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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
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[GMAT math practice question]

(Algebra) If @. is defined as any function, what is the value of 2020 @. 2019?

1) For any x and y, x @ y is defined as x - y.
2) For any x, y and z, x @. x is defined as 0 and x @ (y @ z) is defined as (x @ y) + z.
_________________

Originally posted by MathRevolution on 13 Jan 2020, 00:16.
Last edited by MathRevolution on 15 Jan 2020, 00:21, edited 3 times in total.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Number Properties) $$x, y,$$ and $$z$$ are integers with $$3 ≤ x < y < z ≤ 30$$ and $$y$$ is a prime number. What is the value of $$x + y + z$$?

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

2) $$2xy = z$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Algebra) If @. is defined as any function, what is the value of 2020 @. 2019?

1) For any x and y, x @ y is defined as x - y.
2) For any x, y and z, x @. x is defined as 0 and x @ (y @ z) is defined as (x @ y) + z.

=>

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

Condition 1)
We have 2020 @. 2019 = 2020 – 2019 = 1.
Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
If x = y = z, then x @ (x @ x) = x @ 0, x @ (x @ x) = (x @ x) + x = 0 + x = x and x @ 0 = x.
If y = z, then we have x = x @ 0 = x @ (y @ y) = (x @ y) + y and x @ y = x – y.
Thus we have 2020 @. 2019 = 2020 – 2019 = 1.
Since condition 2) yields a unique solution, it is sufficient.

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).

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.
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(Inequality) Which is greater between $$(a + 2b)^2$$ and $$9ab$$?

1) $$1 < a < 2$$

2) $$\frac{1}{2} < b < 1$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Number Properties) $$x, y,$$ and $$z$$ are integers with $$3 ≤ x < y < z ≤ 30$$ and $$y$$ is a prime number. What is the value of $$x + y + z$$?

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

2) $$2xy = z$$

=>

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)
$$x = 3, y = 5, z = 2*3*5 = 30$$ are unique solutions as it is the only combination of numbers that works within the given conditions of $$3 ≤ x < y < z ≤ 30$$ and $$y$$ is a prime number. If $$x$$ and $$y$$ are larger numbers than $$z$$ is greater than $$30.$$ We then have $$x + y + z = 3 + 5 + 30 = 38.$$

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

Condition 1)
Since $$3 ≤ x < y < z ≤ 30$$, we have $$\frac{1}{30} ≤ \frac{1}{z} < \frac{1}{y} < \frac{1}{x} ≤ \frac{1}{3}$$ when we take reciprocals.

Since we have $$\frac{1}{x} + \frac{1}{y} = \frac{1}{2} + \frac{1}{z}$$, we have $$\frac{1}{2} < \frac{1}{x} + \frac{1}{y} < \frac{1}{x} + \frac{1}{x} = \frac{2}{x}$$ or $$\frac{1}{2} = \frac{2}{4} < \frac{1}{x}.$$

Thus $$x < 4$$ and we have $$x = 3.$$

Since we have $$\frac{1}{2} = \frac{1}{3} + \frac{1}{y}$$, we have $$\frac{1}{6} < \frac{1}{y}$$ or $$y < 6.$$

Since $$3 < y < 6$$ and $$y$$ is a prime number, we have $$y = 5.$$

$$\frac{1}{z} = \frac{1}{x} + \frac{1}{y} – \frac{1}{2} = \frac{1}{3} + \frac{1}{5} – \frac{1}{2} = \frac{10}{30} +\frac{ 6}{30} – \frac{15}{30} = \frac{1}{30}$$ or $$z = 30.$$

Then, $$x + y + z = 3 + 5 + 30 = 38.$$

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

This question is a CMT 4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Algebra) What is $$k$$?

1) $$3x + 5y = k + 1$$ and $$2x + 3y = k$$

2) $$x + y = 2$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Inequality) Which is greater between $$(a + 2b)^2$$ and $$9ab$$?

1) $$1 < a < 2$$

2) $$\frac{1}{2} < b < 1$$

=>

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.

The question is equivalent to the statement $$(a - b)(a - 4b)$$ is greater than or less than $$0$$ for the following reason:

$$(a + 2b)^2 - 9ab > 0$$

=> $$a^2 + 4ab + 4b^2 – 9ab > 0$$

=> $$a^2 - 5ab + 4b^2 > 0$$

=> $$(a - b)(a - 4b) > 0$$

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 $$1 < a < 2$$ and $$\frac{1}{2} < b < 1$$, we have $$\frac{1}{2} < b < 1 < a < 2$$ or $$b < a.$$

Since $$1 < a < 2$$ and $$2 < 4b < 4$$ (by multiplying the equation given in condition 2) by $$4$$), we have $$1 < a < 2 < 4b < 4$$ or $$a < 4b.$$

Then we have $$a – b > 0,$$ and $$a – 4b < 0$$ or $$(a - b)(a - 4b) < 0.$$

Thus, we have $$(a + 2b)^2 - 9ab > 0$$ and $$(a + 2b)^2$$ is greater than $$9ab.$$

Since both conditions together yield a unique solution, they are sufficient.

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: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Algebra) Thomas went to a grocery store and bought some fruit with $$25$$. The prices of the fruit are $$5, 1$$, and $$0.50$$ for each watermelon, pear, and apple, respectively. How many apples did he buy?

1) He bought $$10$$ pieces of fruit, and he didn’t get any change back.

2) He had at least one of each fruit.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Algebra) What is $$k$$?

1) $$3x + 5y = k + 1$$ and $$2x + 3y = k$$

2) $$x + y = 2$$

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

Since we have $$3$$ variables ($$x, y,$$ and $$k$$) 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 $$x + y = 2$$, we have $$y = 2 – x.$$

Substituting $$y = 2 - x$$ into $$3x + 5y = k + 1$$ gives us $$3x + 5(2 - x) = k + 1, 3x + 10 - 5x = k + 1, -2x + 10 = k + 1$$ or $$2x + k = 9.$$

Substituting $$y = 2 - x$$ into $$2x + 3y = k$$ gives us $$2x + 3(2 - x) = k, 2x + 6 = 3x = k, -x + 6 = k$$ or $$x + k = 6.$$

We now have $$2$$ equations: $$2x + k = 9$$ and $$x + k = 6$$. Rewriting the first equation gives us $$k = 9 - 2x$$. Substituting this into the second equation gives us $$x + 9 - 2x = 6, -x = -3$$, and $$x = 3.$$ Then $$x + k = 6$$ becomes $$3 + k = 6,$$ and $$k = 3.$$

Then we have $$x = 3$$ and $$k = 3.$$

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: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Algebra) Thomas went to a grocery store and bought some fruit with $$25$$. The prices of the fruit are $$5, 1$$, and $$0.50$$ for each watermelon, pear, and apple, respectively. How many apples did he buy?

1) He bought $$10$$ pieces of fruit, and he didn’t get any change back.

2) He had at least one of each fruit.

=>

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 $$x, y$$ and $$z$$ are the number of watermelons, pears, and apples, respectively.
Then we have $$5x + y + 0.5z = 25$$ or $$10x + 2y + z = 50.$$

Since we have $$3$$ variables ($$x, y,$$ and $$z$$) and $$1$$ equation, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
We have $$x + y + z = 10$$ and $$5x + y + 0.5z = 25$$ or $$10x + 2y + z = 50$$. Subtracting the first equation from the second equation gives us $$(10x + 2y + z) – (x + y + z) = 50 - 10$$, or $$9x + y = 40.$$

Then $$y = 40 – 9x.$$

Then the possible values of $$(x, y)$$ are $$(1, 31), (2, 22), (3, 13)$$, and $$(4, 4).$$

If $$x = 1,$$ and $$y = 31$$, then $$z = 10 – x – y = -22$$ doesn’t make sense, since $$z$$ must be positive.

If $$x = 2$$, and $$y = 22$$, then $$z = 10 – x – y = -14$$ doesn’t make sense, since $$z$$ must be positive.

If $$x = 3,$$ and $$y = 13$$, then $$z = 10 – x – y = -6$$ doesn’t make sense, since $$z$$ must be positive.

If $$x = 4,$$ and $$y = 4$$, then we have $$z = 10 – x – y = 2.$$

Since both conditions together yield a unique solution, they are sufficient.

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
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GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Geometry) In the figure below, $$AB$$ equals $$7$$. What is the area of the circumscribed circle of $$ABC$$?

1) Point $$O$$ is the circumcenter of $$△ABC.$$

2) The length of the perimeter of $$△AOC$$ is $$19$$.

Attachment: 1.20ds.png [ 8.68 KiB | Viewed 242 times ]

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

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[GMAT math practice question]

(Geometry) Is triangle $$ADE$$ an isosceles triangle?

1) $$AB = AC$$

2) $$BD = CE$$

Attachment: 1.21DS.png [ 7.09 KiB | Viewed 229 times ]

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Status: GMATINSIGHT Tutor
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Schools: IIM (A)
GMAT 1: 750 Q51 V41
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MathRevolution wrote:
[GMAT math practice question]

(Geometry) Is triangle $$ADE$$ an isosceles triangle?

1) $$AB = AC$$

2) $$BD = CE$$

Attachment:
1.21DS.png

The answer would be clearly Option C

but with the information BD = CE we know that triangle ADE also will be Isosceles as per given input that ABC is an isosceles triangle
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Geometry) In the figure below, $$AB$$ equals $$7$$. What is the area of the circumscribed circle of $$ABC$$?

1) Point $$O$$ is the circumcenter of $$△ABC.$$

2) The length of the perimeter of $$△AOC$$ is $$19$$.

Attachment:
1.20ds.png

=>

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 triangle, we have $$3$$ variables and $$1$$ equation, and 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 $$OA + OC + AC = 19$$ and $$AC = 7$$, we have $$OA + OC = 12.$$

Since $$OA = OC$$ is a radius from condition 1), we have the radius $$OA = OC = 6$$.

Then we can figure the area of the circumscribed circle of $$ABC$$ as follows:
$$A = πr^2 = π(6)^2 = 36π.$$

Since both conditions together yield a unique solution, they are sufficient.

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: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

[GMAT math practice question]

(Geometry) In the figure below, is triangle $$AEF$$ an isosceles triangle?

1) $$AB = AC$$

2) $$DF$$ is perpendicular to $$BC$$

Attachment: 1.22ds.png [ 8.57 KiB | Viewed 193 times ]

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Geometry) Is triangle $$ADE$$ an isosceles triangle?

1) $$AB = AC$$

2) $$BD = CE$$

Attachment:
1.21DS.png

=>

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 triangle $$ADE$$ has three sides, we have $$3$$ variables and $$0$$ equations, and 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)

When we consider condition 1), triangle $$ABC$$ is an isosceles triangle, and $$∠B$$ and $$∠C$$ are congruent.

Since $$BD = EC$$ from condition 2) and we have $$AB = AC,$$ and $$∠B = ∠C,$$ triangles $$ABD$$ and $$ACE$$ are congruent to each other using the $$SAS$$ property.

Thus, we have $$AD = AE$$, and the triangle $$ADE$$ is isosceles.

Since both conditions together yield a unique solution, they are sufficient.

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: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

[GMAT math practice question]

(Probability) On each face of a cube, one of $$1, 2$$ or $$3$$ is written. The number of $$1’s$$ on a face is $$a$$, the number of $$2’s$$ is $$b$$, and the number of $$3’s$$ is $$c$$. What is $$c$$?

1) $$a = 2$$ and $$b = 3.$$

2) The probability of throwing the two identical cubes and getting a sum of $$3$$ is $$\frac{1}{3}.$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Geometry) In the figure below, is triangle $$AEF$$ an isosceles triangle?

1) $$AB = AC$$

2) $$DF$$ is perpendicular to $$BC$$

Attachment:
1.22ds.png

=>

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 triangle $$AEF$$ has three sides, we have $$3$$ variables ($$AE, AF,$$ and $$EF$$) and $$0$$ equations, and 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 $$∠B = ∠C$$ since $$AB = AC$$, and the triangle is isosceles.

Assume $$∠B = ∠C = x.$$

Then $$∠DEC = ∠AEF = 90 – x$$ since the triangle is a right triangle, and $$∠DEC$$ is congruent to $$∠AEF.$$

Since triangle $$BDF$$ is a right triangle, we have $$∠AFE = 90 – x.$$

Thus we have $$∠AEF = ∠AFE$$, which means the triangle is isosceles, and we have $$AE = AF.$$

Since both conditions together yield a unique solution, they are sufficient.

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: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

[GMAT math practice question]

(Geometry) The figure shows that $$∠ABC$$ is $$80^o$$. What is $$∠ADC$$?

1) Point $$O$$ is the circumcenter of $$△ABC.$$

2) Point $$O$$ is the circumcenter of $$△ACD.$$

Attachment: 1.24ds.png [ 11.87 KiB | Viewed 179 times ]

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Probability) On each face of a cube, one of $$1, 2$$ or $$3$$ is written. The number of $$1’s$$ on a face is $$a$$, the number of $$2’s$$ is $$b$$, and the number of $$3’s$$ is $$c$$. What is $$c$$?

1) $$a = 2$$ and $$b = 3.$$

2) The probability of throwing the two identical cubes and getting a sum of $$3$$ is $$\frac{1}{3}.$$

=>

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 have $$3$$ variables and $$1$$ equation. However, we should check condition 1) alone first, since it has $$2$$ equations.

Condition 1)
Since we have $$a + b + c = 6, a = 2$$ and $$b = 3$$, we have $$2 + 3 + c = 6, 5 + c = 6,$$ and $$c = 1.$$

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

Condition 2)
Condition 2) tells us that $$\frac{c}{6} + \frac{c}{6} = \frac{1}{3}, \frac{(2c)}{6} = \frac{1}{3}, \frac{c}{3} = \frac{1}{3}, c = \frac{3}{3}.$$ Then we have $$c = 1.$$

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

_________________ Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS   [#permalink] 27 Jan 2020, 05:51

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# Math Revolution DS Expert - Ask Me Anything about GMAT DS  