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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

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.

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

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

Why are you talking about x<0 and x>0 when the transition point is 7/3, not 0?

You should brush up fundamentals on absolute value:

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?[/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[/quote]

we are taking condition x<=7/3 to be equation 3x-7 as <=0 So x=1 is acceptable

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.
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From 1 x=1 or 9. Hence sqrt(x) has 2 values. Not sufficient. From 2 x=0 or 9. Hence sqrt(x) has 2 values. Not sufficient. Both: x=9. Hence sqrt(x) is 3 which is prime. Sufficient. Hence C.

|3x-7|=|3*1-7|=|-4|=4(modulus of any value is +ve)

2x+2=2*1+2=4.

Hence valid

ArunpriyanJ wrote:

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?[/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.

|3x-7|=|3*1-7|=|-4|=4(modulus of any value is +ve)

2x+2=2*1+2=4.

Hence valid

ArunpriyanJ wrote:

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?

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

Statement 1 X = 1 or 9 (Both the values satisfy the equation) If x = 1, then \(\sqrt{x}\) = 1 => Not a prime number. If x = 9, then \(\sqrt{x}\) = 3 => Prime Number 2 different answers, so this is insufficient.

Statement 2 x = 0 or 9 (Both the values satisfy the equation) If x = 0, then \(\sqrt{x}\) = 0 => Not a prime number. If x = 9, then \(\sqrt{x}\) = 3 => Prime Number 2 different answers, so this is insufficient.

Statement 1 + Statement 2 x = 9 (The only common value) => \(\sqrt{x}\) = 3 => Prime Number Unique Answer and hence, Sufficient. C is the answer.

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