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Re: If y = x + 5  x  5, then y can take how many integer vales?
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30 Jul 2017, 05:24
VeritasPrepKarishma wrote: Bunuel wrote: If \(y=x+5x5\), then \(y\) can take how many integer values? A. 5 B. 10 C. 11 D. 20 E. 21 Kudos for a correct solution. Use the number line for mod questions: ___________5 ______________0______________5___________ You want the values of y which is the difference between "distance from 5" and "distance from 5". Anywhere on the left of 5, < < ___________5 ______________0______________5___________ The difference between the two distances will always be 10 (taking difference to mean what it does for GMAT) In between 5 and 5, can y can the value of 9? Sure. Move 0.5 to the right of 5. At x = 4.5, distance from 5 will be .5 and distance from 5 will be 9.5. So y = 0.5  9.5 = 9 Note that x needn't be an integer. Only y needs to be an integer. Similarly, at various points between 5 and 5, y will take all integer values from 9 to 9. To the right of 5, the difference between the two distances will always be 10. So values that y can take: 10, 9, ... 0 ... 9, 10 i.e. a total of 21 values. hi the distance between 2 points is 10 NOT +10 why ...? is this because the key points for x+5 and for x5 are 5 and +5 respectively .. please correct me if I am missing something ... also, if we see the number line we can easily find that y=9 as long as x= 4.5, but if we want to interact this relationship with the equation given, can it be seen as under ..? x+5 = 4.5 + 5 = 0.5 and x  5 = 4.5  5 = 9.5 thus x+5  x5 = .5  9.5 = 9 please correct me if I am missing something... can such problems as this one, however, be solved by recognizing that the number of integer values y can take will be the values that fall within 10<=y<=10 ....? please correct me if I am missing something .... anyway, please shed some light on me, giving some lessons on number lines so that I can solve with the help of number lines virtually any problem pertaining to absolute modulus ... thanks in advance ...



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Re: If y = x + 5  x  5, then y can take how many integer vales?
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31 Jul 2017, 23:52
ssislam wrote: VeritasPrepKarishma wrote: Bunuel wrote: If \(y=x+5x5\), then \(y\) can take how many integer values? A. 5 B. 10 C. 11 D. 20 E. 21 Kudos for a correct solution. Use the number line for mod questions: ___________5 ______________0______________5___________ You want the values of y which is the difference between "distance from 5" and "distance from 5". Anywhere on the left of 5, < < ___________5 ______________0______________5___________ The difference between the two distances will always be 10 (taking difference to mean what it does for GMAT) In between 5 and 5, can y can the value of 9? Sure. Move 0.5 to the right of 5. At x = 4.5, distance from 5 will be .5 and distance from 5 will be 9.5. So y = 0.5  9.5 = 9 Note that x needn't be an integer. Only y needs to be an integer. Similarly, at various points between 5 and 5, y will take all integer values from 9 to 9. To the right of 5, the difference between the two distances will always be 10. So values that y can take: 10, 9, ... 0 ... 9, 10 i.e. a total of 21 values. hi the distance between 2 points is 10 NOT +10 why ...? is this because the key points for x+5 and for x5 are 5 and +5 respectively .. please correct me if I am missing something ... also, if we see the number line we can easily find that y=9 as long as x= 4.5, but if we want to interact this relationship with the equation given, can it be seen as under ..? x+5 = 4.5 + 5 = 0.5 and x  5 = 4.5  5 = 9.5 thus x+5  x5 = .5  9.5 = 9 please correct me if I am missing something... can such problems as this one, however, be solved by recognizing that the number of integer values y can take will be the values that fall within 10<=y<=10 ....? please correct me if I am missing something .... anyway, please shed some light on me, giving some lessons on number lines so that I can solve with the help of number lines virtually any problem pertaining to absolute modulus ... thanks in advance ... Difference between A and B is taken to be A  B in GMAT (At some other places, we take difference to mean Greater  Smaller) Distance from a point is always positive. For any point to the left of 5, say x = 6, Distance from 5 = 1 Distance from 5 = 11 Difference between "distance from 5" and "distance from 5" = 1  11 = 10 Yes, if x = 4.5, then y = 9. We need integer values of y which we get. This is to show that y will take values between 10 and 10. That, x could be a decimal (which is acceptable) but we might still get an integer value for y.
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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30 Sep 2018, 07:51
Bunuel wrote: If \(y=x+5x5\), then \(y\) can take how many integer values?
A. 5 B. 10 C. 11 D. 20 E. 21
For this question, it's useful to know that a  b represents the DISTANCE from point a to point b on the number line. For example, 3  10 = the DISTANCE from 3 to 10 on the number line. Since 3  10 = 7 = 7, we know that 7 is the distance from 3 to 10 on the number line So, in this case, x  5 = the distance from x to 5 Likewise, since x + 5 = x  (5), we know that x + 5 = x  (5) So, x + 5 = the distance from x to 5 The equation y = x + 5  x  5 has two critical points. These are xvalues that MINIMIZE the value of x + 5 and MINIMIZE the value of x  5 First, x + 5 is minimized when x = 5. That is, when x = 5, x + 5 = (5) + 5 = 0 = 0 Next, x  5 is minimized when x = 5. That is, when x = 5, x  5 = 5  5 = 0 = 0 Let's add these critical points ( x = 5 and x = 5) to the number line. Notice that these critical points divide the number line into 3 regions. Let's see what happens to the value of y when x lies in each region. Let's start with region 1 When x lies in region 1 (e.g., x = 7), notice that the blue bar represents the value of x  5When x lies in region 1 (e.g., x = 7), notice that the red bar represents the value of x + 5 (aka x  (5) Since the distance between 5 and 5 is 10, we can see that the length of the blue bar must be 10 units LONGER than the red barSo, it must be true that x + 5  x  5 = 10 Now let's generalize. For ANY value of x in region 1, the length of the blue bar will be 10 units LONGER than the red bar. So, for ANY value of x in region 1, x + 5  x  5 = 10 Now let's see what's going on in region 3 When x lies in region 3 (e.g., x = 8), notice that the blue bar represents the value of x  5When x lies in region 3 (e.g., x = 8), notice that the red bar represents the value of x + 5 (aka x  (5) Since the distance between 5 and 5 is 10, we can see that the length of the red bar must be 10 units LONGER than the blue barSo, it must be true that x + 5  x  5 = 10 To generalize, we can say that, for ANY value of x in region 3, x + 5  x  5 = 10IMPORTANT ASIDE: At this point, we've shown that x + 5  x  5 can equal 10 and 10  Now onto region 2 Let's see what happens when x = 4 If x = 4, then x + 5 (the red bar) = 1 and x  5 = 9 So, x + 5  x  5 = 1  9 = 8 Now let's see what happens when x = 3.5 [img]https://i.imgur.com/1SyK8Ii.png[/img If x = 3.5, then x + 5 = 8.5 and x  5 = 1.5 So, x + 5  x  5 = 8.5  1.5 = 7  CONCLUSION: At this point, we've shown that x + 5  x  5 can equal 10, 10, and all values BETWEEN 10 and 10There are 21 INTEGERS, from 10 to 10 inclusive. So, y can have 21 different integer values Answer: E Cheers, Brent
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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19 Dec 2018, 08:30
Bunuel wrote: If \(y=x+5x5\), then \(y\) can take how many integer values?
A. 5 B. 10 C. 11 D. 20 E. 21
\(?\,\,\,:\,\,\,\# \,\,\,{\rm{integer}}\,\,{\rm{values}}\,\,{\rm{for}}\,\,\,y = \left {x + 5} \right  \left {x  5} \right\) \(\left( {\rm{i}} \right)\,\,\,x \le  5\,\,\,\,\, \Rightarrow \,\,\,\,\,y =  \left( {x + 5} \right)  \left( {5  x} \right) =  10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,1\,\,{\rm{integer}}\) \(\left( {{\rm{ii}}} \right)\,\,\,  5 < x \le 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right)  \left( {5  x} \right) = 2x\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{other}}\,\,10 \cdot 2\,\,{\rm{integers}}\,\,\,\left( * \right)\,\,\,\) \(\left( * \right)\,\,\,x\,\,{\mathop{\rm int}} \,\,\,\left( {  4 \le x \le 5\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\,\,\,\,{\rm{or}}\,\,\,\,\,\left\{ \matrix{ \,x \ne {\mathop{\rm int}} \hfill \cr \,x = {{{\mathop{\rm int}} } \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\,\left( {  9 \le \mathop{\rm int}\,{\rm{odd}}\,\, \le 9\,\,\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\) \(\left( {{\rm{iii}}} \right)\,\,\,x > 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right)  \left( {x  5} \right) = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{already}}\,\,{\rm{obtained}}\,\,{\rm{in}}\,\,\left( {{\rm{ii}}} \right)\) \(? = 1 + 20 = 21\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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19 Dec 2018, 09:36
Bunuel wrote: If anyone interested here is a graph of \(y=x+5x5\): As you can see y is a continuous function from 10 to 10, inclusive. Exactly, Bunuel! This gives us (now explicitly) an immediate alternate solution: There are 21 integers between 10 and 10, both of them included (see blue interval). Regards, Fabio.
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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31 Aug 2019, 04:26
If y=x+5−x−5 then y can take how many integer values? We need to find integer values of y. x can be an integer or noninteger. if you enter in x as 5,4.5,4,3.5,3,2.5,2,1.5,1,0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5 in y=x+5−x−5 then you will get 21 different values of y. Beyond 5 and 5 you will get integer values same as the ones you get by entering the above 21 numbers.
Hence 21.



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Re: If y = x + 5  x  5, then y can take how many integer vales?
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11 Sep 2019, 19:15
For Y =  X+5    X  5 , we can identify critical points as 5 and 5. So X must be in following ranges. X < 5 or 5 <= X < 5 or X > 5
1. Considering X < 5, we can modify modulus as Y = (X + 5) + (X  5) = X 5 + X 5 Y =  10...….(1 integer solution of Y for X < 5)
2. Considering X >= 5, we can modify modulus as Y = (X +5 )  (X  5) = X + 5  X + 5 Y = 10...……(1 integer solution of Y for X >= 5)
3. Considering 3rd and last possible range of X. 5 <= X < 5, Y = (X + 5) + (X 5) Y = 2X...….Thus X can take any value between 5 till 4.9. But need integer values of Y. Y= 2X can give integer values of Y only when X is an integer or X is a fraction with .5 at the end. So the value X can take are ( 5, 4.5, 4, 3.5, 3, 2.5, 2, 1.5, 1, 0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5) so total 20 values. But X = 5 will give Y to be 10. We have already considered Y = 10 in 1st range. hence in this range we get 19 possible integer values of Y.
Thus total possible integer values of Y are 1 + 1 + 19 = 21
hence (E)



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Re: If y = x + 5  x  5, then y can take how many integer vales?
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22 Sep 2019, 07:27
Bunuel wrote: If \(y=x+5x5\), then \(y\) can take how many integer values? A. 5 B. 10 C. 11 D. 20 E. 21 Kudos for a correct solution. Given: \(y=x+5x5\), Asked: \(y\) can take how many integer values? When x = 5; y = 10 when x=5; y = 10 When x <5; y= 10 When x>5; y =10 5<=x<=5; 10<=y<=10; 21 integer values IMO E



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Re: If y = x + 5  x  5, then y can take how many integer vales?
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08 Jan 2020, 00:03
Hello Bunuel, Can you please explain how did you obtain the range. I am getting for Mod (X+5), for positive solution, (X+5)>=0, this implies X>=5 and for negative solution X+5<0, this gives me range X<5. And for Mod(X5) , for positive solution , (X5)>=0, this gives me X>=5 and (X5)<0 gives me X<5.




Re: If y = x + 5  x  5, then y can take how many integer vales?
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