If x and y are numbers such that x < - 10 and y > 6, which of the

As stated, If x and y are numbers such that x < - 10 and y > 6,

if y > 6 , then

| y + 4| = y + 4
| y - 4 | = y - 4

if x < - 10 , then

| x – 4 | = - ( x – 4 )
| x + 4| = - ( x + 4 )

lets solve each
I. |x+4| + | y + 4| + | y-4| + |x – 4| > 32

-(x+4) + (y + 4) + (y - 4) - (x – 4) > 32
-x -4 + y + 4 + y - 4 -x +4 > 32
-2x +2y > 32
-x + y > 16
-(-10.1) + (6.1) > 16 ; True

II. |x – 4| + |y + 4| - | x + 4| - | y – 4| < 0

-(x-4) + (y + 4) - (-(x + 4)) - (y-4) < 0
-x +4 + y + 4 +x -4 -y +4 < 0
8 < 0 ; False

III. |x + 4| + |y + 4| - |x – 4| - |y – 4| = 0

-(x+4) + (y + 4) - (-(x – 4)) - (y - 4) = 0
-x -4 + y + 4 +x -4 -y + 4 = 0
0 = 0 ; True

Ans : D

Official Explanation

Correct Answer: D

First representing the points x and y on the number line: x < -10 and y > 6.

Then, since the expressions talk about |x+4|, |x-4|, |y+4| and |y-4|, representing these distances on the number line as well.

|x+4| represents the distance of x from -4 on the number line
|x-4| represents the distance of x from 4
|y+4| represents the distance of y from -4
|y-4| represents the distance of y from 4



I. |x+4| + | y + 4| + | y-4| + |x – 4| > 32

From the diagram it’s clear that:
|x+4| > 6
|y + 4| >10
|y – 4| > 2
|x – 4| 14

Adding all these we get Expression 1. Therefore, is true always.

II. |x – 4| + |y + 4| - | x + 4| - | y – 4| < 0
From the diagram, we can see that |x – 4| - |x + 4| = 8
And, |y + 4| - | y – 4| = 8
Adding the 2 equations, we get:

|x – 4| + |y + 4| - | x + 4| - | y – 4| = 16

So, expression 2 is NOT TRUE.


III. |x + 4| + |y + 4| - |x – 4| - |y – 4| = 0

From the diagram, we can see that |x + 4| - |x – 4| = -8
And, |y + 4| - | y – 4| = 8
Adding the 2 equations, we get:

|x + 4| + |y + 4| - |x – 4| - |y – 4| = 0

So, expression 3 is TRUE.
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Tried solving the question by substituting values for x and y.
Suppose if we sub. x = -11 and y = 7, and solve the three equations, It's found that statements (i) and (iii) satisfy. Hence D.

How to solve this:

Plug in smart values for x and y. The question tells us that x<-10 and y>6. Therefore I take -11 = x and 7 = y.

I. |x+4| + | y + 4| + | y-4| + |x – 4| > 32
Plug in: 7 + 11 + 3 + 15 = 36 > 32

Be careful now, normally once you have an answer that is true, it doesn't mean that it will be true for all the values of x and y. Here im pretty confident because i took the smallest (integer) numbers for x and y and the result is 36 which is a difference of 4 to 32. Even if you plug in smaller fractional numbers the statement will still be true.

II. |x – 4| + |y + 4| - | x + 4| - | y – 4| < 0
Plug in: 15 + 11 - 7 - 3 > 0, therefore false.

No other plug ins are needed since one false value makes us eliminating it directly.

III. |x + 4| + |y + 4| - |x – 4| - |y – 4| = 0
Plug in: 7 + 11 - 15 - 3 = 0, therefore true.

But is it true for any values of x and y? What about y = 8 and x = -12? Let's try:

Plug in #2: 8 + 12 - 16 - 4 = 0, obviously thats always true.

Answer D like Dragon

Any helpful recommendation how to do this in 2 minutes?

Dear reto

The key to answering this question within 2 minutes is: the ability to represent |x+4|, |x-4|, |y+4|, |y-4| on the number line. Do make sure that you are comfortable with this visual sense of absolute value.

Best Regards

Japinder
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To be honest, I feel more comfortable with plugging in. Just to set up the number line as you did in your example takes me too much time. And I don't feel comfortable with the number line, I assume that the error probability is bigger in my case by visualising the number line than plugging in.

but how do you confirm about this |x – 4| - |x + 4| = 8
And, |y + 4| - | y – 4| = 8 ???

What if non integers are considered? Will you solution work for that as well?

the image is not opening.

Unable to open the image in the solution. Can you please update the image.

Image is not opening.
Can you please upload it again?
EgmatQuantExpert

Can you please share the image of the number line? the one you have shared cant be open

Hey everyone,

We've taken a note of the image opening issue. Since it is an old post, we may take some time to update the diagram. We will update it soon.
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Not sure how the actual image looked like. But I guess it would have looked something like this(attached below). Also, number substitution method of using smart numbers like x=11, y=7(discussed above) works faster than visualizing and over thinking on this number line diagram.

EgmatQuantExpert : Please share the image of the solution

Can we just plug in for this one? x = -20, y = 10.

To me, taking the positive or negative and opening the mod for each one of these is not only extremely time consuming, but error prone considering how many mods there are.

Thoughts please.
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