[GMAT math practice question]

(Geometry) The figure shows right triangle ABC and AB = 5, BC = 4 and CA = 3. What is the area of triangle BDE?

1) AC = AD
2) AB is perpendicular to DE.



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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 (AD and EC) in the original condition, C is most likely the answer. We can figure out where D and E are by AD and EC. 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 triangles AED and AEC are congruent, we have DE = CE and DB = 5 - 3 = 2.
The area of triangle ABC = (1/2)*4*3 = 6, which is also the sum of the areas of triangle ABE and triangle AEC.
Then we have (1/2)*5*DE + (1/2)*3*CE = (5/2)DE + (3/2)DE = (8/2)DE = 4DE = 6.
Thus, we have DE = 6/4, or DE = 3/2.
The area of triangle DBE is (1/2)*DB*DE = (1/2)*2*(3/2) = 3/2.


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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[GMAT math practice question]

(Inequalities) a, b and c are positive and a < 5. a, b and c satisfy a2 - a - 2b - 2c = 0, and a + 2b - 2c + 3 = 0. What is the order among a, b and c?

A. c > a > b
B. a > c > b
C. a > b > c
D. c > b > a
E. b > c > a

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When we add the two equations a^2 - a - 2b - 2c = 0 and a + 2b - 2c + 3 = 0, we have a^2 - a - 2b - 2c + a + 2b - 2c + 3 = 0 + 0, a^2 – 4c + 3 = 0 or 4c = a^2 + 3.
When we subtract 4a from the both sides of the last equation, we have 4c - 4a = a^2 - 4a + 3, 4(c - a) = (a - 1)(a - 3).
Rearranging the equations given in the question gives us a^2 - a = 2(b + c) and a + 3 = 2(c - b) given in the question. When we subtract a + 3 = 2(c - b) from a^2 - a = 2(b + c), we have a^2 - a - (a + 3) = 2(b + c) - 2(c - b), a^2 - a - a - 3 = 2b + 2c - 2c +2b, 4b = a^2 – 2a – 3, 4b = (a - 3)(a + 1) > 0 and a > 3.
We have 4(c - a) = (a - 1)(a - 3) > 0 or c > a since a > 3 and (a - 1)(a - 3) > 0.

We have 4(b - a) = 4b – 4a and 4b – 4a = a^2 – 2a – 3 – 4a (because 4b = a^2 – 2a – 3). Then 4b - 4a = a^2 – 6a – 3 = a(a - 3) – 3(a - 3) – 12 = (a - 3)2 - 12 < 0 or b < a since 3 < a < 5.
Thus, we have c > a > b.

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

(Inequalities) x is an integer. y and z are real numbers with x < 2y < 3z. What is the value of x?

1) x + 2y + 3z = 4
2) 2x + 3y + 4z = 5

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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.

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 + 2y + 3z = 4 and 2x + 3y + 4z = 5.
When we subtract twice the second equation from three times the first equation, we have
3(x + 2y + 3z) - 2(2x + 3y + 4z) = 3(4) - 2(5)
3x + 6y + 9z - 4x - 6y - 8z = 12 - 10
-x + z = 2
z = x + 2.

Substituting z = x + 2 into the first equation gives us:
x + 2y + 3z = 4
x + 2y + 3(x + 2) = 4
x + 2y + 3x + 6 = 4
4x + 2y = -2
2y = -4x - 2
y = -2x – 1

x < 2y < 3z is equivalent to x < 2(-2x – 1) < 3(x + 2), or x < -4x – 2 < 3x + 6.
Then we have x < -2/5 from x < -4x – 2, because:
x < -4x – 2
5x < -2
x < -2/5.

We also have x > -8/7 from -4x - 2 < 3x + 6, because:
-4x - 2 < 3x + 6
-7x < 8
x > -8/7 (the inequality sign changes direction since we divided by a negative)

Thus -8/7 < x < -2/5 and x = -1 since x is an integer.

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

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)

x = -2, y = 1, z = 4/3 and x = -3, y = 0, z = 1 are solutions.
Condition 2)

x = -2, y = 0, z = 9/4 and x = -1, y = 0, z = 7/4 are solutions.
Since condition 2) does not yield a unique solution, it is not 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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[GMAT math practice question]

(Coordinate Geometry) The line y = ax - 1 passes through (1/3, 0). On the line, there is a point whose x-coordinate and y-coordinate are the same. What is this point?

A. (1/5 , 1/5)
B. (1/3 , 1/3)
C. (1/2 , 1/2)
D. (- 1/4 , - 1/4)
E. (- 1/5 , - 1/5)

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Since the line y = ax – 1 passes through (1/3, 0), we have 0 = a/3 – 1 or a = 3.
To figure out the point with the same x-coordinate and y-coordinate on the line y = 3x – 1, we make y = x, which gives us x = 3x – 1 or 2x = 1.
Thus we have x = 1/2, y = 1/2, and the point is (1/2, 1/2).

Therefore, C is the answer.
Answer: C
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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) 1/x + 1/y = 1/2 + 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.

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 1/30 ≤ 1/z < 1/y < 1/x ≤ 1/3 when we take reciprocals.
Since we have 1/x + 1/y = 1/2 + 1/z, we have 1/2 < 1/x + 1/y < 1/x + 1/x = 2/x or 1/2 = 2/4 < 1/x.
Thus x < 4 and we have x = 3.
Since we have 1/2 = 1/3 + 1/y, we have 1/6 < 1/y or y < 6.
Since 3 < y < 6 and y is a prime number, we have y = 5.
1/z = 1/x + 1/y – 1/2 = 1/3 + 1/5 – 1/2 = 10/30 + 6/30 – 15/30 = 1/30 or z = 30.
Then, x + y + z = 3 + 5 + 30 = 38.

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

Therefore, D is the answer.
Answer: D

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.
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[GMAT math practice question]

(Coordinate Geometry) The figure shows that OABC is a parallelogram. What is the equation of the line passing through A and C?

A. y = 1/10x + 3/10
B. y = 5/11x + 20/11
C. y = 5/12x + 3/12
D. y = 1/13x + 3/13
E. y = 5/14x + 3/14





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Since AB = CO = 4, we have point C(-4, 0).

Then the slope of the line AC is (5 - 0) / (7 - (-4)) = 5/11.
The equation of the line AC is y - 0 = (5/11)(x - (-4)) or y = (5/11)x + 20/11.

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

(Probability) What is the probability that a three-digit integer corresponding with the three numbers thrown on three different dice is a multiple of 11?

A. 1/3
B. 2/27
C. 4/27
D. 5/27
E. 2/9

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In order for a three-digit integer abc to be a multiple of 11, a – b + c must be a multiple of 11 since we have 100a + 10b + c = 99a + a + 11b – b + c = (99a + 11b) + a – b + c = 11(9a+b) + a – b +c.

If a – b + c = 0, then the possible cases for (a, b, c) are
(1, 2, 1), (1, 3, 2), (1, 4, 3), (1, 5, 4), (1, 6, 5), (2, 3, 1), (2, 4, 2), (2, 5, 3), (2, 6, 4), (3, 4, 1), (3, 5, 2), (3, 6, 3), (4, 5, 1), (4, 6, 2) and (5, 6, 1), and we have 15 cases.

If a – b + c = 11, then we have only the case (a, b, c) = (6, 1, 6).

Thus the probability is 16/(6^3) = 16/216 = 2/27.

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

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

1) Point O is the circumcenter of △ABC.
2) Point O is the circumcenter of △ACD.



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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 4 angles of a quadrilateral, we have 4 variables (∠A, ∠B, ∠C, and ∠D) and 2 equations (∠B = 80 and ∠A + ∠B + ∠C + ∠D = 360), 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 both conditions together, we have a quadrilateral inscribed by a circle and ∠B + ∠D = 180.
Since we have ∠B = 80, we have ∠D = 100.

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 ∠BOC in the figure?

1) Point O is the circumcenter of triangle ABC.
2) ∠OAC = 23° and ∠OBA = 48°



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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 4 variables (∠BOC, ∠ABC, ∠BCA, and ∠CAB) and 1 equation (∠ABC + ∠BCA + ∠CAB = 180°), 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 point O is the circumcenter of triangle ABC from condition 1), triangles OAB, OBC, and OCA are isosceles and OA, OB, OC are congruent.

Since ∠OAC = 23°, we have ∠OCA = 23° (because it is an isosceles triangle) and ∠AOC = 134° (180° - 23° - 23°).
Since ∠OBA = 48°, we have ∠OAB = 48° and ∠AOB = 84°.
Then we have ∠AOB + ∠BOC + ∠COA = 360°, 134° + 84° + ∠BOC = 360°, 218° + ∠BOC = 360° or ∠BOC = 142°.

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 in which the answer is A, B, C, or D.
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[GMAT math practice question]

(Geometry) Point I is the incenter of triangle ABC in the figure. The figure shows AB = 10, AC = 8, ∠ABI = 22° and ∠ACI = 30°. Line DE is parallel to line BC. What is the length of the perimeter of triangle ADE?



A. 14
B. 16
C. 18
D. 20
E. 22

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Since I is the incenter of the triangle ABC, IB and IC bisect ∠ABC and ∠ACB, respectively. We know ∠IBC = 22° and ∠ICB = 30°. Since ∠DIB and ∠CBI are alternate interior angles, we have ∠BID = 22° and triangle DBI is an isosceles. Since ∠EIC and ∠BCI are alternate interior angles, we have ∠EIC = 30° and triangle CEI is an isosceles. Thus we have the perimeter of triangle ADE = AD + DE + EA = AD + DI + IE + EA = AD + DB + CE + EA = AB + AC = 10 + 8 = 18.

Therefore, C is the answer.
Answer: C
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