GMAT Challenge Series (700+): Q28

Chalkboard-Stats.jpg

Question:

In a set of 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median?

(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10.

(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.

MBA Wisdom's Answer:

$$\text{Let the 5 numbers be: } \textit{a} \leq \textit{b} \leq \textit{c} \leq \textit{d} \leq \textit{e}$$ $$\text{Median} = \textit{c}$$ $$\text{Mean} = {{\textit{a} + \textit{b} + \textit{c} + \textit{d} + \textit{e}} \over {5}} $$ $$$$ $$\text{We know that: } \textit{e} = {\textit{c} + {4}}$$ $$\text{Mean} = {{\textit{a} + \textit{b} + \textit{c} + \textit{d} + \left(\textit{c} + {4}\right)} \over {5}} $$ $$\text{Mean} = {{\textit{a} + \textit{b} + {2}\textit{c} + \textit{d} + {4}} \over {5}} $$ $$$$ $$\text{We need to work out if:}$$ $${{\textit{a} + \textit{b} + {2}\textit{c} + \textit{d} + {4}} \over {5}} > \textit{c}$$ $${\textit{a} + \textit{b} + {2}\textit{c} + \textit{d} + {4}} > {5}\textit{c}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $$$$ $$\text{Statement 1:}$$ $$\textit{e} = {\textit{c} + {4}}$$ $$\textit{e} + \textit{c} = {34}$$ $$\textit{c} = {15} \text{ and }\textit{e} = {19}$$ $$$$ $$\text{We need to work out if:}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {45}$$ $${\textit{a} + \textit{b} + \textit{d}} > {41}$$ $$$$ $$\text{Test at maximum values: } \textit{a} = {15} \textit{ b} = {15} \textit{ d} = {19}$$ $${{15} + {15} + {19}} > {41}$$ $${49} > {41} \text{ TRUE}$$ $$$$ $$\text{Test at non-maximum values: } \textit{a} = {0} \textit{ b} = {0} \textit{ d} = {15}$$ $${{0} + {0} + {15}} > {41}$$ $${15} > {41} \text{ FALSE}$$ $$$$ $$\text{We cannot determine whether the mean is greater than the median}$$ $$\text{Statement 1 is NOT SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\textit{a} = {\textit{c} - {10}}$$ $$$$ $$\text{We need to work out if:}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $${\left(\textit{c} - {10}\right) + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $${\textit{b} + \textit{d}} > {{2}\textit{c} + {6}}$$ $$$$ $$\text{Test at maximum values: } \textit{b} = \textit{c} \text{ and}\textit{ d} = \textit{e} = \textit{c} + {4}$$ $${\textit{c} + \left(\textit{c} + {4}\right)} > {{2}\textit{c} + {6}}$$ $${{2}\textit{c} + {4}} > {{2}\textit{c} + {6}} \text{ FALSE}$$ $$$$ $$\text{LHS will never be greater than RHS}$$ $$\text{Therefore, the mean is NOT greater than the median}$$ $$\text{Statement 2 is SUFFICIENT}$$

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.