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 Ð B = 2 ÐC                           ( \ Ð C = Ð B ) and Ð B = 2 Ð A Hence Ð A = Ð C = x (say ) then ÐB = 2 x and 2x + x + x = 180 gives x = 450 i.e. hence Ð A = Ð C = 450 and ÐB = 2 x = 900 \ side opposite Ð B is the largest and sides opposite Ð A and ÐC are equal. Hence AB = BC \ AB2 = BC2 \ AB2 < BC2 + AC2 Hence B. Two important facts namely (a) 0 < x < 1 implies x to be a positive proper fraction and (b) x  - :     = Since x < 1. \ x2 < x. \ x2 < 1 ( \ x > 1 ) and hence > 1. (Since the reciprocal of proper positive fraction is more than 1). \ > x2 x -2 > x2 Hence A. Given 0 < a < b < c The sum a + b can be small. i.e. 1 + .1 or 1.1 or it can be very large, i.e. 1 + 1000 i.e. 1001. Similarly c - a can be very small. 1.2 - 1 i.e. .2 or can be very large, i.e. 100001 - 1 i.e. 100000. Hence D. OR Mathematically a + b ³ c - a        or     a + b < c - a If a + a ³ c - b     or     a + a < c - b If 2a ³ or a < If c and b are very close then c - b or can be very small but if c is very large compared to b, c - b or can be very large a can be more than or less than or even equal to . Hence D. Test 1 Section 1 : Verbal Section Section 2 : Quantitative Section Section 3 : Analytical Section Section 4 : Quantitative Section Section 5 : Verbal Section Section 6 : Analytical Section Section 7 : Verbal Section Answer Key To Test 1 Answer Explanation To Test 1 Section 1 : Verbal Section Section 2 : Quantitative Section Section 3 : Analytical Section Section 4 : Quantitative Section Section 5 : Verbal Section Section 6 : Analytical Section Section 7 : Verbal Section