This article offers a comprehensive and detailed guide to mastering the fundamental concepts of real numbers through NCERT Solutions for Class 10 Maths Chapter 1. The step-by-step explanations provided in the article, along with essential study materials, aim to facilitate a thorough understanding of the topic. Chapter 1 of Class 8 Maths NCERT, focusing on rational numbers and their applications, is crucial for not only achieving high academic performance but also for building a strong foundation to grasp subsequent concepts introduced in the eighth grade.
Verify that: -(-x) = x for:
(i) x = 11/15
(ii) x = -13/17
(i) x = 11/15
We have, x = 11/15
The additive inverse of x is – x (as x + (-x) = 0).
Then, the additive inverse of 11/15 is – 11/15 (as 11/15 + (-11/15) = 0).
The same equality, 11/15 + (-11/15) = 0, shows that the additive inverse of -11/15 is 11/15.
Or, – (-11/15) = 11/15
i.e., -(-x) = x
(ii) -13/17
We have, x = -13/17
The additive inverse of x is – x (as x + (-x) = 0).
Then, the additive inverse of -13/17 is 13/17 (as 13/17 + (-13/17) = 0).
The same equality (-13/17 + 13/17) = 0, shows that the additive inverse of 13/17 is -13/17.
Or, – (13/17) = -13/17,
i.e., -(-x) = x
Find the multiplicative inverse of the following:
(i) -13
(ii) -13/19
(iii) 1/5
(iv) -5/8 × (-3/7)
(v) -1 × (-2/5)
(vi) -1
(i) -13
Multiplicative inverse of -13 is -1/13.
(ii) -13/19
Multiplicative inverse of -13/19 is -19/13.
(iii) 1/5
Multiplicative inverse of 1/5 is 5.
(iv) -5/8 × (-3/7) = 15/56
Multiplicative inverse of 15/56 is 56/15.
(v) -1 × (-2/5) = 2/5
Multiplicative inverse of 2/5 is 5/2.
(vi) -1
Multiplicative inverse of -1 is -1.
Name the property under multiplication used in each of the following:
(i) -4/5 × 1 = 1 × (-4/5) = -4/5
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
(iii) -19/29 × 29/-19 = 1
(i) -4/5 × 1 = 1 × (-4/5) = -4/5
Here 1 is the multiplicative identity.
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
The property of commutativity is used in the equation.
(iii) -19/29 × 29/-19 = 1
The multiplicative inverse is the property used in this equation.
Multiply 6/13 by the reciprocal of -7/16.
Reciprocal of -7/16 = 16/-7 = -16/7
According to the question,
6/13 × (Reciprocal of -7/16)
6/13 × (-16/7) = -96/91
Tell what property allows you to compute 1/3 × (6 × 4/3) as (1/3 × 6) × 4/3.
1/3 × (6 × 4/3) = (1/3 × 6) × 4/3
Here, the way in which factors are grouped in a multiplication problem supposedly does not change the product. Hence, the Associativity Property is used here.
Is 8/9 the multiplication inverse of –? Why or why not?
– = -9/8
[Multiplicative inverse ⟹ product should be 1]
According to the question,
8/9 × (-9/8) = -1 ≠ 1
Therefore, 8/9 is not the multiplicative inverse of –.
If 0.3 is the multiplicative inverse of? Why or why not?
= 10/3
0.3 = 3/10
[Multiplicative inverse ⟹ product should be 1]
According to the question,
3/10 × 10/3 = 1
Therefore, 0.3 is the multiplicative inverse of.
Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
(I) The rational number that does not have a reciprocal is 0.
Reason:
0 = 0/1
Reciprocal of 0 = 1/0, which is not defined.
(ii) The rational numbers that are equal to their reciprocals are 1 and -1.
Reason:
1 = 1/1
Reciprocal of 1 = 1/1 = 1, similarly, reciprocal of -1 = – 1
(iii) The rational number that is equal to its negative is 0.
Reason:
Negative of 0=-0=0
Fill in the blanks.
(i) Zero has _______reciprocal.
(ii) The numbers ______and _______are their own reciprocals
(iii) The reciprocal of – 5 is ________.
(iv) Reciprocal of 1/x, where x ≠ 0 is _________.
(v) The product of two rational numbers is always a ________.
(vi) The reciprocal of a positive rational number is _________.
(i) Zero has no reciprocal.
(ii) The numbers -1 and 1 are their own reciprocals
(iii) The reciprocal of – 5 is -1/5.
(iv) Reciprocal of 1/x, where x ≠ 0 is x.
(v) The product of two rational numbers is always a rational number.
(vi) The reciprocal of a positive rational number is positive.
Write the additive inverse of each of the following:
(i) 2/8
(ii) -5/9
(iii) -6/-5 = 6/5
(iv) 2/-9 = -2/9
(v) 19/-16 = -19/16
(i) The Additive inverse of 2/8 is – 2/8
(ii) The additive inverse of -5/9 is 5/9
(iii) The additive inverse of 6/5 is -6/5
(iv) The additive inverse of -2/9 is 2/9
(v) The additive inverse of -19/16 is 19/16
Using appropriate properties, find:
(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
= -2/3 × 3/5– 3/5 × 1/6+ 5/2 (by commutativity)
= 3/5 (-2/3 – 1/6)+ 5/2
= 3/5 ((- 4 – 1)/6)+ 5/2
= 3/5 ((–5)/6)+ 5/2 (by distributivity)
= – 15 /30 + 5/2
= – 1 /2 + 5/2
= 4/2
= 2
(ii)
= 2/5 × (- 3/7) + 1/14 × 2/5 – (1/6 × 3/2) (by commutativity)
= 2/5 × (- 3/7 + 1/14) – 3/12
= 2/5 × ((- 6 + 1)/14) – 3/12
= 2/5 × ((- 5)/14)) – 1/4
= (-10/70) – 1/4
= – 1/7 – 1/4
= (– 4– 7)/28
= – 11/28
Represent these numbers on the number line.
(i) 7/4
(ii) -5/6
(i) 7/4
Divide the line between the whole numbers into 4 parts, i.e. divide the line between 0 and 1 to 4 parts, 1 and 2 to 4 parts, and so on.
Thus, the rational number 7/4 lies at a distance of 7 points away from 0 towards the positive number line.
(ii) -5/6
Divide the line between the integers into 4 parts, i.e. divide the line between 0 and -1 to 6 parts, -1 and -2 to 6 parts, and so on. Here, since the numerator is less than the denominator, dividing 0 to – 1 into 6 parts is sufficient.
Thus, the rational number -5/6 lies at a distance of 5 points, away from 0, towards the negative number line.
Represent -2/11, -5/11, -9/11 on a number line.
Divide the line between the integers into 11 parts.
Thus, the rational numbers -2/11, -5/11, and -9/11 lie at a distance of 2, 5, and 9 points away from 0, towards the negative number line, respectively.
Write five rational numbers which are smaller than 2.
The number 2 can be written as 20/10
Hence, we can say that the five rational numbers which are smaller than 2 are:
2/10, 5/10, 10/10, 15/10, 19/10
Find the rational numbers between -2/5 and ½.
Let us make the denominators the same, say 50.
-2/5 = (-2 × 10)/(5 × 10) = -20/50
½ = (1 × 25)/(2 × 25) = 25/50
Ten rational numbers between -2/5 and ½ = ten rational numbers between -20/50 and 25/50.
Therefore, ten rational numbers between -20/50 and 25/50 = -18/50, -15/50, -5/50, -2/50, 4/50, 5/50, 8/50, 12/50, 15/50, 20/50.
Find five rational numbers between:
(i) 2/3 and 4/5
(ii) -3/2 and 5/3
(iii) ¼ and ½
(i) 2/3 and 4/5
Let us make the denominators the same, say 60
i.e., 2/3 and 4/5 can be written as:
2/3 = (2 × 20)/(3 × 20) = 40/60
4/5 = (4 × 12)/(5 × 12) = 48/60
Five rational numbers between 2/3 and 4/5 = five rational numbers between 40/60 and 48/60.
Therefore, five rational numbers between 40/60 and 48/60 = 41/60, 42/60, 43/60, 44/60, 45/60.
(ii) -3/2 and 5/3
Let us make the denominators the same, say 6
i.e., -3/2 and 5/3 can be written as:
-3/2 = (-3 × 3)/(2× 3) = -9/6
5/3 = (5 × 2)/(3 × 2) = 10/6
Five rational numbers between -3/2 and 5/3 = five rational numbers between -9/6 and 10/6.
Therefore, five rational numbers between -9/6 and 10/6 = -1/6, 2/6, 3/6, 4/6, 5/6.
(iii) ¼ and ½
Let us make the denominators the same, say 24
i.e., ¼ and ½ can be written as:
¼ = (1 × 6)/(4 × 6) = 6/24
½ = (1 × 12)/(2 × 12) = 12/24
Five rational numbers between ¼ and ½ = five rational numbers between 6/24 and 12/24.
Therefore, five rational numbers between 6/24 and 12/24 = 7/24, 8/24, 9/24, 10/24, 11/24.
Write five rational numbers greater than -2.
-2 can be written as – 20/10
Hence, we can say that the five rational numbers greater than -2 are
-10/10, -5/10, -1/10, 5/10, 7/10
Find ten rational numbers between 3/5 and ¾.
Let us make the denominators the same, say 80.
3/5 = (3 × 16)/(5× 16) = 48/80
3/4 = (3 × 20)/(4 × 20) = 60/80
Ten rational numbers between 3/5 and ¾ = ten rational numbers between 48/80 and 60/80.
Therefore, ten rational numbers between 48/80 and 60/80 = 49/80, 50/80, 51/80, 52/80, 54/80, 55/80, 56/80, 57/80, 58/80, 59/80.
The NCERT solution for class 8 Chapter 1: Rational is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tacking more difficult concepts in their further education.
Yes, the NCERT solution for class 8 Chapter 1: Rational is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Rational ally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for class 8 Maths Chapter 1 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for class 8 Maths Chapter 1 : Rational as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for class 8 Maths Chapter 1 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution. Also, you can make Rational al notes and jot down the important concepts for your understanding.