Data Sufficiency

[GMAT math practice question]

If x, y, z are positive integers, is xyz ≥ 64?

1) xy ≥ yz ≥ zx ≥ 16
2) x + y + z = 64

=>

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 3 variables (x, y, z) and 0 equations, E is most likely to be the answer. So, we should consider each condition on its own first. As condition 1) includes 3 equations, we should consider it first.

Condition 1)
Since xy ≥ 16, yz ≥ 16, and zx ≥ 16, we have (xyz)^2 ≥ 16^3 or xyz ≥ 64.
Condition 1) is sufficient.

Condition 2)
If x = 20, y = 20 and z = 24, then xyz = 9600 ≥ 64 and the answer is 'yes'.
If x = 1, y = 1 and z = 62, then xyz = 62 < 64 and the answer is 'no'.
Since condition 2) does not yield a unique answer, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

This is a CMT(Common Mistake Type) 4(A) question. If a question is from one of the key question areas and C should be the answer, CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

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

If x, y, z are positive integers, is xyz ≥ 64?

1) xy ≥ yz ≥ zx ≥ 16
2) x + y + z = 64

Excellent problem, Max. Congrats!

$$x,y,z\,\, \ge 1\,\,{\rm{ints}}$$
$$xyz\,\,\mathop \ge \limits^? \,\,64$$

$$\left( 1 \right)\,\,xy \ge yz \ge xz \ge 16$$
$$\left\{ \matrix{
\,\left( {yz} \right)x \ge 16x \hfill \cr
\,\left( {xz} \right)y \ge 16y \hfill \cr
\,\left( {xy} \right)z \ge 16z \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits_{\left( {{\rm{all}}\,\,{\rm{positive}}} \right)}^{multiply\,\,!} \,\,\,\,{\left( {xyz} \right)^3} \ge {16^3}\left( {xyz} \right)\,\,\,\,\,\,\,\left( * \right)$$
$$\left( * \right)\,\,\,\,\mathop \Rightarrow \limits_{\left( {xyz\,\,{\rm{positive}}} \right)}^{:\,\,xyz} \,\,\,\,\,{\left( {xyz} \right)^2} \ge {16^3}\,\,\,\,\mathop \Rightarrow \limits^{xyz\,\, > \,\,0} \,\,\,\,xyz \ge \sqrt {{2^{12}}} = 64\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

$$\left( 2 \right)\,\,\,x + y + z = 64\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( {1,1,62} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( {2,2,60} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \, \hfill \cr} \right.$$


The correct answer is therefore (A).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.