two taps running together can fill a tank in 3 1/13 hours . if one tap takes 3 hours more than the other to fill the tank , then how much time will each tap take to fill the tank?
Answers
Let one tap fill the tank in x hrs.
Therefore, other tap fills the tank in (x + 3) hrs.
Work done by both the taps in one hour is
1/x + 1/(x+3) = 13/40
(2x + 3)40 = 13(x2 + 3x)
13x2 – 41x – 120 = 0
(13x + 24)(x – 5) = 0
x = 5 (rejecting the negative value)
Hence, one tap takes 5 hrs and another 8 hrs separately to fill the tank.
Answer:
Step-by-step explanation:
Solution :-
Let the time taken by 1st tap be x hours.
And the time taken by 2nd tap be x + 3 hours.
According to the Question,
⇒ 1/x + 1/(x + 3) = 13/40
⇒ x + 3 + x/x(x + 3) = 13/40
⇒ 2x + 3/x² - 3x = 13/40
⇒ 80x + 120 = 13x² + 39c
⇒ 13x² - 41x - 120 = 0
⇒ 13x² - 65x + 24x - 120 = 0
⇒ 13x(x - 5) + 24(x - 5) = 0
⇒ (x - 5) (13x + 24) = 0
⇒ x = 5, - 24/13 (x can't be negative)
⇒ x = 5
Time taken by 1st tap = x = 5 hours
Time taken by 2nd tap = x + 3 = 5 + 3 = 8 hours.