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Re: Tricky inequality problem [#permalink]
17 Jul 2010, 03:36

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DenisSh wrote:

Bunuel wrote:

gurpreetsingh's solution is correct.

Is \frac{x}{3}+\frac{3}{x}>2? --> is \frac{(x-3)^2}{x}>0? Answer: C.

How did you get \frac{(x-3)^2}{x}>0?

Step 1: \frac{x}{3}+\frac{3}{x}>2? Step 2 (multiply by 3x): x^2+9>6x Step 3 (move 6x to the left side): x^2-6x+9>0 Step 4 (convert to the compact form): (x-3)^2>0

Please explain...

This is the most common error when solve inequalities. I keep writing this over and over again:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign.

So you CANNOT multiply \frac{x}{3}+\frac{3}{x}>2 by 3x since you don't know the sign of x.

What you CAN DO is: \frac{x}{3}+\frac{3}{x}>2 --> \frac{x}{3}+\frac{3}{x}-2> --> common denominator is 3x --> \frac{x^2+9-6x}{3x}>0 --> multiply be 3 --> \frac{x^2+9-6x}{x}>0 --> \frac{(x-3)^2}{x}>0.

Re: Tricky inequality problem [#permalink]
16 Jul 2010, 06:42

2

This post received KUDOS

Expert's post

DenisSh wrote:

Is \frac{x}{3} + \frac{3}{x} > 2?

(1) x < 3

(2) x > 1

Please outline your approach!

gurpreetsingh's solution is correct.

Is \frac{x}{3}+\frac{3}{x}>2? --> is \frac{(x-3)^2}{x}>0? Now, nominator is non-negative, thus the fraction to be positive nominator must not be zero (thus it'll be positive) and denominator mut be positive --> x\neq{3} and x>0.

Statement (1) satisfies the first requirement and statement (2) satisfies the second requirement, so taken together they are sufficient.

Re: Tricky inequality problem [#permalink]
19 Jul 2010, 08:18

1

This post received KUDOS

Expert's post

ulm wrote:

(x-3)^2/x >0

(x-3)^2 is always >0, therefore we just need x to be >0. The correct answer is (0;3) and (3;+inf) It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.

Generally, unknown (or expression with unknown) in even power is NOT always positive, it's non-negative. Not knowing this is the cause of many mistakes on GMAT.

So, (x-3)^2\geq{0}, because if x=3, then (x-3)^2=0 and \frac{(x-3)^2}{x} also equals to zero (and not more than zero). We need statement (1) to exclude the possibility of x being 3 by saying that x<3. That's why the answer to this question is C, not B.

Re: Tricky inequality problem [#permalink]
17 Jul 2010, 01:09

Bunuel wrote:

gurpreetsingh's solution is correct.

Is \frac{x}{3}+\frac{3}{x}>2? --> is \frac{(x-3)^2}{x}>0? Answer: C.

How did you get \frac{(x-3)^2}{x}>0?

Step 1: \frac{x}{3}+\frac{3}{x}>2? Step 2 (multiply by 3x): x^2+9>6x Step 3 (move 6x to the left side): x^2-6x+9>0 Step 4 (convert to the compact form): (x-3)^2>0

Re: Tricky inequality problem [#permalink]
18 Jul 2010, 09:55

The way I solved it.

1. x < 3 : LT3 - Insuf because with x = -1 the inequality is false and with x = 2 it is true. 2. x > 1 : GT1 - Insuf because with x = 2 the inequality is true and with x = 3 it is false

Combining the two options yeilds 1 < x < 3 which clearly shows that the original expression is > 2.

So the correct answer is C. 1 and 2 together are sufficient.

Re: Tricky inequality problem [#permalink]
19 Jul 2010, 02:20

(x-3)^2/x >0

(x-3)^2 is always >0, therefore we just need x to be >0. The correct answer is (0;3) and (3;+inf) It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.

Re: Tricky inequality problem [#permalink]
19 Jul 2010, 08:48

Expert's post

ulm wrote:

(x-3)^2/x >0

(x-3)^2 is always >0, therefore we just need x to be >0. The correct answer is (0;3) and (3;+inf) It's (B), because if we consider (C), we will loose answers. Try 10, 11, 12 and so on. It fits.

ulm wrote:

I'm completely agree that x not equal 3 (as i wrote earlier;)

But we miss all positive x's that a > 3, don't we?

I'm not sure exactly what you're asking here. You said that the answer to this question should be B: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Statement (2) is: x>1 --> denominator is positive, nominator is also positive EXCEPT for one value of x: when x=3>1, then \frac{(x-3)^2}{x}=0 (not more than zero). Hence we have TWO different answers to the question "is \frac{(x-3)^2}{x}>0": one is NO for x=3>1 and another is YES for all other values of x>1. Two different answers = not sufficient.

Re: Tricky inequality problem [#permalink]
21 Jul 2010, 08:41

Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...

1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient

2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient

When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C _________________

Re: Tricky inequality problem [#permalink]
21 Jul 2010, 09:10

Expert's post

Michmax3 wrote:

Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...

1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient

2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient

When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C

Unfortunately your approach is not correct.

First of all if x=2--> \frac{x}{3} + \frac{3}{x}=\frac{13}{6} > 2, so you made an error in calculations (\frac{2}{3} + \frac{3}{2}=\frac{13}{6}\neq{\frac{5}{6}}). Again \frac{x}{3} + \frac{3}{x}>2 is true for ANY value of x but 3, for which \frac{x}{3} + \frac{3}{x}=2.

Next: you say that "When combined, you know x has to be 2". Not so, as we are not told that x is an integer, hence x<3 and x>1 does not mean that x=2, it can be 2.5 or 1.777, basically ANY number from 1 to 3, not inclusive.

Re: Tricky inequality problem [#permalink]
21 Jul 2010, 09:30

Bunuel wrote:

Michmax3 wrote:

Not sure if I took the long way to do this problem and I just got lucky that it worked out quickly for me, but I took a different approach and started plugging in numbers in case this explanation helps anyone...

1) x<3, so I plugged in 2 for x and got 5/6 which is less than 2; I then plugged in 1 for x trying to find an instance where the opposite was true and got 10/3, which is greater than 2, so statement 1 is insufficient

2) x>1; I already had the result when 2 is plugged in which is 5/6 and less than 2; so I tried plugging in 6, since I knew this would give me a higher result and got 2.5 which is greater than 2, so statement 2 insufficient

When combined, you know x has to be 2 which results in 5/6 and is less than 2, so together they are sufficient. C

Unfortunately your approach is not correct.

First of all if x=2--> \frac{x}{3} + \frac{3}{x}=\frac{13}{6} > 2, so made you an error in calculations (\frac{2}{3} + \frac{3}{2}=\frac{13}{6}\neq{\frac{5}{6}}). Again \frac{x}{3} + \frac{3}{x}>2 is true for ANY value of x but 3, for which \frac{x}{3} + \frac{3}{x}=2.

Next: you say that "When combined, you know x has to be 2". Not so, as we are not told that x is an integer, hence x<3 and x>1 does not mean that x=2, it can be 2.5 or 1.777, basically ANY number from 1 to 3, not inclusive.

Hope it's clear.

Yes, then I guess I got lucky and need a lot more practice with these. Thanks for clarifying _________________

Re: Tricky inequality problem [#permalink]
05 Oct 2010, 21:25

Bunuel wrote:

DenisSh wrote:

Bunuel wrote:

gurpreetsingh's solution is correct.

Is \frac{x}{3}+\frac{3}{x}>2? --> is \frac{(x-3)^2}{x}>0? Answer: C.

How did you get \frac{(x-3)^2}{x}>0?

Step 1: \frac{x}{3}+\frac{3}{x}>2? Step 2 (multiply by 3x): x^2+9>6x Step 3 (move 6x to the left side): x^2-6x+9>0 Step 4 (convert to the compact form): (x-3)^2>0

Please explain...

This is the most common error when solve inequalities. I keep writing this over and over again:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

So you CAN NOT multiply \frac{x}{3}+\frac{3}{x}>2 by 3x since you don't know the sign of x.

Wheat you CAN DO is: \frac{x}{3}+\frac{3}{x}>2 --> \frac{x}{3}+\frac{3}{x}-2> --> common denominator is 3x --> \frac{x^2+9-6x}{3x}>0 --> multiply be 3 --> \frac{x^2+9-6x}{x}>0 --> \frac{(x-3)^2}{x}>0.

Hope it helps.

thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit _________________

Re: Tricky inequality problem [#permalink]
05 Oct 2010, 23:23

hirendhanak wrote:

thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit

Consider the inequality \frac{x^2-12}{x}>-1

Lets say I multiply both sides by 7x without considering the signs of the variable, what happens ?

x^2-12>-x x^2+x-12>0 (x+4)(x-3)>0

Which is true whenever x>3 (both terms positive) or when x<-4 (both terms negative)

But since we haven't kept the Sign of x in mind when we multiplied in step 1, the solution is wrong.

For eg. Take x=-1 which according to us is not a solution. It is easy to see ((-1)^2-12)/(-1)=11>-1. So it should be a solution Similarly take x=-6 which according to us is a solution, but ((-6)^2-12)/-6=-4<-1. So it should not be a solution _________________

Re: Tricky inequality problem [#permalink]
06 Oct 2010, 03:09

Expert's post

hirendhanak wrote:

thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit

Consider a simple inequality 4>3 and some variable x.

Now, you can't multiply (or divide) both parts of this inequality by x and write: 4x>3x, because if x=1>0 then yes 4*1>3*1 but if x=-1 then 4*(-1)=-4<3*(-1)=-3. Similarly, you can not divide an inequality by x not knowing its sign.

is (x/3+3/x) > 2? (1) x < 3 (2) x > 1 This ds has [#permalink]
08 Jun 2011, 02:52

is (x/3+3/x) > 2? (1) x < 3 (2) x > 1

This ds has been discussed thoroughly at http://gmatclub.com/forum/tricky-inequality-problem-97331.html. It inequality simplified there as [(x - 3)^2]/3 > 0. if i do not simplify then i can i solved it as: (1) if x = 2 then the (x/3+3/x) > 2 but if x = negative the (x/3+3/x) > 2 is not true. so Insufficient. (2) if x = 2 then (x/3+3/x) > 2 but if x = 1 then (x/3+3/x) > 2 is true. so insufficient.

for C x =2 and (x/3+3/x) > 2.

so why i will simplify as i am getting direct answer? _________________