Two pipes A and B together can fill a tank in 45/8 minutes. Another two pipes i.e. pipe C having efficiency twice of pipe A and pipe D having efficiency half of pipe B can together fill the same tank in 180/44 minutes. Find the ratio of the time taken by pipe A and the time taken by pipe B alone to fill the tank.
Let time taken by pipe A and pipe B alone to fill the tank be ‘x’ minutes and ‘y’ minutes, respectively. Part of the tank filled by pipe A and pipe B together in one minute = 1/x + 1/y = (x + y)/xy Capacity of tank = [(45/8) × (x + y)]/xy = 45 (x + y)/8xy Part of tank filled by C and D together in one minute = 2/x + 1/2y = (4y + x)/2xy Capacity of tank = [(180/44) × (x + 4y)]/2xy = 180 (x + 4y)/88xy So, 45 (x + y)/8 = 180 (x + 4y)/88 ⇒ 11(x + y) = 4(x + 4y) ⇒ 11x – 4x = 16y – 11y ⇒ 7x = 5y ⇒ x/y =5 /7 ⇒ x : y = 5:7
Time taken by pipe B to fill the tank is 2.8 times of the time taken by the pipe A to fill the tank. Pipe C empties the full tank in 20 minutes while tank is filled completely in (35/3) minutes if all pipes are opened together. If pipe A is opened for first 5 minutes and then pipe B is opened and pipe C is opened after 7 minutes of pipe A, then find the time taken to fill the tank.
Let, time taken by pipe A to fill the tank = x minutes Then, time taken by pipe B to fill the tank = 2.8x minutes So, 1/x + 1/2.8x – 1/20 = 3/35 ⇒ (2.8 + 1)/2.8x = 3/35 + 1/20 ⇒ 3.8/2.8x = (12 + 7)/140 ⇒ x = (140 × 38) /(28 × 19) ⇒ x = 10 Total capacity of tank = LCM of 10, 28 and 20 = 140 units So, part of tank filled by pipe A in 5 minutes = 140/10 x 5 = 70 units And, part of tank filled by pipe A and pipe B in 2 minutes = (140/10) x 2 + (140/28) x 2 = 28 + 10 = 38 units Total part of the tank which is filled = 70 + 38 = 108 units Remaining part = 140 – 108 = 32 units Part of tank emptied by pipe C in one minutes = 140/20 = 7 units Time taken to fill 32 units of tank = 32/(14 + 5 - 7) = 32/12 = 2(2/3) minutes So, total time taken to fill the tank = 7 + 2(2/3) minutes = 9 (2/3) minutes or 9 min 40 seconds
Two pipes P and Q can fill tank A in 36 minutes and 72 minutes respectively and empty pipe M can empty the tank in 54 minutes. Tank A have the capacity of 216 liters. If all three pipes opened in tank B for (x – 48) minutes together they filled 60 liter of the tank which is 25% of the quantity of tank B. Find in x minutes what portion of tank B filled, if all pipe P and Q and M opened alternatively in each minute starting with P, followed by Q and M respectively?
For tank A- Pipes P – 36 min, Q – 72 min & R - 54 min Total quantity = 216 units Efficiency of P = 216/36 = 6 units/min Efficiency of Q = 216/72 = 3 units/min Efficiency of R = 216/54 = (- 4) units/min According to the questions: For tank B — 6 (x – 48) + 3 (x – 48) – 4 (x – 48) = 60 6x – 288 + 3x – 144 – 4x + 192 = 60 5x = (288 + 144 + 60 – 192) x = 300/5 = 60 minutes Total quantity of tank B = 60 × 4 = 240 liter Alternatively (P + Q – M) for 60 minutes, means each pipe for 20 minutes— All three in 14 minutes (P + Q – M) = 20 × 6 + 20 × 3 – 20 × 4 = 100 litres Filled portion = 100/240 = 5/12
Pipe P can fill a tank in 54 hrs, pipe Q is 25% more efficient than P and pipe R can fill the same tank in 7.2 hr less than Q. P and Q opened together for X hr and closed after that and pipe R fill remaining tank in (X + 12) hr, if the ratio between tank filled by (P + Q) together to tank filled by pipe R is 1 : 2. Find the value of X ?
P = 54 hr P : Q = 100 : 125 = 4 : 5 Total capacity of tank = 54 × 4 = 216 litre C = 216/5 – 7.2 = 36 hrs R efficiency = 216/36 = 6 litres/hr According to question ⇒ 9X/[6(X + 12)] = 1/2 ⇒ 18X – 6X = 84 x = 7 hrs
Two pipes A and B together can fill a tank in 45/8 minutes. Another two pipes i.e. pipe C having efficiency twice of pipe A and pipe D having efficiency half of pipe B can together fill the same tank in 180/44 minutes. Find the ratio of the time taken by pipe A and the time taken by pipe B alone to fill the tank.
Let time taken by pipe A and pipe B alone to fill the tank be ‘x’ minutes and ‘y’ minutes, respectively. Part of the tank filled by pipe A and pipe B together in one minute = 1/x + 1/y = (x + y)/xy Capacity of tank = [(45/8) × (x + y)]/xy = 45 (x + y)/8xy Part of tank filled by C and D together in one minute = 2/x + 1/2y = (4y + x)/2xy Capacity of tank = [(180/44) × (x + 4y)]/2xy = 180 (x + 4y)/88xy So, 45 (x + y)/8 = 180 (x + 4y)/88 ⇒ 11(x + y) = 4(x + 4y) ⇒ 11x – 4x = 16y – 11y ⇒ 7x = 5y ⇒ x/y =5 /7 ⇒ x : y = 5:7
Three pipes P, Q and R which are used to fill a tank. Each of the three pipes can be used either as an inlet pipe or as an outlet pipe. Each of the pipes P, Q and R can individually fill/empty the tank in 18 hours, 27 hours and 30 hours respectively. On a given day one pipe becomes outlet pipe and the remaining pipes become inlet and this continues alternatively. If on first day pipe P becomes outlet pipe, next day pipe Q and pipe R the following day, then in how many days will the tank be filled?
Pipe P, Q and R can fill/empty the tank in 18, 27 and 30 hrs respectively. Let the total capacity of the tank = 270 units (LCM of 18, 27 and 30) Pipe P can fill/empty (270/18) = 15 units/hrs Pipe Q can fill/empty (270/27) = 10 units/hrs Pipe R can fill/empty (270/30) = 9 units/hrs Total work done on day 1 = - 15 + 10 + 9 = + 4 units Total work done on day 2 = 15 – 10 + 9 = + 14 units Total work done on day 3 = 15 + 10 – 9 = + 16 units In 3 days, (4 + 14 + 16) = 34 units work is done. In 3 x 7 = 21 days, (34 x 7) = 238 units work is done. On 22nd day, Pipe P works as outlet. Therefore, In 22 days, 238 + 4 = 242 units work is done. Remaining work = 270 – 242 = 28 units On 23rd day, Q works as outlet pipe = 14 units work is done On 24th day, R works as outlet pipe = 14 units work is done Therefore, total time taken = 23 (7/8) days
Pipe L can empty the tank which is 45% filled in 9 hours. The time taken by pipe K and J together to fill an empty tank completely in 7.5 hours. Pipe J alone can fill an empty tank completely in (z-5) hours. The ratio between the efficiencies of pipe J and M is 3:2 respectively. The efficiency of pipe K is one third of the efficiency of pipe J. Find out the value of ‘z’. It is assumed that except pipe L, none of the pipe is used to empty the tank.
Let’s assume the total capacity of the tank is 60 units.
Pipe L can empty the tank which is 45% filled in 9 hours.
Time taken by pipe L alone to empty the tank which is completely filled = (9/45)x100
= 20 hours
Efficiency of pipe L = 60/20 = -3 units/hour (Here negative sign represents that the pipe is used to empty the tank.)
The ratio between the efficiencies of pipe J and M is 3:2 respectively.
Let’s assume the efficiencies of pipe J and M are ‘3a‘ and ‘2a‘ respectively.
The efficiency of pipe K is one third of the efficiency of pipe J.
efficiency of pipe K = ⅓ of 3a = a
The time taken by pipe K and J together to fill an empty tank completely in 7.5 hours.
7.5(a+3a) = 60
(a+3a) = 8
4a = 8
a = 2
Pipe J alone can fill an empty tank completely in (z-5) hours.
efficiency of pipe J = 3a = 3x2 = 6 units/hour
So 6x(z-5) = 60
(z-5) = 10
z = 10+5
Value of z = 15
There are four pipes A, B, C and D where A and B are used to fill the tank and C and D are used to empty the tank. Pipe A and C together can fill the tank in (a+9) hours. Pipe B alone can fill 75% of the tank in (a-1.5) hours. The efficiency of pipe C is 25% less than that of pipe B. Pipe D alone can empty half of the tank in (a-3) hours. The efficiency of pipe A is 50% more than that of pipe B. Find out the time taken by pipe A and D together to fill the empty tank completely.
Let’s assume the total capacity of the tank is 72 units. The efficiency of pipe A is 50% more than that of pipe B. Let’s assume the efficiency of pipe B is ‘4y’. efficiency of pipe A = 4y of (100+50)% = 4y of 150% = (4y x 150)/100 = 600y/100 = 6y units/hour The efficiency of pipe C is 25% less than that of pipe B. efficiency of pipe C = (100-25)% of (efficiency of pipe B) efficiency of pipe C = 75% of (efficiency of pipe B) efficiency of pipe C = 75% of 4y = -3y units/hour [Here the negative sign represents the pipe used to empty the tank.] Pipe A and C together can fill the tank in (a+9) hours. (a+9)x(6y-3y) = 72 (a+9)x3y = 72 Eq.(i) Pipe B alone can fill 75% of the tank in (a-1.5) hours. Time taken by Pipe B alone to fill 100% of the tank = ((a-1.5)/75)x100 = ((a-1.5)/3)x4 So ((a-1.5)/3)x4x4y = 72 Eq.(ii) We can say that Eq.(i) = Eq.(ii). (a+9)x9 = (a-1.5)x16 9a+81 = 16a-24 16a-9a = 81+24 7a = 105 a = 15 Pipe D alone can empty half of the tank in (a-3) hours. Time taken by Pipe D alone to empty the tank = 2x(a-3) So 2x(a-3)x(efficiency of Pipe D) = 72 Put the value of ‘a’ in the above equation. 2x(15-3)x(efficiency of Pipe D) = 72 2x12x(efficiency of Pipe D) = 72 24x(efficiency of Pipe D) = 72 efficiency of Pipe D = -3 units/hour [Here the negative sign represents the pipe used to empty the tank.] Put the value of ‘a’ in Eq.(i) to obtain the value of ‘y’. (15+9)x3y = 72 24x3y = 72 y = 1 Time taken by pipe A and D together to fill the empty tank completely = capacity of tank/(efficiency of pipe A and D together) = 72/(6y-3) Put the value of ‘y’ in the above equation. = 72/(6-3) = 72/3 = 24 hours
