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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 --> \(3.5<x+y<4.5\). Not sufficient. (2) 1 is the integer that is closest to x-y --> \(0.5<x-y<1.5\). 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.

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 x-y --> \(0.5<x-y<1.5\). 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.

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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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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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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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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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 2.5-1.5 or 2.4-1.4, both meet S1 and S2. However, 2.5 = 3 rounded and 2.4 = 2 rounded. So, insufficient

E.
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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

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