Question

Two water taps together can fill a tank in  75/8 hours. The tap of larger diameter takes 10hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

Say the smaller tap fill the tank in x hours
and the bigger tap fill the tank in (x-10) hours

Therefore in 1 hour the small tap can fill 1/x part of the tank.
and, in 1 hour the small tap can fill 1/(x-10) part of the tank.

Two water taps together can fill a tank in  75/8 hours

So,, in one hour the taps fill 8/75 portion of the tank. so the equation will be,

1/(x) + 1/(x-10) = 8/75
=>4x^2 - 115x + 375 = 0 (solving eventually)
=>(4x-15)(x-25)

so, solving the equation we will get X = 15/4 and 25

so, if, x =15/4  than (x-10)= -25/4 (it is not possible because time cant be negative)

If x=25 than (x-10) = 15

thus, the smaller tap fills the tank in 25 hours where the large tap fills it in 15 hours.

Answer 2

Answer:

Answer:

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

Explanation−

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

x

1

hours.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

x−10

1

hours.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

x

1

+

x−10

1

=

75

8

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Taking LCM,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ By cross multiplying,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

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

2

−80

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

2

−230x+750=0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Taking 2 as common

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

: \implies⟹ \rm{4x^2-115x+375=0}4x

2

−115x+375=0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Solving the Quadratic equation, by splitting middle term.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

2

−100x−15x+375=0

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

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

We get,

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

x=

4

15

and25

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

4

15

,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

[/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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

x=25

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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