Question Bank - VIC's question 7

[aaa]

Please answer both questions if possible.

1. Is there another way to do this problem bc i dont understand the answer?
2. Please explain why the auther uses >< 3.

Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|

(2) |x - 3| > 0

We can rephrase the question by opening up the absolute value sign. In other words, we must solve all possible scenarios for the inequality, remembering that the absolute value is always a positive value. The two scenarios for the inequality are as follows:

If x > 0, the question becomes "Is x < 1?"
If x < 0, the question becomes: "Is x > -1?"
We can also combine the questions: "Is -1 < x < 1?"

Since Statement 2 is less complex than Statement 1, begin with Statement 2 and a BD/ACE grid.

(1) INSUFFICIENT: There are three possible equations here if we open up the absolute value signs:

1. If x < -1, the values inside the absolute value symbols on both sides of the equation are negative, so we must multiply each through by -1 (to find its opposite, or positive, value):

|x + 1| = 2|x -1| -(x + 1) = 2(1 - x) x = 3
(However, this is invalid since in this scenario, x < -1.)

2. If -1 < x < 1, the value inside the absolute value symbols on the left side of the equation is positive, but the value on the right side of the equation is negative. Thus, only the value on the right side of the equation must be multiplied by -1:

|x + 1| = 2|x -1| x + 1 = 2(1 - x) x = 1/3

3. If x > 1, the values inside the absolute value symbols on both sides of the equation are positive. Thus, we can simply remove the absolute value symbols:

|x + 1| = 2|x -1| x + 1 = 2(x - 1) x = 3

Thus x = 1/3 or 3. While 1/3 is between -1 and 1, 3 is not. Thus, we cannot answer the question.

(2) INSUFFICIENT: There are two possible equations here if we open up the absolute value sign:

1. If x > 3, the value inside the absolute value symbols is greater than zero. Thus, we can simply remove the absolute value symbols:

|x - 3| > 0 x - 3 > 0 x > 3

2. If x < 3, the value inside the absolute value symbols is negative, so we must multiply through by -1 (to find its opposite, or positive, value).

|x - 3| > 0 3 - x > 0 x < 3

If x is either greater than 3 or less than 3, then x is anything but 3. This does not answer the question as to whether x is between -1 and 1.

(1) AND (2) SUFFICIENT: According to statement (1), x can be 3 or 1/3. According to statement (2), x cannot be 3. Thus using both statements, we know that x = 1/3 which IS between -1 and 1.

The correct answer is C.

[StaceyKoprince]

This is a really difficult problem - I'll give you another option but it may be, on the test, that this is one on which you should just try to make an educated guess and move on before you've gone over 2 minutes.

You do need to be able to figure out that Is |x| < 1 is the same thing as asking whether x is between -1 and 1. Think about actual numbers that would make that statement true. 0 would work. 1/2 would work. -1/2 would work. 1 wouldn't. Neither would -1. Etc. So numbers between -1 and 1 will work, and other numbers won't.

(1)
|x + 1| = 2|x - 1|
So, here, one option is to make each quantity (the stuff in the absolute value signs) positive:
x+1 = 2(x-1)
x+1 = 2x-2
3 = x
The answer to my original question, therefore, is no.

I can also make each quantity negative. Don't actually try this though - in this case, that's the same as multiplying each side by -1, so I should expect the same answer (x=3)

I can also have one be pos and one neg; let's make the first one neg but leave the second one pos:
-(x+1) = 2(x-1)
-x-1 = 2x-2
1 = 3x
x = 1/3
The answer to my original question, therefore, is yes. These are contradictory answers, so statement 1 is NOT sufficient.

(2)
try the pos option first:
x-3 > 0
x > 3
answer to my original question is no

try the neg option now:
-(x-3) > 0
-x + 3 > 0
-x > -3
x < 3
answer to my original question is I don't know. So, statement 2 is NOT sufficient.

(1) and (2) together
From (1): x is either 3 or 1/3
From (2): either x>3 or x<3... which means x does not equal 3
only option when the statements are combined, then, is that x = 1/3

[shaan17]

Hi,
In the first part of the explanation where we get two values for x shouldn't we again substitute these values in the original equation and check it is satisfies ? I know in this particular case both 3 and 1/3 satisfy the equation, but aren't there cases where one value might not satisfy ? Please let me know if this additional step is required .

Thanks in advance.

[StaceyKoprince]

yes, you definitely should check - I should have specified that in my explanation. If one doesn't work, then that one doesn't count and you'd have one (sufficient) solution!

[Guest]

Hi Stacey,

I'm slightly confused my this statement: "I can also have one be pos and one neg; let's make the first one neg but leave the second one pos:"

Why did you not also testing making the first one postive and the second one negative? Why was it done in this order?

Thanks.

[RonPurewal]

Hi Stacey,

I'm slightly confused my this statement: "I can also have one be pos and one neg; let's make the first one neg but leave the second one pos:"

Why did you not also testing making the first one postive and the second one negative? Why was it done in this order?

Thanks.

those two tests are identical.

to see why, consider a simpler problem that just states |x| = |y|. then there are two ways you can express that equation with one positive and one negative:
x = -y
-x = y
...which are of course the same equation.
the same principle applies to the equation above: all you're writing is an equation stating that the two expressions have opposite signs. that can be accomplished regardless of which expression gets the positive sign. (try it yourself: you'll get the same equation, and same solution(s), both ways.)