NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.6

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

(i) \[\frac{1}{2x}+\frac{1}{3y}=2;\text{ }\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}\]

Ans: Taking \[\frac{1}{x}=p\] and \[\frac{1}{y}=q\],

Now, 

\[\frac{p}{2}+\frac{q}{3}=2\]

\[3p+2q-12=0\]           …… (i)

\[\frac{p}{3}+\frac{q}{2}=\frac{13}{6}\]

\[2p+3q-13=0\]        …… (ii)

Using cross multiplication method:

\[\frac{p}{-26-\left( -36 \right)}=\frac{q}{-24-\left( -39 \right)}=\frac{1}{9-4}\]

\[\frac{p}{10}=\frac{q}{15}=\frac{1}{5}\]

\[p=2,q=3\]

\[\frac{1}{x}=2,\frac{1}{y}=3\]

Therefore, \[x=\frac{1}{2}\] and \[y=\frac{1}{3}\].

(ii) \[\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2;\text{}\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\]

Ans: Taking \[\frac{1}{\sqrt{x}}=p\] and \[\frac{1}{\sqrt{y}}=q\],

Now, 

\[2p+3q=2\]           …… (i)

\[4p-9q=-1\]        …… (ii)

Multiplying equation (i) by \[3\], we get

\[6p+9q=6\]          …… (iii)

Adding equation (ii) and (iii), we get

\[10p=5\]

\[p=\frac{1}{2}\]            …… (iv)

Substituting (iv) in equation (i), we get

\[1+3q=2\]

\[q=\frac{1}{3}\]

Now, 

\[\frac{1}{\sqrt{x}}=\frac{1}{2}\]

\[x=4\]

And, \[\frac{1}{\sqrt{y}}=\frac{1}{3}\]

\[y=9\]

Therefore, \[x=4\] and \[y=9\].

(iii) \[\frac{4}{x}+3y=14;\text{ }\frac{3}{x}-4y=23\]

Ans: Taking \[\frac{1}{x}=p\] ,

Now, 

\[4p+3y-14=0\]           …… (i)

\[3p-4y-23=0\]        …… (ii)

Using cross multiplication method:

\[\frac{p}{-69-56}=\frac{y}{-42-\left( -92 \right)}=\frac{1}{-16-9}\]

\[\frac{p}{-125}=\frac{y}{50}=\frac{-1}{25}\]

\[p=5,y=-2\]

\[\frac{1}{x}=5,y=-2\]

Therefore, \[x=\frac{1}{5}\] and \[y=-2\].

(iv) \[\frac{5}{x-1}+\frac{1}{y-2}=2;\text{ }\frac{6}{x-1}-\frac{3}{y-2}=1\]

Ans: Taking \[\frac{1}{x-1}=p\] and \[\frac{1}{y-2}=q\],

Now, 

\[5p+q=2\]           …… (i)

\[6p-3q=1\]        …… (ii)

Multiplying equation (i) by \[3\], we get

\[15p+3q=6\]          …… (iii)

Adding equation (ii) and (iii), we get

\[21p=7\]

\[p=\frac{1}{3}\]            …… (iv)

Substituting (iv) in equation (i), we get

\[\frac{5}{3}+q=2\]

\[q=\frac{1}{3}\]

Now, 

\[\frac{1}{x-1}=\frac{1}{3}\]

\[x=4\]

And, \[\frac{1}{y-2}=\frac{1}{3}\]

\[y=5\]

Therefore, \[x=4\] and \[y=5\].

(v) \[\frac{7x-2y}{xy}=5;\text{ }\frac{8x+7y}{xy}=15\]

Ans: 

\[\frac{7x-2y}{xy}=5\]

\[\frac{7}{y}-\frac{2}{x}=5\]           …… (i)

\[\frac{8x+7y}{xy}=15\]

\[\frac{8}{y}+\frac{7}{x}=15\]         …… (ii)

Taking \[\frac{1}{x}=p\] and \[\frac{1}{y}=q\],

Now, 

\[-2p+7q-5=0\]           …… (iii)

\[7p+8q-15=0\]        …… (iv)

Using cross multiplication method:

\[\frac{p}{-105-\left( -40 \right)}=\frac{q}{-35-30}=\frac{1}{-16-49}\]

\[\frac{p}{-65}=\frac{q}{-65}=\frac{1}{-65}\]

\[p=1,q=1\]

\[\frac{1}{x}=1,\frac{1}{y}=1\]

Therefore, \[x=1\] and \[y=1\].

(vi) \[6x+3y=6xy;\text{ }2x+4y=5xy\]

Ans:

\[6x+3y=6xy\]

\[\frac{6}{y}+\frac{3}{x}=6\]             …… (i)

\[2x+4y=5xy\]

\[\frac{2}{y}+\frac{4}{x}=5\]             …… (ii)

Taking \[\frac{1}{x}=p\] and \[\frac{1}{y}=q\],

Now, 

\[3p+6q-6=0\]           …… (iii)

\[4p+2q-5=0\]        …… (iv)

Using cross multiplication method:

\[\frac{p}{-30-\left( -12 \right)}=\frac{q}{-24-\left( -15 \right)}=\frac{1}{6-24}\]

\[\frac{p}{-18}=\frac{q}{-9}=\frac{1}{-18}\]

\[p=1,q=\frac{1}{2}\]

\[\frac{1}{x}=1,\frac{1}{y}=\frac{1}{2}\]

Therefore, \[x=1\] and \[y=2\].

(vii) \[\frac{10}{x+y}+\frac{2}{x-y}=4;\text{ }\frac{15}{x+y}-\frac{5}{x-y}=-2\]

Ans: Taking \[\frac{1}{x+y}=p\] and \[\frac{1}{x-y}=q\],

Now, 

\[10p+2q-4=0\]           …… (i)

\[15p-5q+2=0\]        …… (ii)

Using cross multiplication method:

\[\frac{p}{4-20}=\frac{q}{-60-20}=\frac{1}{-50-30}\]

\[\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}\]

\[p=\frac{1}{5},q=1\]

Now, 

\[\frac{1}{x+y}=\frac{1}{5}\]

\[x+y=5\]        …… (iii)

And, \[\frac{1}{x-y}=1\]

\[x-y=1\]        …… (iv)

Adding equation (iii) and (iv), we get

\[2x=6\]

\[x=3\]        …… (v)

Substituting (v) in equation (iii), we get

\[y=2\]

Therefore, \[x=3\] and \[y=2\].

(viii) \[\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4};\text{ }\frac{1}{2\left( 3x+y \right)}-\frac{1}{2\left( 3x-y \right)}=\frac{-1}{8}\]

Ans: Taking \[\frac{1}{3x+y}=p\] and \[\frac{1}{3x-y}=q\],

Now, 

\[p+q=\frac{3}{4}\]           …… (i)

\[\frac{p}{2}-\frac{q}{2}=\frac{-1}{8}\]

\[p-q=\frac{-1}{4}\]        …… (ii)

Adding equation (i) and (ii), we get

\[2p=\frac{1}{2}\]

\[p=\frac{1}{4}\]        …… (iii)

Substituting (iii) in equation (ii), we get

\[\frac{1}{4}-q=\frac{-1}{4}\]

\[q=\frac{1}{2}\]

Now, 

\[\frac{1}{3x+y}=\frac{1}{4}\]

\[3x+y=4\]        …… (iv)

And, \[\frac{1}{3x-y}=\frac{1}{2}\]

\[3x-y=2\]        …… (v)

Adding equation (iv) and (v), we get

\[6x=6\]

\[x=1\]        …… (vi)

Substituting (vi) in equation (iv), we get

\[y=1\]

Therefore, \[x=1\] and \[y=1\].

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

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

Ans: Assuming that speed of Ritu in still water is \[\text{x }km/h\] and the speed of stream be \[\text{y }km/h\].

Speed of Ritu while rowing upstream will be \[\text{=}\left( x-y \right)\text{ }km/h\]

Speed of Ritu while rowing downstream will be \[\text{=}\left( x+y \right)\text{ }km/h\]

Writing the algebraic representation using the information given in the question:

\[2\left( x+y \right)=20\]

\[x+y=10\]             …… (i)

\[2\left( x-y \right)=4\]

\[x-y=2\]    …… (ii)

Adding equation (i) and (ii), we get

\[2x=12\]

\[x=6\]     …… (iii)

Substituting (iii) in equation (i), we obtain

\[y=4\]

Hence, speed of Ritu in still water is \[\text{6 }km/h\] and speed of stream is \[\text{4 }km/h\].

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

Ans: Assuming a woman takes \[x\] number of days and a man takes \[\text{y}\] number of days.

So, work done by a women in one day \[=\frac{1}{x}\]

Work done by a man in one day \[=\frac{1}{y}\]

Writing the algebraic representation using the information given in the question:

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

\[\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\]             …… (i)

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

 \[\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\]   …… (ii)

Taking \[\frac{1}{x}=p\] and \[\frac{1}{y}=q\],

Now, 

\[2p+5q=\frac{1}{4}\]

 \[8p+20q=1\]          …… (iii)

\[3p+6q=\frac{1}{3}\]

\[9p+18q=1\]        …… (iv)

Using cross multiplication method:

\[\frac{p}{-20-\left( -18 \right)}=\frac{q}{-9-\left( -8 \right)}=\frac{1}{144-180}\]

\[\frac{p}{-2}=\frac{q}{-1}=\frac{1}{-36}\]

\[p=\frac{1}{18},q=\frac{1}{36}\]

\[\frac{1}{x}=\frac{1}{18},\frac{1}{y}=\frac{1}{36}\]

Therefore, \[x=18\] and \[y=36\].

Hence, a woman takes \[18\] days to finish a work while a man takes \[36\] days to finish a work.

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

Ans: Assuming that the speed of train is \[u\text{ }km/h\] and the speed of bus is \[\text{v }km/h\].

Writing the algebraic representation using the information given in the question:

\[\frac{60}{u}+\frac{240}{v}=4\]             …… (i)

\[\frac{100}{u}+\frac{200}{v}=\frac{25}{6}\]    …… (ii)

Taking \[\frac{1}{u}=p\] and \[\frac{1}{v}=q\], we get

\[60p+240q=4\]     …… (iii)

\[100p+200q=\frac{25}{6}\]

\[600p+1200q=25\]        …… (iv)

Multiplying equation (iii) by \[10\], we get

\[600p+2400q=40\]       …… (v)

Subtracting equation (iv) from equation (v), we get

\[1200q=15\]

\[q=\frac{1}{80}\]        …… (vi)

Substituting (vi) in equation (iii), we get

\[60p=1\]

\[p=\frac{1}{60}\]

So, \[\frac{1}{u}=\frac{1}{60},\frac{1}{v}=\frac{1}{80}\]

\[u=60\] and \[v=80\].

Hence, speed of train is \[\text{60 }km/h\] and speed of bus is \[\text{80 }km/h\].

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

Opting for the NCERT solutions for Ex 3.6 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 3.6 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Pair of Linear Equations in Two Variables textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 3 Exercise 3.6 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 3 Exercise 3.6, there are plenty of exercises in this chapter that contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 3 Exercise 3.6 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

NCERT Solutions for Class 10 Maths  Chapter 3 Exercises