Question

A tank can be filled up by two taps in 6 hours. The smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required by each tap to fill the tank separately.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• Time taken by bigger tap to fill the tank individually be x hours.

So,

• Time taken by smaller tap to fill the tank individually be (x + 5) hours.

Let V be the volume of tank.

So,

• Part of tank filled by bigger tap in 1 hour = V/x

and

• Part of tank filled by smaller tap in 1 hour = V/(x + 5)

As it is given that, A tank can be filled up by two taps in 6 hours.

So, part of tank filled by two taps in 1 hour = V/6

Thus,

$\rm \: \dfrac{V}{x} + \dfrac{V}{x + 5} = \dfrac{V}{6}$

$\rm \: V\bigg(\dfrac{1}{x} + \dfrac{1}{x + 5}\bigg) = \dfrac{V}{6}$

$\rm \: \dfrac{1}{x} + \dfrac{1}{x + 5}= \dfrac{1}{6}$

$\rm \: \dfrac{x + 5 + x}{x(x + 5)} = \dfrac{1}{6}$

$\rm \: \dfrac{2x + 5}{x(x + 5)} = \dfrac{1}{6}$

$\rm \: x(x + 5) = 6(2x + 5)$

$\rm \: {x}^{2} + 5x = 12x + 30$

$\rm \: {x}^{2} + 5x - 12x - 30 = 0$

$\rm \: {x}^{2}- 7x - 30 = 0$

$\rm \: {x}^{2}- 10x + 3x - 30 = 0$

$\rm \: x(x - 10) + 3(x - 10) = 0$

$\rm \: (x - 10)(x + 3) = 0$

$\rm\implies \:x = 10 \: \: or \: \: x = - 3 \: \{rejected \}$

Hence,

Time taken by bigger tap to fill the tank individually = 10 hours.

Time taken by smaller tap to fill the tank individually = (10 + 5) = 15 hours.

$\rule{190pt}{2pt}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac