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Updated on: 07 Apr 2021, 02:40
Originally posted by MathRevolution on 10 Jul 2020, 02:03.
Last edited by MathRevolution on 07 Apr 2021, 02:40, edited 1 time in total.
Last edited by MathRevolution on 07 Apr 2021, 02:40, edited 1 time in total.
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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 (\(P\)) and \(0\) equations, D is most likely the answer. So, we should consider each condition on its own first.
Let’s look at condition 1). It tells us that \(P + 1\) is a multiple of \(6\).
Since \(P\) is a prime number, \(P\) can’t have a remainder of \(0, 2, 3\), or \(4\) when \(P\) is divided by \(6\).
Since \(P + 2\) is a prime number, \(P + 2\) can’t have a remainder \(0, 2, 3\), or \(4\) when \(P + 2\) is divided by \(6\), which means \(P\) can’t have a remainder of \(0, 1, 2\), or \(4.\)
Then, the possible remainder of \(P\) when divided by \(6\) is only \(5\).
Thus, \(P + 1\) has a remainder of \(0\) when divided by \(6\), and \(P + 1\) is a multiple of \(6\).
The answer is unique, yes, and the answer is sufficient according to Common Mistake Type 2, which states that the answer must be a unique yes or no.
Let’s look at condition 2). It tells us that we don’t have a unique solution.
If \(P = 5\), then \(P + 1 = 6\) and \(P + 1\) is a multiple of \(6\) and the answer is ‘yes’.
If \(P = 10\), then \(P + 1 = 11\) and \(P + 1\) is not a multiple of \(6\) and the answer is ‘no’.
The answer is not unique, yes and no, so the condition is not sufficient, according to Common Mistake Type 2, which states that if we get both yes and no as an answer, it is not sufficient.
Condition 1) ALONE is sufficient.
Therefore, A is the correct 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.
10 Jul 2020, 02:04
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[GMAT math practice question]
(Equation) What is the value of \(a\)?
1) The equation \(\frac{x}{3} + a = \frac{x}{2} - \frac{x - 18}{6}\) in terms of \(x\) has no solution.
2) \(a\) is an integer.
(Equation) What is the value of \(a\)?
1) The equation \(\frac{x}{3} + a = \frac{x}{2} - \frac{x - 18}{6}\) in terms of \(x\) has no solution.
2) \(a\) is an integer.
Updated on: 12 May 2021, 02:37
Originally posted by MathRevolution on 12 Jul 2020, 19:57.
Last edited by MathRevolution on 12 May 2021, 02:37, edited 1 time in total.
Last edited by MathRevolution on 12 May 2021, 02:37, edited 1 time in total.
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 \(2\) 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:
If \(a = 4\) and \(b = 3\), we have \((a - b)^2 = (4 - 3)^2 = 1.\)
If \(a = 4\) and \(b = -3\), we have \((a - b)^2 = (4 - (-3))^2 = 7^2 = 49.\)
The answer is not unique, and both conditions 1) and 2) together are not sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Both conditions 1) & 2) together are not sufficient.
Therefore, E is the correct answer.
Answer: E
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.
