NCERT Solutions Class 6 Maths Chapter 10 The Other Side of Zero Exercise 10.1

Class 6's Maths Chapter 10 - The Other Side of Zero- begins with introducing negative numbers and how we show them on the number line. The first exercise (10.1) asks students to understand these numbers in real-life contexts - such as dating something below zero, debts or to indicate something that is notated as being below a certain starting point.

NCERT Solutions Class 6 Maths Chapter 10, The Other Side of Zero, Exercise 10.1, will make learning about negative numbers easy and fun. With caring help from the best educators such as those in Allen, the solutions explain every problem using easy language ingredients and with clear examples to help the student master and get over this brand new concept.

1.0Download NCERT Solutions of Class 6 Maths Chapter 10 The Other Side of Zero Exercise 10.1: Free PDF

Let's access the NCERT Solutions for "Class 6 Maths Chapter 10, The Other Side of Zero, Exercise 10.1." The solutions present clear meaning which will help students grasp the basics of integers and their representation.

2.0Key Concepts of Exercise 10.1

• Introduction to Integers: Integers can be surfaced as a group of whole numbers and negative numbers.
• Real world examples of integers: Depicting situations using positive and negative signs (Profits/Losses, Above/Below Sea level or change in temperature).
• Representation on a Number Line: Locating and identifying integers on a number line, noting that the number line has numbers you represent to the left of zero (Negative integers), and right of zero are positive integers.
• Ordering of Integers: Helps to remember (greater than, less than) via the number line
• Opposites: Every positive integer has a negative integer which is its opposite and vice versa.

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

4.0NCERT Class 6 Maths Chapter 10 Exercise 10.1 : Detailed Solutions

• You start from Floor +2 and press - 3 in the lift. Where will you reach? Write an expression for this movement. Sol. Starting from Floor +2 and pressing -3 means moving down 3 floors. Expression: (2) $+(-3)=-1$ You will reach Floor -1.
• Evaluate these expressions (you may think of them as Starting Floor + Movement by referring to the Building of Fun). a. $(+1)+(+4)=$ b. $(+4)+(+1)=$ c. $(+4)+(-3)=$ d. $(-1)+(+2)=$ e. $(-1)+(+1)=$ f. $0+(+2)=$ g. $0+(-2)=$ Sol. a. $(+1)+(+4)=+5$ b. $(+4)+(+1)=+5$ c. $(+4)+(-3)=+1$ d. $(-1)+(+2)=+1$ e. $(-1)+(+1)=0$ f. $0+(+2)=+2$ g. $0+(-2)=-2$
• Starting from different floors, find the movements required to reach Floor - 5. For example, if I start at Floor +2 , I must press -7 to reach Floor -5 . The expression is $(+2)+(-7)=-5$. Find more such starting positions and the movements needed to reach Floor - 5 and write the expressions. Sol. (a) If I start at floor +1 , I must press ( -6 ) to reach floor -5. The expression is $(1)+(-6)=-5$. (b) If I start at floor +3 , I must press ( -8 ) to reach the floor ( -5 ). The expression is $(+3)+(-8)=-5$.

Evaluate these expressions by thinking of them as the resulting movement of combining button presses: a. $(+1)+(+4)=$ b. $(+4)+(+1)=$ c. $(+4)+(-3)+(-2)=$ d. $(-1)+(+2)+(-3)=$ Sol. a. Target floor $=(+1)+(+4)=+5$ b. Target floor $=(+4)+(+1)=+5$ c. Target floor $=(+4)+(-3)+(-2)$ $=4+(-5)=-1$ d. Target floor $=(-1)+(+2)+(-3)$ $=(-4)+(2)=-2$

• Compare the following numbers using the Building or Fun and fill in the boxes with < or >. a. -2 $+5$ b. -5 $+4$ c. -5 -3 d. +6 -6 e. 0 $-4$ f. 0 $+4$ Notice that all negative number floors are below Floor 0 . So, all negative numbers are less than 0 . All the positive number floors are above 0 . So, all positive numbers are greater than 0 . Sol. a. Here, floor -2 is lower than floor +5 then $-2<+5$. b. Here, floor -5 is lower than floor +4 then $-5<+4$. c. Here, floor -5 is lower than floor -3 then $-5<-3$. d. Here, floor +6 is greater than floor -6 then $+6>-6$. e. Here, floor 0 is greater than floor -4 then $0>-4$. f. Here, floor 0 is lower than floor +4 then $0<+4$.
• Imagine the Building of Fun with more floors. Compare the numbers and fill in the boxes with < or >. a. $-10\square -12$ b. $+17\square -10$ c. $0\square -20$ d. $+9\square -9$ e. -25 $-7$ f. +15 -17 Sol. a. Here, floor -10 is greater than floor -12 then $-10>-12$ b. Here, floor +17 is greater than floor -10 then $+17>-10$ c. Here, floor 0 is greater than floor -20 then $0>-20$ d. Here, floor +9 is greater than floor -9 then $+9>-9$ e. Here, floor -25 is lower than floor -7 then $-25<-7$ f. Here, floor +15 is greater than floor -17 then $+15>-17$
• If Floor $\mathrm{A}=-12$, Floor $\mathrm{D}=-1$ and Floor $\mathrm{E}=+1$ in the building shown on the right as a line, find the numbers of Floors B, C, F, G, and H. Sol. Let's mark the numbers on the line. $0,-1,-2,-3$,............, -12 and $1,2,3$,.........., 12 Now count each floor, and we get the number of floors
  
  Here, $\mathrm{B}=-9,\mathrm{C}=-6,\mathrm{F}=+2,\mathrm{G}=+6$ and $\mathrm{H}=+11$
• Mark the following floors of the building shown on the right. a. -7 b. -4 c. +3 d. -10 Sol. Here, one line is considered as one floor. The floor above floor 0 is marked with positive numbers and the floor below 0 is numbered with negative numbers.. Now let's mark the following numbers ( $-7,-4,+3,-10$ ) in the building by circling the floor.
  

• Complete these expressions. You may think of them as finding the movement needed to reach the Target Floor from the Starting Floor. (a) $(+1)-(+4)=$ (b) $(0)-(+2)=$ (c) $(+4)-(+1)=$ (d) $(0)-(-2)=$ (e) $(+4)-(-3)=$ (f) $(-4)-(-3)=$ (g) $(-1)-(+2)=$ (h) $(-2)-(-2)=$ (i) $(-1)-(+1)=$ (j) $(+3)-(-3)=$ Sol. (a) $(+1)-(+4)=\underline{-3}$ (b) $(0)-(+2)=\underline{-2}$ (c) $(+4)-(+1)=\underline{+3}$ (d) $(0)-(-2)=\underline{+2}$ (e) $(+4)-(-3)=\underline{+7}$ (f) $(-4)-(-3)=\underline{-1}$ (g) $(-1)-(+2)=\underline{-3}$ (h) $(-2)-(-2)=\underline{0}$ (i) $(-1)-(+1)=\underline{-2}$ (j) $(+3)-(-3)=\underline{+6}$
• Complete these expressions. a. $(+40)+$ $=+200\mathrm{b}.(+40)+$ $=-200$ c. $(-50)+$ $=+200\mathrm{d}.(-50)+$ $=-200$ e. $(-200)-(-40)=$ f. $(+200)-(+40)=$ g. $(-200)-(+40)=$ Check your answers by thinking about the movement in the mineshaft. Sol. a. Given (+40) + = +200 Let $(+40)+\mathrm{x}=+200$ $\Rightarrow +\mathrm{x}=200-40=160$ $\therefore (+40)+(+160)=+200$ b. Given (+40) + $=-200$ Let $(+40)+\mathrm{x}=-200$ $\Rightarrow \mathrm{x}=-200-40=-240$ $\therefore (+40)+(-240)=-200$ c. Given (-50) + = +200 Let $(-50)+\mathrm{x}=+200$ $\Rightarrow \mathrm{x}=+200-(-50)=+250$ $\therefore (-50)+(+250)=+200$ d. Given (-50) + $=-200$ Let $(-50)+\mathrm{x}=-200$ $\Rightarrow \mathrm{x}=-200-(-50)=-150$ $\therefore (-50)+(-150)=-200$ e. Given $(-200)-(-40)=$ Let $(-200)-(-40)=x$ $\Rightarrow (-200)-(-40)=-160=\mathrm{x}$ $\therefore (-200)-(-40)=-160$ f. Given (+200) - (+40) = Let (+200) - (+40) = x $\Rightarrow +160=x$ $\therefore (+200)-(+40)=+160$ g. Given (-200) - (+40) = Let $(-200)-(+40)=\mathrm{x}$ $\Rightarrow (-200)+(-40)=-240=\mathrm{x}$ $\therefore (-200)+(-40)=\underline{-240}$

• Mark 3 positive numbers and 3 negative numbers on the number line above. Sol. On the Number Line: Positive numbers: We can mark any three positive numbers, e.g., 3,6 and 9. Negative numbers: We can mark any three negative numbers, e.g., $-2,-5$ and -8 .
  
• Write down the above 3 marked negative numbers in the following boxes: Sol. $-2,-5$ and -8 are three negative numbers.
• Is $2>-3$ ? Why? Is $-2<3$ Why? Sol. Is $2>-3$ Yes, 2 is a positive number and -3 is a negative number. We know that positive numbers are always greater than negative numbers. Hence, 2 is greater than -3 . Is - $2<3$ Why? .: Yes, -2 is less than 3 because -2 is a negative number and 3 is a positive number. Hence -2 < 3
• What are a. $-5+0$ b. $7+(-7)$ c. $-10+20$ d. 10-20 e. 7-(-7) f. $-8-(-10)$ ? Sol. a. $-5+0$ Adding 0 to any number does not change the value of the number. $-5+0=-5$ b. $7+(-7)$ Adding a number to its negative counterpart results in 0 . $7+(-7)=0$ c. $-10+20$ To add numbers with different signs, subtract the smaller absolute value from the larger absolute value and take the sign from the larger absolute value. $-10+20=10$ d. 10-20 Subtracting a larger number from a smaller number gives a negative result. $10-20=-10$ e. 7-(-7) Subtracting a negative number is the same as adding the positive counterpart of the number. $7-(-7)=7+7=14$ f. $-8-(-10)$ Subtracting a negative number is the same as adding the positive counterpart of the number. $-8-(-10)=-8+10=2$

5.0Key Features and Benefits Class 6 Maths Chapter 10 The Other Side of Zero : Exercise 10.1

• Step-by-step NCERT Solutions for clear thinking
• Language appropriate for Class 6 students
• Includes examples relevant to their lives
• Done with guidance provided by Allen to achieve clear concept clarity
• Allow students to see they can successfully work with integers and should
• Support homework, revision and exam preparation