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We have: The one who likes 2 sits 5m from left end, since only such combinations are (2, 3), (1, 1, 3) & (1, 2, 2). Two person sits between the one who likes 2 and F, who likes 8 and sits facing north. F neither sits at end of the row nor sits adjacent to W, that means we have four possible cases, in case (1) the one who likes 2 sits (2 + 3)a from left end, in case (2) the one who likes 2 sits (3 + 2)a from left end, in case (3) the one who likes 2 sits (1 + 3 + 1)a sits from left end, in case (4) the one who likes 2 sits (2 + 1 + 2)a from left end. W sits 4m left of the one who likes 8 and is 1m right of the one who likes 5, since only possible such combination is (1, 4). The one who likes 5 sits three places away from A, who doesn’t like odd number. A sits immediate right of the one who likes 6, thus in case (1a) A sits at right end and the one who likes 5 sits fourth from right end, in case (1b) A sits second from left end and the one who likes 5 sits fourth from right end, in case (2a) A sits at right end and the one who likes 5 sits fourth from right end, in case (2b) A sits second from left end and the one who likes 5 sits fourth from the right end, in case (3) & case (4) A sits third from left end and the one who likes 5 sits third from right end. Based on above given information we have
Again, we have: Z sits third to right of the one who likes 3, since none of the adjacent person likes consecutive numbers. Z neither sits adjacent to A nor likes even number, that means in case (1a), case (1b), case (2a) & case (2b) the one who likes 3 sits at right end facing south and case (3) & case (4) are not valid. The one who likes 4 sits 2m away from X. X sits 7m left of the one who likes 7, thus in case (1a) & case (1b) X sits at left end and W likes 7, in case (2b) X sits at right end and W sits facing south. Based on above given information we have: Case-1a
Case (3) & case (4) are not valid as Z sits third to right of the one who likes 3 and case (2a) is not valid as X sits 7m left of the one who likes 7. Again, we have: Y sits immediate right of the one who likes 1. J sits 8m left of E, who neither likes prime number nor sits adjacent to the one who likes 4, that means in case (1a) & case (1b) J sits third from left end and E sits facing north, case (2b) is not valid. Z sits 6m right of Y, that means distance between E and Y is 1m and case (1a) is not valid. Y and A sits facing in same direction but opposite to the direction of J, that means A sits facing south and J sits facing North. Based on above given information we have final arrangement as follow: Case (2b) is not valid as J sits 8m left of E and case (1a) is not valid as Z sits 7m right of Y. Case-1b
1000÷ 250 = ( 3√? × √1444) ÷ ( 3√512 × √361)
1540 ÷ 7 - 184 ÷ 8 = ?
12.232 + 29.98% of 539.99 = ? × 5.99
√256 * 3 – 15% of 300 + ? = 150% of 160
18 × 15 + 86 – 58 =? + 38
[5 X {(52 X 5) - 10} + 50 of 20] = ?
4(1/3) × 2(11/14) = 50% of ? + 86/11
(15 x 6 + 60% of 500 - 16 x 7) = ?
25% of 140 + 2 × 8 = ? + 9 × 5
If a nine-digit number 389x6378y is divisible by 72, then the value of √(6x + 7y) will be∶