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A Pump Can Be Operated Both For Filling A Tank And For Emptying It. The Capacity Of The Tank Is 2400 M3. The Emptying Capacity Of The Pump Is 10m3 Per Minute Higher Than Its Filling Capacity. Consequently, The Pump Needs 8 Minutes Less To Empty The Tank Than To Fill It. Find The Filling Capacity Of The Pump? |
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Answer» Let the filling CAPACITY of the pump = x M3/ min. Then the emptying capacity of the pump = (x + 10) m3/ min. Time required for filling the TANK = 2400/x MINUTES Time required for emptying the tank = 2400/x+10 minutes Pump needs 8 minutes lesser to EMPTY the tank than it needs to fill it ⇒2400/x−2400/x+10=8 ⇒300/x−300/x+10=1 ⇒300/(x+10)−300/x=x(x+10) ⇒3000=x2+10x ⇒x2+10x−3000=0 (x+60)(x−50)=0 x = 50 or -60 Since x can not be negative, x=50 i.e.,filling capacity of the pump = 50 m3/min. Let the filling capacity of the pump = x m3/ min. Then the emptying capacity of the pump = (x + 10) m3/ min. Time required for filling the tank = 2400/x minutes Time required for emptying the tank = 2400/x+10 minutes Pump needs 8 minutes lesser to empty the tank than it needs to fill it ⇒2400/x−2400/x+10=8 ⇒300/x−300/x+10=1 ⇒300/(x+10)−300/x=x(x+10) ⇒3000=x2+10x ⇒x2+10x−3000=0 (x+60)(x−50)=0 x = 50 or -60 Since x can not be negative, x=50 i.e.,filling capacity of the pump = 50 m3/min. |
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