Inequality and absolute value questions from my collection

Doubt :The question is is |N|<4 ,from statement 2 we can see that n can only be a fraction or its value lies between 0 and 1 like say 1/2 .1/3 etc.same for negative fractions
for example for n =1/2 , 1/(|1/2|)=2 which is greater than n=1/2

Also for n=-1/2,1/(|-1/2|)=2 which is greater than n=-1/2


for negative integers also its true ,say we take n=-3 ,
1/|n|=1/3 which is greater than n=-3



But point to note is that both negative nos and fractions are less than 4 ,so this statement is equally sufficient ,correct me if i am wrong !!!!

Dear Bunuel,

I don't quite understand your explanation on question 4, could you please explain to me again?.
When I solved both statement 1 and 2, the answer can be + or -, therefore, how the answer comes up with only 1 sign?.

Thanks
New member

Dear Bunuel,

I don't quite understand your explanation on question 4, could you please explain to me again?.
When I solved both statement 1 and 2, the answer can be + or -, therefore, how the answer comes up with only 1 sign?.

Thanks
New member

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hi

I am not sure if the answer is C.From my point of view it needs to be E.Reason is even if we combine both I and II ,the answer is insufficient.

Consider putting the values as 1) x=1 y=1/2 ,now x>y and x-y=1/2.In this case both x and y are positive
Now2) x=-1 and y=-1/2,now again x>y and x-y=-1/2.In this case both are negative.

Hence insufficient to deduce whther x and y are both positive.

Please share your inputs on the same.

7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=-2.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.
(...)

Bunuel, how did you figure out that |x+2|=|y+2| can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0
Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

In the first statement from 2x-2y=1 --> we can sat x-y=1/2

So it cud be 8.5-8 or 0.25 - (-0.25)
HOw can we say both x and y are positive?

similarily statement 2

x/y>1
=>x>y

how can we be sure x and y have the same sign
we can have 8>7
or
8>-8

Bunel can you pls xplain..or am i missing sumtin fundamental?

7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

when we have x+y=-4
Acc to the explanation , we have
This will occur when either x or y is less then -2 and the other is more than -2

so x can be be -3,-4,-5..(X<-2)
and y can be -1,0,1...(y>-2)

a solution could that -2-2=-4

in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption

Thanks in advance.

7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below -
I used number picking.

A. xy<0,
x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value
or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

thanks
jay

SOLUTIONS:

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression \(6*x*y = x^2*y + 9*y\):

\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.

Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.

(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.

OR:

\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.

(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.

The answer to this one is C right? B alone is not sufficient.