Question

Two taps A and B can fill a cistern in 6 hours and 9 hours respectively. In the beginning, both taps are opened but after 3 hours first tap is turned off. How much longer will the tank take to fill?

Answer 1

Answer:

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Answer 2

To fill the tank completely additional 1.5 hours will be required

Solution:

According to question there are two taps  ‘A’ and  ‘B’

Tap A can fill the cistern in 6 hours

Tap B can fill the cistern in 9 Hours

So, if we take LCM(6, 9) = 18

So , let us assume the capacity of cistern be 18

$\text { Efficiency of } \mathrm{A}=\frac{\text { Total Capacity of cistern }}{\text { Time taken By } A}=\frac{18}{6}=3$

$\text { Efficiency of } \mathrm{B}=\frac{\text { Total Capacity of cistern }}{\text { Time taken By } B}=\frac{18}{9}=2$

The Tap A and B are kept open for 3 hours simultaneously so total water filled by these in 3 hours  

= (efficiency of A + efficiency of B ) × 3 hour

= 5 x 3 = 15  

Remaining water to be filled = 18 – 15 = 3

Now, Tap A is closed and only Tap B is filling water

So, time taken to fill remaining water is given as:

$=\frac{\text {Remaining water to be filled}}{\text {efficiency of } B}=\frac{3}{2}=1.5 \text { hour }$

So, To fill the tank completely additional 1.5 hours will be required

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