Question

a tank fills completely in 2 hours if both the taps are open. if only one of the tab is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank how much time does each tap take fill the tank completely.


ANSWER :

LET, larger tap take time to fill the tank completely = x hours

therefore, In one hour (1/x) part of tank filled by larger tap only.

Similarly, smaller tap takes time to fill the tank completely = x+3 hours

therefore, In 1 hour(1/x+3) parts of tank filled by smaller tap only.

In 1 hour (1/x+1/x+3)part of tank filled by both tap.
but, According to the given condition.

2 ( 1/x + 1/x+3 ) = 1
[x+3+x/x(x+3)] = 1/2
2x+3/x^2+3x=1/2
4x+6=x^2+3x
x^2-x-6=0
(x-3)(x+2)
x = 3 or x = -2
x = 3 because x = -2 is invalid because time is cannot be negative.

therefore, larger tap take time to filled tank completely = x = 3 hours.

and smaller tap take time to fill the tank completely = (x+3) = (3+3) = 6 hours.

therefore, for larger tap 3 hours and smaller tap 6 hours.
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Answer 1

Answer:

Smaller tap take hours. Both fill the tank in 2 hours if they are open together. Thus, larger tap takes 3 hours while the smaller tap takes 3 + 3 = 6 hours.