Is 10m 5n - k?

Question

Is 10m > 5n - k?

(1) n = 2m
(2) |k| = -k

Kurtosis · Accepted Answer

St1: n = 2m 
Is 5n > 5n - k 
Is 0 > -k ? We do not know this as it depends on the value of k. If k is -ve or 0 then the answer is no. If k is +ve then the answer is yes.
Not sufficient

St2: |k| = -k 
This implies that -k is positive or 0. Therefore k has to be negative value or 0.
But is 10m > 5n - k ? We cannot answer this because we have 3 unknown values.
Not sufficient

Combining St1 and St2 we have,
Is 0 > -k ?
From St2, if k is -ve in value, then is 0 > -(-k) ? Answer is No 
but if k is 0, then is 0 > 0 ? Answer is still No.
Since we get a unique answer 'No' when both the statements are combined, it is sufficient.

Answer: C

MathRevolution · Answer

Forget conventional ways of solving math questions. 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 independent equations ensures a solution.

Is 10m > 5n - k?

(1) n = 2m
(2) |k| = -k

In the original condition, there are 3 variables(m,n,k), which should match with the number of equations. So you need 3 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer.
When 1) & 2), 10m>10m-k? -> -k<0? -> k>0?. In 2), k<=0, which is no and sufficient.
Therefore, the answer is C.

 For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

stonecold · Answer

Here is the My approach 
Here we are not given much info in the question stem apart from 10m>5m-k to be proven
Statement 1 => no clue of k => not sufficient 
statement 2 => it can be seen from this that k is negative but we have no clue of m and n => not sufficient 
combining them => to be proven => 10m>5m-k as n=2m => 10m>10m-k or the question is asking us is k>0 or not which as per statement 2 is false.
hence C is sufficient

infinitemac · Answer

Why does statement 2 imply that k can be 0? I understand that it can imply that k is negative, but not understanding why k can be 0.

Thanks!

Bunuel · Answer

If k =0, then |k| = |0| = 0 and -k = -0 = 0, so |k| = -k = 0.

fskilnik · Answer

10m\,\,\mathop > \limits^? \,\,5n - k

\left( 1 \right)\,\,\,n = 2m\,\,\,\,\left\{ \matrix{
 \,{\rm{Take}}\,\,\,\left( {n,m,k} \right) = \left( {0,0,0} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\, \hfill \cr 
 \,{\rm{Take}}\,\,\,\left( {n,m,k} \right) = \left( {2,1,1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\, \hfill \cr} \right.\,

\left( 2 \right)\,\,\,\,\left| k \right| = - k\,\,\,\, \Leftrightarrow \,\,\,\,k \le 0

\left\{ \matrix{
 \,{\rm{Take}}\,\,\,\left( {n,m,k} \right) = \left( {0,0,0} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\, \hfill \cr 
 \,{\rm{Take}}\,\,\,\left( {n,m,k} \right) = \left( {0,1, - 1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\, \hfill \cr} \right.

\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
 10m\,\,\mathop > \limits^? \,\,5\left( {2m} \right) - k\,\,\,\,\, \Leftrightarrow \,\,\,\,k\,\,\mathop > \limits^? \,\,0 \hfill \cr 
 k \le 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

bumpbot · Answer

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Is 10m > 5n - k?

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