$$let\ the\ numbers\ be\ x_1,\ x_2,\ x_3,\ x_4,x_5,\ x_6$$
range = (greatest - smallest) numbers
range = $$x_6-x_1$$ = 16
$$mean=\text{}\frac{£FX}{£F}=\frac{\left(x_1+x_2+x_3+x_4+x_5+x_6\right)}{6}=7$$
$$Hence\ x_3+x=6\cdot2=12$$
mode of 7 i.e 7 is the most occurring number, it comes at least more than once.

Statement 1
The sum of the two smallest number is $$\frac{1}{5}th\ $$ the sum of the the two greatest numbers
$$Therefore,\ x_1+x_2=\frac{1}{5}\left(x_5+x_6\right)_{ }$$
From question analysis out of six numbers, 3 numbers will be less than 6 the third number must be 5 to make the median =6 an 7 has to appear at least twice.
$$let\ x_1=x$$
$$let\ x_2=y$$
$$x+y=\frac{1}{5}\left(7+\left(16+x\right)\right)$$
$$5x+5y=7+16+x$$
$$4x+5y=23..............eqn\ \left(i\right)$$
$$if\ mean\ =\frac{\left(x_1+x_2+x_3+x_4+x_5+x_6\right)}{6}=7$$
The sum of 6 numbers =6*7= 42
$$x+y+5+7+7+\left(16+x\right)=42$$
$$x+y+19+16+x=42$$ $$2x+y+35=42$$
$$2x+y=42-35=9...................eqn\left(ii\right)$$
multiply eqn(ii) by 2
$$\left(2x\cdot2\right)+\left(y\cdot2\right)=9\cdot2$$
$$4x+2y=14$$
subtracting eqn(ii) from i
$$4x+2y=14$$
$$4x-5y=23$$
$$-\frac{3y}{-3}=\frac{-9}{-3}=3$$
from eqn (ii)
2x +y = 9
2x +3 = 9 $$\frac{2x}{2}=\frac{9-3}{2}=\frac{6}{2}=3$$ $$x=3.65$$
$$x_1=x=2$$
$$x_2=y=3$$
$$x_6=16+x_1=18$$
hence greatest number = 18, statement 1 is SUFFICIENT

Statement 2
The middle two numbers e 5 and 7
$$x_1,x_2,5,7,7,x_6$$
$$x_1,x_2,5,7,7,x_6$$
$$x_1,x_2,5,7,7,x_6$$
$$from\ mean\ x_1+x_2+5+7+7+16+x_1=42$$
$$2x_1+x_2=42-35=7$$
$$if\ x_1=x\ and\ x_2=y$$
$$2\left(x\right)+\left(y\right)=7$$
$$2x+y=7$$
information given is not enough to solve for x and y hence statement 2 is INSUFFICIENT
$$answer\ is\ option\ A$$

BTGmoderatorDC wrote:Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Source: Princeton Review

Very nice problem (in which the "structured candidate" has a considerable advantage)!
$${a_1} \le {a_2} \le \ldots \le {a_6}\,\,\,\,\,\left\{ \matrix{
\,{a_6} = {a_1} + 16\, \hfill \cr
\,{a_3} + {a_4} = 12 \hfill \cr
\,\sum\nolimits_6 {\, = } \,\,\,42 \hfill \cr
\,7{\rm{s}}\,\,{\rm{appear}}\,\,{\rm{most}}\, \hfill \cr} \right.$$
$$? = {a_6}$$

We MUST start with statement (2) : it is easier and it will help us "gain knowledge and sensitivity" to what is really going on!

$$\left( 2 \right)\,\,\,\,{a_1}\,\,\,{a_2}\,\,\,5\,\,\,7\,\,\,7\,\,\,{a_6}\mathop \ne \limits^{\left( {**} \right)} 7\,\,\,\,\,\left\{ \matrix{
\,{a_6} = {a_1} + 16\, \hfill \cr
\,{a_3} + {a_4} = 12\,\,\,\,\left( {{\rm{sure}}} \right) \hfill \cr
\,\sum\nolimits_6 {\, = } \,\,\,42\,\,\,\left( * \right) \hfill \cr
\,7{\rm{s}}\,\,{\rm{appear}}\,\,{\rm{most}}\,\,\,\,\left( {\,\left( {**} \right)\,\,{a_6} = 7\,\,\, \Rightarrow \,\,\,{a_1} = 7 - 16 = - 9\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_2} = 25 > {a_3}\,\,{\rm{impossible}}\,} \right)\, \hfill \cr} \right.$$
$$\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {{a_1},{a_6},{a_2}} \right) = \left( {1,17,5} \right)\,\,\,{\rm{viable}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{a_6} = \,\,17 \hfill \cr
\,{\rm{Take}}\,\,\left( {{a_1},{a_6},{a_2}} \right) = \left( {2,18,3} \right)\,\,\,{\rm{viable}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{a_6} = \,\,18\,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.$$

$$\left( 1 \right)\,\,\,5\left( {{a_1} + {a_2}} \right) = {a_5} + {a_6}\,\,\,\,\left\{ \matrix{
\,{a_6} = {a_1} + 16\, \hfill \cr
\,{a_3} + {a_4} = 12 \hfill \cr
\,\sum\nolimits_6 {\, = } \,\,\,42 \hfill \cr
\,7{\rm{s}}\,\,{\rm{appear}}\,\,{\rm{most}}\, \hfill \cr} \right.$$
$${a_3} = 7\,\,\,\, \Rightarrow \,\,{a_4} = 12 - 7 = 5 < {a_3}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}$$
$${a_{`5}} = {a_6} = 7\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,{a_1} = 7 - 16 = - 9 \hfill \cr
\,{a_2} = \sum\nolimits_6 {\, - \left[ {{a_1} + \left( {{a_3} + {a_4}} \right) + {a_5} + {a_6}} \right] = } \,\,42 - \left( { - 9 + 12 + 2 \cdot 7} \right) = 25\,\,\, > \,\,\,{a_5} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}$$
$${\rm{Hence}}:\,\,\,\,{a_1}\,\,\,{a_2}\,\,\,{a_3}\,\,\,7\,\,\,7\,\,\,{a_6} \ne 7\,\,\,\,\,\left\{ \matrix{
\,{a_6} = {a_1} + 16\, \hfill \cr
\,{a_3} + {a_4} = 12\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{a_3} = 12 - 7 = 5 \hfill \cr
\,\sum\nolimits_6 {\, = } \,\,\,42\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{a_1} + {a_2} + {a_6} = 42 - {a_3} - \left( {{a_4} + {a_5}} \right) = 42 - 5 - 14 = 23 \hfill \cr
\,5\left( {{a_1} + {a_2}} \right) = {a_5} + {a_6}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,5\left( {23 - {a_6}} \right) = 7 + {a_6}\,\,\,\,\, \Rightarrow \,\,\,\,? = {a_6}\,\,\,{\rm{unique}} \hfill \cr} \right.$$

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

Regards,
Fabio.