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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
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The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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

=>

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.

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
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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

=>

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
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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

Attachment: 1.13PS.png [ 12.51 KiB | Viewed 272 times ]

=>

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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[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 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 c/6 + c/6 = 1/3, (2c)/6 = 1/3, c/3 = 1/3, c = 3/3. Then we have c = 1.
Since condition 2) yields a unique solution, it is sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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

=>

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.

_________________
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Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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 227 times ]

=>

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
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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.

Attachment: 1.24DS.png [ 9.51 KiB | Viewed 212 times ]

=>

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.

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
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Probability) There are six cards numbered 0, 0, 1, 2, 3, and 4. How many 3-digit even numbers are possible when picking 3 cards?

A. 30
B. 32
C. 34
D. 36
E. 38

=>

Case 1: X00 (We have two zero digits.)
The number of possible cases is 4.

Case 2: X0Y / XY0 (We have one zero digit.)
X0Y: The number of possible values of y is 2 since y is an even number.
Then the number of possible values of x is 3.
Thus we have 2 * 3 = 6 cases.
XY0: The number of possible cases is 4 * 3 = 12
When we have one zero digit, we have 6 + 12 = 18 cases.

Case 3: XYZ (We don’t have any zero digits.)
The number of possible values of z is 2 since z is an even number.
Then we have 3 * 2 = 6 cases for each even integer z.
Thus we have 6 * 2 = 12 cases.

Then we have 4 + 18 + 12 = 34 cases.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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°

Attachment: 1.27DS.png [ 13.67 KiB | Viewed 173 times ]

=>

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.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Geometry) The figure shows that ∠BAD is 30°, and ∠CAE is 40°. What is the measure of ∠ADE?

1) Point O is the circumcenter of triangle ABC.
2) Point I is the incenter of triangle ABC.

Attachment: 1.30DS.png [ 7.84 KiB | Viewed 154 times ]

=>

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 6 variables (∠ABC, ∠BCA, ∠CAB, ∠ADE, ∠DAE, and ∠DEA) and 2 equations (∠ABC + ∠BCA + ∠CAB = 180°, ∠ADE + ∠DAE + ∠DEA = 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 I is the incenter of triangle ABC, we have ∠IAB = ∠IAC = 40° and ∠DAE = ∠IAB - ∠BAD, ∠DAE = 40° - 30°, ∠DAE = 10° .
Since point O is the circumcenter of triangle ABC, we have ∠BAO = ∠ABO = 30° and ∠OBC = ∠OCB = (1/2)(180° – (2*30° + 2*50°)) = 10°.
Then ∠ABC = ∠ABO + ∠OBC = 30° + 10° = 40°.
Since ∠ADE is an exterior angle of triangle ABD, we have ∠ADE = ∠DAB + ∠ABD = 30° + 40° = 70°.

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 in 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
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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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?

Attachment: 1.27DS.png [ 13.67 KiB | Viewed 141 times ]

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

=>

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.

_________________
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Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Geometry) Point O is the circumcenter of triangle ABC, and the length of AC is 7. The length of the perimeter of triangle ABC is 19. What is the area of the circumscribed circle of triangle ABC?

A. 30π
B. 32π
C. 34π
D. 36π
E. 38π

Attachment: 1.28ps.png [ 8.69 KiB | Viewed 111 times ]

=>

Since point O is the circumcenter of triangle ABC, OA = OB = OC is the radius of the circumscribed circle of the triangle.
Since the perimeter of the triangle ABC is 19, we have OA + OC + 7 = 19, OA + OA + 7 = 19, 2(OA) + 7 = 19, 2(OA) = 12, or the radius OA = 6.
Thus, the area of the circumscribed circle of the triangle is (62)π = 36π.

_________________
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Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Number Properties) How many integers are there satisfying 1 < [3 - x/2] < 4? ([x] means the greatest integer less than or equal to x)

A. 2
B. 3
C. 4
D. 5
E. 6

=>

Since 1 < [3 - x/2] < 4, we have [3 - x/2] = 2 or [3 - x/2] = 3.

If [3 - x/2] = 2, then we have 2 ≤ 3 - x/2 < 3 or -1 ≤ - x/2 < 0 (subtracting 3). This is equivalent to 0 < x ≤ 2 (multiplying by -2, which changes the direction of the inequality signs).
If [3 - x/2] = 3, then we have 3 ≤ 3 - x/2 < 4 or 0 ≤ - x/2 < 1 (subtracting 3). This is equivalent to -2 < x ≤ 0 (multiplying by -2, which changes the direction of the inequality signs).

Thus, we have -2 < x ≤ 2 and the integer values of x are -1, 0, 1, and 2.
We have 4 integer solutions.

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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

(Number Properties) How many integers are there satisfying 1 < [3 - x/2] < 4? ([x] means the greatest integer less than or equal to x)

A. 2
B. 3
C. 4
D. 5
E. 6

=>

Since 1 < [3 - x/2] < 4, we have [3 - x/2] = 2 or [3 - x/2] = 3.

If [3 - x/2] = 2, then we have 2 ≤ 3 - x/2 < 3 or -1 ≤ - x/2 < 0 (subtracting 3). This is equivalent to 0 < x ≤ 2 (multiplying by -2, which changes the direction of the inequality signs).
If [3 - x/2] = 3, then we have 3 ≤ 3 - x/2 < 4 or 0 ≤ - x/2 < 1 (subtracting 3). This is equivalent to -2 < x ≤ 0 (multiplying by -2, which changes the direction of the inequality signs).

Thus, we have -2 < x ≤ 2 and the integer values of x are -1, 0, 1, and 2.
We have 4 integer solutions.

Hi MathRevolution,

This is a good question.

I have 1 query.

Why can't x be 3? Can you please confirm the answer again?

Thank you.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8765
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Geometry) The figure below shows the dimensions of the right triangle ABC with AB = 13, BC = 12, CA = 5 and I is a point inside triangle ABC. Angle C is 90o. What is the minimum distance from the point I to sides AB, BC and CA?

1) Point I is the incenter of △ABC.
2) Line AI bisects angle A, and line BI bisects angle B.

Attachment: 2.3ds.png [ 12.43 KiB | Viewed 437 times ]

=>

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)

Attachment: 2.3DS.A1.png [ 9.44 KiB | Viewed 436 times ]

Since I is the incenter of the triangle, the distances to all sides from point I are equal. Assume the distances are x.

Attachment: 2.3DS(A2).png [ 11.47 KiB | Viewed 436 times ]

The area of triangles IAB, IBC and ICA are (1/2)*13*x + (1/2)*12*x + (1/2)*5*x = (13/2)x + 6x + (5/2)x = (18/2)x + 6x = 9x + 6x = 15x.
The area of triangle ABC = (1/2)*5*12 = 30.
Since the sum of the areas of triangles IAB, IBC and ICA is equal to the area of triangle ABC, we have 15x = 30 or x = 2.

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

Condition 2)

Since we can find an incenter of a triangle by the intersection of lines bisecting interior angles, I is the incenter of the triangle from condition 2).

Thus, condition 2) is sufficient with the previous reasoning in condition 1).

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).
_________________
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Joined: 16 Aug 2015
Posts: 8765
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GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Geometry) What is the length of CD in the figure?

1) Point I is the incenter of triangle ABC, and D, E, and F are the tangential points.
2) The length of BC is 11, and BE is 8.

Attachment: 1.29DS(A).png [ 10.73 KiB | Viewed 412 times ]

=>

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.

Attachment: 1.29DS(A).png [ 10.73 KiB | Viewed 411 times ]

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.

Since we have a triangle, we have 3 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)
Since point I is the incenter of the triangle ABC, AD, AE, BE, BF, CD, and CE are tangent to the same circle, and we have AD = AE, BE = BF and CD = CF.
Then we have CD = FC = BC – BF = BC – BE = 11 – 8 = 3.

Since both conditions together yield a unique solution, they are sufficient.
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[GMAT math practice question]

(Geometry) The figure below shows the dimensions of triangle ABC. What is ∠OBI?

1) AB = AC,
2) Point O is the circumcenter and point I is the incenter of triangle ABC.

Attachment: 2.6DS.png [ 11.66 KiB | Viewed 373 times ]

=>

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 a triangle has 3 variables, 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 triangle ABC is an isosceles, we have interior angles ∠ABC = ∠ACB = (1/2)(180 - ∠A) = (1/2)(180 - 36) = 72.
Since O is the circumcenter of triangle ABC, we have ∠OBA = ∠OCA = (1/2) ∠A = (1/2)36 = 18.
Since I is the incenter of triangle ABC, we have ∠IBA = (1/2) ∠ABC = (1/2)72 = 36.
Thus, we have ∠OBI = ∠IBA - ∠OBA = 36 – 18 = 18.

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 in which the answer is A, B, C, or D.
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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Number Properties) <x> denotes x - 10[x/10] and n is a positive integer. What is the value of <9n - 1>? ([x] means the greatest integer less than or equal to x.)

1) <9^n - 1> is not positive.
2) n is an even number.

=>

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.

<x> means the unit digit of x.
For example, if x = 123, then x - 10[x/10] = 123 - 10[12.3] = 123 - 120 = 3.

We have 9^1 = 9, 9^2 = 81, 9^3 = 729, 9^4 = 6561, ...
Then, 9^1 - 1 = 8, 9^2 - 1 = 80, 9^3 - 1 = 728, 9^4 - 1 = 6560, ....
We notice that if n is an odd number, the unit digit of 9^n - 1 is 8, and if n is an even number, the unit digit of 9^n - 1 is 0.

The question asks what the unit digit of 9^n - 1 is.
Condition 2) tells us that n is an even number. Therefore 9^n - 1 is 0. Since condition 2) yields a unique solution, it is sufficient.

Condition 1)
Since the only possible values of <9^n - 1> are 0 and 8, <9^n - 1> is 0 if <9^n - 1> is not positive.

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

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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(Geometry) As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is the incenter of triangle ABC. What is the length of AC?

Attachment: 2.3ps.png [ 11.56 KiB | Viewed 338 times ]

A. 3
B. 4
C. 5
D. 6
E. 7

=>

Since AD = AF, we have AF = 3.
Since BC = 9 and BD = 5, we have FC = EC = BC – BE = BC – BD = 9 – 5 = 4.

Then we have AC = AF + FC = 3 + 4 = 7.

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Posts: 8765
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[GMAT math practice question]

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

1) Triangle ABC is a right isosceles triangle with AB = AC.
2) BD = 7 and CE = 15.

Attachment: 2.10DS.png [ 5.69 KiB | Viewed 325 times ]

=>

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 a triangle has 3 variables, 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: 2.10(A).png [ 8.97 KiB | Viewed 325 times ]

Since we have AB = CA, ∠ADB = ∠CEA = 90° and ∠BAD = 90° - ∠CAE = ∠ACE, triangles ABD and CAE are congruent.
Then we have BD = AE = 7 and AD = CE = 15.
Thus, DE = AD – AE = 15 – 7 = 8.

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.
_________________ Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 16 Feb 2020, 17:58

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# The Ultimate Q51 Guide [Expert Level]

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