12 Days of Christmas GMAT Competition - Day 12: Is x > 0 ? (1) |x : Data Sufficiency (DS) - Page 2

Is x > 0 ?

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


(1)
x+2 > 3 - x
x > 0.5 > 0

Sufficient

(2)
-2+x = 10 - 2x
x = 4 > 0

2 - x = 10 - 2x, x= 8 solution doesn't work for this equation

Sufficient

D is the correct answer

Is x > 0 ?

(1) |x + 2| > 3 – x
Case 1 : x+2>3-x
2x>1
x> (x > 0)
Case 2 : -x-2>3-x
-x+x>5 (NO)
Sufficient


(2) |2 – x| = 10 – 2x
Case 1 : 2-x=10-2x
x=8
On substituting x=8 LHS not equal to RHS
Case 2 : -2+x=10-2x
3x=12
x=4
On substituting x=4 , LHS = RHS
x=4 (x>0)
Sufficient

Ans : D

Is x > 0 ?

(1) |x + 2| > 3 – x ⟹ Sufficient - Simplifying we get x > 0.5
(2) |2 – x| = 10 – 2x ⟹ Sufficient - Simplifying we get x = 4

Ans D

Is x > 0?

Condition (1):

|x + 2| > 3 – x

Squaring both sides,

(x+2)^2 > (3-x)^2


4x+4 > 9-6x

10x > 5

x > 1/2.

Condition (1) alone is sufficient to prove x > 0.

So eliminate BCE and keep A and D.




Condition (2):


|2 – x| = 10 – 2x

(2-x)^2 = (10 -2x)^2

4+x^2-4x = 4x^2-40x+100

3x^2-36x+96 = 0

x^2-12x+32 = 0
(x-4)(x-8)= 0

Two possible value of x both greater than 0.

So we can say x > 0.

Condition (2) alone is sufficient.

Keep D and eliminate A.

So the best answer choice is D.

Is x > 0 ?

(1) |x + 2| > 3 – x
In this case x > 1/2
Conclusion: Definitely x is positive.
Statement 1 is SUFFICIENT

(2) |2 – x| = 10 – 2x

Case 1:
2 – x = 10 – 2x
X = 8

Case 2:
X – 2 = 10 – 2x
X = 4
In this case x = 8 or 4
Conclusion: Definitely x is positive.
Statement 2 is SUFFICIENT

Answer : Option D

Bunuel wrote:12 Days of Christmas GMAT Competition with Lots of Fun

Is x > 0 ?

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

This question was provided by Experts'Global for the 12 Days of Christmas Competition Win $30,000 in prizes: Courses, Tests & more

 

When you draw the graph of |x+2| it will intersect 3-x at 1/2.

Hence to make statement 1 true x has to be greater than 1/2.

Hence its possible to say x>0.

Similarly using statement 2 the value of x that satisfy the equation is x=4.

Hence we can say x>0.

IMO D

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Bunuel wrote:12 Days of Christmas GMAT Competition with Lots of Fun

Is x > 0 ?

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

This question was provided by Experts'Global for the 12 Days of Christmas Competition Win $30,000 in prizes: Courses, Tests & more

 

st.1)

let x>-2, then
x+2>3-x
or 2x>1 or x>0.5 possible solution

let x<-2, then
-(x+2)>3-x
or x+2<3+x
no possible solution
since we have solution at x>-2

A sufficient

St. 2)
x>2,
-2+x =10-2x
or 3x = 12 or x=4
if x<2
2-x =10-2x
x=8, since x<2, so x=8 is not a valid solution
so possible solution exist at x>2, and solution is x =4,

B sufficient

correct answer should be D

Is x > 0?

(1) |x + 2| > 3 – x
when x>0
x+2>3-x
2x>1
x>1/2
when x<0
Not possible
Sufficient



(2) |2 – x| = 10 – 2x

when 2-x>0
x=4

Sufficient

OA:D

Let's deal with the conditions one by one.

(1)

To open the modulus, there are two options:

• and
  
  Clearly, this is impossible - therefore, this option is invalid.
• and
  
  This actually addresses our question, because X is always above 0.

Therefore, having only one functioning possibility among these two, Condition 1 is sufficient by itself.

(2)

To open the modulus, there are two options again:

• and
  
  This addresses our question, because X is always above 0.
• and
  
  Clearly, this is impossible, because X has to be below 2 and equal to 8 at the same time.

Again, there is only one legit option under this condition, so Condition 2 is sufficient by itself.

Therefore, the correct answer is D.

Is x > 0 ?

(1) |x + 2| > 3 – x, we have to deal with 2 cases here:
i. When x+2>=0,
x+2>3-x
2x>1
x>1/2

ii. When x+2<0
-(x+2)<3-x
In this case the inequality is invalid

So the answer is x>1/2. SUFFICIENT

(2) |2 – x| = 10 – 2x, the same concept works here in our equation

i. When 2-x>0
2-x=10-2x
x=8, we can not take 8 as a root of equation because it does not match the condition of our case (x<2)

ii. When 2-x<0
-(2-x)=10-2x
-2+x=10-2x
x=4
The only root to the equation is 4. SUFFICIENT

Answer D

OA)D
Is x > 0 ?

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

Statement I)

if x<0 then -(x+2) > 3-x ,
-x-2>3-x , -2>3 which cannot be true so x cannot be less than zero

if x=0 2>3, which cannot be true as well, so cannot be zero too

if x>0 , x+2 > 3-x then 2x>1 ,x>1/2 which in line with x>0
so based on that we can answer the question

so statement I is sufficient to answer the question

Statement II)
|2-x|=10-2x
now based in that 2-x=10-2x or 2-x=-(10-2x)
if, 2-x=10-2x then x=8 but |2-x|=10-2x so x can not be 8

now if 2-x=-(10-2x) then x=4 , and |2-x|=10-2x which holds true for x=4
so statement II is also sufficient to answer the question x>0

Asked: Is x > 0 ?

(1) |x + 2| > 3 – x

Case 1: x >=-2
|x+2| = x+2 > 3-x
2x >1
x > 1/2>0
Case 2: x<-2
|x+2| = -x-2 > 3-x
-2 > 3: Not possible
x > 1/2 >0
SUFFICIENT

(2) |2 – x| = 10 – 2x

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 D