Absolute Values

[Uva@90]

if (10-X)/3 < -2X, which of the following must be true ?

1) 2 < X
11) |X-5| >= 7
111) |X-1|/|X| > 1

A) 1 only
B) 11 Only
C) 111 only
D) 11 and 111
E) 1,11 and 111

OA D

[[email protected]]

Hi Uva@90,

This question 'looks' scarier than it really is. With a bit of Algebra and a basic understanding of Absolute Values, you can get to the solution.

First, we're given an inequality:

(10-X)/3 < -2X

We can simplify this inequality with a few Algebra steps...

10 - X < -6X
10 + 5X < 0
5X < -10
X < -2

Now we know that X MUST be less than -2

I) 2 < X

This is NOT true. Eliminate Answers A and E.

II) |X-5| >= 7

Since X < -2, the |X-5| will be greater than OR equal to 7. Plug in ANY values for X < -2 and you'll see. Thus, Roman Numeral 2 IS TRUE. Eliminate Answer C.

III) |X-1| / |X| > 1

This is actually a subtle Number Property rule. Since X < -2, X MUST be NEGATIVE. Knowing that, the |X-1| will always be greater than |X| (since (X-1) will be "more negative", the |X-1| will be MORE positive).

The numerator of this fraction will be positive and GREATER than the denominator (which will also be positive). In that type of fraction, you will ALWAYS have a number that is greater than 1. Thus, Roman Numeral 3 IS TRUE. Eliminate Answer B.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Brent@GMATPrepNow]

If (10-x)/3 < -2x, which of the following must be true ?

1) 2 < x
11) |x-5| >= 7
111) |x-1|/|x| > 1

A) 1 only
B) 11 Only
C) 111 only
D) 11 and 111
E) 1,11 and 111

Start with (10-x)/3 < -2x
Multiply both sides by 3 to get: 10-x < -6x
Add x to both sides: 10 < -5x
Divide both sides by -5 to get: -2 > x
So, we now know that x is LESS THAN -2

At this point, we can use some number sense.

1) 2 < X
Must this be true?
No.
If x is LESS THAN -2, then x is NEGATIVE.
This statement says that x is greater than 2.

11) |x-5| > 7
Must this be true?
Yes!
If x is LESS THAN -2, then if we subtract 5 from x, the result will be less than -7
In other words, x-5 is less than -7, which means |x-5| will definitely be greater than 7

111) |x-1|/|x| > 1
Must this be true?
Yes!
If x is LESS THAN -2, then x - 1 will be negative.
More importantly, x-1 will have a GREATER MAGNITUDE than x.
This means that |x-1| will have a GREATER MAGNITUDE than |x|, which means it must be true that |x-1|/|x| > 1

Answer: D

Cheers,
Brent

[Uva@90]

Hi Uva@90,

This question 'looks' scarier than it really is. With a bit of Algebra and a basic understanding of Absolute Values, you can get to the solution.

First, we're given an inequality:

(10-X)/3 < -2X

We can simplify this inequality with a few Algebra steps...

10 - X < -6X
10 + 5X < 0
5X < -10
X < -2

Now we know that X MUST be less than -2

I) 2 < X

This is NOT true. Eliminate Answers A and E.

II) |X-5| >= 7

Since X < -2, the |X-5| will be greater than OR equal to 7. Plug in ANY values for X < -2 and you'll see. Thus, Roman Numeral 2 IS TRUE. Eliminate Answer C.

III) |X-1| / |X| > 1

This is actually a subtle Number Property rule. Since X < -2, X MUST be NEGATIVE. Knowing that, the |X-1| will always be greater than |X| (since (X-1) will be "more negative", the |X-1| will be MORE positive).

The numerator of this fraction will be positive and GREATER than the denominator (which will also be positive). In that type of fraction, you will ALWAYS have a number that is greater than 1. Thus, Roman Numeral 3 IS TRUE. Eliminate Answer B.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

Thanks a lot, that's a wonderful explanation.

However, I didn't get this part,

10 - X < -6X
10 + 5X < 0
5X < -10
X < -2

In the second line above, how you shifted -6X from RHS to LHS. We dont't know whether X is positive or Negative. So we are not supposed to shift it isn't it ? Question doesn't mention anything about it right ?

We can shift only if it know their sign.

Please correct me if I am wrong ?

Thanks in advance.
Regards,
Uva.

[nikhilgmat31]

Hi Uva@90,

This question 'looks' scarier than it really is. With a bit of Algebra and a basic understanding of Absolute Values, you can get to the solution.

First, we're given an inequality:

(10-X)/3 < -2X

We can simplify this inequality with a few Algebra steps...

10 - X < -6X
10 + 5X < 0
5X < -10
X < -2

Now we know that X MUST be less than -2

I) 2 < X

This is NOT true. Eliminate Answers A and E.

II) |X-5| >= 7

Since X < -2, the |X-5| will be greater than OR equal to 7. Plug in ANY values for X < -2 and you'll see. Thus, Roman Numeral 2 IS TRUE. Eliminate Answer C.

III) |X-1| / |X| > 1

This is actually a subtle Number Property rule. Since X < -2, X MUST be NEGATIVE. Knowing that, the |X-1| will always be greater than |X| (since (X-1) will be "more negative", the |X-1| will be MORE positive).

The numerator of this fraction will be positive and GREATER than the denominator (which will also be positive). In that type of fraction, you will ALWAYS have a number that is greater than 1. Thus, Roman Numeral 3 IS TRUE. Eliminate Answer B.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

Thanks for giving the simple approach to this question.

10 - X < -6X Are you assuming that x < 0 . We don't know the sign of x, 10 + 5X < 0
5X < -10
X < -2

[theCEO]

Hi Uva@90,

This question 'looks' scarier than it really is. With a bit of Algebra and a basic understanding of Absolute Values, you can get to the solution.

First, we're given an inequality:

(10-X)/3 < -2X

We can simplify this inequality with a few Algebra steps...

10 - X < -6X
10 + 5X < 0
5X < -10
X < -2

Now we know that X MUST be less than -2

I) 2 < X

This is NOT true. Eliminate Answers A and E.

II) |X-5| >= 7

Since X < -2, the |X-5| will be greater than OR equal to 7. Plug in ANY values for X < -2 and you'll see. Thus, Roman Numeral 2 IS TRUE. Eliminate Answer C.

III) |X-1| / |X| > 1

This is actually a subtle Number Property rule. Since X < -2, X MUST be NEGATIVE. Knowing that, the |X-1| will always be greater than |X| (since (X-1) will be "more negative", the |X-1| will be MORE positive).

The numerator of this fraction will be positive and GREATER than the denominator (which will also be positive). In that type of fraction, you will ALWAYS have a number that is greater than 1. Thus, Roman Numeral 3 IS TRUE. Eliminate Answer B.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

Thanks a lot, that's a wonderful explanation.

However, I didn't get this part,

10 - X < -6X
10 + 5X < 0
5X < -10
X < -2

In the second line above, how you shifted -6X from RHS to LHS. We dont't know whether X is positive or Negative. So we are not supposed to shift it isn't it ? Question doesn't mention anything about it right ?

We can shift only if it know their sign.

Please correct me if I am wrong ?

Thanks in advance.
Regards,
Uva.

When you are adding or substracting - you can switch sides and not worry about signs
10 - X < -6X
10 + 5X < 0
acceptable

When you are dividing or multiplying - you have to know what the signs are

[nikhilgmat31]

Thanks CEO, It clarifies & clears my doubts.

[[email protected]]

Hi Uva@90,

When ADDING or SUBTRACTING like terms, it does not matter whether the terms are positive or negative.

eg.

X + 2X = 3X no matter what number X actually is.

Including an inequality does not alter this rule...

-X < -6X

We can add 6X to both sides....

-X + 6X < -6X + 6X

5X < 0

GMAT assassins aren't born, they're made,
Rich

[nikhilgmat31]

Thanks Rich,

you, Mitch,Brent are always helping me.

Hope you are with me on GMAT day

[Matt@VeritasPrep]

Another approach here:

1:: Multiply both sides by 3

(10 - x) < -6x

2:: Add x to both sides

10 < -5x

3:: Divide both sides by -5

-2 > x

(I) is obviously false.

(II) gives |x - 5| ≥ 7. This has two solution sets: (a) x - 5 ≥ 7, or x ≥ 12; (b) -(x-5)≥7, or -x ≥ 2, or -2 ≥ x. Success!

(III) |x-1|/|x| > 1. Since |x| must be positive -- it can't be negative, and it's a denominator so it can't be 0 -- we can multiply both sides by |x|. Now we have |x - 1| > |x|, which is a pain to interpret algebraically, so we'll do it intuitively instead.

|x - y| = the distance between x and y

so |x - 1| = the distance between x and 1 and |x| = the distance between x and 0.

|x - 1| > |x| thus means that x is further from 1 than it is from 0 ... which is true of ALL negatives, so this is true for all x less than -2.

Hence II and III and both true, and we're done!

[GMATGuruNY]

|x - 1| > |x|, which is a pain to interpret algebraically

Since both sides of the inequality are enclosed in absolute value symbols, we can square the inequality:

(x - 1)² > x²

x² - 2x + 1 > x²

-2x + 1 > 0

-2x > -1

x < 1/2.

Since the prompt indicates that x<-2 -- and any value less than -2 must be less than 1/2 -- Statement III must be true.