Data Sufficiency

A

Plug in different numbers, we would get that only statement 1 is sufficient. Whereas statement 2 is insufficient

Statement II is not contradicting statement I.....statement I told that X is a perfect square. but according to statement II u cannot find the value of X and thus it is insufficient. the statement II only tells that sqrt 3x+4 is integer. here x could be 4 or 20 to prove that "sqrt 3x+4" is an integer but it doesn't help us to solve the issue. but statement I actually solves for x and is sufficient.

1) sqrt(36) can be simplified to 6*(x)^(1/2). If this is an integer, than x^(1/2) is an integer. SUFFICIENT.

2) You cannot simplify this expression... or isolate x in the sqrt. INSUFFICIENT.

Answer is A.

A the Answer.


Let, 3x+4=64

then, x =20

here, sqrt{20} is not an integer.

OA is A

Good Question.
The point is that to remember that x is an integer and Sqrt (x) can't be a fraction ==> A

DATA SUFFICIENCY QUESTION:
If x is a positive integer, is sqrt(x) an integer?

1. sqrt(36x) is an integer
2. sqrt(3x + 4) is an integer

My answer: A
Explanation:
1. sqrt(36x) is an integer

36 is already a perfect square (6*6). SO x must be 4, 16, ....
Sufficient
2. sqrt(3x + 4) is an integer
x must be 20
and sqrt of 20 is not an integer.

Answer is A even if X =1 for first statement

krusta80 wrote:Interesting side question I posed to myself while looking at part 2 of this question is (and I warn anyone trying to stay on topic to AVOID the rest of this post):

How many square integers, x, exist such that 3x+4 is also a perfect square?

This is a great Q.

Since x is a perfect square, we can write x as k², where k is some integer. In these terms, 3x + 4 is 3k² + 4. This is also a perfect square; call it m². This gives us the equation

3k² + 4 = m²

3k² = m² - 4

3k² - m² + 4 = 0

Let's treat this as a quadratic in k, so that the a-coefficient = 3, the b-coefficient is 0, and the c-coefficient is (4 - m²).

This gives us k = ±√(0² - 4*3*(4-m²)) / 6, or ±√(12m² - 48)/6, or ±2*√(3m² - 12)/6, or ±√(3m² - 12)/3.

So provided that √(3m² - 12)/3 is an integer, we'll have a solution. To find some of these solutions, let's begin with an easy one: m = 2, which gives √(3m² - 12) = 0, an integer multiple of 3. Another easy one is m = 4, which gives √(3m² - 12) = 6, another integer multiple of 3.

In fact -- and I can't find a way to represent this next step in arithmetic appropriate to the GMAT -- we can go further, proceed inductively, and find more of solutions.

Let's start with our two basic solutions, m = 2 and m = 4. Notice that 4*4 - 2 = 14 is another solution: √(3*14² - 12) = 24, again an integer multiple of 3. Now we have m = 14 as a solution. Observe that 4*14 - 4 = 52 is another solution, as is 4*52 - 14 = 194, as is 4*194 - 52 = 724, etc. The inductive step would require us to prove that if r and s are "consecutive" solutions, then (4s - r) is also a solution, but I don't know a nice way of doing this in GMAT-friendly terms.

In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and equations ensures a solution.

=> Since there is only 1 variable in the original condition, it is likely that D is the answer.

In case of 1), sqrt 36x=int ==> 6sqrt x=int, sqrt x=int/6, and since the original condition states that x is an interger, sqrt x is also an interger. Thus this is a sufficient answer.

In case of 2), x=4 yes, x=7 no is not a sufficient answer.

Therefore the best answer is A.


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shubhamkumar wrote:

krusta80 wrote:Interesting side question I posed to myself while looking at part 2 of this question is (and I warn anyone trying to stay on topic to AVOID the rest of this post):

How many square integers, x, exist such that 3x+4 is also a perfect square?

Very True,That made me spend a minute more on this because Statement 2 is actually contradicting statement 1.as per statement 1, x is a square.considering stmt 2, I was not able to get any value from it which makes x a square.
Experts Please correct me if I am wrong.

Put x = 4, then sqrt(36x) is an integer and sqrt(3x+4) is also an integer. The value of sqrt(36x) = 12 and value of sqrt(3x+4) = 4