Two pipes together fill 6(2/3)hours. If the pipe of smaller diameter takes 3 hours more than the pipe of larger diameter to fill the tank separately, find the time taken by each pipe to fill the tank separately /
Answers
||✪✪ QUESTION ✪✪||
Two pipes together fill 6(2/3)hours. If the pipe of smaller diameter takes 3 hours more than the pipe of larger diameter to fill the tank separately, find the time taken by each pipe to fill the tank separately ?
|| ✰✰ ANSWER ✰✰ ||
Lets Assume That, The pipe of Smaller diameter can fill the Tank in = x Hours.
Than, Pipe of Larger diameter will take = (x - 3) hours .
So,
→ Per hour work of Smaller Diameter pipe = (1/x)
→ Per hour work of Larger Diameter pipe = 1/(x-3)
Now, Given That, Both can Fill it in 6(2/3) hours = (20/3) Hours .
So,
→ 1/x + 1/(x-3) = 3/20
→ [ (x-3+x) / x(x-3) ] = 3/20
→ [ (2x - 3) /(x² - 3x) ] = 3/20
→ 20(2x - 3) = 3(x² - 3x)
→ 40x - 60 = 3x² - 9x
→ 3x² - 9x - 40x + 60 = 0
→ 3x² - 49x + 60 = 0
_____________________
Now, we know That,
→ Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,
→ x = [ -b±√(b²-4ac) / 2a ]
or,
→ x = [ - b± √D /2a ] where D(Discriminant)= b²-4ac..
___________________________
From Equation , 3x² - 49x + 60 = 0 we have :-
→ a = 3
→ b = (-49)
→ c = 60
→ D(Discriminant)= b²-4ac
→ D = (-49)² - 4*3*60
→ D = 2401-720
→ D = 1681
→ √D = ± 41
So,
→ x = [ - b± √D /2a ]
→ x = [ 49 ± 41 /2*3 ]
→ x = (90/6) = 15
→ x = (8/6) = (4/3) [ Not Possible. ]
Hence,
→ The pipe of Smaller diameter can fill the Tank in = x Hours. = 15 Hours .
Than, Pipe of Larger diameter will take = (x - 3) hours . = 12 Hours.
_________
Time taken by two pipes together to fill the tank =
volume of water filled in the tank in this time =
Now,
let time taken by smaller diameter pipe = x hrs
then, time taken by larger diameter pipe = x-3 hrs
and
Hence,
If we will take
therefore,
time taken by smaller pipe to fill the tank = x = 15 hrs
and
time taken by larger pipe to
fill the tank = x-3 = 15-3= 12 hrs.