NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.6

Go back to  'Pair of Linear Equations in Two Variables'

Question 1

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) \(\begin{align} \quad \frac{1}{{2x}} + \frac{1}{{3y}} &= 2\\\frac{1}{{3x}} + \frac{1}{{2y}} &= \frac{{13}}{6}\end{align}\) 

(ii) \(\begin{align} \quad \frac{2}{{\sqrt x }} + \frac{3}{{\sqrt y }} &= 2\\\frac{4}{{\sqrt x }} - \frac{9}{{\sqrt y }} &= - 1\end{align}\)

(iii) \(\begin{align}\quad \frac{4}{x} + 3y &= 14\\\frac{3}{x} - 4y &= 23\end{align}\)

(iv) \(\begin{align}\quad \frac{5}{{x - 1}} + \frac{1}{{y - 2}} &= 2\\\frac{6}{{x - 1}} - \frac{3}{{y - 1}}& = 2\end{align}\)

(v) \(\begin{align}\quad \frac{{7x - 2y}}{{xy}} &= 5\\\frac{{8x + 7y}}{{xy}} &= 15\end{align}\)

(vi) \(\begin{align} \quad 6x + 3y &= 6xy\\2x + 4y &= 5xy\end{align}\)

(vii) \(\begin{align}\quad \frac{{10}}{{x + y}} + \frac{2}{{x - y}} &= 4\\\frac{{15}}{{x + y}} - \frac{5}{{x - y}}& = - 2\end{align}\)

(viii) \(\begin{align} \quad \frac{1}{{3x + y}} + \frac{1}{{3x - y}} &= \frac{3}{4}\\\frac{1}{{2(3x + y)}} - \frac{1}{{2(3x - y)}} &= \frac{{ - 1}}{8}\end{align}\)

Solution

Video Solution

Reasoning:

When the variable is in denominator, consider the reciprocal of variable as new variable.

Steps:

(i) 

\(\begin{align} \quad \frac{1}{{2x}} + \frac{1}{{3y}} &= 2\\\frac{1}{{3x}} + \frac{1}{{2y}} &= \frac{{13}}{6}\end{align}\) 

Let \(\frac{1}{x} = p\; {\rm{ and }}\;\frac{1}{y} = q\), then the equations change as follows:

\[\begin{align} & \frac{1}{{2x}} + \frac{1}{{3y}} = 2  \\ & \Rightarrow \frac{p}{2} + \frac{q}{3} = 2 
 \\ \\ & \Rightarrow \,\,3p\, + 2q - 12\, = 0 \qquad (1)\\ \\ & \frac{1}{{3x}} + \frac{1}{{2y}} = \frac{{13}}{6} \\  & \Rightarrow \frac{p}{3} + \frac{q}{2} = \frac{{13}}{6} \\ \\ & \Rightarrow \,\,2p + 3q - 13\, = 0 \quad (2)
\end{align}\]

Using cross-multiplication method, we obtain

\[\begin{align}\frac{p}{{ - 26 - ( - 36)}}& = \frac{q}{{ - 24 - ( - 39)}} \\ & = \frac{1}{{9 - 4}}\\ \frac{p}{{10}} &= \frac{q}{{15}} = \frac{1}{5}\\\frac{p}{{10}} &= \frac{1}{5}\,\,{\rm{and}}\,\,\frac{q}{{15}} = \frac{1}{5}\\p &= 2\,\,{\rm{and}}\,\,q = 3\\\end{align}\]

Therefore, \(\begin{align}\frac{1}{x} = 2 \end{align}\) and \(\begin{align}\frac{1}{y} = 3 \end{align}\)
Hence, \(\begin{align}x = \frac{1}{2}\end{align}\) and  \(\begin{align}y = \frac{1}{3}\end{align}\)

(ii)

\(\begin{align} \quad \frac{2}{{\sqrt x }} + \frac{3}{{\sqrt y }} &= 2\\\frac{4}{{\sqrt x }} - \frac{9}{{\sqrt y }} &= - 1\end{align}\)

Substituting \(\begin{align}\frac{1}{{\sqrt x }} = p\end{align}\)  and \(\begin{align}\frac{1}{{\sqrt y }} = q \end{align}\) in the given equations, we obtain

\[\begin{align}&\frac{2}{{\sqrt x }} + \frac{3}{{\sqrt y }} = 2 \\ \Rightarrow \, & 2p + 3q = 2 \ldots \left( {\rm{1}} \right)\\ & \frac{4}{{\sqrt x }} - \frac{9}{{\sqrt y }} = - 14 \\ \Rightarrow \,  & 4p - 9q = - 14 \ldots \left( {\rm{2}} \right)\end{align}\]

Multiplying equation (1) by 3, we obtain

\[6p + 9q = 6 \ldots \left( {\rm{3}} \right)\]

Adding equation \((2)\) and \((3)\), we obtain

\[\begin{align}10p &= 5\\p &= \,\frac{1}{2}\end{align}\]

Putting \(\begin{align}p = \,\frac{1}{2}\end{align}\) in equation \((1)\), we obtain

\[\begin{align}2 \times \frac{1}{2} + 3q\, &= \,2\\3q &= 2 - 1\\q &= \frac{1}{3}\end{align}\]

\(\begin{align}{\text{Therefore, }}p &= \frac{1}{{\sqrt x }} = \frac{1}{2} & \\ &\Rightarrow \sqrt x \,\, = \,\,2\\& \Rightarrow x\,\, = \,\,4\\{\text{And }}q = \frac{1}{{\sqrt y }} &= \frac{1}{3} & \\ &\Rightarrow \sqrt y = 3\\ &\Rightarrow y = 9\\{\text{Hence, }}\,\,x &= 4\,{\text{ and }}\,y = 9\end{align}\)

\(\begin{align}{\rm{(iii) }}\quad \frac{4}{x} + 3y &= 14\\\frac{3}{x} - 4y &= 23\end{align}\)

Substituting \(\begin{align}\frac{1}{x} = p \end{align}\) in the given equations, we obtain

\[\begin{align}&4p+3y=14 \\ \Rightarrow \, & 4p+3y-14=0 \ldots \left( 1 \right) \\ \\ &3p-4y=23 \\ \Rightarrow \, &3p-4y-23=0 \ldots \left( 2 \right)\end{align}\]

By cross-multiplication, we obtain

\[\begin{align}\frac{p}{\begin{Bmatrix} - 69 \\ - 56\end{Bmatrix}} &= \frac{y}{\begin{Bmatrix} - 42 \\- ( - 92) \end{Bmatrix}} \\ &= \frac{1}{\begin{Bmatrix} - 16 \\- 9\end{Bmatrix}}\\
\frac{p}{{ - 125}} &= \frac{y}{{50}} = \frac{1}{{ - 25}}\\
\frac{p}{{ - 125}} &= \frac{1}{{ - 25}}{\text{  and  }}\frac{y}{{50}} = \frac{1}{{ - 25}}\\
p &= 5{\text{  and    }}y = - 2\end{align}\]

\(\begin{align}{\text{Therefore, }}p &= \,\,\frac{1}{x}\,\,\, = \,\,5\\
 \Rightarrow x &= \frac{1}{5}\\{\text{Hence, }}x &= \frac{1}{5}{\text{ and }}y = - 2
\end{align}\)

\(\begin{align}{\rm{ (iv)}}\quad \frac{5}{{x - 1}} + \frac{1}{{y - 2}} &= 2\\\frac{6}{{x - 1}} - \frac{3}{{y - 1}}& = 2\end{align}\)

Putting \(\begin{align}\frac{1}{{x - 1}} = p\end{align}\) and \(\begin{align}\frac{1}{{y - 2}} = q\end{align}\) in the given equation, we obtain

\[\begin{align} & \frac{5}{{x - 1}} + \frac{1}{{y - 2}} = 2 \\ \Rightarrow \, & 5p + q = 2 \cdots \cdots \left( 1 \right)\\ \\ & \frac{6}{{x - 1}} - \frac{3}{{y - 1}} = 2 \\  \Rightarrow \, & 6p - 3q = 1 \cdots \cdots \left( 2 \right) \end{align}\]

Multiplying equation \((1)\) by \(3\), we obtain

\[15p + 3q = 6\qquad \left( 3 \right)\]

Adding \((2)\) and \((3)\), we obtain

\[\begin{align} 21p&=7 \\ p &=\frac{1}{3} \\\end{align}\]

Putting \(\begin{align}p = \frac{1}{3} \end {align}\) in equation (\(1\)), we obtain

\[\begin{align}5 \times \frac{1}{3} + q &= 2\\\,q &= 2 - \frac{5}{3}\\q &= \frac{1}{3}\\\end{align}\]

\(\begin{align}{\text{Therefore, }}p &= \frac{1}{{x - 1}} = \frac{1}{3}\\
 &\Rightarrow x - 1 = 3\\ &\Rightarrow \,\,x = 4\\{\text{and }}q &= \frac{1}{{y - 2}} = \frac{1}{3}\\ &\Rightarrow y - 2 = 3\\ &\Rightarrow y = 5\\{\text{Hence, }}x &= 4{\text{ and }}y = 5\end{align}\)

\(\begin{align}{\rm{(v)}} \quad \frac{{7x - 2y}}{{xy}} &= 5\\\frac{{8x + 7y}}{{xy}} &= 15\end{align}\)

\[\begin{align}&\frac{{7x - 2y}}{{xy}} = 5\\ \Rightarrow\, & \frac{{7x}}{{xy}} - \frac{{2y}}{{xy}} = 5 \\ \Rightarrow \, & \frac{7}{y} - \frac{2}{x} = 5 \cdots \left( 1 \right)\\ \\ &\frac{{8x + 7y}}{{xy}} = 15 \\ \Rightarrow \, & \frac{{8x}}{{xy}} + \frac{{7y}}{{xy}} = 15\\ \Rightarrow\, & \frac{8}{y} + \frac{7}{x} = 15 \cdots \left( 2 \right)\end{align}\]

Putting \(\begin{align} \frac{1}{x} = p \end{align}\) and  \(\begin{align}\frac{1}{y} = q \end{align}\)in the equations \((1)\) and \((2)\), we obtain

\[\begin{align} & \frac{7}{y} - \frac{2}{x} = 5 \\ \Rightarrow \, & - 2p + 7q - 5 = 0 \cdots \left( 3 \right)\\ \\ 
&\frac{8}{y} + \frac{7}{x} = 15 \\ \Rightarrow \, & 7p + 8q - 15 = 0 \cdots \left( 4 \right)\end{align}\]

By cross-multiplication method, we obtain

\[\begin{align}\frac{p}{\begin{Bmatrix} - 105 \\- ( - 40) \end{Bmatrix}} &= \frac{q}{\begin{Bmatrix} - 35\\ - 30\end{Bmatrix}} \\ &= \frac{1}{{\begin{Bmatrix} - 16 \\ - 49\end{Bmatrix}}}\\
\frac{p}{{ - 65}} &= \frac{q}{{ - 65}} = \frac{1}{{ - 65}}\\ \\ \frac{p}{{ - 65}} &= \frac{1}{{ - 65}} {\text{ and }} \\  \frac{q}{{ - 65}} &= \frac{1}{{ - 65}}\\ \\ p& = 1\quad {\text{ and }}\quad q = 1
\end{align}\]

\(\begin{align}{\text{Therefore, }}p &= \frac{1}{x} = 1\\& \Rightarrow x = 1\\{\text{ and, }}q &= \frac{1}{y} = 1\\& \Rightarrow y = 1\\\\{\text{Hence, }}x &= 1{\text{ and }}y = 1\end{align}\)

\(\begin{align}{\rm{(vi)}} \quad 6x + 3y &= 6xy\\2x + 4y &= 5xy\end{align}\)

By dividing both the given equations by \((xy),\) we obtain

\[\begin{align} & 6x + 3y = 6xy \\ \Rightarrow \, & \frac{6}{y} + \frac{3}{x} = 6 \cdots \left( 1 \right)\\ \\ 
& 2x + 4y = 5xy \\ \Rightarrow \, & \frac{2}{y} + \frac{4}{x} = 5 \cdots \left( 2 \right)\end{align}\]

Substituting \(\begin{align}\frac{1}{x} = p\end{align}\) and \(\begin{align}\frac{1}{y} = q\end{align}\) in the equations \((1)\) and \((2)\), we obtain

\[\begin{align}3p + 6q - 6 &= 0 \cdots \left( 3 \right)\\4p + 2q - 5 &= 0 \cdots \left( 4 \right)
\end{align}\]

By cross-multiplication method, we obtain

\[\begin{align}\frac{p}{{ - 30 - ( - 12)}} &= \frac{q}{{ - 24 - ( - 15)}}\\& = \frac{1}{{6 - 24}}\\ \\
\frac{p}{{ - 18}} &= \frac{q}{{ - 9}} = \frac{1}{{ - 18}}\\ \\ \frac{p}{{ - 18}} &= \frac{1}{{ - 18}}{\text{ and }} \\ \frac{q}{{ - 9}} & = \frac{1}{{ - 18}}\\p &= 1 \quad {\text{ and }}\quad q = \frac{1}{2}\end{align}\]

\(\begin{align}{\text{Therefore, }}p &= \frac{1}{x} = 1\\& \Rightarrow x = 1\\{\text{and, }}q &= \frac{1}{y} = \frac{1}{2}\\& \Rightarrow y = 2\\\\{\text{Hence, }}x &= 1{\text{ and }}y = 2\end{align}\)

\(\begin{align}{\rm{(vii)}}\quad \frac{{10}}{{x + y}} + \frac{2}{{x - y}} &= 4\\\frac{{15}}{{x + y}} - \frac{5}{{x - y}}& = - 2\end{align}\)

Substituting \(\begin{align}\frac{1}{{x + y}} = p \end{align}\) and  \(\begin{align}\frac{1}{{x - y}} = q \end{align}\) in the given equations, we obtain

\[\begin{align} & \frac{{10}}{{x + y}} + \frac{2}{{x - y}} = 4 \\ \Rightarrow \, &10p + 2q = 4 \\ \Rightarrow \, & 5p + q - 2 = 0  \cdots\left( 1 \right)\\ \\ &\frac{{15}}{{x + y}} - \frac{5}{{x - y}} =  - 2 \\ \Rightarrow \, & 15p - 5q =  - 2 \\ \Rightarrow \, &15p - 5q + 2 = 0 \cdots  \left( 2 \right)
\end{align}\]

Using cross-multiplication method, we obtain

\[\begin{align}\frac{p}{{2 - 10}} &= \frac{q}{{ - 30 - 10}} \\ &= \frac{1}{{ - 25 - 15}}\\\frac{p}{{ - 8}} &= \frac{q}{{ - 40}} = \frac{1}{{ - 40}}\\ \\ \frac{p}{{ - 8}} &= \frac{1}{{ - 40}}{\text{ and }} \\ \frac{q}{{ - 40}}& = \frac{1}{{ - 40}}\\ p& = \frac{1}{5} \quad {\text{ and }}\quad q = 1\end{align}\]

\[\begin{align}{\text{Therefore, }}p &= \frac{1}{{x + y}} = \frac{1}{5}\\ &\Rightarrow x + y = 5 \qquad \left( 3 \right)\\{\text{and, }}q &= \frac{1}{{x - y}} = 1\\ &\Rightarrow x - y = 1 \qquad \left( 4 \right)\end{align}\]

Adding equation \((3)\) and \((4)\), we obtain

\[\begin{align}2x &= 6\\x &= 3\end{align}\]

Substituting\(x = 3\) in equation (3), we obtain

\[\begin{align}3 + y &= 5\\y &= 2\end{align}\]

Hence, \(x = 3\) and \(y = 2\)

\(\begin{align}{\rm{(viii)}} \quad \frac{1}{{3x + y}} + \frac{1}{{3x - y}} &= \frac{3}{4}\\\frac{1}{{2(3x + y)}} - \frac{1}{{2(3x - y)}} &= \frac{{ - 1}}{8}\end{align}\)

Substituting \(\begin{align} \frac{1}{{3x + y}} = p \end{align}\) and  \(\begin{align}\frac{1}{{3x - y}} = q \end{align}\) in these equations, we obtain

\[\begin{align}& \frac{1}{{3x + y}} + \frac{1}{{3x - y}} = \frac{3}{4}\\ \Rightarrow \, & p + q = \frac{3}{4} \cdots \left( 1 \right)\\ \\ &\frac{1}{{2(3x + y)}} - \frac{1}{{2(3x - y)}} = \frac{{ - 1}}{8} \\ \Rightarrow \, &\frac{p}{2} - \frac{q}{2} =  - \frac{1}{8} \\ \Rightarrow \, &p - q =  - \frac{1}{4} \cdots \left( 2 \right)
\end{align}\]

Adding \((1)\) and \((2)\), we obtain

\[\begin{align}2p &= \frac{3}{4} - \frac{1}{4}\\2p &= \frac{1}{2}\\p &= \frac{1}{4}\end{align}\]

Substituting \(\begin{align} p = \frac{1}{4} \end{align}\) in \((2)\), we obtain

\[\begin{align}\frac{1}{4} – q &= - \frac{1}{4}\\q &= \frac{1}{4} + \frac{1}{4}\\q& = \frac{1}{2} \end{align}\]

\[\begin{align}{\text{Therefore, }}p &= \frac{1}{{3x + y}} = \frac{1}{4}\\
 &\Rightarrow 3x + y = 4 \cdots \left( 3 \right)\\{\text{and, }}q &= \frac{1}{{3x - y}} = \frac{1}{2}\\
 &\Rightarrow 3x - y = 2  \cdots \left( 4 \right)\end{align}\]

Adding equations \((3)\) and \((4)\), we obtain

\[\begin{align}6x &= 6\\x &= 1\end{align}\]

Substituting\(x = 1\) in \((3)\), we obtain

\[\begin{align}3 \times 1 + y &= 4\\y& = 1\end{align}\]

Hence, \(x = 1\) and \(y = 1\)

Question 2

Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream \(20\,\rm{ km}\) in \(2\) hours, and upstream \(4 \,\rm{km}\) in \(2\) hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and \(5\) men can together finish an embroidery work in \(4\) days, while \(3\) women and 6 men can finish it in \(3\) days. Find the time taken by \(1\) woman alone to finish the work, and also that taken by \(1 \)man alone.

(iii) Roohi travels \(300 \,\rm{km}\) to her home partly by train and partly by bus. She takes \(4\) hours if she travels \(60 \,\rm{km}\) by train and remaining by bus. If she travels \(100\,\rm{ km}\) by train and the remaining by bus, she takes \(10\) minutes longer. Find the speed of the train and the bus separately.

Solution

Video Solution

Reasoning:

Steps:

(i)

Let the Ritu’s speed of rowing in still water and the speed of stream be \(x\,{\rm{ km/h}}\)and \(y\,{\rm{km/h }}\) respectively.

Ritu’s speed of rowing;

Upstream \(= \left( {x – y} \right) \,\rm{km/h}\)

Downstream \(= \left( {x + y} \right)\,\rm{km/h}\)

According to question,

Ritu can row downstream \(20 \,\rm{km}\) in \(2\) hours,

\[\begin{align}2\left( {x + y} \right) &= 20\\x + y& = 10 \qquad\left( 1 \right)\end{align}\]

Ritu can row upstream \(4\,\rm{ km}\) in \(2\) hours,

\[\begin{align}2\left( {x - y} \right) &= 4\\x - y &= 2 \qquad  \left( 2 \right)\end{align}\]

Adding equation (\(1\)) and (\(2\)), we obtain

\[\begin{align} 2x&=12 \\  x&=6 \\\end{align}\]

Putting \(x = 6\) in equation \((1)\), we obtain

\[\begin{align}6 + y &= 10\\y& = 4\end{align}\]

Hence, Ritu’s speed of rowing in still water is \(6 \,\rm{km/h}\) and the speed of the current is \(4 \,\rm{km/h}.\)

(ii)

Let the number of days taken by a woman and a man to finish the work be \(x\) and \(y\) respectively.

Therefore, work done by a woman in \(1\) day \( = \frac{1}{x}\)

and work done by a man in 1 day \( = \frac{1}{y}\)

According to the question,

\(2\) women and \(5\) men can together finish an embroidery work in \(4\) days;

\[\frac{2}{x} + \frac{5}{y} = \frac{1}{4} \qquad \left( 1 \right)\]

\(3\) women and \(6\) men can finish it in \(3\) days

\[\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \qquad \left( 2 \right)\]

Substituting \(\begin{align}\frac{1}{x} = p \end{align}\) and \(\begin{align}\frac{1}{y} = q \end{align}\)in equations \((1)\) and \((2)\), we obtain

\[\begin{align} &\frac{2}{x} + \frac{5}{y} = \frac{1}{4} \\ \Rightarrow\, & 2p + 5q = \frac{1}{4} \\ \Rightarrow \, & 8p + 20q - 1 = 0 \cdots \left( 3 \right)\\ \\ 
&\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \\ \Rightarrow \, & 3p + 6q = \frac{1}{3} \\ \Rightarrow \, & 9p + 18q - 1 = 0 \cdots \left( 4 \right) \end{align}\]

By cross-multiplication, we obtain

\[\begin{align}\frac{p}{{ - 20 - ( - 18)}} &= \frac{q}{{ - 9 - ( - 8)}} \\ & = \frac{1}{{144 - 180}}\\
\frac{p}{{ - 2}} &= \frac{q}{{ - 1}} = \frac{1}{{ - 36}}\\ \\ \frac{p}{{ - 2}}& = \frac{1}{{ - 36}}{\text{ and }} \\ \frac{q}{{ - 1}} & = \frac{1}{{ - 36}}\\p &= \frac{1}{{18}}\quad {\text{ and }} \\  q & = \frac{1}{{36}}\end{align}\]

\(\begin{align}{\text{Therefore, }}p &= \frac{1}{x} = \frac{1}{{18}}\\
& \Rightarrow x = 18\\{\text{and, }}q &= \frac{1}{y} = \frac{1}{{36}}\\& \Rightarrow y = 36\end{align}\)

Hence, number of days taken by a woman is \(18\) and by a man is \(36.\)

(iii)

Let the speed of train and bus be \(u \,\rm{km/h}\) and \(v\,\rm{ km/h}\) respectively.

According to the given information,

Roohi travels \(300\,\rm{ km}\) and takes \(4\) hours if she travels \(60 \,\rm{km}\) by train and the remaining by bus

\[\frac{{60}}{u} + \frac{{240}}{v} = 4 \qquad \left( 1 \right)\]

If she travels \(100\,\rm{ km}\) by train and the remaining by bus, she takes \(10\) minutes longer

\[\frac{{100}}{u} + \frac{{200}}{v} = \frac{{25}}{6} \qquad \left( 2 \right)\]

Substituting \(\begin{align}\frac{1}{u}=p\end{align}\) and \(\begin{align}\frac{1}{v}=q \end{align}\) in equations \((1)\) and \((2)\), we obtain

\[\begin{align} & \frac{{60}}{{u}} + \frac{{240}}{v} = 4 \\ \Rightarrow\, & 60p + 240q = 4 \cdots \left( 3 \right)\\ \\ & \frac{{100}}{u} + \frac{{200}}{v} = \frac{{25}}{6} \\  \Rightarrow \, & 100p + 200q = \frac{{25}}{6}\\ \Rightarrow \, & 600p + 1200q = 25 \cdots \left( 4 \right)\end{align}\]

Multiplying equation \((3)\) by \(10\), we obtain

\[600p + 2400q = 40 \quad \left( 5 \right)\]

Subtracting equation \((4)\) from \((5)\), we obtain

\[\begin{align}1200q &= 15\\q &= \frac{{15}}{{1200}}\\q& = \frac{1}{{80}}\end{align} \]

Substituting \(\begin{align}q = \frac{1}{{80}} \end{align}\) in equation \((3)\), we obtain

\[\begin{align}60p + 240 \times \frac{1}{{80}} &= 4\\60p &= 4 - 3\\p &= \frac{1}{{60}}\end{align}\]

\(\begin{align}{\text{Therefore, }}p &= \frac{1}{u} = \frac{1}{{60}}\\& \Rightarrow u = 60\\{\text{and, }}q &= \frac{1}{v} = \frac{1}{{80}}\\& \Rightarrow v = 80\end{align}\)

Hence, speed of the train \(= 60\,{\rm{ km/h}}\)

And speed of the bus \(= 80\,{\rm{ km/h}}\)

Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.6 for FREE
Ncert Class 10 Exercise 3.6
Ncert Solutions For Class 10 Maths Chapter 3 Exercise 3.6
  
Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.6 for FREE
Ncert Class 10 Exercise 3.6
Ncert Solutions For Class 10 Maths Chapter 3 Exercise 3.6
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