Question

2 water taps together can fill a tank in 75 by 8 hours. the tap of larger diameter take 10 hours less than the smaller one to fill the tank separately. find the time in which each tap can separately fill the tank ​

Answer 1

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.