The 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 x-y

answer: E.

I got c using 5/2 and 3/2 and also 8/3 , 4/3.. maybe i am missing a particular case..

A. 4 is the integer that is closest to x+y

3.5 < X + Y <= 4.5 ( I have taken <= with 4.5 since I observed GMAT rounds down 0.5, contrary to mathematics)


B. 1 is the integer that is closest to x-y

-0.5 < X - Y <= 0.5

using: If A > B and C > D then we can say A + C > B + D (this holds only for addition)

using this on LHS and RHS of the inequalities from A and B

3.5 + -0.5 < (X+Y) + (X-Y) --> 2 < 2 X --> 1 < X

AND

(X+Y) + (X-Y) <= 4.5 + 0.5 --> 2 X <= 5 --> X <= 2.5

1 < X <=2.5 --> X can be close to 1 as well as 2

I hope the steps are correct. (Also as a suggestion, please donot post the answers with the question, our minds tend to fit a solution to the answer, rather than give it a best shot :), maybe its just me)

A. 4 is the integer that is closest to x+y

3.5 < X + Y <= 4.5 ( I have taken <= with 4.5 since I observed GMAT rounds down 0.5, contrary to mathematics)


B. 1 is the integer that is closest to x-y

0.5 < X - Y <= 1.5

using: If A > B and C > D then we can say A + C > B + D (this holds only for addition)

using this on LHS and RHS of the inequalities from A and B

3.5 + 0.5 < (X+Y) + (X-Y) --> 4 < 2 X --> 2 < X

AND

(X+Y) + (X-Y) <= 4.5 + 1.5 --> 2 X <= 6 --> X <= 3

2 < X <=3 --> X can be close to 2 as well as 3 (NOT 3 since X is not Integer)

a couple of notes.

one:


no, the gmat doesn't round 5's down in general. and neither does anyone else.

the biggest thing to notice here is that the issue of "rounding" isn't even a factor in this problem.
in fact, the problem goes out of its way to avoid using the term "rounding", because the issue of how to round EXACT 5's (i.e., 5's that are not followed by anything else, except maybe a bunch of zeroes) is actually a source of some disagreement in the mathematical, scientific, and even banking community.
specifically, "traditional" rounding rounds ALL 5's up, ALL the time, even if they're followed by nothing at all.
there's another method of rounding, though, called "convergent rounding" or "bankers' rounding" (read about it here). this method rounds 5's up if they're followed by any nonzero digits - just as does traditional rounding - but it rounds exact 5's (not followed by any other digits) to the nearest even number.
you may never have heard of this second method, but it's in wide use, especially in the scientific community.

this is why the problem doesn't contain the word "round": according to traditional rounding, 4.5 rounds to 5, but according to bankers' rounding, 4.5 rounds to 4. that's bad.

the wording in the actual problem, though, is completely unambiguous: "4 is the integer that is closest to x + y".
this statement actually rules out BOTH 3.5 and 4.5, because each of those numbers is equidistant from two integers: the former from 3 and 4, and the latter from 4 and 5.

therefore, here are the CORRECT rephrases:
(1) 3.5 < x + y < 4.5
(2) 0.5 < x - y < 1.5

all four of those signs are strict inequalities. there are no <'s or >'s in this problem.


--

two:

there's no reason to separate the "sandwich inequalities" into two inequalities apiece; if you do so, you're merely doing twice the work in order to achieve the same result. not a good thing in general, but especially not on a time-management-intensive test.

you can add all 3 corresponding parts of the inequalities directly:
3.5 < x + y < 4.5
0.5 < x - y < 1.5
_____________________

4 < 2x < 6

therefore
2 < x < 3

notice that all this discussion of <'s, <'s, >'s, and >'s is immaterial in the final analysis, because there are still numbers greater than 2.5 (which are closest to 3) and numbers less than 2.5 (which are closest to 2). therefore, insufficient even if you misinterpret the question prompt as referring to "rounding".
but they could, easily, write a problem that would turn on the inclusion/exclusion of a number such as 4.5. i could write such a problem easily by making minor modifications to this problem (i can do so if any posters here would like to see such a problem).

thanks ron.

could you please show me that kind of an 'inclusion' problem and also the solution to it?

here's an example:

what number results if the number x is rounded to the nearest hundred?
(1) the multiple of 20 that is closest to x is 140.
(2) x is within ten units of 140.

here, statement (1) means that 130 < x < 150. that's a strict inequality, which doesn't apply to 130 and 150 themselves (since 130 is just as close to 120 as to 140, and 150 is just as close to 160 as to 140).
all of these numbers give 100 when rounded to the nearest hundred, so this statement is sufficient.

statement (2), on the other hand, means that 130 < x < 150. this inequality includes 130 and 150.
since 150 rounds to 200, this statement is insufficient.

in this problem, the inclusion vs. exclusion of 150 makes all the difference.

@ Ron: Could you please answer the 2 questions below?

Number 1:
So as per the GMAT, can 3.5 be rounded off to 4. Or, would the GMAT not ask a question where this dilemma is encountered?

Number 2:

Does this condition include 150?
I thought within 10 units of 140 means that it does include 140.99(9) but not 150.

You're asking about distinctions the GMAT won't place any weight on.

Number 1: The GMAT will avoid contentious issues, and this is one such issue.

Number 2: "Within" is inclusive, unless otherwise specified, but that's not unique to the GMAT. (Still, if your boss says that he wants the report within a week, you had better make sure that he's operating with the same understanding as you are.)