Dattu

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2.

• The Triangle and its Properties Class 7 Ex 6.1
• The Triangle and its Properties Class 7 Ex 6.3
• The Triangle and its Properties Class 7 Ex 6.4
• The Triangle and its Properties Class 7 Ex 6.5
• The Triangle and its Properties Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 6
Chapter Name	The Triangle and its Properties
Exercise	Ex 6.2
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2

Question 1.
Find the value of the unknown exterior angle x in the following diagrams:

Solution:

Question 2.
Find the value of the unknown interior angle x in the following figures:

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3.

• Data Handling Class 7 Ex 3.1
• Data Handling Class 7 Ex 3.2
• Data Handling Class 7 Ex 3.4
• Data Handling Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 3
Chapter Name	Data Handling
Exercise	Ex 3.3
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3

Question 1.
Use the bar graph to answer the following questions.
(a) Which is the most popular pet?
(b) How many children have dog as a pet?

Solution:
(a) The most popular pet is ‘Cat’.
(b) 8 children have a dog as a pet.

Question 2.
Read the bar graph which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:

1. About how many books were sold in 1989?1990?1992?
2. ln which year were about 475 books sold? About 225 books sold?
3. In which years were fewer than 250 books sold?
4. Can you explain how you would estimate the number of books sold in 1989?

Solution:
Clearly, from the given graph, we have

1. Number of books sold in the year
   1989: 170 (approx.)
   1990: 475 (approx.)
   1992: 225 (approx.)
2. In the year 1990, about 475 books were sold. In the year 1992, about 225 books were sold.
3. Fewer than 250 books were sold in the years 1989 and 1992.
4. It can be estimated using the height of the bar such that the height of 1 cm = 100 books.

Question 3.
The number of children in six different classes is given below. Represent the data on a bar graph.

(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the students of the class eighth.
Solution:

(a) Start the scale at 0. The greatest value in the data is 135, so end the scale at a value greater than 135, such as 140. Use equal divisions along the axes, such as increments of 10.
We know that all the bars would lie between 0 and 140.
We choose the scale such that the length between 0 and 140 is neither too long nor too small.
Here, we take 1 unit for 10 children.
(b) (i) Fifth class has the maximum number of children. Tenth class has the minimum number of children.
(ii) Ratio of students of class sixth to eighth = 120 : 100 = \(\frac { 12 }{ 100 } \) = \(\frac { 6 }{ 5 } \) = 6 : 5

Question 4.
The performance of a student in 1st Term and, 2nd Term is given. Draw a double bar graph choosing the appropriate scale and answer the following :

1. In which subject, has the child improved his performance the most?
2. In which subject is the improvement the least?
3. Has the performance gone down in any subject?

Solution:

1. The child improved his performance the most in the subject of Maths.
2. The improvement is the least in the subject of S. Science.
3. Yes! The performance has gone down in the subject of Hindi.

Question 5.
Consider this data collected from a survey of a colony.

1. Draw a double graph choosing an appropriate scale.
   What do you infer from the bar graph?
2. Which sport is most popular?
3. Which is more preferred, watching or participating in sports?

Solution:

1. It is inferred that more people prefer cricket and fewer athletics.
2. The most popular sport is cricket.
3. Watching is more preferred than participating.

Question 6.
Take the data giving the minimum and the maximum temperature of various cities given at the beginning of this chapter. Plot a double bar graph using the data and answer the following:

1. Which city has the largest difference in the minimum and maximum temperature on the given date?
2. Which is the hottest city and which is the coldest city?
3. Name two cities where the maximum temperature of one was less than the minimum temperature of the other.
4. Name the city which has the least difference between its minimum and the maximum temperature.

Solution:

1. The city Jammu has the largest difference in the minimum and maximum temperature on the given date.
2. Jammu is the hottest city and Bangalore is the coldest city.
3. The name of the two cities where the maximum temperature of one was less than the minimum temperature of the other is Bangalore and Jaipur or Bangalore and Ahmedabad.
4. Mumbai has the least difference between its minimum and maximum temperature.

We hope the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4.

• Data Handling Class 7 Ex 3.1
• Data Handling Class 7 Ex 3.2
• Data Handling Class 7 Ex 3.3
• Data Handling Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 3
Chapter Name	Data Handling
Exercise	Ex 3.4
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4

Question 1.
Tell whether the following is certain to happen, impossible, can happen but not certain.

1. You are older today than yesterday.
2. A tossed coin will land heads up.
3. A die when tossed shall land up with 8 on top.
4. The next traffic light seen will be green.
5. Tomorrow will be a cloudy day.

Solution:

1. Certain to happen
2. Can happen but not certain
3. Impossible
4. Can happen but not certain
5. Can happen but not certain

Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.

1. What is the probability of drawing a marble with the number 2?
2. What is the probability of drawing a marble with the number 5?

Solution:
Out of 6 marbles, one can be drawn in 6 ways. So, the total number of events = 6

1. The marble with the number 2 can be obtained only in one way.
∴ Required probability = \(\frac { 1 }{ 6 } \)

2. The marble with the number 5 can be obtained only in one way.
∴ Required probability = \(\frac { 1 }{ 6 } \)

Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Solution:
On tossing a coin, the possible outcomes are head (H) or tail (T).
Required probability = \(\frac { 1 }{ 2 } \)

We hope the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7.

• Fractions and Decimals Class 7 Ex 2.1
• Fractions and Decimals Class 7 Ex 2.2
• Fractions and Decimals Class 7 Ex 2.3
• Fractions and Decimals Class 7 Ex 2.4
• Fractions and Decimals Class 7 Ex 2.5
• Fractions and Decimals Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimals
Exercise	Ex 2.7
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

Question 1.
Find:

1. 0.4 ÷ 2
2. 0.35 ÷ 5
3. 2.48 ÷ 4
4. 65.4 ÷ 6
5. 651.2 ÷ 4
6. 114.49 ÷ 7
7. 3.96 ÷ 4
8. 0.80 ÷ 5

Solution:

Question 2.
Find:

1. 4.8 ÷ 10
2. 52.5 ÷ 10
3. 0.7 ÷ 10
4. 33.1 ÷ 10
5. 272.23 ÷ 1o
6. 0.56 ÷ 10
7. 3.97 ÷ 10

Solution:

Question 3.
Find:

1. 2.7 ÷ 100
2. 0.3 ÷ 100
3. 0.78 ÷ 100
4. 432.6 ÷ 100
5. 23.6 ÷ 100
6. 98.53 ÷ 100

Solution:

1. 2.7 ÷ 100 = 0.027
2. 0.3 ÷ 100 = 0.003
3. 0.78 ÷ 100 = 0.0078
4. 432.6 ÷ 100 = 4.326
5. 23.6 ÷ 100 = 0.236
6. 98.53 ÷ 100 = 0.9853.

Question 4.
Find:

1. 7.9 ÷ 1000
2. 26.3 ÷ 2000
3. 38.53 ÷ 1000
4. 128.9 ÷ 2000
5. 0.5 ÷ 2000

Solution:

1. 7.9 ÷ 1000 = 0.0079
2. 26.3 ÷ 1000 = 0.0263
3. 38.53 ÷ 1000 = 0.03853
4. 128.9 ÷ 1000 = 0.1289
5. 0.5 ÷ 1000 = 0.0005

Question 5.
Find:

1. 7 ÷ 3.5
2. 36 ÷ 0.2
3. 3.25 ÷ 0.5
4. 30.94 ÷ 0.7
5. 0.5 ÷ 0.25
6. 7.75 ÷ 0.25
7. 76.5 ÷ 0.15
8. 37.8 ÷ 1.4
9. 2.73 ÷ 1.3

Solution:

Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?
Solution:
∵ Distance covered in 2.4 litres of petrol = 43.2 km
∴ Distance travelled in 1 litre of petrol = 43.2 ÷ 2.4

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6.

• Fractions and Decimals Class 7 Ex 2.1
• Fractions and Decimals Class 7 Ex 2.2
• Fractions and Decimals Class 7 Ex 2.3
• Fractions and Decimals Class 7 Ex 2.4
• Fractions and Decimals Class 7 Ex 2.5
• Fractions and Decimals Class 7 Ex 2.7
• Fractions and Decimals Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimals
Exercise	Ex 2.6
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

Question 1.
Find:

1. 0.2 × 6
2. 8 × 4.6
3. 2.71 × 5
4. 20.1 × 4
5. 0.05 × 7
6. 211.02 × 4
7. 2 × 0.86

Solution:

1. 0.2 × 6 = 1.2
2. 8 × 4.6 = : 36.8
3. 2.71 × 5 = 13.55
4. 20.1 × 4 = 80.4
5. 0.05 × 7 = 0.35
6. 211.02 × 4 = 844.08
7. 2 × 0.86 = 1.72

Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
∴ Area of the rectangle
= Length × Breadth
= 5.7 × 3 = 17.1 cm²

Question 3.
Find:

1. 1.3 × 10
2. 36.8 × 10
3. 153.7 × 10
4. 168.07 × 10
5. 31.1 × 100
6. 156.1 × 100
7. 3.62 × 100
8. 43.07 × 100
9. 0.5 × 10
10. 0.08 × 10
11. 0.9 × 100
12. 0.03 × 1000

Solution:

1. 1.3 × 10 = 13.0
2. 36.8 × 10 = 368.0
3. 153.7 × 10 = 1537.0
4. 168.07 × 10 = 1680.70
5. 31.1 × 100 = = 3110.0
6. 156.1 × 100 = 15610.0
7. 3.62 × 100 = = 362.00
8. 43.07 × 100 = 4307.00
9. 0.5 × 10 = 5 .0
10. 0.08 × 10 = 0.80
11. 0.9 × 100 = 90.0
12. 0.03 × 1000 = 30.0.

Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
Distance covered in one liter of petrol = 55.3 km
Distance covered in 10 litres of petrol = (55.3 × 10) km = 553 km

Question 5.
Find:

1. 2.5 × 0.3
2. 0.1 × 51.7
3. 0.2 × 316.8
4. 1.3 × 3.1
5. 0.5 × 0.05
6. 11.2 × 0.15
7. 1.07 × 0.02
8. 10.05 × 1.05
9. 101.01 × 0.01
10. 100.01 × 1.1

Solution:

1. 2.5 × 0.3 = 0.75
2. 0.1 × 51.7 = 5 .17
3. 0.2 × 316.8 = 63.36
4. 1.3 × 3.1 = 4.03
5. 0.5 × 0.05 = 0 .025
6. 11.2 × 0.15 = 1.680
7. 1.07 × 0.02 = 0.0214
8. 10.05 × 1.05 = 10.5525
9. 101.01 × 0.01 = 1.0101
10. 100.01 × 1.1 = 110.011

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5.

• Fractions and Decimals Class 7 Ex 2.1
• Fractions and Decimals Class 7 Ex 2.2
• Fractions and Decimals Class 7 Ex 2.3
• Fractions and Decimals Class 7 Ex 2.4
• Fractions and Decimals Class 7 Ex 2.6
• Fractions and Decimals Class 7 Ex 2.7
• Fractions and Decimals Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimals
Exercise	Ex 2.5
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Which is greater?

1. 0.5 or 0.05
2. 0.7 or 0.5
3. 7 or 0.7
4. 1.37 or 1.49
5. 2.03 or 2.30
6. 0.8 or 0.88

Solution:

Question 2.
Express as rupees using decimals:

1. 7 paise
2. 7 rupees 7 paise
3. 77 rupees 77 paise
4. 50 paise
5. 235 paise.

Solution:

1. 7 paise = ₹ 0.07
2. 7 rupees 7 paise = ₹ 7.07
3. 77 rupees 77 paise = ₹ 77.77
4. 50 paise – ₹ 0.50
5. 235 paise = ₹ 2.35

Question 3.

1. Express 5 cm in metre and kilometre
2. Express 35 mm in cm, m and km?

Solution:

1. 5 cm = 0.05 m = 0.00005 km
2. 35 mm = 3.5 cm = 0.035 m = 0.000035 km

Question 4.
Express in kg:

1. 200 g
2. 3470 g
3. 4 kg 8 g

Solution:

1. 200 g = 0.200 kg = 0.2 kg
2. 3470 g = 3.470 kg
3. 4 kg 8 g = 4.008 kg.

Question 5.
Write the following decimal numbers in the expanded form:

1. 20.03
2. 2.03
3. 200.03
4. 2.034

Solution:

Question 6.
Write the place value of 2 in the following decimal numbers:

1. 2.56
2. 21.37
3. 10.25
4. 9.42
5. 63.352

Solution:
(i) Place value of 2 in the decimal number 2.56 = 2 × 1 = 2
(ii) Place value of 2 in decimal number 21.37 = 2 × 10 = 20
(iii) Place value of 2 in the decimal number

Question 7.
Dinesh went from place A to place t B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. f Who travelled more and by how much?
Solution:

Question 8.
Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Solution:
For Shyama
Apples bought = 5 kg 300 g = 5.300 kg
Mangoes bought = 3 kg 250 g = 3.250 kg
∴ Fruits bought = Apples bought
+ Mangoes bought = 5.300 kg + 3.250 kg

For Sarala
Oranges bought = 4 kg 800 g = 4.800 kg
Bananas bought = 4 kg 150 g = 4.150 kg
∴ Fruits bought = Oranges bought
+ Bananas bought

Question 9.
How much less is 28 km than 42.6 km?
Solution:

So, 28 km is less than 42.6 km by 14.6 km.

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4.

• Fractions and Decimals Class 7 Ex 2.1
• Fractions and Decimals Class 7 Ex 2.2
• Fractions and Decimals Class 7 Ex 2.3
• Fractions and Decimals Class 7 Ex 2.5
• Fractions and Decimals Class 7 Ex 2.6
• Fractions and Decimals Class 7 Ex 2.7
• Fractions and Decimals Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimals
Exercise	Ex 2.4
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4

Question 1.
Find:

Solution:

Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

Solution:

Question 3.
Find:

Solution:

Question 4.
Find:


Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3.

• Fractions and Decimals Class 7 Ex 2.1
• Fractions and Decimals Class 7 Ex 2.2
• Fractions and Decimals Class 7 Ex 2.4
• Fractions and Decimals Class 7 Ex 2.5
• Fractions and Decimals Class 7 Ex 2.6
• Fractions and Decimals Class 7 Ex 2.7
• Fractions and Decimals Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimals
Exercise	Ex 2.3
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3

Question 1.
Find:

Solution:

Question 2.
Multiply and reduce to lowest form (if possible):


Solution:

Question 3.
Multiply the following fractions:

Solution:

Question 4.
Which is greater:

Solution:

Question 5.
Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is \(\frac { 3 }{ 4 } \) m. Find the distance between the first and the last sapling.
Solution:

Let A, B, C and D be the four saplings planted in a row.
Distance between two adjacent saplings = \(\frac { 3 }{ 4 } \) m
∴ Distance between the first and the last sapling = AD = 3 × AB

Question 6.
Lipika reads a book for 1 \(\frac { 3 }{ 4 } \) hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:
Hours in all required by Lipika to read the book

Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 \(\frac { 3 }{ 4 } \) litres of petrol?
Solution:
Distance covered by the car using 2 \(\frac { 3 }{ 4 } \) litres of petrol

Question 8.

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2.

• Fractions and Decimals Class 7 Ex 2.1
• Fractions and Decimals Class 7 Ex 2.3
• Fractions and Decimals Class 7 Ex 2.4
• Fractions and Decimals Class 7 Ex 2.5
• Fractions and Decimals Class 7 Ex 2.6
• Fractions and Decimals Class 7 Ex 2.7
• Fractions and Decimals Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimals
Exercise	Ex 2.2
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2

Question 1.
Which of the drawings (a) to (d) show:

Solution:
(i) (d)
(ii) (b)
(iii) (a)
(iv) (c)

Question 2.
Some pictures (a) to (c) are given below. Tell which of them show:


Solution:

Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction:

Solution:

Question 4.
Shade

1. \(\frac { 1 }{ 2 } \) of the circles in box (a)
2. \(\frac { 2 }{ 3 } \) of the triangles in box (b)
3. \(\frac { 3 }{ 5 } \) of the squares in box (c)

Solution:

Question 5.
Find:

Solution:

Question 6.
Multiply and express as a mixed fraction:

Solution:

Question 7.
Find

Solution:

Question 8.
Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters of water. Vidya consumed \(\frac { 2 }{ 5 } \) of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Solution:
(i) Quantity of water drank by Vidya

(ii) Quantity of water drank by Pratap
= 5 litres – 2 litres = 3 litres
∴ The fraction of the total quantity of water drank by Pratap

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4.

• Integers Class 7 Ex 1.1
• Integers Class 7 Ex 1.2
• Integers Class 7 Ex 1.3
• Integers Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 1
Chapter Name	Integers
Exercise	Ex 1.4
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4

Question 1.
Evaluate each of the following:

(a) (-30)+ 10
(b) 50 + (-5)
(c) (-36) +(-9)
(d) (-49) + (49)
(e) 13 + [(- 2) + 1]
(f) 0 + (-12)
(g) (-31) + [(-30) + (-1)]
(h) [(-36)+ 12]+3
(i) [(- 6) + 5] + [(- 2) + 1].

Solution:

(a) (- 30) + 10 = – 3
(b) 50 +(-5) = – 10
(c) (-36) +(-9) = 4
(d) (- 49) + (49) = – 1
(e) 13 + [(- 2) + 1] = 13 + (- 1) = – 13
(f) 0 + (- 12) = 0
(g) (- 31) + [(- 30) + (- 1)] = (- 31) + (- 31) = 1
(h) [(- 36) + 12] + 3 = (- 3) + 3 = – 1
(i) [(- 6) + 5] + [(- 2) + 1] = (- 1) + (- 1) = 1.

Question 2.
Verify that
a + (b + c) ≠ (a + b) + (a ÷ c)
for each of the following values of a, b and c.
(a) a = 12, b = – 4, c = 2
(b) a = (- 10), b = 1, c = l.
Solution:
(a) a + (b + c) = 12 ÷ [(- 4) + 2] = 12 + (- 2) = – 6
(a ÷ b) + (a ÷ c) = 12 ÷ (- 4) + 12 ÷ 2 = -3 + 6 = 3
So, a + (b + c) ≠ (a + b) + (a + c)

(b) a ÷ (b + c) = (- 10) + (1 + 1) = (- 10) + 2 = – 5
a ÷ b + a ÷ c = (- 10) ÷ 1 + (- 10) ÷ 1 = (- 10) + (- 10) = – 20
So, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c).

Question 3.
Fill in the blanks:

(a) 369 ÷ …….. = 369
(b) -75 ÷ …….. = – 1
(c) (- 206) ÷ ……. = 1
(d) -87 ÷ …….. = 87
(e) ……. ÷ 1 = -87
(f) ……. ÷ 48 = -1
(g) 20 ÷ …… = -2
(h) …… ÷ (4) = – 3.

Solution:

(a) 369 ÷ 1 = 369
(b) – 75 ÷ 75 = -1
(c) (- 206) ÷ (- 206) = 1
(d) – 87 ÷ – 1 = 87
(e) – 87 ÷ 1 = – 87
(f) – 48 ÷ 48 = – 1
(g) 20 ÷ (-10) = – 2
(h) – 12 ÷ (4) = – 3.

Question 4.
Write five pairs of integers (a, b) such that a + b = -3. One such pair is (6, -2) because 6 +(-2) = (-3).
Solution:
Five pairs of integers (a, b) such that a + b = -3 are (- 6, 2), (-9, 3), (12,- 4), (21, -7), (-24, 8)
Note: We may write many such pairs of integers.

Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until mid-night, at what time would the temperature be 8°C degrees below zero? What would be the temperature at mid night?
Solution:
Difference in temperatures +10 °C and -8
= [10 – (- 8)] °C = (10 + 8)° C = 18 °C
Decrease in temperature in one hour = 2°C
Number of hours taken to have temperature 8 °C below zero \(=\frac { Total\quad decrease }{ Decrease\quad in\quad one\quad hour } \)
\(=\frac { 18 }{ 2 }\)
So, at 9 P.M., the temperature will be 8 °C below zero
Temperature at mid-night = 10 °C – (2 x 12) °C
= 10°C – 24 °C = -14 °C

Question 6.
In a class test (+3) marks are given for every correct answer and (- 2) marks are given for every incorrect answer and no marks for not attempting any question.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores – 5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
(i) Let ‘x’ be the number of incorrect questions attempted by Radhika.
According to the question, we get
(+ 3) × 12 + x × (-2) = 20
⇒ 36 – 2x = 20
⇒ 2x = 36 – 20
⇒ x = \(\frac { 16 }{ 2 } \) = 8
Therefore, Radhika attempted 8 incorrect questions.

(ii) Let ‘x’ be the number of incorrect question attempted by Mohini.
According to the question, we get
(+ 3) × 7 + x × (- 2) = – 5
⇒ 21 – 2x = -5
⇒ 2x = 21 + 5
⇒ x = \(\frac { 26 }{ 2 } \) = 13
Therefore, Mohini attempted 13 incorrect questions.

Question 7.
An elevator descends into a mine shaft at the rate of 6m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Solution:
Difference in heights at two positions = 10 m – (-350 m) = 360 m
Rate of descent = 6 m/minute
∴ Time taken \(=\left( 360 \right) \div \left( 6 \right)\) minutes = 60 minutes = 1 hour
Hence, the elevator will take 1 hour to reach = 350 m.

We hope the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4, drop a comment below and we will get back to you at the earliest.

 

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3.

• Integers Class 7 Ex 1.1
• Integers Class 7 Ex 1.2
• Integers Class 7 Ex 1.4
• Integers Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 1
Chapter Name	Integers
Exercise	Ex 1.3
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

Question 1.
Find each of the following products:
(a) 3 × (- 1)
(b) (- 1) × 225
(c) (-21) × (- 30)
(d) (- 316) × (- 1)
(e) (- 15) × 0 × (- 18)
(f) (- 12) × (- 11) × (10)
(g) 9 × (-3) × (-6)
(h) (- 18) ×(-5)× (- 4)
(i) (- 1) × (-2) × (-3) × 4
(j) (- 3) × (- 6) × (-2) × (- 1).
Solution:
(a) 3 x (- 1) = – (3 x 1) = – 3
(b) (- 1) x 225 = – (1 x 225) = – 225
(c) (- 21) x (- 30) = 21 x 30 = 630
(d) (- 316) x (- 1) = 316 x 1 = 316
(e) (- 15) x 0 x (- 18) = [(- 15) x 0]  x  (- 18) = 0 x (- 18) = 0
(f) (- 12) x (- 11) x (10) = [(- 12) x (- 11)] x (10) = (132) x (10) = 1320
(g) 9 x (- 3) x (- 6) = [9 x (- 3)] x (- 6) = (- 27) x (- 6) = 162
(h) (- 18) x (- 5) x (- 4) = [(- 18) x (- 5)] x (- 4) = 90 x (- 4) = – 360
(i) (- 1) x (- 2) x (- 3) x 4 = [(- 1) x (- 2)] x [(- 3) x 4] = (2)x (- 12) = -24
(j) (- 3) x (- 6) x (- 2) x (- 1) = [(- 3) x (- 6)] x [(- 2) x (- 1)] = (18) x (2) = 36

Question 2.
Verify the following:
(a) 18 × [7 + (- 3)] = [18 × 7] + [18 × (- 3)]
(b) (-21)×[(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)
Solution:
(a) 18 × [7 + (- 3)] = [18 × 7] + [18 × (- 3)]
L.H.S. = 18 × [7 + (- 3)]
= 18 × L(7 – 3)] = 18 × (4) = 18 × 4 = 72
R.H.S. = [18 × 7] + [18 × (- 3)]
= 126 + [- (18 × 3)] = 126 + (- 54) = 126 – 54 = 72
So, 18 × [7 + (- 3)]
= [18 × 7] + [18 × (- 3)]

(b) (- 21) × [(- 4) + (- 6)] = [(- 21) × (- 4)] + [(- 21) × (- 6)]
L.H.S. = (- 21) × [(- 4) + (- 6)]
= (- 21) × (- 10)
= 21 × 10 = 210
R. H.S. = [(- 21) × (- 4)] + [(- 21) × (- 6)]
= (21 × 4) + (21 × 6)
= 84 + 126 = 210
So, (- 21) × [(- 4) + (- 6)]
= [(- 21) × (- 4)] + [(- 21) × (- 6)].

Question 3.
(i) For any integer a, what is (-1)×a equal to?
(ii) Determine the integer whose product with (- 1) is
(a) – 22
(b) 37
(c) 0.
Solution:
(i) For any integer a, (-1) x a = -a.
(ii) We know that the product of any integer and (-1) is the additive inverse of an integer.
The integer whose product with (-1) is
(a) additive inverse of -22, t. e., 22.
(b) additive inverse of 37, i.e., -37.
(c) additive inverse of 0, i.e., 0.

Question 4.
Starting from (- 1) × 5, write various products showing some pattern to show (- 1) × (-1) – 1.
Solution:
(- 1) × 5 = – 5
(- 1) × 4 = – 4 [= (- 5) + 1]
(- 1) × 3 = – 3 [= (- 4) + 1]
(- 1) × 2 = – 2 [= (- 3) + 1]
(- 1) × 1 = – 1 [= (- 2) + 1]
(- 1) × 0 = 0 [= (- 1) + 1]
(- 1) × (- 1) = 1 [= 0 + 1]

Question 5.
Find the product, using suitable properties:
(a) 26 × (- 48) + (- 48) × (- 36)
(b) 8 × 53 × (- 125)
(c) 15×(-25)×(-4)×(- 10)
(d) (-41) × 102
(e) 625 × (-35) + (- 625) × 65
(f) 7 × (50 -2)
(g) (-17) × (-29)
(h) (- 57) ×(-19)+ 57.
Solution:
(a) We have, 26 x (-48) + (- 48) x (- 36)
= (- 48) x 26 + (- 48) x (- 36)
= (- 48) x [26 + (- 36)]
= (- 48) x (26 – 36)
=(- 48) x (- 10)= 480
(b) We have,
8 x 53 x (- 125) = [8 x (- 125)] x 53
= (- 1000) x 53 = – 53000
(c) We have,
15 x (- 25) x (- 4) x (- 10)
=15 x [(- 25) x (-4)] x (- 10)
= 15 x (100) x (- 10)
= (15 x 100) x (- 10)
= 1500 x (- 10) = – 15000
(d) We have,
(- 41) x 102 = (- 41) x (100 + 2)
= (- 41) x 100 + (- 41) x 2 = -4100 – 82 = – 4182
(e) We have, 625 x (- 35) + (- 625) x 65
= 625 x (- 35) + (625) x (- 65)
= 625 x [(- 35)+ (- 65)]
= 625 x (- 100) = – 62500
(f) 7 x (50 – 2) = 7 x 50 – 7 x 2
= 350 -14 =336
(g) (-17) x (- 29) = (-17) x [(- 30) + 1]
= (- 17) x (- 30) + (- 17) x 1 = 510 – 17 = 493
(h) (- 57) x (-19)+ 57 =57 x 19 + 57 x 1
= 57 x (19 +1)
= 57 x 20 = 1140

Question 6.
A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5° C every hour. What will be the room temperature 10 hours after the process begins?
Solution:
Room temperature 10 hours after the process begins
= 40°C – 10 × 5°C
= 40°C – 50°C
= – (50 – 40)°C = – 10°C

Question 7.
In a class test containing 10 questions, 5 marks are awarded for every correct answer and (-2) marks are awarded for every incorrect answer and 0 for questions not attempted.
(i) Mohan gets four correct and si× incorrect answers. What is his score?
(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
Solution:
(i) Mohan gets for four correct answers 4 × 5 = 20 marks
He also gets for si× incorrect answers. 6 × (- 2) = – 12 marks.
Therefore, Mohan’s score = 20 + (- 12) = 20-12 = 8 marks.

(ii) Reshma gets for five correct answers 5 × 5 = 25 marks
She also gets for five incorrect answers 5 × (- 2) = – 10 marks Therefore, Reshma’s score = 25 + (- 10) = 25-10 = 15 marks.

(iii) Heena gets for two correct answers
2 × 5 = 10 marks.
She also gets for five incorrect answers 5 × (- 2) = – 10 marks
She didn’t attempt three questions. For these, she gets 3×0 = 0 marks
Therefore, Heena’s score = 10 + (- 10) + 0 = 10 – 10 + 0 = 0 marks.

Question 8.
A cement company earns a profit of ₹ 8 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold.
(a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?
(b) What is the number of white cement bags it must sell to have neither profit nor loss if the number of grey bags sold is 6,400 bags.
Solution:
(a) The company sells 3,000 bags of white cement. So her profit = 3,000 × 8 = ₹ 24,000
Also, the company sells 5,000 bags of grey cement. So her loss = 5,000 × 5 = ₹ 25,000
Since 25,000 > 24,000
Therefore, the company is at a loss and the loss is = 25000 – 24000 = ₹ 1000

(b) Let ‘×’ be the number of white cement bags sold.
According to the question, we get
x × 8 = 6400 × 5
⇒ x = \(\frac { 6400\times 5 }{ 8 }\) = 800 × 5 = 4,000 bags.
Therefore, 4,000 bags of white cement must be sold to have neither profit nor loss.

Question 9.
Replace the blank with an integer to make it a true statement.

1. (a) (- 3) × …….. = 27
2. (b) 5 × …….. = -35
3. (c) …….. × (- 8) = – 56
4. (d) …….. × (- 12) = 132.

Solution:

1. (a) (-3) x (- 9) = 27
2. (b) 5 x (-7) = (-35)
3. (c) 7 x (-8) = (-56)
4. (d) (-11) x (-12) = 132

We hope the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3, drop a comment below and we will get back to you at the earliest.