EXERCISE 1.1
Introduction - "Rational Numbers" Chapter 1 - NCERT Class 8th Maths Solutions
Q 1(i),(ii) - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Using appropriate properties find .
(i) $ -\dfrac{2}{3}\times\dfrac{3}{5}+\dfrac{5}{2}-\dfrac{3}{5}\times\dfrac{1}{6} $
sol :
$\\=-\dfrac{2}{3}\times\dfrac{3}{5}+\dfrac{5}{2}-\dfrac{3}{5}\times\dfrac{1}{6}$
$=-\dfrac{2}{3}\times\dfrac{3}{5}-\dfrac{3}{5}\times\dfrac{1}{6}+\dfrac{5}{2}$ $\text{[Using commutativity of rational number]}$
$=\Bigg(-\dfrac{3}{5}\Bigg)\times\Bigg(\dfrac{2}{3}+\dfrac{1}{6}\Bigg)+\dfrac{5}{2}$ $\text{[Distributivity]}$
$=\Bigg(-\dfrac{3}{5}\Bigg)\times\Bigg(\dfrac{2\times{2}+1}{6}\Bigg)+\dfrac{5}{2}$
$=\Bigg(-\dfrac{3}{5}\Bigg)\times\Bigg(\dfrac{5}{6}\Bigg)+\dfrac{5}{2}$
$=\Bigg(-\dfrac{3}{6}\Bigg)+\dfrac{5}{2}$
$=\Bigg(\dfrac{-3+5\times3}{6}\Bigg)$
$=\dfrac{12}{6}$
= 2
(ii) $ \dfrac{2}{5}\times\Bigg(-\dfrac{3}{7}\Bigg)-\dfrac{1}{6}\times\dfrac{3}{2}+\dfrac{1}{14}\times\dfrac{2}{5} $
sol :
$=\dfrac{2}{5}\times\Bigg(-\dfrac{3}{7}\Bigg)+\dfrac{1}{14}\times\dfrac{2}{5}-\dfrac{1}{6}\times\dfrac{3}{2}$ $\text{[By commutativity of rational number]}$
$=\dfrac{2}{5}\times\Bigg(-\dfrac{3}{7}+\dfrac{1}{14}\Bigg)-\dfrac{1}{4}$ $\text{[by~distributivity]}$
$=\dfrac{2}{5}\times\Bigg(\dfrac{-3\times2+1}{14}\Bigg)-\dfrac{1}{4}$
$=\dfrac{2}{5}\times\Bigg(\dfrac{-5}{14}\Bigg)-\dfrac{1}{4}$
$=-\dfrac{1}{7}-\dfrac{1}{4}$
$=\dfrac{-4-7}{28}$
$=-\dfrac{11}{28}$
Q 2 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Question 2 :
Write the additive inverse of each of the following :
(i) $ \dfrac{2}{8} $
sol :
Additive inverse= $-\dfrac{2}{8}$
(ii) $ -\dfrac{5}{9} $
sol :
Additive inverse= $ \dfrac{5}{9} $
(iii)
$ \dfrac{-6}{-5} $
sol :
$ \dfrac{-6}{-5}=\dfrac{6}{5} $
Additive inverse=$ -\dfrac{6}{5} $
(iv)
$\dfrac{~~2}{-9}$
sol :
$\dfrac{~~2}{-9}=\dfrac{-2}{~~~9}$
Additive inverse=$\dfrac{2}{9}$
(v)
$\dfrac{~~19}{-6}$
sol :
$\dfrac{~~2}{-9}=\dfrac{-19}{~~~6}$
Additive inverse =$\dfrac{19}{6}$
Q 3 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Question 3 :
Verify that -(-x)=x for
(i) $x=\dfrac{11}{15}$
sol :
The additive inverse of $ x=\dfrac{11}{15}$ is $ -x=-\dfrac{11}{15} $ as $ \dfrac{11}{15}+\Bigg(-\dfrac{11}{15}\Bigg)=0 $
This equality $ \dfrac{11}{15}+\Bigg(-\dfrac{11}{15}\Bigg)=0 $ represents that the additive inverse of is $ x=\dfrac{11}{15} $ or it can be said that $ -\bigg(-\dfrac{11}{15}\bigg)=\dfrac{11}{15} $ i.e., -(-x)=x
(ii) $ x=-\dfrac{13}{17} $
sol :
The additive inverse of $x=-\dfrac{13}{17}$ is $-x=\dfrac{13}{17}$ as $-\dfrac{13}{17}+\Bigg(\dfrac{13}{17}\Bigg)=0$
This equality $ -\dfrac{13}{17}+\Bigg(\dfrac{13}{17}\Bigg)=0 $ represents that the additive inverse of $ x=-\dfrac{13}{17} $ is $ -x=\dfrac{13}{17} $ or it can be said that $ -\bigg(-\dfrac{13}{17}\bigg)=\dfrac{13}{17} $ i.e., -(-x)=x
Q 4 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Question 4 :
Find the multiplicative inverse of the following.
(i) -13
sol :
Multiplicative inverse =$-\dfrac{1}{13}$
(ii) $-\dfrac{13}{19}$
sol :
Multiplicative inverse = $-\dfrac{19}{13}$
(iii) $\dfrac{1}{5}$
sol :
Multiplicative inverse =5
(iv) $\dfrac{-5}{~~~8}\times\dfrac{-3}{~~~7}$
sol :
$\dfrac{-5}{~~~8}\times\dfrac{-3}{~~~7}=\dfrac{15}{56}$
Multiplicative inverse =$\dfrac{56}{15}$
(v) $-1\times\dfrac{-2}{~~~5}$
sol :
$-1\times\dfrac{-2}{~~~5}=\dfrac{2}{5}$
Multiplicative inverse =$\dfrac{5}{2}$
(vi) -1
sol :
Multiplicative inverse =-1
Q 5 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Name the property under multiplication used in each of the following :
(i) $\dfrac{-4}{5}\times{1}=1\times\dfrac{-4}{5}=-\dfrac{4}{5}$
sol : 1 is a multiplicative identity.
(ii) $-\dfrac{13}{17}\times\dfrac{-2}{7}=\dfrac{-2}{7}\times\dfrac{-13}{17}$
sol : Commutativity
(iii) $\dfrac{-19}{29}\times\dfrac{~~~29}{-19}=1$
sol : Multiplicative inverse
Q 6 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Question 6 :
Multiply $\dfrac{6}{13}$ by the reciprocal of $-\dfrac{7}{16}$.
sol :
$
\\=\dfrac{6}{13}\times\Bigg(Reciprocal~of~-\dfrac{7}{16}\Bigg)\\=\dfrac{6}{13}\times-\dfrac{16}{7}\\=-\dfrac{96}{91}
$
Q 7 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Question 7 :
Tell what property allow you to compute
$\dfrac{1}{3}\times\Bigg(6\times\dfrac{4}{3}\Bigg)$ as $\Bigg(\dfrac{1}{3}\times6\Bigg)\times\dfrac{4}{3}$
sol : Associativity
Q 8 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Is $\dfrac{8}{9}$ the multiplicative inverse of $-1\dfrac{1}{8}$ ?Why or why not ?
sol : If it is the multiplicative inverse , then the product should be 1. However, here the product is not 1 as
$\dfrac{8}{9}\times\Bigg(-1\dfrac{1}{8}\Bigg)=\dfrac{8}{9}\times\Bigg(-\dfrac{9}{8}\Bigg)=-1\not=1$
Q 9 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Is 0.3 the multiplicative inverse of $3\dfrac{1}{3}$ ? Why or Why not ?
sol :
$
\\=3\dfrac{1}{3}=\dfrac{10}{3}\\=0.3\times3\dfrac{1}{3}=0.3\times\dfrac{10}{3}\\=\dfrac{3}{10}\times\dfrac{10}{3}\\=1
$
Here is the product is 1 . Hence , 0.3 is the multiplicative inverse of $3\dfrac{1}{3}$
Q 10 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
Write :
(i) The rational number that does not have a reciprocal .
sol : 0 is a rational number but its reciprocal is not defined.
(ii) The rational numbers that are equal to their reciprocals .
sol : 1 and -1 are the rational numbers that are equal to their reciprocals .
(iii) The rational number that is equal to its negative .
sol : 0 is the rational number that is equal to its negative.
Q 11 - Ex 1.1 - Rational Numbers - NCERT Maths Class 8th - Chapter 1
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 $-\dfrac{1}{5}$
(iv) Reciprocal of $\dfrac{1}{x}$ , where $x\not=0~is~x$
(v) The product of two rational number is always a rational number .
(vi) The reciprocal of a positive rational number is positive rational number .
0 comments:
Post a Comment