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If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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14 May 2017, 09:26
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If y = x2 + x  x+2 where x is an integer, then y can take how many nonzero integral values between 10 and 10, exclusive? A.10 B.11 C.12 D.13 E.14
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Re: If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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15 May 2017, 00:45
I am not sure about how to do this... Can anybody please suggest a process?



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If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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15 May 2017, 02:51
msk0657 wrote: If y = x2 + x  x+2 where x is an integer, then y can take how many nonzero integral values between 10 and 10, exclusive?
A.10 B.11 C.12 D.13 E.14 Start substituting values for the simplest approach to this question. y = x2 + x  x+2 @x = 0, y = 0 @x = 1, y = 1 @x = 2, y = 2 @x = 3, y = 1 @x = 4, y = 0 @x = 5, y = 1 @x = 6, y = 2 @x = 7, y = 3 @x = 8, y = 4 @x = 9, y = 5 and so on so all positive values till 10 so total Integer values of x = 12 {2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Answer: Option C
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If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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15 May 2017, 03:11
mine is A, only 10 values, after breaks the absolute sign, y can only takes range from 2 to 9, and except 0 => 10 values 2 0 2 x2  0 + x  0 + x+2  0 + /x+2/ x+2 x2
x >=2 y = x 4 => 10 values



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Re: If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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15 May 2017, 03:25
chesstitans wrote: mine is A, only 10 values, after breaks the absolute sign, y can only takes range from 2 to 9, and except 0 => 10 values 2 0 2 x2  0 + x  0 + x+2  0 + /x+2/ x+2 x2
x >=2 y = x 4 => 10 values I see two mistakes here 1) 2 to 9 there are 12 values 2) y may be 0 @x=0
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Re: If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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15 May 2017, 05:04
msk0657 wrote: If y = x2 + x  x+2 where x is an integer, then y can take how many nonzero integral values between 10 and 10, exclusive?
A.10 B.11 C.12 D.13 E.14 Use the number line to figure this out. ................. 5 ........ 4 ....... 3 ....... 2 ....... 1 ........ 0 ........ 1 ........ 2 ........ 3 ......... 4 ........ 5 .............. We need the sum of "distance from 0" + "distance from 2" and from that we need to subtract the "distance from 2". This will be y. A few things to note right away: At 0, distance from 2 = distance from 2 and hence y = 0 At the right of 0, distance from 2 and distance from 0 effectively will always add up to 2. We will just keep adding distance from 2 to this. At x = 1, this will give us y = 1. At x = 2, this will give us y = 2. At x = 3, this will give us y = 1. and now the distance from 2 will keep increasing so we will get all values from 0 to infinity. In our desired range, we get 0 to 9. At the left of 0, the distance from 2 will always get cancelled out by distance from 0 and hence y will always be positive. We have already got all positive values of y. So all values of y from 2 to 9 are possible i.e. 12 values. Answer (C)
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Re: If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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15 May 2017, 05:54
Hi, As explained above, the values y can take is from 2 to 9... When x is 0, y=0... At all values of x above 2, an increase of 1 in value of x, increase in x2 and x will also be 1 each and x+2 will increase by 1.. Thus effect will be 1+11=1.. At x as 2, y=22+22+2=2.. so y can take all values of 2 and above it.. So 12 values.. But be careful, inspite of doing everything correct, you will land up with wrong answer as we are looking for nonzero.. There are 12 values including 0.. Ans=121=11
B
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Re: If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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15 May 2017, 08:11
msk0657 wrote: If y = x2 + x  x+2 where x is an integer, then y can take how many nonzero integral values between 10 and 10, exclusive?
A.10 B.11 C.12 D.13 E.14 If \(x<2\) then \(y=(2x)x+(x+2)=4x\)\(x=3 \implies y=7\) \(x=4 \implies y=8\) If \(x\) decreases then \(y\) increases. If \(2 \leq x < 0\) then \(y=(2x)x(x+2)=3x\)\(x=2 \implies y=6\) \(x=1 \implies y=3\) If \(0 \leq x < 2\) then \(y=(2x)+x(x+2)=x\)\(x=0 \implies y=0\) \(x=1 \implies y=1\) If \(x \geq 2\) then \(y=(x2)+x(x+2)=x4\)\(x=2 \implies y=2\) \(x=3 \implies y=1\) \(x=4 \implies y=0\) ... \(x=13 \implies y=9\) \(x=14 \implies y=10\) If \(x\) increases then \(y\) increases. Based on cases above, we see that \(y\) could receive value from \(2\) to positive infinitive. Hence, y could receive 13 values between 10 and 10 (from 10 to 10). However, the question asks for nonzero integral values, hence we need to exclude 0 and 10. Hence the answer will be 11. 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9 The answer is B
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Re: If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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16 May 2017, 04:52
GMATinsight wrote: msk0657 wrote: If y = x2 + x  x+2 where x is an integer, then y can take how manynonzero integral values between 10 and 10, exclusive?
A.10 B.11 C.12 D.13 E.14 Start substituting values for the simplest approach to this question. y = x2 + x  x+2 @x = 0, y = 0@x = 1, y = 1 @x = 2, y = 2 @x = 3, y = 1 @x = 4, y = 0 @x = 5, y = 1 @x = 6, y = 2 @x = 7, y = 3 @x = 8, y = 4 @x = 9, y = 5 and so on so all positive values till 10 so total Integer values of x = 12 {2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Answer: Option C The question asks fro Nonzero integer values.



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Re: If y = x2 + x  x+2 where x is an integer, then y can take how [#permalink]
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17 May 2017, 10:55
Bunuel plz can u explain this qn... i m not very confident on how the sign changes for range related values as above soln does.




Re: If y = x2 + x  x+2 where x is an integer, then y can take how
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17 May 2017, 10:55






