answer the above question
Answers
Let the tap with smaller diameter fills the tank alone in x hours
Let the tap with larger diameter fills the tank alone in (x – 10) hours.
In 1 hour, the tap with a smaller diameter can fill 1/x part of the tank.
In 1 hour, the tap with larger diameter can fill 1/(x – 10) part of the tank.
The tank is filled up in 75/8 hours.
Thus, in 1 hour the taps fill 8/75 part of the tank.
1/x + 1/(x-10) = 8/75
(x-10) + x / x(x-10) = 8/75
2x – 10/x(x-10) = 8/75
75 (2x-10) = 8(x2-10x) by cross multiplication
150x – 750 = 8x2 – 80x
8x2 − 230x + 750 = 0
4x2−115x + 375 = 0
4x2 − 100x −15x + 375 = 0
4x(x−25)−15(x−25) = 0
(4x−15)(x−25) = 0
4x−15 = 0 or x – 25 = 0
x = 15/4 or x = 25
Case 1: When x = 15/4
Then x – 10 = 15/4 – 10
⇒ 15-40/4
⇒ -25/4
Time can never be negative so x = 15/4 is not possible.
Case 2: When x = 25 then
x – 10 = 25 – 10 = 15
∴ The tap of smaller diameter can separately fill the tank in 25 hours and the time taken by the larger tap to fill the tank = ( 25 – 10 ) = 15 hours.
Answer:
15 hours is write answer
Step-by-step explanation:
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