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Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8

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Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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Is |x| ≤ 4 ?

(1) |x + 2| + |x − 4| > 2|x − 1|
(2) |(x + 4)(x − 3)| > 8

Originally posted by Ayush1692 on 20 Mar 2018, 10:18.
Last edited by Bunuel on 22 Mar 2018, 02:12, edited 1 time in total.
Edited the question.
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Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 20 Mar 2018, 19:17
2
Ayush1692 wrote:
Is |x| ≤ 4 ?

1. |x + 2| + |x − 4| > 2|x − 1|
2. |(x + 4)(x − 3)| > 8




ans A..


1. |x + 2| + |x − 4| > 2|x − 1|
statement I tells us ..
a) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6...
here RHS is 2|x-1| is always less than 6..
b) Any value above 4 and below -2 will have similar effect on LHS and RHS as the increase / decrease will have SAME effect on both sides since there are 2xs on each side..
so -2<x<4, which is a subset of |x| ≤ 4
sufficient

2. |(x + 4)(x − 3)| > 8
when x = 0, |(x + 4)(x − 3)| = 4*3=12
when x = 10, |(x + 4)(x − 3)| = 14*3
so |x| ≤ 4 is NOT necessary
insuff

A
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Re: Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 22 Mar 2018, 02:16
Ayush1692 wrote:
Is |x| ≤ 4 ?

(1) |x + 2| + |x − 4| > 2|x − 1|
(2) |(x + 4)(x − 3)| > 8


9. Inequalities



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Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 22 Mar 2018, 03:16
chetan2u wrote:
Ayush1692 wrote:
Is |x| ≤ 4 ?

1. |x + 2| + |x − 4| > 2|x − 1|
2. |(x + 4)(x − 3)| > 8




ans A..


1. |x + 2| + |x − 4| > 2|x − 1|
statement I tells us ..
a) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6...
here RHS is 2|x-1| is always less than 6..
b) Any value above 4 and below -2 will have similar effect on LHS and RHS as the increase / decrease will have SAME effect on both sides since there are 2xs on each side..
so-2≤x≤4, which is a subset of |x| ≤ 4
sufficient

2. |(x + 4)(x − 3)| > 8
when x = 0, |(x + 4)(x − 3)| = 4*3=12
when x = 10, |(x + 4)(x − 3)| = 14*3
so |x| ≤ 4 is NOT necessary
insuff

A


Hi chetan2u,

Please correct me if I am wrong.

For Statement I, the Range should be \(-2≤x<4\) but not \(-2≤x≤4\) as you have mentioned above. When \(x = 4,\) \(L.H.S = R.H.S\). So, Can we say that \(-2≤x<4\) is a sub-set of \(x≤4\).?? Please advise.
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Re: Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 22 Mar 2018, 03:29
rahul16singh28 wrote:
chetan2u wrote:
Ayush1692 wrote:
Is |x| ≤ 4 ?

1. |x + 2| + |x − 4| > 2|x − 1|
2. |(x + 4)(x − 3)| > 8




ans A..


1. |x + 2| + |x − 4| > 2|x − 1|
statement I tells us ..
a) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6...
here RHS is 2|x-1| is always less than 6..
b) Any value above 4 and below -2 will have similar effect on LHS and RHS as the increase / decrease will have SAME effect on both sides since there are 2xs on each side..
so-2≤x≤4, which is a subset of |x| ≤ 4
sufficient

2. |(x + 4)(x − 3)| > 8
when x = 0, |(x + 4)(x − 3)| = 4*3=12
when x = 10, |(x + 4)(x − 3)| = 14*3
so |x| ≤ 4 is NOT necessary
insuff

A


Hi chetan2u,

Please correct me if I am wrong.

For Statement I, the Range should be \(-2≤x<4\) but not \(-2≤x≤4\) as you have mentioned above. When \(x = 4,\) \(L.H.S = R.H.S\). So, Can we say that \(-2≤x<4\) is a sub-set of \(x≤4\).?? Please advise.


Hi..

The actual range is -2<x<4, as at -2 and 4 both sides are equal to 6..
But still it is a subset of |x|<=4..
|x|<=4 means -4<=x<=4...
-2<x<4 is a subset of -4<=x<=4..
And will remain A
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 22 Mar 2018, 07:44
chetan2u wrote:
Ayush1692 wrote:
Is |x| ≤ 4 ?

1. |x + 2| + |x − 4| > 2|x − 1|
2. |(x + 4)(x − 3)| > 8




ans A..


1. |x + 2| + |x − 4| > 2|x − 1|
statement I tells us ..
a) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6...
here RHS is 2|x-1| is always less than 6..
b) Any value above 4 and below -2 will have similar effect on LHS and RHS as the increase / decrease will have SAME effect on both sides since there are 2xs on each side..
so -2<x<4, which is a subset of |x| ≤ 4
sufficient

2. |(x + 4)(x − 3)| > 8
when x = 0, |(x + 4)(x − 3)| = 4*3=12
when x = 10, |(x + 4)(x − 3)| = 14*3
so |x| ≤ 4 is NOT necessary
insuff

A


Hello chetan2u,

I understand that by solving statement 1, we get 1<=x<4; sufficient

However, please point out the mistake in my statement 2 analysis:

|x^2+x-12| > 8

if x^2+x-12 > 0; then -->
(x+5)(x-4)>0
Thus range of x is -->
x>-5 and x>4 if both are +ve
x<-5 and x<4 if both are -ve
We get, x<-5 and x>4 combining above statements -------------------------B

if x^2+x-12 < 0; then -->
x^2+x-4<0
(x+2.5)(x-1.5)<0
Thus range of x is -->
x>-2.5 and x<1.5 if (x+2.5)>0
x<-2.5 and x>1.5 if (x+2.5)<0
We get, x<-2.5 and x>1.5 combining above statements ----------------------A

Combining A and B we get x<-5 and x>4 --- which is sufficient to answer the question.

Please point out where am I going wrong?

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Re: Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 22 Mar 2018, 08:47
I find this question tough.
It would be very helpful if some can share a trick, how to select numbers to plug in and check range fast

Posted from my mobile device
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Re: Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 22 Mar 2018, 08:58
chetan2u wrote:
Ayush1692 wrote:
Is |x| ≤ 4 ?

1. |x + 2| + |x − 4| > 2|x − 1|
2. |(x + 4)(x − 3)| > 8




ans A..


1. |x + 2| + |x − 4| > 2|x − 1|
statement I tells us ..
a) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6...
here RHS is 2|x-1| is always less than 6..
b) Any value above 4 and below -2 will have similar effect on LHS and RHS as the increase / decrease will have SAME effect on both sides since there are 2xs on each side..
so -2<x<4, which is a subset of |x| ≤ 4
sufficient

2. |(x + 4)(x − 3)| > 8
when x = 0, |(x + 4)(x − 3)| = 4*3=12
when x = 10, |(x + 4)(x − 3)| = 14*3
so |x| ≤ 4 is NOT necessary
insuff

A



Please can you explain me the reasoning to analyze this type of questions? How did you know that for (1) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6...?

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Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8  [#permalink]

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New post 23 Mar 2018, 04:15
Ayush1692 wrote:
Is |x| ≤ 4 ?

(1) |x + 2| + |x − 4| > 2|x − 1|
(2) |(x + 4)(x − 3)| > 8


I will use other visual approach

(1) |x + 2| + |x − 4| > 2|x − 1|

Critical points -2, 1, 4

Representing on number line

-----Invalid-----2--+++----1----+++-------4----Invalid--------

Now let's examine each region

x\leq{-2} :

Let x = -10..........|-8| + |-14| > 2|-11|..........22 > 22.......Invalid.............Hence NO number satisfy the region.

-2<x<1:

Let x = 0..............|2| + |-4| > 2|-1|..........6 > 4.........Hence every Number satisfies the region...........Answer Is Yes

1<x<4:

Let x = 3..............|5| + |-1| > 2|2|..........6 > 4.........Hence every Number satisfies the region............Answer Is Yes

x\geq{4}

Let x = 10..........|12| + |6| > 2|-9|..........18 > 18.......Invalid.............Hence NO number satisfy the region.

From above:

Every x in region -2<x<4 satisfies the statement 1 and |x| also less than 4...........Answer is always yes

Sufficient

(2) |(x + 4)(x − 3)| > 8

critical points: -4 & 3

------++++-------4--++++-------3----+++--------

x\leq{-4} :

Let x = -10..........|6 * -13 | > 8....................Answer is NO

-4<x<3:

Let x = 0..........|4 * -3 | > 8........................Answer is Yes

We can stop here as two different answers but I will continue the last region for clarification

X >3

Let x = 10..........|14* 7 | > 8......................Answer is NO.

Insufficient

Answer: A
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Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8 &nbs [#permalink] 23 Mar 2018, 04:15
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