Ahloha!!
Two pipes running together can fill a cistern in 6 minutes.If one pipe takes 5 minutes more than the other to fill the cistern. Find the time in which each pipe will fill the cistern individually .
ALL THE BEST!
Answers
SOLUTION
Let the time taken by 1st tap to fill the cistern completely = x minutes.
Thus,
let the time taken by 2nd tap to fill the cistern= (x+5) minutes.
Work done by 1st tap (in 1 min)= 1/x
Work done by 2nd tap(in 1min)=1/(x+5)
Work done by 1st tap+ work done by 2nd tap in 1 minute
Work done by 1st tap+ work done by 2nd tap in 1 minute=)Cistern filled in 1 minute= 1/6.
Work done by 1st tap+ work done by 2nd tap in 1 minute=)Cistern filled in 1 minute= 1/6.Thus, the equation is 1/x +1/(x+5)= 1/6
=) 6(2x+5) =x^2 +5x
=) 12x +30= x^2 +5x
=) x^2 -7x-30=0
=) x^2+3x-10x-30= 0
=) x(x+3)-10(x+3)=0
=) (x+3) (x-10) =0
=) x+3= 0 or x-10=0
=) x= -3 or x= 10
Time can't be negative.
Thus, x= 10
Time taken by 1st tap= 10 minutes.
Time taken by 2nd tap= x+5= 10+5= 15 minutes.
hope it helps ☺️⬆️
Solution
Let the volume of cistern = V litre (L)
let the flow rate of one pipe be x L/min and that of other pipe be y L/min
Since 6 minute are needed to fill the pipe
6x+6y = V = total volume (1)
V/x = time taken by one pipe and V/y = time taken by other pipe
V/x + V/y = 5 min
1/x - 1/y = 5/V (2)
Solve
Time taken by one pipe = 15 min
and Time taken by other pipe = 10 min
=>15 min and 10 min