# NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3

## 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

What is the unknown?

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

What is known?

What is Unknown?

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

What is known?

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

What is Unknown?

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

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

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

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:

${x^2} - 60x - 2700 = 0$

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

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

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}

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

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}

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

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

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 $$= 12 \,\rm{m}.$$

Ncert Class 10 Exercise 4.3
Ncert Solutions For Class 10 Maths Chapter 4 Exercise 4.3

Ncert Class 10 Exercise 4.3
Ncert Solutions For Class 10 Maths Chapter 4 Exercise 4.3
Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school