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Math Revolution GMAT Instructor
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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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17 Jan 2020, 00:05
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. 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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Math Revolution GMAT Instructor
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17 Jan 2020, 00:06
[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.
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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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19 Jan 2020, 06:22
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 \(k = 3.\) 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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19 Jan 2020, 06:25
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. 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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19 Jan 2020, 23:07
[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 225 times ]
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21 Jan 2020, 00:20
[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 212 times ]
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21 Jan 2020, 00:31
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 CAB = AC doesn't give us any information about triangle ADE 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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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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22 Jan 2020, 17:16
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. 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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Math Revolution DS Expert  Ask Me Anything about GMAT DS
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22 Jan 2020, 17:17
[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 176 times ]
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22 Jan 2020, 23:50
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. 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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22 Jan 2020, 23:52
[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 \(\frac{1}{3}.\)
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24 Jan 2020, 01:58
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. 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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24 Jan 2020, 02:00
[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 162 times ]
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27 Jan 2020, 05:51
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. Therefore, D is the answer. Answer: D
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27 Jan 2020, 05:52
MathRevolution wrote: [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 => 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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27 Jan 2020, 05:55
[GMAT math practice question] (Geometry) What is the measure of \(∠BOC\) in the figure? 1) Point \(O\) is the circumcenter of triangle \(ABC.\) 2) \(∠OAC = 23^o\) and \(∠OBA = 48^o\) Attachment:
1.27ds.png [ 17.07 KiB  Viewed 129 times ]
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28 Jan 2020, 00:04
[GMAT math practice question] (Geometry) What is the measure of \(∠BIC\) in the figure? Attachment:
1.28ds.png [ 4.2 KiB  Viewed 115 times ]
1) Point \(I\) is the incenter (the point where the three angle bisectors meet) of triangle \(ABC.\) 2) \(∠BAC = 50^o\)
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28 Jan 2020, 00:37
1) Point \(I\) is the incenter (the point where the three angle bisectors meet) of triangle \(ABC.\) No other info, NOT Sufficient2) \(∠BAC = 50^o\) No other info, NOT SufficientCombining St 1 & 2 we get \(∠ABC +∠BCA = 180  ∠BAC\) As Sum of angles in a triangle = 180So ∠ABC +∠BCA = 130 hence ∠IBC +∠BCI = (∠ABC +∠BCA)/2 = 65 in triangle IBC, \(∠IBC +∠BCI \)+\(∠BIC\) = 180 Hence, \(∠BIC\) = 18065  115 Sufficient , Answer = C MathRevolution wrote: [GMAT math practice question] (Geometry) What is the measure of \(∠BIC\) in the figure? Attachment: 1.28ds.png 1) Point \(I\) is the incenter (the point where the three angle bisectors meet) of triangle \(ABC.\) 2) \(∠BAC = 50^o\)
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28 Jan 2020, 01:53
MathRevolution wrote: [GMAT math practice question] (Geometry) What is the measure of \(∠BIC\) in the figure? Attachment: 1.28ds.png 1) Point \(I\) is the incenter (the point where the three angle bisectors meet) of triangle \(ABC.\) 2) \(∠BAC = 50^o\) BunuelHave GMAT questions ever used terms like Incenter, Circumcenter, Orthocenter etc.??? If not (as I believe), then I consider the language of these questions inappropriate for GMAT questions practice.
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28 Jan 2020, 01:57
GMATinsight wrote: MathRevolution wrote: [GMAT math practice question] (Geometry) What is the measure of \(∠BIC\) in the figure? Attachment: 1.28ds.png 1) Point \(I\) is the incenter (the point where the three angle bisectors meet) of triangle \(ABC.\) 2) \(∠BAC = 50^o\) BunuelHave GMAT questions ever used terms like Incenter, Circumcenter, Orthocenter etc.??? If not (as I believe), then I consider the language of these questions inappropriate for GMAT questions practice. I cannot recall any official question using this term. The term is defined there so it makes things easier.
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