Ncert Maths Class 8 Try These Solutions Chapter 1

NCERT solutions for class 8 maths chapter 1 Rational Numbers- It is a very important chapter for the students from the examination point of view and also to build basic knowledge about the numbers. Rational numbers is the subpart of the unit numbers system. Rational numbers are those numbers which you can represent in the form of a fraction or numerator(p) upon a denominator(q) where the denominator can be any value except 0 or we can say that rational numbers are those numbers which can be represented in p/q form where q ≠0 and p & q are integers. From this definition, we can conclude that all the integers come under the category of a rational number. In this chapter, you will learn about rational numbers, real numbers, whole numbers, integers, and natural numbers and also study their properties like closure, commutativity, associativity. In this particular chapter, there are a total of 24 questions in 2 exercises. Students must complete the NCERT Class 8 Maths syllabus to the earliest. Apart from this, students must also refer to the NCERT Class 8 Maths books . NCERT solutions for class 8 maths chapter 1 Rational Numbers are covering detailed answers to all the 24 questions. NCERT solutions are downloadable for free through the link.

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

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

(i) \left \{ \frac{7}{5}\times \left ( \frac{-3}{12} \right ) \right \}+\left \{ \frac{7}{5}\times \frac{5}{12} \right \} (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

\left \{ \frac{7}{5}\times \left ( \frac{-3}{12} \right ) \right \}+\left \{ \frac{7}{5}\times \frac{5}{12} \right \} = \frac{7}{5}\left ( \frac{-3}{12}+\frac{5}{12} \right )\\\\\\=\frac{7}{5}\times\frac{1}{6} = \frac{7}{30}

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

\left \{ \frac{9}{16}\times \frac{4}{12} \right \}+\left \{ \frac{9}{16}\times \frac{-3}{9} \right \} = \frac{9}{16}\left ( \frac{4}{12}-\frac{3}{9} \right )\\\\\\=\frac{9}{16}\times0 = 0

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 the commutativity property of numbers, we get,

-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times\frac{1}{6} = -\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times\frac{1}{6}+\frac{5}{2}

(Now we will use distributivity of numbers)

= = \frac{3}{5}\left ( \frac{-2}{3}+\frac{-1}{6} \right )+\frac{5}{2}\\\\\\=\frac{3}{5}\times \frac{-5}{6}+\frac{5}{2}\\\\\\=\frac{-1}{2}+\frac{5}{2}\\\\\\=\frac{-1+5}{2} = \frac{4}{2}\\\\=2

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 = \frac{-13}{17} 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

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.

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}

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 solutions for class 8 maths: Chapter-wise

NCERT solutions for class 8: Subject-wise

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  • NCERT solutions for class 8 science

How to use NCERT solutions for class 8 maths chapter 1 Rational Numbers?

  • Clear your concepts about various types of numbers using the text given in the textbook.
  • Learn the proper usage of those concepts while solving a particular problem.
  • Start with the practice exercises implement the learnt concepts.
  • During the practice, keep using NCERT solutions for class 8 maths chapter 1 Rational Numbers to boost your preparation.

Keep working hard and happy learning!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

Question: What are the important topics of chapter Rational Numbers ?

Answer:

 Properties of rational numbers like commutativity and associativity, negative of a number, reciprocal, and distributivity of multiplication over addition for rational numbers are the important topics of this chapter.

Question: How many chapters are there in the CBSE class 8 maths ?

Answer:

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

Question: Where can I find the complete solutions of NCERT for class 8 ?

Question: Where can I find the complete solutions of NCERT for class 8 maths ?

Question: How does the NCERT solutions are helpful ?

Answer:

NCERT solutions are helpful for the students if they are not able to NCERT problems on their own. These solutions are provided in a very detailed manner which will give them conceptual clarity.

Question: Which is the official website of NCERT ?

Answer:

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

Ncert Maths Class 8 Try These Solutions Chapter 1

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