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