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[GMAT math practice question]

(Inequalities) Is \(b-a\) positive?

1) The solution set of \((a - 3b)x + (b - 3a) < 0\) is \(x > \frac{5}{3}\)

2) \(ab < 0\)
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[GMAT math practice question]

(Inequalities) What is the minimum value of \(a\)?

1) The solution of \(3 - 3x ≥ 2x - 7\) and \(x + 3 > a\) is \(ø\).

2) \(a\) is an integer.
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[GMAT math practice question]

(Inequalities) Is \(b-a\) positive?

1) The solution set of \((a - 3b)x + (b - 3a) < 0\) is \(x > \frac{5}{3}\)

2) \(ab < 0\)

=>

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)
\((a - 3b)x + (b - 3a) < 0\)

\(=> (a - 3b)x < (3a - b)\)

\(=> x > \frac{(3a - b) }{ (a - 3b)}\) under the assumption \(a - 3b < 0.\)

Then, we have \(\frac{(3a - b)}{(a - 3b)} = \frac{5}{3}\) or \(3(3a - b) = 5(a - 3b).\)

It is equivalent to \(9a – 3b = 5a – 15b, 4a = -12b\) or \(a = -3b.\)

Since we have the assumption \(a – 3b < 0\), we have \((-3b) - 3b = -6b < 0\) or \(b > 0.\) Since b is positive, \(a = -3b\) is negative.

Thus \(b – a\) is positive and condition 1) is sufficient, since it yields a unique solution.

Condition 2)
If \(a = 1\) and \(b = -1\), then \(b – a = (-1) - 1 = -2\) is negative and the answer is ‘no’.

If \(a = -1\) and \(b = 1\), then \(b – a = 1 - (-1) = 2\) is positive and the answer is ‘yes’.

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

Therefore, A is the answer.
Answer: A
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[GMAT math practice question]

(Inequalities) Is \(xy + 1\) greater than \(x + y\)?

1) \(0 ≤ x < 1\) and \(0 ≤ y < 1\).

2) \(x + y\) is negative.
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[GMAT math practice question]

(Inequalities) What is the minimum value of \(a\)?

1) The solution of \(3 - 3x ≥ 2x - 7\) and \(x + 3 > a\) is \(ø\).

2) \(a\) is an integer.

=>

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

Condition 1)
\(3 - 3x ≥ 2x - 7\)

=> \(10 ≥ 5x\)

=> \(x ≤ 2\)

\(x + 3 > a\)
=> \(x > a – 3\)

Looking at the intersection of \(x ≤ 2\) and \(x > a - 3\), we have \(a - 3 ≥ 2\) or \(a ≥ 5.\)

Thus, the minimum value of \(a\) is \(5\).



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

Condition 2)
Condition 2) tells us that a is an integer, which gives us an infinite number of possibilities.

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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[GMAT math practice question]

(Algebra) \(a, b\), and \(c\) are integers greater than \(1\) with \(a < b < c\). What is the value of \(a + b + c\)?

1) \((ab - 1)(bc - 1)(ca - 1)\) is divisible by \(abc\).

2) \(a, b,\) and \(c\) are prime numbers less than \(6.\)
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[GMAT math practice question]

(Inequalities) Is \(xy + 1\) greater than \(x + y\)?

1) \(0 ≤ x < 1\) and \(0 ≤ y < 1\).

2) \(x + y\) is negative.

=>

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.

The question \(xy + 1 > x + y\) is equivalent to \((x - 1)(y - 1)>0\) for the following reason.
\(xy + 1 > x + y\)

=> \(xy + 1 – x – y > 0\)

=> \(xy - x - y + 1 > 0\) (rearranging the equation)

=> \(x(y - 1) - 1(y - 1) > 0\)

=> \((x - 1)(y - 1) > 0\)

=> \(x > 1, y > 1\) or \(x < 1, y < 1\)

Condition 1)
Condition 1) tells us that \(0 ≤ x < 1\) and \(0 ≤ y < 1\). This tells us that \(x < 1\), and \(y < 1\), which fits our modified condition. Therefore, the answer is unique, 'yes,' and the condition is sufficient.

Condition 2)
If \(x = -1\), and \(y = -1\), then \(xy + 1 = (-1)(-1) + 1 = 1 + 1 = 2, x + y = (-1) + (-1) = -2\). In this case, \(xy + 1\) is greater than \(x + y\) and the answer is ‘yes’.

If \(x = 2,\) and \(y = -3\), then \(xy + 1 = (2)(-3) + 1 = -6 + 1 = -5, x + y = 2 + (-3) = -1.\) In this case, \(xy + 1\) is less than \(x + y\) 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
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[GMAT math practice question]

(Number Properties) \(a, b\) and \(c\) are positive integers with \(a ≤ b ≤ c\). What is the value of \(a + b + c\)?

1) \(abc = a + b + c.\)

2) \(abc = 6.\)
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[GMAT math practice question]

(Algebra) \(a, b\), and \(c\) are integers greater than \(1\) with \(a < b < c\). What is the value of \(a + b + c\)?

1) \((ab - 1)(bc - 1)(ca - 1)\) is divisible by \(abc\).

2) \(a, b,\) and \(c\) are prime numbers less than \(6.\)

=>

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.

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)
Since \(a, b\) and \(c\) are prime numbers less than \(6\) with \(a < b < c\), we have \(a = 2, b = 3\) and \(c = 5.\)

Thus we have \(a + b + c = 10.\)

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

Condition 1)
\((ab - 1)(bc - 1)(ca - 1)\)

\(= (ab^2c – ab – bc + 1)(ca - 1)\)

\(= a^2b^2c^2 – ab^2c – a^2bc – abc^2 + ab + bc + ca – 1\)

\(= a^2b^2c^2 – abc(a + b + c) + (ab + bc + ca) – 1\)

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

Since \((ab - 1)(bc - 1)(ca - 1)\) is divisible by \(abc\), we notice that \(ab + bc + ca – 1\) is divisible by \(abc.\)

Then we have \(ab + bc + ca – 1 = abc*n\) for some integer.

When we divide both sides of the equation by \(abc,\) we have \(n = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} – \frac{1}{abc} < \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2},\) since \(2 ≤ a < b < c\) or \(\frac{1}{c} > \frac{1}{b} > \frac{1}{a} ≥ \frac{1}{2}.\)

Then the positive integer \(n\) equals \(1.\)

When we divide both sides of the equation \(ab + bc + ca – 1 = abc\) by \(bc\), we have \(a = 1 + \frac{a}{c} + \frac{a}{b} – \frac{1}{bc} < 1 + 1 + 1 – \frac{1}{bc} < 3.\)

Then we have \(a = 2.\)

When we substitute a with \(2\) in the equation \(ab + bc + ca - 1 = abc,\) we have \(2b + bc + 2c - 1 = 2bc\) or \(bc – 2b – 2c + 1 = 0.\)

Then we have \(bc – 2b – 2c + 4 = 3\) or \((b - 2)(c - 2) = 3.\)

Then we have \(b – 2 = 1, c – 2 = 3\) or \(b = 3, c = 5.\)

Thus, we have \(a + b + c = 10.\)

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

Therefore, D is the answer.
Answer: D

This question is a CMT4 (B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT4 (B) questions, D is most likely to be the answer.
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[GMAT math practice question]

(Number Properties) \(a, b\) and \(c\) are positive integers with \(a ≤ b ≤ c\). What is the value of \(a + b + c\)?

1) \(abc = a + b + c.\)

2) \(abc = 6.\)

=>

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.

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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[GMAT math practice question]

(Functions) A function \(f(x) = -3x + 16\) is a linear function and \(f(a+b) = c.\) What is \(f(|c|)\)?

1) \(f(a) = -a\)

2) \(f(b) = b\)
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[GMAT math practice question]

(Geometry) There is a point \(P(a, b).\) What is the value of \(a+b\)?

1) \(P\) is on the line \(\frac{x}{3} + \frac{y}{4} = 1 \)

2) The line \(\frac{x}{3} + \frac{y}{4} = 1\) is parallel to the line \(\frac{a}{3}x + \frac{b}{4}y = 1 \)
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[GMAT math practice question]

(Functions) A function \(f(x) = -3x + 16\) is a linear function and \(f(a+b) = c.\) What is \(f(|c|)\)?

1) \(f(a) = -a\)

2) \(f(b) = b\)

=>

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.
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.

\(f(a+b) = c\)

=> \(-3(a+b) + 16 = -3c + 16\)

=> \(-3a - 3b + 16 = -3c + 16\)

=> \(-3a - 3b = -3c\) (by subtracting \(16\) on both sides)

=> \(a + b = c\) (by dividing by \(-3\) on both sides)

Since we have \(3\) variables (\(a, b\), and \(c\)) and \(1\) equation (\(a + b = c\)), 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 \(a = 8\) from condition 1) for the following reason.

\(f(a) = -a\)

=> \(-3a + 16 = -a\)

=> \(2a = 16\)

=> \(a = 8\)

We have \(b = 4\) from condition 2) for the following reason.

\(f(b) = b\)

=> \(-3b + 16 = b\)

=> \(4b = 16\)

=> \(b = 4.\)

Then we have \(c = a + b = 8 + 4 = 12.\)

\(f(|c|) = f(|12|) = f(12) = -3(12) + 16 = -36 + 16 = -20.\)

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.
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[GMAT math practice question]

(Geometry) What is the measure of the angle \(∠EAC\)?

1) \(BD = DE = EA = AC\)

2) \(∠ACE = ∠DBE + 40^O\)

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1.1 ds.png [ 11.79 KiB | Viewed 1450 times ]
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[GMAT math practice question]

(Geometry) There is a point \(P(a, b).\) What is the value of \(a+b\)?

1) \(P\) is on the line \(\frac{x}{3} + \frac{y}{4} = 1 \)

2) The line \(\frac{x}{3} + \frac{y}{4} = 1\) is parallel to the line \(\frac{a}{3}x + \frac{b}{4}y = 1 \)

=>

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.

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)

We have \(\frac{a}{3} + \frac{b}{4} = 1\) since the point \(P(a, b)\) is on the line \(\frac{x}{3} + \frac{y}{4} = 1.\)

We have \(a = b\) since \(\frac{x}{3} + \frac{y}{4} = 1\) is parallel to the line \((\frac{a}{3})x + (\frac{b}{4})y = 1\) or \(\frac{1}{3} : \frac{a}{3} = \frac{1}{4} : \frac{b}{4}.\)

Then, we have \(\frac{a}{3} + \frac{a}{4} = 1, \frac{4a}{12} + \frac{3a}{12} = 1, (\frac{7}{12})a = 1\) or \(a = \frac{12}{7}.\)

Thus we have \(a + b = \frac{12}{7} + \frac{12}{7} = \frac{24}{7}.\)

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.
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[GMAT math practice question]

(Geometry) As the figure below shows, line \(l\) is perpendicular to \(BD\) and \(CE\). What is the length of \(DE\)?

1) \(△ABC\) is a right isosceles triangle with \(∠BAC = 90^o\).

2) \(BD = 3\), and \(CE = 4.\)

Attachment:
1.2ds.png
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[GMAT math practice question]

(Geometry) What is the measure of the angle \(∠EAC\)?

1) \(BD = DE = EA = AC\)

2) \(∠ACE = ∠DBE + 40^O\)

Attachment:
The attachment 1.1 ds.png is no longer available

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Since we have \(9\) variables from \(3\) triangles and \(5\) equations, \(∠BDE + ∠EDA = 180, ∠BED + ∠DEA + ∠EAD = 180, ∠ABC + ∠BCA + ∠CAB = 180, ∠EBD + ∠EDB + ∠BED = 180\) and \(∠EAC + ∠ECA + ∠AEC = 180\), 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)

Attachment:
1.1ds(a).png
1.1ds(a).png [ 19.24 KiB | Viewed 1434 times ]

Since we have \(BD = DE = EA = AC\) from condition 1), we have the measures of the interior angles, as shown in the above figure.

Since \(∠ACE = ∠DBE + 40°\) from condition 2), we have \(3x = x + 40°, 2x = 40°\) or \(x = 20°.\)

Thus, we have \(∠EAC = 180° – 6x = 180° – 6(20), ∠EAC = 180° = 120 = 60°.\)

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.
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