| Direction: In the following questions two quantities I and II are given. Solve both the quantities and choose the correct option accordingly. |
| I. Two pipes A and B can fill a tank in 18 minutes and 24 minutes respectively. Both pipes are opened together and 3 minutes before the tank is filled completely, pipe B is closed. Calculate the total time required to fill the tank. |
| II. Pipe A and Pipe B can fill a tank in 12 hours and 15 hours respectively while Pipe C can empty the full tank in 18 hours. If Pipe A is opened at 7 am. Pipe B opened at 8 : 30 am and Pipe C at 10 am, then after 10 am, how much more time will be taken by all three pipes together to fill the tank? |
A) Quantity I > Quantity II
B) Quantity II > Quantity I
C) Quantity I > Quantity II
D) Quantity I < Quantity II
E) Quantity I = Quantity II or relation can't be established.
Correct Answer: B
Solution :
Quantity I. Let total capacity = 72 litres (LCM of 18 and 24) A + B fill 72 litres (if both worked till end). B can fill 3\[\times \]3 = 9 litres in 3 min But B was closed 3 min before so A + B together filled (72 - 9 =) 63 litres in \[\frac{63}{4+3}\]= 9 min So, total = 9 + 3 = 12 minutes Quantity II. Let capacity of tank = 180 lit (LCM of 12, 15, 18) A fills \[\frac{180}{12}\]= 15 litre/hour B fills \[\frac{180}{15}\]= 12 litre/hour C empties \[\frac{180}{18}\]= 10 litre/hours Till 10 am 15 \[\times \] 3 + 12 \[\times \] 1.5 = 45 + 18 = 63 litres Remaining volume to be filled = 180 - 63 = 117 litres After 10 am, when all pipes are opened. 15 + 12 - 10 = 17 litres/hr is filled. \[\therefore \] Reqd time \[=\frac{117}{17}=6\frac{15}{17}\] hours Hence, Quantity I < Quantity II Note: We can easily solve without calculation. QI is given in minutes white Q is given in hours. Thus, QI < QII
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