Overview of GMAT Math Question Types and Patterns on the GMAT

=>

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 \(4\) variables (\(a, b, x\), and \(y\)) 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)

If \(a = √3, b = 0, x = 2\), and \(y = 0,\) then \((ax + by)(bx + ay) = (√3*2 + 0*0)(0*2 + √3*0) = 2√3*0 = 0.\)

If \(a = √2, b = 1, x = √3\), and \(y = 1\), then \((ax + by)(bx + ay) = (√2*√3 + 1*1)(1*√3 + √2*1) = (1+√6)(√2√+3) ≠ 0.\)

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.

=>

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.

Condition 1) means the average of \(a, b\) and \(24\) is \(\frac{(a + b + 24 )}{3} = 24\) since it implies that \(\frac{( a + b )}{2} = 24\). Since one of the three data points is their average, it is also their median. Condition 1) is sufficient.

Condition 2)
If \(a > 24\) and \(b < 24\), the median of \(a, b\) and \(24\) is \(24\).
If \(a < 24\) and \(b > 24\), the median of \(a, b\) and \(24\) is \(24\).
Thus, we have a unique value for the median, and condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

Note: Condition 1) is easy to understand and condition 2) is hard to understand. Thus, this question is a CMT4(B) question. When one condition is easy to understand, and the other is hard to understand, D is most likely to be the answer.

=>

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 original condition \(a>b>c>d>0\) is equivalent to \(0 < \frac{1}{a} < \frac{1}{b} < \frac{1}{c} < \frac{1}{d}\). The question asks if \(d < 4\). This is equivalent to asking if \(\frac{1}{d} > \frac{1}{4}\).

By condition \(1, \frac{1}{d} > \frac{1}{4}\) since \(\frac{1}{c} < \frac{1}{d}\). So, \(d < 4\). Condition 1) is sufficient.

Condition 2)
Since \(0 < \frac{1}{a} < \frac{1}{b} < \frac{1}{c} < \frac{1}{d}\) and \((\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})=1\), we have \(\frac{1}{a}<\frac{1}{d}\), \(\frac{1}{b}<\frac{1}{d}, \frac{1}{c}<\frac{1}{d}\) and \(\frac{1}{a} +\frac{1}{b} + \frac{1}{c} + \frac{1}{d} < \frac{1}{d} +\frac{1}{d} +\frac{1}{d} + \frac{1}{d} = \frac{4}{d}.\)
Therefore, \(1 < \frac{4}{d}\) and \(d < 4\).
Condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

Note: This question is a CMT4(B) question: condition 1) is easy to work with and condition 2) is hard. For CMT4(B) questions, D is most likely to be the answer.

=>

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.

Asking if “\(n^2+2n+44\) is divisible by \(4\)” is equivalent to asking if \(n\) is an even integer, since \(44\) is a multiple of \(4\) and \(n^2+2n = n(n+2)\) is a multiple of \(4\) when \(n\) is an even number.
Thus, condition 1) is sufficient.

Condition 2)
Since \(n^2\) is divisible by \(144\), \(n\) is divisible by \(12\) and \(n\) is an even number.
Thus, condition 2) is also sufficient.

Therefore, D is the answer.
Answer: D

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).

=>

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 (\(x, y\) and \(z\)) 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)
If \(x = 2, y = 3, z = 5\), then \(x\) and \(z\) are relatively prime, and the answer is ‘yes’.

If \(x = 2, y = 3, z = 2,\) then \(x\) and \(z\) are not relatively prime, and the answer is ‘no’.

Both conditions together are not sufficient, since they don’t yield a unique answer.

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.

=>

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

Condition 1)
Since the sum of \(2, 3, 4, 5\) and \(6\) is \(20\), which is an even number, and \(n=5\) is an odd number, the answer is sometimes ‘no’.
Since the sum of \(1, 2, 3\), and \(4\) is \(10\), which is an even number, and \(n=4\) is an even number, the answer is sometimes ‘yes’.
Condition 1) is not sufficient since it does not yield a unique answer.

Condition 2)
If \(n = 4\), then \([\frac{n}{2}] = 2\) is even, and \(n\) is even.

If \(n = 5\), then \([\frac{n}{2}] = 2\) is also even, but \(n\) is not even.
Condition 2) is not sufficient since it does not yield a unique answer.

Conditions 1) & 2)
Since the sum of \(2, 3, 4, 5\) and \(6\) is \(20\), which is an even number, and \(n=5\) is an odd number such that \([\frac{n}{2}] = 2\) is even, the answer is sometimes ‘no’.

Since the sum of \(1, 2, 3\), and \(4\) is \(10\), which is an even number, and \(n=4\) is an even number such that \([\frac{n}{2}] = 2\) is even, the answer is sometimes ‘yes’.

Conditions 1) & 2) together are not sufficient, since they do not yield a unique answer.

Therefore, E is the answer.
Answer: E

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

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

Condition 1)
\(n\) is a prime factor of \(21 = 3*7\) and \(n\) is \(3\) or \(7\).
Since it does not give a unique answer, condition 1) is not sufficient.

Condition 2)
If \(n\) is a factor of \(49 = 7^2\), then \(n\) is \(1, 7\) or \(49\).
Since it does not give a unique answer, condition 2) is not sufficient.

Conditions 1) & 2)
The unique integer satisfying both conditions is \(n = 7.\)
Both conditions are sufficient, when taken together.

Therefore, C is the answer.
Answer: C

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The expression from the question is equivalent to \(\frac{xy}{( x^2 + xy + y^2)}\) as shown below, which is the same as condition 1):
\(\frac{( x^2y – xy^2 )}{( x^3 – y^3 )}\)
\(= \frac{xy(x-y)}{(x-y)(x^2+xy+y^2)}\)
\(= \frac{xy}{( x^2 + xy + y^2 )}\)
Thus, condition 1) is sufficient.

Condition 2)
If \(x = 3, y = 1\), then \(\frac{( x^2y – xy^2 )}{( x^3 – y^3 )} = \frac{( 9 – 3 )}{( 27 – 1)} = \frac{6}{26} = \frac{3}{13}.\)
If \(x = -3, y = 1\), then \(\frac{( x^2y – xy^2 )}{( x^3 – y^3 )} = \frac{( 9 + 3 )}{( - 27 – 1)} = \frac{-12}{28} = \frac{-3}{7}.\)
Since it does not yield a unique solution, condition 2) 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.

Since we have \(3\) variables (\(x, y\), and \(z\)) 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)

We have \(x + 2y + 3z = 4\) and \(2x + 3y + 4z = 5.\)

When we subtract twice the second equation from three times the first equation, we have
\(3(x + 2y + 3z) - 2(2x + 3y + 4z) = 3(4) - 2(5)\)

\(3x + 6y + 9z - 4x - 6y - 8z = 12 - 10\)

\(-x + z = 2\)

\(z = x + 2.\)

Substituting \(z = x + 2\) into the first equation gives us:

\(x + 2y + 3z = 4\)

\(x + 2y + 3(x + 2) = 4\)

\(x + 2y + 3x + 6 = 4\)

\(4x + 2y = -2\)

\(2y = -4x - 2\)

\(y = -2x – 1\)

\(x < 2y < 3z\) is equivalent to \(x < 2(-2x – 1) < 3(x + 2)\), or \(x < -4x – 2 < 3x + 6\).

Then we have \(x < \frac{-2}{5}\) from \(x < -4x – 2, \)because:

\(x < -4x – 2\)

\(5x < -2\)

\(x < \frac{-2}{5}.\)

We also have \(x > \frac{-8}{7}\) from \(-4x - 2 < 3x + 6\), because:

\(-4x - 2 < 3x + 6\)

\(-7x < 8\)

\(x > \frac{-8}{7}\) (the inequality sign changes direction since we divided by a negative)

Thus \(\frac{-8}{7} < x < \frac{-2}{5}\) and \(x = -1\) since \(x\) is an integer.

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)

\(x = -2, y = 1, z = \frac{4}{3}\) and \(x = -3, y = 0, z = 1\) are solutions.
Condition 2)

\(x = -2, y = 0, z = \frac{9}{4}\) and \(x = -1, y = 0, z = \frac{7}{4}\) are solutions.

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.

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 put \(x\) after \(y\), we have \(S = 100y + x.\)

And we have \(13 ≤ x ≤ 9\) and \(13 ≤ x ≤ y ≤ 99\) since \(x\) and \(y\) are two-digit numbers.

Since we have \(3\) variables (\(x, y\), and \(S\)) 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)
We can rearrange the equation \(S - (y - x) = 4289\) from condition 1) to get \(S = 4289 + (y - x).\) Substituting \(y - x = 27\) from condition 2) we get \(S = 4289 + 27 = 4316.\) Then \(y = 43\) and \(x = 16\).

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)
We have many possible pairs of \(x\) and \(y\) from \(y – x = 27.\)

Since condition 1) does not yield a unique solution, it is not sufficient

Condition 2)
Substituting \(S = 100y + x\), into \(S – (y-x)\) we get \(100y + x – (y - x) = 99y + 2x = 4289.\)

If we make \(y = 10a + b\) and \(x = 10 + c\) where \(1 ≤ a ≤ 9, 0 ≤ b ≤ 9\) and \(3 ≤ c ≤ 9.\)

Then we have \(S = 99y + 2x = 990a + 99b + 20 + 2c = 4289\) or \(990a + 99b + 2c = 4269.\)

When we substitute integers between \(1\) and \(9\) for the variable a one by one, we notice \(a = 4\) is the unique value in order to have units digits \(b\) and \(c\).

Then we have \(99b + 2c = 4269 – 3960 = 309.\)

When we substitute integers between \(1\) and \(9\) for the variable \(b\) one by one, we notice \(b = 3\) is the unique value in order to have a units digit \(c\).
Then we have \(c = 6.\)

Thus the boy’s age is \(16\) and his father’s age is \(43.\)

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

Therefore, B is the answer.
Answer: B

This question is a CMT 4 (A) question: When we easily get C as an answer, consider A and B as an answer.
If the question has both C and B as its answer, then B is an answer rather than C by the definition of DS questions. Also this question is a 50/51 level question and can be solved by using the relationship between the Variable Approach and Common Mistake Type 3 and 4 (A or B).

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) tells us that ( x1 + x2 + … + x8 ) / 8 = 5 or x1 + x2 + … + x8 = 40.
Condition 2) tells us that x9 + x10 = 9.
Thus, when we consider both conditions together, we have ( x1 + x2 + … + x8 + x9 + x10 ) / 10 = (40 + 9) / 10 = 49/10 = 4.9.

Therefore, C is the answer.
Answer: C

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

Condition 1)
x^3+1<0
⇔ (x+1)(x^2-x+1) < 0
⇔ x+1<0 since x^2-x+1 > 0
⇔ x < -1
In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
Thus, since the solution set of the question, ‘x<0’, includes that of condition 1), ‘x<-1’, condition 1) is sufficient.

Condition 2)
x^3+2x^2+x+2=0
⇔ x^2(x+2)+(x+2)=0
⇔ (x^2+1)(x+2)=0
⇔ x+2=0 since x^2+1 > 0
⇔ x = -2
Thus, x < 0 and condition 2) is sufficient.

Therefore, D is 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. 

=>

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 original condition \(|x+1|=2|x-1|\) is equivalent to \(x=\frac{1}{3}\) or \(x=3\) as shown below:

\(|x+1|=2|x-1|\)
\(=> |x+1|^2=(2|x-1|)^2\)
\(=> (x+1)^2=4(x-1)^2\)
\(=> x^2+2x+1=4(x^2-2x+1)\)
\(=> x^2+2x+1=4x^2-8x+4\)
\(=> 3x^2-10x+3 = 0\)
\(=> (3x-1)(x-3) = 0\)
\(=> 3x-1=0\) or \(x-3 = 0\)
\(=> x=\frac{1}{3}\) or \(x=3\)

Thus, condition 1) is sufficient since it gives a unique solution.

Condition 2) is not sufficient, since it allows both possible solutions.

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.

Since we have many variables (x1, x2, …, x7) 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)
We write the numbers as x1≤x2≤x3≤x4≤x5≤x6≤x7. Then x4 is their median.
From condition 1), x4 = 9 and the smallest possible number is 1 + 2 + 3 + 9 + 10 + 11 + 12 = 48. Therefore, the answer is ‘yes’.
Both conditions 1) & 2) together 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 we didn’t use condition 2) in the above argument, condition 1) is sufficient on its own.

By condition 1), x4 = 9 and the smallest possible number is 1 + 2 + 3 + 9 + 10 + 11 + 12 = 48. Therefore, the answer is ‘yes’.


Condition 2)
When the numbers are 6,7,8,9,10,11,12, their sum is 6 + 7 + 8 + 9 + 10 + 11 + 12 = 63 > 48, and the answer is ‘yes’.
When the numbers are 6,7,8,9,10,11,12, their sum is 1 + 2 + 3 + 4 + 5 + 6 + 12 = 33 < 48, and the answer is ‘no’.
Since we don’t obtain a unique answer, condition 2) is not 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.

=>

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 question is equivalent to asking if \(xy < 0\). This can be seen by multiplying both sides of the inequality by \(y^2.\)

Condition 1) is equivalent to \(xy < 0\) as shown below:
\(|x+y|<|x|+|y|\)
\(=> |x+y|^2<(|x|+|y|)^2\)
\(=> (x+y)^2<|x|^2+2|x||y|+|y|^2\)
\(=> x^2+2xy+y^2<x^2+2|xy|+y^2\)
\(=> 2xy<2|xy|\)
\(=> xy<|xy|\)
\(=> xy<0\)
Thus, condition 1) is sufficient.

Condition 2)
If \(x = -2\) and \(y = 1\), then the answer is ‘yes’.
If \(x = -1\) and \(y = -1\), then the answer is ‘no’.
Since it does not give a unique answer, condition 2) is not sufficient.

Therefore, the correct answer is A.
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.

Since we have \(3\) variables (\(x, y\), and \(a\)) 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 we have \(2^x = a\) and \(3^y=a\), we have \(2=a^{\frac{1}{x}}\) and \(3=a^{1/y}.\)

Then \(2*3 = a^{\frac{1}{x}}a^{\frac{1}{y}}=a^{\frac{1}{x}+\frac{1}{y}}=a^{\frac{1}{2}}\) or \(a^{\frac{1}{2}}=6.\)
So we have \(a = 6^2 = 36.\)

Since we have 2 equations from condition 1) and 1 equation from condition 2), even though we have three variables, both conditions together are 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 when 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.

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.

Let’s look at the condition 1). It tells us that the average of \(1\) and \(3\) is \(2\). If \(x\) is between \(1\) and \(3\) inclusive, then the standard deviation of \(1, 3\), and \(x\) is less than that of \(1\) and \(2\).

The answer is not unique, and the condition is not sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one.

Condition 2) is not sufficient, obviously.

When we consider both conditions 1) & 2) together, \(1, 2\), and \(3\) are possible values of \(x\).

The answer is not unique, and both conditions are not sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one.

Both conditions 1) and 2) together are not sufficient

Therefore, E is the answer.
Answer: E

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.