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You guys have taken Sqrt(9) as 3 where as it should be plus or minus 3. In this case, statements (1) and (2) taken together wont be sufficient hence the answer should be E. Please explain. I know OA is C but it may be wrong.

Note that from statement 1: x>7/3 but root of x is not required to be greater than 7/3 or in fact sqrt(x) has no conditions on it so that logic wont work too.

The red part is not correct.

The point here is that square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

1) \(|3x-7|=2x+2\) --> we have one check point 7/3 (check point - the value of x for which an expression in absolute value equals to zero):

A. \(x\leq{\frac{7}{3}}\) --> \(3x-7\leq{0}\) hence \(|3x-7|=-(3x-7)\) --> \(-(3x-7)=2x+2\) --> \(x=1\) --> \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) --> \(3x-7>0\) hence \(|3x-7|=3x-7\) --> \(3x-7=2x+2\) --> \(x=9\) --> \(\sqrt{9}=3=prime\). Two different answer. Not sufficient.

2) \(x^2=9x\) --> \(x(x-9)=0\) --> \(x=0\) or \(x=9\) --> \(\sqrt{0}=0\neq{prime}\) or \(\sqrt{9}=3=prime\). Not sufficient.

(1)+(2) Intersection of values from (1) and (2) is \(x=9\) --> \(\sqrt{9}=3=prime\). Sufficient.

You guys have taken Sqrt(9) as 3 where as it should be plus or minus 3. In this case, statements (1) and (2) taken together wont be sufficient hence the answer should be E. Please explain. I know OA is C but it may be wrong.

Note that from statement 1: x>7/3 but root of x is not required to be greater than 7/3 or in fact sqrt(x) has no conditions on it so that logic wont work too.

Please correct me if i am wrong x\sqrt{2}=9x if we divide both sides by x then we get x= 9 which makes B sufficient. isnt it??

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) \(x^2=9x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution (notice that both x=9 AND x=0 satisfy the equation).

As from the first statement x >-1 so it can be 0 too. Second statement gives values 0 and 3

Even after combining both statements, As we are not sure of answer ( 0 or 3 ) i.e why I concluded E.

I know am skipping few imp. concepts here please help.

any explanation for this?

x > -1 implies that whatever value of x will satisfy this equation, it will be greater than -1. It does not mean that every value greater than -1 will satisfy it. You cannot take one part of an equation in isolation and solve from it.

|3x-7|=2x+2

Point is that no value of x less than -1 can satisfy this equation. But, it doesn't mean that every value greater than or equal to -1 will satisfy it. When you solve this equation, you get x = 1 or 9 (both greater than -1). No other value of x satisfies this equation. If you put x = 0, you get 7 = 2 which is not true. So x cannot be 0.
_________________

1) \(|3x-7|=2x+2\) --> we have one check point 7/3 (check point - the value of x for which an expression in absolute value equals to zero):

A. \(x\leq{\frac{7}{3}}\) --> \(3x-7\leq{0}\) hence \(|3x-7|=-(3x-7)\) --> \(-(3x-7)=2x+2\) --> \(x=1\) --> \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) --> \(3x-7>0\) hence \(|3x-7|=3x-7\) --> \(3x-7=2x+2\) --> \(x=9\) --> \(\sqrt{9}=3=prime\). Two different answer. Not sufficient.

2) \(x^2=9x\) --> \(x(x-9)=0\) --> \(x=0\) or \(x=9\) --> \(\sqrt{0}=0\neq{prime}\) or \(\sqrt{9}=3=prime\). Not sufficient.

(1)+(2) Intersection of values from (1) and (2) is \(x=9\) --> \(\sqrt{9}=3=prime\). Sufficient.

Answer: C.

Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done

ohh yes. Thanks Bunuel. I read other articles and realized what you have said is followed by GMAT. Anyways, i appreciate the explanation, thanks for the information.

Please correct me if i am wrong x\sqrt{2}=9x if we divide both sides by x then we get x= 9 which makes B sufficient. isnt it??

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) \(x^2=9x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution (notice that both x=9 AND x=0 satisfy the equation).

There is one confusion .In many of your posts you have suggested whenever we have modulus at one side (Foreg: LHS in the first statement here). Why can't we compare the RHS as below

2x +2 >=0(LHS absolute value so always above or equal zero) x >-1 In this case answer could be different ie E

Please suggest where I am doing wrong
_________________

There is one confusion .In many of your posts you have suggested whenever we have modulus at one side (Foreg: LHS in the first statement here). Why can't we compare the RHS as below

2x +2 >=0(LHS absolute value so always above or equal zero) x >-1 In this case answer could be different ie E

Please suggest where I am doing wrong

How do you get that the answer is E from x>-1?
_________________

As from the first statement x >-1 so it can be 0 too. Second statement gives values 0 and 3

Even after combining both statements, As we are not sure of answer ( 0 or 3 ) i.e why I concluded E.

I know am skipping few imp. concepts here please help.

any explanation for this?

x > -1 implies that whatever value of x will satisfy this equation, it will be greater than -1. It does not mean that every value greater than -1 will satisfy it. You cannot take one part of an equation in isolation and solve from it.

|3x-7|=2x+2

Point is that no value of x less than -1 can satisfy this equation. But, it doesn't mean that every value greater than or equal to -1 will satisfy it. When you solve this equation, you get x = 1 or 9 (both greater than -1). No other value of x satisfies this equation. If you put x = 0, you get 7 = 2 which is not true. So x cannot be 0.

simply... for condition 1...square both sides... x=9 or x=1 we get two solutions

for cndition 2..we get x=9 therefore, both statements are reqd.

1) \(|3x-7|=2x+2\) --> we have one check point 7/3 (check point - the value of x for which an expression in absolute value equals to zero):

A. \(x\leq{\frac{7}{3}}\) --> \(3x-7\leq{0}\) hence \(|3x-7|=-(3x-7)\) --> \(-(3x-7)=2x+2\) --> \(x=1\) --> \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) --> \(3x-7>0\) hence \(|3x-7|=3x-7\) --> \(3x-7=2x+2\) --> \(x=9\) --> \(\sqrt{9}=3=prime\). Two different answer. Not sufficient.

2) \(x^2=9x\) --> \(x(x-9)=0\) --> \(x=0\) or \(x=9\) --> \(\sqrt{0}=0\neq{prime}\) or \(\sqrt{9}=3=prime\). Not sufficient.

(1)+(2) Intersection of values from (1) and (2) is \(x=9\) --> \(\sqrt{9}=3=prime\). Sufficient.

Answer: C.

Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done

Both x=1 and x=9 are valid for (1). Please elaborate what you mean?
_________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done[/quote]

Both x=1 and x=9 are valid for (1). Please elaborate what you mean?[/quote]

Bunel, I too have the same doubt in my mind. I will try to explain it.

By taking condition, X>0, i got the value of X as 9

And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.

So i took X=9 only and got the answer as A.

Pls explain how X=1 with the condition X<0 is valid.

Thanks in advance

gmatclubot

Re: Is root{x} a prime number?
[#permalink]
25 Dec 2015, 00:24

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