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

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

NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations are created here for boosting your preparation of the exam.NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations will clear your basic concepts on linear equations in two variables. Exercise 3.6 has different structures of linear equations which are solved here in the easiest ways. These NCERT solutions are not only helpful to the class 10 maths students but also for the candidates who are going to appear in the competitive exams.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Download the pdf of the NCERT Solutions of Class 10 Chapter 3 Pair of Linear Equations in Two Variables

PDF -NCERT Solutions of Linear Equations in Two Variables

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

Exercise 3.1 -Linear Equations in Two Variables

Exercise 3.2 -Linear Equations in Two Variables

Exercise 3.3-Linear Equation in Two Variables

Exercise 3.4-Pair of Linear Equations

Exercise 3.5-Pair of Linear Equations

Class 10 maths NCERT solutions of important questions of chapter 3 Pair of Linear Equations

Exercise 3.7 – Linear Equations in Two Variables

Latest Sample paper Class 10 maths for Term 1 2021 CBSE board

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

Q1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Ans.The given pair of linear equations is

1/2x + 1/3y = 2…..(i)

1/3x + 1/2y = 13/6….(ii)

Let 1/x = a and 1/y = b,then we get equation (iii) and equation (iv)

a/2  + b/3 = 2

3a + 2b =12…….(iii)

a/3  + b/2 =13/6

2a + 3b =13/6

12a +18b = 78….(iv)

Multiplying equation (iii) by 4,we get equation (v)

12a + 8b = 48……(v)

Substracting equation (v) from equation (iv)

10b = 30

b = 3

Putting the value of b in equation (iii)

3a + 2×3 = 12

3a +6 =12

3a = 12-6 =6

a = 2

Since 1/x = a and 1/y = b

Substituting the value a and b ,we get x = 1/2 and y = 1/3

(ii) The given pair of linear equation is

2/√x + 3/√y = 2……(i)

4/√x – 9/√y = -1…..(ii)

Let 1/√x = a and b = 1/√y, we get the equations (iii) and equation (iv)

2a + 3b = 2…….(iii)

4a – 9b =-1……(iv)

Multiplying equation (iii) by 2,we get equation (v)

4a + 6b = 4…..(v)

Substracting equation (v) from equation (iv)

-15b = -5

b = 1/3

Putting the value of b in equation (iii)

2a + 3(1/3) =2

2a +1 =2

2a =1

a =1/2

Since 1/√x = a and 1/√y = b

Substituting the value of a and b

1/√x = 1/2 and 1/√y = 1/3

Squaring both equations

1/x = 1/4 and 1/y = 1/9

x =4 and y = 9

(iii) The given equations are

4/x + 3y = 14….(i)

3/x -4y = 23……(ii)

Let 1/x =z

4z + 3y =14……(iii)

3z – 4y =23……(iv)

Multiplying equation (iii) by 3 and equation (iv) by 4,we get equation (v) and equation (vi)

12z + 9y = 42…..(v)

12z – 16y = 92……(vi)

Substracting equation (vi) from equation (v)

25 y = -50

y = -2

Putting the value of y in equation (v)

12z +9×-2 =42

12z -18 =42

12z = 42 +18 =60

z = 5

Since 1/x = z

Substituting the value of z

x = 1/5

Therefore the value of x is 1/5 and the value of y is -2

(iv) The given equations are

5/(x-1) + 1/(y-2) = 2…..(i)

6/(x-1) – 3/(y-2) = 1….(ii)

Let 1/(x -1) = a and 1/(y-2) = b

5a + b = 2……(iii)

6a – 3b = 1……(iv)

Multiplying equation (iii) by 3,we get equation (v)

15a + 3b = 6….(v)

Adding equation (iv) and (v)

21a = 7

a = 7/21 = 1/3

Putting the value of a in equation (iii)

5×1/3 +b =2

5/3 + b =2

b = 2 -(5/3)

b = 1/3

Putting the value of a and b in 1/(x -1) = a and 1/(y-2) = b

1/(x -1) = 1/3 and 1/(y-2) = 1/3

x -1 =3 and y-2 =3

x = 4 and y =5

Amazon online Cloths

(v)  The given pair of linear equations is

(7x-2y)/ xy = 5

(8x + 7y)/xy = 15

Simplifying the given equations

7x/xy – 2y/xy = 5

7/y  – 2/x = 5……(i)

8x/xy +7y/xy = 15

8/y  +7/x = 15….(ii)

Let 1/y =a and 1/x = b

7a -2b = 5……(iii)

8a +7b = 15…..(iv)

Multiplying equation (iii) by 8 and equation (iv) by 7,we get equation (v) and equation (vi)

56a – 16b = 40….(v)

56a + 49b = 105…(vi)

Substracting equation (vi) from equation (v)

-65b = -65

b = 1

Putting the value of b in equation (iii)

7a -2×1 = 5

7a -2 = 5

7a = 7

a = 1

Since 1/y =a and 1/x = b

∴ 1/y = 1 and 1/x = 1

y = 1 and x = 1

(vi) The given equations are

6x + 3y = 6xy

2x + 4y = 5xy

Simplifying the given equations

6x/xy +3y/xy = 6xy/xy

6/y + 3/x = 6….(i)

2x/xy + 4y/xy = 5xy/xy

2/y  + 4/x = 5…(ii)

Let 1/y =a and 1/x = b

6a +3b = 6……(iii)

2a + 4b =5….(iv)

Multiplying equation (iii) by 2 and equation (iv) by 6,we get equation (v) and equation (vi)

12a +6b = 12….(v)

12a + 24b = 30….(vi)

Substracting equation (vi) from equation (v)

-18b = -18

b = 1

Putting the value of b in equation (iii)

6a + 3×1=6

6a + 3 = 6

6a = 3

a = 3/6 = 1/2

Since 1/y =a and 1/x = b

Substituting the value of a and b

1/y =1/2 and 1/x = 1

y =2 and x =1

(vii) The given pair of linear equations is

10/(x+y) + 2/(x-y) = 4…..(i)

15/(x+y) – 5/(x-y) = -2….(ii)

Let 1/(x+y) = a and 1/(x-y) = b

10a + 2b = 4…….(iii)

15a – 5b = = -2….(iv)

Multiplying equation (iii) by 5 and equation (iv) by 2,we get equation (v) and equation (vi)

50a + 10b = 20…….(v)

30a – 10b = = -4….(vi)

Adding both of the equation (v) and equation (vi)

80a = 16

a = 16/80 = 1/5

Putting the value of a in equation (v)

50 ×1/5 +10b =20

10 + 10b =20

10b = 20-10 = 10

b = 1

Since 1/(x+y) = a and 1/(x-y) = b

Substituting the value of a and b ,we get equation (vii) and equation (viii)

1/(x+y) = 1/5 and 1/(x-y) = 1

x + y = 5…..(vii) and x -y =1(viii)

Adding both equations (vii) and (viii)

2x  = 6

x = 3

Putting the value of x in equation (vii)

3 + y = 5

y = 5-3 = 2

Hence the value of x is 3 and value of y is 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4….(i)

1/2(3x+y) – 1/2(3x-y) = -1/8…..(ii)

Let 1/(3x+y) =a and  1/(3x-y) = b

a  + b = 3/4

4a + 4b = 3…..(iii)

a/2  – b/2 = -1/8

4a – 4b =-1…..(iv)

Adding both equations (iii) and (iv)

8a = 2

a = 2/8 = 1/4

Putting the value of a in equation (iv)

4×1/4 + 4b =3

1 + 4b = 3

4b = 3-1 = 2

b = 2/4 = 1/2

Since 1/(3x+y) =a and  1/(3x-y) = b

1/(3x+y) =1/4 and  1/(3x-y) = 1/2

3x + y = 4 ….(v) and 3x -y = 2….(vi)

Adding the equations (v) and equation (vi)

6x = 6

x = 1

Putting the value of x in equation (v)

3 ×1+ y =4

3 + y = 4

y = 4 -3 = 1

Hence the value of x is and value of y is 1

Q2.Formulate the following problems as a pair of equations, and hence find their solutions:

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

Ans. Let the speed of the boat in still water is x km/h and the speed of the current is y km/h

The speed of boat in down stream is = x + y

Since we are given distance covered by the boat in downstream is 20 km in time 2 hour

Therefore speed of boat in downstream = Distance/Time = 20/2 = 10 km/h

According to first condition

x + y = 10…….(i)

The speed of boat in down stream is = x – y

Since we are given distance covered by the boat in upstream is 4 km in time 2 hour

Therefore speed of boat in upstream = Distance/Time = 4/2 = 2 km/h

According to second condition

x – y = 2……(ii)

Adding both equations (i) and (ii)

2x = 8

x = 8/2 = 4

Putting the value of x in equation (ii)

4 -y = 2

y = 4 -2 = 2

Hence the speed of the boat in still water is 4 km/h and the speed of the current is 2 km/h

(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.

Ans.Let 1 woman alone can finish the work in x days and 1 man alone can finish the work in y days

Since in x days   1 woman do 1 work

In 1 day a woman will do = 1/x work

In 1 day 2 women will do = 2/x work

Since in y days, 1 man do 1 work

In 1-day a man will do = 1/y work

In 1 day 5 men will do = 5/y work

2 woman and 5 men do a work in 4 days

In 1 day they will work =1/4 work

Therefore according to first condition

2/x + 5/y = 1/4….(i)

Similarly according to second condition,we can creat second equation

3/x + 6/y = 1/3….(ii)

Let 1/x =a and 1/y = b

2a + 5b = 1/4

8a + 20b =1…….(iii)

3a +6b =1/3

9a + 18b =1……(iv)

Multiplying equation (iii) by 9 and equation (iv) by 8

72a + 180b = 9…(v)

72a +144b =8….(vi)

36b = 1

b = 1/36

Putting the value of b in equation (iii)

8a + 20×1/36 =1

8a + 5/9 = 1

8a = 1-5/9 = 4/9

a = 1/18

Since 1/x =1/18 and 1/y = 1/36

x =18 and y = 36

Hence a woman alone can do the work in 18 days and a man alone can do the given work in 36 days.

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

Ans. Let the speed of the train is x km/h and speed of the bus is y km/h

According to first condition Roohi travels 60 km by train and the remaining (300 -60 = 240 km) by bus and total time taken is 4 hours.

Time taken to cover the distance 60 km + Time taken to cover the distance 240 km/h = 4 hours

Since time = Distance/Speed

60/x  + 240/y = 4……(i)

According to second condition Roohi travels 100 km by train and the remaining (300 -100 = 200 km) by bus and total time taken is 4 hours +10 minutes

Time taken to cover the distance 100 km + Time taken to cover the distance 200 km/h = (4+10/60)=4+1/6 =(25/6)h

Since time = Distance/Speed

100/x  + 200/y = 25/6……(ii)

Let a = 1/x and b = 1/y

60a + 240b = 4…….(iii)

100a + 200b = 25/6

600a + 1200b = 25….(iv)

Multiplying the equation (iii) by 10,we get equation (v)

600a + 2400b = 40…(v)

Substracting equation (v) from equation (iv)

-1200b = -15

b = 15/1200 = 1/80

Putting the value of b in equation

60a + 240×1/80 =4

60a +3 = 4

60a = 4-3 =1

a = 1/60

Since a = 1/x and b = 1/y

1/60 = 1/x and 1/80 = 1/y

x = 60 and y =80

Hence the speed of train is 60 km/h and speed of the bus is 80 km/h

You can compensate us by donating any amount of money for our survival

## NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions of class 9 science

 Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

### NCERT Solutions of class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT solutions of class 10 science

 Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world Chapter 4- Carbon and its Compounds Chapter 12- Electricity Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current Chapter 6- Life Process Chapter 14-Sources of Energy Chapter 7-Control and Coordination Chapter 15-Environment Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

### Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

### NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Scroll to Top
The text passage is a set of NCERT solutions for Class 10 Maths, specifically for Exercise 3.6 of Chapter 3, which deals with pairs of linear equations in two variables. The passage contains multiple questions and their corresponding solutions.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations in Two Variables - Future Study Point

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

NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations are created here for boosting your preparation of the exam.NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations will clear your basic concepts on linear equations in two variables. Exercise 3.6 has different structures of linear equations which are solved here in the easiest ways. These NCERT solutions are not only helpful to the class 10 maths students but also for the candidates who are going to appear in the competitive exams.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Download the pdf of the NCERT Solutions of Class 10 Chapter 3 Pair of Linear Equations in Two Variables

PDF -NCERT Solutions of Linear Equations in Two Variables

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

Exercise 3.1 -Linear Equations in Two Variables

Exercise 3.2 -Linear Equations in Two Variables

Exercise 3.3-Linear Equation in Two Variables

Exercise 3.4-Pair of Linear Equations

Exercise 3.5-Pair of Linear Equations

Class 10 maths NCERT solutions of important questions of chapter 3 Pair of Linear Equations

Exercise 3.7 – Linear Equations in Two Variables

Latest Sample paper Class 10 maths for Term 1 2021 CBSE board

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

Q1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Ans.The given pair of linear equations is

1/2x + 1/3y = 2…..(i)

1/3x + 1/2y = 13/6….(ii)

Let 1/x = a and 1/y = b,then we get equation (iii) and equation (iv)

a/2  + b/3 = 2

3a + 2b =12…….(iii)

a/3  + b/2 =13/6

2a + 3b =13/6

12a +18b = 78….(iv)

Multiplying equation (iii) by 4,we get equation (v)

12a + 8b = 48……(v)

Substracting equation (v) from equation (iv)

10b = 30

b = 3

Putting the value of b in equation (iii)

3a + 2×3 = 12

3a +6 =12

3a = 12-6 =6

a = 2

Since 1/x = a and 1/y = b

Substituting the value a and b ,we get x = 1/2 and y = 1/3

(ii) The given pair of linear equation is

2/√x + 3/√y = 2……(i)

4/√x – 9/√y = -1…..(ii)

Let 1/√x = a and b = 1/√y, we get the equations (iii) and equation (iv)

2a + 3b = 2…….(iii)

4a – 9b =-1……(iv)

Multiplying equation (iii) by 2,we get equation (v)

4a + 6b = 4…..(v)

Substracting equation (v) from equation (iv)

-15b = -5

b = 1/3

Putting the value of b in equation (iii)

2a + 3(1/3) =2

2a +1 =2

2a =1

a =1/2

Since 1/√x = a and 1/√y = b

Substituting the value of a and b

1/√x = 1/2 and 1/√y = 1/3

Squaring both equations

1/x = 1/4 and 1/y = 1/9

x =4 and y = 9

(iii) The given equations are

4/x + 3y = 14….(i)

3/x -4y = 23……(ii)

Let 1/x =z

4z + 3y =14……(iii)

3z – 4y =23……(iv)

Multiplying equation (iii) by 3 and equation (iv) by 4,we get equation (v) and equation (vi)

12z + 9y = 42…..(v)

12z – 16y = 92……(vi)

Substracting equation (vi) from equation (v)

25 y = -50

y = -2

Putting the value of y in equation (v)

12z +9×-2 =42

12z -18 =42

12z = 42 +18 =60

z = 5

Since 1/x = z

Substituting the value of z

x = 1/5

Therefore the value of x is 1/5 and the value of y is -2

(iv) The given equations are

5/(x-1) + 1/(y-2) = 2…..(i)

6/(x-1) – 3/(y-2) = 1….(ii)

Let 1/(x -1) = a and 1/(y-2) = b

5a + b = 2……(iii)

6a – 3b = 1……(iv)

Multiplying equation (iii) by 3,we get equation (v)

15a + 3b = 6….(v)

Adding equation (iv) and (v)

21a = 7

a = 7/21 = 1/3

Putting the value of a in equation (iii)

5×1/3 +b =2

5/3 + b =2

b = 2 -(5/3)

b = 1/3

Putting the value of a and b in 1/(x -1) = a and 1/(y-2) = b

1/(x -1) = 1/3 and 1/(y-2) = 1/3

x -1 =3 and y-2 =3

x = 4 and y =5

Amazon online Cloths

(v)  The given pair of linear equations is

(7x-2y)/ xy = 5

(8x + 7y)/xy = 15

Simplifying the given equations

7x/xy – 2y/xy = 5

7/y  – 2/x = 5……(i)

8x/xy +7y/xy = 15

8/y  +7/x = 15….(ii)

Let 1/y =a and 1/x = b

7a -2b = 5……(iii)

8a +7b = 15…..(iv)

Multiplying equation (iii) by 8 and equation (iv) by 7,we get equation (v) and equation (vi)

56a – 16b = 40….(v)

56a + 49b = 105…(vi)

Substracting equation (vi) from equation (v)

-65b = -65

b = 1

Putting the value of b in equation (iii)

7a -2×1 = 5

7a -2 = 5

7a = 7

a = 1

Since 1/y =a and 1/x = b

∴ 1/y = 1 and 1/x = 1

y = 1 and x = 1

(vi) The given equations are

6x + 3y = 6xy

2x + 4y = 5xy

Simplifying the given equations

6x/xy +3y/xy = 6xy/xy

6/y + 3/x = 6….(i)

2x/xy + 4y/xy = 5xy/xy

2/y  + 4/x = 5…(ii)

Let 1/y =a and 1/x = b

6a +3b = 6……(iii)

2a + 4b =5….(iv)

Multiplying equation (iii) by 2 and equation (iv) by 6,we get equation (v) and equation (vi)

12a +6b = 12….(v)

12a + 24b = 30….(vi)

Substracting equation (vi) from equation (v)

-18b = -18

b = 1

Putting the value of b in equation (iii)

6a + 3×1=6

6a + 3 = 6

6a = 3

a = 3/6 = 1/2

Since 1/y =a and 1/x = b

Substituting the value of a and b

1/y =1/2 and 1/x = 1

y =2 and x =1

(vii) The given pair of linear equations is

10/(x+y) + 2/(x-y) = 4…..(i)

15/(x+y) – 5/(x-y) = -2….(ii)

Let 1/(x+y) = a and 1/(x-y) = b

10a + 2b = 4…….(iii)

15a – 5b = = -2….(iv)

Multiplying equation (iii) by 5 and equation (iv) by 2,we get equation (v) and equation (vi)

50a + 10b = 20…….(v)

30a – 10b = = -4….(vi)

Adding both of the equation (v) and equation (vi)

80a = 16

a = 16/80 = 1/5

Putting the value of a in equation (v)

50 ×1/5 +10b =20

10 + 10b =20

10b = 20-10 = 10

b = 1

Since 1/(x+y) = a and 1/(x-y) = b

Substituting the value of a and b ,we get equation (vii) and equation (viii)

1/(x+y) = 1/5 and 1/(x-y) = 1

x + y = 5…..(vii) and x -y =1(viii)

Adding both equations (vii) and (viii)

2x  = 6

x = 3

Putting the value of x in equation (vii)

3 + y = 5

y = 5-3 = 2

Hence the value of x is 3 and value of y is 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4….(i)

1/2(3x+y) – 1/2(3x-y) = -1/8…..(ii)

Let 1/(3x+y) =a and  1/(3x-y) = b

a  + b = 3/4

4a + 4b = 3…..(iii)

a/2  – b/2 = -1/8

4a – 4b =-1…..(iv)

Adding both equations (iii) and (iv)

8a = 2

a = 2/8 = 1/4

Putting the value of a in equation (iv)

4×1/4 + 4b =3

1 + 4b = 3

4b = 3-1 = 2

b = 2/4 = 1/2

Since 1/(3x+y) =a and  1/(3x-y) = b

1/(3x+y) =1/4 and  1/(3x-y) = 1/2

3x + y = 4 ….(v) and 3x -y = 2….(vi)

Adding the equations (v) and equation (vi)

6x = 6

x = 1

Putting the value of x in equation (v)

3 ×1+ y =4

3 + y = 4

y = 4 -3 = 1

Hence the value of x is and value of y is 1

Q2.Formulate the following problems as a pair of equations, and hence find their solutions:

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

Ans. Let the speed of the boat in still water is x km/h and the speed of the current is y km/h

The speed of boat in down stream is = x + y

Since we are given distance covered by the boat in downstream is 20 km in time 2 hour

Therefore speed of boat in downstream = Distance/Time = 20/2 = 10 km/h

According to first condition

x + y = 10…….(i)

The speed of boat in down stream is = x – y

Since we are given distance covered by the boat in upstream is 4 km in time 2 hour

Therefore speed of boat in upstream = Distance/Time = 4/2 = 2 km/h

According to second condition

x – y = 2……(ii)

Adding both equations (i) and (ii)

2x = 8

x = 8/2 = 4

Putting the value of x in equation (ii)

4 -y = 2

y = 4 -2 = 2

Hence the speed of the boat in still water is 4 km/h and the speed of the current is 2 km/h

(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.

Ans.Let 1 woman alone can finish the work in x days and 1 man alone can finish the work in y days

Since in x days   1 woman do 1 work

In 1 day a woman will do = 1/x work

In 1 day 2 women will do = 2/x work

Since in y days, 1 man do 1 work

In 1-day a man will do = 1/y work

In 1 day 5 men will do = 5/y work

2 woman and 5 men do a work in 4 days

In 1 day they will work =1/4 work

Therefore according to first condition

2/x + 5/y = 1/4….(i)

Similarly according to second condition,we can creat second equation

3/x + 6/y = 1/3….(ii)

Let 1/x =a and 1/y = b

2a + 5b = 1/4

8a + 20b =1…….(iii)

3a +6b =1/3

9a + 18b =1……(iv)

Multiplying equation (iii) by 9 and equation (iv) by 8

72a + 180b = 9…(v)

72a +144b =8….(vi)

36b = 1

b = 1/36

Putting the value of b in equation (iii)

8a + 20×1/36 =1

8a + 5/9 = 1

8a = 1-5/9 = 4/9

a = 1/18

Since 1/x =1/18 and 1/y = 1/36

x =18 and y = 36

Hence a woman alone can do the work in 18 days and a man alone can do the given work in 36 days.

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

Ans. Let the speed of the train is x km/h and speed of the bus is y km/h

According to first condition Roohi travels 60 km by train and the remaining (300 -60 = 240 km) by bus and total time taken is 4 hours.

Time taken to cover the distance 60 km + Time taken to cover the distance 240 km/h = 4 hours

Since time = Distance/Speed

60/x  + 240/y = 4……(i)

According to second condition Roohi travels 100 km by train and the remaining (300 -100 = 200 km) by bus and total time taken is 4 hours +10 minutes

Time taken to cover the distance 100 km + Time taken to cover the distance 200 km/h = (4+10/60)=4+1/6 =(25/6)h

Since time = Distance/Speed

100/x  + 200/y = 25/6……(ii)

Let a = 1/x and b = 1/y

60a + 240b = 4…….(iii)

100a + 200b = 25/6

600a + 1200b = 25….(iv)

Multiplying the equation (iii) by 10,we get equation (v)

600a + 2400b = 40…(v)

Substracting equation (v) from equation (iv)

-1200b = -15

b = 15/1200 = 1/80

Putting the value of b in equation

60a + 240×1/80 =4

60a +3 = 4

60a = 4-3 =1

a = 1/60

Since a = 1/x and b = 1/y

1/60 = 1/x and 1/80 = 1/y

x = 60 and y =80

Hence the speed of train is 60 km/h and speed of the bus is 80 km/h

You can compensate us by donating any amount of money for our survival

## NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions of class 9 science

 Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

### NCERT Solutions of class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT solutions of class 10 science

 Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world Chapter 4- Carbon and its Compounds Chapter 12- Electricity Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current Chapter 6- Life Process Chapter 14-Sources of Energy Chapter 7-Control and Coordination Chapter 15-Environment Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

### Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

### NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Scroll to Top
user
The text passage is a set of NCERT solutions for Class 10 Maths, specifically for Exercise 3.6 of Chapter 3, which deals with pairs of linear equations in two variables. The passage contains multiple questions and their corresponding solutions.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS