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OE is S1 is true even if G =1/10 and K=1/10 ; therefore, it is not sufficient. S2 is obviously insufficient, but together they are sufficient, because then G> K^2 > 1 and this implies G> K

My answer is A

When 1. G = 2, K =1 G >K^2 is TRUE, Is G<K - No/False - Suff 2. G = 1/2, K =1/2 G >K^2 is TRUE, Is G<K - No/False - Suff 3. G = 1/2, K =-1/2 G >K^2 is TRUE, Is G<K - No/False - Suff 4. G = 9, K = -2 G >K^2 is TRUE, Is G<K - No/False - Suff 5. G = 1/3, K = -1/3 G >K^2 is TRUE, Is G<K - No/False - Suff 6. G = 1/3, K = 0 G >K^2 is TRUE, Is G<K - No/False - Suff

I could not think of an example which will the contradict the answer

Am I missing something here I am completely confused

Livestronger, you were close enough on your number substitution sample #2. But why did you chose the same numbers for both G and K.

If you had chosen G = 1/3 and K = 1/2. It satisfies, G > K^2. And clearly answers if G < K. it appears sufficient. But let's try with another substitution. G = 1 and K = 1/2. Again G > K^2 is satisfied. But this time G < K gives you a different answer. Now that's the crucial part(IMHO) of DS. If you can come up with 2 different results for the original question stem, then that particular choice is not Sufficient. Safely POE on that..so A and D are gone.

statement #2, seems obvious. It is definitely not sufficient to answer whether G < K. All we know if G > 0 and K > 0 and that they are integers. So B is gone.

Now together, they tell a story. Statement 2 affirms us that G and K are not fractions. So our only other contradiction about statement 1 is eliminated. Which leaves us with the fact that if G > 0 and K > 0 and both are integers, then if G > K^2, then always G is > K.

HTH...
_________________

excellence is the gradual result of always striving to do better

From statement 1: 1. If G = 0.5 and K = 0.7, G > K^2. here G < K. 2. If G = 5 and K = 2, G > K^2. Here G > K.

NSF.

From statement 2:

Either one is greater or smaller than the other. NSF.

From Statement 1 and 2:

G > K or G is not smaller than K. So Suff....

//C//

LiveStronger wrote:

Is G< K ? 1. G>K^2 2. G and K are positive integers

OA is C OE is S1 is true even if G =1/10 and K=1/10 ; therefore, it is not sufficient. S2 is obviously insufficient, but together they are sufficient, because then G> K^2 > 1 and this implies G> K

My answer is A

When 1. G = 2, K =1 G >K^2 is TRUE, Is G<K - No/False - Suff 2. G = 1/2, K =1/2 G >K^2 is TRUE, Is G<K - No/False - Suff 3. G = 1/2, K =-1/2 G >K^2 is TRUE, Is G<K - No/False - Suff 4. G = 9, K = -2 G >K^2 is TRUE, Is G<K - No/False - Suff 5. G = 1/3, K = -1/3 G >K^2 is TRUE, Is G<K - No/False - Suff 6. G = 1/3, K = 0 G >K^2 is TRUE, Is G<K - No/False - Suff

I could not think of an example which will the contradict the answer

Am I missing something here I am completely confused

Livestronger, you were close enough on your number substitution sample #2. But why did you chose the same numbers for both G and K.

If you had chosen G = 1/3 and K = 1/2. It satisfies, G > K^2. And clearly answers if G < K. it appears sufficient. But let's try with another substitution. G = 1 and K = 1/2. Again G > K^2 is satisfied. But this time G < K gives you a different answer. Now that's the crucial part(IMHO) of DS. If you can come up with 2 different results for the original question stem, then that particular choice is not Sufficient. Safely POE on that..so A and D are gone.

statement #2, seems obvious. It is definitely not sufficient to answer whether G < K. All we know if G > 0 and K > 0 and that they are integers. So B is gone.

Now together, they tell a story. Statement 2 affirms us that G and K are not fractions. So our only other contradiction about statement 1 is eliminated. Which leaves us with the fact that if G > 0 and K > 0 and both are integers, then if G > K^2, then always G is > K.

HTH...

Ahhh, I just simply couldn't think of that example Thanks masuhari and GMAT TIGER

Is G < K ? (1) G > K^2 (2) G and K are positive integers.

Statement 2: G and K are positive integers. G = 1 and K = 2 => G < K is True. G = 2 and K = 1 => G < K is False. Statement 2 is Clearly not sufficient by itself.

This eliminates answers B and D.

Remember this statement says G and K are positive and G and K are integers. So, G and K are not fractions.

Statement 1: G > K^2 G = 5 and K = 2: 5 > 4, so G > K^2 is True, but G < K is False. G = 1/3 and K = 1/2: 1/3 > 1/4, so G > K^2 is True, but G < K is True. Statement 2 is Clearly not sufficient by itself.

This eliminates the answer A.

Take Both Statement 1 and 2: G > K^2 and G and K are positive integers. In this case, G and K are position and are integers, so cannot be fractions.

Examples: G = 2 and K = 1: 2 > 1, so G > K^2 is True, but G < K is False. G = 5 and K = 2: 5 > 4, so G > K^2 is True, but G < K is False. G = 101 and K = 10: 101 > 100, so G > K^2 is True, but G < K is False.

So, the result is consistent and both statement 1 and 2 jointly are sufficient to answer is G < K.

The answer choice is C.

Do not forget to award me with Kudos+1, if this helps you!! Cheers!!!

The text in red contains an error. If you use information from both statements, you should take such G that follows S1 (G>K^2). The text in red clearly contradicts S1, so this is not a valid proof. S1+S2 is sufficient, OA is C.

I hope it helps you.

lnarayanan wrote:

We know that -

is G<K

Opt - A) G>K^2 - alone not sufficient

Opt - B) G and K are +ve integers - alone not sufficient

Combaining together, (two possibilities ) ex) G=3 , K = 10 implies 3>100 = False G=20 , K = 3 implies 20>9 = True. So two answers are contradicting each other.

it took me so much time, and so many examples to get this one right! Lesson learnt: use fractions when it is NOT given that the numbers are "integers".
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1) G > K to power 2 2) G and K are positive integers.

Lets take (1)

Both G and K can be positive or negative integers or even fractions. (a)If both are positive, when G >K^2, => G > K (b)If both are negative, when G cannot be greater than K^2 which will then become positive (c)When both are fractions, we can get all kinds of results.

So, (1) is not enough for us to get an answer - Choce (A) and (D) are ruled out.

Now lets take (2) Both G and K are postive integers, so G < K only when exact values of G and K are known, which we dont know so (2) is not enough and option (B) is ruled out.

Now its either (C) or (E).

Lets evaluate C -

If both are positive integers and G>K^2, as explained in (a) above, this means that G > K always. Hence (C) is the choice.