Question

1) two pipes together can fill a tank in 6 hours 20 minutes . one tap takes 3 hours more than the other to fill the tank separately, find the time in which each tap can separately fill the tank​

Answer 1

Step-by-step explanation:

one tap takes X hours,

another takes X+3 hours

1. + 1. = 3.

X. X+3. 19

X+3+X. = 3.

X^2+3X. 19

(2x+3) 19 = 3(X^2+3X)

38X+57 = 3X^2+9X

3X^2+9X-38X-57 = 0

3X^2-29X-57 =0

try your self to solve the Quadratic equatoion

Answer 2

Given,

Two pipes can fill a tank in 6 hours 20 minutes together.

Separately, one tap takes 3 hours more than the other to fill the tank.

To Find,

The time in which each tap can separately fill the tank.

Solution,

Suppose in x hours the first tap can fill the tank completely.

And in x+3 the second tap can fill the tank completely.

So in  1 hour, the first tap can fill the total $\frac{1}{x}$ part of the tank.

And the second tap can fill totally $\frac{1}{x+3}$ part of the tank.

We know 1 total part means a full tank.

If two taps are open, Total ($\frac{1}{x} +\frac{1}{x+3}$) =$\frac{2x+3}{x^{2} +3x}$ part of the tank can be filled in 1 hour.

∴total 1 part of the tank can be filled in $\frac{1}{\frac{2x+3}{x^{2} +3x}}$ hours.=$\frac{x^{2} +3x}{2x+3}$ hours.

Given together two taps can fill the tank in 6 hours 20 minutes = 6$\frac{20}{60}$=$\frac{19}{3}$ hours.

So $\frac{x^{2} +3x}{2x+3}$ =$\frac{19}{3}$ .

⇒$3x^{2} -29x-57=0$.

Use the rule of shridharacharya to know the value of x.

∴x=$\frac{29+\sqrt{29^{2} -4.3.(-57)} }{2.(3)}$=$\frac{29+5\sqrt{61} }{6}$=11.34.

So the first tap can fill the tank in 11.34 hours alone.

The second tap can do the so in (11.34+3)=14.34 hours,

Hence, The first and second tap can fill the tank separately in respectively 11.34 and 14.34 hours.