Question 511 Mark
For every rational numbers $x, y$ and $z, x + (y × z) = (x + y) × (x + z).$
Answer
View full question & answer→For all rational numbers $a, b$ and $c.$
$a(b + c) = ab + ac.$
$a(b + c) = ab + ac.$
Questions · Page 2 of 2
1 Marks Question
Question 521 Mark
If $\text{a}\neq0,$ the multiplicative inverse of $\frac{\text{a}}{\text{b}}$ is $\frac{\text{b}}{\text{a}}.$
Answer
View full question & answer→True. Solution: If $\text{a}=0,$ then multiplicative inverse of $\frac{\text{a}}{\text{b}}$ is not difined. So, if $\text{a}\neq0,$ then multiplicative inverse of $\frac{\text{a}}{\text{b}}$ is $\frac{\text{b}}{\text{a}}.$
Question 531 Mark
The negative of a negative rational number is a positive rational number.
Answer
View full question & answer→True.
Solution:
Let be a positive rational number.
Then, $-x$ be the negative rational number.
Hence, negative of negative rational number $= -(-x) = x =$ Positive rational number.
Solution:
Let be a positive rational number.
Then, $-x$ be the negative rational number.
Hence, negative of negative rational number $= -(-x) = x =$ Positive rational number.
Question 541 Mark
If $\frac{\text{x}}{\text{y}}$ is the additive inverse of $\frac{\text{c}}{\text{d}},$ then $\frac{\text{x}}{\text{y}}+\frac{\text{c}}{\text{d}}=0.$
Answer
View full question & answer→ True.
Solution:
If $\frac{\text{x}}{\text{y}}$ is the additive inverse of $\frac{\text{c}}{\text{d}}.$
i.e. $\frac{\text{x}}{\text{y}}+\frac{\text{c}}{\text{d}}=0$
Solution:
If $\frac{\text{x}}{\text{y}}$ is the additive inverse of $\frac{\text{c}}{\text{d}}.$
i.e. $\frac{\text{x}}{\text{y}}+\frac{\text{c}}{\text{d}}=0$
Question 551 Mark
The rational number $\frac{-8}{-3}$ lies neither to the right nor to the left of zero on the number line.
Answer
View full question & answer→ False.
Solution:
$-\frac{-8}{-3}=-\frac{8}{3}$ is a positive rational number.
Hence, it lies on the right of zero on the number line.
Solution:
$-\frac{-8}{-3}=-\frac{8}{3}$ is a positive rational number.
Hence, it lies on the right of zero on the number line.
Question 561 Mark
The additive inverse of $\frac{1}{2}$ is $-2.$
Answer
View full question & answer→Let additive inverse of $\frac{1}{2}$ be $x.$
i.e. $\frac{1}{2}+\text{x}=0$
$\Rightarrow\text{x}=\frac{-1}{2}$
Hence, additive inverse of $\frac{1}{2}$ is $\frac{-1}{2}.$
i.e. $\frac{1}{2}+\text{x}=0$
$\Rightarrow\text{x}=\frac{-1}{2}$
Hence, additive inverse of $\frac{1}{2}$ is $\frac{-1}{2}.$
Question 571 Mark
There are countless rational numbers between $\frac{5}{6}$ and $\frac{8}{9}.$
Answer
View full question & answer→True. Solution: $\frac{5}{6}$ and $\frac{8}{9}$ are rational number and there are infinite (countiess) rational numbers lie between $\frac{5}{6}$ and $\frac{8}{9}.$ Note: We know that there infinite rational numbers lie between two rational numbers.
Question 581 Mark
$(213 × 657)^{-1}= 213^{-1}× \_\_\_\_\_.$
Answer
View full question & answer→$(213 × 657)^{-1}= 213^{-1}×$ $\frac{1}{657}$
Solution:
Suppose, $(213 × 657)^{-1}= 213^{-1}× x$ $\Rightarrow\frac{1}{213\times657}=\frac{1}{213}\times\text{x}$ $\Rightarrow\text{x}=\frac{213}{213\times657}$ $\Rightarrow\text{x}=\frac{1}{657}$
Solution:
Suppose, $(213 × 657)^{-1}= 213^{-1}× x$ $\Rightarrow\frac{1}{213\times657}=\frac{1}{213}\times\text{x}$ $\Rightarrow\text{x}=\frac{213}{213\times657}$ $\Rightarrow\text{x}=\frac{1}{657}$
Question 591 Mark
Rational numbers can be added (or multiplied) in any order $\frac{-4}{5}\times\frac{-6}{5}=\frac{-6}{5}\times\frac{-4}{5}$
Answer
View full question & answer→We know, $\frac{-4}{5}\times\frac{-6}{5}=\frac{-6}{5}\times\frac{-4}{5}$
$\Rightarrow\frac{24}{25}=\frac{24}{25}$
So, rational mumber can be added (or multiplied) in any order.
Then, $ab = ba [$Commutative under multiplication$] a + b = b + a$
$[$Commutative under addition$]$
Hence, rational numbers can be added (or multiplied) in any order.
$\Rightarrow\frac{24}{25}=\frac{24}{25}$
So, rational mumber can be added (or multiplied) in any order.
Then, $ab = ba [$Commutative under multiplication$] a + b = b + a$
$[$Commutative under addition$]$
Hence, rational numbers can be added (or multiplied) in any order.
Question 601 Mark
If $y$ be the reciprocal of $x,$ then the reciprocal of $y^2$ in terms of $x$ will be _____.
Answer
View full question & answer→If $y$ be the reciprocal of $x,$ then the reciprocal of $y^2$ in terms of $x$ will be $x^2$
Solution:
Given, $\frac{1}{\text{x}}=\text{y}$
Now, reciprocal of $\text{y}^2=\frac{1}{\text{y}^2}=\frac{1}{\Big(\frac{1}{\text{x}}\Big)^2}$
$=\text{x}^2$
Solution:
Given, $\frac{1}{\text{x}}=\text{y}$
Now, reciprocal of $\text{y}^2=\frac{1}{\text{y}^2}=\frac{1}{\Big(\frac{1}{\text{x}}\Big)^2}$
$=\text{x}^2$
Question 611 Mark
The rational numbers $\frac{1}{2}$ and $-1$ are on the opposite sides of zero on the number line.
Answer
View full question & answer→Since, positive rational number and negative rational number are on the opposite sides of zero on the number line. Hence, $-\frac{1}{2}$ and $-1$ are on the opposite sides of zero on the number line.
Question 621 Mark
$\frac{9}{6}$ lies between $1$ and $2.$
Answer
View full question & answer→First, we convert the given rational numbers with denominator as $6,$
we get $1=1\times\frac{6}{6}=\frac{6}{6}$
$2=2\times\frac{6}{6}=\frac{12}{6}$
$\because\frac{6}{6}<\frac{9}{6}<\frac{12}{6}$
$\therefore1<\frac{9}{6}<2$ Therefore, $\frac{9}{6}$ lies between $1$ and $2.$
we get $1=1\times\frac{6}{6}=\frac{6}{6}$
$2=2\times\frac{6}{6}=\frac{12}{6}$
$\because\frac{6}{6}<\frac{9}{6}<\frac{12}{6}$
$\therefore1<\frac{9}{6}<2$ Therefore, $\frac{9}{6}$ lies between $1$ and $2.$
Question 631 Mark
If $x$ and $y$ are two rational numbers such that $x > y,$ then $x - y$ is always a positive rational number.
Answer
View full question & answer→ If $x$ and $y$ are two rational numbers such that $x > y.$ Then, there are three possible cases, i.e. Case $I$ $x$ and $y$ both are positive. Case $II\ x$ is positive and y is negative. Case $III\ x$ and $y$ both are negative. In all three cases, $x - y$ is always a positive rational number.
Question 641 Mark
Identify the rational number that does not belong with the other three. Explain your reasoning. $\frac{-5}{11},\frac{-1}{2},\frac{-4}{9},\frac{-7}{3}$
Answer
View full question & answer→Does not belong with the other three. Since, $\frac{-7}{3}$ as it is smaller than $-1$ whereas rest of the numbers are greater than $-1.$
Question 651 Mark
The multiplicative inverse of $\frac{4}{3}$ is _____.
Answer
View full question & answer→The multiplicative inverse of $\frac{4}{3}$ is $\frac{3}{4}.$
Solution:
Let $x$ be the multiplicative inverse of $\frac{4}{3}.$
By the definition, i.e. $\text{x}\times\frac{4}{3}=1$
$\Rightarrow\text{x}=\frac{3}{4}$
Hence, the multiplication inverse of $\frac{4}{3}$ is $\frac{3}{4}.$
Solution:
Let $x$ be the multiplicative inverse of $\frac{4}{3}.$
By the definition, i.e. $\text{x}\times\frac{4}{3}=1$
$\Rightarrow\text{x}=\frac{3}{4}$
Hence, the multiplication inverse of $\frac{4}{3}$ is $\frac{3}{4}.$
Question 661 Mark
The numbers _____ and _____ are their own reciprocal.
Answer
View full question & answer→The numbers $1$ and $-1$ are their own reciprocal.
Solution:
The reciprocal of $1$ and $-1$ are $\frac{1}{1}$ and $\frac{1}{-1},$ i.e. $1$ and $-1$ respectively.
Solution:
The reciprocal of $1$ and $-1$ are $\frac{1}{1}$ and $\frac{1}{-1},$ i.e. $1$ and $-1$ respectively.
Question 671 Mark
The population of India in $2004 - 05$ is a rational number.
Answer
View full question & answer→The population of India in $2004-05$ is a rational number.
Question 681 Mark
The rational number $10.11$ in the from $\frac{\text{p}}{\text{q}}$ is _____.
Answer
View full question & answer→The rational number $10.11$ in the from $\frac{\text{p}}{\text{q}}$ is $\frac{1011}{100}.$
Solution: Let, $\text{x}=10.11$
$\Rightarrow100\text{x}=10.11\times100 [$multiplying both sides by $100]$
$\Rightarrow100\text{x}=1011$
$\Rightarrow\frac{100\text{x}}{100}=\frac{1011}{100} [$dividing both sides by $100]$
$\Rightarrow\text{x}=\frac{1011}{100}$
Hence, the rational number $10.11$ in the form $\frac{\text{p}}{\text{q}}$ is $\frac{1011}{100}.$
Solution: Let, $\text{x}=10.11$
$\Rightarrow100\text{x}=10.11\times100 [$multiplying both sides by $100]$
$\Rightarrow100\text{x}=1011$
$\Rightarrow\frac{100\text{x}}{100}=\frac{1011}{100} [$dividing both sides by $100]$
$\Rightarrow\text{x}=\frac{1011}{100}$
Hence, the rational number $10.11$ in the form $\frac{\text{p}}{\text{q}}$ is $\frac{1011}{100}.$
Question 691 Mark
The rational numbers $\frac{1}{2}$ and $-\frac{5}{2}$ are on the opposite sides of $0$ on the number line.
Answer
View full question & answer→True.
Solution:
Positive rational number and negative rational number remain on opposite sides of zero on the number line.
Solution:
Positive rational number and negative rational number remain on opposite sides of zero on the number line.
Question 701 Mark
Can you find a rational number whose multiplicative inverse is $-1?$
Answer
View full question & answer→No, we cannot find a rational number whose multiplicative inverse is $-1.$
Question 711 Mark
$\frac{-5}{7}$ is _____ than $-3.$
Answer
View full question & answer→$\frac{-5}{7}$ is $ > $ than $-3.$
Solution: First we convert the given rational number into like denominator.
Now, $LCM$ of and $1 = 7.$
$-3=\frac{-3\times7}{7}$
$[$On multiolying and dividing by $7]$
$=\frac{-21}{7}$ As, $\frac{-5}{7}>\frac{-21}{7}$ i.e. $\frac{-5}{7}>-3$
Hence, $-\frac{5}{7}$ is greater than $-3.$
Solution: First we convert the given rational number into like denominator.
Now, $LCM$ of and $1 = 7.$
$-3=\frac{-3\times7}{7}$
$[$On multiolying and dividing by $7]$
$=\frac{-21}{7}$ As, $\frac{-5}{7}>\frac{-21}{7}$ i.e. $\frac{-5}{7}>-3$
Hence, $-\frac{5}{7}$ is greater than $-3.$
Question 721 Mark
If $\frac{\text{p}}{\text{q}}$ is a rational number, then $p$ cannot be equal to zero.
Answer
View full question & answer→If $\frac{\text{p}}{\text{q}}$ is a rational number. Then, $p$ can be equal to any integer. i.e. $p$ can be zero.
Question 731 Mark
Between the numbers $\frac{15}{20}$ and $\frac{35}{40},$ the greater number is _____.
Answer
View full question & answer→Between the numbers $\frac{15}{20}$ and $\frac{35}{40},$
the greater number is $\Big(\frac{35}{40}\Big).$
Solution: Given number are $\frac{15}{20}$ and $\frac{35}{40}.$
$LCM$ of $20$ and $40 = 2 \times 2 \times 2 \times 5 = 40$
Now, $\frac{15}{20}=\frac{15}{20}\times\frac{2}{2} [$On multiplying and dividing by $2]$
$\begin{array}{c|c}2& 20,\ \ \ 40\ \\\hline2&\ \ \ \ 10, \ \ \ 20\ \ \ \ \ \\ \hline2&\ 5, \ \ 10\\\hline5&5,\ \ 5\\ \hline&1,\ \ 1 \end{array}$
$=\frac{30}{40}$ On comparing,
$\frac{35}{40}>\frac{30}{40}$
$\Rightarrow\frac{35}{40}>\frac{15}{20}$
Hence, $\frac{35}{40}$ is greater.
the greater number is $\Big(\frac{35}{40}\Big).$
Solution: Given number are $\frac{15}{20}$ and $\frac{35}{40}.$
$LCM$ of $20$ and $40 = 2 \times 2 \times 2 \times 5 = 40$
Now, $\frac{15}{20}=\frac{15}{20}\times\frac{2}{2} [$On multiplying and dividing by $2]$
$\begin{array}{c|c}2& 20,\ \ \ 40\ \\\hline2&\ \ \ \ 10, \ \ \ 20\ \ \ \ \ \\ \hline2&\ 5, \ \ 10\\\hline5&5,\ \ 5\\ \hline&1,\ \ 1 \end{array}$
$=\frac{30}{40}$ On comparing,
$\frac{35}{40}>\frac{30}{40}$
$\Rightarrow\frac{35}{40}>\frac{15}{20}$
Hence, $\frac{35}{40}$ is greater.
Question 741 Mark
Rational numbers can be added or multiplied in any _____.
Answer
View full question & answer→Rational numbers can be added or multiplied in any order. Solution: Rational numbers can be added or multiplied in any order and this concept is known as commutative property.
Question 751 Mark
Name the property used in each of the following. $\frac{-2}{7}+0=0+\frac{-2}{7}=-\frac{2}{7}$
Answer
View full question & answer→Existence of additive identity.
Question 761 Mark
The reciprocal of a non-zero rational number $\frac{\text{q}}{\text{p}}$ is the rational number $\frac{\text{q}}{\text{p}}.$
Answer
View full question & answer→False. Solution: The reciprocal of a non-zero rational number $\frac{\text{q}}{\text{p}}.$ is the rational number $\frac{\text{p}}{\text{q}}.$
Question 771 Mark
$0$ is whole number but it is not a rational number.
Answer
View full question & answer→$0$ is a whole number and also a rational number.
Question 781 Mark
Find the multiplicative inverse of: $3\frac{1}{3}.$
Answer
View full question & answer→Given number is $3\frac{1}{3},$ i.e. $\frac{10}{3}.$The multiplicative inverse of $\frac{10}{3}$ is $\frac{3}{10}.$
Question 791 Mark
Simplify each of the following by using suitable property. Also name the property. $\Big[\frac{1}{2}\times\frac{1}{4}\Big]+\Big[\frac{1}{2}\times6\Big]$
Answer
View full question & answer→Given, $\Big[\frac{1}{2}\times\frac{1}{4}\Big]+\Big[\frac{1}{2}\times6\Big]$ $=\frac{1}{2}\Big[\frac{1}{4}+6\Big]$ $=\frac{1}{2}\Big[\frac{1+24}{4}\Big]$ [Using distributive property over addition] $=\frac{25}{8}$
Question 801 Mark
$-1$ is not the reciprocal of any rational number.
Answer
View full question & answer→$-1$ is the reciprocal of $-1.$
Question 811 Mark
The negative of $1$ is _____.
Answer
View full question & answer→The negative of $1$ is $-1$
Solution: $-1$ The negative of $1$ is $-1.$
Solution: $-1$ The negative of $1$ is $-1.$
Question 821 Mark
The negative of $1$ is $1$ itself
Answer
View full question & answer→The negative of $1$ is $-1.$
Question 831 Mark
Simplify: $\frac{3}{7}+\frac{-2}{21}\times\frac{-5}{6}$
Answer
View full question & answer→Given, $\frac{3}{7}+\frac{-2}{21}\times\frac{-5}{6}$$=\frac{3}{7}+\frac{5}{63}$
$=\frac{27+5}{63}$
$=\frac{32}{63}$
$=\frac{27+5}{63}$
$=\frac{32}{63}$
Question 841 Mark
If $\frac{\text{r}}{\text{s}}$ is a rational number, then $s$ cannot be equal to zero.
Answer
View full question & answer→If $\frac{\text{r}}{\text{s}}$ is a rational number, Then, $s$ can be any non-zero integer. Hence, $s$ cannot be equal to zero.
Question 851 Mark
The equivalent of $\frac{5}{7},$ whose numerator is $45$ is _____.
Answer
View full question & answer→The equivalent of $\frac{5}{7},$ whose numerator is $45$ is $\Big(\frac{45}{63}\Big).$
Solution:
Take $\frac{5}{7},\frac{5}{7}\times\frac{9}{9} [$On multiplying numberator and denominator by denominator by $9]$
$=\frac{45}{63}$
Hence, $\frac{45}{63}$ is equivalent to $\frac{5}{7}.$
Solution:
Take $\frac{5}{7},\frac{5}{7}\times\frac{9}{9} [$On multiplying numberator and denominator by denominator by $9]$
$=\frac{45}{63}$
Hence, $\frac{45}{63}$ is equivalent to $\frac{5}{7}.$
Question 861 Mark
Name the property used in each of the following.
$-\frac{1}{3}+\bigg[\frac{4}{9}+\Big(\frac{-4}{3}\Big)\bigg]=\Big[\frac{1}{3}\times\frac{4}{9}\Big]+\Big[\frac{-4}{3}\Big]$
$-\frac{1}{3}+\bigg[\frac{4}{9}+\Big(\frac{-4}{3}\Big)\bigg]=\Big[\frac{1}{3}\times\frac{4}{9}\Big]+\Big[\frac{-4}{3}\Big]$
Answer
View full question & answer→ Associative property over addition.
Question 871 Mark
Verify -(-x) = x for: $\text{x}=\frac{-7}{9}$
Answer
View full question & answer→Given, $\text{x}=\frac{-7}{9}$ $\Rightarrow-\text{x}=-\Big(\frac{-7}{9}\Big)$ $\Rightarrow-\text{x}=\frac{7}{9}$ $\Rightarrow-(-\text{x})=\frac{-7}{9}$ $=\text{x}$
Question 881 Mark
The equivalent rational number of $\frac{7}{9},$ whose denominator is $45$ is _____.
Answer
View full question & answer→The equivalent rational number of $\frac{7}{9},$ whose denominator is $45$ is $\Big(\frac{35}{45}\Big).$
Solution: Take $\frac{7}{9},\frac{7}{9}\times\frac{5}{5}$
$ [$On multiplying numberator and denominator by $5]$
$=\frac{35}{45}$
Hence, $\frac{35}{45}$ is equivalent to $\frac{7}{9}.$
Solution: Take $\frac{7}{9},\frac{7}{9}\times\frac{5}{5}$
$ [$On multiplying numberator and denominator by $5]$
$=\frac{35}{45}$
Hence, $\frac{35}{45}$ is equivalent to $\frac{7}{9}.$
Question 891 Mark
$-\frac{3}{4}$ is smaller than $-2.$
Answer
View full question & answer→Here, $\frac{-3}{4}$ and $-2$ (like) First,
we do same denominator.
We get, $\frac{-3}{4}$ and $\frac{-2\times4}{1\times4}$
$\Rightarrow\frac{-3}{4}$ and $\frac{-8}{4}$
Now, comparing both numbers, $\frac{-3}{4}>\frac{-8}{4}$
$\Rightarrow\frac{-3}{4}>-2$ So, $-\frac{3}{4}$ is greater than $-2.$
we do same denominator.
We get, $\frac{-3}{4}$ and $\frac{-2\times4}{1\times4}$
$\Rightarrow\frac{-3}{4}$ and $\frac{-8}{4}$
Now, comparing both numbers, $\frac{-3}{4}>\frac{-8}{4}$
$\Rightarrow\frac{-3}{4}>-2$ So, $-\frac{3}{4}$ is greater than $-2.$
Question 901 Mark
The rational numbers $\frac{1}{3}$ and $\frac{-1}{3}$ are on the _____ sides of zero on the number line.
Answer
View full question & answer→The rational numbers $\frac{1}{3}$ and $\frac{-1}{3}$ are on the opposite sides of zero on the number line. Solution: 
Question 911 Mark
Name the property used in each of the following. $-\frac{7}{11}\times\frac{-3}{5}=\frac{-3}{5}\times\frac{-7}{11}$
Answer
View full question & answer→Commutative property over multiplication.
Question 921 Mark
There are _____ rational numbers between any two rational numbers.
Answer
View full question & answer→There are infinite rational numbers between any two rational numbers.
Question 931 Mark
For rational numbers $\frac{\text{a}}{\text{b}},\frac{\text{c}}{\text{d}}$ and $\frac{\text{e}}{\text{f}}$ we have $\frac{\text{a}}{\text{b}}\times\Big(\frac{\text{c}}{\text{d}}+\frac{\text{e}}{\text{f}}\Big)=$ _____ + _____.
Answer
View full question & answer→For rational numbers $\frac{\text{a}}{\text{b}},\frac{\text{c}}{\text{d}}$ and $\frac{\text{e}}{\text{f}}$ we have $\frac{\text{a}}{\text{b}}\times\Big(\frac{\text{c}}{\text{d}}+\frac{\text{e}}{\text{f}}\Big)=$ $\frac{\text{ac}}{\text{bd}}+\frac{\text{ae}}{\text{bf}}.$ Solution: If $\frac{\text{a}}{\text{b}}\times\Big(\frac{\text{c}}{\text{d}}+\frac{\text{e}}{\text{f}}\Big)$ $=\frac{\text{a}}{\text{b}}\times\frac{\text{c}}{\text{d}}+\frac{\text{a}}{\text{b}}\times\frac{\text{e}}{\text{f}}$ $=\frac{\text{ac}}{\text{bd}}+\frac{\text{ae}}{\text{bf}}$
Question 941 Mark
The rational number $\frac{7}{-4}$ lies to the right of zero on the number line.
Answer
View full question & answer→False. Solution: Since, $-\frac{7}{-4}$ is a negative rational number. So, it lies on the left of zero on the number line.
Question 951 Mark
For all rational numbers $x$ and $y, x × y = y × x.$
Answer
View full question & answer→For all rational numbers $x$ and $y, x × y = y × x.$
Question 961 Mark
Simplify each of the following by using suitable property. Also name the property. $\Big[\frac{1}{5}\times\frac{2}{15}\Big]-\Big[\frac{1}{5}\times\frac{2}{5}\Big]$
Answer
View full question & answer→Given, $\Big[\frac{1}{5}\times\frac{2}{15}\Big]-\Big[\frac{1}{5}\times\frac{2}{5}\Big]$ $=\frac{1}{5}\Big[\frac{2}{15}-\frac{2}{5}\Big]$ $=\frac{1}{5}\Big[\frac{2-6}{15}\Big]$ [Using distributive property over addition] $=\frac{-4}{75}$
Question 971 Mark
Simplify: $\frac{32}{5}+\frac{23}{11}\times\frac{22}{15}$
Answer
View full question & answer→Given, $\frac{32}{5}+\frac{23}{11}\times\frac{22}{15}$$=\frac{32}{5}+\frac{46}{15}$
$=\frac{96+46}{15}$
$=\frac{142}{15}$
$=\frac{96+46}{15}$
$=\frac{142}{15}$