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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(Arithmetic) $$a, b, c, d$$, and $$e$$ are integers with $$a < b < c < d < e$$. What is the maximum value of $$(a + e)$$?

1) $$a, b, c, d$$, and $$e$$ are consecutive integers.

2) The sum of $$a, b, c, d$$, and $$e$$ is negative.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Inequalities) $$m$$ is an integer where $$m = x + y.$$ What is the minimum value of $$m$$?

1) $$5 < x < 8.$$

2) $$-7 < y < -4.$$

=>

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 $$m$$) 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) together give us:
When we add the inequalities $$5 < x < 8$$ and $$-7 < y < -4$$, we have $$-2 < x + y < 4$$ or $$-2 < m < 4.$$

Since $$m$$ is an integer, we have $$-1 ≤ m ≤ 3.$$

Thus, the minimum value of $$m$$ is $$-1.$$

The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Since this question is an inequality 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.

Let’s look at the condition 1). Since it doesn’t have any information regarding the variable y, 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.

Let’s look at the condition 2). Since it doesn’t have any information regarding the variable x, 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.

Both conditions 1) and 2) together are sufficient.
Therefore, C is the correct answer.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Rate Problems) How far can Adam walk by talking $$18$$ steps?

1) The distance Adam travels in $$4$$ steps is equivalent to the distance Ben travels in $$5$$ steps.

2) The length of Ben’s stride is $$36 cm$$.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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1
MathRevolution wrote:
[GMAT math practice question]

(Arithmetic) $$a, b, c, d$$, and $$e$$ are integers with $$a < b < c < d < e$$. What is the maximum value of $$(a + e)$$?

1) $$a, b, c, d$$, and $$e$$ are consecutive integers.

2) The sum of $$a, b, c, d$$, and $$e$$ 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.

Since we have $$5$$ variables ($$a, b, c, d,$$ and $$e$$) 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) together give us that $$a + e = -2:$$

When we put five consecutive integers on the number line, the following case has the maximum negative value of $$a + b + c + d + e.$$

Attachment: 6.25(A).png [ 3.33 KiB | Viewed 293 times ]

Then $$a + e = (-3) + 1 = -2.$$

The answer is unique, and conditions 1) and 2) together are 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 sufficient.

Therefore, C is the correct answer.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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1
MathRevolution wrote:
[GMAT math practice question]

(Rate Problems) How far can Adam walk by talking $$18$$ steps?

1) The distance Adam travels in $$4$$ steps is equivalent to the distance Ben travels in $$5$$ steps.

2) The length of Ben’s stride is $$36 cm$$.

=>

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.

Assume $$a$$ and $$b$$ are the stride lengths of Adam and Ben, respectively.
The question asks for the value of $$18a.$$

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) together give us:

Condition 1) tells us that $$4a = 5b$$ and condition 2) tells us that $$b = 36.$$

Then we have $$4a = 5b = 5·36 = 180$$ or $$a = 45.$$

Thus, we have $$18a = 18·45 = 810.$$

The answer is unique, and conditions 1) and 2) together are 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 sufficient.

Therefore, C is the correct answer.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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1
[GMAT math practice question]

(Algebra) $$A○B$$ denotes the midpoint of $$A$$ and $$B$$ where $$A$$ and $$B$$ are points on the number line. $$[A]$$ denotes the coordinates of $$A. [C] = 92.$$ What is the value of $$[X]$$?

1) $$[A] = -5$$, and $$[B] = 7.$$

2) $$[A○B] = [C○X].$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(Statistics) a1, a2, …., an are n integers. What is the average of a1, a2, …., an?

1) The minimum value of possible averages of $$n-1$$ integers among a1, a2, …., an is $$60.$$

2) a1, a2, …., an are consecutive integers and a1 = 21.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Algebra) $$A○B$$ denotes the midpoint of $$A$$ and $$B$$ where $$A$$ and $$B$$ are points on the number line. $$[A]$$ denotes the coordinates of $$A. [C] = 92.$$ What is the value of $$[X]$$?

1) $$[A] = -5$$, and $$[B] = 7.$$

2) $$[A○B] = [C○X].$$

=>

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 ($$A, B, C,$$ and $$X$$) and $$1$$ equation, E is most likely the answer. So, we should consider conditions 1) & 2) together first After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2) together give us $$x = \frac{-5}{2}.$$

Assume $$x = [X].$$

We have $$[A○B] = \frac{[(-5) + 7] }{ 2} = \frac{2}{2} = 1$$, since we have $$[A] = -5$$ and $$[B] = 7.$$

We have $$[C○X] = \frac{[(9}{2) + x] / 2} = \frac{9}{4} + \frac{x}{2}.$$

Thus, we have $$\frac{9}{4} + \frac{x}{2} = 1$$ or $$\frac{x}{2} = 1 – (\frac{9}{4}) = \frac{-5}{4}.$$

Then we have $$x = \frac{-5}{2.}$$

The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Both conditions 1) & 2) together are sufficient.

Therefore, C is the correct answer.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Algebra) There are $$c$$ couches, and $$p$$ people sit on the couches. What is the value of $$p$$?

1) $$5$$ people sit on each couch first, and $$3$$ people sit on the last settee.

2) $$p$$ is a two-digit prime number.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Statistics) a1, a2, …., an are n integers. What is the average of a1, a2, …., an?

1) The minimum value of possible averages of $$n-1$$ integers among a1, a2, …., an is $$60.$$

2) a1, a2, …., an are consecutive integers and a1 = 21.

=>

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 many variables 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) together give us:

We have a1 = 20 + 1, a2 = 20 + 2, … , an = 20 + n from condition 2).
(a1 + a2 + … + an-1 ) / (n - 1) = [(20 + 1) + (20 + 2) + … + (20 + (n - 1))] / (n - 1) = [20(n - 1) + n(n - 1) / 2] / (n-1) = 20 + n/2 = 60.
We have n/2 = 40 or n = 80.

Thus, the maximum value of possible averages of a1, a2, … , an is
(21 + 22 + 23 + … + 100) / 80
= [80·(21 + 100) / 2] / 80, (21 + 100) / 2, or 121/2.

The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

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.

Let’s look at condition 1).
If a1 = 100 and a2 = 60, the average of a1 and a2 is 80.
If a1 = 80 and a2 = 60, the average of a1 and a2 is 70.

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.

Let’s look at condition 2).
If a1 = 21, a2 = 22 and a3 = 23, then the average of a1, a2 and a3 is (21 + 22 + 23) / 3 = 66/3 = 22.
If a1 = 21, a2 = 22, a3 = 23, a4 = 24 and a5 = 25, then the average of a1, a2 and a3 is (21 + 22 + 23 + 24 + 25) / 5 = 115/5 = 23.

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.

Both conditions 1) & 2) together are sufficient.

Therefore, C is the correct answer.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Algebra) $$A○B$$ denotes $$\frac{1}{A + B}-\frac{1}{A}$$ ($$A ≠ -B$$ and $$A ≠ 0$$). What is the value of $$xy$$?

1) $$x = (1 + A) ○ (1 - A).$$

2) $$y = (1 - A) ○ (1 + A).$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Algebra) There are $$c$$ couches, and $$p$$ people sit on the couches. What is the value of $$p$$?

1) $$5$$ people sit on each couch first, and $$3$$ people sit on the last settee.

2) $$p$$ is a two-digit prime number.

=>

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

Since we have $$2$$ variables ($$c$$ and $$p$$) 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) together give us:
We have $$5(c - 1) + 3 = p$$ or $$p = 5c – 2.$$

If $$c = 3$$, then $$p = 5·3 – 2 = 15 – 2 = 13.$$

If $$c = 5$$, then $$p = 5·5 – 2 = 25 – 2 = 23.$$

The answer is not unique, and both conditions 1) and 2) together are not sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Both conditions 1) & 2) together are not sufficient.

Therefore, E is the correct answer.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

[GMAT math practice question]

(Algebra) $$a$$ and $$x$$ are real numbers. What is the value of $$a$$?

1) $$3x - [7x - {2x - 4(5 - 6x)}] = -10x + 4.$$

2) $$–a + 5 = 11x.$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Algebra) $$A○B$$ denotes $$\frac{1}{A + B}-\frac{1}{A}$$ ($$A ≠ -B$$ and $$A ≠ 0$$). What is the value of $$xy$$?

1) $$x = (1 + A) ○ (1 - A).$$

2) $$y = (1 - A) ○ (1 + A).$$

=>

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

Since we have $$2$$ variables ($$x$$ and $$y$$) and $$0$$ equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2) together give us:
$$x = (1 + A) ○ (1 - A) = [\frac{1 }{ (1 + A + 1 - A)}] – [\frac{1 }{ (1 + A)}] = [\frac{1}{2}] – [\frac{1 }{ (1 + A)}] = [\frac{(1 + A) }{ 2(1 + A)}] – [\frac{2 }{ 2(1 + A)}] = [\frac{(1 + A - 2) }{ 2(1 + A)}] = [\frac{(A - 1) }{ 2(1 + A)}].$$

$$y = (1 - A) ○ (1 + A) = [\frac{1 }{ (1 – A + 1 + A)}] – [\frac{1 }{ (1 - A)}] = [\frac{1}{2}] – [\frac{1 }{ (1 - A)}] = [\frac{(1 - A) }{ 2(1 - A)}] – [\frac{2 }{ 2(1 - A)}] = [\frac{(1 – A - 2) }{ 2(1 - A)}] = \frac{-(A + 1) }{ 2(1 - A)}.$$

Then, we have $$xy = [\frac{(A - 1) }{ 2(1 + A)}] · [\frac{-(A + 1) }{ 2(1 - A)}] = \frac{[(A – 1)·(-A – 1)] }{ [4·(1 + A)·(1 – A)]} = \frac{[-A^2 – A + A + 1] }{ [4·(1 – A + A - A^2]} = \frac{[-A^2 + 1] }{ [4·(1 – A^2]} = \frac{[1 – A^2] }{ [4·(1 - A^2]} = \frac{1}{4}.$$

The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Both conditions 1) & 2) together are sufficient.

Therefore, C is the correct answer.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Algebra) $$a$$ and $$x$$ are real numbers. What is the value of $$a$$?

1) $$3x - [7x - {2x - 4(5 - 6x)}] = -10x + 4.$$

2) $$–a + 5 = 11x.$$

=>

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 $$x$$) 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) together give us:

From condition 1) we get:
$$3x - [7x - {2x - 4(5 - 6x)}] = -10x + 4$$

$$3x – 7x + (2x – 20 + 24x) = -10x + 4$$

$$3x - 7x + 26x – 20 = -10x + 4$$

$$22x – 20 = -10x + 4$$

$$32x = 24$$

$$x= \frac{24}{32} = \frac{3}{4}$$

From condition 2) we get:
$$–a + 5 = 11x$$

$$a = 5 – 11x$$

Thus, we have $$x = \frac{3}{4}$$ and $$a = 5 – 11x$$. So, $$a = 5 – 11(\frac{3}{4}) = 5 – \frac{33}{4} = \frac{20}{4} - \frac{33}{4} = \frac{-13}{4}.$$

Since 'no' is also a unique answer, according to CMT (Common Mistake Type) 1, both conditions are sufficient.

The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Both conditions 1) & 2) together are sufficient.

Therefore, C is the correct 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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(Geometry) The figure shows square $$ABCD$$ and $$2$$ lines, $$BP$$ and $$BQ$$. Are triangles $$ABP$$ and $$CBQ$$ congruent?

Attachment: 7.6DS.png [ 5.77 KiB | Viewed 179 times ]

1 )$$PD=DQ.$$

2) $$BP=BQ.$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Set) Set $$A$$ is given as {$$1, 2, 3, … , n$$} and $$n$$ is a positive integer. What is the value of $$n$$?

1) The number of subsets of $$A$$ containing both $$1$$ and $$n$$ is $$16.$$

2) $$n$$ is less than $$8.$$
_________________

Originally posted by MathRevolution on 07 Jul 2020, 02:15.
Last edited by MathRevolution on 09 Jul 2020, 02:20, edited 1 time in total.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Geometry) The figure shows square $$ABCD$$ and $$2$$ lines, $$BP$$ and $$BQ$$. Are triangles $$ABP$$ and $$CBQ$$ congruent?

Attachment:
7.6DS.png

1 )$$PD=DQ.$$

2) $$BP=BQ.$$

=>

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 a square, we have $$1$$ variable and $$0$$ equations, and D is most likely the answer. So, we should consider each condition on its own first.

Let’s look at condition 1). It tells us that triangles $$ABP$$ and $$CBQ$$ are congruent to each other.

Since $$PD = DQ$$, we have $$AP = AD – PD$$, and $$CQ = CD – DQ$$ are congruent and $$∠A$$ and $$∠C$$ are congruent as well. We can prove triangles $$ABP$$ and $$CBQ$$ are congruent with the SAS (Side-Angle-Side) triangle congruence property.

The answer is unique, and the condition is sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Let’s look at condition 2). It tells us that triangles $$ABP$$ and $$CBQ$$ are congruent.

Since $$BP$$ and $$BQ$$ are congruent hypotenuses of two right triangles, and $$AB$$ and $$BC$$ are two congruent legs, triangles $$ABP$$ and $$CBQ$$ are congruent according to the $$HL$$ right triangle congruence property.

The answer is unique, and the condition is sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Each condition alone is sufficient.

Therefore, D is the correct answer.

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

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]

(Number Properties) $$P$$ is a positive integer greater than $$3$$. Is $$P + 1$$ a multiple of $$6$$?

1) Both $$P$$ and $$P+2$$ are prime numbers.

2) $$P$$ is a multiple of $$5$$.
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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MathRevolution wrote:
[GMAT math practice question]

(Set) Set $$A$$ is given as {$$1, 2, 3, … , n$$} and $$n$$ is a positive integer. What is the value of $$n$$?

1) The number of subsets of $$A$$ containing both $$1$$ and $$n$$ is $$16.$$

2) $$n$$ is less than $$8.$$

=>

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

Let’s look at condition 1). It tells us that set $$A$$ has four elements.

Remember that the number of subsets of a set with $$n$$ elements is $$2n$$.
Since $$2n = 24 = 16$$, we have $$n = 4$$.

The answer is unique, and the condition is sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Let’s look at condition 2). It tells us that we don’t have a unique solution since $$n = 6$$ and $$n = 7$$ are possible values.

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 1) ALONE is sufficient.

Therefore, A is the correct answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations,” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C, or E.
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# Math Revolution DS Expert - Ask Me Anything about GMAT DS  