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Is x > k?

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by Nijeesh » Mon Mar 22, 2010 10:09 am
Is x > k?
(1) 2^x "¢ 2^k = 4
(2) 9^x "¢ 3^k = 81
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Source: — Data Sufficiency |
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by Stuart@KaplanGMAT » Mon Mar 22, 2010 10:29 am
Nijeesh wrote:Is x > k?

(1) 2^x "¢ 2^k = 4

(2) 9^x "¢ 3^k = 81
Step 1: Analyze the Problem

We see "is", we think "yes/no question: if I can answer with a definite yes or definite no, sufficient; if I answer with a maybe or sometimes or not sure, insufficient".

What do I know? Absolutely nothing! We have 2 variables and 0 equations, so 2 equations would be nice.

Step 2: Evaluate the Statements

(1) 2^x * 2^k = 4

Using the rules of exponents, we now know that:

2^(x+k) = 2^2

so:

x + k = 2

No clue which one is bigger, so insufficient. Eliminate A and D.

(2) 9^x * 3^k = 81

By the same process as above, we know that:

(3^2)^x * 3^k = 3^4

3^2x * 3^k = 3^4

3^(2x + k) = 3^4

2x + k = 4

Again, no clue which one is bigger, so insufficient. Eliminate B.

Together:

2 equations, 2 unknowns, we can solve for x and k: sufficient. Choose C.

* * *

This question is a great illustration of the power of the "number of equations for number of unknowns" rule; the better you understand and apply that rule, the less math you'll have to do.

For example, if on this question we quickly realized that each statement would give us one distinct linear equation, and that each equation contained our two variables (and no others), we could have gone directly to "C" without actually solving each statement.
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by akahuja143 » Mon Mar 22, 2010 8:35 pm
Nice breakdown.. two equations two unknowns ..

Thanks for the explanation !!
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by tgou008 » Sun Feb 20, 2011 8:21 am
I got to C, but used a different technique; testing numbers.

Is X>K? = Y/N question

(1) 2^x.2^k = 4

Possible value combinations for x & k that total to four include:
x = 1, k = 1. Here x = k
x = 0, k = 2. Here x < k
x = 2, k = 0 . Here x > k

As there are multiple possibilities for X>K, we can eliminate answers A and D.

(2) 9^x.3^k = 81

Possible value combinations for x & k that total to 81 include:
x = 2, k = 0. Here x > k
x = 0, k = 4. Here x < k
x = 1, k = 2 . Here x < k

Once again there are multiple possibilities for X>K, so we can eliminate B. Must now choose between C and E.

For it to be C, there must be 1 X vs K combination that is common across both (1) and (2). We can see that there is, the only combination common across both (1) and (2) is x= 2, k=0, which is sufficient as we can now see that x > k.

Therefore answer = C

That took me ~2mins to do. Not sure whether this or the other approach is quicker. What level difficultly would you say this question is?

Thanks
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by Night reader » Sun Feb 20, 2011 9:08 am
@tgou008, way to go BUT I think Stuart has hastened himself
st(1) 2^x "¢ 2^k = 4 --> 2^x "¢ 2^k = |2|^2 AND x+k=2 isn't a unique solution, Not Sufficient
st(2) 9^x "¢ 3^k = 81 --> again :( |3|^2x "¢ 3^k = 3^4, so obviously 2x+k=4 isn't a unique solution here, Not Sufficient

Only after combining st(1&2) we can assume that x and k have the unique linear relationship, hence x+k=2 And 2x+k=4, x=2 and k=0, choose C
Nijeesh wrote:Is x > k?
(1) 2^x "¢ 2^k = 4
(2) 9^x "¢ 3^k = 81
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by pankajks2010 » Wed Apr 06, 2011 10:28 pm
tgou008 wrote:I got to C, but used a different technique; testing numbers.

Is X>K? = Y/N question

(1) 2^x.2^k = 4

Possible value combinations for x & k that total to four include:
x = 1, k = 1. Here x = k
x = 0, k = 2. Here x < k
x = 2, k = 0 . Here x > k

As there are multiple possibilities for X>K, we can eliminate answers A and D.

(2) 9^x.3^k = 81

Possible value combinations for x & k that total to 81 include:
x = 2, k = 0. Here x > k
x = 0, k = 4. Here x < k
x = 1, k = 2 . Here x < k

Once again there are multiple possibilities for X>K, so we can eliminate B. Must now choose between C and E.

For it to be C, there must be 1 X vs K combination that is common across both (1) and (2). We can see that there is, the only combination common across both (1) and (2) is x= 2, k=0, which is sufficient as we can now see that x > k.

Therefore answer = C

That took me ~2mins to do. Not sure whether this or the other approach is quicker. What level difficultly would you say this question is?

Thanks
Hi tgou008,

Its good that you have considered all the possible integers as solution for x & k. However, as it is not specified in the question whether x & k are integers or not, it would be worthwhile considering values of x & k in fractions. Although, going to that detail is not required in this question. However, it would really help in GMAT if we do not just consider integers as solutions.

Thanks!!
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by pankajks2010 » Wed Apr 06, 2011 10:52 pm
Night reader wrote:@tgou008, way to go BUT I think Stuart has hastened himself
st(1) 2^x "¢ 2^k = 4 --> 2^x "¢ 2^k = |2|^2 AND x+k=2 isn't a unique solution, Not Sufficient
st(2) 9^x "¢ 3^k = 81 --> again :( |3|^2x "¢ 3^k = 3^4, so obviously 2x+k=4 isn't a unique solution here, Not Sufficient

Only after combining st(1&2) we can assume that x and k have the unique linear relationship, hence x+k=2 And 2x+k=4, x=2 and k=0, choose C
Nijeesh wrote:Is x > k?
(1) 2^x "¢ 2^k = 4
(2) 9^x "¢ 3^k = 81
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Hey Night reader,
Your post definitely helps me in considering the negative square roots for a number (like 2 & -2 for 4). Although, in this question, it was not required as its the power which we are equating to get the equations. Nonetheless, its good to accustom one's mind to consider both positive & negative roots.

Thanks!!
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by sureng » Thu Apr 14, 2011 8:53 pm
BTW when i received email i got it like below:


DATA SUFFICIENCY QUESTION:
Is x > r?

1. 2^x * 2^k = 4
2. 9^x * 3^k = 81


Instead of k i see r and thought answer as E:)
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by piyush2112 » Sun May 15, 2011 1:36 am
tgou008 wrote:I got to C, but used a different technique; testing numbers.

Is X>K? = Y/N question

(1) 2^x.2^k = 4

Possible value combinations for x & k that total to four include:
x = 1, k = 1. Here x = k
x = 0, k = 2. Here x < k
x = 2, k = 0 . Here x > k

As there are multiple possibilities for X>K, we can eliminate answers A and D.

(2) 9^x.3^k = 81

Possible value combinations for x & k that total to 81 include:
x = 2, k = 0. Here x > k
x = 0, k = 4. Here x < k
x = 1, k = 2 . Here x < k

Once again there are multiple possibilities for X>K, so we can eliminate B. Must now choose between C and E.

For it to be C, there must be 1 X vs K combination that is common across both (1) and (2). We can see that there is, the only combination common across both (1) and (2) is x= 2, k=0, which is sufficient as we can now see that x > k.

Therefore answer = C

That took me ~2mins to do. Not sure whether this or the other approach is quicker. What level difficultly would you say this question is?

Thanks
Just wondering why we cant assume negative values for x and k?
x=2+n and k=0-n; will satisfy x+k=2 for all integers.
so please let me know why we are choosing only values equal or greater than 0.
Am i missing something here?

Thanks in advance.

Regards
Piyush
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by sayanchakravarty » Mon Jun 06, 2011 8:09 pm
C

x + k = 2

2x + k = 4

2 equations. 2 unknowns. Therefore, solvable.
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by cans » Mon Jun 06, 2011 8:31 pm
Is x > k?
(1) 2^x "¢ 2^k = 4
(2) 9^x "¢ 3^k = 81
a)2^(x+k) = 2^2
x+k=2
insufficient.
b)3^(2x+k) = 3^4
2x+k=4
Insufficient
both a&b) two equations, 2 variables. x=2 and k=0
Sufficient
IMO C
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by aftableo2006 » Mon Jun 06, 2011 8:45 pm
C is the correct answer
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by d.maguy » Fri Jun 10, 2011 1:59 am
My answer is C.
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by abhisays » Tue Aug 02, 2011 11:50 pm
d.maguy wrote:My answer is C.
Yes, C is the right answer.
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by Deependra1 » Wed Aug 31, 2011 11:30 am
ANSWER: C
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