Question

two water taps can filla tank in 75/8 hours tap of larger diameter takes 10 hours less than the tap of smaller diameter to fill the tank separately find the time taken by each tap to fill the tank separately

Answer 1

Sol :
Let the tap of the larger diameter fills the tank alone in (x – 10) hours.
 
In 1 hour, the tap of the  smaller diameter can fill 1/x part of the tank.
 
In 1 hour, the tap of the  larger diameter can fill 1/(x – 10) part of the tank.
 
Two water taps together can fii a tank in 75 / 8 hours.
 
But  in 1 hour the taps fill 8/75 part of the tank.
 
1 / x  +  1 / (x – 10) = 8 / 75.
 
( x – 10 + x ) / x ( x – 10) =  8 / 75.
 
2( x – 5) / ( x2 – 10 x) = 8 / 75.
 
4x2 – 40x = 75x – 375.
 
4x2 – 115x + 375 = 0
 
4x2 – 100x – 15x + 375 = 0
 
4x ( x – 25) – 15( x – 25) = 0
 
( 4x -15)( x – 25) = 0.
 
x = 25, 15/ 4.
 
But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time
 
But x = 25 then x – 10 = 15.

Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill
the tank in 25 hours

Answer 2

Answer:

Answer:

$\huge\underline\mathrm\green{Answer-}$

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Tap of smaller diameter can fill the tank in 25 hours separately.

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$\huge\underline\mathrm\red{Explanation-}$

Explanation−

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Let tap of smaller diameter can fill the tank in x hours.

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So, tap of larger diameter can fill the tank in (x - 10) hours ( as mention in the question )

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In 1 hour, tap of smaller diameter can fill tank in \dfrac{1}{x}

x

1

hours.

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Similarly, In 1 hour, tap of larger diameter can fill tank in $\dfrac{1}{x-10}$

x−10

1

hours.

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It is given that, both taps can fill the tank in $\dfrac{75}{8}$

8

75

hours. So, in hour, both taps can fill tank in $\dfrac{8}{75}$

75

8

hours.

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$\therefore∴ \large{\boxed{\rm{\dfrac{1}{x}\:+\:\dfrac{1}{x-10}\:=\:\dfrac{8}{75}}}}$

x

1

+

x−10

1

=

75

8

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★ Taking LCM,

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$:\implies⟹ \rm{\dfrac{x-10+x}{x(x-10)}\:=\:\dfrac{8}{75}}$

x(x−10)

x−10+x

=

75

8

$:\implies⟹ \rm{\dfrac{2x-10}{x^2-10x}\:=\:\dfrac{8}{75}}$

x

2

−10x

2x−10

=

75

8

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★ By cross multiplying,

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$:\implies⟹ \rm{75(2x-10)\:=\:8(x^2-10)}75(2x−10)=8(x </p><p>2</p><p> −10)$

$implies⟹ \rm{150x-750=8x^2-80}$ 150x−750=8x

2

−80

$</p><p>:\implies⟹ \rm{8x^2-230x+750=0}$

2

−230x+750=0

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★ Taking 2 as common

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: \implies⟹ \rm{4x^2-115x+375=0}4x

2

−115x+375=0

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★ Solving the Quadratic equation, by splitting middle term.

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: \implies⟹ \rm{4x^2-100x-15x+375=0}4x

2

−100x−15x+375=0

$:\implies⟹ \rm{4x(x-25)-15(x-25)=0}4x(x−25)−15(x−25)=0$

$:\implies⟹ \rm{(4x-15)(x-25)=0}(4x−15)(x−25)=0$

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We get,

$\large{\boxed{\rm{\pink{x=\dfrac{15}{4}\:and\:25}}}}$

x=

4

15

and25

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$\bold{x=\dfrac{15}{4}}$

4

15

,

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[/tex]:\implies⟹ \rm{x-10\:=\:\dfrac{15}{4}\:-\:10}x−10=[/tex]

4

15

−10

$:\implies⟹ \rm{x-10=\dfrac{-25}{4}}$ x−10=

4

−25

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Since time can't be negative, so we reject this value.

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\therefore∴ $\huge{\boxed{\rm{\blue{x=25}}}}$

x=25

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Therefore, tap of smaller diameter can fill the tank in 25 hours separately.