Question

One pipe empties a pool in one whole 1 by 2 hours. A second pipe empties it 1 whole 3 by 4 hours. How long will it take to empty the pool if both the pipes are opened simultaneously?​

Answer 1

Given :-

• Pipe A can empty the whole pool in = 1(1/2) = (3/2) hours.
• Pipe B can Empty the whole Pool in = 1(3/4) = (7/4) hours.

To Find :-

• In How many Hours Both Pipe Will Empty the full tank Together ?

Solution :-

Basic Method :-

→ Pipe A total Time to Empty pool = (3/2) hours.

→ So, Efficiency of Pipe A = (2/3) per hours.

Similarly,

→ Efficiency of Pipe B = (4/7) per hours.

So,

→ Efficiency of Both Pipe Per hour = (2/3) + (4/7) = (14+12)/21 = (26/21) unit of water per hour.

So, Time Taken By Them to Empty Whole Tank = (21/26) Hours. (Ans).

Hence, Both Pipe can Empty the pool in (21/26) Hours.

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LCM Method :-

→ LCM of (3/2) & (7/4) = (21/2) units = Let Capacity of Pool.

So,

→ Efficiency of Tank A per Hour = (Total Capacity) / (Total Time) = (21/2) / (3/2) = (21/2) * (2/3) = 7 unit / hour.

Similarly,

→ Efficiency of Tank B per Hour = (21/2) / (7/4) = (21/2) * (4/7) = 6 unit / hour.

So,

→ Efficiency of Both Pipe = (6 + 7) = 13 unit / Hour.

Hence,

→ Time Taken By Them to Empty Full tank = (Total Capacity of Tank) / (Total Efficiency of Both) = (21/2) / (13) = (21/2*13) = (21/26) Hours.

Hence, Both Pipe can Empty the pool in (21/26) Hours.

______________________

Answer 2

Given:-

Pipe A can empty the pool in  $1\frac{1}{2} \ hours.$

Pipe B can empty the pool in $1\frac{3}{4} \ hours.$

To find:-

Total time taken to empty the pool if both the pipes are opened simultaneously.

Solution:-

Pipe A takes = $\frac{(2*1)+1}{2}\ hours$ = $\frac{3}{2}\ hours$

Pipe B takes = $\frac{(4*1)+3}{4} \ hours$ = $\frac{7}{4}\ hours$

$LCM\ of\ \frac{3}{2}\ and\ \frac{7}{4}\ = \frac{21}{2}$.

Let 21/2 be the capacity of the pool.

Now,

efficiency of pipe A = Total capacity / Time taken to fill the capacity = $\boxed{\frac{\frac{21}{2}}{\frac{3}{2}} =\frac{21}{2}*\frac{2}{3} = 7\ units/hour}$

efficiency of pipe B = Total capacity / Time taken to fill the capacity =

$\boxed{\frac{\frac{21}{2}}{\frac{7}{4}} =\frac{21}{2}*\frac{4}{7} = 6\ units/hour}$

Efficiency of both the pipes = efficiency of pipe A + efficiency of pipe B =

7 + 6 = 13 units/hour

So, total time taken by both the pipes to empty the full tank =

Capacity of the tank/Efficiency of both the pipes = (21/2)/13 = (21/2)*(1/13)

= $\boxed{21/26\ hours}$

∴So, the total time taken to empty the pool if both the pipes are opened simultaneously is 21/26 hours.