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27 Mar 2021, 01:25
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MathRevolution
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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.
Visit 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 \(2\) variables (\(a\) and \(b\)) and \(0\) equations, C 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) together give us:
Since \(a\) and \(b\) have the same parities in order for \(y = ax\) and \(y = \frac{b}{x}\) to have an intersection, we have \(ab > 0\) from condition 1).
Since we have \(a + b > 0\) from condition 2), we have \(a > 0\) and \(b > 0\) when we consider both conditions together.
Then point (\(a, b\)) is in the first quadrant.
The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Both conditions 1) & 2) together are sufficient.
Therefore, C is the correct answer.
Answer: C
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.
30 Mar 2021, 02:55
Kudos
Bookmarks
MathRevolution
=>
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 \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C 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) together give us:
\(x = (1 + A) ○ (1 - A) = [\frac{1 }{ (1 + A + 1 - A)}] – [\frac{1 }{ (1 + A)}] = [\frac{1}{2}] – [\frac{1 }{ (1 + A)}] = [\frac{(1 + A) }{ 2(1 + A)}] – [\frac{2 }{ 2(1 + A)}] = [\frac{(1 + A - 2) }{ 2(1 + A)}] = [\frac{(A - 1) }{ 2(1 + A)}].\)
\(y = (1 - A) ○ (1 + A) = [\frac{1 }{ (1 – A + 1 + A)}] – [\frac{1 }{ (1 - A)}] = [\frac{1}{2}] – [\frac{1 }{ (1 - A)}] = [\frac{(1 - A) }{ 2(1 - A)}] – [\frac{2 }{ 2(1 - A)}] = [\frac{(1 – A - 2) }{ 2(1 - A)}] = \frac{-(A + 1) }{ 2(1 - A)}.\)
Then, we have \(xy = [\frac{(A - 1) }{ 2(1 + A)}] · [\frac{-(A + 1) }{ 2(1 - A)}] = \frac{[(A – 1)·(-A – 1)] }{ [4·(1 + A)·(1 – A)]} = \frac{[-A^2 – A + A + 1] }{ [4·(1 – A + A - A^2]} = \frac{[-A^2 + 1] }{ [4·(1 – A^2]} = \frac{[1 – A^2] }{ [4·(1 - A^2]} = \frac{1}{4}.\)
The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Both conditions 1) & 2) together are sufficient.
Therefore, C is the correct answer.
Answer: C
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.
01 Apr 2021, 02:53
Kudos
Bookmarks
MathRevolution
=>
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.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have \(1\) variable (\(N\)) and \(0\) equations, D is most likely the answer. So, we should consider each condition on its own first.
Condition 1)
Assume \(N\) is \(1\) greater than a product of four consecutive integers, \(x, x+1, x+2,\) and \(x+3\) where \(x\) is an integer.
We have
\(N = x(x + 1)(x + 2)(x + 3) + 1\)
\(N = x(x + 3)(x + 1)(x + 2) + 1 \)
\(N = (x^2 + 3x)(x^2 + 3x + 2) + 1 \)
\(N = (x^2 + 3x)^2 + 2(x^2 + 3x) + 1 \)
\(N = (x^2 + 3x + 1)^2\)
Thus, \(N\) is a perfect square.
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
If \(N = 1 + 3 + 5 + 7 = 16\), then \(N\) is a perfect square and the answer is ‘yes’.
If \(N = 3 + 5 + 7 + 9 = 24\), then \(N\) is not a perfect square and the answer is ‘no’.
Since condition 2) does not yield a unique solution, it is not sufficient.
Therefore, A is the answer.
Answer: A
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations,” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C, or E.
