The first question looks incomplete. Could you clarify what the question is?

For the second question:

2- If X is a positive integers, is square root X an integer?

(1) square root 4x is an integer

Plug in x=1, which works because sqrt(4*1) is an integer

Next number that works is x=4, which works because sqrt(4*4) is an integer.

Next number that works is x=9, which works because sqrt(4*9) is an integer.

If you plug in x=1,4,9 back into the prompt, you'd find that sqrt(x) will always be an integer.

But the faster way to realize this is to notice that you can break sqrt(4x) down:

sqrt(4x) = sqrt(4) * sqrt(x) = 2*sqrt(x)

So (1) basically tells us that 2*sqrt(x) is an integer. If x is an integer, then the only way 2*sqrt(x) could be an integer is if sqrt(x) is an integer. SUFFICIENT.



(2) Square root 3x is not an integer

Plug in x=1, which works because sqrt(3*1) is not an integer.

If we plug x=1 back into the prompt, we find that sqrt(x) is an integer.

Plug in x=2, which works because sqrt(3*2) is not an integer.

If we plug x=2 back into the prompt, we find that sqrt(x) is not an integer.

INSUFFICIENT

raz1024 wrote:The first question looks incomplete. Could you clarify what the question is?

For the second question:

2- If X is a positive integers, is square root X an integer?

(1) square root 4x is an integer

Plug in x=1, which works because sqrt(4*1) is an integer

Next number that works is x=4, which works because sqrt(4*4) is an integer.

Next number that works is x=9, which works because sqrt(4*9) is an integer.

If you plug in x=1,4,9 back into the prompt, you'd find that sqrt(x) will always be an integer.

But the faster way to realize this is to notice that you can break sqrt(4x) down:

sqrt(4x) = sqrt(4) * sqrt(x) = 2*sqrt(x)

So (1) basically tells us that 2*sqrt(x) is an integer. If x is an integer, then the only way 2*sqrt(x) could be an integer is if sqrt(x) is an integer. SUFFICIENT.



(2) Square root 3x is not an integer

Plug in x=1, which works because sqrt(3*1) is not an integer.

If we plug x=1 back into the prompt, we find that sqrt(x) is an integer.

Plug in x=2, which works because sqrt(3*2) is not an integer.

If we plug x=2 back into the prompt, we find that sqrt(x) is not an integer.

INSUFFICIENT

For question 1 it says [yes there was a typo]

1- If R,S, and T are nonzero integers, is R^5 S^3 T^4 Negative?

(1) R T is negative

(2) S is Negative

1- If R,S, and T are nonzero integers, is R^5 S^3 T^4 Negative?

T^4 will never be negative, since an even exponent is involved. R^5 and S^3 are either positive or negative depending on the signs of R and S themselves. R will have the same sign as R^5, since an odd exponent is involved. Same for S and S^3.

For the entire product to be negative, either R^5 must be positive and S^3 negative, or vice versa. If both R^5 and S^3 have the same sign, then the entire product will be positive.

So the question really becomes: Do R and S have opposite signs?

(1) R T is negative

Nothing about S. INSUFFICIENT

(2) S is Negative

Nothing about R. INSUFFICIENT

Together, (1) and (2) tell us that S is negative and RT is negative. We have the sign of S, but do we necessarily have the sign of R?

Well, we know RT is negative. That means R and T have opposite signs. We know that T^4 is positive, but T itself could be either positive or negative. Therefore, R could be either positive or negative.

INSUFFICIENT

1- If R,S, and T are nonzero integers, is R^5 S^3 T^4 ?

(1) R T is negative

(2) S is Negative

Lets see statement 2, S in Negative.

S^3 will be negative.
T^4 will always be positive.
R^5 will be either positive or negative, depending on the value of R which is unknown.
Hence its INSUFFICIENT.

Statement 1, RT is negative. This implies either R or T is negative. Since it doesn't touch S we, in anycase, can't say what will be S^3, hence even this is INSUFFICIENT.

Now taking both together.
S^3 is negative (from statement 1).
One out of R or T is also negative (which one is not known.) Hence its NOT POSSIBLE to find an answer using both the statements as well.