The numbers x and y are not integers. The value of x is closest to which integer?
1) 4 is the integer that is closer to x+y.
2) 1 is the integer that is closer to x-y
x and y
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Answer is E.CITI29 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 closer to x+y.
2) 1 is the integer that is closer to x-y
It should be fairly obvious why neither of these individually would work. But what about together? Let's assume that we weren't rounding for a second and solve for x and y:
x-y = 1
x+y =4
2x =5 (add two previous equations together)
x = 2.5
y = 1.5
So x is just straddling the line here (more on this in a sec)... we need to see if it's possible for x to be a little less than 2.5 and a little more than 2.5. Sure enough it can be. For example both these scenarios work:
x = 2.4, y= 1.5 so: x+y = 3.9 and x-y = .9
or
x = 2.6, y = 1.4 so: x+ y = 4 and x-y = 1.2
Since both of these scenarios work, the answer is E.
There's also another problem with the question. It doesn't tell us how to treat a number like 2.5... It's equally close to both 2 and 3. We aren't told to round, where the definition would require us to change 2.5 to 3... So we don't know what to do with it. In any case, I doubt this is a real gmat question...
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Closer is different from rounded.
In my view, x+y and x-y cannot have a tenth digit which is 5
3,5 is not closer to 3 than to 4, so I eliminated the 0.5
I wrote these inequalities,
3,5<x+y<4,5
0,5<x-y<1,5
I summed the two inequalities.
4<2x<6
2<x<3
So x can be closer either to 2 or to 3
I would like to know if it is permitted to add 2 inequalities like that, I try to find maths principles about that, but I have nothing.
Well, I think
A<B<C
D<E<F
A+D<B+E<C+F only if we know everything is positive
In my view, x+y and x-y cannot have a tenth digit which is 5
3,5 is not closer to 3 than to 4, so I eliminated the 0.5
I wrote these inequalities,
3,5<x+y<4,5
0,5<x-y<1,5
I summed the two inequalities.
4<2x<6
2<x<3
So x can be closer either to 2 or to 3
I would like to know if it is permitted to add 2 inequalities like that, I try to find maths principles about that, but I have nothing.
Well, I think
A<B<C
D<E<F
A+D<B+E<C+F only if we know everything is positive
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While you should definitely be very careful about adding inequalities, you can certainly do it. What you've done above is correct, except for the part I've highlighted in red- what you've done is true even if some of the numbers are negative. In brief, if you know:pepeprepa wrote: I would like to know if it is permitted to add 2 inequalities like that, I try to find maths principles about that, but I have nothing.
Well, I think
A<B<C
D<E<F
A+D<B+E<C+F only if we know everything is positive
a > b
c > d
Then a+c > b+d. Just be sure your inequalities are going in the same direction before you try to do this (i.e., be sure both are '>' or '<'). While this should probably make intuitive sense if you think about what we're doing here, there are many ways to see why it is permissible to combine inequalities in this way- you could take a formal approach:
if a > b, this just means that a = b + x, where x is positive.
if c > d, then c = d + y, where y is positive.
Then a + c = b + d + x + y, and since x+y is positive, a+c must be larger than b+d. And it doesn't matter if a, b, c and d are positive or negative.
I said above that you should be very careful when doing this, and you should only do what you know is allowed. You cannot, for example, subtract inequalities- you can only add them as above, when the inequalities are in the same direction. You can see why subtraction leads to inconsistent answers:
3 > 1
3 > 0
and if you thought you could subtract here, you'd conclude that 0 > 1, which is clearly nonsense. I'd also warn against trying to combine inequalities in more exotic ways- e.g. by multiplying them- unless you really know what you're doing. In some circumstances you can do such things, but in many situations you cannot.
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