Question

a tank can empty in 20 hrs while another tank empty in 15hrs.if both tap open simultaneously how long it take to empty the full
tank

Answer 1

Correct Question :-

• a tap can empty the full tank in 20 hrs while another tap can empty in 15hrs. if both tap open simultaneously how long it take to empty the full
• tank ?

Solution :-

→ LCM of 20 & 15 = 60 unit = Let Total Capacity of Tank.

So,

→ Tap one Efficiency = (Total Capacity of Tank ) / ( Total Time) = 60/20 = 3 unit/Hour.

Similarly,

→ Another Tap Efficiency = (Total Capacity of Tank ) / ( Total Time) = 60/15 = 4 unit/Hour.

Hence,

→ Efficiency of Both Tank = 3 + 4 = 7unit/hour.

So,

→ Time Taken by Both Tank to Empty Full tank = (60/7) Hours = 8(4/7) Hours. (Ans).

ஃ Both tap open simultaneously take (60/7) Hours to empty the full tank...

Answer 2

Correct Question

Tap A can empty a tank in 20hours while Tap B empty tank in 15hours .both tap open simultaneously how long it take to empty the full

tank

Given

• Tap A Can empty tank in 20h
• Tap B Can empty tank in 15h

To find

• The time taken to empty the tank if both tap is open

Solution

For Tap A

In 1 hour the tank fill = $\frac{1}{20}$

For Tap B

In 1 hour the tank fill = $\frac{1}{15}$

If both open

Let time be taken for empty the both tap be x

$\implies\sf\frac{1}{x}=\frac{1}{20}+\frac{1}{15}$

$\implies\sf\frac{1}{x}=\frac{3+4}{60}$ Taking lcm

$\implies\sf\frac{1}{x}=\frac{7}{60}$

$\therefore\sf\X=\frac{60}{7}$

Hence,the time taken by both tap to empty y tank is.

$\therefore\sf\X=\frac{60}{7}$