Question

two water taps together can fill a tank in 75 by 8 hours the tap of larger diameter takes 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

Answer:

Time taken by smaller one is 25 minutes and the larger one 15 minutes.

Step-by-step explanation:

Time taken by the smaler pipe is = x hrs

Time taken by the larger pipe is =  x - 10 hrs

Let us take the reciprocols of them, i.e.,

1/x + 1/x - 10 = 8/75

x - 10 + x / x(x - 10) = 8/75

(2x - 10)75 = 8(x² - 10x)

150x - 750 = 8x² - 80x

8x² - 80x - 150x + 750 = 0

8x² - 230x + 750 = 0

Divide the eqn by 2,

4x² - 115x + 375 = 0

(4x - 15) (x - 25) = 0

4x - 15 = 0           x - 25 = 0

x = 15/4                x = 25

There are two possibilities, in which second one is correct, i.e., x = 25.

Time taken by the smaller pipe to fill is 25 hrs.

Time taken by the larger pipe to fill is 25 - 10 = 15 hrs

Hope its helpful!!!

Answer 2

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.