NCERT-class-8-maths-chapter-1-exercise-1.1

Ncert Class 8 Maths Chapter 1 exercise 1.1

NCERT solutions for class 8 maths| Chapter 1 Rational numbers 

A  detailed and step-wise solutions to all the questions at the end of the chapter from the NCERT Maths book   are given below:

Ncert-class-8-maths-chapter-1-exercise-1.1

1. Using appropriate properties find

(i) \mathbf{-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}}

Sol.  -\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}       (by commutativity)

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

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

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

\frac{4}{2}=2

(ii) \mathbf{\frac{2}{5}\times (-\frac{3}{7})-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}} 

Sol.  \frac{2}{5}\times (-\frac{3}{7})-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}=\frac{2}{5}\times (-\frac{3}{7})+\frac{1}{14} \times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}                     (by commutativity)

=\frac{2}{5}\times \left ( (-\frac{3}{7}) +\frac{1}{14}\right )-\frac{1}{4}

\frac{2}{5 }\times \left ( \frac{-6+1}{14} \right )-\frac{1}{4} = \frac{2}{5 }\times \left ( \frac{-5}{14} \right )-\frac{1}{4}

= -\frac{1}{7}-\frac{1}{4} = \frac{-4-7}{28}

= \frac{-11}{28}

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

(i) \frac{2 }{8}

Sol.  -\frac{2}{8}  is the additive inverse of  \frac{2 }{8}  because -\frac{2}{8}+\frac{2}{8} =\frac{-2+2}{8}=\frac{0}{8} =0 .

(ii)  -\frac{5}{9}  

Sol.  \frac{5}{9}  is the additive inverse of  -\frac{5 }{9}  because -\frac{5}{9}+\frac{5}{9} =\frac{-5+5}{9}=\frac{0}{9} =0 .

(iii) \frac{-6}{-5}  

Sol.  \frac{-6}{-5}=\frac{6}{5} . So additive inverse of \frac{6}{5}   is  -\frac{6}{5} .

(iv)  \frac{2}{-9}  

Sol. Since -\frac{2}{9}+\frac{2}{9} =0 , so  additive inverse of \frac{2}{-9}   is  \frac{2}{9} .

(v)  \mathbf{\frac{19}{-6}} 

Sol. Since -\frac{19}{6}+\frac{19}{6} =0 , so  additive inverse of \frac{19}{-6}   is  \frac{19}{6} .

3. Verify that - (- x) = x  for  (i) x=\frac{11}{15}        (ii) x=-\frac{13}{17}   

Sol. (i) We have x=\frac{11}{15} ,

The additive inverse of x=\frac{11}{15}  is -x=-\frac{11}{15}  since -\frac{11}{15}+\frac{11}{15} =0 .

The same equality  -\frac{11}{15}+\frac{11}{15} =0 , shows that the additive inverse of -\frac{11}{15}  is  \frac{11}{15}  or -(-\frac{11}{15}) =\frac{11}{15} , i.e.  - (- x) = x

(ii) We have x=-\frac{13}{17} ,

The additive inverse of x=-\frac{13}{17}  is -x=\frac{13}{17}  since -\frac{13}{17}+\frac{13}{17} =0 .

The same equality  -\frac{13}{17}+\frac{13}{17} =0 , shows that the additive inverse of \frac{13}{17}  is  -\frac{13}{17}   , i.e.  - (- x) = x .

4. Find the multiplicative inverse of the following.

(i) – 13 

Sol.  We say that a rational number \frac{c}{d} is called the reciprocal or multiplicative inverse of another non-zero rational number \frac{a}{b}
if \frac{a}{b}\times \frac{c}{d} =1 .

Since  (-13)\times \frac{1}{(-13)} =1 , so  multiplicative inverse of  -13 is  -\frac{1}{13} .

(ii)\frac{-13}{19}

Sol. Since  \frac{(-13)}{19}\times \frac{19}{(-13)} =1 , so  multiplicative inverse of  –\frac{(-13)}{19} is  -\frac{19}{13} .

(iii)\mathbf{\frac{1}{5}}  

Sol. Multiplicative inverse of \frac{1}{5}   is 5 , because \frac{1}{5}\times 5=1.

(iv)\mathbf{\frac{-5}{8}}\times \frac{-3}{7}

Sol. \frac{-5}{8}\times \frac{-3}{7} =\frac{15}{56}

Since \frac{15}{56}\times \frac{56}{15}=1,so multiplicative inverse of \frac{15}{56}  is \frac{56}{15}.

(v) -1\times \frac{-2}{5}

Sol -1\times \frac{-2}{5} =\frac{2}{5}

Since \frac{2}{5}\times \frac{5}{2}=1,so multiplicative inverse of \frac{2}{5}  is \frac{5}{2}  .

(vi) – 1 

Sol. Multiplicative inverse of -1 is  -1 , because (-1) x (-1) =1 .

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

(i) \mathbf{\frac{-4}{5}\times 1= 1\times \frac{-4}{5} =\frac{-4}{5}}                 

Sol.  1 is the multiplicative identity.

(ii) \mathbf{-\frac{13}{17}\times \frac{-2}{7} = \frac{-2}{7}\times -\frac{13}{17} }

Sol. Commutativity of rational numbers.

(iii) \mathbf{\frac{-19}{29}\times \frac{29}{-19}=1}

Sol. Multiplicative inverse .

6.  Multiply \mathbf{\frac{6}{13}} by the reciprocal of \mathbf{\frac{-7}{16}}  . 

Sol.  The reciprocal of \mathbf{\frac{-7}{16}}  is \frac{16}{-7}  because \frac{-7}{16}\times \frac{16}{-7}=1 .

Now the product of  \mathbf{\frac{6}{13}}  and  \frac{16}{-7}  = \frac{6}{13}\times \frac{16}{-7}= -\frac{96}{91} .

7. Tell what property allows you to compute  \mathbf{\frac{1}{3}\times (6\times \frac{4}{3})} as \mathbf{(\frac{1}{3}\times 6)\times \frac{4}{3}} .

Sol. Associativity of rational numbers.

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

Sol.  No,  \frac{8}{9}  is not  the multiplicative inverse of -1\frac{1}{8}  because their product is not 1.

9. Is 0.3 the multiplicative inverse of 3\frac{1}{3} ? Why or why not?  

Sol.  Yes, 0.3  is the multiplicative inverse of 3\frac{1}{3}  because 0.3\times 3\frac{1}{3} = \frac{3}{10} \times \frac{10}{3}=1 .

10. Write.

(i) The rational number that does not have a reciprocal.    Ans. 0

(ii) The rational numbers that are equal to their reciprocals.   Ans. 1 and -1

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

11. Fill in the blanks.

(i) Zero has ____no____ reciprocal.

(ii) The numbers ___1_____ and __-1______ are their own reciprocals

(iii) The reciprocal of – 5 is ___\frac{-1}{5}_____.

(iv) Reciprocal of  \frac{1}{x}where x\neq 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______.

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