Question

Ellow
1) A tank fills completely in 2 hours if both the taps are open.if only one tabs is open at the given time,the smaller tap taken 3 hours more than the larger one to fill the tank. How much time does each tap taken to fill the tank completely?☺​

Answer 1

Answer:

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Let larger tap can fills the tank in = x hrs.

then smaller tap will fill the tank in = (x+3) hrs

Part of the tank filled in an hr by both taps =1/x+1/(x+3) acordingly:-

1/x + 1/(x+3) = 1/2

(x+3+x)/x.(x+3) =1/2

or (2x+3)/(x^2+3x) = 1/2

or x^2+3x = 4x+6

or x^2 - x - 6 = 0

or (x-3) (x+2) = 0

x= 3 , -2(Negligenle )

Larger tap can fills the tank in = x hrs = 3 hrs.

Smaller tap can fills in = (x+3 )=(3+3)= 6 hrs.

Let the larger tap, A, take t hours to fill the tank, alone.

The smaller tap, B, takes (t+3) hours to fill the tank, alone.

A fills (1/t)th of the tank in 1 hour.

B fills [1/(t+3)]th of the tank in 1 hour.

So A and B take (1/t) + [1/(t+3)] = (t+3+t)/[t(t+3)] th part of the tank in one hour which is half the tank. Thus

(t+3+t)/[t(t+3)] = 1/2 or

2(2t+3) = t^2+3t, or

4t+6 = t^2+3t, or

t^2-t+6 = 0

(t-3)(t+2) = 0

t = 3 hours. [-2 hours has no meaning and so rejected]’

So the larger tap takes 3 hours to fill the tank, alone, while the smaller tap takes 6 hours to fill the tank, alone.

Check: (1/3)+(1/6) = (2/6)+(1/6) = (3/6) = 1/2 of the tank in one hour. Correct.

Answer 2

Answer:

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