exercise 1.1 class 8

Question 1:

Using appropriate properties find:$$

1. ${\displaystyle \frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}}$

2. ${\displaystyle \frac{2}{5}\times [\frac{-3}{7}]-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}}$

Solution:

1). ${\displaystyle \frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}}$ 

using commutative property of rational numbers

= ${\displaystyle \frac{-2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}}$

using distributive property of rational numbers

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

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

= ${\displaystyle (\frac{-3}{5})(\frac{5}{6})+\frac{5}{2}}$

= ${\displaystyle (\frac{-3}{6})+\frac{5}{2}}$

= ${\displaystyle (\frac{-3+5\times 3}{6})}$

=  ${\displaystyle \frac{-3+15}{6}}$

=  ${\displaystyle \frac{12}{6}}$

=  $2$

2). ${\displaystyle \frac{2}{5}\times [\frac{-3}{7}]-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}}$

using commutative property of rational numbers

= ${\displaystyle \frac{2}{5}\times [\frac{-3}{7}]+\frac{1}{14}\times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}}$

= ${\displaystyle \frac{2}{5}\times [\frac{-3}{7}+\frac{1}{14}]-\frac{1}{4}}$

= ${\displaystyle \frac{2}{5}\times [\frac{-3\times 2+1}{14}]-\frac{1}{4}}$

= ${\displaystyle \frac{2}{5}\times [\frac{-5}{14}]-\frac{1}{4}}$

= ${\displaystyle \frac{-1}{7}-\frac{1}{4}}$

= ${\displaystyle \frac{-4-7}{28}}$

=${\displaystyle \frac{-11}{28}}$

Question 2:

Write the additive inverse of each of the following:

${\displaystyle (i)\frac{2}{8}(ii)\frac{-5}{9}(iii)\frac{-6}{-5}(iv)\frac{2}{-9}(v)\frac{19}{-6}}$

Solution:

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

Additive inverse = ${\displaystyle \frac{-2}{8}}$

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

Additive inverse = ${\displaystyle \frac{5}{9}}$

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

Additive inverse = ${\displaystyle \frac{-6}{5}}$

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

Additive inverse = ${\displaystyle \frac{2}{9}}$

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

Additive inverse = ${\displaystyle \frac{19}{6}}$

Question 3:

Verify that −(−x) = x for.

(i) ${\displaystyle x=\frac{11}{15}}$ (ii) ${\displaystyle x=\frac{-13}{17}}$

Solution:

(i)${\displaystyle x=\frac{11}{15}}$

The additive inverse of ${\displaystyle x=\frac{11}{15}}$ is ${\displaystyle -x=-\frac{11}{15}}$ as ${\displaystyle \frac{11}{15}+\frac{-11}{15}=0}$

This equality ${\displaystyle \frac{11}{15}+\frac{-11}{15}=0}$ represents that the additive inverse of ${\displaystyle \frac{-11}{15}}$ is ${\displaystyle \frac{11}{15}}$ or it can be said that ${\displaystyle -(-\frac{11}{15})=\frac{11}{15}}$ i.e., −(−x) = x.

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

The additive inverse of ${\displaystyle x=\frac{-13}{17}}$ is ${\displaystyle -x=\frac{13}{17}}$ as ${\displaystyle \frac{-13}{17}+\frac{13}{17}=0}$

This equality ${\displaystyle \frac{-13}{17}+\frac{13}{17}=0}$ represents that the additive inverse of ${\displaystyle \frac{13}{17}}$ is ${\displaystyle \frac{-13}{17}}$ i.e., −(−x) = x.

Question 4:

Find the multiplicative inverse of the following.

(i)${\displaystyle -13}$

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

(iii)${\displaystyle \frac{1}{5}}$

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

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

(vi)${\displaystyle -1}$

Solution:

(i) ${\displaystyle -13}$

Multiplicative inverse = ${\displaystyle -\frac{1}{13}}$

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

Multiplicative inverse =$-\frac{19}{13}$

(iii)${\displaystyle \frac{1}{5}}$

Multiplicative inverse = ${\displaystyle 5}$

(iv)${\displaystyle \frac{-5}{8}\times \frac{-3}{7}=\frac{15}{56}}$

Multiplicative inverse=${\displaystyle \frac{56}{15}}$

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

Multiplicative inverse=${\displaystyle \frac{5}{2}}$

(vi) ${\displaystyle -1}$

Multiplicative inverse = ${\displaystyle -1}$

Question 5:

Name the property under multiplication used in each of the following:

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

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

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

Answer:

(i)Multiplicative identity.

(ii) Commutativity

(iii) Multiplicative inverse

Question 6:

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

Answer:

${\displaystyle \frac{6}{13}\times (}$ the reciprocal of ${\displaystyle \frac{-7}{16})}$

=${\displaystyle \frac{6}{13}\times \frac{-16}{7}=\frac{-96}{91}}$

Question 7:

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

Answer:

Associativity

Question 8:

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

Answer:

If it is the multiplicative inverse, then the product should be ${\displaystyle 1}$.

However, here, the product is not ${\displaystyle 1}$ as

${\displaystyle \frac{8}{9}\times (-1\frac{1}{8})=\frac{8}{9}\times (-\frac{9}{8})=-1\ne 1}$

Question 9:

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

Answer:

${\displaystyle 3\frac{1}{3}=\frac{10}{3}}$

${\displaystyle 0.3\times 3\frac{1}{3}=0.3\times \frac{10}{3}=\frac{3}{10}\times \frac{10}{3}=1}$

Here, the product is 1. Hence, 0.3 is the multiplicative inverse of ${\displaystyle 3\frac{1}{3}}$.

Question 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.

Answer:

(i) $0$ is a rational number but its reciprocal is not defined.

(ii) $1$ and $-1$ are the rational numbers that are equal to their reciprocals.

(iii) $0$ is the rational number that is equal to its negative.

Question 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 ${\displaystyle x\ne 0}$ is __________.

(v) The product of two rational numbers is always a __________.

(vi) The reciprocal of a positive rational number is __________.

Answer:

(i) No

(ii) 1, −1 

(iii)${\displaystyle -\frac{1}{5}}$

(iv)  x 

(v) Rational number

(vi) Positive rational number