Problem Solving

If x/lxl<x, which of the following must be true about x
(Note: Read lxl as modulus x)

A) x>1
B) x>-1
C) lxl<1
D) lxl = 1
E) lxl^2 > 1

This question has been discussed previously and popular answer was B. But I am not able to understand why it was not A. Pls someone help me...

mathewmithun wrote:If x/lxl<x, which of the following must be true about x
(Note: Read lxl as modulus x)

A) x>1
B) x>-1
C) lxl<1
D) lxl = 1
E) lxl^2 > 1

This question has been discussed previously and popular answer was B. But I am not able to understand why it was not A. Pls someone help me...

Case 1 :- X is +ive; then

X/X < X => 1 < X.

Case 2 :- X is -ive; then

X/X > -X => 1 > -X => -1 < X.

Combining both the case, we can say that Case 1 range for value of X is included in case 2 range for value of X. Hence Answer should be B.

You can also try with few values of X for -1 < X say X = -1/2, -1/3 etc.. These values satisfy inequality.

Hi,
I am totally on your side.
The question is NOT asking us the complete solution of x, but is only asking which of the values of x among the options will satisfy the inequality.
Here goes my solution:

x/|x| < x
Multiplying both sides by |x|, we have
x < x.|x|
x(|x| - 1) > 0

I)
When x > 0, |x| = x and
x(x - 1) > 0

Using Critical Points' method,
x > 1

So, [spoiler](A)[/spoiler] is correct. We don't need to go further. The next case is of x < 0 so x > 1 will always satisfy the given inequality.
But we will, for the sake of more analysis.

(B) cannot be correct; x = 1/2 does not satisfy the inequality.
----------------------------------------
II)
When x < 0, |x| = -x
x(-x - 1) > 0
x(x + 1) < 0

Using Critical Points' method,
-1 < x < 0

Combining I) and II),
-1 < x < 0 OR x > 1

What is the Critical Points' method?
Read here: https://www.beatthegmat.com/critical-poi ... 10450.html

mathewmithun wrote:If x/lxl<x, which of the following must be true about x
(Note: Read lxl as modulus x)

A) x>1
B) x>-1
C) lxl<1
D) lxl = 1
E) lxl^2 > 1

An efficient approach is to COMPARE THE ANSWER CHOICES and to plug in values that are included in some ranges but not in others.

A: x > 1
B: x > -1

x=-1/2 is included in the range of answer choice B but not in the range of answer choice A.
Plugging x = -1/2 into x/lxl<x, we get:
(-1/2)/|1/2| < -1/2
-1 < -1/2.
This works.
Eliminate A and any other answer choice that does not include x=-1/2 within its range.
Eliminate A, D and E.

B: x > -1
C: |x| < 1.

x=2 is included in the range of answer choice B but not in the range of answer choice C.
Plugging x=2 into x/lxl<x, we get:
(2)/|2| < 2
1 < 2.
This works.
Eliminate C.

The correct answer is B.

Alternate Method:

x/|x| < x

When x > 0,
x/x < x
1 < x

And that is enough.
(A) is correct.

(B) is ruled out for many reasons:
(i) As I said, the problem is not asking for a complete solution. If it were, none of the options would be correct.
(ii) x > 1 satisfies the inequality as we just saw. You can also plug-in any value of x > 1 and notice the same.
(iii) x = 1/3, 1/2 do not satisfy the inequality.

Lets find the complete solution.

When x < 0,
(x/-x) < x
x > -1
But we're dealing with x < 0,
So -1 < x < 0 from this case.

Complete solution:
-1 < x < 0 OR x > 1

aneesh.kg wrote:Hi,
I am totally on your side.
The question is NOT asking us the complete solution of x, but is only asking which of the values of x among the options will satisfy the inequality.
Here goes my solution:

x/|x| < x
Multiplying both sides by |x|, we have
x < x.|x|
x(|x| - 1) > 0

I)
When x > 0, |x| = x and
x(x - 1) > 0

Using Critical Points' method,
x > 1

So, [spoiler](A)[/spoiler] is correct. We don't need to go further. The next case is of x < 0 so x > 1 will always satisfy the given inequality.
But we will, for the sake of more analysis.

(B) cannot be correct; x = 1/2 does not satisfy the inequality.
----------------------------------------
II)
When x < 0, |x| = -x
x(-x - 1) > 0
x(x + 1) < 0

Using Critical Points' method,
-1 < x < 0

Combining I) and II),
-1 < x < 0 OR x > 1

What is the Critical Points' method?
Read here: https://www.beatthegmat.com/critical-poi ... 10450.html

Dear Aneesh,

This is a Must Be True kind of question. Agreed that Inequality does not satisfy for X being between 0 & 1.

Imagine 2 scenarios...

Scenarios 1:- X is say 3, Inequality satisfies, and options A & B both satisfy.

Scenarios 2:- X is say -1/2, option B satisfies, option A does not.

So we cannot necessarily state that X > 1 only, whereas we can necessarily state that X > -1. I think Answer is B.

Can we have your opinion Mitch?

Thank you for the response...
I still go with A as the answer and this is the reason:

After finding the constraints, the question boils down to -1<x<0 and x>1.
There is a shady region between 0 and 1 that does not support x/|x|<x. to prove the point, plug 1/2 as x. (1/2)/(1/2)=1 which is not less than 1/2. This proves question wrong.

Since x>-1 also covers this shady region 0 to 1 which does not fall into our solution, for the equation to be true (must be true question for all values of x), x>1 and hence A.

I feel this question is classic example of plugging values not working our way.

And Aneesh, Critical points method is one cool method, thanks for posting it.

Shalabh's Quants wrote:

aneesh.kg wrote:Hi,
I am totally on your side.
The question is NOT asking us the complete solution of x, but is only asking which of the values of x among the options will satisfy the inequality.
Here goes my solution:

x/|x| < x
Multiplying both sides by |x|, we have
x < x.|x|
x(|x| - 1) > 0

I)
When x > 0, |x| = x and
x(x - 1) > 0

Using Critical Points' method,
x > 1

So, [spoiler](A)[/spoiler] is correct. We don't need to go further. The next case is of x < 0 so x > 1 will always satisfy the given inequality.
But we will, for the sake of more analysis.

(B) cannot be correct; x = 1/2 does not satisfy the inequality.
----------------------------------------
II)
When x < 0, |x| = -x
x(-x - 1) > 0
x(x + 1) < 0

Using Critical Points' method,
-1 < x < 0

Combining I) and II),
-1 < x < 0 OR x > 1

What is the Critical Points' method?
Read here: https://www.beatthegmat.com/critical-poi ... 10450.html

Dear Aneesh,

This is a Must Be True kind of question. Agreed that Inequality does not satisfy for X being between 0 & 1.

Imagine 2 scenarios...

Scenarios 1:- X is say 3, Inequality satisfies, and options A & B both satisfy.

Scenarios 2:- X is say -1/2, option B satisfies, option A does not.

So we cannot necessarily state that X > 1 only, whereas we can necessarily state that X > -1. I think Answer is B.

Can we have your opinion Mitch?

Correct.
If x>1, then it must be true that x/|x| < x.
But the question here is the reverse:
If x/|x| < x, then what must be true about x?
Since any value between -1 and 0 satisfies x/|x| < x, we cannot say that it must be true that x > 1.
Hence, we can eliminate A (and any other answer choice that does not include -1<x<0 within its range).
What must be true about x is that x > -1, since no value less than or equal to -1 satisfies x/|x| < x.
Hence, the correct answer is B.

Okay, Another reason why B (x>-1) is not correct option is, let x=0. It falls in the interval x>-1.

Substituting in our equation x/|x|<x it leads to division by 0 which is not defined. Hence A is the answer.

mathewmithun wrote:Thank you for the response...
I still go with A as the answer and this is the reason:

After finding the constraints, the question boils down to -1<x<0 and x>1.
There is a shady region between 0 and 1 that does not support x/|x|<x. to prove the point, plug 1/2 as x. (1/2)/(1/2)=1 which is not less than 1/2. This proves question wrong.

Since x>-1 also covers this shady region 0 to 1 which does not fall into our solution, for the equation to be true (must be true question for all values of x), x>1 and hence A.

I feel this question is classic example of plugging values not working our way.

And Aneesh, Critical points method is one cool method, thanks for posting it.

'Must Be True' kinds are surely tricky. Pl. have a look at this post Be Careful When Interpreting "Must Be True" Questions by Knewton. You will gain a lot of understanding through this & will change your mindset.

https://www.beatthegmat.com/mba/2011/10/ ... -questions

Shalabh's Quants wrote:

mathewmithun wrote:Thank you for the response...
I still go with A as the answer and this is the reason:

After finding the constraints, the question boils down to -1<x<0 and x>1.
There is a shady region between 0 and 1 that does not support x/|x|<x. to prove the point, plug 1/2 as x. (1/2)/(1/2)=1 which is not less than 1/2. This proves question wrong.

Since x>-1 also covers this shady region 0 to 1 which does not fall into our solution, for the equation to be true (must be true question for all values of x), x>1 and hence A.

I feel this question is classic example of plugging values not working our way.

And Aneesh, Critical points method is one cool method, thanks for posting it.

'Must Be True' kinds are surely tricky. Pl. have a look at this post Be Careful When Interpreting "Must Be True" Questions by Knewton. You will gain a lot of understanding through this & will change your mindset.

https://www.beatthegmat.com/mba/2011/10/ ... -questions

Thanks Shalabh for the link. I am just pasting the important line form this link:
So the big takeaway here is: On "must be true" questions, just one example of "false" is enough to demonstrate that a statement does not have to be true. But just one example of "true" is not enough to show that the statement must be true in all cases.

By using the above method, to prove a must be true question, I just have to prove a condition is false to strike it off from question options. in B, x>-1; when x lies between 0 and 1 or when x is 0 the question x/|x|<x is not true. So we can eliminate option B. While option A,x>1, according to me, satisfies this equation and is true.

mathewmithun wrote:

Shalabh's Quants wrote:

mathewmithun wrote:Thank you for the response...
I still go with A as the answer and this is the reason:

After finding the constraints, the question boils down to -1<x<0 and x>1.
There is a shady region between 0 and 1 that does not support x/|x|<x. to prove the point, plug 1/2 as x. (1/2)/(1/2)=1 which is not less than 1/2. This proves question wrong.

Since x>-1 also covers this shady region 0 to 1 which does not fall into our solution, for the equation to be true (must be true question for all values of x), x>1 and hence A.

I feel this question is classic example of plugging values not working our way.

And Aneesh, Critical points method is one cool method, thanks for posting it.

'Must Be True' kinds are surely tricky. Pl. have a look at this post Be Careful When Interpreting "Must Be True" Questions by Knewton. You will gain a lot of understanding through this & will change your mindset.

https://www.beatthegmat.com/mba/2011/10/ ... -questions

Thanks Shalabh for the link. I am just pasting the important line form this link:
So the big takeaway here is: On "must be true" questions, just one example of "false" is enough to demonstrate that a statement does not have to be true. But just one example of "true" is not enough to show that the statement must be true in all cases.

By using the above method, to prove a must be true question, I just have to prove a condition is false to strike it off from question options. in B, x>-1; when x lies between 0 and 1 or when x is 0 the question x/|x|<x is not true. So we can eliminate option B. While option A,x>1, according to me, satisfies this equation and is true.

Dear Mithun,

Lets take an interesting example...

Say in a forgotten match Messy scored either 'More than 1 but less than 5 goals" or "More than 7 goals". Which of the following must be true for Messy?

A. Messy scored more than 7 goals.
B. Messy scored more than 1 goal.

Option 1...

If you choose option 1, it means you categorically state that Messy Must Not score 2, 3 or 4 goals. Which is wrong to say.

Option 2...

It takes into account all possible no. of goals options.

I think you are mixing it with solution set.
Stating messy must have scored more than 1 goal does not comply that No. of goal be necessarily all the elements of solution set.

Messy's solution set is {2, infinity} with drop-out intervels, which is {5,7}. Question is not asking for solution set.

Hope this makes sense.

Shalabh's Quants wrote:

mathewmithun wrote:

Shalabh's Quants wrote:

mathewmithun wrote:Thank you for the response...
I still go with A as the answer and this is the reason:

After finding the constraints, the question boils down to -1<x<0 and x>1.
There is a shady region between 0 and 1 that does not support x/|x|<x. to prove the point, plug 1/2 as x. (1/2)/(1/2)=1 which is not less than 1/2. This proves question wrong.

Since x>-1 also covers this shady region 0 to 1 which does not fall into our solution, for the equation to be true (must be true question for all values of x), x>1 and hence A.

I feel this question is classic example of plugging values not working our way.

And Aneesh, Critical points method is one cool method, thanks for posting it.

'Must Be True' kinds are surely tricky. Pl. have a look at this post Be Careful When Interpreting "Must Be True" Questions by Knewton. You will gain a lot of understanding through this & will change your mindset.

https://www.beatthegmat.com/mba/2011/10/ ... -questions

Thanks Shalabh for the link. I am just pasting the important line form this link:
So the big takeaway here is: On "must be true" questions, just one example of "false" is enough to demonstrate that a statement does not have to be true. But just one example of "true" is not enough to show that the statement must be true in all cases.

By using the above method, to prove a must be true question, I just have to prove a condition is false to strike it off from question options. in B, x>-1; when x lies between 0 and 1 or when x is 0 the question x/|x|<x is not true. So we can eliminate option B. While option A,x>1, according to me, satisfies this equation and is true.

Dear Mithun,

Lets take an interesting example...

Say in a forgotten match Messy scored either 'More than 1 but less than 5 goals" or "More than 7 goals". Which of the following must be true for Messy?

A. Messy scored more than 7 goals.
B. Messy scored more than 1 goal.

Option 1...

If you choose option 1, it means you categorically state that Messy Must Not score 2, 3 or 4 goals. Which is wrong to say.

Option 2...

It takes into account all possible no. of goals options.

I think you are mixing it with solution set.
Stating messy must have scored more than 1 goal does not comply that No. of goal be necessarily all the elements of solution set.

Messy's solution set is {2, infinity} with drop-out intervels, which is {5,7}. Question is not asking for solution set.

Hope this makes sense.

Thank you Shalabh. At this point, I am just confused... a small break from this question will definitely help me. But I am giving extra time on must be true question and for than I thank you once again for the link you provided. If you have some more material/links on must be true DS question, pls post them for me.

Ah, very interesting!
I switch to (B). Thanks Shalabh and Mitch.

Dear Mithun,

See this one from Master GMAT.

https://www.beatthegmat.com/mba/2012/04/ ... perts-miss