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If x < y < –1, which of the following must be true?

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

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New post 24 Jun 2017, 07:30
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

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

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

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New post 24 Jun 2017, 11:39
\(x < y < –1\)

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

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

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New post 24 Jun 2017, 11:50
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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New post 25 Jun 2017, 09:00
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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New post 14 Aug 2018, 04:11
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]


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

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New post 14 Aug 2018, 04:22
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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New post 14 Aug 2018, 05:40
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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Re: If x < y < –1, which of the following must be true?   [#permalink] 14 Aug 2018, 05:40
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