Question

Two pipes A and B can fill a tank in 9 hours and 18 hours respectively. If both pipes
are opened together in an empty tank, how much time will it take to fill it?​

Answer 1

Answer:

Step-by-step explanation:

A can do 1/9th work in 1hour

B can do 1/18 work in 1hour

working together they take

1/9+1/18=3/18=6hours

Answer 2

$\huge{\underline{\underline{\mathtt{Solution}}}}$

Given :-

• Pipe A can fill tank = 9 hours
• Pipe B can fill tank = 18 hours

To find :-

• Together both can fill tank in how many hours .

_____________________

Tap A can fill units in 1 hour = $\frac{1}{9}$ part of total .

Tap B can fill units in 1 hour = $\frac{1}{18}$. part of total.

Let's find out the total capacity of tank . ( LCM of Units filled by each tap in 1 hour )

$\implies \: \frac{1}{9} \: + \: \frac{1}{18} \: = lcm \: = \: 18$

So , Total capacity of tank = 18 units .

⇝ Capacity of Tap A = $\frac{1}{9}$ × 18

⇝ 2 units /. hour .

⇝ Capacity of Tap B = $\frac{1}{18}$ × 18

⇝ 1 units / hour .

So together in 1 hour taps fill tank = 1+2 = 3 units .

So 18 units will be filled in = $\frac{18}{3}$

⇝ 6 hours .

So both tap A and B can fill the tank in 6 hours if opened together .