MathRevolution wrote:
[GMAT math practice question]
(number properties) \(a\) and \(b\) are positive integers. What is the value of \(b-a\)?
\(1) \frac{a}{b} = \frac{2}{7}\)
\(2) a+b\) is a two-digit integer greater than \(20\)
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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.
Since we have \(2\) variables (\(a\) and \(b\)) and \(0\) equations, C is most likely to be 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)
Since \(\frac{a}{b}= \frac{2}{7}\), we have \(7a = 2b.\)
If \(a = 4\) and \(b = 14\), then \(\frac{a}{b} = \frac{4}{14} = \frac{2}{7}, a+ b = 21 > 20\), and \(b – a = 10.\)
If \(a = 6\) and \(b = 21\), then \(\frac{a}{b} = \frac{6}{21} = \frac{2}{7}, a + b = 27 > 20,\) and \(b – a = 15.\)
Since both conditions together don’t yield a unique solution, they are not sufficient.
Therefore, E is the answer.
Answer: E
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.
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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