Data Sufficiency

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

Is x^3-y^3 > x^2+xy+y^2?

1) x > y + 1
2) 0 < y < x

High-level question. Congrats, Max!
$${x^2} + xy + {y^2} = \underbrace {{x^2} + 2 \cdot x \cdot {y \over 2} + {{{y^2}} \over 4}}_{{{\left( {x + {y \over 2}} \right)}^{\,2}}} + {{3{y^2}} \over 4} \ge 0\,\,\,\,\,\,\,\left[ {\,\, \Rightarrow \,\,\,\,\,{x^2} + xy + {y^2} = 0\,\,\, \Leftrightarrow \,\,\left\{ \matrix{
\,{y^2} = 0 \hfill \cr
\,x + {y \over 2} = 0 \hfill \cr} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {x,y} \right) = \left( {0,0} \right)\,\,\,\,\,\,\left( * \right)\,\,} \right]$$
[The "completing the squares" technique (above) is carefully explained in our course!]
$${x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\,\,\,\,\left( {**} \right)$$
$${x^3} - {y^3}\,\,\,\mathop > \limits^? \,\,\,{x^2} + xy + {y^2}\,$$
$$\left( 1 \right)\,\,x > y + 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{
\,\left( {x,y} \right) \ne \left( {0,0} \right)\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{x^2} + xy + {y^2} > 0\,\,\,\,\left( {***} \right) \hfill \cr
\,\,x - y > 1\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,\,\,\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\,\,\, > \,\,\,1 \cdot \left( {{x^2} + xy + {y^2}} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\,$$
$$\left( 2 \right)\,\,0 < y < x\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {3,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

The correct answer is therefore (A).


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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The original condition x^3-y^3 > x^2+xy+y^2 is equivalent to x > y + 1 as shown below:
x^3-y^3 > x^2+xy+y^2
=> (x-y)(x^2+xy+y^2) > x^2+xy+y^2
=> x - y > 1 after dividing both sides by x^2+xy+y^2, since x^2+xy+y^2 > 0.

Since the final inequality is equivalent to x > y + 1, condition 1) is sufficient.

Condition 2)
If x = 3 and y = 1, then x - y = 2 > 1, and the answer is 'yes'.
If x = 1 and y = 1/2, then x - y = 1/2 < 1, and the answer is 'no'.
Since it doesn't give a unique answer, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A