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# Is x > 0?

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16 May 2013, 13:00
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Is x > 0?

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

I understand #1 but I am getting tripped up with #2

X2 - 10x + 21 >/= (greater than or equal to) 0.
(x-7)(x-3) >/= 0
So I get at least two results (this is the type of question that asks whether you can narrow it down to one definitive result or not) that make the question valid.
X=0 (0-7)(0-3) = 21 21>0 and X=0
X= 3 (3-7)(3-3) = 0 0=0 and X=3
So basically, the question asks whether X>0. I have X=0 that satisfies the equation and X=3 that satisfies the equation (it asks if the equation is greater than or equal to 0) So the problem cannot be answered with certainty.
However, the explanation in the book uses a number line to explain which looks something like this:

ee <---------- eeeeeeeeeeeee ----------->

|__________3___________7___________|

I get that with a typical absolute value problem, say |x| >/= 3 you can have cases where x >/= 3 and x </= -3 but with (x-7) and (x-3) why do I know that it's 3 that goes left on the # line and 7 that goes right on the # line.
Thanks!
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16 May 2013, 13:07
2
Is $$x>0$$?

1.) $$|2x-12| < 10$$
$$1<x<11$$
Sufficient

2.) $$x^2 - 10x \geq{} -21$$

$$x^2 - 10x +21\geq{}0$$
This equation has two roots x=3 and x=7
So $$x<3$$ or $$x>7$$
Not sufficient to say that x is positive

A

If you have doubts on inequalities, check my post tips-and-tricks-inequalities-150873.html to define the interval of the inequality
Should remove your doubts (I hope!)
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16 May 2013, 13:19
2
WholeLottaLove wrote:
Is x>0

1.) |2x-12| < 10
2.) x2 - 10x >/= -21

I understand #1 but I am getting tripped up with #2

X2 - 10x + 21 >/= (greater than or equal to) 0.
(x-7)(x-3) >/= 0
So I get at least two results (this is the type of question that asks whether you can narrow it down to one definitive result or not) that make the question valid.
X=0 (0-7)(0-3) = 21 21>0 and X=0
X= 3 (3-7)(3-3) = 0 0=0 and X=3
So basically, the question asks whether X>0. I have X=0 that satisfies the equation and X=3 that satisfies the equation (it asks if the equation is greater than or equal to 0) So the problem cannot be answered with certainty.
However, the explanation in the book uses a number line to explain which looks something like this:

eeee <---------- eeeeeee ----------->

|__________3___________7___________|

I get that with a typical absolute value problem, say |x| >/= 3 you can have cases where x >/= 3 and x </= -3 but with (x-7) and (x-3) why do I know that it's 3 that goes left on the # line and 7 that goes right on the # line.
Thanks!

hey WholeLottaLove!

The answer [A] is accurate. I understand the confusion. Lemme describe my way of approaching the problem.

Statement 1 clearly says |2x-12| < 10, so you can directly put ranges for x > 6 and x < 6 and derive at the conclusion that x always lies between the region {1,11}. Hence cl;early x > 0.

Statement 2 on the other hand, presents an inequality $$x^2 - 10x + 21 >=0$$
Solving the same we arrive at (x-7)(x-3) >=0. Now we can test values between the region, {3,7} and find that the values appear to be -ve. For example when x = 5, f(x) = -4 < 0. Hence we can conclude that for all values between {3,7}, the function will have -ve values. When we test the values for regions {-infinity, 3} and {7, infinity} we realize that all the values are +ve. Hence we conclude from the above, that the above range {-infinity, 3} and {7, infinity} satisfies our equation.
Now since the range {-infinity, 3} includes -ve values as well, we'd have to rule it out as a valid answer i.e. for such values x < 0.

You can visualize the change in values of $$x^2 - 10x + 21 >=0$$ once you plot the same. Please refer to my *not-geometrically-accurate* graph for the same!

Hope my explanation satisfies your query!

Regards,
Arpan
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16 May 2013, 13:27
Thanks!

I understand the bit about plugging in values, but what I don't understand is the explanation given in the book. What I don't understand is why it's [- infinity to 3] and [7 to infinity] in the inequality we have (x-7)(x-3)>/=0 so wouldn't we test for all cases greater than or equal to 3 and greater than or equal to 7? Why does the number line have x is less than or equal to 3 (i.e. [-infinity to 3]?
arpanpatnaik wrote:
WholeLottaLove wrote:
Is x>0

1.) |2x-12| < 10
2.) x2 - 10x >/= -21

I understand #1 but I am getting tripped up with #2

X2 - 10x + 21 >/= (greater than or equal to) 0.
(x-7)(x-3) >/= 0
So I get at least two results (this is the type of question that asks whether you can narrow it down to one definitive result or not) that make the question valid.
X=0 (0-7)(0-3) = 21 21>0 and X=0
X= 3 (3-7)(3-3) = 0 0=0 and X=3
So basically, the question asks whether X>0. I have X=0 that satisfies the equation and X=3 that satisfies the equation (it asks if the equation is greater than or equal to 0) So the problem cannot be answered with certainty.
However, the explanation in the book uses a number line to explain which looks something like this:

eeee <---------- eeeeeee ----------->

|__________3___________7___________|

I get that with a typical absolute value problem, say |x| >/= 3 you can have cases where x >/= 3 and x </= -3 but with (x-7) and (x-3) why do I know that it's 3 that goes left on the # line and 7 that goes right on the # line.
Thanks!

hey WholeLottaLove!

The answer [A] is accurate. I understand the confusion. Lemme describe my way of approaching the problem.

Statement 1 clearly says |2x-12| < 10, so you can directly put ranges for x > 6 and x < 6 and derive at the conclusion that x always lies between the region {1,11}. Hence cl;early x > 0.

Statement 2 on the other hand, presents an inequality $$x^2 - 10x + 21 >=0$$
Solving the same we arrive at (x-7)(x-3) >=0. Now we can test values between the region, {3,7} and find that the values appear to be -ve. For example when x = 5, f(x) = -4 < 0. Hence we can conclude that for all values between {3,7}, the function will have -ve values. When we test the values for regions {-infinity, 3} and {7, infinity} we realize that all the values are +ve. Hence we conclude from the above, that the above range {-infinity, 3} and {7, infinity} satisfies our equation.
Now since the range {-infinity, 3} includes -ve values as well, we'd have to rule it out as a valid answer i.e. for such values x < 0.

You can visualize the change in values of $$x^2 - 10x + 21 >=0$$ once you plot the same. Please refer to my *not-geometrically-accurate* graph for the same!

Hope my explanation satisfies your query!

Regards,
Arpan
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16 May 2013, 13:34
1
When you have an inequality such as

$$x^2 - 10x +21\geq{}0$$

replace the > with an = and find its roots

x=3 and x=7 are your results

Now the interval must be defined by those values, and will be either $$3<x<7$$ (first case) or $$x<3$$ , $$x>7$$ (second case)
so x will be "inside" (first case) the interval or "outside" (second case) -no other possible solutions-

to pick the right one refer to my tips and tricks post on inequalities

in this case the right one is x<3 or x>7 (sing of x^2 is + and the operator is >)

Hope this helps
_________________

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16 May 2013, 13:36
I'm running on next to no sleep so I apologize if I am a bit dense

So basically, when you have greater than/less than, you test values less than the smaller value (in this case, 3) and greater than the larger value (in this case, 7)?

Zarrolou wrote:
When you have an inequality such as

$$x^2 - 10x +21\geq{}0$$

replace the > with an = and find its roots

x=3 and x=7 are your results

Now the interval must be defined by those values, and will be either $$3<x<7$$ (first case) or $$x<3$$ , $$x>7$$ (second case)
so x will be "inside" (first case) the interval or "outside" (second case) -no other possible solutions-

to pick the right one refer to my tips and tricks post on inequalities

in this case the right one is x<3 or x>7 (sing of x^2 is + and the operator is >)

Hope this helps
VP
Status: Far, far away!
Joined: 02 Sep 2012
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16 May 2013, 13:43
1
WholeLottaLove wrote:
I'm running on next to no sleep so I apologize if I am a bit dense

So basically, when you have greater than/less than, you test values less than the smaller value (in this case, 3) and greater than the larger value (in this case, 7)?

Generally speaking an inequality of second degree has 2 roots

$$ax^2+bx+c=0$$ gives us two solutions $$x_1, x_2$$ (where$$x_2$$ is bigger than $$x_1$$)

when you have to define the interval, it will be $$x_1<x<x_2$$ (x is between the smaller and the bigger root)
OR
$$x>x_2$$ , $$x<x_1$$ (x bigger than the bigger root or less than the smaller)

Now everything comes down to choose the right one
As I said before refer to my post on Inequalities where I explain which and why pick one over the other

Hope this makes sense, let me know
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16 May 2013, 13:43
1
WholeLottaLove wrote:
Thanks!

I understand the bit about plugging in values, but what I don't understand is the explanation given in the book. What I don't understand is why it's [- infinity to 3] and [7 to infinity] in the inequality we have (x-7)(x-3)>/=0 so wouldn't we test for all cases greater than or equal to 3 and greater than or equal to 7? Why does the number line have x is less than or equal to 3 (i.e. [-infinity to 3]?

If you'd look at the graph I plotted for the function, you'd realize that for [- infinity to 3] and [7 to infinity], the function satisfies the condition of being > 0 i.e. our Statement 2. The values 3 and 7 lie on the function where the value is 0. If we were to use any value greater than equal to 3 but less than 7, the function will no longer be > 0. For such values between {3,7} the values of f(x) < 0. Hence we cannot use them as valid values of x satisfying the inequality!

ee <---------- eeeeeeeeeeeee ----------->

|__________3___________7___________|

Now if you look at the number line mentioned in the book, for all values less than 3, f(x) > 0. Similarly for all values of x > 7, f(x) > 0. Now since for the region x < 3, we have values like {-1,-2.....} -ve values for x. We can not use the same as a solution for x>0, our question! This is the reason why the above number line has arrows starting from 3 towards -ve infinity and 7 to positive infinity to represent the solution of f(x) > 0.

Regards,
Arpan
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16 May 2013, 13:51
1
Zarrolou wrote:
WholeLottaLove wrote:
I'm running on next to no sleep so I apologize if I am a bit dense

So basically, when you have greater than/less than, you test values less than the smaller value (in this case, 3) and greater than the larger value (in this case, 7)?

Generally speaking an inequality of second degree has 2 roots

$$ax^2+bx+c=0$$ gives us two solutions $$x_1, x_2$$ (where$$x_2$$ is bigger than $$x_1$$)

when you have to define the interval, it will be $$x_1<x<x_2$$ (x is between the smaller and the bigger root)
OR
$$x>x_2$$ , $$x<x_1$$ (x bigger than the bigger root or less than the smaller)

Now everything comes down to choose the right one
As I said before refer to my post on Inequalities where I explain which and why pick one over the other

Hope this makes sense, let me know

Totally agree with Zarrolou!

Would like to add an observation which would help you over such inequalities. In general, when you look at functions, such as

F(x) > 0 where F(x) may be a mod function or quadratic equation, you would always arrive at an open-ended solution set i.e.
{-ve infinity, p1} and {+ve infinity, p2}, where p1 and p2 are roots of the equation F(x) = 0.

So whenever you see such an quadratic equation, and you calculate the roots, you can plug them in the above range and determine a solution set!

Similarly, for functions of the form, F(x) < 0 the solution set is always of the form {p1,p2} and is a closed-set! You can go ahead and test the theory over as many equations you like! The above idea is extremely useful in determining quick answers to such inequality questions!

Hope it helps!

Regards,
Arpan
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16 May 2013, 14:40
arpanpatnaik wrote:

Totally agree with Zarrolou!

Would like to add an observation which would help you over such inequalities. In general, when you look at functions, such as

F(x) > 0 where F(x) may be a mod function or quadratic equation, you would always arrive at an open-ended solution set i.e.
{-ve infinity, p1} and {+ve infinity, p2}, where p1 and p2 are roots of the equation F(x) = 0.

So whenever you see such an quadratic equation, and you calculate the roots, you can plug them in the above range and determine a solution set!

Similarly, for functions of the form, F(x) < 0 the solution set is always of the form {p1,p2} and is a closed-set! You can go ahead and test the theory over as many equations you like! The above idea is extremely useful in determining quick answers to such inequality questions!

Hope it helps!

Regards,
Arpan

I am afraid but I have to disagree with you.

"F(x) > 0 where F(x) may be a mod function or quadratic equation, you would always arrive at an open-ended solution set"

This is not true: Consider $$F(x)=-x^2+x$$ for example
if you study F(x)>0 you will arrive at a closed solution

In this particular case is $$F(x)=-x^2+x>0$$ or $$0<x<1$$ => closed interval

Also "F(x) < 0 the solution set is always of the form {p1,p2} and is a closed-set" is wrong.

Do you agree?
_________________

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16 May 2013, 20:47
WholeLottaLove wrote:

I understand the bit about plugging in values, but what I don't understand is the explanation given in the book. What I don't understand is why it's [- infinity to 3] and [7 to infinity] in the inequality we have (x-7)(x-3)>/=0 so wouldn't we test for all cases greater than or equal to 3 and greater than or equal to 7? Why does the number line have x is less than or equal to 3 (i.e. [-infinity to 3]?

Just another way to look at it, which might help:

We have (x-1)(x-3)$$\geq{0}$$

Taking the inequality first, we have (x-1)(x-3)>0. Thus, both (x-1) and (x-7) have to be positive or both have to be negative.

When both are positive, we have x>3 AND x>7--> If x>7 it will automatically ensure that x>3. Thus x>7 and extends till positive infinity.

When both are negative, we have x<3 and x<7--> If x<3, we will have x<7. Thus x<3 and extends till negative infinity.

And obviously, the equality gives the values as x=3 OR x=7.
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Re: Is x > 0?  [#permalink]

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16 May 2013, 22:23
WholeLottaLove wrote:
Is x > 0?

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

I understand #1 but I am getting tripped up with #2

X2 - 10x + 21 >/= (greater than or equal to) 0.
(x-7)(x-3) >/= 0
So I get at least two results (this is the type of question that asks whether you can narrow it down to one definitive result or not) that make the question valid.
X=0 (0-7)(0-3) = 21 21>0 and X=0
X= 3 (3-7)(3-3) = 0 0=0 and X=3
So basically, the question asks whether X>0. I have X=0 that satisfies the equation and X=3 that satisfies the equation (it asks if the equation is greater than or equal to 0) So the problem cannot be answered with certainty.
However, the explanation in the book uses a number line to explain which looks something like this:

ee <---------- eeeeeeeeeeeee ----------->

|__________3___________7___________|

I get that with a typical absolute value problem, say |x| >/= 3 you can have cases where x >/= 3 and x </= -3 but with (x-7) and (x-3) why do I know that it's 3 that goes left on the # line and 7 that goes right on the # line.
Thanks!

This post might help: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476
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17 May 2013, 00:40
Zarrolou wrote:
I am afraid but I have to disagree with you.

"F(x) > 0 where F(x) may be a mod function or quadratic equation, you would always arrive at an open-ended solution set"

This is not true: Consider $$F(x)=-x^2+x$$ for example
if you study F(x)>0 you will arrive at a closed solution

In this particular case is $$F(x)=-x^2+x>0$$ or $$0<x<1$$ => closed interval

Also "F(x) < 0 the solution set is always of the form {p1,p2} and is a closed-set" is wrong.

Do you agree?

I'd slightly modify the above equation Zarrolou!

For F(x) = $$-x^2 + x$$> 0
=> $$-x(x+1)$$ > 0
Now when we multiply minus sign on both the sides,
F(x) = $$x(x+1) < 0$$

The above equation translates to a less than value equation, Hence can be used over my asseration that its a closed set soultion set i.e {0,1}

But I do see your point that my assertion was ambiguous! I should specify that the sign of the x^2 term should be +ve and the equation F(x) should not be further reducable to any other form. The observation I mentioned is purely something that came up after looking at several problems of inequalities. You can check them for mod functions like $$|x-4| < 3$$ whose values lie betwee {1,7} and is a closed set, whereas $$|x-4| > 3$$ has a solution set {-infinity,1} and {7, infinity}, which is an open set! The same can also be observed for quadratic equations!

Please correct me if I am wrong!

Regards,
Arpan
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17 May 2013, 00:55
1
arpanpatnaik wrote:
I'd slightly modify the above equation Zarrolou!

For F(x) = $$-x^2 + x$$> 0
=> $$-x(x+1)$$ > 0
Now when we multiply minus sign on both the sides,
$$F(x) = x(x+1) < 0$$

The above equation translates to a less than value equation, Hence can be used over my asseration that its a closed set soultion set i.e {0,1}

But I do see your point that my assertion was ambiguous! I should specify that the sign of the x^2 term should be +ve and the equation F(x) should not be further reducable to any other form. The observation I mentioned is purely something that came up after looking at several problems of inequalities. You can check them for mod functions like $$|x-4| < 3$$ whose values lie betwee {1,7} and is a closed set, whereas $$|x-4| > 3$$ has a solution set {-infinity,1} and {7, infinity}, which is an open set! The same can also be observed for quadratic equations!

Please correct me if I am wrong!

Regards,
Arpan

As I explained above, you method is not correct.

"Now when we multiply minus sign on both the sides" if you do this you are studying not $$F(x)$$ but $$-F(x)$$
$$F(x) = x(x+1) < 0$$ is wrong
its
$$-F(x) = x(x+1) < 0$$ note the minus
$$-F(x)<0$$ which according to your methos gives us external values (f(x)>0)=> INCORRECT in this case the solution is internal

I know what you mean, and I don't want to confuse you BUT your method is mathematically incorrect.

Hope it's clear now
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05 Jun 2013, 12:01
Hi there.

I revisited this problem and attacked it in the way you have taught me to go about solving problems. Namely, for #1 I checked for which values would 2x-12 be negative and positive and then went about establishing the positive and negative cases as follows:

2x-12 is positive when x>6.
2x-12 is negative when x<6.

Therefore,

Positive: 2x-12<10 ==> 2x<22 ==> x<11
Negative: -2x+12<10 ==> -2x<-2 ==> x>1

Here is my problem. For example, regarding the positive case. 2x-12 is positive when x>6 and indeed for the positive case we find that x<11. Therefore, x may or may not be greater than 6. I see that for this problem you combine the statements into 1<x<11 but even then, couldn't x be less than 1? I understand the concepts used to solve this problem but I'm not sure I would be able to differentiate between two (seemingly) different problems that demand (seemingly) different approaches to solve them.

Thanks!

Zarrolou wrote:
Is $$x>0$$?

1.) $$|2x-12| < 10$$
$$1<x<11$$
Sufficient

2.) $$x^2 - 10x \geq{} -21$$

$$x^2 - 10x +21\geq{}0$$
This equation has two roots x=3 and x=7
So $$x<3$$ or $$x>7$$
Not sufficient to say that x is positive

A

If you have doubts on inequalities, check my post tips-and-tricks-inequalities-150873.html to define the interval of the inequality
Should remove your doubts (I hope!)
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05 Jun 2013, 12:21
WholeLottaLove wrote:
Hi there.

I revisited this problem and attacked it in the way you have taught me to go about solving problems. Namely, for #1 I checked for which values would 2x-12 be negative and positive and then went about establishing the positive and negative cases as follows:

2x-12 is positive when x>6.
2x-12 is negative when x<6.

Therefore,

Positive: 2x-12<10 ==> 2x<22 ==> x<11
Negative: -2x+12<10 ==> -2x<-2 ==> x>1

Here is my problem. For example, regarding the positive case. 2x-12 is positive when x>6 and indeed for the positive case we find that x<11. Therefore, x may or may not be greater than 6. I see that for this problem you combine the statements into 1<x<11 but even then, couldn't x be less than 1? I understand the concepts used to solve this problem but I'm not sure I would be able to differentiate between two (seemingly) different problems that demand (seemingly) different approaches to solve them.

Thanks!

Your method is almost correct, remember to include an $$=$$ sign (as I told you before).

Here we'll work on the intervals, so

Positive $$x\geq{6}$$: 2x-12<10 ==> 2x<22 ==> x<11
You have found the range of solutions $$x<11$$, correct. But we are considering values $$x\geq{6}$$, so we have to MERGE the intervals into
$$6\leq{x}<11$$ (just united them). This right here is the result of the analysis, so the equation $$|2x-12|<10$$ (in its positive case) holds true into that range of values.
This is an important concept to understand.

Negative $$x<6$$: -2x+12<10 ==> -2x<-2 ==> x>1
Same thing here, merge the intervals into $$1<x<6$$. So the function $$|2x-12|<10$$ (in its negative case) holds true into that range of values. Same thing as above.

But we are studying the overall function $$|2x-12|<10$$, so we have to merge the intervals in which its separated cases hold true, so
$$1<x<6$$+$$6\leq{x}<11$$ = $$1<x<11$$
And here is my result

The question asks is x positive? YES! It's in the range 1-11 so it's positive

When you study and abs value, first and foremost pay attention to the interval you are considering.
If you are asked a specific value of $$x=?$$, then just check weather x is in the interval you are considering.
If you have an inequality, MERGE the intervals you have => so you'll find the range into which the equation holds true.
If the intervals have NO COMMON VALUES example x>5 and x<2, that equation NEVER holds true.
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05 Jun 2013, 12:33
Ahhh....I see. So for say 6≤x<11 x is indeed positive and that case is true. Then we could test the negative case 1<x<6 in which case x is positive so for both the positive and negative cases, x is a positive #.

Thanks!

Zarrolou wrote:
WholeLottaLove wrote:
Hi there.

I revisited this problem and attacked it in the way you have taught me to go about solving problems. Namely, for #1 I checked for which values would 2x-12 be negative and positive and then went about establishing the positive and negative cases as follows:

2x-12 is positive when x>6.
2x-12 is negative when x<6.

Therefore,

Positive: 2x-12<10 ==> 2x<22 ==> x<11
Negative: -2x+12<10 ==> -2x<-2 ==> x>1

Here is my problem. For example, regarding the positive case. 2x-12 is positive when x>6 and indeed for the positive case we find that x<11. Therefore, x may or may not be greater than 6. I see that for this problem you combine the statements into 1<x<11 but even then, couldn't x be less than 1? I understand the concepts used to solve this problem but I'm not sure I would be able to differentiate between two (seemingly) different problems that demand (seemingly) different approaches to solve them.

Thanks!

Your method is almost correct, remember to include an $$=$$ sign (as I told you before).

Here we'll work on the intervals, so

Positive $$x\geq{6}$$: 2x-12<10 ==> 2x<22 ==> x<11
You have found the range of solutions $$x<11$$, correct. But we are considering values $$x\geq{6}$$, so we have to MERGE the intervals into
$$6\leq{x}<11$$ (just united them). This right here is the result of the analysis, so the equation $$|2x-12|<10$$ (in its positive case) holds true into that range of values.
This is an important concept to understand.

Negative $$x<6$$: -2x+12<10 ==> -2x<-2 ==> x>1
Same thing here, merge the intervals into $$1<x<6$$. So the function $$|2x-12|<10$$ (in its negative case) holds true into that range of values. Same thing as above.

But we are studying the overall function $$|2x-12|<10$$, so we have to merge the intervals in which its separated cases hold true, so
$$1<x<6$$+$$6\leq{x}<11$$ = $$1<x<11$$
And here is my result

The question asks is x positive? YES! It's in the range 1-11 so it's positive

When you study and abs value, first and foremost pay attention to the interval you are considering.
If you are asked a specific value of $$x=?$$, then just check weather x is in the interval you are considering.
If you have an inequality, MERGE the intervals you have => so you'll find the range into which the equation holds true.
If the intervals have NO COMMON VALUES example x>5 and x<2, that equation NEVER holds true.
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Re: Is x > 0?  [#permalink]

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06 Jun 2013, 00:46
Is x > 0?

(1) |2x-12| < 10 --> reduce by 2: $$|x-6| < 5$$. So, we are told that the distance between x and 6 is less than 5: --1-----6-----11. Thus we have that $$1<x<11$$ --> $$x>0$$. Sufficient.

(2) x^2 - 10x >= -21 --> $$x^2 - 10x+21\geq{0}$$ --> $$(x-3)(x-7)\geq{0}$$. The "roots" are 3 and 7. "$$\geq$$" sign indicates that the solutions lies to the left of the smaller root and to the right of the greater root (check here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus $$x\leq{3}$$ OR $$x\geq{7}$$. Not sufficient.

Hope it's clear.
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Re: Is x > 0?  [#permalink]

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14 Mar 2017, 02:48
(1) SUFFICIENT: Manipulate the absolute value expression so we can represent this inequality on a number line:
|2x – 12| < 10
|2(x – 6)| < 10
2x|x – 6| < 10
|x – 6| < 5

This can be interpreted as “The distance between x and 6 is less than 5.” On a number line, this is the region between 1 and 11:
All possible values for x are positive, so the answer to the question is a definite “Yes.”

(2) INSUFFICIENT: Manipulate the inequality to get 0 on one side, then factor the resulting
x^2 – 10x ≥ –21
x^2 – 10x + 21 ≥ 0
(x – 7)(x – 3) ≥ 0
The factored quadratic on the left side will equal 0 when x = 3 and 7. These are the boundary points. On a number line, we can check the regions on either side of these boundary points to determine the valid region(s) for x:
Note that when 3 < x < 7, (x – 7)(x – 3) = (neg)(pos) = neg.
Since both positive and negative values are possible for x, the answer is “Maybe.”
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Re: Is x > 0?  [#permalink]

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15 Mar 2017, 04:19
St 1: 2x-12 will be 10 at x=11 and -10 at 1. there fore for this inequality 1<x<11. therefore x>0 ANSWER
St 2: for the inequality x<3 and x>7. cannot say is x>0 or not. INSUFFICIENT

Option A
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Re: Is x > 0? &nbs [#permalink] 15 Mar 2017, 04:19

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