12 Days of Christmas GMAT Competition - Day 12: Is x > 0 ? (1) |x

Asked: Is x > 0 ?


Case 1: x >=-2
|x+2| = x+2 > 3-x
2x > -1
x > -1/2
Case 2: x<-2
|x+2| = -x-2 > 3-x
-2 > 3: Not possible
x > -1/2
If x = -1/4 ; x < 0
But if x=2; x>0
NOT SUFFICIENT


Case 1: x>=2
|2-x| = x-2 = 10-2x
x = 4
Case 2: x<2
|2-x| = 2-x = 10-2x
x = 8: not possible since x<2
x=4>0
SUFFICIENT

IMO B
_________________

S1)

solving graphically:
|x + 2| > 3 – x

in x>0 region
=> x+2=3-x;
=> x=1/2

when x<1/2, |x+2| is below 3-x
when x>1/2, |x+2| is above 3-x (x>0)


Hence suff

S2)
|2 – x| = 10 – 2x

Graphically both intersects at x-2=10-2x;
=>x=4
x>0

(Suff)

hence D

Is x > 0 ?

(1) |x + 2| > 3 – x
(2) |2 – x| = 10 – 2x

1. Concept: > equality opens as: x+2 >3-x and x+2< -(3-x)

solving this gives x>0.5 only solution. Second x cancels so no solution.

1 is sufficient

2. |2 – x| = 10 – 2x

10-2x = 2-x and 10-2x= x+2
solving this gives x=4,8

Sufficient as x>0

Ans: D

target is x>0

#1
|x + 2| > 3 – x
case 1 :
x+2>3-x
2x>1
x >1/2
for all -ve values #1 will not hold true as RHS will always > LHS
sufficient that x>0
#2
l2 – x| = 10 – 2x
case 1
2-x= 10-2x
x = 8
case 2
-2+x = 10-2x
3x = 12
x = 4
as at x=8 LHS is +ve but RHS is -ve
x>0
sufficient
OPTION D is correct

1) abs(x+2)>3-x
We can solve for the inequality
Case 1:
x+2>3-x
2x>1
x>1/2

Case 2:
-x-2>3-x
-2>3
This is always false, so we can ignore case 2. That leaves us with x>1/2. Clearly, X>0. Suff.

2)
abs(2-x)=10-2x
Case 1:
2-x=10-2x
x=8
But this is not a valid solution, since 10-2(8)<0, and abs cannot be less than 0. discard x=8
Case 2:
-2+x=10-2x
3x=12
x=4
This is a valid solution:
abs(2-4)=abs(-2)=2
10-2(4)=2.

therefore, x=2 which is clearly>0.

Both are suff, therefore D.

Solution:

Statement (1) |x + 2| > 3 – x
Taking cases:
Case 1: x<-2
Then, -x-2 > 3-x
or -2 > 3, which is never possible.
Hence, x<-2 is not possible. Case rejected.

Case 2: x>=-2
Then, x+2 > 3-x
or 2x > 1
or x > 1/2.
This is an acceptable value as it lies in the range x>= -2. Also, x>1/2 implies, x is always positive.
So, x>0.
Sufficient.

Statement (2) |2 – x| = 10 – 2x
or |x - 2| = 10 – 2x
Taking cases:
Case 1: x<2
Then, -x + 2 = 10 - 2x
or x=8.
But x<2 so x cannot be 8. Case rejected.

Case 2: x>=2
Then, x - 2 = 10 - 2x
or 3x=12
or x=4.
As x=4 lies in the range x>=2, thus it is an acceptable value. Also, x=4 implies that x is always positive.
SO, x>0.
Sufficient.

ANSWER D.

(1) |x + 2| > 3 – x
for x>=-2; |x+2|=x+2
x+2>3-x
x>1/2
and
for x<-2
|x+2|=-(x+2)
-x-2>3-x
No such value of x exists which can satisfy this equation.
Therefore, this statement is sufficient
(2) |2 – x| = 10 – 2x
x>=2; |2 – x|=-(2-x)
-2+x=10-2x
3x=12
x=4
x<2; |2 – x|= 2-x
2-x=10-2x
3x=8
x=8/3 (not possible since x<2)
Therefore, this statement is also sufficient
Therefore, option D

Statement 1: |x+2|>3-x
x+2>3-x or x+2<x-3
2x>1 or 2<-3 (not possible)
x>1/2
i.e. x is always +ve.
Statement 1 alone is sufficient.

Statement 2: |2-x|=10-2x
Analysis: 10-2x>=0
10>=2x
5>=x
x<=5

Case 1: 2-x=10-2x
x=8 which is not possible basis above analysis.

Case 2: x-2=10-2x
3x=12
x=4 is possible basis above analysis.
x is +ve and we get a unique answer.
Statement 2 alone is sufficient.

Answer is D.
_________________

Is x > 0 ?

(1) |x + 2| > 3 – x

On solving this, we get x>1/2

Hence, x > 0

Sufficient

(2) |2 – x| = 10 – 2x

only one value satisfies ie, x = 4

Hence, x > 0

Sufficient

Hence, (D) is the correct answer
_________________

Is x > 0 ?

statement 1) |x + 2| > 3 – x --> sufficient
solving equation x>1/2
Statement 2) |2 – x| = 10 – 2x --> sufficient
solving equation x=4
ans D

Is x > 0 ?

Case (1) |x + 2| > 3 – x
If x>-2 then x>1/2
x<-2 then -2>3 not possible
hence Case-1 is sufficient

(2) |2 – x| = 10 – 2x

X>2 then x-2=10-2x, then x=4
x<2 then 2-x= 10-2x then x=8 Not possible
hence Case-2 is sufficient

Hence Option D

Is x > 0 ?

(1) |x + 2| > 3 – x
x+2>3-x
x>1/2;
|-1+2|>3-(-1) so x can't be negative number Sufficient
(2) |2 – x| = 10 – 2x
2-x=10-2x
x=8 yes
-2+x=10-2x
x=4 yes Sufficient

IMO D

Is x > 0 ?

YES/NO Data Sufficiency question

(1) |x + 2| > 3 – x

For x>= -2;

x + 2 > 3 – x
x> 1/2 leads to x > 1/2

X<2;

-x - 2 > 3 – x
-2>3 (Inconsistent)

overall; implies x> 0..YES

SUFFICIENT


(2) |2 – x| = 10 – 2x

x<2;

2-x = 10-2x
x = 8..Inconsistent

x>=2;

-2+x = 10-2x
x=4

overall; implies x> 0..YES

SUFFICIENT

IMO, (D) will be the answer

Is x > 0 ?

(1) |x + 2| > 3 – x Suff cuz x>1/2 when x+2>0 otherwise there is no solution when x<-2
(2) |2 – x| = 10 – 2x Suff because x=8 or 4 which are >0

Ans D

Is x > 0 ?

(1) |x + 2| > 3 – x

It will solve to lead only to x > 0.5

Thus, X > O

YES is the solution.....So, Sufficient


(2) |2 – x| = 10 – 2x

It will solve to lead only to x = 4

Thus, X > 0

YES is the solution.....So, Sufficient

Thus, (D) will be correct

Taking statement (1),
First case, x+2>3-x
x>(1/2). So, x is definitely positive.

Second case, -x-2<3-x. Hence, it is not true for any negative value of x.

So, (1) is enough.

Taking statement (2),
First case, 2-x=10-2x
x=8.

Second case, -2+x=10-2x
3x=12
x=4.

So, (2) is enough.

Hence, IMO D.

Is x > 0 ?

Answer is D

(1) Sufficient. we get x > 1/2
(2) Sufficient. We get x has two roots, which are 4 & 8

Statment 1: For x < 0 this equation will not hold e.g. x = -1 1> 4(not possible) => x > 0 this statement works. -> Sufficient
Statment 2: This equation will hold only for x = 4 i.e x > 0 => Sufficient

Answer should be D

Is x > 0 ?

(1) |x + 2| > 3 – x
x+2>3-x or x+2<-(3-x)
2x>1; x>1/2 or 0<-5 (which is not true)
Hence we get x>1/2 therefore it is greater than zero hence A is sufficient --> check II to verify whether it is A or D
(2) |2 – x| = 10 – 2x
2-x = 10-2x or 2-x = -10+2x
x=8 or 12=3x (x=4)
but x=8 cannot be the answer b/c it yields the answer in negative result
so we get only x=4 which is >0

Hence both 1 and 2 is sufficient --> D