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The numbers x and y are NOT integers. The value of x is [#permalink]
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04 Jan 2011, 13:52
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The numbers x and y are NOT integers. The value of x is closest to which integer? (1) 4 is the integer that is closest to x+y (2) 1 is the integer that is closest to xy
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Last edited by Bunuel on 09 Apr 2012, 12:37, edited 1 time in total.
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Re: The numbers x and y are not integers ... [#permalink]
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04 Jan 2011, 14:05
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Re: DS  Integers [#permalink]
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Numbers X and Y are not integers. the value of x is closest to which integer? A. 4 is the integer that is closest to X+Y B. 1 is the integer that is closest to XY  Answer: I dont see any helpful rephrasing. Concepts tested are decimals. 1) so x + y equals 3.51 to 4.49 and x can be 0.52 and close to one or could be equal 3.48 and close to 3. Insuff. AD out 2) so x  y can be equal to 0.51 or 1.49 for example and x can be 1.01 and close to 1 or x could be 0.48 and x close to 0 Insuff B out 1+2) we know 3.51 < x + y < 4.49 0.51 < x  y < 1.49 If we add both y cancels out and we get 4.02 < 2x < 5.98 and so 2.01 < x < 2.99 so x could be close to 2 or 3 Insuff Answer is E.
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Re: The numbers x and y are not integers ... [#permalink]
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04 May 2011, 23:31
1+2 gives , 2.1 < x < 2.9 thereby giving either 2 or 3 as value. Hence E.
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Re: The numbers x and y are not integers ... [#permalink]
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07 May 2011, 15:50
1. Not sufficient
3.5<= x+y <= 4.4
we can choose different x values that satisfy the above expression
2. Not sufficient
0.5<=xy<=1.4
we can choose different x values that satisfy the above expression
1&2 2<=x<=2.9. still not sufficient as x can be close to 2 or 3.
Answer is E.



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Re: The numbers x and y are not integers ... [#permalink]
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21 Nov 2011, 18:50
Bunuel wrote: The numbers x and y are NOT integers. The value of x is closest to which integer? (1) 4 is the integer that is closest to x+y > \(3.5<x+y<4.5\). Not sufficient. (2) 1 is the integer that is closest to xy > \(0.5<xy<1.5\). Not sufficient. (1)+(2) Sum above inequalities: \(4<2x<6\) > \(2<x<3\) > so \(x\) can be closer to 2 (for example if \(x=2.1\)) as well as to 3 (for example if \(x=2.9\)). Not sufficient. Answer: E. Check similar question: if500isthemultipleof99421.htmlHope it helps. How do we know we need to take \(3.5<x+y<4.5\) or \(3.5<=x+y<=4.5\) ??????



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Re: The numbers x and y are not integers ... [#permalink]
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24 Nov 2011, 22:35
siddhans wrote: How do we know we need to take \(3.5<x+y<4.5\) or \(3.5<=x+y<=4.5\) ?????? 4 is the integer that is closest to x+y i.e. there is a single integer that is closest to (x+y) If (x+y) = 3.5, which integer is closest to it? Both 3 and 4 are at equal distance i.e. they are both 0.5 away from (x+y). But then, we cannot say that 4 is the integer closest to x+y. Hence, x+y must be greater than 3.5. It must also be less than 4.5 due to the same reason. Note: 3.5 is rounded up to 4 instead of 3 only because we generally follow round up convention. If we follow 'round down' convention, 3.5 will be rounded off to 3. 3.5 is equidistant from both 3 and 4.
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Re: The numbers x and y are not integers ... [#permalink]
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25 Nov 2011, 17:27
VeritasPrepKarishma wrote: siddhans wrote: Note: 3.5 is rounded up to 4 instead of 3 only because we generally follow round up convention. If we follow 'round down' convention, 3.5 will be rounded off to 3. 3.5 is equidistant from both 3 and 4. If on another question, we knew that x when rounded is equal to 4 then : \(3,5\leq x <4,5\). Correct?



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Re: The numbers x and y are not integers ... [#permalink]
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26 Nov 2011, 05:22
If you are not good with inequalities, you can also do this with selecting values to see if you can come up with values that satisfy both 1. & 2. but give different answers for which integer x is closest to.
e.g. from 1. & 2. you can see that x is around 2.5 and y around 1.5.
if x is 2.49 and y is 1.5 you can see that both statements hold (x is closest to 2). if x is 2.51 and y is 1.5 you can see that both statements hold (x is closest to 3).
so even together, there is INSUFFICIENT information to solve.
Same as the answers above, but a different way of approaching it.



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Re: The numbers x and y are not integers ... [#permalink]
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04 Jan 2012, 12:26
E is the answer. The statements are insufficient individually and also when combined. The value of x is between 2 and 3 but a definite answer to a single integer cannot be obtained. Hence, INSUFFICIENT.
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Re: The numbers x and y are not integers ... [#permalink]
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04 Jan 2012, 13:57
Clearly solving S1 and S2 does not make any sense since they could be any fractional combination. So, it is C or E. Together, we could have 2.51.5 or 2.41.4, both meet S1 and S2. However, 2.5 = 3 rounded and 2.4 = 2 rounded. So, insufficient E.
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The numbers x and y are NOT integers. The value of x is closest [#permalink]
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21 Jan 2013, 23:54
amitdgr wrote: The numbers x and y are not integers. The value of x is closest to which integer ?
(1) 4 is the integer that is closest to x+y (2) 1 is the integer that is closest to xy Statement 1: 3.5 < x+y < 4.5 We could also have multiple values of possible y which would either give us an x closest to 0 to 4. INSUFFICIENT! Statement 2: 0.5 < xy < 1.5 This will also be INSUFFICIENT like Statement 1.Together: Let x+y = 4.4 and xy = 1.4 ==> 2x = 3 ==> x= 1.5, closest to 2 Let x+y = 4.1 and xy = 1.4 ==> 2x= 2.7 ==> x=1.35, closest to 1 .. and there are many more possible combinations...Answer: E
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