Data Sufficiency

[Math Revolution GMAT math practice question]

If m and n are positive integers, is m + n an odd number?

1) m/n is an even number
2) m or n is an even number

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If m and n are positive integers, is m + n an odd number?

1) m/n is an even number
2) m or n is an even number

\[m,n\,\,\, \geqslant 1\,\,\,{\text{ints}}\,\,\,\,\left( * \right)\]
$$m + n\,\,\,\,\mathop = \limits^? \,\,{\text{odd}}\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\boxed{\,\,\,?\,\,\,:\,\,\,\left( {m\,\,{\text{odd}}\,,\,\,n\,\,{\text{even}}} \right)\,\,\,{\text{or}}\,\,\,{\text{vice - versa}\,\,}\,\,}$$
\[\left( 1 \right)\,\,\,\frac{m}{n} = {\text{even}}\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {4,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,m\,\,{\text{even}}\,\,\,{\text{or}}\,\,\,n\,\,{\text{even}}\,\,\,\,\left\{ \begin{gathered}
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,\left( {m,n} \right) = \left( {4,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\text{E}} \right)\]

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

Regards,
Fabio.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (m and n) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since m/n is even, m/n = 2k and m = (2k)n for some positive integer k.
If m = 2 and n = 1, then m + n = 3, is an odd integer and the answer is 'yes'.
If m = 4 and n = 2, then m + n = 6 is an even integer and the answer is 'no'.
Since we do not obtain a unique answer, conditions 1) & 2) are not sufficient when considered together.

Therefore, E is the answer.
Answer: E

In cases where 3 or more additional equations are required, such as for original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

For m+n = odd number either m or n must be odd number e.g 5+4=9 ; 2+3=5, Even +Odd =Odd

Statement 1
$$\frac{m}{n}is\ an\ Even\ \ number$$
$$division\ of\ two\ even\ number=Even\ number$$
$$sumof\ two\ even\ number=Even\ number$$
$$dividing\ even\ number\ by\ some\ odd\ numbers\ also=Even\ number$$
$$but\ sum\ of\ even\ and\ odd\ number\ =\ odd\ number$$
Hence statement 1 is INSUFFICIENT.

Statement 2
m or n is an even number
m = even , n= odd
If m=odd , n= even
Let m=odd, n= even
Sum of m+n such that m=odd and n = even becomes
$$2n+\left(2m+1\right)$$
$$2n+2m+1$$
$$2\left(n+m\right)+1$$
but m and n is just some integer k, hence sum of an even plus an odd number is always = 2k+1
for some integers k which is m+n and it is always = odd number
Thus m+n = Odd number if m or n is an even number
$$Statement\ 2\ is\ INSUFFICIENT$$