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Math Expert V
Joined: 02 Sep 2009
Posts: 56302
If x < y < –1, which of the following must be true?  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 76% (01:27) correct 24% (01:39) wrong based on 75 sessions

### HideShow timer Statistics If $$x < y < –1$$, which of the following must be true?

(A) $$\frac{x}{y} > xy$$

(B) $$\frac{y}{x} > x + y$$

(C) $$\frac{y}{x} > xy$$

(D) $$\frac{y}{x} < x + y$$

(E) $$\frac{y}{x} > \frac{x}{y}$$

Source: Nova GMAT
Difficulty Level: 600

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Re: If x < y < –1, which of the following must be true?  [#permalink]

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$$x < y < –1$$

This means that both x and y are NEGATIVE numbers

(A) $$\frac{x}{y} > xy$$

Take $$x = -3$$ and $$y = -2$$ - when we add these values in the above equation we get

$$x/y = 1.5 xy = 6$$

So $$x/y < xy$$ ====> Hence, FALSE

(B) $$\frac{y}{x} > x + y$$

Take $$x = -3$$ and $$y = -2$$ - when we add these values in the above equation we get

$$y/x = -2/-3 = 0.66$$
$$x+y = -3+(-2) = -3-2 = -5$$

So, $$y/x > x+y$$ ===> TRUE

(C) $$\frac{y}{x} > xy$$

Take $$x = -3$$ and $$y = -2$$ - when we add these values in the above equation we get

$$y/x = -2/-3 = 0.66$$
$$xy = 6$$

So$$y/x < xy$$ ===> Hence, FALSE

(D) $$\frac{y}{x} < x + y$$

Take $$x = -3$$ and $$y = -2$$ - when we add these values in the above equation we get

$$y/x = -2/-3 = 0.66$$
$$x+y = -3+(-2) = -3-2 = -5$$

So, $$y/x > x+y$$ ===> Hence, FALSE

(E) $$\frac{y}{x} > \frac{x}{y}$$

Take $$x = -3$$ and $$y = -2$$ - when we add these values in the above equation we get

$$y/x = -2/-3 = 0.66$$
$$x/y = -3/-2 = 1.5$$

So, $$y/x < x/y$$ ===> Hence, FALSE

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Re: If x < y < –1, which of the following must be true?  [#permalink]

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Bunuel wrote:
If $$x < y < –1$$, which of the following must be true?

(A) $$\frac{x}{y} > xy$$

(B) $$\frac{y}{x} > x + y$$

(C) $$\frac{y}{x} > xy$$

(D) $$\frac{y}{x} < x + y$$

(E) $$\frac{y}{x} > \frac{x}{y}$$

it is clear that x & y are negative numbers. So summation of x & y will be negative (for eg. -3+-2=-5)
So, x+y<0 and

y/x will be positive (as negative signs will cancel out. for eg. -3/-2 = 3/2 = 1.5)
So, y/x >0

Obviously Positive number is greater than negative number

So, $$\frac{y}{x} > x + y$$

Hence Option B
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Re: If x < y < –1, which of the following must be true?  [#permalink]

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Best way to solve such questions is take a value for x and y and put these values in all the options.

Let y be -5 and x be -10.
Now, on solving, clearly B must be true.
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Re: If x < y < –1, which of the following must be true?  [#permalink]

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Bunuel wrote:
If $$x < y < -1$$, which of the following is true?

A. $$\frac{x}{y} > xy$$

B. $$\frac{y}{x} > x+y$$

C. $$\frac{y}{x} > xy$$

D. $$\frac{y}{x} < x+y$$

E. $$\frac{y}{x} > \frac{x}{y}$$

We should staraightaway check with option by assuming several values of x and y and judging the property of outcome

x < y < -1

Let, x = -3 and y = -2

A. $$\frac{x}{y} > xy$$ x/y = 1.5 and xy = 6 hence Incorrect option

B. $$\frac{y}{x} > x+y$$ y/x = 2/3 (Always Positive) and x+y = -5 (Always Negative) hence Correct option

C. $$\frac{y}{x} > xy$$

D. $$\frac{y}{x} < x+y$$

E. $$\frac{y}{x} > \frac{x}{y}$$[/quote]

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Re: If x < y < –1, which of the following must be true?  [#permalink]

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Bunuel wrote:
If $$x < y < -1$$, which of the following is true?

A. $$\frac{x}{y} > xy$$

B. $$\frac{y}{x} > x+y$$

C. $$\frac{y}{x} > xy$$

D. $$\frac{y}{x} < x+y$$

E. $$\frac{y}{x} > \frac{x}{y}$$

Here x<y and both are negative.

option (B) LHS : y/x >0 , RHS: x+y <0
So, $$\frac{y}{x} > x+y$$

Ans. (B)
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Re: If x < y < –1, which of the following must be true?  [#permalink]

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If x<y<−1x<y<−1, which of the following is true?

A. xy>xyxy>xy

B. yx>x+yyx>x+y

C. yx>xyyx>xy

D. yx<x+yyx<x+y

E. yx>xyyx>xy

+ B. consider values for x= -4, y=-2 check for all options only B suffices
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If you liked my solution then please give Kudos. Kudos encourage active discussions. Re: If x < y < –1, which of the following must be true?   [#permalink] 14 Aug 2018, 05:40
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