NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3

Go back to  'Quadratic Equations'

Question 1

Find the roots of the following quadratic equations, if they exist, by the method of completing the square.

(i) \(2x^2-7x+3=0\)

(ii) \(2x^2+x-4=0\)

(iii) \(4x^2+ 4\sqrt {\left( 3 \right)} x +3=0\)

(iv) \(2x^{2}+x+4=0\)

Solution

Video Solution

What is the unknown?

Roots of the quadratic equation.

Reasoning:

Steps required to solve a quadratic equation by applying the ‘completing the square’ method are given below:

Let the given quadratic equation be: \(ax{}^\text{2}+bx+c=0\)

(i) Divide all the terms by \(a\)

\[\begin{align}\frac{{a{x^2}}}{a} + \frac{{bx}}{a} + \frac{c}{a} &= 0\\{x^2} + \frac{{bx}}{a} + c &= 0\end{align}\]

(ii) Move the constant term \(\begin{align}\frac{c}{a}\end{align}\) to the right side of the equation:

\[{x^2} + \frac{b}{a}x = - \frac{c}{a}\]

(iii) Complete the square on the left side of the equation by adding \(\begin{align} \frac{{{b^2}}}{{4{a^2}}}.\end{align}\) Balance this by adding the same value to the right side of the equation.

Steps:

(i) \(2x^2-7x+3=0\)

Divide \(2x{}^\text{2}-7x+3=0\) by \(2:\)

\[\begin{align}{x^2} - \frac{7}{2}x + \frac{3}{2}& = 0\\{x^2} - \frac{7}{2}x &= - \frac{3}{2}\\\end{align}\]

Since \(\begin{align}\frac{7}{2} \div 2 = \frac{7}{4} \end{align}\), \(\begin{align}{\left( {\frac{7}{4}} \right)^2} \end{align}\) should be added to both sides of the equation:

\[\begin{align}{x^2} - \frac{7}{2}x + {\left( {\frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + {\left( {\frac{7}{4}} \right)^2}\\{\left( {x - \frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + \frac{{49}}{{16}}\\ &= \frac{{ - 24 + 49}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} & = \frac{{25}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} &= {\left( { \pm \frac{5}{4}} \right)^2}\\x - \frac{7}{4} = \frac{5}{4} &\qquad \!\!x - \frac{7}{4} = \frac{{ - 5}}{4}\\x = \frac{5}{4} + \frac{7}{4} &\qquad \!\!x = \frac{{ - 5}}{4} + \frac{7}{4}\\
x = \frac{{12}}{4} &\qquad  x = \frac{2}{4}\\x = 3&\qquad x = \frac{1}{2}\end{align}\]

Roots are \(\begin{align}3, \;\frac{1}{2}.\end{align}\)

(ii) \(2x^2+x-4=0\)

\[\begin{align}{x^2} + \frac{x}{2} - 2 &= 0\\{x^2} + \frac{x}{2} &= 2\\{x^2} + \frac{x}{2} &= 2\end{align}\]

Since, \(\begin{align} \frac{{\rm{1}}}{{\rm{2}}} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4},\;
{\left( {\frac{1}{4}} \right)^2} \end{align}\) should be added on both sides

\[\begin{align}{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= 2 + {\left( {\frac{1}{4}} \right)^2}\\
{\left( {{x^2} + \frac{1}{4}} \right)^2} &= 2 + \frac{1}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{32 + 1}}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{33}}{{16}}\\\left( {x + \frac{1}{4}} \right) &= \pm \frac{{\sqrt {33} }}{4}\end{align}\]

\( \begin{align} \left( {x + \frac{1}{4}} \right) \!\!= \frac{{\sqrt {33} }}{4} ,& \quad \!\!\!\!\!\!\left( \!{x \!+\! \frac{1}{4}} \!\!\right) \!\!= - \frac{{\sqrt {33} }}{4}\\ x = \frac{{\sqrt {33} }}{4} - \!\! \frac{1}{4}, &\quad \!\!x \!\!= - \!\! \frac{{\sqrt {33} }}{4} - \! \frac{1}{4}\\x = \frac{{\sqrt {33} - 1}}{4}, &\quad \!\!x \! = \frac{{ - \sqrt {33} - 1}}{4}\end{align}\)

Roots are \(\begin{align} \frac{{\sqrt {33} - 1}}{4},\frac{{ - \sqrt {33} - 1}}{4}\end{align}\)

iii) \(4x+ 4\sqrt {\left( 3 \right)} x +3=0\)

\[\begin{align}x^2+ \sqrt 3 x +\frac{3}{4}&= 0\\{x^2} + \sqrt 3 x& = - \frac{3}{4}\\  x^2 \!\! + \!\! \sqrt 3 x \!\!+ \!\!  {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} & \!\! =  - \frac{3}{4} \!\!+\!\! {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} 
 \end{align}\]

\(\begin{align}{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \end{align}\) is added on both sides

\[\begin{align}{{\left( {x + \frac{{\sqrt 3 }}{2}} \right)}^2} &= - \frac{3}{4} + \frac{3}{4}\\{\left( {x + \frac{{\sqrt 3 }}{2}} \right)^2} &= 0\\x = - \frac{{\sqrt 3 }}{2} &\quad x= - \frac{{\sqrt 3 }}{2}\end{align}\]

Roots are \(\begin{align} - \frac{{\sqrt 3 }}{2},\, - \frac{{\sqrt 3 }}{2}\end{align}\)

iv) \(2x^{2}+x+4=0\)

\[\begin{align}{x^2} + \frac{x}{2} + 2 &= 0\\{x^2} + \frac{x}{2}& = - 2\\{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= - 2 + {\left( {\frac{1}{4}} \right)^2}\\{\left( {x + \frac{1}{4}} \right)^2} &= - 2 + \frac{1}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 32 + 1}}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 31}}{{16}} < 0\end{align}\]

Square of any real number can’t be negative. 

\(\therefore\;\)Real roots don’t exist.

Question 2

Find the roots of the quadratic equations given in Q1 above by applying quadratic formula.

Solution

Video Solution

What is known?

A quadratic equation.

What is Unknown?

Roots of the quadratic equation.

Reasoning:

If the given quadratic equation is: \(ax{}^{2}+bx+c=0\), then:

If \(b^{2}-4ac\ge 0\) then the roots are \(\begin{align} x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \end{align}\)

If \(b{}^{2}-4ac<0\) then no real roots exist.

Steps:

 i)

\[\begin{align}a &= 2,\;b = - 7,\;c = 3\\{b^2} - 4ac &= {\left( { - 7} \right)^2} - 4\left( 2 \right)\left( 3 \right)\\&= 49 - 24\\{b^2} - 4ac &= 25 > 0\end{align}\]

The Roots are,

\[\begin{align} x&=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ x &= \frac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4(2)(3)} }}{{2(2)}}\\x &= \frac{{ - ( - 7) \pm \sqrt {49 - 24} }}{{2(2)}}\\x &= \frac{{7 \pm 5}}{4}\\x &= \frac{{7 + 5}}{4}\qquad x = \frac{{7 - 5}}{4}\\x &= \frac{{12}}{4}\qquad \quad x = \frac{2}{4}\\x &= 3 \qquad \qquad x = \frac{1}{2}\end{align}\]

The Roots are,

\(\begin{align}3,\frac{1}{2} \end{align}\)

ii)

\[\begin{align}{{a}} &= 2,\;{{b}} = 1,\;{{c}} =  - 4\\{{{b}}^2} - 4{{ac}} &= {{(1)}^2} - 4(2)( - 4)\\ &= 1 + 32\\& = 33 > 0\end{align}\]

The Roots are,

\[\begin{align} x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - 1 \pm \sqrt {33} }}{{2(2)}}\\ &= \frac{{ - 1 \pm \sqrt {33} }}{4}\\x& = \frac{{ - 1 + \sqrt {33} }}{4}\\x &= \frac{{ - 1 - \sqrt {33} }}{4}\end{align}\]

The Roots are,

\(\begin{align}\frac{{ - 1 + \sqrt {33} }}{4},\;\frac{{ - 1 - \sqrt {33} }}{4} \end{align}\)

iii)

\[\begin{align}a &= 4,b = (4\sqrt 3 ),c = 3\\{b^2} - 4ac& = {(4\sqrt 3 )^2} - 4(4)(3)\\ &= (16 \times 3) - (16 \times 3)\\{b^2} - 4ac &= 0\end{align}\]

The Roots are,

\[\begin{align}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
 &= \frac{{ - b \pm 0}}{{2a}}\\&= \frac{{ - b}}{{2a}}\\&= \frac{{ - 4\sqrt 3 }}{{2(4)}}\\x &= \frac{{ - \sqrt 3 }}{2}\end{align}\]

The Roots are,

\(\begin{align}\frac{{ - \sqrt 3 }}{2},\;\frac{{ - \sqrt 3 }}{2}.\end{align}\)

iv)

\[\begin{align}a &= 2, b = 1, c = 4\\{b^2} - 4ac &= {{(1)}^2} - 4(2)(4)\\ &= 1 - 32\\&=  - 31 < 0\\{b^2} - 4ac &< 0\end{align}\]

\(\therefore\) No real roots are exist.

Question 3

Find the roots of the following equations:

(i) \(\begin{align}x - \frac{1}{x} = 3,x \ne 0\end{align}\)

(ii)

\(\begin{align} & \frac{1}{{x + 4}} - \frac{1}{{x - 7}} = \frac{{11}}{{30}}, \\ \\ & x \ne - 4,\,7\end{align}\)

Solution

Video Solution

What is known?

Quadratic equation, which is not in the form of \(ax^\text{2}+bx+c=0\)

What is Unknown?

Roots of a quadratic equation.

Reasoning:

Convert the given equation in the form of \(\begin{align}ax^\text{2}+bx+c=0 \end{align}\) and by using the quadratic formula, find the roots.

Steps:

(i) \(\begin{align}x - \frac{1}{x} = 3,x \ne 0\end{align}\)

\(x - \frac{1}{x} = 3,x \ne 0\) can be rewritten as (multiplying both sides by \(x\)):

\[\begin{align}{x^2} - 1 &= 3x\\{x^2} - 3x - 1 &= 0\end{align}\]

Comparing this against the standard form \(ax^\text{2}+bx+c=0\), we find that:

\[a = 1,\;b = - 3,\;c = - 1\]

\[\begin{align}{{{b}}^2} - 4{{ac}}& = {( - 3)^2} - 4(1)( - 1)\\&= 9 + 4\\&= 13 > 0\end{align}\]

\[\begin{align}\therefore x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\&= \frac{{ - ( - 3) \pm \sqrt {13} }}{{2(1)}}\\x &= \frac{{3 \pm \sqrt {13} }}{2}\end{align}\]

The roots are \(\begin{align}\frac{{3 + \sqrt {13} }}{2},\frac{{3 - \sqrt {13} }}{2}.\end{align}\)

(ii) \(\begin{align}\frac{1}{{x + 4}} - \frac{1}{{x - 7}} = \frac{{11}}{{30}},\,x \ne - 4,\,7\end{align}\)

By cross multiplying we get:

\[\begin{align}\frac{{(x - 7) - (x + 4)}}{{(x + 4)(x - 7)}} &= \frac{{11}}{{30}}\\\frac{{x - 7 - x - 4}}{{{x^2} + 4x - 7x - 28}} &= \frac{{11}}{{30}} \\ \frac{{ - 11}}{{{x^2} - 3x - 28}} &= \frac{{11}}{{30}}\end{align}\] \[\begin{align} - 11 \times 30\! &\!= \!\!11 \! \left({x^2} - 3x - 28 \right) \\- 30 &=  {x^2} - 3x  - 28 \end{align}\] \[\begin{align} {x^2} - 3x - 28 + 30 &= 0\\{x^2} - 3x + 2 &= 0\end{align}\]

Comparing this against the standard form \(ax^\text{2}+bx+c=0\), we find that:

\[a = 1\;b = - 3,\;c = 2\]

\[\begin{align}{{b^2} - 4ac}& = {{( - 3)}^2} - 4(1)(2)\\&= 9 - 8\\&= 1 > 0\end{align}\]

 \(\therefore \;\)Real roots exist for this quadratic equation.

\[\begin{align}x&=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\{x}&=\frac{-(-3)\pm \sqrt{{{(-3)}^{2}}-4(1)(2)}}{2(1)} \\{x}&=\frac{3\pm 1}{2} \\{x} &=\frac{3+1}{2}\quad {x}=\frac{3-1}{2} \\
 x &=\frac{~4~}{2} \qquad \;x=\frac{2}{2} \\ x&=2 \qquad \;\;\;\;x=1 \\\end{align}\]

Roots are \(2, \,1.\)

Question 4

The sum of the reciprocals of Rehman’s age (in years) \(3\) years ago and \(5\) years from now is \(\begin{align}\frac{1}{3} \end{align}.\) Find his present age.

Solution

Video Solution

What is Known?

i) Sum of reciprocals of Rehman’s age (in years) \(3\) years ago and \(5\) years from now is \(\begin{align}\frac{1}{3} \end{align}.\)

What is Unknown?

Rehman’s age.

Reasoning:

Let the present age of Rehman be \(x\) years.

\(3\) years ago, Rehman’s age was \(= x - 3\)

\(5\) years from now age will be \(=x+5\)

Using this information and the given condition, we can form the following equation:

\[\begin{align}\frac{1}{{x - 3}} + \frac{1}{{x + 5}} = \frac{1}{3} \end{align}\]

Steps:

\[\begin{align}\frac{1}{{x - 3}} + \frac{1}{{x + 5}} = \frac{1}{3} \end{align}\]

 By cross multiplying we get:

\[\begin{align}\frac{{(x + 5) + (x - 3)}}{{(x - 3)(x + 5)}} &= \frac{1}{3}\\\frac{{2x + 2}}{{{x^2} + 2x - 15}} &= \frac{1}{3} \end{align}\] \[\begin{align} (2x + 2)(3) &=\!\! {x^2} + 2x  - 15  \\6x + 6 &=\!\! {x^2} + 2x - 15  \\{x^2} + 2x - 15 &= 6x + 6\\ {x^2} + 2x - 15\\ - 6x - 6 &= 0\\{x^2} - 4x - 21 &= 0\end{align}\]

Finding roots by factorization:

\[\begin{align}{x^2} - 7x + 3x - 21& = 0\\x(x - 7) + 3(x - 7) &= 0\\ (x - 7)(x + 3)& = 0\end{align}\]

\[ \begin{align} x - 7 &= 0 \quad x + 3 = 0\\x &= 7 \quad x = - 3\end{align}\]

Age can’t be a negative value.

\(\therefore \;\) Rehman’s present age is \(7\) year.

Question 5

In a class test the sum of Shefali’s marks in Mathematics and English is \(30.\) Had She got \(2\) marks more in Mathematics and \(3\) marks less in English, the product of their marks would have been \(210.\) Find her marks in the two subjects.

Solution

Video Solution

What is known?

(i) Sum of Shefali’s marks in Mathematics and English is \(30.\)

(ii) Had she got \(2\) marks more in Mathematics and \(3\) marks less in English, product of marks would have been \(210 .\)

What is Unknown?

Marks of Shefali in two subjects.

Reasoning:

Let the marks Shefali scored in mathematics be \(x.\)

i) Then, marks scored by her in English\( \Rightarrow 30 - \) Marks scored in Mathematics \(= 30 – x\)

ii) \(2\) more marks in Mathematics \(= x + 2\)

3 marks less in English 
\[\begin{align}  &= 30-x-3\\&= 27-x \end{align}\]

Product of these two
\[\begin{align}\left( {x + 2} \right)\left( {27 - x} \right) &= 210 \end{align}\]

Steps:

\[\begin{align}({{x}} + 2)(27 - {{x}}) &= 210\\27{{x}} - {{{x}}^2} + 54 - 2{{x}} &= 210\\ - {{{x}}^2} + 25{{x}} + 54 &= 210\\ - {{{x}}^2} + 25{{x}} + 54 - 210 &= 0\\ - {{{x}}^2} + 25{{x}} - 156 &= 0\end{align}\]

Multiplying both sides by \( -1:\)

\[{{{x}}^2} - 25{{x + }}156 = 0\]

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

\[\begin{align}a& = 1,{b} = - 25,c = 156\\{{b^2}} - 4ac &= {( - 25)^2} - 4(1)(156)\\&= 625-624\\ &= 1\\{b^2} - 4ac&> 0\end{align}\]

\[\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - ( - 25) \pm \sqrt {{{( - 25)}^2} - 4(1)(156)} }}{{2(1)}}\\ &= \frac{{ - ( - 25) \pm \sqrt 1 }}{{2(1)}}\\
 &= \frac{{25 \pm 1}}{2}\\x &= \frac{{25 + 1}}{2}\quad x = \frac{{25 - 1}}{2}\\x &= \frac{{26}}{2}\qquad \;x = \frac{{24}}{2}\\x &= 13 \qquad\;\; x = 12\end{align}\]

Two possible answers for this given question:

If Shefali scored \(13\) mark in mathematics, then mark in English \( = \left( {30 - 13} \right) = 17.\)

If Shefali scored \(12\) mark in mathematics, then mark in English \( = \left( {30 - 12} \right) = 18.\)

Question 6

The diagonal of a rectangular field is \(60\) meters more than the shorter side. If the longer side is \(30\) meters more than the shorter side, find the sides of the fields.

Solution

Video Solution

What is known?

i) The diagonal of the rectangular field is \(60\) meters more than the shorter side.

ii) The longer side is \(30\) meters more than the shorter side.

What is Unknown?

Sides of rectangular field.

Reasoning:

Let the shorter side be \(x\) meter. Then the length of diagonal of field will be \(x+60\) and length of longer side will be \(x+30.\) Using Pythagoras theorem, value of \(x\) can be found.

By applying Pythagoras theorem:

\[\begin{align}\text{Hypotenuse }^2&= \text{ Side 1}^{2} + \text{Side 2 }\!\!{}^\text{2}\\(60 + x)^2 &= {x^2} + (30 + x)^2\end{align}\]

Steps:

\[\begin{align}{{(60 + x)}^2} &= {x^2} + {{(30 + x)}^2} \\60 + 2(60)x + {x^2} &=\begin{bmatrix} {x^2} + {{30}^2} +\\ 2(30)x + {x^2}\end{bmatrix}\\\quad\because {{\left( {a + b} \right)}^2}& = {a^2} + 2ab + {b^2} \\3600 + 120x + {x^2} &= \begin{bmatrix}{x^2} + 900 +\\ 60x + {x^2}\end{bmatrix} \\\begin{bmatrix}3600 \!+ \!120x \!+\! {x^2}\! -\! \\{x^2}\! -\! 900 \!-\! 60x\! -\! {x^2}\end{bmatrix}&= 0 \\2700 + 60x - {x^2} &= 0\end{align}\]

Multiplying both sides by \(-1:\)

\[{x^2} - 60x - 2700 = 0\]

Solving by quadratic formula:

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

\[a =1,\; b=  - 60,\; c = -2700\]

\[\begin{align}{b^2} - 4ac&= {{( - 60)}^2} - 4(1)( - 2700)\\& = 3600 + 10800\\{b^2} - 4ac& = 14400 > 0\end{align}\]

\(\therefore\;\)Roots exist.

\[\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - ( - 60) \pm \sqrt {14400} }}{2}\\ &= \frac{{60 \pm \sqrt {14400} }}{2}\\x &= \frac{{60 \pm 120}}{2}\\x &= \frac{{60 + 120}}{2} \quad x = \frac{{60 - 120}}{2}\\x &= \frac{{180}}{2}\qquad \;\;x = \frac{{ - 60}}{2}\\x &= 90 \qquad \;\;x = - 30\end{align}\]

Length can’t be a negative value. 

\(\therefore x = 90\)

Length of shorter side \(x = 90 \,\rm{m}\)

Length of longer side

 \(=\text{ }30+x= 30 + 90 = 120\,\rm{ m.}\)

Question 7

The difference of squares of two numbers is \(180.\) The square of smaller number is \(8\) times the larger number. Find the two numbers.

Solution

Video Solution

What is known?

i) Difference of squares of two numbers is \(180.\)

ii) The square of the smaller number is \(8\) times the larger number.

What is Unknown?

Two numbers.

Reasoning:

Let the larger number be \(x.\)

Square of smaller number is \(= 8x.\)

Difference of squares of the two numbers is \(180.\)

Square of larger number \(-\) Square of smaller number \(= 180\)

\[{x^2} - 8x - 180 = 0\]

Steps:

\[\begin{align}{x^2} - 8x - 180 &= 0\\{x^2} - 18x + 10x - 180 &= 0\\x(x - 18) + 10(x - 18) &= 0\\(x - 18)(x + 10) &= 0\\x - 18 &= 0 \quad x + 10 \!=\! 0\\x &= 18  \;\;  x = - 10\end{align}\]

If the larger number is \(18,\) then square of smaller number \( = 8 \times 18\)

Therefore, the smaller number
\[\begin{align}&{ = \pm \sqrt {8 \times 18} }\\{}&{ = \pm \sqrt {2 \times 2 \times 2 \times 2 \times 3 \times 3} }\\{}&{ = \pm 2 \times 2 \times 3|}\\{}&{ = \pm 12}\end{align}\]

If larger number is \(– 10,\,\) then square of smaller number \( = 8 \times ( - 10) = - 80\)

Square of any number cannot be negative.

\(\therefore x = - 10\) is not applicable.

The numbers are \(18, \,12\) (or) \(18, -12.\)

Question 8

A train travels \(360\,\rm{ km}\) at a uniform speed. If the speed had been \(5\,\rm{ km /hr}\) more, it would have taken \(1\) hour less for the same journey. Find the speed of the train.

Solution

Video Solution

What is known?

i) Distance covered by the train at a uniform speed \(= 360\,\rm{ km} \)

ii) If the speed had been \(5 \,\rm{km/hr}\) more, it would have taken \(1\) hour less for the same journey.

What is Unknown?

Speed of the train.

Reasoning:

Let the speed of the train be \(s\; \rm{km/hr}\) and the time taken be \(t\) hours.

\[\begin{align}\rm{Distance} &= \rm{Speed} \times \rm{Time}\\360&=s\times t\\360 &= s \times t\\t &= \left( {\frac{{360}}{s}} \right)\end{align}\]

Increased speed of the train: \(s + 5\)

New time to cover the same distance: \(t – 1\)

\[(s + 5)(t - 1) = 360\,\,\,\,\,\,\,\,\,\, \ldots (2)\]

Steps:

\[\begin{align}(s + 5)(t - 1) &= 360\\st - s + 5t - 5 &= 360\\360 - s + 5\left( {\frac{{360}}{s}} \right) - 5 &= 360\\ - s + \frac{{1800}}{s} - 5 &= 0 \\ - {s^2} + 1800 - 5s &= 0\\{s^2} + 5s - 1800 &= 0\end{align}\]

Solving by quadratic formula:

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

\[a{\rm{ }} = {\rm{ }}1,{\rm{ }}b{\rm{ }} = {\rm{ }}5,{\rm{ }}c{\rm{ }} = {\rm{ }} - {\rm{ }}1800\]

\[\begin{align}{{{b}}^2} - 4{\rm{ac}}& = {{(5)}^2} - 4(1)( - 1800)\\&= 25 + 7200\\& = 7225 > 0\end{align}\]

\(\therefore\) Real roots exist.

\[\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\{{s}} &= \frac{{ - 5 \pm \sqrt {7225} }}{2}\\{{s}} &= \frac{{ - 5 \pm 85}}{2}\\{{s}} &= \frac{{ - 5 + 85}}{2},\quad {{s}} = \frac{{ - 5 - 85}}{2}\\{{s}}& = \frac{{80}}{2}\qquad\qquad  {{s}} = \frac{{ - 90}}{2}\\{{s}} &= 40 \qquad\qquad\;  {{s}} = - 45\end{align}\]

Speed of the train cannot be a negative value.

\(\therefore\) Speed of the train is \(40 \,\rm{km /hr.}\)

Question 9

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?

Solution

Video 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.

Question 10

An express train takes \(1\) hour less than a passenger train to travel \(132 \,\rm{km}\) between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of express train is \(11\, \rm{km /hr}\) more than that of passenger train, find the average speed of the two trains.

Solution

Video Solution

What is Known?

i) Express train takes \(1\) hour less than a passenger train to travel \(132\,\rm{ km.}\)

ii) Average speed of express train is \(11\,\rm{km/hr}\) more than that of passenger train.

What is Unknown?

Average speed of express train and the passenger train.

Reasoning:

Let the average speed of passenger train

\(= x\,\rm{km/hr}\)

Average speed of express train

\(= (x + 11)\, \rm{km /hr}\)

\[\begin{align}\rm{Distance} &= \rm{Speed} \times \rm{Time}\\\rm{Time}& = \frac{{{\rm{Distance}}}}{{\rm{Speed}}}\end{align}\]

Time taken by passenger train to travel

\( \begin{align}132\,{\rm{km}}= \frac{{132}}{x}\end{align}\)

Time taken by express train to travel

\(\begin{align}132 \,{\rm{km}} =\frac{{132}}{{x + 11}}\end{align}\)

Difference between the time taken by the passenger and the express train is \(1\) hour. Therefore, we can write:

\[\frac{{132}}{x} - \frac{{132}}{{x + 11}} = 1\]

Steps:

Solving \(\begin{align} \frac{{132}}{x} - \frac{{132}}{{x + 11}} = 1 \end{align}\) by taking the LCM on the LHS:

\[\begin{align}\frac{{132\left( {x + 11} \right) - 132x}}{{x\left( {x + 11} \right)}}& = 1\\\frac{{132x + 1452 - 132x}}{{{x^2} + 11x}} &= 1\\1452 &= {x^2} + 11x\\{x^2} + 11x - 1452 &= 0\end{align}\]

By comparing \({x^2} + 11x - 1452 = 0\) with the general form of a quadratic equation

\(ax² + bx + c = 0:\)

\[a = 1,\; b = 11,\; c = - 1452\]

\[\begin{align} {{b}^{2}}-4ac&={{11}^{2}}-4\left( 1 \right)\left( -1452 \right) \\ & =121+5808 \\
 & =5929>0 \\ b{}^\text{2}\text{ }-\text{ }4ac &>0 \\\end{align}\]

\(\therefore\) Real roots exist.

\[\begin{align}{\rm{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
&= \frac{{ - 11 \pm \sqrt {5929} }}{{2(1)}}\\&= \frac{{ - 11 \pm 77}}{2}\\
x &= \frac{{ - 11 + 77}}{2} \qquad x = \frac{{ - 11 - 77}}{2}\\&= \frac{{66}}{2} \qquad \qquad \quad\;\;  = \frac{{ - 88}}{2}\\&= 33 \qquad \qquad \quad\;\;\;\;= - 44\\\\&x=33 \qquad \qquad \qquad -x=44\end{align}\]

\(x\) can’t be a negative value as it represents the speed of the train.

Speed of passenger train \(= 33\,\rm{ km/hr}\)

Speed of express train

\(=x+11 \\ = 33 + 11 \\= 44\;\rm{km/hr.}\)

Question 11

Sum of the areas of two squares \(468 \,\rm{m}^2.\) If the difference of perimeters is \(24\,\rm{m,}\) find the sides of two squares.

Solution

Video Solution

What is Known?

i) Sum of the areas of two squares is \(468\,\rm{m}^2.\)

ii) The difference of perimeters is \(24\,\rm{m.}\)

What is Unknown?

Sides of two squares.

Reasoning:

Let the side of first square is \(x\) and side of the second square is \(y.\)

\(\text{Area of the square}= \rm{Side} \times \rm{Side}\)

\(\text{Perimeter of the square} = 4 \times \rm{Side}\)

Therefore, the area of the first and second square are \({x^2}\) and \({y^2}\) respectively. Also, the perimeters of the first and second square are \(4x\) and \(4y\) respectively. Applying the known conditions:

\(\begin{align}&{\rm{(i)}}\quad{x^2} + {y^2} = 468 \quad \ldots ..\left( 1 \right)\\\\
&{\rm{(ii)}}\quad 4x - 4y = 24 \quad \ldots .{\rm{ }}\left( 2 \right)\end{align}\)

Steps:

\[\begin{align}{x^2} + {y^2} &= 468\\4x - 4y &= 24\\4(x - y) &= 24\\x - y &= 6\\x &= 6 + y\end{align}\]

Substitute \(x = y + 6\) in equation (1)

\[\begin{align}{(y + 6)^2} + {y^2} &= 468\\{y^2} + 12y + 36 + {y^2} &= 468\\2{y^2} + 12y + 36 &= 468\\2({y^2} + 6y + 18)& = 468\\
{y^2} + 6y + 18 &= 234\\{y^2} + 6y + 18 - 234 &= 0\\{y^2} + 6y - 216 &= 0
\end{align}\]

Solving by factorization method

\[\begin{align}{y^2} + 18y - 12y - 216 &= 0\\y\left( {y + 18} \right) - 12\left( {y + 18} \right)& = 0\\\left( {y + 18} \right)\left( {y - 12} \right) &= 0\\y + 18 = 0 &\qquad y - 12= 0\\y = - 18 &\qquad y = 12\end{align}\]

\(y\) can’t be negative value as it represents the side of the square.

Side of the first square

\(x = y + 6 = 12 + 6 = 18\,\rm{m}\)

Side of the second square \(= 12 \,\rm{m}.\)

Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 for FREE
Ncert Class 10 Exercise 4.3
Ncert Solutions For Class 10 Maths Chapter 4 Exercise 4.3
  
Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 for FREE
Ncert Class 10 Exercise 4.3
Ncert Solutions For Class 10 Maths Chapter 4 Exercise 4.3
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