1 | 13 | hours |
17 |
2 | 8 | hours |
11 |
3 | 9 | hours |
17 |
4 | 1 | hours |
2 |
3 | 9 | hours |
17 |
Net part filled in 1 hour | ❨ | 1 | + | 1 | - | 1 | ❩ | = | 17 | . |
5 | 6 | 12 | 60 |
∴ The tank will be full in | 60 | hours i.e., 3 | 9 | hours. |
17 | 17 |
4 | 1 | hours |
3 |
Work done by the leak in 1 hour = | ❨ | 1 | - | 3 | ❩ | = | 1 | . |
2 | 7 | 14 |
∴ Leak will empty the tank in 14 hrs.
Part filled in 4 minutes = 4 | ❨ | 1 | + | 1 | ❩ | = | 7 | . |
15 | 20 | 15 |
Remaining part = | ❨ | 1 - | 7 | ❩ | = | 8 | . |
15 | 15 |
Part filled by B in 1 minute = | 1 |
20 |
∴ | 1 | : | 8 | :: 1 : x |
20 | 15 |
x = | ❨ | 8 | x 1 x 20 | ❩ | = 10 | 2 | min = 10 min. 40 sec. |
15 | 3 |
∴ The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.
Part filled by (A + B) in 1 minute = | ❨ | 1 | + | 1 | ❩ | = | 1 | . |
60 | 40 | 24 |
Suppose the tank is filled in x minutes.
Then, | x | ❨ | 1 | + | 1 | ❩ | = 1 |
2 | 24 | 40 |
⟹ | x | x | 1 | = 1 |
2 | 15 |
⟹ x = 30 min.
Then, faster pipe will fill it in | x | minutes. |
3 |
∴ | 1 | + | 3 | = | 1 |
x | x | 36 |
⟹ | 4 | = | 1 |
x | 36 |
⟹ x = 144 min.
Part filled A in 1 hr= (1/45)
Part filled B in 1 hr= (1/36)
Part filled by (A+B) together in 1 hr=(1/45)+(1/36)=1/20
So, The tank will be full in 20 hr.
We know that, when a pipe fills a tank in m h, then the part of tank filled in 1 h = 1/m
Here, m = 6
? Required part of the tank to be filled in 1h = 1/6 part
Work done by leak in 1 hour = (1/8 - 1/10) = 1/40
? The leak will empty the cistern in 40 hours.
Required answer = (12 x 16) /(12 + 16) = 48 / 7 = 66/7 minutes.
Here x = 8 hrs. and y = 8 + 2 = 10 hrs.
Now, applying the given rule, we have the
Required answer = (8 x 10) /(10 - 8) = 40 hrs.
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