51. Find the area of a rhombus one side of which measures 20cm and one diagonal 24cm.

Sol:Let, other diagonal = 2x cm,
      Since halves of diagonals and one side of rhombus form a right angled triangle
      with side as hypotenuse, we have:
      (20)^2 =(12)^2+x^2 or x=Ö(20)^2-(12)^2 =Ö256=16 cm.
      Therefore, other diagonal = 32 cm.

52. A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty the tank can empty it in 6 hours. If all the pipes are kept open when the tank is full, it will take exactly 6 hours for the tank to empty. How many of these are fill pipes?

Sol:  Let the number of fill pipes be ‘n'. Therefore, there will be 8-n, waste pipes.
       Each of the fill pipes can fill the tank in 8 hours. Therefore, each of the fill pipes will fill 1/8th
       of the tank in an hour.Hence, n fill pipes will fill n/8th of the tank in an hour.
       Similarly, each of the waste pipes will drain the full tank in 6 hours. That is, each of the     
       waste pipes will drain 1/6th of the tank in an hour.
       Therefore, (8-n) waste pipes will drain ((8-n)/6)th of the tank in an hour.
       Between the fill pipes and the waste pipes, they drain the tank in 6 hours. That is, when all  
       8 of them are opened, 1/6th of the tank gets drained in an hour.
       (Amount of water filled by fill pipes in 1 hour - Amount of water drained by waste pipes 1
                                           hour)
                                          = 1/6th capacity of the tank drained in 1 hour.
                        

53. A pump can be used either to fill or to empty a tank. The capacity of the tank is 3600 m3. The emptying capacity of the pump is 10 m3/min higher than its filling capacity. What is the emptying capacity of the pump if the pump needs 12 more minutes to fill the tank than to empty it?

Sol.   Let ‘f’ m3/min be the filling capacity of the pump. Therefore, the emptying capacity of the
         pump will be = (f + 10 ) m3 / min.
        The time taken to fill the tank will be = minutes
       And the time taken to empty the tank will be = .
       We know that it takes 12 more minutes to fill the tank than to empty it
       i.e => 3600 f + 36000 - 3600 f = 12 (f2 + 10 f)
       => 36000 = 12 (f2 + 10 f) => 3000 = f2 + 10 f => f2 + 10 f - 3000 = 0.
         Solving for positive value of ‘f’ we get, f = 50.
         Therefore, the emptying capacity of the pump = 50 + 10 = 60 m3 / min

54. X alone can do a piece of work in 15 days and Y alone can do it in 10 days. X and Y undertook to do it for Rs. 720. With the help of Z they finished it in 5 days. How much is paid to Z?

Sol.  In one day X can finish 1/15th of the work.
         In one day Y can finish 1/10th of the work.
         Let us say that in one day Z can finish 1/Zth of the work.
         When all the three work together in one day they can finish 1/15 + 1/10 + 1/Z = 1/5th of          the work.
         Therefore, 1/Z = 1/30.
         Ratio of their efficiencies = 1/15: 1/10: 1/30 = 2: 3: 1.Therefore Z receives 1/6th of the          total
         money.
         According to their efficiencies money is divided as 240: 360: 120.
         Hence, the share of Z = Rs. 120.

55. Pipe A usually fills a tank in 2 hours. On account of a leak at the bottom of the tank, it takes pipe A 30 more minutes to fill the tank. How long will the leak take to empty a full tank if pipe A is shut?