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JEE Main & Advanced AIEEE Solved Paper-2010

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    Directions: Questions number 86 are Assertion - Reason type questions. Each of these questions contains two statements: Statement - 1 (Assertion) and Statement - 2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ....., 20}. Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is\[\frac{1}{85}\] Statement - 2: If the four chosen numbers form an AP, then the set of all possible values of common difference is \[(\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\}\] Statement - 1 (Assertion) and Statement - 2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.       AIEEE  Solved  Paper-2010

    A) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1

    B) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1.

    C) Statement -1 is true, Statement -2 is false.

    D) Statement -1 is false, Statement -2 is true.  

    Correct Answer: C

    Solution :

    S : 1 required no of groups (1, 2, 3, 4) ??????.. (17, 18, 19, 20) = 17 ways (1, 3, 5, 7) ???????.. (14, 16, 18, 20) = 14 ways (1, 4, 7, 10) ?????.. (11, 14, 17, 20) = 11 ways (1, 5, 9, 13) ???????.. (8, 12, 16, 20) = 8 ways (1, 6, 11, 16) ???????.. (5, 10, 15, 20) = 5 ways (1, 7, 13, 19) ??????? (2, 8, 14, 20) = 2 ways required arability\[=\frac{(17+14+11+8+5+2)4!}{^{20}{{C}_{4}}4!}\] \[=\frac{57\,4!}{20.19.18.17}=\frac{3.4.3.2.1}{20.18.17}\] \[=\frac{1}{85}\] \[S:1\]is true. \[S:2\] possible cases of common difference are \[[\pm 1,\pm \text{ }2,\pm 3,\pm \text{ }4,\pm 5,\pm 6]\] \[S:2\]is false


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