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# How many values of x satisfy the equation ||x-7|-9|=11?

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How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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07 Feb 2018, 00:49
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[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8261 GMAT 1: 760 Q51 V42 GPA: 3.82 How many values of x satisfy the equation ||x-7|-9|=11? [#permalink] ### Show Tags 09 Feb 2018, 01:42 3 2 => $$||x-7|-9| = 11$$ $$|x-7|-9 = ±11$$ $$|x-7| = 9 ±11$$ $$|x-7| = 20$$ or $$|x-7| = -2$$ $$|x-7| = 20$$ since $$|x-7| ≥ 0$$ $$x-7 = ±20$$ $$x = 7±20$$ $$x = 27$$ or $$x = -13$$ Therefore, the answer is B. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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07 Feb 2018, 05:07
If |x-7| = a, the equation will become |a - 9| = 11

Squaring on both sides,
$$a^2 - 18a + 81 = 121$$
$$a^2 - 18a - 40 = 0$$
Solving for a, a = 20 or -2

If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)
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How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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07 Feb 2018, 08:14
pushpitkc wrote:
If |x-7| = a, the equation will become |a - 9| = 11

Squaring on both sides,
$$a^2 - 18a + 81 = 121$$
$$a^2 - 18a - 40 = 0$$
Solving for a, a = 20 or -2

If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)

Hi pushpitkc

Normally valus in modulus have two solutions (one is positive and another one is negative) one cant have 5 or 4 solutions No ?

Here is my solution

ok so we have $$||x-7| -9|= 11$$

open modulus:)
step one

1.) $$x-7-9= 11$$ (rewrite as it is)
$$x = 27$$

step two
2.) $$| (-1) |x-7| (-1)-9|= 11$$ here i mulptiply by -1

so i get $$-x+7+9=11$$

$$-x= -13$$

is my solution/approach correct ?

I wonder how did you get this and why are you squaring both sides....no square roots, or powers are mentioned....just simple modulus in question stem .. please explain

$$a^2 - 18a + 81 = 121$$
$$a^2 - 18a - 40 = 0$$
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Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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07 Feb 2018, 09:03
1
MathRevolution wrote:
[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

$$||x-7|-9|=11$$

$$=> |x-7|-9=11$$ or $$|x-7|-9=-11$$

$$=>|x-7|=20$$ or $$|x-7|=-2$$. as Mod cannot be negative, hence this equation is not valid

so we have $$x-7=20$$ or $$x-7=-20$$

$$=>x=27$$ or $$x=-13$$. We have two solutions

Option B
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Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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07 Feb 2018, 09:07
1
dave13 wrote:
pushpitkc wrote:
If |x-7| = a, the equation will become |a - 9| = 11

Squaring on both sides,
$$a^2 - 18a + 81 = 121$$
$$a^2 - 18a - 40 = 0$$
Solving for a, a = 20 or -2

If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)

Hi pushpitkc

Normally valus in modulus have two solutions (one is positive and another one is negative) one cant have 5 or 4 solutions No ?

Here is my solution

ok so we have $$||x-7| -9|= 11$$

open modulus:)
step one

1.) $$x-7-9= 11$$ (rewrite as it is)
$$x = 27$$

step two
2.) $$| (-1) |x-7| (-1)-9|= 11$$ here i mulptiply by -1

so i get $$-x+7+9=11$$

$$-x= -13$$

is my solution/approach correct ?

I wonder how did you get this and why are you squaring both sides....no square roots, or powers are mentioned....just simple modulus in question stem .. please explain

$$a^2 - 18a + 81 = 121$$
$$a^2 - 18a - 40 = 0$$

Hi dave13,

your method is incorrect. You have two methods to solve the question in this forum, so as a learning exercise, try to find out where you made mistake and what is the correct approach.

Also check your ADDITION in the highlighted portion. how can you get -13 with this approach?
VP
Joined: 09 Mar 2016
Posts: 1228
Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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09 Feb 2018, 14:21
niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

$$||x-7|-9|=11$$

$$=> |x-7|-9=11$$ or $$|x-7|-9=-11$$

$$=>|x-7|=20$$ or $$|x-7|=-2$$. as Mod cannot be negative, hence this equation is not valid

so we have $$x-7=20$$ or $$x-7=-20$$

$$=>x=27$$ or $$x=-13$$. We have two solutions

Option B

Hello niks18 :-)

Why did you remove ONLY one bracket after "9" $$||x-7|-9|=11$$ $$=> |x-7|-9=11$$ or $$|x-7|-9=-11$$

Shouldnt we open modulus as whole by removing all brackets :? ? like here in the first sample question ? https://www.veritasprep.com/blog/2014/0 ... -the-gmat/

do like this

CASE ONE: x is positive or ZERO :) $$||x-7|-9|=11$$ open modulus / remove all brackets $$x-7-9=11$$ $$x = 11+7+9$$ ---> $$x = 27$$:) Really ? answering mysellf :) well yes, its valid because our initial condition was x is positive HENCE we get x= 27 VALID :)

CASE TWO: x is negative $$||x-7|-9|=11$$ open modulus by PUTTING minus sign in front of FIRST MODULUS $$|-|x-7|-9|=11$$

so we get $$-x+7+9=11$$ --->$$-x = 11-7-9$$---- > $$-x = -5$$ $$-x = -5$$ ??? but i followed this rule ? whats wrong ? how should i understand $$-x = -5$$:?
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How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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09 Feb 2018, 16:01
3
MathRevolution wrote:
[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

To simplify brackets within brackets, let $$u$$ stand for |x - 7|

Then |u - 9| = 11
Case 1:
u - 9 = 11
u = 20

Case 2
u - 9 = -11
u = -2

We now use the values just calculated and plug in u = |x - 7|
|x - 7| = 20
|x - 7| = -2
An absolute value is nonnegative. |x - 7| cannot = -2. We are left with:

|x - 7| = 20
Case 1
x - 7 = 20
x = 27

Case 2
x - 7 = -20
x = -13

Two solutions, 27 and -13.

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Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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09 Feb 2018, 19:22
1
dave13 wrote:
niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

$$||x-7|-9|=11$$

$$=> |x-7|-9=11$$ or $$|x-7|-9=-11$$

$$=>|x-7|=20$$ or $$|x-7|=-2$$. as Mod cannot be negative, hence this equation is not valid

so we have $$x-7=20$$ or $$x-7=-20$$

$$=>x=27$$ or $$x=-13$$. We have two solutions

Option B

Hello niks18 :-)

Why did you remove ONLY one bracket after "9" $$||x-7|-9|=11$$ $$=> |x-7|-9=11$$ or $$|x-7|-9=-11$$

Shouldnt we open modulus as whole by removing all brackets :? ? like here in the first sample question ? https://www.veritasprep.com/blog/2014/0 ... -the-gmat/

do like this

CASE ONE: x is positive or ZERO :) $$||x-7|-9|=11$$ open modulus / remove all brackets $$x-7-9=11$$ $$x = 11+7+9$$ ---> $$x = 27$$:) Really ? answering mysellf :) well yes, its valid because our initial condition was x is positive HENCE we get x= 27 VALID :)

CASE TWO: x is negative $$||x-7|-9|=11$$ open modulus by PUTTING minus sign in front of FIRST MODULUS $$|-|x-7|-9|=11$$

so we get $$-x+7+9=11$$ --->$$-x = 11-7-9$$---- > $$-x = -5$$ $$-x = -5$$ ??? but i followed this rule ? whats wrong ? how should i understand $$-x = -5$$:?

Hi dave13

There are TWO mod equations here and not one. you need to solve for both mod function one-by-one. Opening the bracket doesn't mean you open all the bracket at once. Even for your simple simplification questions where you use BODMAS rule you solve for different bracket taking one bracket at a time i.e you solve for small bracket first and then for big bracket.
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Posts: 334
Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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09 Feb 2018, 23:17
||x-7|-9|=11

This gives us 2 options , the final number can be 11 or -11 before taking absolute values.

Before calculating X, one thing to consider is that |x-7| will always a positive value. So any positive value that yields 11 or -11 when subtracting 9 from it gives us the answer.

The |x-7| is either 20 or -2. Based on which the X is either 27 or -13.

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Joined: 09 Mar 2016
Posts: 1228
Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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10 Feb 2018, 06:27
generis wrote:
MathRevolution wrote:
[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

To simplify brackets within brackets, let $$u$$ stand for |x - 7|

Then |u - 9| = 11
Case 1:
u - 9 = 11
u = 20

Case 2
u - 9 = -11
u = -2

We now use the values just calculated and plug in u = |x - 7|
|x - 7| = 20
|x - 7| = -2
An absolute value is nonnegative. |x - 7| cannot = -2. We are left with:

|x - 7| = 20
Case 1
x - 7 = 20
x = 27

Case 2
x - 7 = -20
x = -13

Two solutions, 27 and -13.

Greetings generis :)

you are writing solution in sui generis style :) i thought |x - 7| has two numbers so replacing |x - 7| with $$u$$ made it clear :)

Just one question :-)

when you write case Case 2

$$x - 7 = -20$$ ( why didnt you write like this putting minus in front of x ? --> $$-(x - 7) = 20$$ ? $$-x+7 =20$$ but in this case i get $$-x = 13$$ ? why ?

And one more question ibidem

you say |x - 7| = -2 absolute cant be negative

but in case TWO you got negative value $$-13$$ why

have a great weekend
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Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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17 Feb 2018, 05:53
MathRevolution wrote:
=>

$$||x-7|-9| = 11$$
$$⇔ |x-7|-9 = ±11$$
$$⇔ |x-7| = 9 ±11$$
$$⇔ |x-7| = 20$$ or $$|x-7| = -2$$
$$⇔ |x-7| = 20$$ since $$|x-7| ≥ 0$$
$$⇔ x-7 = ±20$$
$$⇔ x = 7±20$$
$$⇔ x = 27$$ or $$x = -13$$

Why mod of X-7 is greater tha equal to 0 {[m]|x-7| ≥ 0}?
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How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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17 Feb 2018, 08:36
1
Salamander911 wrote:
MathRevolution wrote:
=>

$$||x-7|-9| = 11$$
$$⇔ |x-7|-9 = ±11$$
$$⇔ |x-7| = 9 ±11$$
$$⇔ |x-7| = 20$$ or $$|x-7| = -2$$
$$⇔ |x-7| = 20$$ since $$|x-7| ≥ 0$$
$$⇔ x-7 = ±20$$
$$⇔ x = 7±20$$
$$⇔ x = 27$$ or $$x = -13$$

Why mod of X-7 is greater tha equal to 0 {$$|x-7| ≥ 0}? Hi Salamander911 Mod means magnitude or distance from 0 on a number line. Distance is never negative, it is either positive or 0. Hence mod is never negative it is either positive or 0, that's why [m]|x-7|≥0$$
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Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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18 Feb 2018, 01:50
Ix-7I-9=11 or Ix-7I-9=-11
Ix-7I=20 or Ix-7I=-2
x-7=20 or x-7=-20 , Ix-7I cannot take a negative value because modulus of a number is non negative
x=27 or x=-13

So x can have two values

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Joined: 09 Mar 2016
Posts: 1228
Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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18 Feb 2018, 05:30
pushpitkc wrote:
If |x-7| = a, the equation will become |a - 9| = 11

Squaring on both sides,
$$a^2 - 18a + 81 = 121$$
$$a^2 - 18a - 40 = 0$$
Solving for a, a = 20 or -2

If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)

H pushpitkc its me again

i have one question regarding this $$a^2 - 18a - 40 = 0$$

you say now we need to solve for a.

did you solve it using discriminant ?

or through expansion ? like this $$a(a-18)-40= 0$$

i remember for example if equation were of this form $$5a^2 - 18a - 40 = 0$$ then i wouldnt be able to take $$a$$ from the brackets --- right ? and would need to solve it through discriminant

thank you
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How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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18 Feb 2018, 05:43
1
dave13 wrote:
pushpitkc wrote:
If |x-7| = a, the equation will become |a - 9| = 11

Squaring on both sides,
$$a^2 - 18a + 81 = 121$$
$$a^2 - 18a - 40 = 0$$
Solving for a, a = 20 or -2

If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)

H pushpitkc :) its me again

i have one question regarding this $$a^2 - 18a - 40 = 0$$

you say now we need to solve for a.

did you solve it using discriminant ?

or through expansion ? like this $$a(a-18)-40= 0$$

i remember for example if equation were of this form $$5a^2 - 18a - 40 = 0$$ then i wouldnt be able to take $$a$$ from the brackets --- right ? and would need to solve it through discriminant

thank you :)

Hey dave13

In order to solve the equation a quadratic equation $$a^2 - 18a - 40 = 0$$

we need to understand that the product of the roots are -40 and sum is -18
A combination of numbers which gives us this is -20 and +2

Hence, the equation can be re-written as $$a^2 - 20a + 2a - 40 = 0$$
$$a(a - 20) + 2(a - 20) = 0$$
$$(a + 2)(a - 20) = 0$$ -> a = -2 or 20

Hope this helps you!
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Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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18 Feb 2018, 22:16
dave13 wrote:
niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

$$||x-7|-9|=11$$

$$=> |x-7|-9=11$$ or $$|x-7|-9=-11$$

$$=>|x-7|=20$$ or $$|x-7|=-2$$. as Mod cannot be negative, hence this equation is not valid

so we have $$x-7=20$$ or $$x-7=-20$$

$$=>x=27$$ or $$x=-13$$. We have two solutions

Option B

Hello niks18

Why did you remove ONLY one bracket after "9" $$||x-7|-9|=11$$ $$=> |x-7|-9=11$$ or $$|x-7|-9=-11$$

Shouldnt we open modulus as whole by removing all brackets ? like here in the first sample question ? https://www.veritasprep.com/blog/2014/0 ... -the-gmat/

do like this

CASE ONE: x is positive or ZERO $$||x-7|-9|=11$$ open modulus / remove all brackets $$x-7-9=11$$ $$x = 11+7+9$$ ---> $$x = 27$$:) Really ? answering mysellf well yes, its valid because our initial condition was x is positive HENCE we get x= 27 VALID

CASE TWO: x is negative $$||x-7|-9|=11$$ open modulus by PUTTING minus sign in front of FIRST MODULUS $$|-|x-7|-9|=11$$

so we get $$-x+7+9=11$$ --->$$-x = 11-7-9$$---- > $$-x = -5$$ $$-x = -5$$ ??? but i followed this rule ? whats wrong ? how should i understand $$-x = -5$$:?

We can't remove all brackets at once.
We can remove brackets one by one with carefulness.
If $$b ≥0$$, $$|x-a| = b ⇔ x-a = ±b$$
If $$c ≥0$$, $$||x-a| - b| = c ⇔ |x-a| - b = ±c ⇔ |x-a| = b ±c ⇔ x-a = ±(b ±c) ⇔ x = a ±(b ±c)$$
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Re: How many values of x satisfy the equation ||x-7|-9|=11?  [#permalink]

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Re: How many values of x satisfy the equation ||x-7|-9|=11?   [#permalink] 26 Nov 2019, 12:16
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