NCERT Solutions for Class 7 Maths Chapter 1
Class 7 Maths Chapter 1 all Exercises Solution
| Class: 7 | Maths (English and Hindi Medium) |
| Chapter 1: | Integers |
NCERT Solutions App for Class 7
7 Maths Chapter 1 Integers Solutions
NCERT Books for class 7 and CBSE Solutions 2021-22 for other subjects are also available for free download. We have prepared the Solutions in the simplified format so that students can understand easily. 7 Maths Chapter 1 solutions are given below in Hindi and English Medium.
7 Maths Chapter 1 All Exercises in English Medium
7 Maths Chapter 1 All Exercises in Hindi Medium
Class 7 Maths Chapter 1 Solution in Video
Class 7 Maths Exercise 1.1 Solution in VideoClass 7 Maths Exercise 1.2 Solution in VideoClass 7 Maths Exercise 1.3 Solution in VideoClass 7 Maths Exercise 1.4 Solution in Video
About NCERT Solutions for Class 7 Maths Chapter 1
In 7 Maths Chapter 1 Integers, we will explore all the operations based on integer properties. Properties of integers on the operations like addition, subtraction, multiplication and division. We have to learn about all the properties like closure, commutative, associative, etc. Integers are Closure under Addition but not under subtraction.
Class 7 Maths Chapter 1 Important Questions for Practice
- When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient. That is, we get a negative integer.
- The product of three integers does not depend upon the grouping of integers and this is called the associative property for multiplication of integers.
- When we divide a negative integer by a positive integer, we divide them as whole numbers and then put a minus sign (–) before the quotient. We, thus, get a negative integer.
- Addition is associative for integers, i.e., (a + b) + c = a + (b + c) for all integers a, b and c.
- Integer 0 is the identity under addition. That is, a + 0 = 0 + a = a for every integer a.
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We have prepared Hindi Medium NCERT Solutions for class 7 Maths on the demand of students. Now it is online on website to view online as well as for download. Time to time we modify our website on the basis of student’s suggestions. That is why Class 7 Maths App in English or Kaksha 7 Ganit App in Hindi Medium are developed for offline use.
Class 7 Maths Chapter 1 Important Questions
At Srinagar temperature was C on Monday and then it dropped by C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by C. What was the temperature on this day?
On Monday, temperature at Srinagar = –5 oC
On Tuesday, temperature dropped = 2 oC
Temperature on Tuesday = –5 oC – 2 oC = –7 oC
On Wednesday, temperature rose up = 4oC
Temperature on Wednesday = –7 oC + 4 oC = –3 oC
Thus, temperature on Tuesday and Wednesday was –7 oC and –3 oC respectively.
A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?
Height of a place above the sea level = 5000 m
Floating a submarine below the sea level = 1200 m
The vertical distance between the plane and the submarine = 5000 + 1200 = 6200 m
Thus, the vertical distance between the plane and the submarine is 6200 m.
Mohan deposits ₹2,000 in his bank account and withdraws ₹1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s accounts after the withdrawal?
Deposit amount = ₹2,000 and
Withdrawal amount = ₹1,642
Balance = 2,000 – 1,642 = ₹358
Thus, the balance in Mohan’s account after withdrawal is ₹ 358.
A certain freezing process requires that room temperature be lowered from 40 oC at the rate of 5 oC every hour. What will be the room temperature 10 hours after the process begins?
Given:
Present room temperature = 40 oC
Decreasing the temperature every hour = 5 oC
Room temperature after 10 hours = 40 oC + 10 x (–5 oC ) = 40 oC – 50 oC = – 10 oC
Thus, the room temperature after 10 hours is – 10 oC after the process begins.
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. 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?
Given:
Profit of 1 bag of white cement = ₹ 8
And Loss of 1 bag of grey cement = ₹ 5
Profit on selling 3000 bags of white cement = 3000 x ₹ 8 = ₹ 24,000
Loss of selling 5000 bags of grey cement = 5000 x ₹ 5 = ₹ 25,000
Since Profit
On Monday, temperature at Srinagar = –5 oC
On Tuesday, temperature dropped = 2 oC
Temperature on Tuesday = –5 oC – 2 oC = –7 oC
On Wednesday, temperature rose up = 4oC
Temperature on Wednesday = –7 oC + 4 oC = –3 oC
Thus, temperature on Tuesday and Wednesday was –7 oC and –3 oC respectively. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “suggestedAnswer”: {“@type”: “Answer”, “upvoteCount”: “1”, “text”: “
On Monday, temperature at Srinagar = –5 oC
On Tuesday, temperature dropped = 2 oC
Temperature on Tuesday = –5 oC – 2 oC = –7 oC
On Wednesday, temperature rose up = 4oC
Temperature on Wednesday = –7 oC + 4 oC = –3 oC
Thus, temperature on Tuesday and Wednesday was –7 oC and –3 oC respectively. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}]}]]>
Height of a place above the sea level = 5000 m
Floating a submarine below the sea level = 1200 m
The vertical distance between the plane and the submarine = 5000 + 1200 = 6200 m
Thus, the vertical distance between the plane and the submarine is 6200 m. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “suggestedAnswer”: {“@type”: “Answer”, “upvoteCount”: “1”, “text”: “
Height of a place above the sea level = 5000 m
Floating a submarine below the sea level = 1200 m
The vertical distance between the plane and the submarine = 5000 + 1200 = 6200 m
Thus, the vertical distance between the plane and the submarine is 6200 m. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}]}]]>
Deposit amount = ₹2,000 and
Withdrawal amount = ₹1,642
Balance = 2,000 – 1,642 = ₹358
Thus, the balance in Mohan’s account after withdrawal is ₹ 358. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “suggestedAnswer”: {“@type”: “Answer”, “upvoteCount”: “1”, “text”: “
Deposit amount = ₹2,000 and
Withdrawal amount = ₹1,642
Balance = 2,000 – 1,642 = ₹358
Thus, the balance in Mohan’s account after withdrawal is ₹ 358. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}]}]]>
Given:
Present room temperature = 40 oC
Decreasing the temperature every hour = 5 oC
Room temperature after 10 hours = 40 oC + 10 x (–5 oC ) = 40 oC – 50 oC = – 10 oC
Thus, the room temperature after 10 hours is – 10 oC after the process begins. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “suggestedAnswer”: {“@type”: “Answer”, “upvoteCount”: “1”, “text”: “
Given:
Present room temperature = 40 oC
Decreasing the temperature every hour = 5 oC
Room temperature after 10 hours = 40 oC + 10 x (–5 oC ) = 40 oC – 50 oC = – 10 oC
Thus, the room temperature after 10 hours is – 10 oC after the process begins. “, “dateCreated”: “January 28, 2020”, “url”: “https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/”, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}, “author”: {“@type”:”Person”,”name”:”Rakesh Tiwari” }}]}]]>
Given:
Profit of 1 bag of white cement = ₹ 8
And Loss of 1 bag of grey cement = ₹ 5
Profit on selling 3000 bags of white cement = 3000 x ₹ 8 = ₹ 24,000
Loss of selling 5000 bags of grey cement = 5000 x ₹ 5 = ₹ 25,000
Since Profit < Loss Therefore, his total loss on selling the grey cement bags = Loss – Profit = ₹ 25,000 – ₹ 24,000 = ₹ 1,000 Thus, he has lost of `₹1,000 on selling the grey cement bags. ", "dateCreated": "January 28, 2020", "url": "https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/", "author": {"@type":"Person","name":"Rakesh Tiwari" }}, "suggestedAnswer": {"@type": "Answer", "upvoteCount": "1", "text": "
Given:
Profit of 1 bag of white cement = ₹ 8
And Loss of 1 bag of grey cement = ₹ 5
Profit on selling 3000 bags of white cement = 3000 x ₹ 8 = ₹ 24,000
Loss of selling 5000 bags of grey cement = 5000 x ₹ 5 = ₹ 25,000
Since Profit
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