Question

Answer this :-

Two water taps together can fill a tank in $\dfrac{75}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time(in hrs.) in which tap of smaller diameter can separately fill the tank.

Thank you ❤​

Answer 1

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

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

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

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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 $\dfrac{1}{x}$ hours.

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

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

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

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★ Taking LCM,

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

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

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★ By cross multiplying,

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: $\implies$ $\rm{75(2x-10)\:=\:8(x^2-10)}$

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

: $\implies$ $\rm{8x^2-230x+750=0}$

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★ Taking 2 as common

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: $\implies$ $\rm{4x^2-115x+375=0}$

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★ Solving the Quadratic equation, by splitting middle term.

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

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

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

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We get,

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

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

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: $\implies$ $\rm{x-10\:=\:\dfrac{15}{4}\:-\:10}$

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

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

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

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

Answer 2

Step-by-step explanation:

SOLUTION :

Given : Two water taps together can fill a tank in 9 ⅜ = 75/8 h

Let the smaller tap fill the tank in ‘x’ hours

The tap of larger diameter fills the tank in ‘x – 10’ hours.

In 1 hour the smaller tap fills the portion of the tank : 1/x

In 75/8 hour the smaller tap fills the portion of the tank : 75/8 × 1/x = 75/8x

In 1 hour the tap of larger diameter fills the portion of the tank : 1/(x - 10)

In 75/8 hour the tap of larger diameter fills the portion of the tank : 75/8 × 1/(x - 10) = 75/8(x - 10)

A.T.Q

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

75/8[1/x + 1/(x - 10)] = 1

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

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

[By taking LCM]

{x - 10 + x}/(x² - 10x) = 8/75

75(x + x – 10) = 8(x² – 10x)

75(x + x – 10) = 8x² – 80x

75(2x– 10) = 8x² – 80x

150x – 750 = 8x² – 80x

8x² – 80x - 150x + 750 = 0

8x² – 230x + 750 = 0

2(4x² – 115x + 375) = 0

4x² – 115x + 375 = 0

4x² – 100x – 15x + 375 = 0

[By middle term splitting]

4x(x – 25) – 15(x – 25) = 0

(4x – 15)(x – 25) = 0

(4x – 15) = 0 (x – 25) = 0

x = 15/4 or x = 25

Value of x can’t be 15/4 as (x – 10) will be negative. Since, time cannot be negative,

Therefore, x = 25

The smaller tap fill the tank = x = 25 hours

The tap of larger diameter fills the tank = (x – 10) = 25 - 10 = 15 h

Hence, the time taken by the tap of larger diameter be 15 h & time taken by smaller tap be 25 hours.

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