Latest Elitmus Aptitude Question SOLUTION: A large tank is filled by a large drainage pipe. when the tank is filled all the outer pipes of the tank that is 10 are opened and tank gets empty in 2.5 hours. When only 5 outer

Let the volume of tank be : x lit
Let the tank be filled by drainage pipe at the rate of : y lit/hr

So, when the tank is fully filled and then all the 10 outer pipes of tank are opened (where as the drainage is still active and it is being filled with a rate of y lit/hr) , the tank gets empty in 2.5 hrs.
So, the total volume of water drain out by 10 pipes in 2.5 hr is : x + 2.5y
So the rate of drain out per pipe is : (x + 2.5y)/10 x 2.5 .....(i)

When 5 pipes are used to drain out the water , the tank gets empty in 5.5 hrs,
So, the total volume of water drain out by 5 pipes in 5.5 hr is : x + 5.5y
So the rate of drain out per pipe is : (x + 5.5y)/5 x 5.5 .....(ii)

Equating equation (i) & (ii),
(x + 2.5y)/25 = (x + 5.5y)/27.5
or, 11(x + 2.5y) = 10(x + 5.5y)
or, 11x + 27.5y = 10x + 55y
or, x = 27.5y

Therefore when 3 pipes are used to empty the tank,
let the time taken be : t hrs
So, the total volume of water drain out by 3 pipes in t hr is : x + ty
So the rate of drain out per pipe is : (x + ty)/3t

Therefore; (x + ty)/3t = (x + 2.5y)/25
or, (27.5y + ty)/3t = (27.5y + 2.5y)/25
or, (27.5 + t)/3t = 30/25 = 1.2
or, 27.5 = t = 3.6t
or, 2.6t = 27.5
or, t = 27.5/2.6 = 10.57hrs

one pipe can empty tank in 2.5*10=25h
so 5 pipes can empty tank in 25/5=5h since total 10 pipes so remaining 5 pipes fill the tank
so some additional work should be done by 5 pipes used to empty tank
given total required if 5 outer pipes are opened is 5.5h=5+0.5=(5+1/2)h
additional time=10/5=1/2........
if 3 outer pipes are opened then time to empty tank is (25/3)+(10/3)= 11.66h

1)part of water drained in an hour by a pipe=2.5/10=0.25;2)part of water drained by a pipe=5.5/5=1.1;then the series as AP with d as 0.17 is 0.25,0.42,0.59,0.76,0.93,1.1,1.27,1.44====>x/3=1.44=>x=4.32
10 9 8 7 6 5 4 3

10 pipes work-hour=1/2.5
5 pipes work-hour=1/5.5
Therefore,(10-5)pipes work-hour=(1/2.5)-(1/5.5)
= 60/275
Then, 3 pipes work-hour=(60/275)*(3/5) ------------(Unitary method)
= 36/275
So,time required to empty the tank by 3 pipes is= 275/36
=7.63 Hours

Let the volume of tank be : x lit
Let the tank be filled by drainage pipe at the rate of : y lit/hr

So, when the tank is fully filled and then all the 10 outer pipes of tank are opened (where as the drainage is still active and it is being filled with a rate of y lit/hr) , the tank gets empty in 2.5 hrs.
So, the total volume of water drain out by 10 pipes in 2.5 hr is : x + 2.5y
So the rate of drain out per pipe is : (x + 2.5y)/10 x 2.5 .....(i)

When 5 pipes are used to drain out the water , the tank gets empty in 5.5 hrs,
So, the total volume of water drain out by 5 pipes in 5.5 hr is : x + 5.5y
So the rate of drain out per pipe is : (x + 5.5y)/5 x 5.5 .....(ii)

Equating equation (i) & (ii),
(x + 2.5y)/25 = (x + 5.5y)/27.5
or, 11(x + 2.5y) = 10(x + 5.5y)
or, 11x + 27.5y = 10x + 55y
or, x = 27.5y

Therefore when 3 pipes are used to empty the tank,
let the time taken be : t hrs
So, the total volume of water drain out by 3 pipes in t hr is : x + ty
So the rate of drain out per pipe is : (x + ty)/3t

Therefore; (x + ty)/3t = (x + 2.5y)/25
or, (27.5y + ty)/3t = (27.5y + 2.5y)/25
or, (27.5 + t)/3t = 30/25 = 1.2
or, 27.5 = t = 3.6t
or, 2.6t = 27.5
or, t = 27.5/2.6 = 10.57hrs