ePracCAT # 13 ANALYSIS About PracCATs Hi! Welcome to PT’s trendsetting ePracCAT series that provides the cutting edge to all test takers with its innovative approach towards paper setting styles, question patterns, and inductive & deductive reasoning skill requirements. The ePracCAT series has all the features that make it unique in content and approach, thus making it the best tool in the country for students aiming at the best B Schools. The ePracCATs by PT Education will provide you various possible simulations of the actual CAT. The multiple paper patterns, differential marking, and different negative markings will give you an actual feel of the unpredictable CAT.
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Solutions 1.
The first number in the given range that is divisible by 23 is 805 and the last one is 1196. Common difference = 23 Tn = a + (n 1) d Where a = first term n = number of terms d = common difference Tn = n th term So, 1196 = 805 + (n – 1) ´ 23 Or n = 18 Therefore, sum of all the 18 terms by the formula: Sn = n/2 [2a + (n – 1) d] = 18/2 [(2 × 805) + (18 – 1) 23] = 18009 Ans. (1)
2.
The distance AC = 55 ´ 5 = 275 km Similarly, BC=150 km Since, D is the midpoint of BC, BD = 75 km The distance from A to D can be obtained from the median theorem: AB 2 +AC 2 =2(BD 2 +AD 2 ) Hence, AD=228.4 or approximately 228 km. Ans. (1)
3.
Solution: G(x) = x 2 + nx – 24 = 0 x(x + n) = 24 x can take the values 1, 1, 2, 2, 3, 3, 4, 4, 6, 6, 8, 8, 12, 12, 24, 24 Thus, n can take the corresponding values: 23, 23, 10, 10, 5, 5, 2, 2, 2, 2, 5, 5, 10, 10, 23, 23 But, 10 0 (x 1) 2 is already greater than zero. Hence, x must be greater than zero. Also, x can neither be 1. Hence, x > 0 and x ¹ 1 Case II: Taking x 2x 2 + x or x (x + 1) 2
