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M03 Q14

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M03 Q14 [#permalink] New post 08 Nov 2008, 08:28
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Is G < K ?

(1) G>K^2
(2) G and K are positive integers

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[Reveal] Spoiler:

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 :roll:
I am completely confused :?
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Re: M03 Q14 [#permalink] New post 08 Nov 2008, 10:54
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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...
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Re: M03 Q14 [#permalink] New post 08 Nov 2008, 14:42
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 :roll:
I am completely confused :?

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Re: M03 Q14 [#permalink] New post 09 Nov 2008, 02:52
masuhari wrote:
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 :oops:
Thanks masuhari and GMAT TIGER :)
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Re: M03 Q14 [#permalink] New post 13 May 2010, 05:15
Option C... easy one.... But easy to miss (and mark A incorrectly ) if you do not consider equal fractions for both numbers.
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Re: M03 Q14 [#permalink] New post 13 May 2010, 05:53
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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.

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Re: M03 Q14 [#permalink] New post 13 May 2010, 18:25
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.

Both A&B together are insufficient.
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Re: M03 Q14 [#permalink] New post 13 May 2010, 23:34
Hi,

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.

Both A&B together are insufficient.

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Re: M03 Q14 [#permalink] New post 15 May 2010, 04:22
I also got C. But missed to consider 1/2 and 1/3 option :(
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Re: M03 Q14 [#permalink] New post 03 Jan 2011, 00:33
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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Re: M03 Q14 [#permalink] New post 17 May 2011, 04:09
(1)

G > K^2

G = 1/2

K = 1/2

G = 1

K = 1/2

Insufficient

(2)

Not Sufficient

(1) + (2)

G = 4, K = 1

G = 10, K = 3

Sufficient

Answer - C
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Re: M03 Q14 [#permalink] New post 17 May 2011, 05:37
C is the answer,
always consider fractions when nit mentioned integers.
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Re: M03 Q14 [#permalink] New post 18 May 2011, 21:40
I got C thinking about + and - cases, but I think that will not work and we have to consider fractions.
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Re: M03 Q14 [#permalink] New post 19 May 2011, 21:36
Is G < K

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.
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Re: M03 Q14 [#permalink] New post 22 May 2011, 08:31
C since (2) eliminates the fractions that were available in (1)
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Re: M03 Q14 [#permalink] New post 21 Sep 2011, 07:33
Consider case : g=1/3, k=1/2 therefore k^2 = 1/4. Here g>k^2 but g<k; Hence Answer should be E.
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Re: M03 Q14 [#permalink] New post 07 May 2014, 07:51
Um, you guys, where can I find the answer options? I'm new and feel really stupid as obviously everyone else can find them but me...

Thanks!
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Re: M03 Q14 [#permalink] New post 07 May 2014, 07:58
Expert's post
financebunny wrote:
Um, you guys, where can I find the answer options? I'm new and feel really stupid as obviously everyone else can find them but me...

Thanks!


This is a data sufficiency question. Options for DS questions are always the same.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Hope this helps.
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Re: M03 Q14 [#permalink] New post 07 May 2014, 07:59
Expert's post
LiveStronger wrote:
Is G < K ?

(1) G>K^2
(2) G and K are positive integers

Source: GMAT Club Tests - hardest GMAT questions

[Reveal] Spoiler:

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 :roll:
I am completely confused :?


m03 q14

Is x < y ?

(1) x>y^2 --> if x=2 and y=1 then the answer is NO but if x=\frac{1}{3} and y=\frac{1}{2} then the answer is YES. Not sufficient.

(2) x and y are positive integers. Not sufficient on its own.

(1)+(2) Since from (2) x and y are positive integers then from (1) x>y^2\geq{1}, which means that x>y, so the answer to the question is NO. Sufficient.

Answer: C.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M03 Q14 [#permalink] New post 07 May 2014, 08:38
Bunuel: Yes, thank you so much! :)

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Re: M03 Q14   [#permalink] 07 May 2014, 08:38
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