Question

Two taps P and Q can fill a tank in 20 minutes and 30 minutes respectively. Both the taps are opened together but after 3 minutes tap Q is closed. How long will the tap P takes to fill the remaining portion of the tank?​

Answer 1

Given:

• P can fill tank in 20 minutes
• Q can fill tank in 30 minutes
• Both tap are opened for 3 minutes together
• After 3 minutes Q is closed.

To Find:

• How long will the tap P takes to fill the remaining portion of the tank?

Solution:

We have given that tap P and Q can fill tank in 20 minutes and 30 minutes respectively.

✪ 1 minute work of P = 1/20

✪ 1 minute work of Q = 1/30

Therefore,

◉ 1 minute work of P and Q = $\frac{1}{20} + \frac{1}{30}$

$\implies\frac{3 + 2}{60}$

$\implies\dfrac{5}{60}~=~\dfrac{1}{12}$

Hence,

★ 3 minutes work of P and Q = 1/12 × 3 = 1/4

→ Remaining work = 1 - 1/4 = 3/4

￣￣￣￣￣￣￣￣￣￣￣￣￣￣

Now, we found that after working 3 minutes together, remaining work in 3/4. We know that P is working 1/20 part in 1 minute. We have to find that how much time it will take to finish 3/4 work.

→ 1/20 part = 1 minute

→ 3/4 part = 3/4 ÷ 1/20

→ 3/4 part = 3/4 × 20 = 15 minutes

Therefore

• Tap P will take 15 minutes.

Answer 2

Question:

• Two taps P and Q can fill a tank in 20 minutes and 30 minutes respectively. Both the taps are opened together but after 3 minutes tap Q is closed. How long will the tap P takes to fill the remaining portion of the tank?

Find:

• How long will the tap P takes to fill the remaining portion of the tank?

Solution:

P=10

Q=20

Take LCM of both Nd it is 20 so now unit

P = 2 unit

Q=1 unit

LCM is the capacity of the tank

P+Q = 2*3 = 6 unit , both will fill in the 2 mint

20–6 = 14 unit left

So 14+ 1 = 15 minute.

Note:

• Tap P will take 15 minutes.