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If y = x + 5  x  5, then y can take how many integer vales?
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01 Jul 2014, 04:06
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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 M3021
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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01 Jul 2014, 04:07
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. Try NEW absolute value DS question. Attachment:
WolframAlphayx5x5_20140708_0728.png [ 12.01 KiB  Viewed 32409 times ]
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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01 Jul 2014, 13:32
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 + 5  x  5, then y can take how many integer vales?
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01 Jul 2014, 07:14
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 + 5  x  5, then y can take how many integer vales?
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01 Jul 2014, 10:39
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 + 5  x  5, then y can take how many integer vales?
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01 Jul 2014, 19:36
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 + 5  x  5, then y can take how many integer vales?
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08 Jul 2014, 04:33
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 + 5  x  5, then y can take how many integer vales?
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07 May 2015, 23:18
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 + 5  x  5, then y can take how many integer vales?
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20 May 2015, 08:17
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 + 5  x  5, then y can take how many integer vales?
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20 May 2015, 09:14
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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Re: If y = x + 5  x  5, then y can take how many integer vales?
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21 May 2015, 11:24
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:
Unbenannt.jpg [ 38.03 KiB  Viewed 18636 times ]
So just tick a box, but don't take it too serious



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Re: If y = x + 5  x  5, then y can take how many integer vales?
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21 May 2015, 20:23
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 + 5  x  5, then y can take how many integer vales?
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21 May 2015, 20:27
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 + 5  x  5, then y can take how many integer vales?
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04 Jul 2015, 12:13
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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Re: If y = x + 5  x  5, then y can take how many integer vales?
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09 Jul 2015, 21:17
ankushbagwale wrote: 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, Yes, when you are adding the two terms, there will no limit on the values y can take. Keep increasing x and y will keep increasing. When x = 10, y = 20 When x = 20, y = 30 and so on... y can take infinite integer values.
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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01 Sep 2016, 23:42
In the range 5<x<=5, there are only 11 integers. So, if we are solving for x using 10<=2x<10 and it gives us 20 integers,then only 11 of these are possible within the range 5<x<=5. Can someone please help me understand as to why we are going with 21 as the answer?



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Re: If y = x + 5  x  5, then y can take how many integer vales?
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02 Sep 2016, 00:35
La1yaMalhotra wrote: In the range 5<x<=5, there are only 11 integers. So, if we are solving for x using 10<=2x<10 and it gives us 20 integers,then only 11 of these are possible within the range 5<x<=5. Can someone please help me understand as to why we are going with 21 as the answer? Check Bunuel's solution here: ifyx5x5thenycantakehowmanyinteger173626.html#p1378885In the range x <= 5, you have 1 solution. In the range x < x < 5, you have 19 solutions. In the range x >= 5, you have 1 solution. Total 21 solutions.
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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12 Mar 2017, 19:12
Bunuel wrote: SOLUTION
If \(y=x+5x5\), 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. Dear Bunuel, If the question changed to \(y=x+5+x5\). When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5+x5=x5+(5x)=2x\). When \(5<x<5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5+x5=x+5+(5x)=10\). 1 integer values of \(y\) for this range. When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5+x5=x+5+(x5)=2x\). Does the answer equal to infinite?
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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12 Mar 2017, 23:26
ziyuen wrote: Bunuel wrote: SOLUTION
If \(y=x+5x5\), 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. Dear Bunuel, If the question changed to \(y=x+5+x5\). When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5+x5=x5+(5x)=2x\). When \(5<x<5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5+x5=x+5+(5x)=10\). 1 integer values of \(y\) for this range. When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5+x5=x+5+(x5)=2x\). Does the answer equal to infinite? Yes, in this case y can take infinitely many integer values. You can see this directly from the graph: when x<5 or when x>5, the value of y increases along with increase in absolute value of x.
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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22 May 2017, 09:40
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. There are a lot of good mathematical solutions here, but I'll be honest  even as somebody with a 790 GMAT, that's absolutely not what I'd do. I'm not confident in my ability to consistently draw correct graphs for complex absolute value functions like this one. Especially not within 2 minutes. I would (and I did) actually just start plugging in numbers to see what happens. The biggest answer choice is 21, so there can't possibly be more than 21 values to count. That really isn't that many, if I work quickly. The numbers in the problem aren't tough either (just adding and subtracting 5). I might be able to find some patterns along the way that would speed it up as well. Plus, if I come up with an easy strategy for a problem that I know will eventually work, that means I can safely spend a little extra time on the problem. I know that I'll figure it out, so I might spend as much as 3 minutes on this one if I really had to. Okay, let's try it: x = 0: y = 5  5 = 0 x = 1: y = 6  4 = 2 x = 2: y = 7  3 = 4 Interesting  I notice that there's a pattern there. To save time, I'm going to guess that the values keep increasing until I hit x = 5, because x + 5 will keep getting bigger, and x  5 will keep getting smaller. I'm going to assume that x = 3, x = 4, and x = 5 give me the values 6, 8, and 10. If I go bigger than x = 5, what happens? How about x = 6? 6 + 5  6  5 = 10  interesting  that's a value I already got. 7 + 5  7  5 = 10  same value again. Boring. I'm going to stop testing bigger numbers. Right now, I've got on my paper: 0, 2, 4, 6, 8, 10. Now let's test negative numbers, since we're dealing with absolute values. x = 1: 1 + 5  1  5 = 4  6 = 2 x = 2: 2 + 5  2  5 = 3  7 = 4 Okay, I'm going to guess that it's symmetrical, and that I can get the values 2, 4, 6, 8, and 10. If I had extra time, I could check that. I'm almost done, but let me think just a little bit more. Is there any way I could possibly get a different value? Maybe I should try some fractions, like x = 1/2... just to see what happens. x = 1/2: 1/2 + 5  1/2  5 = 5.5  4.5 = 1 Interesting! That gave me a value I didn't have before. How about x = 1/3? x = 1/3: 1/3 + 5  1/3  5 = 5.333  4.6666 = not an integer I'm thinking right now that I have to work with halves, since when I subtract, it has to come out to an integer. So x = 1/2 gives me a value of 1. x = 3/2 gives a value of 3, x = 5/2 gives a value of 5... I can probably also get 1, 3, 5, ... just by dealing with negative numbers instead. Finally, how big can these numbers get? What about x = 9/2? x = 9/2: 9/2 + 5  9/2  5 = 9.5  0.5 = 9 x = 11/2: 11/2 + 5  11/2  5 = 10.5  0.5 = 10 x = 13/2: 13/2 + 5  13/2  5 = 11.51.5 = 10 Okay, it gets 'stuck' at 10 again. Looks like my values are: 0, 2, 4, 6, 8, 10 2, 4, 6, 8, 10 1, 3, 5, 7, 9 1, 3, 5, 7, 9 For a total of 21 values. I did a lot of writing and a lot of arithmetic, but I didn't take that much time on this one. That's because I went with a plan I felt pretty confident about right from the beginning. I took the 'easy way out'  that's what you want to do on the GMAT!
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Re: If y = x + 5  x  5, then y can take how many integer vales?
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