A water tab A takes 7 min more than water tab B for filling up a tank with water.the tabA takes 16 min more than the time taken by both the tap together to fill the tank. find the time each tap alone would take to fill the tank
Answers
Let the time taken by A is x minutes and B is y minutes,
Hence from the 1st statement we have,
x-y = 7 = > y = x-7..........................eq1
Now, in one minute tank filled by A = 1/x
and in one minute tank filled by B = 1/y
So combine together both can fill = 1/x + 1/y tank in one minute,
Hence time taken by both the taps to fill the tank
= 1/(1/x + 1/y)
= xy/x+y
Now from the second statement,
x- xy/x+y = 16
=> x² + xy - xy = 16(x+y)
=> x² -16x - 16y = 0
Putting the value of y from eq1
x²-16x - 16(x-7) = 0
=> x² - 32x + 112 = 0.
Solving the above quadratic eqn, we get
x = 28 or 4
rejecting 4 as x can't be less than 7
x = 28
y = x-7 = 28-7 = 21
Hence time taken by both the taps are 28 and 21 minutes alone.
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!