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If 0 < x < y, then which of the following MUST be true

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If 0 < x < y, then which of the following MUST be true [#permalink] New post 12 May 2017, 06:58
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If 0 < x < y, then which of the following MUST be true?

A) \(\frac{x + 2}{y + 2} > x/y\)

B) \(\frac{x - y}{x} < 0\)

C) \(\frac{2x}{x + y} < 1\)

Answer:
[Reveal] Spoiler:
A, B and C
[Reveal] Spoiler: OA

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Re: If 0 < x < y, then which of the following MUST be true [#permalink] New post 12 Apr 2018, 18:06
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Statement A:
There are two ways to solve this one. Firstly, if you substitute various values for x and y, you'll find that it must be true. It can be a bit haphazard sometimes to substitute values, however, since if you don't really understand the math, you're never quite positive that another set of values wouldn't give you a different answer.

In this case, using algebraic knowledge is a bit more surefire. Since y is larger than x, we know that x/y must be smaller than 1. It's often useful to know the following rule: if you ever add the same amount to both the numerator and denominator of any fraction, it will approach 1. In this case, since we've added 2 to both the top and bottom of the fraction, it must be getting closer to one, and since x/y is less than one, (x+2)/(y+2) must be greater. So A must be true.

Statement B:
In general, in these three part types of questions, you must do something different for each answer choice. So the trick from last time won't work here. Here's the trick for this one: if you ever have a fraction whose numerator is a few terms being added or subtracted, it's a good idea to split it into several fractions. In this case, we can rewrite

(x-y)/x

as

x/x - y/x

which cancels to

1 - y/x

Just as we know that x/y must be less than one, we know that y/x must be greater than one. So if you subtract something greater than one from one, you must get a negative. Thus B must also be true.

Statement C
Let's multiply both sides of this inequality by the denominator, (x+y), to simplify. Is that safe? With inequalities you should always be careful about multiplying or dividing by variables if you don't know whether they're positive or negative, since you therefore won't know whether to flip the inequality. In this case, however, we know that both x and y are positive, so (x+y) must be positive, so it's safe to multiply by it. Thus we get

2x < x + y

Subtract x from both sides:

x < y

Since we're told this in the question, C must also be true.
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Re: If 0 < x < y, then which of the following MUST be true [#permalink] New post 24 May 2018, 10:09
All options are true..
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Re: If 0 < x < y, then which of the following MUST be true [#permalink] New post 25 May 2018, 09:53
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GreenlightTestPrep wrote:
If 0 < x < y, then which of the following MUST be true?

A) \(\frac{x + 2}{y + 2} > x/y\)

B) \(\frac{x - y}{x} < 0\)

C) \(\frac{2x}{x + y} < 1\)

Answer:
[Reveal] Spoiler:
A, B and C


Let's examine each statement individually:

A) (x + 2)/(y + 2) > x/y
Since y is POSITIVE, we can safely take the given inequality and multiply both sides by y to get: (y)(x+2)/(y+2) > x
Also, if y is POSITIVE, then (y+2) is POSITIVE, which means we can safely multiply both sides by (y+2) to get: (y)(x+2) > x(y+2)
Expand: xy + 2y > xy + 2x
Subtract xy from both sides: 2y > 2x
Divide both sides by 2 to get: y > 2
Perfect! This checks out with the given information that says 0 < x < y
So, statement A is TRUE

B) (x - y)/x < 0
Let's use number sense here.
If x < y, then x - y must be NEGATIVE
We also know that x is POSITIVE
So, (x - y)/x = NEGATIVE/POSITIVE = NEGATIVE
In other words, it's TRUE that (x - y)/x < 0
Statement B is TRUE


C) 2x/(x + y) < 1
More number sense...
If x is positive, then 2x is POSITIVE
If x and y are positive, then x + y is POSITIVE
If x < y, then we know that x + x < x + y
In other words, we know that 2x < x + y
If 2x < x + y, then the FRACTION 2x/(x + y) must be less than 1
Statement C is TRUE


Answer: A, B, C

Cheers,
Brent
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Re: If 0 < x < y, then which of the following MUST be true   [#permalink] 25 May 2018, 09:53
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