ncert solutions for class 8 maths
ncert maths class 8
class 8 maths solutions
class 8th maths

Ncert solutions for class 8 maths chapter 1 Rational Numbers. Here We learn what is in ncert maths class 8 EXERCISE 1 and how to solve questions with easiest method.  In this chapter we solve the question of NCERT class 8th maths EXERCISE 1. NCERT class 8 maths solutions Rational Numbers are part of NCERT Solutions for class 8 maths chapter 1 solution PDF. Ncert solutions for class 8 EXERCISE 1 with formula and solution. 8th class maths

Ncert solutions for class 8 maths Here we solve class 8th maths ncert solutions chapter 1 concepts all questions with easy method with expert solutions. It help students in their study, home work and preparing for exam. Soon we provide ncert solutions for class 8 maths EXERCISE 1 Rational Numbers question and answers. Soon we provided ncert solutions for class 8th maths chapter 1 Rational Numbers in free PDF here. ncert solutions for class 8 maths EXERCISE 1 pdf will be provide soon. 8th class maths chapter 1 NCERT Solution and ncert solutions for class 8 maths chapter 1 pdf download book PDF.

NCERT Solutions for Class 8 Maths EXERCISE 1

Rational Numbers

NCERT Solutions for Class 8 Maths chapter 1
class 8th maths
Ex 1.1 

EXERCISE 1.1

1. Using appropriate properties find.
(i) \(\displaystyle -\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}\)
(ii) \(\displaystyle \left( {\frac{2}{3}} \right)\times \left( {-\frac{3}{7}} \right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{{14}}\times \frac{2}{5}\)

Solution : 
(i) \(\displaystyle -\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}\)
\(\displaystyle \begin{array}{l}=-\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}\\=\left( {-\frac{3}{5}} \right)\times \left( {\frac{2}{3}+\frac{1}{6}} \right)+\frac{5}{2}\\=\left( {-\frac{3}{5}} \right)\times \left( {\frac{{4+1}}{6}} \right)+\frac{5}{2}\\=\left( {-\frac{3}{5}} \right)\times \left( {\frac{5}{6}} \right)+\frac{5}{2}\\=\left( {-\frac{3}{6}} \right)+\frac{5}{2}\\=\frac{{-3+15}}{6}\\=\frac{{12}}{6}\end{array}\)
= 2

(ii) \(\displaystyle \left( {\frac{2}{3}} \right)\times \left( {-\frac{3}{7}} \right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{{14}}\times \frac{2}{5}\)
\(\displaystyle \begin{array}{l}=\left( {-\frac{6}{{35}}} \right)-\frac{3}{{12}}+\frac{2}{{70}}\\=-\frac{6}{{35}}-\frac{1}{4}+\frac{1}{{35}}\\=\frac{{-24-35+4}}{{140}}\\=\frac{{-59+4}}{{140}}\\=\frac{{-55}}{{140}}\\=\frac{{-11}}{{28}}\end{array}\)

2. Write the additive inverse of each of the following.

(i) \(\displaystyle \frac{2}{8}\)

(ii) \(\displaystyle \frac{{-5}}{9}\)

(iii) \(\displaystyle \frac{{-6}}{{-5}}\)

(iv) \(\displaystyle \frac{2}{{-9}}\)

(v) \(\displaystyle \frac{{19}}{{-6}}\)

Solution : 
(i) \(\displaystyle \frac{2}{8}\)
additive inverse = \(\displaystyle \frac{-2}{8}\)

(ii) \(\displaystyle \frac{{-5}}{9}\)
additive inverse = \(\displaystyle \frac{{5}}{9}\)

(iii) \(\displaystyle \frac{{-6}}{{-5}}\)
additive inverse = \(\displaystyle \frac{{-6}}{{5}}\)

(iv) \(\displaystyle \frac{2}{{-9}}\)
additive inverse = \(\displaystyle \frac{2}{{9}}\)

(v) \(\displaystyle \frac{{19}}{{-6}}\)
additive inverse = \(\displaystyle \frac{{19}}{{6}}\)

3. Verify that − (− x) = x
(i) \(\displaystyle x=\frac{{11}}{{15}}\)    (ii) \(\displaystyle x=-\frac{{13}}{{17}}\)
Solution :  
(i) \(\displaystyle x=\frac{{11}}{{15}}\)
given \(\displaystyle x=\frac{{11}}{{15}}\)
additive inverse of \(\displaystyle x=\frac{{11}}{{15}}\) is \(\displaystyle -x=\frac{{-11}}{{15}}\) ,
because \(\displaystyle \frac{{11}}{{15}}+\left( {\frac{{-11}}{{15}}} \right)=0\)
the same equality \(\displaystyle \frac{{11}}{{15}}+\left( {\frac{{-11}}{{15}}} \right)=0\) , shows that the additive inverse of \(\displaystyle -x=\frac{{-11}}{{15}}\) is \(\displaystyle x=\frac{{11}}{{15}}\) 
so, \(\displaystyle -\left( {\frac{{-11}}{{15}}} \right)=\frac{{11}}{{15}}\)
so, − (− x) = x

(ii) \(\displaystyle x=-\frac{{13}}{{17}}\)
given \(\displaystyle x=-\frac{{13}}{{17}}\)
additive inverse of \(\displaystyle x=-\frac{{13}}{{17}}\) is \(\displaystyle x=\frac{{13}}{{17}}\) है, 
because \(\displaystyle \left( {\frac{{-13}}{{17}}} \right)+\frac{{13}}{{17}}=0\) है।
समिका \(\displaystyle \left( {\frac{{-13}}{{17}}} \right)+\frac{{13}}{{17}}=0\) , shows that the additive inverse of \(\displaystyle x=\frac{{13}}{{17}}\) is \(\displaystyle x=-\frac{{13}}{{17}}\) 
so, \(\displaystyle -\left( {\frac{{-13}}{{17}}} \right)=\frac{{13}}{{17}}\)
so, − (− x) = x

4. Find the multiplicative inverse of the following.

(i) − 13	(ii) \(\displaystyle \frac{{-13}}{{19}}\)	(iii) \(\displaystyle \frac{{1}}{{5}}\)	(iv) \(\displaystyle \frac{{-5}}{8}\times \left( {\frac{{-3}}{7}} \right)\)
(v) \(\displaystyle -1\times \left( {\frac{{-2}}{5}} \right)\)	(vi) − 1

Solution :
(i) − 13
multiplicative inverse = \(\displaystyle \frac{{-1}}{{13}}\)

(ii) \(\displaystyle \frac{{-13}}{{19}}\)
multiplicative inverse = \(\displaystyle \frac{{-19}}{{13}}\)

(iii) \(\displaystyle \frac{{1}}{{5}}\)
multiplicative inverse = \(\displaystyle \frac{{5}}{{1}}\) या 5

(iv) \(\displaystyle \frac{{-5}}{8}\times \left( {\frac{{-3}}{7}} \right)\)
multiplicative inverse = \(\displaystyle \frac{{-56}}{{15}}\)

(v) \(\displaystyle -1\times \left( {\frac{{-2}}{5}} \right)\)
multiplicative inverse = \(\displaystyle \frac{{-5}}{{2}}\)

(vi) − 1
multiplicative inverse = − 1

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

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

Solution : 
(i) \(\displaystyle \frac{{-4}}{5}\times 1=1\times \left( {\frac{{-4}}{5}} \right)=-\frac{4}{5}\)
Answer – 1 is the multiplicative identity

(ii) \(\displaystyle -\frac{{13}}{{17}}\times \left( {-\frac{2}{7}} \right)=\frac{{-2}}{7}\times \left( {\frac{{-13}}{{17}}} \right)\)
Answer – Commutativity

(iii) \(\displaystyle \frac{{-19}}{{29}}\times \frac{{29}}{{-19}}=1\)
Answer – Multiplicative inverse

6. Multiply \(\displaystyle \frac{6}{{13}}\) by the reciprocal of \(\displaystyle \frac{{-7}}{{16}}\)
Solution : 
Multiplicative inverse \(\displaystyle \frac{{-7}}{{16}}\) is \(\displaystyle \frac{{-16}}{{7}}\)
so, \(\displaystyle \frac{6}{{13}}\times \frac{{-16}}{7}\)
\(\displaystyle =\frac{{-96}}{{91}}\)

7. Tell what property allows you to compute \(\displaystyle \frac{1}{3}\times \left( {\frac{{-2}}{7}} \right)=\frac{{-2}}{7}\times \left( {\frac{{-13}}{{17}}} \right)\)
\displaystyle \frac{{-7}}{{16}}$
Solution : Associativity property

8. Is \(\displaystyle -1\frac{1}{8}\) the multiplicative inverse of \(\displaystyle \frac{{8}}{{9}}\) ? Why or why not?
Solution : 
converting the \(\displaystyle -1\frac{1}{8}\) in simple fraction we get \(\displaystyle \frac{{-9}}{8}\)
Multiplicative inverse of \(\displaystyle \frac{{-9}}{8}\) is \(\displaystyle \frac{{-8}}{9}\) ,
which does not correspond to the given number.
so, Multiplicative inverse of \(\displaystyle -1\frac{1}{8}\) is \(\displaystyle \frac{{-8}}{9}\) , not \(\displaystyle \frac{{8}}{{9}}\)

9. Is 0.3 the multiplicative inverse of \(\displaystyle 3\frac{1}{3}\) ? Why or why not?
Solution :
converting the \(\displaystyle 3\frac{1}{3}\) in simple fraction we get \(\displaystyle \frac{{10}}{3}\) 
so, Multiplicative inverse of \(\displaystyle \frac{{10}}{3}\) is \(\displaystyle \frac{{3}}{10}\) 
\(\displaystyle \frac{{3}}{10}\) = 0.3
So, multiplicative inverse of \(\displaystyle 3\frac{1}{3}\) is 0.3

10. 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.
Solution :
(i) The rational number that does not have a reciprocal.
Answer – 0

(ii) The rational numbers that are equal to their reciprocals.
Answer – 1 and − 1

(iii) The rational number that is equal to its negative.
Answer – 0

11. 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 \(\displaystyle \frac{{1}}{x}\) , where (x ≠ 0) is _______________ .
(v) The product of two rational numbers is always a _______.
(vi) The reciprocal of a positive rational number is ________.

Solution :
(i) Zero has no reciprocal.
(ii) The numbers 1 and − 1 are their own reciprocals.
(iii) The reciprocal of – 5 is \(\displaystyle \underline{{\frac{{-1}}{5}}}\) .
(iv) Reciprocal of \(\displaystyle \frac{{1}}{x}\) , 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.