Question

two water taps can together fill a tank in 9 3/8 hours the tap of larger diameter takes 10 hours less than smaller one to fill the tank separately​

Answer 1

Step-by-step explanation:

hope it helps...........

Answer 2

Let the time taken by the larger diameter tap be x hours.

The time taken by the smaller diameter tap = (x+10) hours.

Let us assume the volume of the tank to be 1 unit.

(you can take any value for the volume of x, I have taken 1 unit because it is a small value and easy will be easy to solve)

The amount of water filled in hour by smaller diameter tap = 1/(x+10) units.

The amount of water filled in 1 hour by larger diameter tap= (1/x) units.

{1/(x+10)*75/8} + {1/x*75/8)= 1

75/8{{1/(x+10)} + {1/x} = 1

75/8{(x+x+10)/(x^2 + 10x)} = 1

(2x+10)/(x^2+10x) = 8/75

75(2x+10) = 8(x^2+10x)

150x+750 = 8x^2+ 80x

8x^2 + 80x - 150x - 750 = 0

8x^2 - 70x - 750 = 0

2(4x^2 - 35x - 375) = 0

4x^2 - 35x - 375 = 0

4x^2 - 60x + 25x - 375 = 0 (by splitting method)

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

x - 15 = 0

x= 15

4x + 25 = 0

x = -25/4

Time can never be calculated in negative values.

So,

x = 15,

x + 10 = 15 + 10 = 25

Therefore,

Time taken by larger diameter tap = 15 hours

Time taken by smaller diameter tap = 25 hours.

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