gmat prep integers

[vinviper1]

Thanks!

[vinviper1]

bumpy bump

[Stuart@KaplanGMAT]

We really need to do this one by picking numbers.

It's easy to eliminate (1) and (2) individually, since each one gives us an infinite number of possible values for x. So, we're at the combination stage and know the answer must be either (c) or (e).

Let's pick numbers that work for both statements and see if we can narrow down x enough.

(1) x + y is closest to 4
(2) x - y is closest to 1

Well, we could pick x = 2.6 and y = 1.4. In this case, the closest integer to x is 3.

However, we could also pick x = 2.4 and y = 1.6. In this case, the closest integer to x is 2.

Since we can get more than 1 possible answer, the statements taken together are insufficient: choose (e).

[leovonp]

I tried to approach it analytically:

4 ~ x + y
1 ~ x - y

If you subtract the second statement from the first statement you end up with

3 ~ 2y

Which means y ~ 1.5 and x ~ 2.5

Now you are stuck in the middle and none of the statements really help you to solve the issue so you pick (e)

[Ian Stewart]

Yes, it's certainly possible to do this without picking numbers.

From Statement 1:
3.5 < x+y < 4.5

and from Statement 2:
0.5 < x-y < 1.5

Combining these,

4 < 2x < 6
2 < x < 3

And thus not enough information to answer the question.

[netigen]

Yes, it's certainly possible to do this without picking numbers.

From Statement 1:
3.5 < x+y < 4.5

and from Statement 2:
0.5 < x-y < 1.5

Combining these,

4 < 2x < 6
2 < x < 3

And thus not enough information to answer the question.

Just a quick note.

The upper bounds will not be 4.5 and 1.5 but 4.49 and 1.49. Though this will not affect the reasoning and final answer.

[Ian Stewart]

Yes, it's certainly possible to do this without picking numbers.

From Statement 1:
3.5 < x+y < 4.5

and from Statement 2:
0.5 < x-y < 1.5

Combining these,

4 < 2x < 6
2 < x < 3

And thus not enough information to answer the question.

Just a quick note.

The upper bounds will not be 4.5 and 1.5 but 4.49 and 1.49. Though this will not affect the reasoning and final answer.

No, the upper bounds will be as I've written them. I'm guessing you're assuming that we're writing things with two decimal place precision, but if x+y = 4.49999, it is still closer to 4 than it is to 5 (or any other integer, for that matter).