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Bunuel
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Hi Bunuel

I have a question for the number 3:
What if we select x= 10; y = -5 or x= -5; y= 10. For the first par we can say that x/y = -2, which is less than -1 or, on the other hand y=10 / x=-5 which is equal to -2 also.. Please help me understand. Thanks

Your pairs do not satisfy the condition x + y < 0; therefore, they are not valid.
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Sorry Bunuel , I meant (-10,5) or (5, -10) then, x/y would be -2 which is less than -1 or y/x = -2.. Thanks in advance!
Bunuel
Javmex
Hi Bunuel

I have a question for the number 3:
What if we select x= 10; y = -5 or x= -5; y= 10. For the first par we can say that x/y = -2, which is less than -1 or, on the other hand y=10 / x=-5 which is equal to -2 also.. Please help me understand. Thanks

Your pairs do not satisfy the condition x + y < 0; therefore, they are not valid.
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Javmex
Sorry Bunuel , I meant (-10,5) or (5, -10) then, x/y would be -2 which is less than -1 or y/x = -2.. Thanks in advance!
Bunuel
Javmex
Hi Bunuel

I have a question for the number 3:
What if we select x= 10; y = -5 or x= -5; y= 10. For the first par we can say that x/y = -2, which is less than -1 or, on the other hand y=10 / x=-5 which is equal to -2 also.. Please help me understand. Thanks

Your pairs do not satisfy the condition x + y < 0; therefore, they are not valid.

If x = -10 and y = 5, then y/x = -1/2, and in this case \(\frac{y}{x} > -1\) is true.

If x = 5 and y = -10, then x/y = -1/2, and in this case \(\frac{x}{y} > -1\) is true.

So, in any case either \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\) is true, making III true.

Hope it's clear.
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Third inequality is tricky.

x/y>-1 or y/x>-1. Don't multiply each side to reach conclusion. Rather bring the terms on same side of inequality.

So lets say y was -ve and x was +ve from xy<0. and magnitude of y is greater than x as x+y<0,

therefore, x/y>-1 >>>>>>bringing to same side x/y+1>0 OR (x+y)/y>0. Since x+y<0 "given" the numerator is always -ve and denominator was also -ve (y we assumed to be negative) So (-ve)N/(-veD) becomes positive and hence rightly greater than 0.

Apply same to the viceversa case, x-ve, y+ve, Mag(x)>(Mag(y). Here, y/x>-1 inequaility holds true.
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Bunuel
Official Solution:


If \(xy < 0\) and \(x + y < 0\), then which of the following must be true?

I. \(x - y > 0\)

II. \(x > 0\) or \(y > 0\)

III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)


A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III


\(xy < 0\) implies that \(x\) and \(y\) have opposite signs. Hence, in conjunction with \(x + y < 0\), either \(x > 0 > y\) and \(y\) is further from 0 than \(x\) (to make \(x + y\) negative, for example \(x = 1\) and \(y = -2\)), or \(y > 0 > x\) and \(x\) is further from 0 than \(y\) (to make \(x + y\) negative, for example \(x = -2\) and \(y = 1\)).

Let's examine the options:

I. \(x - y > 0\)

The above implies that \(x > y\). This option is not always true because we know that either \(x > 0 > y\) or \(y > 0 > x\). Discard.

II. \(x > 0\) or \(y > 0\)

This option must be true. Since from \(xy < 0\), it follows that one of them must be positive.

III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)

This one is indeed tricky.

If \(x > 0 > y\) and \(y\) is further from 0 than \(x\), then \(\frac{x}{y} > -1\) must be true. We can derive this by dividing the inequality \(x + y < 0\) by the negative value \(y\), and then flipping the inequality sign to get \(\frac{x}{y} + 1 > 0\), which leads to \(\frac{x}{y} > -1\).

On the other hand, if \(y > 0 > x\) and \(x\) is further from 0 than \(y\), then \(\frac{y}{x} > -1\) must be true. We can derive this by dividing the inequality \(x + y < 0\) by the negative value \(x\), and then flipping the inequality sign to get \(1 + \frac{y}{x} > 0\), which leads to \(\frac{y}{x} > -1\). Therefore, in either case, either \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\) is always true.

Hence, only options II and III are always true.


Answer: D
Hi Bunuel,
I have a doubt for second statement. i understand x>0 or y>0 is true but for both the condition to satisfy should also when x>0, y<-x and for y>0, x<-y, these should happen together or x>0 and y>0 satisfies both condition so this must be true. What am i missing here?
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btsaami

Hi Bunuel,
I have a doubt for second statement. i understand x>0 or y>0 is true but for both the condition to satisfy should also when x>0, y<-x and for y>0, x<-y, these should happen together or x>0 and y>0 satisfies both condition so this must be true. What am i missing here?

Statement II does not need to include the full condition from the stem. Since xy < 0, x and y must have opposite signs, so exactly one of them is positive. Therefore, x > 0 or y > 0 must be true.
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I like the solution - it’s helpful.
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Can i substitute values for option 3, say -3 and 2. Are putting values reliable for must be true questions?
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Can i substitute values for option 3, say -3 and 2. Are putting values reliable for must be true questions?


"MUST BE TRUE" ("ALWAYS TRUE"/"IS TRUE") questions:


These questions ask which statement is always true for every valid set of numbers. If you can find just one valid set of numbers where a statement is not true, it means the statement is not always true and therefore not the correct answer. So, for 'MUST BE TRUE' questions, the plug-in method is good for eliminating an option, but it does not provide a 100% guarantee that an option is always true. For "MUST BE TRUE" questions, when using the plug-in method, if you find that more than one option appears to be correct for a particular number or set of numbers, try using different numbers to double-check. Reevaluate only those options that were previously considered correct.

"COULD BE TRUE" questions:


The questions that ask which of the following statements could be true are different. If you can demonstrate that a statement is true for a specific set of numbers, it implies that the statement could be true and therefore is a correct answer.

You can practice more questions of this type HERE.

Hope it helps.­
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