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If x + x + y = 7 and x + y  y =6 , then x + y =
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Updated on: 13 Aug 2012, 23:31
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Originally posted by Hussain15 on 13 Jun 2010, 04:41.
Last edited by Bunuel on 13 Aug 2012, 23:31, edited 1 time in total.
Edited the question.




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Re: Interesting Absolute value Problem
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13 Jun 2010, 06:01
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.
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Re: Interesting Absolute value Problem
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13 Jun 2010, 05:42
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.




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Re: Interesting Absolute value Problem
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18 Jun 2010, 06:23
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?



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Re: Interesting Absolute value Problem
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18 Jun 2010, 06:28
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: \(xyy=6\) > \(x2y=6\).
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Re: Interesting Absolute value Problem
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18 Jun 2010, 07:17
^ 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...



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Re: Interesting Absolute value Problem
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14 Jun 2011, 02:25
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



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Re: If x + x + y = 7 and x + y  y =6 , then x + y =
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02 Jul 2013, 05:38
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



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Re: If x + x + y = 7 and x + y  y =6 , then x + y =
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02 Jul 2013, 10:22
I solved in 1.36 mins
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Re: Interesting Absolute value Problem
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05 Jul 2013, 10:19
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?



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Re: Interesting Absolute value Problem
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05 Jul 2013, 10:30
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.
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Re: If x + x + y = 7 and x + y  y =6 , then x + y =
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09 Jul 2013, 15:24
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)



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Re: If x + x + y = 7 and x + y  y =6 , then x + y =
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27 Sep 2015, 23:10
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
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Re: If x + x + y = 7 and x + y  y =6 , then x + y =
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24 Oct 2015, 06:05
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?
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If x + x + y = 7 and x + y  y =6 , then x + y =
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24 Oct 2015, 06:55
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.



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Re: If x + x + y = 7 and x + y  y =6 , then x + y =
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24 Oct 2015, 08:52
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!
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Re: If x + x + y = 7 and x + y  y =6 , then x + y =
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12 Jun 2017, 12:55
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?



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