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Re: If y = x + 5x  5, then y can take how many integer
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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 + 5x  5, then y can take how many integer
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31 Aug 2016, 00:42
I'd like to know if i did this correctly.
y=x+5−x−5
1. y = (x+5)  (x  5) = 10 2. y = (x+5)  ((x5)) = 10
Thus, 10 <= y <= 10
10  (10) = 20 + including integer zero = 21 integers



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Re: If y = x + 5x  5, then y can take how many integer
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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 + 5x  5, then y can take how many integer
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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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If y = x + 5x  5, then y can take how many integer
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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 + 5x  5, then y can take how many integer
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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 + 5x  5, then y can take how many integer
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22 May 2017, 08:28
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. Hi Bunuel, in many a question I have seen this graph view for abs questions. Could you please explain how to draw a graph from the abs values equations.



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Re: If y = x + 5x  5, then y can take how many integer
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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 + 5x  5, then y can take how many integer
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09 Jul 2017, 16:24
Can we say that in order to get the number of integer values that Y can take, we can take test cases where we will see the that maximum Y can be is 10 and minimum it can be is 10. If we test cases such as x=5, 5, and x=6 , 6 we will see the the graph will not expand nor will it converge.
Therefore the range is 20 but we need to count zero as an integer therefore, the integer values will be 21.
had it been how many integer values will x take in such a way that y will be different, then we would have 11integer values including zero.



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Re: If y = x + 5x  5, then y can take how many integer
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29 Jul 2017, 13:04
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 mam Thank you for your wonderful solution to the problem... Anyway, do you believe  on the basis of number line knowhow you shared here  this problem can be solved by solving only "10<= y <=10" ..? thanks in advance...



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Re: If y = x + 5x  5, then y can take how many integer
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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 + 5x  5, then y can take how many integer
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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 + 5x  5, then y can take how many integer
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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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If y = x + 5x  5, then y can take how many integer
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30 Sep 2018, 20:33
I attempted this by taking the maximum and minimum values of y. The minimum value is possible when x+5 is least and x5 Is maximum. The least possible value of an absolute is 0. Hence X+5 = 0 X=5
Therefore, y = 0  10 Y = 10
Similarly , maximum value of y is when x+5 is maximum and x5 is least. The least values of an absolute is 0. Therefore y = 10 + 0 y = 10.
Now between the minimum and maximum value of 10 and 10; their are 21 values.



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If y = x + 5x  5, then y can take how many integer
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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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If y = x + 5x  5, then y can take how many integer
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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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Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our highlevel "quant" preparation starts here: https://gmath.net




If y = x + 5x  5, then y can take how many integer &nbs
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