NCERT Solutions Class 8 Maths Chapter 1 Rational Numbers Ex 1.1

NCERT Solutions Class 8 Mathematics 
Chapter – 1 (Rational Numbers) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 1 Rational Numbers Exercise 1.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter 1: Rational Numbers

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}}

Solution –

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

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

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

= \frac{-15}{30} +\frac{5}{2}

= -\frac{1}{2} + \frac{5}{2}

= \frac{4}{2}

= 2

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

Solution –

= \frac{2}{5}\times \left ( -\frac{3}{7} \right )-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}

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

= \frac{2}{5} \times \left ( \frac{-3}{7}+\frac{1}{14} \right)-\frac{3}{12}

= \frac{2}{5} \left ( \frac{-6+1}{14} \right ) -\frac{3}{12}

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

= -\frac{10}{70} -\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)  \mathbf{\frac{2}{8}}          (ii) \mathbf{\frac{-5}{9}}             (iii) \mathbf{\frac{-6}{-5}}                 (iv) \mathbf{\frac{2}{-9}}          (v) \mathbf{\frac{19}{-6}}

Solution –

(i)  \mathbf{\frac{2}{8}}
The Additive inverse of  \frac{2}{8}  = – \frac{2}{8}  

(ii) \mathbf{\frac{-5}{9}}
The additive inverse of  \frac{-5}{9} =  \frac{5}{9}

(iii) \mathbf{\frac{-6}{-5}} or \mathbf{\frac{6}{5}}
The additive inverse of  \frac{6}{5}  = – \frac{6}{5} 

(iv) \mathbf{\frac{2}{-9}} or \mathbf{-\frac{2}{9}}
The additive inverse of  – \frac{2}{9} = \frac{2}{9} 

(v) \mathbf{\frac{19}{-6}} or \mathbf{-\frac{19}{6}}

The additive inverse of  – \frac{19}{6} = \frac{19}{6} 

3. Verify that: -(-x) = x for:
(i) x = \mathbf{\frac{11}{15}}

(ii) x = \mathbf{\frac{-13}{17}}

Solution –

(i) x = \mathbf{\frac{11}{15}}

∴ – x = – \frac{11}{15}
– (– x) = – \left (-\frac{11}{15} \right ) = \frac{11}{15} = x (verified)

(ii) x = \mathbf{\frac{-13}{17}}

∴ – x = – \left ( - \frac{13}{17} \right ) = \frac{13}{17}

– (– x) = – \left ( - \frac{13}{17} \right ) = x (verified)

4. Find the multiplicative inverse of the following:
(i) -13

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

(iii) \mathbf{\frac{1}{5}}

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

(v) -1 × \mathbf{\frac{-2}{5}}
(vi) -1

Solution –

(i) -13
Multiplicative inverse of -13 = \frac{-1}{13}

(ii) \mathbf{\frac{-13}{19}}
Multiplicative inverse of \frac{-13}{19} = \frac{-19}{13}

(iii) \mathbf{\frac{1}{5}}
Multiplicative inverse of \frac{1}{5} = 5

(iv) \mathbf{\frac{-5}{8}} × \mathbf{\frac{-3}{7}} = \mathbf{\frac{15}{56}}
Multiplicative inverse of \frac{15}{56}  =  \frac{56}{15}

(v) -1 × \mathbf{\frac{-2}{5}} = \mathbf{\frac{2}{5}}
Multiplicative inverse of \frac{2}{5} = \frac{5}{2}

(vi) -1
Multiplicative inverse of -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}}

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

(iii) \mathbf{\frac{-19}{20}\times \frac{29}{-19} = 1}

Solution –

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

Here 1 is the multiplicative identity.

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

The property of commutativity is used in the equation.

(iii) \mathbf{\frac{-19}{20}\times \frac{29}{-19} = 1}

The multiplicative inverse is the property used in this equation.

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

Solution –
Reciprocal of \frac{-7}{16} = \frac{16}{-7} = -\frac{16}{7}
According to the question,

\frac{6}{13} × (Reciprocal of \frac{-7}{16})

\frac{6}{13} × -\frac{16}{7} = \frac{-96}{91}

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

Solution –
\frac{1}{3}\times \left ( 6\times \frac{4}{3} \right )  = \left ( \frac{1}{3} \times 6\right )\times \frac{4}{3}

Hence, the Associativity Property of Multiplication is used here.

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

Solution –
-1\frac{1}{8} = \frac{-9}{8}

Since multiplicative inverse of \frac{8}{9} is \frac{9}{8} but not − \frac{9}{8}.

\frac{8}{9} is not the multiplicative inverse of -1\frac{1}{8}

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

Solution –
3\frac{1}{3}= \frac{10}{3} and 0.3 = \frac{3}{10}

Multiplicative inverse of 0.3 or \frac{3}{10} = \frac{10}{3}.

Thus, 0.3 is the multiplicative inverse of 3\frac{1}{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.
The rational number that does not have a reciprocal is 0.
∵ 0 = \frac{0}{1} is not defined

(ii) The rational numbers that are equal to their reciprocals.
The rational numbers that are equal to their reciprocals are 1 and -1.
∵ 1 = \frac{1}{1}
Reciprocal of 1 = \frac{1}{1}  = 1, similarly, reciprocal of -1 = – 1
Thus, 1 and -1 are the required rational numbers.

(iii) The rational number that is equal to its negative.
The rational number that is equal to its negative is 0.
∵ Negative of 0 = – 0 = 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 \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 \mathbf{-\frac{1}{5}}
(iv) Reciprocal of \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.

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