ncert-solutions-class-8-maths-ch-1-rational-numbers

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

NCERT solutions for class 8 maths chapter 1 Rational Numbers Topic: Properties of Rational Numbers

Q1 Fill in the blanks in the following table.

Addition	Subtraction	Multiplication	Division
Rational Numbers	Yes	Yes	...	No
Integers	...	Yes	...	No
Whole Numbers	...	...	Yes	...
Natural Numbers	...	No	...	...

Answer:

It can be seen that rational numbers, integers, whole numbers, natural numbers are not closed under division because of Zero is included in these numbers. Any number divided by zero is not defined.

Addition	Subtraction	Multiplication	Division
Rational Numbers	Yes	Yes	Yes	No
Integers	Yes	Yes	Yes	No
Whole Numbers	Yes	No	Yes	No
Natural Numbers	Yes	No	Yes	Yes

Q2 Complete the following table:

Commutative for

Addition	Subtraction	Multiplication	Division
Rational Numbers	Yes	..	...	...
Integers	...	No	...	...
Whole Numbers	...	...	Yes	...
Natural Numbers	...	...	...	No

Answer:

In rational numbers, a $\div$ b $\neq$ b $\div$ a

also a-b $\neq$ b-a

Addition	Subtraction	Multiplication	Division
Rational Numbers	Yes	No	Yes	No
Integers	Yes	No	Yes	No
Whole Numbers	Yes	No	Yes	No
Natural Numbers	Yes	No	Yes	No

Q3 Complete the following table:

Associative for

Addition	Subtraction	Multiplication	Division
Rational Numbers	...	...	...	No
Integers	...	...	Yes	...
Whole Numbers	Yes	...	...	...
Natural Numbers	...	No	...	...

Answer:

For associative in multiplication:- a $\times$ (b $\times$ c) = (a $\times$ b) $\times$ c

Addition	Subtraction	Multiplication	Division
Rational Numbers	Yes	No	Yes	No
Integers	Yes	No	Yes	No
Whole Numbers	Yes	No	Yes	No
Natural Numbers	Yes	No	Yes	No

Q4 Find using distributivity:

(i) (ii) $\left \{ \frac{9}{16}\times \frac{4}{12} \right \}+\left \{ \frac{9}{16}\times \frac{-3}{9} \right \}$

Answer:

(i) Using distributivity, a(b+c) = ab + ac

(ii) Using distributivity of multiplication over addition and subtraction,

Q5 Write the rational number for each point labeled with a letter:

Answer:

(i) In this, we can see that 1 is divided into 5 parts each, so when we are moving from zero to the right-hand side, it is easy to observe that

All the numbers should contain 5 in their denominator. Thus, A is equal to $\frac{1}{5}$ , B is equal to $\frac{4}{5}$ , C is equal to $\frac{5}{5} = 1$ , D is equal to $\frac{8}{5}$ , E is equal to $\frac{9}{5}$

(ii) Here we see that 1 is divided in 6 parts each. So when we move from zero towards left we observe that

All the numbers should contain 6 in their denominator. Thus, F is equal to $\frac{-2}{6}$ , G is equal to $\frac{-5}{6}$ , H is equal to $\frac{-7}{6}$ , I is equal to $\frac{-8}{6}$ , J is equal to $\frac{-11}{6}$

NCERT solutions for class 8 maths chapter 1 Rational Numbers Excercise: 1.1

Q1 (i) Using appropriate properties find. $(i)\:\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$

Answer:

By using commutativity property of numbers, we get,

(Now we will use distributivity of numbers)

Q1 (ii) Using appropriate properties find. $(ii)\frac{2}{5}\times \frac{-3}{7}- \frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

Answer:

By using commutativity, we get

Now by distributivity,

Q2 Write the additive inverse of each of the following:

(i) $\frac{2}{8}$ (ii) $\frac{-5}{9}$ (iii) $\frac{-6}{-5}$ (iv) $\frac{2}{-9}$ (v) $\frac{19}{-6}$

Answer:

(i) The additive inverse of $\\ \\ \frac{2}{8}$ is $\frac{-2}{8}$ because $\frac{2}{8}+\frac{-2}{8} = \frac{2-2}{8} = 0$

(ii) The additive inverse of $\frac{-5}{9}$ is $\frac{5}{9}$ because $\frac{-5}{9}+\frac{5}{9} = \frac{-5+5}{9} = 0$

(iii) The additive inverse of $\frac{-6}{-5}$ is $\frac{6}{-5}$ because $\frac{-6}{-5}+\frac{6}{-5} = \frac{-6+6}{-5} = 0$

(iv) The additive inverse of $\frac{2}{-9}$ is $\frac{-2}{-9}$ because $\frac{2}{-9}+\frac{-2}{-9} = \frac{2-2}{-9} = 0$

(v) The additive inverse of $\frac{19}{-6}$ is $\frac{-19}{-6}$ because $\frac{19}{-6}+\frac{-19}{-6} = \frac{19-19}{-6} =0$

Q3 Verify that – (– x) = x for (i) x = $\frac{11}{15}$ (ii) x = $\frac{-13}{17}$

Answer:

(i) We have x = $\frac{11}{15}$

The additive inverse of x = $\frac{11}{15}$ is -x = $\frac{-11}{15}$

The same equality $\frac{11}{15}+\left ( \frac{-11}{15} \right ) = 0$

which implies $-\left ( \frac{-11}{15} \right ) = \frac{11}{15}$ shows that -(-x) = x

(ii) Additive inverse of x = <img alt="\frac{-13}{17}" height="38"

src="https://lh3.googleusercontent.com/GGb5ZP8AglOdWLxrbQP32LviEN6eci17sbaxYTUiO5o7zqZvCF5ajywF9kcovuhhro1WyxHhJEGyxGYCb0KgU83fm1N8cgtaQ6gcBUcNV-lGTc3o5eO_zUPJzydIwL-SD6e1gOE" style="margin-left: 0px; margin-top: 0px;" width="33" /> is -x = $\frac{13}{17}$ (since $\frac{-13}{17}+\frac{13}{17} = 0$ )

The same quality shows that the additive inverse of $\frac{13}{17}$ is $\frac{-13}{17}$

i.e., -(-x) = x

Q4 Find the multiplicative inverse of the following. (i) - 13 (ii) $\frac{-13}{19}$ (iii) $\frac{1}{5}$ (iv) $\frac{-5}{8}\times \frac{-3}{7}$ (v) $-1\times \frac{-2}{5}$ (vi) - 1

Answer:

(i) The multiplicative inverse of -13 is $\frac{-1}{13}$ because $-13\times \frac{-1}{13} = 1$

(ii) The multiplicative inverse of $\frac{-13}{19}$ is $\frac{-19}{13}$ because of i

(iii) The multiplicative inverse of $\frac{1}{5}$ is 5 because $\frac{1}{5}\times 5 = 1$

(iv) The multiplicative inverse of $\frac{-5}{8}\times \frac{-3}{7}$ is $\frac{56}{15}$ because $\frac{15}{56}\times \frac{56}{15} = 1$

(v) The multiplicative inverse of $-1\times \frac{-2}{5}$ is $\frac{5}{2}$ because $\frac{2}{5}\times \frac{5}{2} = 1$

(vi) The multiplicative inverse of -1 is -1 because $-1\times -1 = 1$

Q5 Name the property under multiplication used in each of the following.

(i) $\frac{-4}{5}\times 1= 1\times \frac{-4}{5} = \frac{-4}{5}$ (ii) $-\frac{13}{17}\times \frac{-2}{7} = \frac{-2}{7}\times \frac{-13}{17}$ (iii) $\frac{-19}{29}\times \frac{29}{-19} = 1$

Answer:

(i) Multiplying any number with 1 we get the same number back.

i.e., a $\times$ 1 = 1 $\times$ a = a

Hence 1 is the multiplicative identity for rational numbers.

(ii) Commutavity property states that a $\times$ b = b $\times$ a

(iii) It is the multipicative inverse identity, i.e., $a\times\frac{1}{-a} = 1$

Q6 Multiply $\frac{6}{13}$ by the reciprocal of $\frac{-7}{16}$ .
Answer:
We know that the reciprocal of $\frac{-7}{16}$ is $\frac{16}{-7}$ .
Now, $\frac{6}{13}\times \frac{16}{-7} = \frac{-96}{91}$

Q7 Tell what property allows you to compute $\frac{1}{3}\times \left ( 6\times \frac{4}{3} \right )$ as $\left ( \frac{1}{3}\times 6 \right )\times \frac{4}{3}$

Answer:

By the Associativity

property for multiplication, we know that

a × (b × c) = (a × b) × c

Thus property used here is the associativity.

Q8 Is $\frac{8}{9}$ the multiplicative inverse of $-1\frac{1}{8}$ ? Why or why not?

Answer:

We know that A is the multiplicative inverse of B if B $\times$ A = 1

Applying this in given question, we get,

$\frac{-9}{8}\times \frac{8}{9} = -1$ Thus $\frac{8}{9}$ is not the multiplicative inverse of $\frac{-9}{8}$ .

Q9 Is 0.3 the multiplicative inverse of $3\frac{1}{3}$ ? Why or why not?
Answer:

We know that if A is multiplicative inverse of B then B $\times$ A = 1

In this question, $\frac{10}{3}\times\frac{3}{10} = 1$ (Since 0.3 = $\frac{3}{10}$ )

This 0.3 is the multiplicative inverse of $\frac{10}{3}$ .

Q10 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.

Answer:

(i) Zero(0). We know that reciprocal of A is $\frac{1}{A}$ . So for 0, its reciprocal is not defined.

(ii) 1 and -1 . (Since $\frac{1}{1} = 1$ and $\frac{1}{-1} = -1$ )

(iii) Zero,0. (as -0=0)

Q11 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 $\frac{1}{x}$ , where x ≠ 0 is ________.

(v) The product of two rational numbers is always a _______.

(vi) The reciprocal of a positive rational number is ________.

Answer:

(i) Zero has no reciprocal as it's reciprocal is not defined.

(ii) The numbers 1 and -1 are their own reciprocals as $\frac{1}{1} = 1$ and $\frac{1}{-1} = -1$

(iii) $\frac{1}{-5}$ . We know that reciprocal of A is $\frac{1}{A}$ .

(iv) Since $\frac{1}{\frac{1}{x}} = x$ .

(v) rational number. We know that if p and q are 2 rational numbers then pq is also a rational number.

(vi) Positive. Since reciprocal of A is $\frac{1}{A}$ , now if A is positive then reciprocal is also positive.

NCERT solutions for class 8 maths chapter 1 Rational Numbers Excercise: 1.2

Q1 Represent these numbers on the number line. (i) $\frac{7}{4}$ (ii) $\frac{-5}{6}$

Answer:

(i) To represent $\frac{7}{4}$ on a number line, firstly we will divide 1 in 4 parts and draw it on a line such as 1/4, 2/4, 3/4, ........, 9/4. Then will mark the required number.

(ii) To represent $\frac{-5}{6}$ on the number line, firstly we will divide 1 in 6 parts and draw it on the left side of zero on number line such as -1/6, -2/6, .......,-9/4. Then mark the required number on the number line.

Q2 Represent $\frac{-2}{11}$ , $\frac{-5}{11}$ , $\frac{-9}{11}$ on the number line

Answer:

We will divide 1 into 11 parts, then start marking numbers on left side of zero such as -1/11, -2/11, -3/11,.........,-12/11. Mark the required numbers on

the drawn number line.

Q3 Write five rational numbers which are smaller than 2.
Answer:

The 5 rational numbers smaller than 2 can be any number in the form of p/q where q $\not\equiv$ 0. Hence infinite numbers are possible.

Examples of 5 such numbers are 1, 1/3, 0, -1, -2

Q4 Find ten rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$
Answer:

Rational numbers between any 2 numbers can easily find out by taking their means.

i.e., For $\\\frac{-2}{5}$ and $\\\frac{1}{2}$

Their mean is $\left ( \frac{-2}{5}+\frac{1}{2} \right )\div 2 = \frac{1}{20}$ . Hence 1 rational number between $\frac{-2}{5}$ and $\frac{1}{2}$ is $\frac{1}{20}$ .

Now we will find the mean between $\frac{-2}{5}$ and $\frac{1}{20}$ .

This implies a new required rational number is $\left ( \frac{-2}{5}+\frac{1}{20} \right )/2 = \frac{-7}{40}$ .

Similarly, we will find a mean between $\frac{1}{20}$ and $\frac{1}{2}$

New required rational number is $\left ( \frac{1}{20}+\frac{1}{2} \right ) /2 = \frac{11}{40}$

Similarly, we will take means of new numbers generated between $\frac{-2}{5}$ and $\frac{1}{2}$ .

Q5 Find five rational numbers between.(i) $\frac{2}{3}$ and $\frac{4}{5}$ (ii) $\frac{-3}{2}$ and $\frac{5}{3}$ (iii) $\frac{1}{4}$ and $\frac{1}{2}$
Answer:

(i) For finding rational numbers between 2 numbers one method is to find means between the numbers repeatedly.

Another method is:- For $\frac{2}{3}$ and $\frac{4}{5}$

$\frac{2}{3}$ can be written as $\frac{10}{15}$ $\left ( \frac{2}{3}\times \frac{5}{5} = \frac{10}{15}\right )$

and $\frac{4}{5}$ can be written as $\frac{12}{15}$ $\left ( \frac{4}{5}\times \frac{3}{3} = \frac{12}{15}\right )$

Thus numbers between $\frac{10}{15}$ and $\frac{12}{15}$ are the required numbers.

Now since we require 5 numbers in between, thus we multiply numerator and denominator both by 4.

It becomes numbers between $\frac{40}{60}$ and $\frac{48}{60}$ .

Thus numbers are $\frac{41}{60}, \frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60}$ .

(ii) Similarly for $\frac{-3}{2}$ and $\frac{5}{3}$

Required numbers fall between $\frac{-9}{6}$ and $\frac{10}{6}$ $\left \{ \left ( \frac{-3}{2}\times \frac{3}{3} \right )= \frac{-9}{6} \right \}$

Thus numbers are $\frac{-8}{6}, \frac{-7}{6}, \frac{-6}{6}, \frac{-5}{6}, \frac{-4}{6}$

(iii) For $\frac{1}{4}$ and $\frac{1}{2}$

Required numbers lie between $\frac{1}{4}$ and $\frac{2}{4}$ or we can say between $\frac{8}{32}$ and $\frac{16}{32}$

Thus numbers are $\frac{9}{32}, \frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}$

Q6 Write five rational numbers greater than –2.
Answer:

There exist infinitely many rational numbers (can be expressed in the form of p/q where q $\not\equiv$ 0) greater than -2 .

Few such examples are -1, -1/2, 0, 1, 1/3 etc.

Q7 Find ten rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$ .
Answer: Finding rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$ is equivalent to find rational numbers between rational numbers between $\frac{12}{20}$ and $\frac{15}{20}$ ,since these numbers are obtained by just making their denominators equal.

Further, it is equivalent to find rational number between $\frac{96}{160}$ and $\frac{120}{160}$

(We obtained above numbers by multiplying and dividing numbers by 8 to create a difference of at least 10 numbers).

Thus required numbers are $\frac{97}{160},\frac{98}{160},\frac{99}{160},........,\frac{106}{160}$

Alternate:- Rational numbers can also be found by taking mean of the given numbers and the newly obtained number.

NCERT Class 8 Mathematics Solutions

Chapter 01 - Rational Numbers

Chapter 02 - Linear Equations in One Variable

Chapter 03 -Understanding Quadrilaterals

Chapter 04 - Practical Geometry

Chapter 05 - Data Handling

Chapter 06 - Squares and Square Roots

Chapter 07 - Cubes and Cube Roots

Chapter 08 - Comparing Quantities

Chapter 09 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Indirect proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

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NCERT solutions for class 8 maths chapter 1 Rational Numbers Topic: Properties of Rational Numbers

NCERT solutions for class 8 maths chapter 1 Rational Numbers Excercise: 1.1

NCERT solutions for class 8 maths chapter 1 Rational Numbers Excercise: 1.2

NCERT Class 8 Mathematics Solutions

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