A farmer divides his herd of n cows among his four sons so that the first son gets one – third the herd, the second son gets one – fifth, the third son gets one – sixth and the fourth son gets 42 cows. The value of n is
According to the question, (n/3) + (n/5) + (n/6) + 42 = n (10n + 6n + 5n)/30 + 42 = n 21n/30 + 42 = n 42 = n – (21n/30) 42 = 9n/30 n = (30 x 42)/9 = 140
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