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If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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20 May 2015, 10:12
2x – 15 < 2 So 6.5 < x < 8.5 In y = 2 + x  2 – x for above values of X , 2+x will be equal to 2+X and 2 – x will be equal to x2 Substituting y=2+xx+2=4 Answer 1
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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20 May 2015, 13:30
EgmatQuantExpert wrote: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values can y take? (A) 0 (B) 1 (C) 2 (D) 4 (E) Cannot be determined This is Question 4 for the eGMAT Question Series on Absolute Value.Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 22nd May. Till then, Happy Solving! Best Regards The eGMAT Team Here's how to solve it:First find the range of values for x:Inequalities involving absolute values  here you have a number case again. First solve like this without the brackets: 2x – 15 < 2 2x < 17 x < 8.5 Second case (FLIP the sign and put a negative sign around the other side of the inequality): 2x – 15 > 2 x>6.5 Hence: 6.5<x<8.5 >>> since x is an integer, only 7 is possible as a value. Now put in 7 in If y = 2 + x  2 – x to see that there is only 1 solution. ANSWER CHOICE A
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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20 May 2015, 21:12
Solving 2x – 15 < 2 first to get range of x (2x – 15) < 2 x < 8.5 (2x – 15) < 2 x > 6.5 Range of x 6.5<x<8.5y = 2 + x  2 – x to see that there is only 1 solution. \(2 + x = 2 + x\) ...... is always +ve , x > 2 as (6.5<x<8.5) \(2 – x = (2  x)\) ...... range of x > 2 , so 2x will be negative \(y = 2 + x  2 – x\) \(y = 2 + x ((2  x))\) \(y = 4 ;\) constant value
Ans : A
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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20 May 2015, 21:20
reto wrote: Hence: 6.5<x<8.5 >>> since x is an integer, only 7 is possible as a value.
Now put in 7 in If y = 2 + x  2 – x to see that there is only 1 solution. ANSWER CHOICE A reto, want to correct one thing in your solution, there are more possible values for x that will result in y as integer for Example : integer value 8 , and you can take x = 15/2 , 6.9, 7.2 any fraction between (6.5  8.5) all result in integer value for Y
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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21 May 2015, 03:21
UJs wrote: reto wrote: Hence: 6.5<x<8.5 >>> since x is an integer, only 7 is possible as a value.
Now put in 7 in If y = 2 + x  2 – x to see that there is only 1 solution. ANSWER CHOICE A reto, want to correct one thing in your solution, there are more possible values for x that will result in y as integer for Example : integer value 8 , and you can take x = 15/2 , 6.9, 7.2 any fraction between (6.5  8.5) all result in integer value for YDear retoPlease note that the question doesn't mention that x is an integer. So, it would be wrong to assume that. In this case, you were able to answer the question correctly even with this wrong assumption. But in a different question, this wrong assumption would have led you to an incorrect answer. For example, try this variation of the above question: If y = 2 + x  2 – x and x – 1 < 1, how many integer values can y take?
(A) None (B) 1 (C) 2 (D) 3 (E) Cannot be determinedFirst solve it without using the constraint that x is an integer. Then solve the question using the constraint that x is an integer. How does your answer change? Best Regards Japinder
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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21 May 2015, 10:33
Lucky2783 wrote: 2x – 15 < 2 So 6.5 < x < 8.5 In y = 2 + x  2 – x for above values of X , 2+x will be equal to 2+X and 2 – x will be equal to x2
Substituting y=2+xx+2=4
Answer 1 Also using the graphs we can immediately see that y=4 is only solution.
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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22 May 2015, 03:21
X is between 0 and 2. y=2x Does this mean the number of integer values of y can't be determined ? Please suggest. EgmatQuantExpert wrote: UJs wrote: reto wrote: Hence: 6.5<x<8.5 >>> since x is an integer, only 7 is possible as a value.
Now put in 7 in If y = 2 + x  2 – x to see that there is only 1 solution. ANSWER CHOICE A reto, want to correct one thing in your solution, there are more possible values for x that will result in y as integer for Example : integer value 8 , and you can take x = 15/2 , 6.9, 7.2 any fraction between (6.5  8.5) all result in integer value for YDear retoPlease note that the question doesn't mention that x is an integer. So, it would be wrong to assume that. In this case, you were able to answer the question correctly even with this wrong assumption. But in a different question, this wrong assumption would have led you to an incorrect answer. For example, try this variation of the above question: If y = 2 + x  2 – x and x – 1 < 1, how many integer values can y take?
(A) None (B) 1 (C) 2 (D) 3 (E) Cannot be determinedFirst solve it without using the constraint that x is an integer. Then solve the question using the constraint that x is an integer. How does your answer change? Best Regards Japinder



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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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22 May 2015, 06:40
Official ExplanationCorrect Answer: B2x – 15 < 2 Can also be written as x – 7.5 < 1 The expression x  7.5 represents the distance between x and 7.5 on the number line. The inequality x  7.5 < 1 means that the distance between x and 7.5 on the number line is less than 1. This means, x lies between 6.5 and 8.5, exclusive. Now, y = 2+x   2 – x  Rewriting this as y = x + 2   x – 2 Here x+2 represents the distance of x from 2 on the number line And, x2 represents the distance of x from 2 on the number line. Representing these 2 distances on number line: It’s easy to see that y = 4. So, only 1 possible value of y. Note: the question is asking about HOW MANY values y can have, not WHAT values y can have.
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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22 May 2015, 06:49
csirishac wrote: X is between 0 and 2. y=2x Does this mean the number of integer values of y can't be determined ? Please suggest. EgmatQuantExpert wrote: For example, try this variation of the above question:
If y = 2 + x  2 – x and x – 1 < 1, how many integer values can y take?
(A) None (B) 1 (C) 2 (D) 3 (E) Cannot be determined
Dear csirishacYou are right that the highlighted part in the quote above gives us: 0 < x < 2 You're also right that for this range of x, y = 2x Now, the question is: how many integer values can y have? Note that we are not given that x is an integer. So, x can take any decimal values as well. When x = 0.5, y = 1 When x = 1, y = 2 When x = 1.5, y = 3 Thus, y can take 3 integer values only. So, for this question, the correct answer will be D. Does this clarify your doubt? Best Regards, Japinder
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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02 Dec 2015, 15:24
EgmatQuantExpert wrote: Official ExplanationCorrect Answer: B2x – 15 < 2 Can also be written as x – 7.5 < 1 The expression x  7.5 represents the distance between x and 7.5 on the number line. The inequality x  7.5 < 1 means that the distance between x and 7.5 on the number line is less than 1. This means, x lies between 6.5 and 8.5, exclusive. Now, y = 2+x   2 – x  Rewriting this as y = x + 2   x – 2 Here x+2 represents the distance of x from 2 on the number line And, x2 represents the distance of x from 2 on the number line. Representing these 2 distances on number line: It’s easy to see that y = 4. So, only 1 possible value of y. Note: the question is asking about HOW MANY values y can have, not WHAT values y can have. Hi Japinder, Everything is clear to me untill the point you say that clearly y=4. Can you pls show me how we derive that y=4? Thanks. Regards.
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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02 Dec 2015, 17:25
elisabettaportioli wrote:
Hi Japinder,
Everything is clear to me untill the point you say that clearly y=4. Can you pls show me how we derive that y=4?
Thanks.
Regards. Let me try to answer your question. I am assuming that you were able to understand how you got 6.5<x<8.5 from 2x15<2. Once you get this, you realize that for the interval 6.5 < x < 8.5, 2 + x = 2+x and 2x = (2x) = 2+x ....as xa = xa for x \(\geq\) a and = (xa) for x < a So putting the above values in the equation for y, you get, y = 2+x2x = 2+x  (x2) = 2+x x +2 = 4 Hope this helps.



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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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04 Dec 2015, 01:23
Engr2012 wrote: elisabettaportioli wrote:
Hi Japinder,
Everything is clear to me untill the point you say that clearly y=4. Can you pls show me how we derive that y=4?
Thanks.
Regards. Let me try to answer your question. I am assuming that you were able to understand how you got 6.5<x<8.5 from 2x15<2. Once you get this, you realize that for the interval 6.5 < x < 8.5, 2 + x = 2+x and 2x = (2x) = 2+x ....as xa = xa for x \(\geq\) a and = (xa) for x < a So putting the above values in the equation for y, you get, y = 2+x2x = 2+x  (x2) = 2+x x +2 = 4 Hope this helps. ok, let me see If I got it right... If 6.5<x<8.5, then x is positive. Now, with this understanding we try to calculate the following equation y=2+x2x and to do so we need to know the value of 2+x and 2x but since we know that x>0, we can consider only the positive case for 2+x that leads to 2+x and the positive case for 2x that leads to 2x. Now we can substitute the values for the mods in the equation and we get y=2+x(2x)=2+x2+x=2x The outcome I get is not 4 as it you should be. I only get 4 if I rearrange the sign inside the second mod, and precisely: 2x=x2 indeed y=2+x(x2)=2+xx+2=4 Now my question is WHY SHOULD WE ASSUME THAT TO GET THE CORRECT ANSWER WE NEED TO ARRANGE THE SIGN IN 2x?? Please advise. Your contitube is pure gold! Thanks.
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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04 Dec 2015, 04:42
elisabettaportioli wrote: Engr2012 wrote: Let me try to answer your question.
I am assuming that you were able to understand how you got 6.5<x<8.5 from 2x15<2.
Once you get this, you realize that for the interval 6.5 < x < 8.5, 2 + x = 2+x and 2x = (2x) = 2+x ....as xa = xa for x \(\geq\) a and = (xa) for x < a
So putting the above values in the equation for y, you get,
y = 2+x2x = 2+x  (x2) = 2+x x +2 = 4
Hope this helps. ok, let me see If I got it right... If 6.5<x<8.5, then x is positive. Now, with this understanding we try to calculate the following equation y=2+x2x and to do so we need to know the value of 2+x and 2x but since we know that x>0, we can consider only the positive case for 2+x that leads to 2+x and the positive case for 2x that leads to 2x.Now we can substitute the values for the mods in the equation and we get y=2+x(2x)=2+x2+x=2xThe outcome I get is not 4 as it you should be. I only get 4 if I rearrange the sign inside the second mod, and precisely: 2x=x2 indeed y=2+x(x2)=2+xx+2=4 Now my question is WHY SHOULD WE ASSUME THAT TO GET THE CORRECT ANSWER WE NEED TO ARRANGE THE SIGN IN 2x?? Please advise. Your contitube is pure gold! Thanks. Ok. Couple of things. Text in red above is not correct. Second, you need to brush up your knowledge of absolute values. Look here mathabsolutevaluemodulus86462.htmlOnce you get that 6.5<x<8.5, this means that x is not just positive but this range of 'x' will also give you 2x = (2x) and NOT 2x This is because, ab = (ab) if a < b. So in our example as all values of 'x' will be >2 , 2x = (2x). Text in greegn above "WHY SHOULD WE ASSUME THAT TO GET THE CORRECT ANSWER WE NEED TO ARRANGE THE SIGN IN 2x??" is absolutely the wrong way of looking at the correct logic behind this question. We are not assuming anything. We are going by the rules of absolute values. Hope this helps.



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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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16 May 2017, 09:32
EgmatQuantExpert wrote: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values can y take? (A) 0 (B) 1 (C) 2 (D) 4 (E) Cannot be determined This is Question 4 for the eGMAT Question Series on Absolute Value.Provide your solution below. Kudos for participation. Happy Solving! Best Regards The eGMAT Team Hey..i got it in 1:49:) from mod(2x+15)<2 we get lx+7.5l<1 i.e. distance of x from 7.5 or 7.5 is less than 1 therefore 8.5<x<6.5 or 6.5<x<8.5 ....as the question is screaming integer...test first the integer value say 7 wwe get y = 4 now test any decimal value say 8.4 and 6.6...we get non integer value just to be sure try another non integer value say 8.3 and 6.6 we get no integer vakue so only one integer value... But i need a concept clarity it is said that one should interpret lxl as distance of x from zero i.e. lx0l..so drawing a parallel we can interpret lx7l as distance of x from 7 my question, so should we interpret lx+7l as distance of x from () 7??...had i been sure about it..i would not have tried above question from both positive and negative values.. Pl help:)



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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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13 Jul 2017, 07:25
reto wrote: EgmatQuantExpert wrote: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values can y take? (A) 0 (B) 1 (C) 2 (D) 4 (E) Cannot be determined This is Question 4 for the eGMAT Question Series on Absolute Value.Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 22nd May. Till then, Happy Solving! Best Regards The eGMAT Team Here's how to solve it:First find the range of values for x:Inequalities involving absolute values  here you have a number case again. First solve like this without the brackets: 2x – 15 < 2 2x < 17 x < 8.5 Second case (FLIP the sign and put a negative sign around the other side of the inequality): 2x – 15 > 2 x>6.5 Hence: 6.5<x<8.5 >>> since x is an integer, only 7 is possible as a value. Now put in 7 in If y = 2 + x  2 – x to see that there is only 1 solution. ANSWER CHOICE A integer values possible are 7 and 8.
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Re: If y = 2 + x  2 – x and 2x – 15 < 2, how many integer values
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26 Jul 2017, 12:59
elisabettaportioli wrote: EgmatQuantExpert wrote: Official ExplanationCorrect Answer: B2x – 15 < 2 Can also be written as x – 7.5 < 1 The expression x  7.5 represents the distance between x and 7.5 on the number line. The inequality x  7.5 < 1 means that the distance between x and 7.5 on the number line is less than 1. This means, x lies between 6.5 and 8.5, exclusive. Now, y = 2+x   2 – x  Rewriting this as y = x + 2   x – 2 Here x+2 represents the distance of x from 2 on the number line And, x2 represents the distance of x from 2 on the number line. Representing these 2 distances on number line: It’s easy to see that y = 4. So, only 1 possible value of y. Note: the question is asking about HOW MANY values y can have, not WHAT values y can have. Hi Japinder, Everything is clear to me untill the point you say that clearly y=4. Can you pls show me how we derive that y=4? Thanks. Regards. hi as far as the number line is concerned, just subtract 2 from 2, a difference revealing the number of numbers y can take... hope this helps ...



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