NCERT Solutions Class 6 Maths Chapter 10 The Other Side of Zero Exercise 10.2

Class 6 Maths Chapter 10 – The Other Side of Zero introduces positive and negative and how to compare and place them in some order.Exercise 10.2 is the exercise that focuses on comparing integers, and with this help students will develop confidence placing integers on the number line in their correct places.

NCERT Solutions Class 6 Maths Chapter 10 The Other Side of Zero Exercise 10.2, prepared with guidance from institutions like Allen, give you a step-by-step highlight to show how the writing of examples can make learning to work with integers easy and fun. So go on and learn with confidence when using NCERT Solutions for strengthening your understanding and making corpus, no more likely than they would make per region, to do well in term exams of mathematics.

1.0Download NCERT Solutions of Class 6 Maths Chapter 10 The Other Side of Zero Exercise 10.2: Free PDF

Download NCERT Solutions for Class 6 Maths Chapter 10 The Other Side of Zero Exercise 10.2 in free PDF file format.

2.0Key Concepts of Exercise 10.2

• Ordering and Comparing Integers: Identifying what integer is greater than or less than by using the number line (numbers furthest to the right are greater).
• Adding Integers on a Number Line: When adding integers, on a number line, you visualize the addition process by moving right (for positive integers) or left (for negative integers) on the number line.
• Subtracting Integers on a Number Line: Subtraction is addition, but instead of moving right along the number line, you are adding the opposite (the negative), also visualized on the number line.
• Additive Inverse: Reinforcing that opposites (additive opposite) sum to zero.
• Absolute Value (Distance from Zero): The distance from zero (the number line) is what is important to understand when we talk about the value of the integer (whether it is positive or negative).

3.0NCERT Exercise Solutions Class 6 Maths Chapter 10 The Other Side of Zero : All Exercises 

4.0NCERT Class 6 Maths Chapter 10 Exercise 10.2 : Detailed Solutions

• Complete the additions using tokens. a. $(+6)+(+4)$ b. $(-3)+(-2)$ c. $(+5)+(-7)$ d. $(-2)+(+6)$ Sol. a. To represent (+6), we use 6 positive tokens
  
  and to represent (+4), we use 4 positive tokens
  
  Hence, by combining them, we get
  
  $\Rightarrow (+6)+(+4)=(+10)$ $\therefore$ Counting all the positive tokens, we get $(+10)$. b. To represent (-3), we use 3 negative tokens
  
  To represent ( -2 ), we use 2 negative tokens
  
  Combining them, we get
  
  $\Rightarrow (-3)+(-2)=-5$ $\therefore$ Counting all the negative tokens, we get (-5). c. To represent (+5), we use 5 positive tokens
  
  To represent (-7), we use 7 negative tokens
  
  $\therefore$ Remaining tokens $=(+5)+(-7)=-2$
  
  d. To represent (-2), we use 2 negative tokens
  
  To represent (+6), we use 6 positive tokens
  
  Hence,
  
  $(-2)+(+6)=+4$ Counting the remaining tokens, we get
  
  $(+4)$ tokens
• Cancel the zero pairs in the following two sets of tokens. On what floor is the lift attendant in each case? What is the corresponding addition statement in each case?
  
  Sol. a. From the picture we see that we can remove three pairs.
  
  This cancels out to -2 . Remaining tokens =
  
  Since two negative tokens is remaining, the lift attendant is on the second floor below the ground floor. The corresponding addition statement is $(+3)+(-5)=(-2)$ b. From the picture we see that we can remove three pairs.
  

This cancels out to +3. Remaining tokens =

• Evaluate the following differences using tokens. Check that you get the same result as with other methods you now know: a. $(+10)-(+7)$ b. $(-8)-(-4)$ c. $(-9)-(-4)$ d. $(+9)-(+12)$ e. $(-5)-(-7)$ f. $(-2)-(-6)$

Sol. a. Here, from 10 positives take away 7 positives.

• Complete the subtractions: a. $(-5)-(-7)$ b. $(+10)-(+13)$ c. $(-7)-(-9)$ d. $(+3)-(+8)$ e. $(-2)-(-7)$ f. $(+3)-(+15)$ Sol. a. There are not enough tokens to take out -7 negatives from -5 negatives. So, we put an extra two zero pairs (1 pair = a positive and a negative). Now we can take out -7 negatives.
  
  Hence, $(-5)-(-7)=+2$ b. There are not enough tokens to take out 13 positives from 10 positives. So, we put an extra 3 zero pairs ( 1 pair = a positive and a negative). Now we can take out 13 positives.
  
  Hence, (+10) - (+13) = -3 c. There are not enough tokens to take out -9 negatives from -7 negatives. So, we put an extra two zero pairs (1 pair = a positive and a negative). Now we can take out -9 negatives.
  
  Hence, (-7) - (-9) = +2 d. There are not enough tokens to take out 8 positives from 3 positives. So, we put an extra 5 zero pairs ( 1 pair = a positive and a negative). Now we can take out 8 positives.
  
  Hence, $(+3)-(+8)=-5$ e. There are not enough tokens to take out -7 negatives from -2 negatives. So, we put an extra two zero pairs (1 pair = a positive and a negative). Now we can take out -7 negatives.
  
  Hence, (-2) - (-7) = +5 f. There are not enough tokens to take out 15 positives from 3 positives. So, we put an extra 12 zero pairs ( 1 pair = a positive and a negative). Now we can take out 15 positives.
  
  Hence, (+3) - (+15) = -12
• Try to subtract: $-3-(+5)$. How many zero pairs will you have to put in? What is the result? Sol. We have to take out 5 positives from -3 negatives. But there are not enough positives. So, we put down an extra 5 zero pairs ( 1 pair = a positive and a negative). Now we can take out 5 positives.
  
  Hence, $-3-(+5)=-8$
• Evaluate the following using tokens. a. $(-3)-(+10)$ b. $(+8)-(-7)$ c. $(-5)-(+9)$ d. $(-9)-(+10)$ e. $(+6)-(-4)$ f. $(-2)-(+7)$ Sol. a. $(-3)-(+10)$ We have to take out 10 positives from -3 negatives. But there are not enough positives. So, we put down extra 10 zero pairs ( 1 pair = a positive and a negative). Now we can take out 10 positives.
  
  Hence, (-3) - (+10) = -13 b. $(+8)-(-7)$ We have to take out -7 negatives from 8 positives. But there are not enough negatives. So, we put down an extra 7 zero pairs (1 pair = a positive and a negative). Now we can take out -7 negatives.
  
  Hence, $(+8)-(-7)=15$ c. $(-5)-(+9)$ We have to take out 9 positives from -5 negatives. But there are not enough positives. So, we put down an extra 9 zero pairs (1 pair = a positive and a negative). Now we can take out 9 positives.
  
  Hence, $-5-(+9)=-14$ d. $(-9)-(+10)$ We have to take out 10 positives from -9 negatives. But there are not enough positives. So, we put down an extra 10 zero pairs ( 1 pair $=$ a positive and a negative). Now we can take out 10 positives.
  
  Hence, $-9-(+10)=-19$ e. $(+6)-(-4)$ We have to take out -4 negatives from 6 positives. But there are not enough negatives. So, we put down an extra 4 zero pairs (1 pair = a positive and a negative). Now we can take out -4 negatives.
  
  Hence, (+6) - (-4) = 10 f. $(-2)-(+7)$ We have to take out 7 positives from -2 negatives. But there are not enough positives. So, we put down an extra 7 zero pairs (1 pair = a positive and a negative). Now we can take out 7 positives.
  
  Hence, (-2) - (+7) = -9

5.0Key Features and Benefits Class 6 Maths Chapter 10 The Other Side of Zero : Exercise 10.2

• More In-Depth Understanding of Integer Operations: The provided solutions go a long way to clarifying the steps we take when we add and subtract integers, especially with the context of using number lines.
• Better Comparison Skills: Students become skilled in being able to compare and order both positive and negative integers in a correct manner.
• Learning Aid: Using the number lines as models in the proposed solutions allows for a better conceptual and visual understanding of how integer operations work.
• Problem Solving Application: Students demonstrate their application of integers when solving real-life problems they encounter in this exercise.
• Foundation for Algebra: Command of the operations on integers will be the foundation for the next couple of topics, namely applying algebraic properties and rules.
• Exam Preparation: The solutions offered by trusted sources, like the Allen solutions, are meticulously prepared with the CBSE curriculum in mind to aid in students' effective exam preparation.