The Ultimate Q51 Guide [Expert Level]

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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 \(3\) variables (\(p, q\) and \(r\)) and \(0\) equations, E 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):
\(3, 5\) and \(7\) is the unique triplet of three consecutive odd integers which are prime numbers, and \(3*5*7\) is a multiple of \(5\).
Thus, both conditions together 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)
If \(p = 3, q = 5\) and \(r = 7\), then \(pqr = 105\) is a multiple of \(5\), and the answer is ‘yes’.
If \(p = 7, q = 9\) and \(r = 11\), then \(pqr = 693\) is not a multiple of \(5\), and the answer is ‘no’.
Condition 1) is not sufficient since it doesn’t yield a unique answer.

Condition 2)
If \(p = 3, q = 5\) and \(r = 7\), then \(pqr = 105\) is a multiple of \(5\), and the answer is ‘yes’.
If \(p = 3, q = 7\) and \(r = 11\), then \(pqr = 231\) is not a multiple of \(5\), and the answer is ‘no’.
Condition 2) is not sufficient since it doesn’t yield a unique solution.

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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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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The possible pairs \((m,n)\) are \((1,1),(1,3),(2,1),(2,3)\) and \((5,5)\).

Since we have \(2\) variables (\(m\) and \(n\)) 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)
There is a unique pair of integers which satisfies both conditions.
This is \(m = 2\) and \(n = 3\).
So, \(mn = 6\).
Conditions 1) & 2) are sufficient, when applied together.

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)
If \(m = 2\) and \(n = 3\), then \(mn = 6\).
If \(m = 5\) and \(n = 5\), then \(mn = 25\).
Condition 1) is not sufficient since it does not yield a unique solution.

Condition 2)
If \(m = 2\) and \(n = 3\), then \(mn = 6\).
If \(m = 1\) and \(n = 1\), then \(mn = 1\).
Condition 2) is not sufficient since it does not yield a unique solution.

Therefore, C is the 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.

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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 \(3\) variables (\(a, b\), and \(c\)) 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)
\(2^a + 2^b + 2^c \)

\(≤1 + 2^{a+b} + 2^c \)

\(≤1 + 1 + 2^{a+b+c},\) since \(2^{a+b} + 2^c ≤ 1 + 2^{a+b+c}\)

\(≤1 + 1 + 2^5 = 34\)

So, the maximum value of \(2^a + 2^b + 2^c\) is \(34.\)

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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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.

Condition 2)
The minimum possible value of \(x\) is \(3\), and the minimum possible value of \(y\) is \(5\) since \(y\) is a prime number.

Then we have \(z = 2*3*5 = 30\), which is the possible maximum value of \(z\).

\(x = 3, y = 5\) and \(z = 30\) are the unique tuple of solutions.

Since condition 2) yields a unique solution, it is sufficient.

Condition 1)
When we take reciprocals, we have \(\frac{1}{30} < \frac{1}{z} < \frac{1}{y} < \frac{1}{x} ≤ \frac{1}{3}\)

We have \(\frac{1}{2} < \frac{1}{x} + \frac{1}{y} < \frac{1}{x} + \frac{1}{x} = \frac{2}{x}\) from \(\frac{1}{x} + \frac{1}{y} = \frac{1}{2} + \frac{1}{z}.\)

Then we have \(x = 3\) since \(x < 4\) and \(x ≥ 3\).

Since \(\frac{1}{x} + \frac{1}{y} = \frac{1}{2} + \frac{1}{z} > \frac{1}{2},\) we have \(\frac{1}{3} + \frac{1}{y} > \frac{1}{2} or \frac{1}{y} > \frac{1}{6}.\)

Therefore, we have \(y < 6.\)

Since we have \(3 = x < y < 6\) and \(y\) is a prime number, we have \(y = 5.\)

Then, we have \(\frac{1}{z} = (\frac{1}{x} + \frac{1}{y}) – \frac{1}{2} = (\frac{1}{3} + \frac{1}{5}) - \frac{1}{2} = \frac{8}{15} - \frac{30}{60} = \frac{1}{15}\) or \(z = 15.\) Since condition 1) yields a unique solution, it is sufficient.

Since each condition yields a unique solution, the answer is D.
Answer: D

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.

Note: Since this question is a CMT 4(B) question because condition 2) is easy to understand and condition 1) is hard. When one condition is easy to understand, and the other is hard, D is most likely the answer.

Sorry if this is a silly question or if it has already been answered, but what exactly is Murphy's and Sally's law?