Question
A factory has three belt conveyors fitted in its
loading station. Conveyors 'A' and 'B' operating simultaneously lift all the goods in the same time during which the goods can be lifted by the conveyor 'C' operating alone. Conveyor 'B' lifts all the goods five hours faster than conveyor 'A' but four hours slower than the conveyor 'C' The durations of time required to lift all the goods, individually by the conveyors 'A', 'B' and 'C' areSolution
Let conveyor B lifts all the goods in X hours. Conveyor A lifts all goods in X +5 Conveyor C lifts all goods in X – 4 hours. So, conveyor B takes 1/x time to lift in 1 hour Conveyor A takes 1/(x+5) hours to lift in 1 hour Conveyor C takes 1/(x-4) hours to lift in 1 hour. Now, according to the question, to lift all goods, Conveyor A + Conveyor B = Conveyor C 1/(x+5) + 1/x = 1/(x-4) Take LCM on the left side (x+5+x)/x(x+5)= 1/(x-4) 2x+5/x²+5x = 1/(x-4) (2x+5) *(x-4) = x² + 5x 2x²-8x+5x-20 = x² + 5x x²-8x-20 =0 x² - 10x + 2x - 20 = 0 x(x-10) + 2(x-10) = 0 (x-10)*(x+2) = 0 So, x= 10, -2 But x cannot be negative here. So, X = 10 Hence Conveyor B takes X 10 hours. Conveyor A take X + 5 = 10 +5 15 hours Conveyor C takes X-4 hours 10-4 = 6 hours Â
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