Inequality Question MGMAT

[krnverma]

Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?

[sumgb]

Negative solution
-2x<-2
x>2

This is incorrect. This translates into x > 1. (although this doesn't change the outcome, one must be careful )

stmnt 2:

As you have said, stmnt 2 translates into
(x-7)(x-3) >= 0

"these are boundary points, on a number line."

These are indeed boundary conditions. The boundary conditions are -
either 1. x >= 7 or 2. x <= 3

If you notice, when x >= 7 the inequality holds true (if x = 7 then 0 and if x > 7 both values positive so result > 0)

when x <= 3 again the inequality holds tru (if x = 3 then 0 and if x < 3 both values are negative and the result > 0)

hence from stmnt 2 it follows x can take negative value.

try x = -1 in inequality (x-7)(x-3) >= 0 you get
(-8) * (-4) > 0 true. so x can take negative values so you cant conclude if x definitely greater than 0. Hence insuff.

Thus, answer A

Hope this helps.

[gmatboost]

You can visualize this with three number lines

One shows where x-7 is positive and negative

----------------------7++++++++

One shows where x-3 is positive and negative

--------------3++++++++++++++++

One shows the product of these two, which is determined by the signs on the number lines above

++++++++++++++3-------7+++++++

When x < 3, neg * neg = pos
When 3 < x < 7, pos * neg = neg
When x > 7, pos * pos = pos

This approach can be helpful anytime you have an inequality with several expressions on one side and zero on the other.

[krnverma]

Thank you everyone.

[AVbyT]

krnverma - My answer is the for

2) (x^2) - 10x >= -21 => (x^2) -10x + 21 >=0 => (x-3)(x-7)=>0 => x >= 3 OR x>=7 which brings us to x >=7

AND Also, (7-x)(3-x) > 0 => 7 >= x OR 3 >=x Which brings us to 3 >=x

Finally, x >= 7 OR 3 >=x which means the stem is insufficient

[prateek_guy2004]

Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?

Well to correct answer aint D.

Because Statement 1 is sufficient X<11.

statement 2 Gives 2 answers + and - so its may be hence insufficient.

Hence Answer is A.

[[email protected]]

Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?

please tell why statement 2 is not sufficient. HELP HELP

[krnverma]

3 >=x gives us 3 positive values (3, 2, 1) and infinity negative values. So statement 2 insufficient. Read earlier posts under this thread for more details.

Thanks

[GMATGuruNY]

Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?

Statement 1: |2x-12| < 10.
-10 < 2x-12 < 10
2 < 2x < 22
1 < x < 11.
Sufficient.

Statement 2: x²-10x ≥ -21.
It's possible that x=0:
(0)²-10(0)≥-21.
0≥-21.

It's possible that x=1:
1² - 10(1)≥-21.
-9≥-21.

Thus, we cannot determine whether x>0.
Insufficient.

The correct answer is A.

Here's an algebraic way to evaluate statement 2:

x²-10x+21 ≥ 0.
(x-3)(x-7)≥0.

The critical points are x=3 and x=7.
These are the only values where x²-10x+21=0.
When x is any other value, x²-10x+21>0 or x²-10x+21<0.
To determine the range of x, test one value to the left and right of each critical point.

x<3:
Plug x=0 into x²-10x+21 ≥ 0:
0²-10(0)+21≥ 0
21≥0.
This works.
x<3 is part of the range.

3<x<7:
Plug x=5 into x²-10x+21 ≥ 0:
5²-10(5)+21 ≥ 0.
-4 ≥ 0.
Doesn't work.
3<x<7 is not part of the range.

x>7:
Plug x=10 into x²-10x+21≥0:
10²-10(10)+21≥0
21≥0.
This works.
x>7 is part of the range.

Thus, x≤3 OR x≥7.

Using algebra for statement 2 makes the problem HARDER.
I would use algebra to evaluate statement 1 but would plug in values to evaluate statement 2.
Don't limit yourself to algebra.
Many DS questions can be solved more easily -- and more QUICKLY -- by using a COMBINATION of techniques.

Other posts about critical points:

https://www.beatthegmat.com/inequality-c ... 89518.html

https://www.beatthegmat.com/inequalities ... 89788.html

https://www.beatthegmat.com/knewton-q-t89317.html

https://www.beatthegmat.com/which-is-true-t89111.html

Determining the critical points to solve a problem from GMATPrep:
https://www.beatthegmat.com/need-explanation-t85192.html

[sl750]

Statement 1 can be written as,

-10 < 2x -12 < 10, after rearranging the terms, we have 1<x<11. Sufficient

In statement 2, we have a quadratic equation

(x-3)(x-7)>=0, we have two solutions to this, either x>=7 or x<=3. The latter doesn't satisfy x>0. So it is insufficient