A water tap 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 taps together to fill the tank.Find the time each tap alone would take to fill the tank.
Answers
Answer:
21 & 28 minutes
Step-by-step explanation:
Let say b takes B min
1 min work by B = 1/B
A takes B +7 min
1 min work by A = 1/(B+7)
1 min work by A&B together
1/A + 1/B = 1/(B+7) + 1/(B)
=( 2B+7)/{(B)((B+7)}
Time taken by A& B together
= B(B+7)/(2B+7) = B+7 - 16
B^2 +7B = (2B+7)(B-9)
B^2 + 7B = 2B^2 -18B +7B -63
B^2-18B-63=0
B^2-21B+3B-63=0
(B-21)(B+3)=0
B = 21 min
A = B+7 = 28 min
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!