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Let x and y be the respective times in which the tap fills the tank
Now this is work and efficiency question
Before solving this u need to remember that linear operations can be performed only on work done and the rate of doing work
So if a tap is filling a tank in x time then it is doing one work in a designated amount of time
So,
Rate of doing work=(1/x)
Now as per question if both the taps are open then the tank is filled in 2 hours
So
(1/x)+(1/y)=(1/2)
Now it is further stated as
smaller tap alone takes 3 hours more to fill the tank in the same amount
So it can derived that,
(1/x)+(1/(x-3))=(1/2)
Solving this equation ,
(2x+3)/(x^2 -3x)=(1/2)
Rearranging the equation,
x^2-7x+6=0
The following quadratic equation can be solved as,
(x-1)(x-6)=0
x=1,x=6
But x=1 is not possible as x-3 cannot be negative
So
x=6
y=x-3=6-3=3
So the individual time taken by the taps are 3 hours and 6 hours respectively
Now this is work and efficiency question
Before solving this u need to remember that linear operations can be performed only on work done and the rate of doing work
So if a tap is filling a tank in x time then it is doing one work in a designated amount of time
So,
Rate of doing work=(1/x)
Now as per question if both the taps are open then the tank is filled in 2 hours
So
(1/x)+(1/y)=(1/2)
Now it is further stated as
smaller tap alone takes 3 hours more to fill the tank in the same amount
So it can derived that,
(1/x)+(1/(x-3))=(1/2)
Solving this equation ,
(2x+3)/(x^2 -3x)=(1/2)
Rearranging the equation,
x^2-7x+6=0
The following quadratic equation can be solved as,
(x-1)(x-6)=0
x=1,x=6
But x=1 is not possible as x-3 cannot be negative
So
x=6
y=x-3=6-3=3
So the individual time taken by the taps are 3 hours and 6 hours respectively
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