A water tap A takes 7minutesmore than water tap B for filling up a tank with water the tap A takes 16minutes more than the taken by both the tap together to fill the tank find the time each tap alone would take to fill the tank???????
please anwer
your answer is correct I will give५stars
Answers
Answer:
28 minutes, 21 minutes
Explanation:
Let the time taken by tap B to fill the tank is 'x' minutes.
∴ Part filled by tank B in 1 minute = (1/x).
Given that A takes 7 minutes more than tap B.
Then, the time take by tap A is 'x + 7' minutes.
∴ Part filled by tan A in 1 minute = (1/x + 7).
Part of the tank filled by (A + B) in 1-minute = (1/x) + (1/x + 7)
= [x + 7 + x]/[x(x + 7]
= [2x + 7]/[x² + 7x]
∴ Total time = [x² + 7x]/[2x + 7]
Given that Tap A takes 16 minutes more than the time taken by both.
=> x + 7 = {[x² + 7x]/[2x + 7]} + 16
=> (2x + 7)(x + 7) = (x² + 7x) + 16(2x + 7)
=> 2x² + 14x + 7x + 49 = x² + 7x + 32x + 112
=> 2x² + 21x + 49 = x² + 39x + 112
=> x² - 18x - 63 = 0
=> x² - 21x + 3x - 63 = 0
=> x(x - 21) + 3(x - 21) = 0
=> (x - 21)(x + 3) = 0
=> x = 21, -3{∴Cannot be negative}
=> x = 21.
Now:
=> x + 7
=> 28.
Therefore:
→ Time taken by tap A = 28 minutes.
→ Time taken by tap B = 21 minutes.
Hope it helps!