BTGmoderatorDC wrote:If x, y, and z are positive numbers, Is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

Statement 1: x < 2z < y
Let x=2 and y=10.
In this case:
2 < 2z < 10
1 < z < 5.
If z=4, then it is between x=2 and y=10.
If z=1.5, then it is not between x=2 and y=10.
INSUFFICIENT.

Statement 2: 2x < z < 2y
Let x=2 and y=10.
In this case:
2*2 < z < 2*10
4 < z < 20
If z=5, then it is between x=2 and y=10.
If z=19, then it is not between x=2 and y=10.
INSUFFICIENT.

Statements combined:
Inequalities can be ADDED TOGETHER.
Adding together x < 2z < y and 2x < z < 2y, we get:
(x+2x) < (2z+z) < (y+2y)
3x < 3z < 3y
x < z < y.
Thus, z is between x and y.
SUFFICIENT.

The correct answer is C.

BTGmoderatorDC wrote:If x, y, and z are positive numbers, Is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y
Source: GMAT Prep

(At least to my students) There is no need to add both statements to deal with (1+2), although this is a very good idea used by all other experts above.
Could you find another approach? (One possibility is presented in my solution.)
\[x,y,z\,\,\, > 0\,\,\,\left( * \right)\]
\[?\,\,\,\,:\,\,\,\,z\,\,{\text{between}}\,\,x\,\,{\text{and}}\,\,y\,\,\,\]
\[\left( 1 \right)\,\,\,x < 2z < y\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,2,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,2x < z < 2y\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,4} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\,x\,\,\mathop < \limits^{\left( * \right)} \,\,\,2x\,\,\mathop < \limits^{\left( 2 \right)} \,\,z\,\,\mathop < \limits^{\left( * \right)} \,\,\,2z\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x < z < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]

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

Regards,
Fabio.