MathRevolution wrote:
[GMAT math practice question]
(Inequalities) \(x\) is an integer. \(y\) and \(z\) are real numbers with \(x < 2y < 3z.\)
What is the value of \(x\)?
1) \(x + 2y + 3z = 4\)
2) \(2x + 3y + 4z = 5\)
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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.
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https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since we have \(3\) variables (\(x, y\), and \(z\)) and \(0\) equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
We have \(x + 2y + 3z = 4\) and \(2x + 3y + 4z = 5.\)
When we subtract twice the second equation from three times the first equation, we have
\(3(x + 2y + 3z) - 2(2x + 3y + 4z) = 3(4) - 2(5)\)
\(3x + 6y + 9z - 4x - 6y - 8z = 12 - 10\)
\(-x + z = 2\)
\(z = x + 2.\)
Substituting \(z = x + 2\) into the first equation gives us:
\(x + 2y + 3z = 4\)
\(x + 2y + 3(x + 2) = 4\)
\(x + 2y + 3x + 6 = 4\)
\(4x + 2y = -2\)
\(2y = -4x - 2\)
\(y = -2x – 1\)
\(x < 2y < 3z\) is equivalent to \(x < 2(-2x – 1) < 3(x + 2)\), or \(x < -4x – 2 < 3x + 6\).
Then we have \(x < \frac{-2}{5}\) from \(x < -4x – 2, \)because:
\(x < -4x – 2\)
\(5x < -2\)
\(x < \frac{-2}{5}.\)
We also have \(x > \frac{-8}{7}\) from \(-4x - 2 < 3x + 6\), because:
\(-4x - 2 < 3x + 6\)
\(-7x < 8\)
\(x > \frac{-8}{7}\) (the inequality sign changes direction since we divided by a negative)
Thus \(\frac{-8}{7} < x < \frac{-2}{5}\) and \(x = -1\) since \(x\) is an integer.
Since both conditions together yield a unique solution, they are sufficient.
Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
\(x = -2, y = 1, z = \frac{4}{3}\) and \(x = -3, y = 0, z = 1\) are solutions.
Condition 2)
\(x = -2, y = 0, z = \frac{9}{4}\) and \(x = -1, y = 0, z = \frac{7}{4}\) are solutions.
Since condition 2) does not yield a unique solution, it is not sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
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