if a farmer sells 20 of his chickens , his stock of feed will last for 3 more days than planned , but if he buys 15 more chickens , he will run out of feed 4 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ?
let x = total feed required for the planned period n = number of chicken t = total time of the planned feed x = nt 1 ) x = ( n - 15 ) * ( t + 4 ) 2 ) x = ( n + 20 ) * ( t - 3 ) equating 1 & 2 ( n - 15 ) * ( t + 4 ) = ( n + 20 ) * ( t - 3 ) 7 n = 35 t n = 5 t x = n * n / 5 substituting this value in 1 n * n / 5 = ( n - 15 ) * ( n / 5 + 4 ) 5 n = 300 n = 60 b
multiply(add(const_2, const_3), multiply(4, 3))