Question

Two taps together can fill a tank completely in 3 1/13minutes. The smaller 13

tap takes 3 minutes more than the bigger tap to fill the tank. How much time does each tap take to fill the tank completely ?​

Answer 1

Given :-

• Two taps together can fill tank completely in 3 1/13 minutes.
• The smaller tap takes 3 minutes more than the bigger tap to fill the tank.

To Find :-

• How much time does each tap take to fill the tank completely?

Solution :-

• Let the Capacity of the tank to be filled be A and the time taken by the Bigger tap to fill the Tank be B minutes. As per question, smaller tap takes 3 minutes more than bigger tap to fill the tank.Therefore, Time taken by the smaller tap to fill the tank = (B + 3) minutes.

In One Minute:-

• The Bigger Tap fills the Tank by

$\sf :\implies \dfrac{A}{B}$

• The Smaller Tap fills the Tank by

$\sf :\implies \dfrac{A}{B + 3}$

Hence, the two taps fill the Tank by :-

$\mathsf{\implies \bigg(\dfrac{A}{B} + \dfrac{A}{B + 3}\bigg)}$

⠀⠀⠀⠀⠀$\small\underline{\pmb{\sf According \: to \: the \: question :-}}$

$\pink{\mathsf{ \: \: \: \: \: \: \: \: \: \:\::\implies \dfrac{40}{13}\bigg(\dfrac{A}{B} + \dfrac{A}{B + 3}\bigg) = A}}$

$\mathsf{\: \: \: \: \: \: \: \: \: \:\::\implies \dfrac{40}{13}\bigg(\dfrac{1}{B} + \dfrac{1}{B + 3}\bigg) = 1}$

$\mathsf{\: \: \: \: \: \: \: \: \: \:\::\implies \dfrac{40}{13}\bigg(\dfrac{B + 3 + B}{B(B + 3)}\bigg) = 1}$

$\mathsf{\: \: \: \: \: \: \: \: \: \:\::\implies 40(2B + 3) = 13(B^2 + 3B)}$

$\mathsf{\: \: \: \: \: \: \: \: \: \:\::\implies 13B^2 + 39B = 80B + 120}$

$\: \: \: \: \: \: \: \: \: \:\::\implies \mathsf{13B^2 - 41B - 120 = 0}$

$\: \: \: \: \: \: \: \: \: \:\::\implies \mathsf{13B^2 - 65B + 24B - 120 = 0}$

$\: \: \: \: \: \: \: \: \: \:\::\implies \mathsf{13B(B - 5) + 24(B - 5) = 0}$

$\: \: \: \: \: \: \: \: \: \:\::\implies \mathsf{(B - 5)(13B + 24) = 0}$

$\: \: \: \: \: \: \: \: \: \:\::\pink{\mathsf{\implies B = 5\;\;\;(or)\;\;\;B = \dfrac{-24}{13}}}$

(As time cannot be negative)

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies{\underline{\boxed{\frak{\pink{B =5}}}}}\:\bigstar$

$\therefore\:\underline{\textsf{Time taken by Bigger tap = \textbf{5 minutes }}}.$

$\dag\underline{\sf{\sf Time \: taken\: by\: Smaller\: tap:- }}$

$\sf:\implies B+3 = 5+3$

$\sf:\implies \boxed{ \sf Time\: taken = \pink{\pmb {8 \: minutes .}}}$