a water tap take A takes 7 minutes more than water tap B for filling up a tank with water.The tap A takes 16 minutes more than the time taken by both the taps together to fill the tank . Find the time each tap alone would take to fill the tank.
Answers
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Answer:
28 minutes, 21 minutes
Step-by-step 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!