1.

A takes 7 minutes more than water tap B for filling up a tank with water.A takes 16 minutes more than the time taken by both the taps together to filltank. Find the time each tap alone would take to fill the tankthe

Answer»

Let Tap B takes x mins

Tap B 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



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