In bucket A and B, the quantity of mixture are (3z-25) litre and (5z+25) litre respectively. If (y-15) and (y+60) litres of water is mixed in bucket A and B respectively, then in each bucket the quantity of milk will be 20% more than the quantity of water. In each bucket the initial quality of milk is three times of the initial quantity of water. Find out the value of ‘z’.
[(3/4)x(3z-25)] /[(1/4)x(3z-25)+(y-15)] = 120/100 By solving the above expression. 24y = 27z+135 Eq.(i) [(3/4)x(5z+25)] /[(1/4)x(5z+25)+(y+60)] = 120/100 By solving the above expression. 24y = 45z−1215 Eq.(ii) So Eq.(i) = Eq.(ii). 27z+135 = 45z-1215 45z-27z = 1215+135 18z = 1350 value of ‘z’ = 75
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