Question

Two taps can fill a tub 90 and
60 min respectively
And
water A drain can empty
it in 60 mins. if all
3 kept simultaneausly
open for 30 min. after drain
close How long it will take
Per full tap​

Answer 1

Step-by-step explanation:

Time required={(1/90+1/60)-1/60}30.

so,

1/90×30

=1/3 work done

*work left=1-1/3=2/3

then, time required for filling the remaining tub by tap A and B=2/3×[1/60+1/90(which will be resiprocled)]

therefore,2/3×180/5

=2/3×36

=2×12

=24mins

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Answer 2

Time required=${( \frac{1}{90 } + \frac{1}{60} ) - \frac{1}{60} } \times 30$

So,

$\frac{1}{90} \times 30 \\ = \frac{1}{3}$

Then, time required for filing the remaining tub by tap A and B$\frac{2}{3} \times( \frac{1}{60} + \frac{1}{90} )$

Therefore,

$\frac{2}{3} \times \frac{180}{5}$

$= \frac{2}{3} \times 36$

$= 2 \times 12$

$= 24 \: min$