# Ex.4.3 Q9 Quadratic Equations Solutions - NCERT Maths Class 10

## Question

Two water taps together can fill a tank in \(\begin{align} 9\frac{3}{8} \end{align}\) hours. The tap of larger diameter takes \(10\) hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank?

## Text Solution

**What is known?**

i) Two water taps together can fill the tank in \(\begin{align} 9\frac{3}{8} \end{align}\) hours.

ii) The tap of larger diameter takes \(10\) hours less than the smaller one to fill the tank separately.

**What is the Unknown?**

Time taken by smaller tap and larger tap to fill the tank separately.

**Reasoning:**

Let the tap of smaller diameter fill the tank in *\(x\)* hours.

Tap of larger diameter takes \(\left( {x - 10} \right)\) hours.

In* \(x\)* hours, smaller tap fills the tank.

In one hour, part of tank filled by the smaller tap \(\begin{align} =\frac{1}{x} \end{align}\)

In \(\left( {x - 10} \right)\) hours, larger tap fills the tank.

In one hour, part of tank filled by the larger tap\(\begin{align} = \frac{1}{{\left( {x - 10} \right)}} \end{align}\)

In \(1\) hours, the part of the tank filled by the smaller and larger tap together:

\[\begin{align}\frac{1}{x} + \frac{1}{{x - 10}}\end{align}\]

\[\begin{align}\therefore \quad \frac{1}{x} + \frac{1}{{x - 10}} = \frac{1}{{9\frac{3}{8}}} \end{align}\]

**Steps:**

\[\begin{align}\frac{1}{x} + \frac{1}{{x - 10}} = \frac{1}{{\frac{{75}}{8}}}\end{align}\]

By taking LCM and cross multiplying:

\[\begin{align}\frac{{x - 10 + x}}{{x(x - 10)}} &= \frac{8}{{75}}\\

\frac{{2x - 10}}{{{x^2} - 10x}} &= \frac{8}{{75}}\\75\left( {2x - 10} \right) &= 8\left( {{x^2} - 10x} \right)\\150x - 750x &= 8{x^2} - 80x\\8{x^2} - 80x - 150x + 750& = 0\\8{x^2} - 230x + 750& = 0\\4{x^2} - 115x + 375& = 0\end{align}\]

Solving by quadratic formula:

Comparing with \(ax^\text{2}+bx+c=0\)

\[a = 4,\;b = - 115,\;c = 375\]

\[\begin{align} b{}^{2}-4ac&={{\left( -115 \right)}^{2}}-4\left( 4 \right)\left( 375 \right) \\ & =13225-6000 \\ & =7225 \\ b{}^{2}-4ac&>0\end{align}\]

\(\therefore\) Real roots exist.

\[\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\

{{x}} &= \frac{{115 \pm \sqrt {7225} }}{8}\\{{x}} &= \frac{{115 + 85}}{8} \qquad {{x}} = \frac{{115 - 85}}{8}\\{{x}} &= \frac{{200}}{8} \qquad \;\qquad {{x}} = \frac{{30}}{8}\\{{x}} &= 25 \qquad\;\;\; \qquad {{x}} = 3.75\end{align}\]

*\(x\)* cannot be \(3.75\) hours because the larger tap takes \(10\) hours less than *\(x\) *

Time taken by smaller tap \(x = 25\) hours

Time taken by larger tap \((x - 10) =15\) hours.