Ans 1:
Class : Class 10
simple solutionSol: so, in the first row we have 23 15 9 6423 15 = 38 (applied rule 5)38 - 9 = 29 (applied rule 2)29 64 = 93 (applied rule 4)Hence, First row result is 93 which is equivalent to xSo, now in the second row we have 93 31 15 393 divided by 31 = 3 (applied rule 3)3 15 = 18 (applied rule 5)18 X 3 = 54 (applied rule 1)Hence, resultant of second row is 54 which is answer option A
Ans 5:
Class : Class 10
simple solution Sol: so, in the first row we have 23 15 9 64 23 plus 15 = 38 (applied rule 5) 38 - 9 = 29 (applied rule 2) 29 plus 64 = 93 (applied rule 4) Hence, First row result is 93 which is equivalent to x So, now in the second row we have 93 31 15 3 93 divided by 31 = 3 (applied rule 3) 3 plus 15 = 18 (applied rule 5) 18 X 3 = 54 (applied rule 1) Hence, resultant of second row is 54 which is answer option A
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if l is tens place digit of 6th multiple of 36, m is ones place digit of H.C.F. of 72 and 192, n is ones place digit of the least common multiple of 6 and 42.
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