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Question Stats:
40% (01:34) correct
60% (00:54) wrong based on 3 sessions
If A , B , and C are points on the plane, is AB \gt 15 ? 1. BC + AC \gt 142. Area of triangle ABC \lt 1Source: GMAT Club Tests - hardest GMAT questions OA does not explain how you could get an area < 1, with side > 15?
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ConkergMat wrote: A, B, and C are points on the plane. Is AB > 15 ?
1) BC + AC > 14
2) Area of triangle ABC < 1
OA does not explain how you could get an area < 1, with side > 15? 1) If BC + AC > 14, AB could be 0.1, 1 or 5 or 10 or or > 15. nsf.. 2) Ok, area of triangle ABC < 1. This is possible when, either height or base of the triangle is negligibly a fraction of 1. Suppose AB is base and is = 16, height is 0.01, area is < 1. Suppose AB is a base = 0.01, bc = ac = 8, then hieght is 8, then area is < 1. So Not suff.. Togather also not suff. So E.
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Thanks for the helpful explanations. I was thrown off on this question in consideration of the triangle inequality rule, which states: the length of a side of a triangle is less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides.
but the example GMAT Tiger shows us above also fits the rule:
bc - ac < ab < bc + ac
8 - 8 < .01 < 8 + 8
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Given: A,B and C are points in the plane. We don't know wether those 3 points in a straight line or can be used to form a traingle. (1) BC + AC > 14 We are not sure here whether A,B and C are in a straight line or can be used to form a traingle. Hence, Not sufficient (2) Area of triangle ABC < 1 This means - A, B and C are forming a traingle. We can't get anything out of it even. Hence, Not Sufficient (1) + (2) ==> We can think of applying traingle sides rule as below: When BC + AC = 20, AB can take values from >0 to <20. We can't definetly say AB > 15 or not. Hence, Insufficient.
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I hate myself when i cant explain myself, and this has gotta be one of those occassions. On exam day i'd have gone for E.
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E is correct. Although, I must admit that at first I thought there was something wrong with the question.
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initially i thought A is correct,but later realized my mistake.
Ans should be E.
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Is it right on "GMAT-land" not to suppose that ABC is triangle? I mean, from the question stem you can think not only that points A, B and C are all three in the same line, but also that they are not in the same line. However, if GMAT tells us in S2 that the "area of the triangle ABC is less than 1", is it possible that the area is 0?
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From statement 1, we can't deduce about AB.why? A-------------C---------------B . 8. 8 AC=8, BC=8. AB=16 TRUE A----------B-----------C . 8 8 AB=8, BC=8 AND BC+AC>14 BUT AB<15. Hence insufficient. Area of ABC <1. Clearly insufficient. Just the area is given and hence we can't infer individual lengths. On combining: we know that in a triangle, length of any side must be smaller than the sum of lengths of other two sides. But we don't know which side is greater. Hence insufficient.
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ConkergMat wrote: If A , B , and C are points on the plane, is AB \gt 15 ? 1. BC + AC \gt 142. Area of triangle ABC \lt 1Source: GMAT Club Tests - hardest GMAT questions OA does not explain how you could get an area < 1, with side > 15? This problem was replaced by the following question: The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?
I. c>a+b
II. c/a/b=10/6/2
III. c^2>a^2+b^2A. I and III only B. II and III only C. I only D. II only E. III only According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that II is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them). Answer: E. Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.
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