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How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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12 Sep 2016, 11:02
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How many solutions are possible for the inequality  x  1  +  x  6  < 2? A. 3 B. 2 C. 1 D. 0 E. 4 Make sure that you understand the number line approach to solve this question rather than the algebraic approach. Learning the number line approach helped me understand absolute values better. Thanks to the team at ManhattanPrep for this post that explains the concept very well https://www.manhattanprep.com/gmat/tuto ... value.cfm
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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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12 Sep 2016, 12:00
hdwnkr wrote: How many solutions are possible for the inequality  x  1  +  x  6  < 2?
A. 3 B. 2 C. 1 D. 0 E. 4
The left side of the equation is obviously not negative and can only be \(0\) or \(1\) and plugin the values in and around the boundary in \( x  1  +  x  6 \) and none satisfies. Answer is D.



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How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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Updated on: 12 Sep 2016, 12:39
hdwnkr wrote: How many solutions are possible for the inequality  x  1  +  x  6  < 2? A. 3 B. 2 C. 1 D. 0 E. 4 Make sure that you understand the number line approach to solve this question rather than the algebraic approach. Learning the number line approach helped me understand absolute values better. Thanks to the team at ManhattanPrep for this post that explains the concept very well https://www.manhattanprep.com/gmat/tuto ... value.cfmI tried the way what Bunuel taught us and I got it correct. Absolute value properties:
When x≤0 then x=−x, or more generally when some expression≤0 then some expression=−(some expression). For example: −5=5=−(−5);
When x≥0 then x=x, or more generally when some expression≥0 then some expression=some expression. For example: 5=5. x  1  +  x  6  < 2 ... Then we have x = 1 and 6. We can write case 1: x < 1 Case 2: 1<x<6 case 3: x>6 case 1: (x  1)  (x  6) < 2 , we get x > 2.5 and then substitute this value in the equation , suppose if x is 3 then equation fits, if x is 7 then it is greater than 2. not useful. We reject this solution because our condition is not satisfied. ( 2.5 is not less than 1) case 2: (x  1)  (x  6) < 2 , for ex: if we take x as 5 then we get 5+1 not less than 2... or for ex if we take x as 2 then we get 1+4 not less than 2...We reject this solution because our condition is not satisfied. case 3: x1+x6 < 2 , if x is 7 then we get 6+1 not less than 2.We reject this solution because our condition is not satisfied. None satisfy our condition...so option D. Bunuel... please let me know if I am missing anything.
Originally posted by msk0657 on 12 Sep 2016, 12:05.
Last edited by msk0657 on 12 Sep 2016, 12:39, edited 1 time in total.



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How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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12 Sep 2016, 12:22
For example,  x  1  = 3 This means all the values of x whose distance from 1 on the number line is 3 units. So, the two points are 4 and 2 Similarly, if  x  6  = 3 This means all values of x whose distance from 6 on the number line is 3 units. So, the two points are 9 and 3. In the question, we need to find a common point which is placed on the number line such that it's distance from 1 and 6 should add up to less than 2. The difference between 1 and 6 is 5. Any point between the two on a number line would divide the line segment between 1 and 6 into two parts whose sum would be 5, which is not less than 2. If we take the negative value for  x  1 , the distance further increases. Hence, no solution. Option D msk0657
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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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09 Oct 2016, 11:29
hdwnkr wrote: How many solutions are possible for the inequality  x  1  +  x  6  < 2?
A. 3 B. 2 C. 1 D. 0 E. 4
i actually tested all 4 options: + + +   +   1st one: x1 + x6 < 2 2x7 < 2 2x < 9 x < 4.5 suppose x=4 3 + 2 = 5 5>2 not working. E is out. 2nd one: x1 x +6 < 2 x cancels, and we get 5<2  nonsense. A is out. 3rd one: x+1 +x6 <2 x cancels, and we get an inequality that is always true > 5<2. but we don't have an answer for X.... so this one is kind of blurry... 4th one: x+1 x+6 < 2 2x+7 < 2 5<2x 2.5<x so x>2.5 suppose x=3 31 = 2. 36 = 3, absolute value is 3. 2+3 = 5 5 is not < 2. so out. in this case, it's either D or C. since the 3rd case doesn't actually provide an exact solution...I would go with D...



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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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08 Aug 2017, 10:42
msk0657 wrote: hdwnkr wrote: How many solutions are possible for the inequality  x  1  +  x  6  < 2? A. 3 B. 2 C. 1 D. 0 E. 4 Make sure that you understand the number line approach to solve this question rather than the algebraic approach. Learning the number line approach helped me understand absolute values better. Thanks to the team at ManhattanPrep for this post that explains the concept very well https://www.manhattanprep.com/gmat/tuto ... value.cfmI tried the way what Bunuel taught us and I got it correct. Absolute value properties:
When x≤0 then x=−x, or more generally when some expression≤0 then some expression=−(some expression). For example: −5=5=−(−5);
When x≥0 then x=x, or more generally when some expression≥0 then some expression=some expression. For example: 5=5. x  1  +  x  6  < 2 ... Then we have x = 1 and 6. We can write case 1: x < 1 Case 2: 1<x<6 case 3: x>6 case 1: (x  1)  (x  6) < 2 , we get x > 2.5 and then substitute this value in the equation , suppose if x is 3 then equation fits, if x is 7 then it is greater than 2. not useful. We reject this solution because our condition is not satisfied. ( 2.5 is not less than 1) case 2: (x  1)  (x  6) < 2 , for ex: if we take x as 5 then we get 5+1 not less than 2... or for ex if we take x as 2 then we get 1+4 not less than 2...We reject this solution because our condition is not satisfied. case 3: x1+x6 < 2 , if x is 7 then we get 6+1 not less than 2.We reject this solution because our condition is not satisfied. None satisfy our condition...so option D. Bunuel... please let me know if I am missing anything. hi I have also learnt this technique from Bunuel. its simply great, no doubt out there ... you can, however, apply to this problem the following technique...I hope you will also like it... 2 < x  1 + x  6 < 2 solve this inequality within a moment to get 5/2 <x <7/5 here you can see that virtually there is no solution for x.. hope this helps ...



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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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16 Aug 2017, 08:50
gmatcracker2017 wrote: msk0657 wrote: hdwnkr wrote: How many solutions are possible for the inequality  x  1  +  x  6  < 2? A. 3 B. 2 C. 1 D. 0 E. 4 Make sure that you understand the number line approach to solve this question rather than the algebraic approach. Learning the number line approach helped me understand absolute values better. Thanks to the team at ManhattanPrep for this post that explains the concept very well https://www.manhattanprep.com/gmat/tuto ... value.cfmI tried the way what Bunuel taught us and I got it correct. Absolute value properties:
When x≤0 then x=−x, or more generally when some expression≤0 then some expression=−(some expression). For example: −5=5=−(−5);
When x≥0 then x=x, or more generally when some expression≥0 then some expression=some expression. For example: 5=5. x  1  +  x  6  < 2 ... Then we have x = 1 and 6. We can write case 1: x < 1 Case 2: 1<x<6 case 3: x>6 case 1: (x  1)  (x  6) < 2 , we get x > 2.5 and then substitute this value in the equation , suppose if x is 3 then equation fits, if x is 7 then it is greater than 2. not useful. We reject this solution because our condition is not satisfied. ( 2.5 is not less than 1) case 2: (x  1)  (x  6) < 2 , for ex: if we take x as 5 then we get 5+1 not less than 2... or for ex if we take x as 2 then we get 1+4 not less than 2...We reject this solution because our condition is not satisfied. case 3: x1+x6 < 2 , if x is 7 then we get 6+1 not less than 2.We reject this solution because our condition is not satisfied. None satisfy our condition...so option D. Bunuel... please let me know if I am missing anything. hi I have also learnt this technique from Bunuel. its simply great, no doubt out there ... you can, however, apply to this problem the following technique...I hope you will also like it... 2 < x  1 + x  6 < 2 solve this inequality within a moment to get 5/2 <x <7/5 here you can see that virtually there is no solution for x.. hope this helps ... :lol: Why 7/5? For me is 5/2 < x < 9/2 where my mistake?



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How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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16 Aug 2017, 16:41
hdwnkr wrote Quote: How many solutions are possible for the inequality  x  1  +  x  6  < 2?
A. 3 B. 2 C. 1 D. 0 E. 4 gmatcracker2017 wrote: Quote: 2 < x  1 + x  6 < 2
solve this inequality within a moment to get 5/2 <x <7/5
here you can see that virtually there is no solution for x..
hope this helps ... :lol: guilherme28 wrote: Why 7/5?
For me is 5/2 < x < 9/2 where my mistake? guilherme28 , if you are asking about 7/5, see my question to gmatcracker2017 below. I do not think you made a mistake on the arithmetic. I got 9/2, too, see below. I am not sure, but maybe also you forgot to test your answers? I think you removed absolute value bars and did the math for the four cases. FT = first term and ST = second term 1st: +FT + ST < 2 2nd: +FT  ST < 2 3rd: FT + ST < 2 4th: FT  ST < 2 1st: x  1 + x  6 < 2 2x < 9 x < \(\frac{9}{2}\) 2nd: x  1  x + 6 < 2 NO SOLUTION, LHS > x  x = 0 3rd: x + 1 + x  6 < 2 NO SOLUTION, LHS > x + x = 0 4th: x + 1  x + 6 < 2 2x < 5 x > \(\frac{5}{2}\), or \(\frac{5}{2}\) < x Put 1st and 4th cases' solutions together \(\frac{5}{2}\) < x < \(\frac{9}{2}\) Just one problem: with absolute value inequalities, you should check every single solution. Test \(\frac{5}{2}\) < x . Okay, suppose x is 3. Plug 3 into original equation 3  1 + 3  6 < 2 ??? 2 + 3 = 5, and 5 is NOT less than 2. That means the solution is not valid. Eliminate it. Now try x < \(\frac{9}{2}\). So x could be 4. Plug in: 4  1 + 4  6 < 2 ??? 3 + 2 = 5, and 5 is NOT less than 2. That solution is not valid either. Eliminate it. There are ZERO solutions. gmatcracker2017 , you wrote: Quote: 2 < x  1 + x  6 < 2
solve this inequality within a moment to get 5/2 < x < 7/5 I'm not sure how you got the value in bold from the inequality you posited. I think RHS should be my 1st case, above, x < \(\frac{9}{2}\) You are correct; there is no solution. But how would we know that from the inequality you wrote? Am I missing something? Conclusion: if you remove absolute value bars, you should test every solution. guilherme28 , does that help? :)
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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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17 Aug 2017, 06:44
genxer123 wrote: hdwnkr wrote Quote: How many solutions are possible for the inequality  x  1  +  x  6  < 2?
A. 3 B. 2 C. 1 D. 0 E. 4 gmatcracker2017 wrote: Quote: 2 < x  1 + x  6 < 2
solve this inequality within a moment to get 5/2 <x <7/5
here you can see that virtually there is no solution for x..
hope this helps ... :lol: guilherme28 wrote: Why 7/5?
For me is 5/2 < x < 9/2 where my mistake? guilherme28 , if you are asking about 7/5, see my question to gmatcracker2017 below. I do not think you made a mistake on the arithmetic. I got 9/2, too, see below. I am not sure, but maybe also you forgot to test your answers? I think you removed absolute value bars and did the math for the four cases. FT = first term and ST = second term 1st: +FT + ST < 2 2nd: +FT  ST < 2 3rd: FT + ST < 2 4th: FT  ST < 2 1st: x  1 + x  6 < 2 2x < 9 x < \(\frac{9}{2}\) 2nd: x  1  x + 6 < 2 NO SOLUTION, LHS > x  x = 0 3rd: x + 1 + x  6 < 2 NO SOLUTION, LHS > x + x = 0 4th: x + 1  x + 6 < 2 2x < 5 x > \(\frac{5}{2}\), or \(\frac{5}{2}\) < x Put 1st and 4th cases' solutions together \(\frac{5}{2}\) < x < \(\frac{9}{2}\) Just one problem: with absolute value inequalities, you should check every single solution. Test \(\frac{5}{2}\) < x . Okay, suppose x is 3. Plug 3 into original equation 3  1 + 3  6 < 2 ??? 2 + 3 = 5, and 5 is NOT less than 2. That means the solution is not valid. Eliminate it. Now try x < \(\frac{9}{2}\). So x could be 4. Plug in: 4  1 + 4  6 < 2 ??? 3 + 2 = 5, and 5 is NOT less than 2. That solution is not valid either. Eliminate it. There are ZERO solutions. gmatcracker2017 , you wrote: Quote: 2 < x  1 + x  6 < 2
solve this inequality within a moment to get 5/2 < x < 7/5 I'm not sure how you got the value in bold from the inequality you posited. I think RHS should be my 1st case, above, x < \(\frac{9}{2}\) You are correct; there is no solution. But how would we know that from the inequality you wrote? Am I missing something? Conclusion: if you remove absolute value bars, you should test every solution. guilherme28 , does that help? :) of course you will have to test the viability of every solution..... and yes, x is less than 9/2 .... thanks



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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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17 Aug 2017, 08:04
Ans is D case1 x>6 then (x1)+(x6)<2 gives x<9/2 hence no soln case2 x~(1,6) then (x1)+(6x)<2 gives no soln case3 x<1 then (1x) + (6x) <2 gives x>5/2 hence no soln therefore no solutions are possible
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How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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17 Aug 2017, 09:52
gmatcracker2017 wrote: of course you will have to test the viability of every solution..... and yes, x is less than 9/2 ....
thanks gmatcracker2017  Hmm, hope I didn't offend. I certainly did not intend offense. I did not suggest that you failed to test viability of solutions. I simply could not figure out how you got 7/5. I also thought: if 7/5 were correct (which if true = zero solutions b/c no number is greater than 2.5 and simultaneously less than 1.4), no need to test four cases as I had and as guilherme28 might have. That's all. I rarely use the critical points method. There was a decent chance I was missing something. And just in case there is seeming offense elsewhere: Neither did I say that guilherme28 failed to test solutions. I wondered whether or not that might be the case. If it were, I decided it was better to cover the bases than to leave the impression that the inequality in question led to valid solutions. IMO, better safe than sorry for those like I who aren't 100 percent fluent in absolute value inequalities. Regards.
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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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17 Aug 2017, 11:22
genxer123 wrote: gmatcracker2017 wrote: of course you will have to test the viability of every solution..... and yes, x is less than 9/2 ....
thanks gmatcracker2017  Hmm, hope I didn't offend. I certainly did not intend offense. I did not suggest that you failed to test viability of solutions. I simply could not figure out how you got 7/5. I also thought: if 7/5 were correct (which if true = zero solutions b/c no number is greater than 2.5 and simultaneously less than 1.4), no need to test four cases as I had and as guilherme28 might have. That's all. I rarely use the critical points method. There was a decent chance I was missing something. And just in case there is seeming offense elsewhere: Neither did I say that guilherme28 failed to test solutions. I wondered whether or not that might be the case. If it were, I decided it was better to cover the bases than to leave the impression that the inequality in question led to valid solutions. IMO, better safe than sorry for those like I who aren't 100 percent fluent in absolute value inequalities. Regards. hey man Its okay... I am okay with the matter. You know, its possible to solve a PS in many ways. Like you, I also work with critical point method or squaring to help me solve inequalities. Truth be told, I have learnt these methods from gmatclub, from you people. Side by side, however, I was also finding some other way available. So far I know inequality is widely tested on the GMAT. So, I am planing a dexterity on the topic, if not such by now ... Finally, hope to discuss such issues from time to time in upcoming days. Again I aver I am okay with your kudos ... thanks



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Re: How many solutions are possible for the inequality  x  1  +  x  6 [#permalink]
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16 May 2018, 10:56
hdwnkr wrote: How many solutions are possible for the inequality  x  1  +  x  6  < 2?
A. 3 B. 2 C. 1 D. 0 E. 4 Notice that x  1 = 0 means x = 1 and x  6 = 0 means x = 6, so we have 3 cases: (1) x ≥ 6, (2) 1 ≤ x < 6, and (3) x < 1. We need to see what happen in each of these cases. Let’s do that now: Case 1: If x ≥ 6, then both x  1 and x  6 are positive, so x  1 = x  1 and x  6 = x  6, so x  1 + x  6 < 2 becomes: x  1 + x  6 < 2 2x  7 < 2 2x < 9 x < 9/2 However, x < 9/2 contradicts our assumption that x ≥ 6, so there are no solutions when x ≥ 6. Case 2: If 1 ≤ x < 6, then x  1 is positive but x  6 is negative, so x  1 = x  1 and x  6 = (x  6) = x + 6, so x  1 + x  6 < 2 becomes: x  1 + (x + 6) < 2 5 < 2 We see that 5 < 2 is a false statement. Therefore, there are no solutions when 1 ≤ x < 6, either. Case 3: If x < 1, then both x  1 and x  6 are negative, so x  1 = (x  1) = x + 1 and x  6 = (x  6) = x + 6, so x  1 + x  6 < 2 becomes: x + 1 + (x + 6) < 2 2x + 7 < 2 2x < 5 x > 5/2 However, x > 5/2 contradicts our assumption that x < 1, so there are no solutions when x < 1. We see that there no solutions that satisfy the given inequality. Answer: D
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Re: How many solutions are possible for the inequality  x  1  +  x  6
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