Overview of GMAT Math Question Types and Patterns on the GMAT

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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.

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.

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

Condition 1)

\(x^3-x = 0\)
\(=> x(x^2-1) = 0\)
\(=> x(x+1)(x-1) = 0\)
\(=> x = 0, x = -1\) or \(x = 1.\)
Since it does not yield a unique solution, condition 1) is not sufficient.

Condition 2)
\(x = -x\)
\(=> 2x = 0\)
\(=> x = 0.\)
Since it gives a unique solution, condition 2) is sufficient.

Therefore, B is the answer.
Answer: B

=>

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 parity of \(x^2+x+y = x(x+1) + y\) is same as the parity of \(y\), since \(x^2+x = x(x+1)\) is the product of two consecutive integers and so it is always an even integer.
Thus, asking whether \(x^2+x+y = x(x+1) + y\) is odd is equivalent to asking whether \(y\) is odd.

Therefore, B is the answer.
Answer: B

=>

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. We should simplify the conditions if necessary.

Condition 1)
\(m\) and \(n\) have a unique common divisor since \(1\) and \(m\) are the only factors of \(m\) and, \(1\) and \(n\) are the only factors of \(n\). This tells us that \(gcd(m,n)=1\) and condition 1) is sufficient.

Condition 2)
Since the greatest common divisor of consecutive integers is \(1\), condition 2) is 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.

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 asks for the value of \(\frac{( x^2 + y^2 )}{( x^2 - y^2 )}= ( (\frac{x}{y})^2 + 1 ) / (\frac{x}{y})^2 – 1 ).\)

When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient.

Condition 1)

Since \(|\frac{x}{y}|=\frac{1}{3}, \frac{x}{y} = ±(\frac{1}{3})\), and \(\frac{( x^2 + y^2 )}{( x^2 - y^2 )}= ( (\frac{x}{y})^2 + 1 ) / ( (\frac{x}{y})^2 – 1 ) = ( (\frac{1}{3})^2 + 1 ) / ( (\frac{1}{3})^2 – 1) = (\frac{1}{9} + 1)/(\frac{1}{9}-1) = (\frac{10}{9})/(-\frac{8}{9}) = -\frac{10}{8} = -\frac{5}{4}.\)
Condition 1) is sufficient since it gives a unique solution.


Condition 2)
Since \(y = -3x, \frac{x}{y} = -\frac{1}{3}\), and \(\frac{( x^2 + y^2 )}{( x^2 - y^2 )}= ( (\frac{x}{y})^2 + 1 ) / ( (\frac{x}{y})^2 – 1 ) = ( (-\frac{1}{3})^2 + 1 ) / ( (-\frac{1}{3})^2 – 1) = (\frac{1}{9} + 1)/(\frac{1}{9}-1) = (\frac{10}{9})/(-\frac{8}{9}) = -\frac{10}{8} = -\frac{5}{4}.\)
Condition 2) is sufficient since it gives a unique solution.

Therefore, the answer is D.
Answer: D

FYI: Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.

=>

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.

If the maximum or minimum value of a data set is the same as its average, then all data points are equal and their standard deviation is 0.
Thus, each of the conditions is sufficient.

Therefore, D is the answer.
Answer: D

Recall the property that the SD is zero if the maximum value or the minimum value is the same as the average, as this implies that all data points are equal.
Questions related to the above property have appeared frequently on recent GMAT exams.

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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.

Suppose \(x, y\) and \(z\) are different integers with \(x < y < z\).
For their average \(\frac{( x + y + z )}{3}\) to be equal to their median \(y\),
we must have \(z – y = y – x\), and so their range is \(z – x = z – y + y – x = 2(y-x).\)
This implies that \(z – x\) is an even integer.

Condition 1)
Since condition 1) gives an odd value for the range, the answer is ‘no’. Thus, condition 1) is sufficient by CMT (Common Mistake Type) 1.

Condition 2)
If \(x = 10, y = 11\) and \(z = 12\), then the average and the median are the same, and the answer is ‘yes’
If \(x = 10, y = 11\) and \(z = 15\), then the average \(12\) is different from the median \(11\), and the answer is ‘no’.
Thus, condition 2) is not sufficient, since it does not yield a unique solution.

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.

Note: Even though C is most likely to be an answer since we have two variables, D is likely to be the answer by Tip 1) because conditions 1) and 2) provide the same information.

Condition 1)
\((x^2-6xy+9y^2)=0\)
\(=> (x-3y)^2 = 0\)
\(=> x -3y = 0\)
\(=> x = 3y\)
\(=> \frac{x}{y} = 3\)
Thus, condition 1) is sufficient.

Condition 2)
\(|x-3|+|y-1|=0\)
\(=> |x-3|+|y-1|=0\)
\(=> x=3\) and \(y=1\)
So, \(\frac{x}{y} = \frac{3}{1} =3.\)
Thus, condition 2) 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.

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)
By condition 1),
\(m!*n! = (n+1)!*n! = (n+1)(n!)(n!) = (n+1)(n!)^2 = m(n!)^2.\)
Since \((n!)^2\) is an integer squared, and \(m\) is an integer squared by condition 2), \(m!*n!\) is an integer squared.
Thus, both conditions 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)
If \(m = 4\) and \(n = 3\), then \(4!*3! = 4*3!*3! = (2(3!))^2\) and the answer is ‘yes’.
If \(m = 3\) and \(n = 2\), then \(3!*2! = 6*2 = 12\), and the answer is ‘no’.
Since it does not give a unique answer, condition 1) is not sufficient on its own.

Condition 2)
If \(m = 4\) and \(n = 3\), then \(4!*3! = 4*3!*3! = (2(3!))^2\) and the answer is ‘yes’.
If \(m = 4\) and \(n = 2,\) then \(4!*2! = 24*2 = 48\) and the answer is ‘no’.
Since it does not give a unique answer, condition 2) is not sufficient on its own.

Therefore, the answer is C.
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.

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)
\((n-17)(n-21) = 0\) is equivalent to the statement \(n = 17\) or \(n =21\)
If \(n = 17\), then \(17 = 19 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
If \(n = 21\), then \(21 = 23 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
Since it gives a unique answer, condition 1) is sufficient.

Condition 2)
\((n-15)(n-17) = 0\) is equivalent to the statement \(n = 15\) or \(n = 17\)
If \(n = 15\), then \(15 = 17 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
If \(n = 17\), then \(17 = 19 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
Since it gives a unique answer, condition 2) is sufficient.

Therefore, D is the answer.
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.

=>

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. We should simplify conditions if necessary.

Since \((\frac{7}{15})x+\frac{1}{3} = \frac{4}{3}\) and \(7x + 5 = 20\) by definition of set \(A\), we must have \(x = \frac{15}{7}\) and \(A = { \frac{15}{7} }.\)

Since \(2m-(\frac{1}{15})x = 3\) and \(30m – x = 45\) by definition of set \(B\), \(x = 30m – 45\) and \(B = { 30m – 45 }.\)

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

Condition 1)
Since \(A∩B≠Ø\), we must have \(30m – 45 = \frac{15}{7}\) and condition 1) yields a unique solution. It is sufficient.

Condition 2)
Since m can be any value and condition 2) doesn’t 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.

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

Asking whether \(x^2-y^2\) is even is equivalent to asking whether \((x+y)(x-y)\) is an even integer.

Condition 2) is sufficient since \(x^2-y^2 = (x+y)(x-y)\) is an even integer if \(x + y\) is an even integer.

Condition 1)
In order for \(x^3-y^3\) to be an even integer, \(x\) and \(y\) must both have the same parity.
There are only two cases to consider: both \(x\) and \(y\) are even integers or both are odd integers.
Since \(x\) and \(y\) have the same parity, \(x – y\) is always an even integer.
Thus, \(x^2-y^2\) is an even integer. Condition 1) 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.

Since we have \(4\) variables (\(a, b, x\) and \(y\)) 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 \(a = 1, b = 1, x = 1\) and \(y = 1\), then \(ax+by = 2\), and the answer is ‘yes’.
If \(a = -1, b = -1, x = 1\) and \(y = 1\), then \(ax+by = -2\), and the answer is ‘no’.

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

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.

Solution: The resultant percentage will be the weighted average of the percentages of the above two stocks.

=> \(\frac{[(1,200 * 0.25) + (400 * 0.20)]}{ [(1,200 + 400)]}\) * 100

=> \(\frac{(300 + 80) }{ (1,600)}\) * 100

=> (\(\frac{380}{1,600}\)) * 100 => 23.75%

Therefore, B is the correct answer.

Answer B

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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 (\(A\) and \(B\)) 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)
Suppose \(A=aG\) and \(B=bG\) for some integers \(a, b\) and \(G,\) where \(a\) and \(b\) are relatively prime.

Then
\(\frac{G}{A} + \frac{G}{B} = \frac{7}{10}\)
\(=> \frac{G}{(aG)} + \frac{G}{(bG)} = \frac{7}{10}\), since \(A=aG\) and \(B = bG\)
\(=> \frac{bG}{(abG)} + \frac{aG}{(abG)} = \frac{7}{10}\), taking a common denominator
\(=> \frac{(aG+bG)}{(abG)} = \frac{7}{10}\)
\(=> \frac{(A+B)}{L} = \frac{7}{10}\)
\(=> \frac{(A+B)}{70} = \frac{7}{10}\), since \(L = 70\)
\(=> A+B = 49\)
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 = 2 and B = 5, then A + B = 7.
If A = 6 and B = 15, then A + B = 21.
Since condition 1) doesn’t yield a unique solution, it is not sufficient.

Condition 2)
If A = 14 and B = 35, then A + B = 49.
If A = 2 and B = 35, then A + B = 37.
Since condition 2) doesn’t yield a unique solution, it is not 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.

Let \(x\) be the lost weight of \(b%\) saline solution by boiling.

The question asks the value of \(x\) such that \((\frac{a}{100})500 = (\frac{b}{100})(500-x).\)

\((\frac{a}{100})500 = (\frac{b}{100})(500-x)\)

\(=> 500a = b(500-x)\)

\(=> 500a = 500b – bx\)

\(=> bx = 500(b-a)\)

\(=> x = \frac{500(b-a)}{b}\)

\(=> x = (500)(1-\frac{a}{b})\)

Thus, condition 2) is sufficient.

When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient. This tells us that B is most likely to be the answer to this question.

Condition 1) is not sufficient obviously, since we don’t have any information about the variable a.

Therefore, B is the answer.
Answer: B

=>

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.