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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since a function has many variables to determine, 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) together give us:

Assume \(t = 2020x + f(0).\)

Then we have \(x = \frac{(t - f(0))}{2020}.\)

\(f(t) = 2020[\frac{(t - f(0))}{2020}]^2 = \frac{[t - f(0)]^2}{2020}.\)

When we replace \(t\) by \(0\), we have \(f(0) = \frac{(f(0))^2}{2020}\) or \((f(0))^2 – 2020f(0) = 0\)

Then we have \(f(0)(f(0) - 2020) = 0.\)

Thus \(f(0) = 0\) or \(f(0) = 2020.\)

Both conditions 1) and 2) together 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.

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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 \(1\) variable (\(x\)) and \(0\) equations, D is most likely the answer. So, we should consider each condition on its own first.

Condition 1)

Assume \(x = k(k + 1)(k + 2)(k + 3) + 1.\)

Then \(x = (k^2 + 3k)(k^2 + 3k + 2) + 1 = A(A + 2) + 1 = A^2 + 2A + 1 = (A + 1)^2\) for \(A = k^2 + 3k\) for an integer \(k\).

Thus, \(x\) is a perfect integer, and the answer is ‘yes’.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

If \(x = 1 + 3 + 5 + 7 + 9 = 25\), then \(x\) is a perfect integer and the answer is ‘yes’.
If \(x = 3 + 5 + 7 + 9 + 11 = 35\), then \(x\) is not a perfect integer 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.

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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.

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)

We have \((x - y)^2 = (x + y)^2 - 4xy \)

Because
\((x + y)^2 - 4xy\)

\((x + y)(x + y) – 4xy\)

\(x^2+ xy + xy + y^2- 4xy\)

\(x^2 - 2xy + y^2\)

\((x – y)(x – y)\)

\((x – y)^2 \)

Then, we substitute
\((x + y)^2 - 4xy = 9^2 – 4*2 = 81 – 8 = 73.\)

Then, we have \(x – y = ±√73.\)

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the 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.

=>

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 \(3\) variables (\(a, b\) and \(c\)) 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)
Since \(gcd(a,b) = 18 = 2^1*3^2\) and \(gcd(b,c) = 24 =2^3*3^1\),\(gcd(a,b,c) = gcd(18,24) = gcd( 2^1*3^2 , 2^3*3^1 ) = 2^1*3^1 = 6\) by choosing the minimum exponents from the prime factorizations. Since both conditions together yield a unique solution, they are sufficient.

Since this question is a statistics 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)
Since it provides no information about the value of c, condition 1) is not sufficient.

Condition 2)
Since it provides no information about the value of a, condition 2) 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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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.
Since \(4^n\) is a multiple of \(2\), we only need to look at \(n^2+1\).
If \(n\) is an odd number, \(4^n+n^2+1\) is divisible by \(2\).
If \(n\) is an even number, \(4^n+n^2+1\) is not divisible by \(2\).
The question asks if \(n\) is an odd number.
Thus, each of the conditions is sufficient.

Therefore, D is the answer.
Answer: D

=>

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)

Since we have \(\frac{1}{a} – \frac{1}{b} = 2\)

\(\frac{b}{ab} – \frac{a}{ab} = 2\) (common denominator)

\(\frac{(b – a)}{ab} = 2\) (subtracting fractions)
\(a – b = -2ab\) (multiplying both sides by \(ab\))

Since we have \(\frac{1}{a^2} + \frac{1}{b^2} = 3 \)

\(\frac{b^2}{a^2b^2} + \frac{a^2}{a^2b^2} = 3\) (common denominator)

\(\frac{(b^2 + a^2)}{a^2b^2} = 3\) (adding fractions)
\(a^2 + b^2 = 3a^2b^2\) (multiplying both sides by \(a^2b^2\))

Then we have \(a^2 + b^2 = (a-b)^2 + 2ab\)
\(a^2 + b^2 = (-2ab)^2 + 2ab\) (substituting \(-2ab\) in for \(a – b\))

\(a^2 + b^2 = 3a^2b^2\) or \(a^2b^2 + 2ab = 0.\)
We have \(ab = -2\) and \(a – b= -2ab = 4\) from \(ab(ab+2)=0\), since \(ab ≠ 0.\)
\(a^3-b^3 = (a-b)^3 + 3ab(a-b) = 4^3 + 3(-2)4 = 64 – 24 = 40.\)

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

Normally, in problems that 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 a probability of 70%), and E is the answer (with a probability of 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.

=>

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 (\(x\) 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)

We have \(x = 1, y = 2\) or \(x = 2, y = 1.\)

Then we have \(x^5 + y^5 = 33\) for both cases.

Since both conditions together yield a unique solution, they are sufficient.

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.

=>

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 (\(x\) 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)

When we subtract \(√3\) times the equation of condition 1) from the equation of condition 2), we have
\(√3x - √2y - √3(x + y) = 5 - √3(2√3)\)

\(√3x - √2y - √3x - √3y = 5 – 6\)

\(√2y - √3y = -1\)

\(√3y - √2y = 1\)

\((√3 + √2)y = 1 \)

\(y = \frac{1}{(√3 + √2)}\)

\(y = √3 - √2.\)

Then we have
\(x + y = 2√3\)

\(x = 2√3 – y\)

\(x = 2√3 - (√3 - √2)\)

\(x = 2√3 - √3 + √2 \)

\(x = √3 + √2.\)

\(x^2 + y^2 = (x + y)^2 – 2xy \)

\(= (√3 + √2 + √3 - √2)^2 – 2(√3 + √2)(√3 -√2)\)

\(= (2√3)^2 – 2(3 - √6 + √6 - 2)\)

\(= 4(3) - 2(1)\)

\(= 12 – 2\)

\(= 0\)

Since both conditions together yield a unique solution, they are sufficient.

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.

=>

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.

Condition 1)
If we have \(abc = 1\), then we have

\(\frac{a}{ab+a+1} + \frac{ab}{a(bc+b+1)} + \frac{abc}{ab(ca+c+1)}\)

\(= \frac{a}{ab+a+1} + \frac{ab}{abc+ab+a} + \frac{abc}{abca+abc+ab}\)

\(=\frac{a}{ab+a+1} + \frac{ab}{ab+a+1} + \frac{1}{ab+a+1} \) (substituting \(1\) for \(abc\))

\(=\frac{ab+a+1}{ab+a+1} = 1.\)

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
Since we don’t have any information about \(a, b\) and \(c,\) it is not sufficient.

Therefore, A is the answer.
Answer: A

=>

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.

Condition 2)
\(x = 3, y = 5, z = 2*3*5 = 30\) are unique solutions as it is the only combination of numbers that works within the given conditions of \(3 ≤ x < y < z ≤ 30\) and \(y\) is a prime number. If \(x\) and \(y\) are larger numbers than \(z\) is greater than \(30.\) We then have \(x + y + z = 3 + 5 + 30 = 38.\)

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

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

Since we have \(\frac{1}{x} + \frac{1}{y} = \frac{1}{2} + \frac{1}{z}\), we have \(\frac{1}{2} < \frac{1}{x} + \frac{1}{y} < \frac{1}{x} + \frac{1}{x} = \frac{2}{x}\) or \(\frac{1}{2} = \frac{2}{4} < \frac{1}{x}.\)

Thus \(x < 4\) and we have \(x = 3.\)

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

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

\(\frac{1}{z} = \frac{1}{x} + \frac{1}{y} – \frac{1}{2} = \frac{1}{3} + \frac{1}{5} – \frac{1}{2} = \frac{10}{30} +\frac{ 6}{30} – \frac{15}{30} = \frac{1}{30}\) or \(z = 30.\)

Then, \(x + y + z = 3 + 5 + 30 = 38.\)

Since condition 1) yields a unique solution, it is sufficient.

Therefore, D is the answer.
Answer: D

This question is a CMT 4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.

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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)
\(a, b,\) and \(c\) are integers with \(1 ≤ a ≤ 9, 0 ≤ b ≤ 9\), and \(0 ≤ c ≤ 9\).

If \(a ≥ 5,\) then \(b > 10 + c > 10\) never happens and we have \(a ≤ 4\) from condition 1).

Since condition 2) tells us that the minimum value of \(a\) is \(4, a ≥ 4,\) and the unique value of a is \(4.\)

Since \(b > 8 + c\) and \(c ≥ 0\), we have \(b = 9\) and \(c = 0.\)
Therefore, \(N\) is \(490\), and it is a unique answer.

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)
If \(a = 4, b = 9,\) and \(c = 2,\) then \(N=492\) is an answer.
If \(a = 3, b = 9,\) and \(c = 2\), then \(N=392\) is also an answer.
Since condition 1) does not yield a unique solution, it is not sufficient.

Condition 2)
If \(a = 4, b = 9\), and \(c = 2\), then \(N=492\) is an answer.
If \(a = 5, b = 9,\) and \(c = 2\), then \(N=592\) is also an answer.
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.

=>

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.

We have \(3a + 2b + c = 48\) from the original condition.

Since we have \(3\) variables (\(a, b\) and \(c\)) and \(1\) equation, 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)
If \(a = 2\), then we have \(2b + c = 42\) and we have \(c = 34\) when we have \(b = 4.\)

If \(a = 4\), then we have \(2b + c = 36\) and we have \(c = 24\) when we have \(b = 6.\)

Thus, \(a =2, b = 4, c = 34\) and \(a = 4, b = 6, c = 24\) are solutions.

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the 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.

=>

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.

Assume \(f, u\) and \(d\) are distances on the flat sections, uphill sections and downhill sections when she travels from home to school.

Then we have\( f + u + d = 24.\)

Even though we have \(3\) variables and \(1\) equation, condition 1) has \(3\) equations. So we should check condition 1) first.

Condition 1)
The time she travels from home to school is \(\frac{f}{5} + \frac{u}{4} + \frac{d}{6} = 4(\frac{50}{60}) = \frac{29}{6}.\)

The time she travels from school to home is \(\frac{f}{5} + \frac{u}{6} + \frac{d}{4} = 5\) since uphill sections becomes downhill sections and downhill sections becomes uphill when she travels back.

Adding the equations together, we have \((\frac{2}{5})f + \frac{(5u + 5d)}{12} = (\frac{2}{5})f + (\frac{5}{12})(u+d) = \frac{29}{6} + 5 = \frac{59}{6}.\)

Substituting in \(u + d = 24 – f,\) we have \((\frac{2}{5})f +(\frac{5}{12})(24 - f) = \frac{59}{6}, (\frac{2}{5})f + 10 - (\frac{5}{12})f = \frac{59}{6},\) or \((\frac{24}{60})f - (\frac{25}{60})f = \frac{59}{6} - \frac{60}{6}.\) Then we have \((\frac{-1}{60})f = \frac{-1}{6}\), or \(f = 10.\)

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
Since the speed on the flat section is the arithmetic average of the speeds on the uphill and downhill sections, we can assume that s is the speed on the flat sections, \(s – a\) is the speed on the uphill sections and \(s + a\) is the speed on the downhill sections.

Then we have the time she takes when she travels from home to school, \(\frac{f}{s} + \frac{u}{(s-a)} + \frac{d}{(s+a)} = \frac{29}{6}\) and we have the time she takes when she travels from home to school, \(\frac{f}{s} + \frac{u}{(s+a)} + \frac{d}{(s-a)} = 5.\)

When we add those equations, we have \((\frac{2f}{s}) + (u+d)(\frac{1}{(s-a)}+\frac{1}{(s-b)}) = \frac{59}{6}.\)

We can notice that there must be many possibilities for solutions.

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.

=>

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.
\(ax + 2y - 3 = 4x + by + 5\)

\(⇔ (a - 4)x + (2 - b)y – 8 = 0.\)

If we have \(a = 4\) and \(b = 2\), then the equation is equivalent to \(-8 = 0\), which is not an equation of a line.

So, each condition alone is sufficient.

Therefore, D is the answer.
Answer: D

=>

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)
Since \(a, b,\) and \(c\) are positive integers and we have \(abc = 6\) with \(a ≤ b ≤ c,\) we have a unique solution of \(a = 1, b = 2, c = 3.\)

Therefore, we have \(a + b + c = 6.\)

Since both conditions together yield a unique solution, they are sufficient.

Since this question is a statistics 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 2)
If \(a = 1, b = 2\) and \(c = 3\), then we have \(a + b + c = 1 + 2 + 3 = 6.\)

If \(a = 1, b = 1\) and \(c = 6\), then we have \(a + b + c = 1 + 1 + 6 = 8.\)

Since condition 2) does not yield a unique solution, it is not sufficient.

Condition 1)
Since \(abc = a + b + c\) and \(1 ≤ a ≤ b ≤ c,\) we have \(c ≤ abc = a + b + c ≤ 3c\) or \(c ≤ abc ≤ 3c.\)

Therefore, we have \(1 ≤ ab ≤ 3.\)

If \(ab = 1,\) then we have \(a = b = 1\) and \(c = 2 + c,\) which doesn’t make sense.

If \(ab = 2,\) then we have \(a = 1, b = 2\) and \(2c = c + 3\) or \(c = 3,\) which tells us that \(a + b + c = 6.\)

If \(ab = 3,\) then we have \(a = 1, b = 3\) and \(3c = 4 + c\) or \(c = 2,\) which doesn’t make sense since \(c < b.\)

Therefore, \(a = 1, b = 2\) and \(c = 3\) is the unique case and we have \(a + b + c = 6.\)

Since condition 1) yields a unique solution, it is sufficient.

Therefore, A is the answer.
Answer: A

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.

This question is a CMT 4(B) question: We easily figured out condition 2) is not sufficient, and condition 1) is difficult to work with. For CMT 4(B) questions, we assume condition 1) is sufficient. Then A is most likely an answer.

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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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Since we have \(2\) variables (\(x\) and \(y\)) and each condition has \(2\) equations, D is most likely the answer. So, we should consider each condition on its own first.

Condition 1)
We have\( x + y = 7\) and \(x – y = 1.\)

Adding the \(2\) equations together gives us \((x + y) + (x – y) = 7 + 1, 2x = 8\), or \(x = 4.\)

When we substitute \(x\) with \(4\) in the first equation, we have \(4 + y = 7\) or \(y = 3.\)

Therefore, we have \(xy = 4*3 = 12.\)

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
Case 1: \(x ≥ y\)

Then we have \(2x + 3y - 13 = x\) and \(3x - y - 6 = y.\)

Then we have \(x + 3y - 13 = 0\) and \(3x - 2y - 6 = 0.\)

When we subtract three times the first equation from the second equation, we have \((3x - 2y - 6) - 3(x + 3y - 13) = 3x - 2y - 6 - 3x - 9y + 39 = -11y + 33 = 0\) or \(y = 3.\)

When we substitute \(y\) with \(3\) in the first equation, we have \(x + 3*3 - 13 = 0\) or \(x = 4.\)

Then, we have \(xy = 4*3=12.\)

Case 2: \(x < y\)

Then we have \(2x + 3y - 13 = y\) and \(3x - y - 6 = x.\)

Then we have \(2x + 2y - 13 = 0\) and \(2x - y - 6 = 0.\)

When we subtract the second equation from the first equation, we have \((2x - y - 6) - (2x + 2y - 13) = 2x - y - 6 - 2x - 2y + 13 = -3y + 7 = 0\) or \(y = \frac{7}{3}.\)

When we substitute \(y\) with \(\frac{7}{3}\) in the second equation, we have \(2x - \frac{7}{3} - 6 = 2x - \frac{7}{3} - \frac{18}{3} = 2x – \frac{25}{3} = 0, 2x = \frac{25}{3}\), or \(x = \frac{25}{6}.\)

However, we have \(x > y\) in this case so we don’t have a solution in this case.

Therefore, we have a unique solution of \(x = 4\) and \(y = 3.\)

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

Therefore, D is the answer.
Answer: D

This question is a CMT 4(B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.

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.

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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 \(A^2 + 2B^2 = C^2 \), we have \(2B^2 = C^2 - A^2\) or \(2B^2 = (C + A)(C - A)\).

\(C^2 - A^2 = (C + A)(C - A)\) is an even number, both \(A\) and \(C\) must either odd numbers or even numbers.

Then \(2B^2\) is a product of even numbers, and \(2B^2\) is a multiple of \(4\).
\(B\) is an even number.

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)
\(A, B\), and \(C\) are positive integers less than or equal to \(10\).

Since we have \(0 < C^2 - A^2 = 2B^2 < 10^2\) or \(0 < C^2 - A^2 = 2B^2 < 100\), we have \(0 < B < √50\) and \(B = 2, 4 \)or \(6.\)
We have \(C + A > C – A.\)

Case 1: \(B = 2\)
Since we have \((C + A)(C - A) = 8\), we have \(C + A = 4\) and \(C – A = 2.\)
Then we have \(C = 3, A = 1\) and \(A + B + C = 1 + 2 + 3 = 6.\)

Case 2: \(B = 4\)
Since we have \((C + A)(C - A) = 32\), we have \(C + A = 16, C - A = 2\) or \(C + A = 8, C – A = 4.\)
If \(C + A = 16, C – A = 2\), then we have \(A = 7, C = 9\) and \(A + B + C = 7 + 4 + 9 = 20.\)

Since both conditions together do not 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.