# NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6

NCERT Solutions for class 10 Maths Chapter 3 Exercise 3.6 (Class 10 Ex. 3.6) Pair of linear equations in two variables in Hindi Medium and English Medium. 10th Maths CBSE Solutions are also available in Video Format for all board Gujrat Board, UP Board and CBSE board based NCERT Books. UP Board High School students are also using NCERT Textbooks for their course, so they also can use these solutions for their help in Maths of class 10. Download UP Board Solutions for Class 10 Maths Chapter 3 Exercise 3.6 in Hindi Medium free from this page. Download Online as well as Offline Apps for session 2020-21 and CBSE Solutions updated on the basis of CBSE Curriculum 2020-21.

## NCERT Solutions for class 10 Maths Chapter 3 Exercise 3.6

 Class: 10 Maths (English and Hindi Medium) Chapter 3: Exercise 3.6

### 10 Maths Chapter 3 Exercise 3.6 Solutions

NCERT Solutions for class 10 Maths Chapter 3 Exercise 3.6 in English Medium and Hindi medium given below updated for new academic year 2020-2021. In case you want to download these solutions in PDF, visit to NCERT Solutions for Class 10 Maths Chapter 3 main page to get all exercises.

• ### Class 10 Maths Exercise 3.6 Solutions

#### Class 10 Maths Chapter 3 Exercise 3.6 Solution in Videos

Class 10 Maths Exercise 3.6 Question 1(i) Solution
Class 10 Maths Exercise 3.6 Question 1(ii) Solution
Class 10 Maths Exercise 3.6 Question 1(iii) Solution
Class 10 Maths Exercise 3.6 Question 1(iv) Solution

Class 10 Maths Exercise 3.6 Question 1(v) Solution
Class 10 Maths Exercise 3.6 Question 1(vi) Solution
Class 10 Maths Exercise 3.6 Question 1(vii) Solution
Class 10 Maths Exercise 3.6 Question 1(viii) Solution
Class 10 Maths Exercise 3.6 Question 2(i) Solution
Class 10 Maths Exercise 3.6 Question 2(ii) Solution
Class 10 Maths Exercise 3.6 Question 2(iii) Solution
Class 10 Maths Chapter 3 Exercise 3.6 Solution

#### Important Questions – Answers for practice

1. Sunita has some ₹ 50 and ₹ 100 notes amounting to a total of ₹15,500. If the total number of notes is 200, the find how many notes of ₹50 and ₹100 each, she has. [Answer: ₹50 notes = 90, ₹100 notes = 110]
2. The electric bill of a certain establishment is partly fixed and partly varies as the number of units of electricity consumed. When in a certain month 200 units are consumed, the bill is ₹107. In another month, 300 units are consumed and the bill is ₹154. If yet another month, 500 units are consumed, find the bill for the month. [Answer: ₹248]
3. Solve graphically the pair of linear equations 3x – 4y + 3 = 0 and 3x + 4y – 21 = 0. Find the co-ordinates of vertices of triangular region formed by these lines and x-axis. Also calculate the area of this triangle. [Answer: Solutions (3, 3). Vertices: (-1, 0), (7, 0) and (3, 3). Area = 12 square units]
##### Questions on Linear Equation from Board Papers
• A railway half ticket costs half the full fare and the reservation charges is same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs ₹216 and one full and one half reserved first class ticket costs ₹327. What is the basic first class full fare and what is the reservation charge? [Answer: Fare = ₹210 reservation charge = ₹6]
• A man travels 600 km to his home partly by train and partly by bus. He takes 8 hours, if he travels 120 km by train and rest by bus. Further, it takes 20 minute longer, if he travels 200 km by train and rest by bus. Find the speeds of the train and the bus. [Answer: 60 km/h, 80 km/h]

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##### 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.

Let the speed of rowing in still water = x km/h
Let the speed of water = y km/h
Towards downstream, speed = x+y km/h, distance = 20 km, time = 2 hours
According to question, speed = distance/time
x + y = 20/2
⇒ y = 10 – x … (1)
Towards upstream, speed = x-y km/h, distance = 4 km, time = 2 hours
x – y = 4/2
⇒ x – y = 2
Putting the value of y form equation (1), we get
x – (10 – x) = 2
⇒ 2x = 12
⇒ x = 6
Putting the value of x in equation (1), we get
y= 10 – 6 = 4
Hence, speed of rowing in still water is 6 km/h and speed of water 4 km/h.

##### 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.

Let the number of days taken by 1 woman to complete the work = x
Let the number of days taken by 1 man to complete the work = y
So, the work done by 1 woman in 1 day = 1/x
And the work done by 1 man in 1 day = 1/y
According to first condition,
4(2/x + 5/y) = 1
⇒8/x + 20/y = 1 … (1)
According to second condition,
3(3/x + 6/y) = 1
⇒ 9/x + 18/y = 1 … (2)
Let, 1/x = a and 1/y=b … (3)
From equation (1), we get
8a + 20b = 1
⇒ a = (1 – 20b)/8 … (4)
From equation (2), we get
9a + 18b = 1
Putting the value of a form equation (4), we get
9((1 – 20b)/8) + 18b = 1
⇒ 9 – 180b + 144b = 8
⇒ – 36b = – 1
⇒ b = 1/36
Putting the value of b in equation (4), we get
a = (1 – 20(1/36))/8 = 4/72 =1/18
From the equation (3), we get
1/x = 1/18 and 1/y = 1/36
⇒ x = 18 and y = 36
Hence, 1 woman takes 18 days and 1 man takes 36 days to complete the work.

##### 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.

Let the speed of train = x km/h
Let the speed of bus = y km/h
If she travels 60 km by train and the remaining by bus, she take 4 hours, so, according to question,
60/x + 240/y = 4 … (1)
If she travels 100 km by train and the remaining by bus, she take 10 minutes more, so, according to question,
100/x + 200/y = 4+10/60
⇒ 100/x + 200/y = 25/6 … (2)
Let, 1/x = a and 1/y=b … (3)
From the equation (1), we get
60a + 240b = 4
⇒ a = (4 – 240b)/60 … (4)
From the equation (2), we get
100a + 200b = 25/6
Putting the value of a from equation (4), we get
100((4 – 240b)/60) + 200b = 25/6
⇒ 40 – 2400b + 1200b = 25
⇒ – 1200b = -15
⇒ b = 1/80
Putting the value of b in equation (4), we get
a = (4 – 240(1/80))/60 = 1/60
From the equation (3), we get
1/x = 1/60 and 1/y = 1/80
⇒ x = 60 and y = 80
Hence, the speed of train is 60 km/h and the speed of bus is 80 km/h.

##### Which is the most important question in 10th Maths Exercise 3.6?

Most of the questions of Ex. 3.6 Class 10 Maths are important for Board Examination. So, practice all the questions properly to score good marks.               