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If y = x + 5x  5, then y can take how many integer [#permalink]
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01 Jul 2014, 05:06
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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SOLUTIONIf \(y=x+5x5\), then \(y\) can take how many integer values?A. 5 B. 10 C. 11 D. 20 E. 21 When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x5(5x)=10\). 1 integer value of \(y\) for this range. When \(5<x<5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x+5(5x)=2x\). Therefore for this range \(10<(y=2x)<10\). 19 integer values of \(y\) for this range (from 9 to 9, inclusive). When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5x5=x+5(x5)=10\). 1 integer value of \(y\) for this range. Total = 1 + 19 + 1 = 21. Answer: E. If anyone interested here is a graph of \(y=x+5x5\): Attachment:
WolframAlphayx5x5_20140708_0728.png [ 12.01 KiB  Viewed 15690 times ]
As you can see y is a continuous function from 10 to 10, inclusive. Try NEW absolute value DS question.
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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If \(y=x+5x5\), then \(y\) can take how many integer values? A. 5 B. 10 C. 11 D. 20 E. 21 Sol :\(y=x+5x5\), then \(y\) We see that there is a following range of values x<5 5< x<5 x>5 Now for any value of x<5 the expression will give the value 10... Consider x=6 so we have 111= 111=10 Consider x=10, we have 515= 515=10 Similarly when x>5 the expression will give value of 10 Consider x=7, 122=10 X=100,10595=10 So we have 2 values of y possible 10, 10 Consider the range 5<x<5...At x=5 and 5 value is 10 and 10 respectively Consider x=4, 19=8 x=3,28=6 x=2, 37=4 Similarly x=1 will give you different value Like wise x=0 will give you one value and value of x from 1 to 4 you get different values of y. So total possible values 10,10,0, + 4 values 1 to4 and +4 values for range of x 1 to 4= 11 Ans C
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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01 Jul 2014, 08:49
If \(y=x+5x5\), then \(y\) can take how many integer values?
A. 5 B. 10 C. 11 D. 20 E. 21
X>=5 : Y= X+5X+5=10; X=<5 : Y=X5+X5=10; 5<X<5: Y=2X ; we have 9 values here hence C, y can take 11 integer values.



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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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01 Jul 2014, 11:39
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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.
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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01 Jul 2014, 14:32
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y=x+5  x5 Case 1 : x>5 y=x+5 (x5) =10 Case 2: 5<= x <5 y=(x+5)+(x5) =2x Now x could assume 10 values in this range 5, 4, 3, 2, 1, 0, 1, 2, 3, 4 for which y could assume 10 values 10, 8, 6, 4, 2, 0, 2, 4, 6, 8 However since x could be a fraction , x could also assume values like: 4.5, 3.5, 2.5, 1.5, 0.5, 0.5, 1.5, 2.5, 3.5, 4.5 for which y could take values like: 9,7,5,3,1,1,3,5,7,9 Case 3: x< 5 y=(x+5)+(x5) =10 Therefore combining all the unique values of y we see there are 21 values of y(E)



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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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01 Jul 2014, 20:36
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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.
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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01 Jul 2014, 21:55
Plastelin wrote: proposed solution was wrong. del. You solution i presume was correct...No need to delete your post just because someone else did it differently. Your post will be helpful for other members in the forum as well as one can learn from different ways of attempting a question... Good luck
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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01 Jul 2014, 23:13
Just test the number. Alway start with one of the integer given in the question. Let start with 5 we get 10. Now test 5 gives us 10. Now test 0 gives us 0. Now test 6 gives us 10 or 6 gives us 9. So the pattern is as we increase beyond 6 or 6 value is constant at 10. So the total is 11 including 0 between 5 and 5.
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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08 Jul 2014, 05:33
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SOLUTIONIf \(y=x+5x5\), then \(y\) can take how many integer values?A. 5 B. 10 C. 11 D. 20 E. 21 When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x5(5x)=10\). 1 integer value of \(y\) for this range. When \(5<x<5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x+5(5x)=2x\). Therefore for this range \(10<(y=2x)<10\). 19 integer values of \(y\) for this range (from 9 to 9, inclusive). When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5x5=x+5(x5)=10\). 1 integer value of \(y\) for this range. Total = 1 + 19 + 1 = 21. Answer: E. 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. Kudos points given to correct solutions above.Try NEW absolute value DS question.
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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12 Sep 2014, 13:58
Bunuel wrote: SOLUTIONIf \(y=x+5x5\), then \(y\) can take how many integer values?A. 5 B. 10 C. 11 D. 20 E. 21 When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x5(5x)=10\). 1 integer value of \(y\) for this range. When \(5<x<5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x+5(5x)=2x\). Therefore for this range \(10<(y=2x)<10\). 19 integer values of \(y\) for this range (from 9 to 9, inclusive).
When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5x5=x+5(x5)=10\). 1 integer value of \(y\) for this range. Total = 1 + 19 + 1 = 21. Answer: E. If anyone interested here is a graph of \(y=x+5x5\): Attachment: WolframAlphayx5x5_20140708_0728.png As you can see y is a continuous function from 10 to 10, inclusive. Try NEW absolute value DS question. I think the answer is 11. Options C. The highlighted section should have 9 values instead of 19.



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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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12 Sep 2014, 14:47
dransa wrote: Bunuel wrote: SOLUTIONIf \(y=x+5x5\), then \(y\) can take how many integer values?A. 5 B. 10 C. 11 D. 20 E. 21 When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x5(5x)=10\). 1 integer value of \(y\) for this range. When \(5<x<5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x+5(5x)=2x\). Therefore for this range \(10<(y=2x)<10\). 19 integer values of \(y\) for this range (from 9 to 9, inclusive).
When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5x5=x+5(x5)=10\). 1 integer value of \(y\) for this range. Total = 1 + 19 + 1 = 21. Answer: E. If anyone interested here is a graph of \(y=x+5x5\): Attachment: WolframAlphayx5x5_20140708_0728.png As you can see y is a continuous function from 10 to 10, inclusive. Try NEW absolute value DS question. I think the answer is 11. Options C. The highlighted section should have 9 values instead of 19. Have you tried to write down possible values? I guess you are missing 0 and 9 negative values.
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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08 May 2015, 00:18
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Here's a visual way of thinking through this question.First of all, we see that the given expression for y contains two modulus expressions: x + 5 and x  5 What do these expressions mean? x5 represents the distance of point x from the point 5 on the number line. x+5 can be written as x  (5). So, x+5 represents the distance of point x from the point 5 on the number line. So, let's plot the points 5 and 5 on the number line, and consider the different possible values of x. Case 1: x < 5Depicting this case on the number line: It's clear that in this case, x(5)  x5 = (distance between points  5 and 5 on the number line) = 10. The minus sign here comes because the bigger distance (x5) is subtracted from the smaller distance (x+5) Case 2: 5 < = x <= 5Depicting this case on the number line: Let's say x = 5. In that case, it's easy to see that x(5)  x5 = 10 If x = 4.5, then x(5)  x5 = 0.5  9.5 = 9 . . . If x = 5, then x(5)  x5 = 10 Thus, we see that in Case 2, all integral values between 10 and 10, inclusive, are possible for y. Case 3: x > 5Depicting this case on the number line: It's easy to see that in this case, x(5)  x5 = +10 Combining all casesWe see that the following integral values of y are possible: 10, 9, 8 . . . 8, 9, 10 That is, 21 values in all. Hope this visual way of thinking through this and other similar absolute value questions was helpful for you! Japinder
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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20 May 2015, 09:17
I have been taught that if we have a x+b = c. The midpoint is always b.
I tried applying it here: so x+5 gives us midpoint 5. and x5 gives us midpoint 5.
However  how do I know the actual range? I'm not actually given something like "x+5 < 2" or whatever.



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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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20 May 2015, 10:14
erikvm wrote: I have been taught that if we have a x+b = c. The midpoint is always b.
I tried applying it here: so x+5 gives us midpoint 5. and x5 gives us midpoint 5.
However  how do I know the actual range? I'm not actually given something like "x+5 < 2" or whatever. Dear erikvmThe way to read the equation 'x+b = c is: "the distance of x from the point b on the number line is c'. So 2 values of x are possible: one that is c units to the left of b and the second that is c units to the right of b. As you can see, 'b' is indeed the midpoint between the two possible values of x. But note that I didn't start my analysis of x+b = c from this result. I started from the fact that x+b represents the distance of x from the point b on the number line. Let's now come to the question at hand: Start your analysis by considering that x+5 represents the distance of x from 5 and x5 represents the distance of x from 5. Try solving the question from here on. Go through the solution I've posted above, and if something is still not clear, I'll be happy to help! Best Regards Japinder
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If y = x + 5x  5, then y can take how many integer [#permalink]
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20 May 2015, 12:00
EgmatQuantExpert wrote: erikvm wrote: I have been taught that if we have a x+b = c. The midpoint is always b.
I tried applying it here: so x+5 gives us midpoint 5. and x5 gives us midpoint 5.
However  how do I know the actual range? I'm not actually given something like "x+5 < 2" or whatever. Dear erikvmThe way to read the equation 'x+b = c is: "the distance of x from the point b on the number line is c'. So 2 values of x are possible: one that is c units to the left of b and the second that is c units to the right of b. As you can see, 'b' is indeed the midpoint between the two possible values of x. But note that I didn't start my analysis of x+b = c from this result. I started from the fact that x+b represents the distance of x from the point b on the number line.Let's now come to the question at hand: Start your analysis by considering that x+5 represents the distance of x from 5 and x5 represents the distance of x from 5.
Try solving the question from here on. Go through the solution I've posted above, and if something is still not clear, I'll be happy to help! Best Regards Japinder Hey, thanks for trying to explain, but I still don't understand this logic :[ I do understand that we get two different midpoints, somewhere on the number line, but I don't understand how we can decide on their range. How do I know that the range for for the two values is 5 steps to the right/left of x+5 and x5



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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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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. This is how you solve such questions: Attachment:
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So just tick a box, but don't take it too serious
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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21 May 2015, 21:23
erikvm wrote: I have been taught that if we have a x+b = c. The midpoint is always b.
I tried applying it here: so x+5 gives us midpoint 5. and x5 gives us midpoint 5.
However  how do I know the actual range? I'm not actually given something like "x+5 < 2" or whatever. Actually, don't think of it from the midpoint perspective because then it is not useful in many circumstances. The actual logic of absolute value is this: x  a gives you the "distance of x from point a on the number line" You might want to check out this post to understand this perspective: http://www.veritasprep.com/blog/2011/01 ... edoredid/So y = x+5−x−5 gives you the difference between "distance of x from 5" and "distance of x from 5" So if x = 0, difference between "distance of x from 5" and "distance of x from 5" will be "5"  "5" = 0. This gives you y = 0. Similarly, you can handle something like this: x+5 + x−5 This is sum of "distance of x from 5" and "distance of x from 5". Using this method, you can solve this question like this: ifyx5x5thenycantakehowmanyinteger173626.html#p1379135
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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21 May 2015, 21:27
reto wrote: This is how you solve such questions: Attachment: Unbenannt.jpg So just tick a box, but don't take it too serious Amusing! Though on a serious note, try to understand the concept behind it and the HARD question will be bumped up to at least MEDIUM in no time!
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Re: If y = x + 5x  5, then y can take how many integer [#permalink]
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04 Jul 2015, 13:13
VeritasPrepKarishma wrote: erikvm wrote: I have been taught that if we have a x+b = c. The midpoint is always b.
I tried applying it here: so x+5 gives us midpoint 5. and x5 gives us midpoint 5.
However  how do I know the actual range? I'm not actually given something like "x+5 < 2" or whatever. Actually, don't think of it from the midpoint perspective because then it is not useful in many circumstances. The actual logic of absolute value is this: x  a gives you the "distance of x from point a on the number line" You might want to check out this post to understand this perspective: http://www.veritasprep.com/blog/2011/01 ... edoredid/So y = x+5−x−5 gives you the difference between "distance of x from 5" and "distance of x from 5" So if x = 0, difference between "distance of x from 5" and "distance of x from 5" will be "5"  "5" = 0. This gives you y = 0. Similarly, you can handle something like this: x+5 + x−5 This is sum of "distance of x from 5" and "distance of x from 5". Using this method, you can solve this question like this: ifyx5x5thenycantakehowmanyinteger173626.html#p1379135 VeritasPrepKarishmaI am thinking the number of integer solutions to y=  x+5 +  x5 In this case we will have three cases: 1. When x <5 here y can will have 2x values. 2. when 5<=x <= +5 then we have only one integral solution. 3. when x > 5 y will have 2x values So essentially it will have infinite integral values. Please let me know if my understanding is correct,




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