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

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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?

 Video Solution
Quadratic Equations
Ex 4.3 | Question 9

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.