NCERT Solutions for Class 7 Maths Chapter 1 Integers in Video

NCERT Solutions for Class 7 Maths Chapter 1 Integers in Video format is given below updated for new academic session 2024-25. Description of all questions given below the videos. We are using the latest NCERT Books for current academic session 2024-25.

7 Maths Chapter 1 Exercise 1.1 Question 1

Here, using this video, we shall learn how to use number line. First of all, we shall observe the temperature of different cities given on number line. Just by subtracting the temperatures of Lahulspiti and Benguluru, we can find the temperature difference between the hottest and the coldest places. Similarly using the same method, the temperature difference between Lahulspiti and Srinagar can be obtained. After adding the temperatures of Srinagar and Shimla together, we find that it is not less than the temperature at Shimla. The video describes this completely this fact.

7 Maths Chapter 1 Exercise 1.1 Question 2, 3, 4, 5 & 6

Class 7 Maths Exercise 1.1 Questions 2

We shall add negative and positive marks separately. It means we shall add positive numbers 25 + 15 + 10 = 50 and negative numbers – 5 + (– 10) = – 15. Now to get his total at the end subtract 15 form 50 that is 50 – 15 = 35

Class 7 Maths Exercise 1.1 Questions 3

The temperature is dropped by 2°C on Tuesday. So, the temperature will reduce by 2°C from Monday. Therefore, the temperature on Tuesday = – 5°C – 2°C = – 7°C. But on Wednesday, it rose by 4°C, so the temperature on Wednesday = – 7°C + 2°C = – 5°C.

Class 7 Maths Exercise 1.1 Questions 4

To get the vertical distance between plane and submarine, we add the two distances. It means the distance between them = 5000 + 1200 = 6200 m.

Class 7 Maths Exercise 1.1 Questions 5

The balance in Mohan’s account after the withdrawal will the difference between deposits and withdrawal = ₹2000 – ₹1642 = ₹358.

Class 7 Maths Exercise 1.1 Questions 6

According to number line take East direction positive and west direction as negative. Rita goes 20 km towards east from a point A to the point B. It means Rita travels +20 km. Now from B, she moves 30 km towards west along the same road. It shows that she has travelled – 30 km. The final position of Rita = +20 km – 30 km = – 10 km. Therefore, she is now 10 km towards east from A.

Class 7 Maths Exercise 1.1 Questions 7 and 8

Class 7 Maths Exercise 1.1 Questions 7

Taking the box 1, if we add all the integers of row or column, we get the total as 0. In the second box, doing the same thing, we get the total of integers is – 9.

Class 7 Maths Exercise 1.1 Questions 8

In Part (i), to verify the equation, put a = 21 and b = 18 on LHS and RHS. Both the sides, we receive the total of 39. This shows that it is verified. Similarly, in Part (ii), put a = 118 and b = 125, we get LHS and RHS both are same and equal to 243. In Part (iii), put a = 75 and b = 84, we observe that LHS = RHS = 159. In the last Part (iv), put a = 28, b = 11 and repeat the same process. Here, also the LHS = RHS = 39. Hence, by equating LHS to RHS, we can verify all such equation.

Class 7 Maths Exercise 1.1 Questions 9

If LHS is less than RHS, we have to use < sign or if LHS is greater than RHS use > sign. When LHS and RHS are equal, = sign in Question 9 of Exercise 1.1 Class 7 Maths. In Part (i), LHS is – 12 and RHS is – 4. So LHS < RHS. Similarly, we can do the other parts just by comparing LHS and RHS. Always remember that the number which is left side on number line will be lesser.

Class 7 Maths Exercise 1.1 Questions 10

In Part (i), he jumps 3 steps down and then jumps back 2 steps up. It means in two jumps; he moves down just one step only. So, in this way he will move down only five steps in next 10 jumps and will reach to the 6th step. Now in next jump he will reach to the water. Therefore, he will take eleven jumps to reach the water.
In Part (ii), after drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. It means in two jumps; he moves up just two steps. So, in this way he will move up only 4 steps in next four jumps. Now he is at step 5. Therefore, in the fifth jump he will reach to the top.
In Part (iii), if the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, then the first part will be as follows: – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = – 8. Second part will be as follows: 4 – 2 + 4 – 2 + 4 = 8.

7 Maths Chapter 1 Exercise 1.2 Question 1, 2, 3, 4

In Exercise 1.2 of Class 7 Maths Question 1, we have to write down a pair of integers whose sum is – 7. There are so many such pairs. Like 2 + (– 9), 5 + (– 12), – 3 + (– 4), 0 + (– 7), etc. Similarly, to get the difference – 10, we can take integers like 1 – 11, 5 – 15, 3 – 13, etc. Now to get the sum zero, we have to add equal negative and positive number, for example, – 5 + 5, – 2 + 2, – 8 + 8, etc.
In Exercise 1.2 of Class 7 Maths Question 2, we have to write a pair of negative integers whose difference gives 8. There are infinite such pairs whose difference is 8. For example, 10 – 2, 12 – 4, 18 – 10, 9 – 1, etc. In case of second part, we have to write a negative integer and a positive integer whose sum is –5. For this we must take negative number greater than positive number. For example, – 10 + 5, – 12 + 7, – 7 + 2, – 6 + 1, etc. In the third part, we have to write a negative integer and a positive integer whose difference is – 3. For this, we can take + 10 – 13, + 7 – 10, + 2 – 5, + 12 – 15, etc.
In Exercise 1.2 of Class 7 Maths Question 3, we have to find that which team scored more in the quiz. Adding the total scores of Team-A is – 40 + 10 + 0 = – 30 whereas the total score of Team-B is 10 + 0 – 40 = – 30. Finally, both the team have same score. Here, both the teams are scoring same integer points in different order but there total is same. So, we can say that integers can be added in any order.
In Exercise 1.2 of Class 7 Maths Question 4, we have to make the statements true by filling the blanks using suitable integers. Get the total of integers on both the sides and then compare for decision.

7 Maths Chapter 1 Exercise 1.3 Question 1, 2, 3

In Exercise 1.3 of Class 7 Maths Question 1, we have to find the product of two integers. If we are multiplying two negative integers, the result will be positive. It means multiplication of 2 or 4 (even numbers) of negative integers gives positive result and multiplication of 1 or 3 (odd numbers) of negative integers give result negatives irrespective of the number of positive integers.
In Exercise 1.3 of Class 7 Maths Question 2, we have to find LHS and RHS separately and then show that both are equal. For example, the LHS of first part is 18 x 4 = 72 and the RHS is 126 – 54 = 72. Similarly, we can do the second part also.
In Exercise 1.3 of Class 7 Maths Question 3, we have to find the result of (–1) × a. If a is negative integer, the result will be positive value of a, if a is zero, the result also will be zero and otherwise in case a is positive integer the result will be negative.

7 Maths Chapter 1 Exercise 1.3 Question 4, 5

In Exercise 1.3 of Class 7 Maths Question 4, we have to multiply (–1) with 5, 4, 3, 2, 1, 0 and (–1). Here we will observe that the product of one negative and one positive number results in negative whereas the product of two negative number is positive.
In Exercise 1.3 of Class 7 Maths Question 5, we have to apply distributive property of multiplication over addition and commutative of multiplication to simplify the results.

7 Maths Chapter 1 Exercise 1.3 Question 6, 7, 8, 9

In Exercise 1.3 of Class 7 Maths Question 6, the room temperature after 1 hours be 40°C – 5°C = 35°C. Similarly, after 10 hours, the temperature = 40°C – 10 × (5°C) = – 10°C.
In Exercise 1.3 of Class 7 Maths Question 7, here 5 marks are awarded for every correct answer and (–2) marks are awarded for every incorrect answer, therefore Mohan’s score = 4 × 5 – 6 × 2 = 20 – 12 = 8. Similarly, Reshma’s score = 5 × 5 – 5 × 2 = 25 – 10 = 15, and Heena’s score = 2 × 5 – 5 × 2 = 10 – 10 = 0.
In Exercise 1.3 of Class 7 Maths Question 8, we have total profit = ₹8 × 3000 = ₹24000 and total loss = ₹5 × 5000 = ₹25000. Here, loss is more than profit. So, finally company has loss of ₹25000 – ₹24000 = ₹1000. Now loss on 6400 bags of grey cement = ₹5 × 6400 = ₹32000. For no profit no loss, he has to get profit of ₹32000. So, the number of bags of white cement = 32000 ÷ 8 = 4000.
In Exercise 1.3 of Class 7 Maths Question 9, we have to divide the RHS, from the integer given on LHS to get the integer for the blank.