Question

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 the taps together to fill the tank. Find the time each tap alone would take to fill the tank.

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

Let Tap B takes x mins

Tap A takes (x + 7) mins

Tap B takes x mins

⇒ 1 min = 1/x of the tank filled

Tap A takes (x + 7) mins

⇒ 1 min = 1/(x + 7) of the tank filled

Together:

1 min = 1/x + 1/(x + 7)

1 min = [ x + 7 + x] /x(x + 7)

1 min = (2x + 7) /x(x + 7) of the tank

Total time needed to fill the tank together:

mins needed = x(x + 7)/(2x + 7)

The tap A takes 16 minutes more than the time taken by both the taps together to fill the tank:

Tap A = Tap A and Tap B + 16 mins

(x + 7)  = x(x + 7)/(2x + 7) + 16

x + 7 - 16 = x(x + 7)/(2x + 7)

x - 9 = x(x + 7)/(2x + 7)

x(x + 7) = (x - 9)(2x + 7)

x² + 7x = 2x²  + 7x - 18x - 63

x² - 18x - 63 = 0

(x - 21)(x + 3) = 0

x = 21 or x = - 3 (rejected, time cannot be negative)

Find the time needed:

Tap B = x = 21 mins

Tap A = x + 7 = 21 + 7 = 28 mins

Answer: Tap A takes 28 mins and Tap B takes 21 mins