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If Y = X + 1  X2, then [#permalink]
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06 Feb 2017, 14:42
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If \(Y = X + 1  X2\), then (a) \(3 \leq Y \leq 0\) (b) \(3 \leq Y \leq 3\) (c) \(Y \leq 3\) (d) \(Y \geq 3\) (e) No Solution
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Re: If Y = X + 1  X2, then [#permalink]
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06 Feb 2017, 19:22
quantumliner wrote: If Y = X + 1  X2, then
(a) 3 <= Y <=0 (b) 3 <= Y <=3 (c) Y <=3 (d) Y >=3 (e) No Solution 3 ranges to define 1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3 as second condition dependent on value of x thus from range 1 & 2 3 <= Y <=3 Ans B



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Re: If Y = X + 1  X2, then [#permalink]
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09 Feb 2017, 06:30
rohit8865 wrote: 3 ranges to define
1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3
as second condition dependent on value of x thus from range 1 & 2 3 <= Y <=3
Ans B
How did you came up with the 3 ranges and how to define them? I dont understand what you are doing. Thanks



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Re: If Y = X + 1  X2, then [#permalink]
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09 Feb 2017, 07:08
maxschmid wrote: rohit8865 wrote: 3 ranges to define
1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3
as second condition dependent on value of x thus from range 1 & 2 3 <= Y <=3
Ans B
How did you came up with the 3 ranges and how to define them? I dont understand what you are doing. Thanks maxschmidfor such questions always do the below suggested Ix+1I will be 0 when x= 1similarly Ix2I will be 0 when x= 2thus we have boundaries defined as 1 and 2 so i have checked on both sides of boundaries defined in three steps for value y 1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3 hope now u can understand



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Re: If Y = X + 1  X2, then [#permalink]
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13 Feb 2017, 17:05
quantumliner wrote: If Y = X + 1  X2, then
(a) 3 <= Y <=0 (b) 3 <= Y <=3 (c) Y <=3 (d) Y >=3 (e) No Solution Since the expressions in the absolute value will equal zero when X = 1 and X = 2, we need to investigate the following three cases: i) X < 1, ii) 1 < X < 2, and iii) X > 2 Now let’s analyze each case. i) If X < 1, we see that both of the expressions inside the absolute values are negative, and thus: Y = (X  1)  [(X  2)] Y =  X 1 + X  2 Y = 3 ii) If 1 < X < 2, then the first expression is positive but the second expression is negative; thus: Y = X + 1  [(X  2)] Y = X + 1 + X  2 Y = 2X 1 Now, recall that 1 < X < 2, so 2 < 2X < 4 3 < 2X  1 < 3 Since Y = 2X  1, 3 < Y < 3. iii) If X > 2, then both expressions inside the absolute value are positive; thus: Y = X + 1  (X  2) Y = X + 1  X + 2 Y = 3 Combining the results of the three cases, we see that 3 ≤ Y ≤ 3. Answer: B
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Re: If Y = X + 1  X2, then [#permalink]
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15 Feb 2017, 04:45
JeffTargetTestPrep wrote: quantumliner wrote: If Y = X + 1  X2, then
(a) 3 <= Y <=0 (b) 3 <= Y <=3 (c) Y <=3 (d) Y >=3 (e) No Solution Since the expressions in the absolute value will equal zero when X = 1 and X = 2, we need to investigate the following three cases: i) X < 1, ii) 1 < X < 2, and iii) X > 2 Now let’s analyze each case. i) If X < 1, we see that both of the expressions inside the absolute values are negative, and thus: Y = (X  1)  [(X  2)] Y =  X 1 + X  2 Y = 3 ii) If 1 < X < 2, then the first expression is positive but the second expression is negative; thus: Y = X + 1  [(X  2)] Y = X + 1 + X  2 Y = 2X 1 Now, recall that 1 < X < 2, so 2 < 2X < 4 3 < 2X  1 < 3 Since Y = 2X  1, 3 < Y < 3. iii) If X > 2, then both expressions inside the absolute value are positive; thus: Y = X + 1  (X  2) Y = X + 1  X + 2 Y = 3 Combining the results of the three cases, we see that 3 ≤ Y ≤ 3. Answer: B I have a query regarding the range. Shouldn't it be 1 <= X < 2 and X >= 2



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Re: If Y = X + 1  X2, then [#permalink]
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15 Feb 2017, 06:04
warriorguy wrote: I have a query regarding the range. Shouldn't it be 1 <= X < 2 and X >= 2
Dear warriorguy, Graphical illustration will clear your doubt. When \(x = 2\) \(Y = x+1  x2 = 2+1  22 = 3 + 0 = 3\) Coordinate 1 \((2, 3)\) When \(x = 1\) \(Y = x+1  x2 = 1+1  12 = 0  3 = 3\) Coordinate 2 \((1, 3)\) Therefore, the solution is \(3 ≤ y ≤ 3\)
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Re: If Y = X + 1  X2, then [#permalink]
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Updated on: 24 Sep 2017, 13:27
Hi All, This question can be solved by TESTing VALUES. We're told that Y = X + 1  X2. We're asked to find the range of value for Y. Let's start with something easy.... IF.... X = 0, then Y = 1  2 = 1, so Y can equal 1. Eliminate Answers C and E. Considering the three remaining answers, we know that Y either has an "upper limit" or it does not, so let's see what happens if we make X really big or really small.... IF... X = 100, then Y = 101  98 = 3. This is interesting, since we've appeared to now randomly hit the 'upper limit' in Answer B. Eliminate Answer A. What if we try something even bigger.... X = 1000, then Y = 1001  998 = 3. This is the exact SAME value... It certainly looks like we've found the upper limit, but just to be sure, I'll do one more Test... IF.... X = 50, then Y = 49  52 = 3. We clearly have the range now. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If Y = X + 1  X2, then [#permalink]
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23 Sep 2017, 13:15
EMPOWERgmatRichC wrote: ...
IF.... X = 0, then Y = 1  2 = 1, so Y can equal 1.
... Rich How so? How does one approach this via a number line?



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Re: If Y = X + 1  X2, then [#permalink]
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23 Sep 2017, 13:23
mikemcgarry  Could you help me with plotting a solution on the number line, please?



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Re: If Y = X + 1  X2, then [#permalink]
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24 Sep 2017, 13:29
Hi Blackbox, Thanks for catching the error; I've updated my explanation. GMAT assassins aren't born, they're made, Rich
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Re: If Y = X + 1  X2, then
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