Question

Two water taps together can fill a tank 9 and 3/8 hours. The tap of larger diameter takes 10 hour less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.’

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Answer 1

Given:-

• Two taps together take $\sf{9\dfrac{3}{8}}$ hours to fill a tank.
• The tap of larger diameter takes 10 hours let than the smaller one to fill the tank separately.

To Find:-

The time in which each tap can separately fill the tank.

Assumption:-

Let the time taken by smaller tap to fill the tank be x

Time taken by larger tap to fill the tank = $\sf{x-10}$

Solution:-

In one hour tank filled by smaller tap = $\sf{\dfrac{1}{x}}$

In one hour tank filled by larger tap = $\sf{\dfrac{1}{x-10}}$

Total time taken by both the taps to fill the tank = $\sf{9\dfrac{3}{8} = \dfrac{75}{8}}$

In one hour tank filled by both the taps = $\sf{\dfrac{1}{\dfrac{75}{8}} = \dfrac{8}{75}}$

ATQ,

$\sf{\dfrac{1}{x} + \dfrac{1}{x-10} = \dfrac{8}{75}}$

= $\sf{\dfrac{x-10 + x}{x(x-10)} = \dfrac{8}{75}}$

= $\sf{\dfrac{2x-10}{x^2-10x} = \dfrac{8}{75}}$

By Cross multiplication,

$\sf{75(2x-10) = 8(x^2-10x)}$

= $\sf{150x - 750 = 8x^2 - 80x}$

Taking all the variables on LHS,

The sign will change as follows:-

(-) into (+)

(+) into (-)

= $\sf{150x - 750 - 8x^2 + 80x = 0}$

= $\sf{230x - 750 - 8x^2 = 0}$

= $\sf{-8x^2 + 230x - 750 = 0}$

Taking -2 as common,

$\sf{-2(4x^2 - 115x + 375) = 0}$

= $\sf{4x^2 - 115x + 375 = \dfrac{0}{-2}}$

= $\sf{4x^2 - 115x + 375 = 0}$

By splitting the middle term,

= $\sf{4x^2 - 100x - 15x + 375 = 0}$

= $\sf{4x(x-25)-15(x-25) = 0}$

= $\sf{(x-25)(4x-15) = 0}$

Either,

$\sf{x-25 = 0}$

= $\sf{x = 25}$

Or,

$\sf{4x - 15 = 0}$

= $\sf{4x = 15}$

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

Taking x = 25

Time taken by smaller tap to fill the tank = x = 25 hours

Time taken by larger tap to fill the tank = x-10 = 25 - 10 = 15 hours.

Taking x = $\sf{\dfrac{15}{4}}$

Time taken by smaller tap to fill the tank = x = $\sf{\dfrac{15}{4}\:hours}$

Time taken by larger tap to fill the tank = x - 10 = $\sf{\dfrac{15}{4} - 10}$

= $\sf{\dfrac{15-40}{4}}$

= $\sf{\dfrac{-25}{4}}$

As negative time cannot be taken,

Hence,

Time taken by smaller tap = 25 hours

Time taken by larger tap = 15 hours.

______________________________________

Answer 2

Answer:

ANSWER

Consider that the tap with smaller diameter fills the tank in x hours.

Then, the tap with larger diameter fills the tank in x−10 hours.

This shows that the tap with a smaller diameter can fill

x

1

part of the tank in 1 hour. Similarly, the tap with larger diameter can fill

x−10

1

part of the tank in 1 hour.

It is given that the tank is filled in

8

75

hours that is, the taps fill

75

8

part of the tank in 1 hour. Then,

x

1

+

x−10

1

=

75

8

x(x−10)

x−10+x

=

75

8

x

2

−10x

2x−10

=

75

8

75(2x−10)=8(x

2

−10x)

150x−750=8x

2

−80x

8x

2

−230x+750=0

4x

2

−115x+375=0

4x

2

−100x−15x+375=0

4x(x−25)−15(x−25)=0

(4x−15)(x−25)=0

4x−15=0

x=

4

15

Or,

x−25=0

x=25

When x=

4

15

, then, x−10=

4

15

−10

=

4

15−40

=−

4

25

This cannot be possible because time can never be negative.

When x=25, then,

x−10=25−10

x=25

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