Math, asked by asjish4276, 9 months ago

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

Answered by RvChaudharY50
68

||✪✪ 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.

_________

Answered by Anonymous
46

Time taken by two pipes together to fill the tank =

6  \frac{2}{3}  =  \frac{20}{3} hours

volume of water filled in the tank in this time =

 \frac{3}{20}

Now,

let time taken by smaller diameter pipe = x hrs

then, time taken by larger diameter pipe = x-3 hrs

volume \: of \: water \: filled \: by \: smaller \: pipe =  \frac{1}{x}

and

volume \: of \: water \: filled \: by \: larger \: pipe \:  =  \frac{1}{x - 3}

Hence,

 \frac{1}{x}   +  \frac{1}{x  -  3}  =  \frac{3}{20}  \\  \\  (taking \: lcm) \\  \frac{x + x - 3}{ {x}^{2} - 3x }  =  \frac{3}{20}  \\  \\ (cross \: multiplying) \\  40x - 60 = 3 {x}^{2} - 9x \\  \\ 3 {x}^{2}   - 49x  + 60 = 0 \\  \\ (splitting \: the \: middle \: term) \\  {3}^{2}  -45 x -4 x + 60 = 0 \\  \\ 3x(x - 15) - 4(x - 15) = 0 \\  \\ (3x - 4)(x - 15) = 0 \\  \\ so... \\x = 15 \: and \: x =  \frac{4}{3}

If we will take

x =  \frac{4}{3} then \: x - 3 \: will \: become \: negative \: while \: negative \: time \: is \: not \: possible \: hence \:  \\  \\we \: will \: take \:  x = 15

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.

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