5 Marks Questions

Question 15 Marks

Put the $(\checkmark ),$ wherever applicable.

Number	Natural Number	Whole Number	Integer	Fraction	Rational Number
$(a)$	$-114$
$(b)$	$\frac{19}{27}$
$(c)$	$\frac{623}{1}$
$(d)$	$-19\frac{3}{4}$
$(e)$	$\frac{73}{71}$
$(f)$	$0$

Answer

We know that, Natural numbers are $1, 2, 3, 4, ..$
Whole numbers are $0, 1, 2, 3, ...$
Integers are $-2, -1, 0, 1, 2, ...$
Fraction numbers are $\frac{1}{2},\frac{-1}{2},\frac{-7}{8}, \ ...$
Rational numbers are $\frac{3}{2},\frac{-1}{2},\frac{-7}{8},\ ...$
So, acoording to the numbrs systems,
$-114\rightarrow$ Integer and rational number
$a. \frac{19}{27}→$ Fraction and rational number
$b. \frac{623}{1}\rightarrow$ natture number, whole numbers, integer, fraction and rational numbers
$c. -19\frac{3}{4}=-\frac{79}{4}\rightarrow$ Rational number
$d. \frac{73}{71}\rightarrow$ Fraction and rational numbers
$e. 0\rightarrow$ Whole number, integer, fraction and rational number
Hence, the table is,

Number	Natural Number	Whole Number	Integer	Fraction	Rational Number
$(a)$	$-114$		$\checkmark$		$\checkmark$
$(b)$	$\frac{19}{27}$			$\checkmark$	$\checkmark$
$(c)$	$\frac{623}{1}$	$\checkmark$	$\checkmark$	$\checkmark$	$\checkmark$
$(d)$	$-19\frac{3}{4}$				$\checkmark$
$(e)$	$\frac{73}{71}$			$\checkmark$	$\checkmark$
$(f)$	$0$	$\checkmark$	$\checkmark$	$\checkmark$	$\checkmark$

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Question 25 Marks

If $\text{x}=\frac{1}{10}$ and $\text{y}=\frac{-3}{8},$ then evaluate $x ÷ y, x - y, xxy$ and $x ÷ y.$

Answer

Given, $\text{x}=\frac{1}{10}$ and $\text{y}=\frac{-3}{8}$
Now, $\text{x+y}=\frac{1}{10}\div\frac{(-3)}{8}=\frac{1}{10}-\frac{3}{8}$
$=\frac{1\times4}{10\times4}-\frac{3\times5}{8\times5}$
$=\frac{4}{40}-\frac{15}{40}=\frac{4-15}{40}$
$=-\frac{11}{40}$ and $\text{x-y}=\frac{1}{10}-\frac{(-3)}{8}=\frac{1}{10}\div\frac{3}{8}$
$=\frac{1\times4}{10\times4}\div\frac{3\times5}{8\times5}$
$=\frac{4}{40}\div\frac{15}{410}=\frac{4\div15}{40}$
$=\frac{19}{40}$
$\therefore$ Product of rational numbers $=\frac{\text{Product of numerators}}{\text{Product of denominators}}$
$\Rightarrow\text{x}\times\text{y}=\frac{1}{10}\times\frac{(-3)}{8}$
$=\frac{1\times(-3)}{10\times8}=\frac{-3}{80}$ and $\text{x}\div\text{y}=\frac{1}{10}\div\Big(\frac{-3}{8}\Big)$
The rexiprocal of $\Big(\frac{-3}{8}\Big)$ is $\frac{8}{-3}.$
So, $\text{x}\div\text {y}=\frac{1}{10}\times\frac{8}{-3}$ [dividing numerator and denominator by $2]$
$=\frac{1\times8}{10\times-3}=\frac{-8}{30}=\frac{-8\div2}{30\div2}$
$=\frac{-4}{15}$

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Question 35 Marks

Write each of the following rational numbers with positive denomi-natirs: $\frac{5}{-8}, \frac{15}{-28},\frac{-17}{-13}.$

Answer

We can write, $\frac{5}{-8}=\frac{5\times(-1)}{-8\times(-1)}=\frac{-15}{28}$
$\frac{15}{-28}$ can be written as $=\frac{5\times(-1)}{-8\times(-1)}=\frac{-15}{28}$ and $\frac{-17}{-13}$
can be written as $=\frac{-17\times(-1)}{-13\times(-1)}=\frac{17}{13},$ as both negative sings are cancelled.

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Question 45 Marks

Complete the following table by finding the sums.

Answer

Let

Here,
$a. =\frac{2}{3}+\Big(-\frac{1}{9}\Big)$
$=\frac{2}{3}-\frac{1}{9}$
$=\frac{2\times3}{3\times3}-\frac{1\times1}{9\times1}$
$=\frac{6}{9}-\frac{1}{9}$
$=\frac{6-1}{9}$
$=\frac{5}{9}$
$b. =\frac{2}{3}+\frac{4}{11}$
$=\frac{2\times11}{3\times11}+\frac{4\times3}{11\times3}$
$=\frac{22}{33}+\frac{12}{33}$
$=\frac{22+12}{33}$
$=\frac{34}{33}$
$c. =\frac{2}{3}+\Big(-\frac{5}{6}\Big )$
$=\frac{2}{3}-\frac{5}{6}$
$=\frac{2\times2}{3\times2}-\frac{5\times1}{6\times1}$
$=\frac{4}{6}-\frac{5}{6}$
$=\frac{4-5}{6}$
$=-\frac{1}{6}$
$d. =-\frac{5}{4}+\Big(-\frac{1}{9}\Big) $
$=\frac{-5}{4}-\frac{1}{9}$
$=\frac{-5\times9}{4\times9}-\frac{1\times4}{9\times4}$
$=\frac{-45}{36}-\frac{4}{36}$
$=\frac{-45-4}{36}$
$=\frac{-49}{36}$
$e. =-\frac{5}{4}+\Big(-\frac{5}{6}\Big)$
$=-\frac{5}{4}-\frac{5}{6}$
$=\frac{-5\times3}{4\times3}-\frac{5\times2}{6\times2}$
$=\frac{-15}{12}-\frac{10}{12}$
$=\frac{-15-10}{12}$
$=-\frac{25}{12}$
$f. =-\frac{1}{3}+\Big(-\frac{1}{9}\Big)$
$=-\frac{1}{3}-\frac{1}{9}$
$=\frac{-1\times3}{3\times3}-\frac{1\times1}{9\times1}$
$=\frac{-3}{9}-\frac{1}{9}$
$=\frac{-3-1}{9}$
$=\frac{-4}{9}$
$g. =-\frac{1}{3}+\frac{4}{11}$
$=\frac{-1\times11}{3\times11}+\frac{4\times3}{11\times3}$
$=\frac{-11}{33}+\frac{12}{33}$
$h. =-\frac{1}{3}+\Big(\frac{-5}{6}\Big)$
$=-\frac{1}{3}-\frac{5}{6}$
$=\frac{-1\times2}{3\times2}-\frac{5\times1}{6\times1}$
$=\frac{-2}{6}-\frac{5}{6}$
$=\frac{-2-5}{6}$
$=\frac{-7}{6}$
Hence, the complete table is:

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