going wrong here? Or this approach incorrect? Let me try to answer. You are making a mistake with using 6+2y After you get x=6+2y and substitute in equation (1), you get 6+2y+6+2y+y=7 >6+3y+6+2y = 7 > 3y+6+2y = 1. Now the point to note is that the 'nature' of 6+2y changes at y=3 and thus you need to evaluate 6+2y for values smaller than 3 and for values greater than 3. You have already assumed that y<0 in order to get x=6+2y, so your ranges to consider become 3 \(\le\)q y < 0 and y < 3 Case 1: 3 \(\le\) y < 0, giving you 6+2y \(\geq\) 0 > 3y+6+2y = 1 > 3y+6+2y = 1 > 5y=5 > y=1. Acceptable value giving you x=6+2y = 4 > x+y = 3 Case 2: y<3 > 6+2y = (6+2y) > 3y+6+2y = 1 > 3y62y = 1 > y=7 this contradicts the assumption that y<3 , making this out of scope. Thus the only value of x+y = 3. Excellent. Thank you dude!
_________________
+1 Kudos if you liked my post! Thank you!



Intern
Joined: 12 May 2017
Posts: 13

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
12 Jun 2017, 13:55
Can someone help me please.
I understand that if you first that x= pos. and y= neg. you get x=6 and y=5.
But then this wouldn't be right, because it would contradict the assumptions that x should be neg. and y should be pos.
I don't understand, where do i get these assumptions from?



NonHuman User
Joined: 09 Sep 2013
Posts: 11667

Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
Show Tags
12 Aug 2018, 10:03
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If x + x + y = 7 and x + y  y =6 , then x + y =
[#permalink]
12 Aug 2018, 10:03






