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# If x/y >1and x and y are integers, is x>1?

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If x/y >1and x and y are integers, is x>1?  [#permalink]

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26 Jun 2018, 09:26
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If $$\frac{x}{y}>1$$ and x and y are integers, is x>1?
1) $$y^2+5y=6$$
2) $$x^2+x=6$$

New tricky question

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If x/y >1and x and y are integers, is x>1?  [#permalink]

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26 Jun 2018, 11:19
chetan2u wrote:
If $$\frac{x}{y}>1$$ and x and y are integers, is x>1?
1) $$y^2+5y=6$$
2) $$x^2+x=6$$

New tricky question

x/y > 1 and x, y are both integers. Now both x & y have to have the same sign (both positive or both negative) for x/y to be > 1.
If x & y are both positive, then x has to be > y (then only x/y will be > 1, eg, x= 3, y= 2)
If x & y are both negative, then x has to be < y (then only x/y will be > 1, eg, x= -3, y= -2)
Basically for x/y > 1, x & y must have same signs and magnitude of x has to be greater than magnitude of y, i.e., |x| > |y|.

(1) y*(y+5) = 6.
Here y can be either 1 or -6. If y=1, then x > 1. But if y= -6, then x is negative.
So this statement is not sufficient.

(2) x*(x+1) = 6.
Here x can be either 2 or -3. So x can be > 1 or < 1.
This statement is also not sufficient.

Combining the two statements, we know that y can take only two values: 1 or -6. And x can also take only two values: 2 or -3.
Now if y= 1, then x can only be = 2 (because x/y has to be > 1).
If y= -6, then neither value of x will be ok here. Because if x= 2, then x/y will become negative; and if x= -3, then x/y will become < 1.
This means the only acceptable value of y is = 1, and thus the only acceptable value of x is = 2, which is > 1.
Thus sufficient.

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Re: If x/y >1and x and y are integers, is x>1?  [#permalink]

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26 Jun 2018, 11:41
chetan2u wrote:
If $$\frac{x}{y}>1$$ and x and y are integers, is x>1?
1) $$y^2+5y=6$$
2) $$x^2+x=6$$

New tricky question

Interesting question! What makes this one tricky is the extra information in the question. My first thought on seeing it: I really need to remember that x/y is bigger than 1! I'm not trying to figure out whether x/y is bigger or smaller than 1. Instead, that's a fact I already know, and I'll need to combine it with the facts in the statements.

I don't know whether y is positive or negative, so I can't simplify the question by multiplying by y. So, I'll leave it how it is for now.

Statement 1: This simplifies, using quadratic rules, to 'y = -6 or y = 1'. At this point, I know two facts: I know that y is one of those two numbers, -6 or 1, although I don't know which one it is. I also know that x/y > 1, but I don't know what x is.

Given those two facts, is it possible for x to be greater than 1? Is it possible for x to be less than 1? If both of those are possible, then this statement is insufficient.

Well, since y might be 1, we could say that x = 100. That follows all of the rules (y is one of those two values, and x/y > 1), and x is greater than 1.

And since y might be -6, we could say that x = -600. That follows all of the rules (y is one of those two values, and x/y > 1), and x is less than 1.

Since x could be either bigger or smaller than 1, the statement is insufficient. Eliminate A and D.

Statement 2: This simplifies using quadratic rules to 'x = -3 or x = 2'. Again, I know two facts: I know that x is one of those two numbers (although I don't know which one). I also know that x/y > 1. That's not a very interesting fact, since I have no way of figuring out what y is.

It's possible for x to be less than 1, since x could equal -3.

It's also possible for x to be greater than 1, since x could equal 2.

So, the statement doesn't tell us whether x is less than or greater than 1. Insufficient. Eliminate B.

Both statements: Now, we know three facts! Here they are:

x = -3 or 2
y = 1 or -6
x/y > 1

And we're trying to figure out:

is x > 1?

The first thing to figure out is whether x and y could be any of those numbers, or whether we can eliminate some possibilities because they break the rules. It's not possible for x to be -3 and y to be 1, since if that's true, we'd be breaking the third rule, that says x/y > 1. Similarly, it's not possible for x to be -3 and y to be -6. It's also not possible for x to be 2 and y to be -6. The only set of numbers that follows all three of our rules is x = 2 and y = 1.

So basically, we can translate the two statements plus the question as giving us this information:

x = 2 and y = 1.

If we know that, do we know the answer to the question ('is x > 1?')? Yes, we do, since x is definitely 2 and 2 is definitely bigger than 1.
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If x/y >1and x and y are integers, is x>1?  [#permalink]

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26 Jun 2018, 15:39
x and y are integers.
1. We get y is 1 or -6
X can take any value +ve or -ve and accordingly be greater or smaller than y
Not Suff

2. we get x is 2 or -3
Y can take any value +ve or -ve and accordingly be greater or smaller than X
Not suff

Both together
we know $$\frac{X}{Y}$$ >1
So X=2 and Y=1

If x/y >1and x and y are integers, is x>1? &nbs [#permalink] 26 Jun 2018, 15:39
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# If x/y >1and x and y are integers, is x>1?

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