Question

A water tap A takes 7 minutes more than water tap B for filling up a tank with water .Tge 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.

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Answer 1

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 2

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!