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[GMAT math practice question]

(statistics) \(160\) students from School \(A\) and School \(B\) are chosen to take the mathematics test and the average of the test is \(66.5\). How many students from School \(A\) are chosen?

1) The average of School \(A\) is \(1.5\) points higher than the total average.

2) The average of School \(B\) is \(2.5\) points lower than the total average.

[GMAT math practice question]

(number properties) \(x, y,\) and \(z\) are integers and \(30 ≥ z > y > x ≥ 3.\) Also, \(y\) is a prime number. What are the values of \(x, y,\) and \(z\)?

1) \(\frac{1}{x} + \frac{1}{y} = \frac{1}{2} + \frac{1}{z}\)

2) \(2xy = z\)

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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.

We assume that \(a\) and \(b\) are the numbers of students in School \(A\) and School \(B\), and \(x\) and \(y\) are the average scores of Schools \(A\) and \(B\), respectively.



\(a + b = 160\)

Their average is \(\frac{(ax + by) }{ (a + b)} = \frac{(ax + by) }{ 160} = 66.5\) and it is equivalent to \(ax + by = 10640.\)

The question asks for the value of \(a\).

Since we have \(4\) variables and \(2\) 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 \(x = 66.5 + 1.5 = 68\) from condition 1).

We have \(y = 66.5 - 2.5 = 64\) from condition 2).

Then we have \(68a + 64b = 10640.\)

Since we have \(a + b = 160\), which we can multiply by \(64\) to get \(64a + 64b = 64*160 = 10240.\)

Then we can say \(4a = (68a + 64b) - (64a + 64b) = 10640 - 10240 = 400\) or \(a = 100.\)

Therefore, 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)
We have \(x = 66.5 + 1.5 = 68\) from condition 1).

The original condition tells us that \(a + b = 160\) and \(ax + by = 10640.\)

\(a = 100, b = 60, x = 68\) and \(y = 64\) and \(a = 120, b = 40, x = 68, y = 62\) are solutions.

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

Condition 2)
We have \(y = 66.5 - 2.5 = 64\) from condition 2).

The original condition tells us that \(a + b = 160\) and \(ax + by = 10640.\)

\(a = 100, b = 60, x = 68\) and \(y = 64\) and \(a = 64, b = 96, x = 70.25, y = 64\) are solutions.

Since condition 2) does not 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.

[GMAT math practice question]

(geometry) \(ABCD\) and \(AFGE\) are rectangles, and the area of \(ABCD\) is \(120\). What is the area of the rectangle \(AFGE\)?



1) The area of \(GBC\) is \(24.\)

2) The area of \(EGD\) is \(9.\)

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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.

[GMAT math practice question]

(inequality) \(a, b,\) and \(c\) are the lengths of the sides of an obtuse triangle. What is the maximum value of \(a\)?

1) \(a < b < c = 20.\)

2) \(a, b,\) and \(c\) are integers.

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We can draw an additional line \(GH\) on the figure.



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 the area of rectangle \(BFHC\) is \(48\) since it is two times greater than the area of triangle \(GBC\).

Condition 2) tells us that the area of rectangle \(EGHD\) is \(18\) since it is two times greater than the area of triangle \(EGD\).

Then we have\( AFGE = ABCD – (FBCH + EGHD).\)

Since we have \(4\) variables (\(AFGE, ABCD, FBCH\) and \(EGHD\)) and \(2\) equations (\(ABCD = 120\) and \(AFGE = ABCD – (FBCH + EGHD)\)), 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)
\(AFGE = ABCD – (FBCH + EGHD) = 120 – (48 + 18) = 120 – 66 = 54.\)

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.

[GMAT math practice question]

(inequality) Which one of \((a + 2b)^2\) and \(9ab\) is greater?

1) \(1 < a < 2\).

2) \(12 < b < 1.\)

=>

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 the triangle is obtuse, we have \(c^2 > a^2 + b^2\) or \(400 > a^2 + b^2\) from condition 1). Since \(a < b\) or \(a^2 < b^2\) and \(a\) is an integer, we have \(a^2 < 200\) or \(a ≤ 13\) from condition 2).

The maximum value of \(a\) is \(13\).

Since both conditions together yield a unique solution, they 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 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.

\((a + 2b)^2 - 9ab = a^2 + 4ab + 4b^2 - 9ab = a^2 – 5ab + 4b^2 = (a - b)(a - 4b)\)

If \((a - b)(a - 4b) > 0\), then \((a + 2b)^2\) is greater than \(9ab.\)

If \((a - b)(a - 4b) < 0,\) then \(9ab\) is greater than \((a + 2b)^2.\)

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 \(b < 1 < a,\) we have \(b < a\) or \(a – b > 0.\)

Since \(a < 2 < 4b,\) we have \(a < 4b\) or \(a – 4b < 0.\)

Then we have \((a - b)(a - 4b) < 0\) and \((a + 2b)^2\) is greater than \(9ab.\)

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.

[GMAT math practice question]

(algebra) \(x, y\) and \(z\) are the number of \($2\) stamps, \($3\) stamps and \($10\) stamps, respectively. What is the value of \(xyz\)?

1) The total number of stamps is \(20\).

2) The total value of all the stamps is \($100\).

[GMAT math practice question]

(algebra) What is the value of \(xy\)? (max(\(x, y\)) denotes the maximum between \(x\) and \(y,\) and min(\(x, y\)) denotes the minimum between \(x\) and \(y\))

1) \(x + y = 7\) and \(x - y = 1.\)

2) max(\(x, y\)) = \(2x + 3y - 13\) and min(\(x, y\)) = \(3x - y - 6.\)

=>

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 \(z\)) 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 \(x + y + z = 20\) from condition 1) and \(2x + 3y + 10z = 100\) from condition 2).

When we subtract two times the first equation from the second equation, we have \(y + 8z = (2x + 3y + 10z) - 2(x + y + z) = 100 – 2*20, 2x + 3y + 10z - 2x - 2y - 2z = 100 - 40, y + 8z = 60, and y = 60 - 8z.\)

Then we notice that \(x = 9, y = 4, z = 7\) and \(x = 2, y = 12, z = 6\) are solutions of the equation system with \(x + y + z = 20\) and \(2x + 3y + 10z = 100.\)

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.

[GMAT math practice question]

(algebra) There are \(3\) kinds of gifts \(A, B\) and \(C\) in a box. The number of gifts \(A, B\) and \(C\) are \(a, b\) and \(c,\) respectively. The prices of \(A, B\) and \(C\) are \($3, $2\) and \($1.\) The total price of all gifts in the box is \($48\). What is the total price of gift \(A\)?

1) \(a < b < c.\)

2) \(a, b\) and \(c\) are all even numbers.

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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 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.

[GMAT math practice question]

(inequalities) There is an inequality \(ax > b\) in terms of \(x\). Does it have a solution?

1) \(a = 0\)

2) \(b ≥ 0\)

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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.

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.