Question 511 Mark
Write $‘T’$ for true and $‘F’$ for false for the following: Rational numbers are always closed under division.
AnswerFalse.Solution:
$\frac{\text{a}}{0}=\infty$
Hence, Rational numbers are not always closed under division.
View full question & answer→Question 521 Mark
Find the multiplicative inverse (i.e., reciprocal) of: $\Big(\frac{5}{8}\Big)^{-1}$
Answer$\Big(\frac{5}{8}\Big)^{-1}=\frac{8}{5}$
View full question & answer→Question 531 Mark
Fill in the blanks: $\Big(\frac{15}{7}\times\frac{-21}{10}\Big)\times\frac{-5}{6}=(....)\times\Big(\frac{-21}{10}\times\frac{- 5}{6}\Big)$
Answer$\Big(\frac{15}{7}\times\frac{-21}{10}\Big)\times\frac{-5}{6}=\Big(\frac{15}{7}\Big)\times\Big(\frac{-21}{10}\times\frac{- 5}{6}\Big)$
View full question & answer→Question 541 Mark
Are rational numbers always closed under division?
AnswerNo, not always closed under division.
View full question & answer→MCQ 551 Mark
Tick $(\checkmark)$ the correct answer the following: The reciprocal of a negative rational number:
- A
Is a positive rational number.
- ✓
Is a negative rational number.
- C
Can be either a positive or a negative rational number.
- D
AnswerCorrect option: B. Is a negative rational number.
The reciprocal of a negative rational number is also a negative rational number.
View full question & answer→Question 561 Mark
Find the multiplicative inverse (i.e., reciprocal) of: $\frac{13}{25}$
AnswerMultiplicative inverse of $\frac{13}{25}=\frac{25}{13}$
View full question & answer→Question 571 Mark
Fill in the blank: $(.....)\div\Big(\frac{-7}{5}\Big)=\frac{10}{19}$
AnswerLet required number $= x$ Then, $\text{x}\div\frac{-7}{5}=\frac{10}{19}$
$\Rightarrow\text{x}=\frac{10}{19}\div\frac{5}{-7}$
$\Rightarrow\text{x}=\frac{10}{19}\times\frac{-7}{5}$
$=\frac{-70}{95}$
$=\frac{-70\div5}{95\div5}$
$=\frac{-14}{19}$
$\therefore$ Required number $=\frac{-14}{19}$
View full question & answer→Question 581 Mark
Find the multiplicative inverse (i.e., reciprocal) of: $\frac{0}{2}$
AnswerMultiplicative inverse of $\frac{0}{2}$ does not exists.
View full question & answer→Question 591 Mark
Write $‘T’ $for true and $‘F’ $ for false for the following:$-\Big(\frac{-7}{8}\Big)$
Answer$-\Big(\frac{-7}{8}\Big)$
$=1\times\Big(\frac{-7}{8}\Big)$
$=\frac{-1\times-7}{8}$
$=\frac{7}{8}$
View full question & answer→Question 601 Mark
Add the following rational numbers.
$\frac{-11}{8}\ \text{and}\ \frac{5}{8}$
Answer $\frac{-11}{8}+\frac{5}{8}$
$=\frac{-11+5}{8}$
$=\frac{-6}{8}=\frac{-6\div2}{8\div2}=\frac{-3}{4}$
View full question & answer→Question 611 Mark
Is subtraction commutative on rational numbers?
AnswerNo, subtraction is not commutative.
View full question & answer→Question 621 Mark
Add the following rational numbers. $\frac{5}{6}\ \text{and}\ \frac{-1}{6}$
Answer$\frac{5}{6}+\frac{-1}{6}$
$=\frac{5-1}{6}$
$=\frac{4}{6}$
$=\frac{4\div2}{6\div2}=\frac{2}{3}$
View full question & answer→Question 631 Mark
Fill in the blanks. $-12+\Big(\frac{7}{12}+\frac{-9}{11}\Big)=\Big(-12+\frac{7}{12}\Big)+(.....)$
Answer$-12+\Big(\frac{7}{12}+\frac{-9}{11}\Big)=\Big(-12+\frac{7}{12}\Big)+\Big(\frac{-9}{11}\Big)$ (By Associative Law of addition)
View full question & answer→Question 641 Mark
Fill in the blank:
$(-12)\div(.....)=\frac{-6}{5}$
AnswerLet required number $= x$
Then,
$(-12)\div(.....)=\frac{-6}{5}$
$\Rightarrow\frac{-12}{1}\times\frac{1}{\text{x}}=\frac{-6}{5}$
$\Rightarrow\frac{1}{\text{x}}=\frac{-6}{5}\div\frac{12}{1}$
$\Rightarrow\frac{-6}{5}\times\frac{1}{-12}$
$=\frac{-6}{-60}=\frac{6}{60}$
$\therefore\text{x}=\frac{60}{6}$
$=10$
$\therefore$ Required number $= 10$
View full question & answer→Question 651 Mark
Fill in the blank. $\frac{25}{8}\div(......)=-10$
Answer$\frac{25}{8}\div\frac{-5}{16}=-10$Solution:
$\frac{25}{8}\div\text{x}=-10$
$\Rightarrow\text{x}=\frac{25}{8}\div-10$
$\Rightarrow\text{x}=\frac{25}{8}\times\frac{1}{-10}$
$\Rightarrow\text{x}=\frac{25\times1}{8\times-10}$
$\Rightarrow\text{x}=\frac{25}{-80}$
$\Rightarrow\text{x}=\frac{25\times-1}{-80\times-1}$
$\Rightarrow\text{x}=\frac{-25}{80}$
$\Rightarrow\text{x}=\frac{-25\div5}{80\div5}$
$\Rightarrow\text{x}=\frac{-5}{16}$
View full question & answer→Question 661 Mark
Find the multiplicative inverse (i.e., reciprocal) of: $18$
AnswerMultiplicative inverse of $18=\frac{1}{18}$
View full question & answer→Question 671 Mark
Add the following rational numbers. $\frac{-17}{15}\ \text{and}\ \frac{-1}{15}$
Answer$\frac{-17}{15}+\frac{-1}{15}$
$=\frac{-17-1}{15}$
$=\frac{-18}{15}$
$=\frac{-18\div3}{15\div3}=\frac{-6}{5}$
View full question & answer→Question 681 Mark
Fill in the blanks: $\frac{-12}{5}\times\Big(\frac{4}{15}\times\frac{25}{-16}\Big)=\Big(\frac{-12}{5}\times\frac{4}{15}\Big)\times(....)$
Answer$\frac{-12}{5}\times\Big(\frac{4}{15}\times\frac{25}{-16}\Big)=\Big(\frac{-12}{5}\times\frac{4}{15}\Big)\times\frac{25}{-16}$
View full question & answer→Question 691 Mark
Add the following rational numbers. $\frac{-5}{16}\ \text{and}\ \frac{7}{24}$
AnswerThe denominators of the given rational number are $16$ and $24.$
$LCM$ of $16, 24$ is $48$
Now,
$\frac{-5}{16}=\frac{-5\times3}{16\times3}$
$=\frac{-15}{48}$
and $\frac{7}{24}=\frac{7\times2}{24\times2}$
$=\frac{14}{48}$
$\therefore\frac{-5}{16}+\frac{7}{24}$
$=\frac{-15}{48}+\frac{14}{48}$
$=\frac{-15+14}{48}$
$=\frac{-1}{48}$
View full question & answer→Question 701 Mark
Fill in the blank: The reciprocal of $\frac{1}{\text{a}}$, where $a ≠ 0,$ is _____.
AnswerThe reciprocal of $\frac{1}{\text{a}}$ where $a ≠ 0$ is $a.$
View full question & answer→Question 711 Mark
Are rational numbers always commutative under division?
AnswerNo, not always commutative.
View full question & answer→Question 721 Mark
Name the property of multiplication illustrated of the following statement: $\frac{-11}{15}\times\frac{15}{-11}=\frac{15}{-11}\times\frac{-11}{15}=1$
Answer$\frac{-11}{15}\times\frac{15}{-11}=\frac{15}{-11}\times\frac{-11}{15}=1$ It is existance of multiplicative inverse.
View full question & answer→Question 731 Mark
Is addition associative on rational numbers?
Answer Yes, addition associative.
View full question & answer→Question 741 Mark
Add the following rational numbers. $\frac{-6}{11}\ \text{and}\ \frac{-4}{11}$
Answer$\frac{-6}{11}+\frac{-4}{11}$
$=\frac{-6-4}{11}$
$=\frac{-10}{11}$
View full question & answer→Question 751 Mark
Add the following rational numbers. $\frac{-2}{5}\ \text{and}\ \frac{4}{5}$
Answer$\frac{-2}{5}+ \frac{4}{5}$
$=\frac{-2+4}{5}$
$=\frac{2}{5}$
View full question & answer→Question 761 Mark
Are rational numbers always associative under division?
AnswerNo, not always associative.
View full question & answer→Question 771 Mark
Write $‘T’$ for true and $‘F’$ for false for the following:
Rational numbers are always closed under subtraction.
AnswerLet there be two rational numbers $\frac{\text{a}}{\text{b}}$ and $\frac{\text{c}}{\text{d}}$ Then, $\frac{\text{a}}{\text{b}}-\frac{\text{c}}{\text{d}}=\frac{\text{ad}-\text{bc}}{\text{bd}}$ Which is also a rational number Hence, Rational numbers are always closed under subtraction.
View full question & answer→Question 781 Mark
Write $‘T’$ for true and $‘F’$ for false for the following: Subtraction is commutative on rational numbers.
AnswerLet $\frac{\text{a}}{\text{b}}$ and $\frac{\text{c}}{\text{d}}$ represent rational numbers.
Now, we have:$\frac{\text{a}}{\text{b}}-\frac{\text{c}}{\text{d}}=\frac{\text{ad}-\text{bc}}{\text{bd}}$
$\frac{\text{c}}{\text{d}}-\frac{\text{a}}{\text{b}}=\frac{\text{bc}-\text{ad}}{\text{bd}}$
$\therefore\frac{\text{a}}{\text{b}}-\frac{\text{c}}{\text{d}}\neq\frac{\text{c}}{\text{d}}-\frac{\text{a}}{\text{b}}$
View full question & answer→Question 791 Mark
Fill in the blanks.
$-9+\frac{-21}{8}=(......)+(-9)$
Answer $-9+\frac{-21}{8}=\frac{-21}{8}+(-9)$
(By commutative Law of addition)
View full question & answer→Question 801 Mark
Fill in the blank. $(-1)+(......)=\frac{-2}{9}$
AnswerLet the blank space be $x$ Now, we have:
$(-1)\times\text{x}=\frac{-2}{9}$
$\Rightarrow\text{x}=\frac{-2}{9}+1$
$\Rightarrow\text{x}=\frac{-2+9}{9}$
$\Rightarrow\text{x}=\frac{7}{9}$
View full question & answer→Question 811 Mark
Fill in the blanks with the correct symbol out of $>, =$ and $ <.$ $\frac{-3}{7}\ ....\ \frac{6}{-13}$
AnswerBetween $\frac{-3}{7}\ \text{and}\ \frac{6}{-13}$ or $\frac{-3}{7}\ \text{and}\ \frac{-6}{13}$
$LCM$ of $7$ and $13 = 91$
$\therefore\frac{-3}{7}=\frac{-3\times13}{7\times13}=\frac{-39}{91}$ and
$\frac{-6}{13}=\frac{-6\times7}{13\times7}=\frac{-42}{91}$
It is clear $\frac{-39}{91}$ is less than $\frac{-42}{91}$
$\therefore\frac{-3}{7}>\frac{6}{-13}$
View full question & answer→Question 821 Mark
Fill in the blanks: $-38\times\frac{-7}{19}\times=\frac{-7}{19}\times(.....)$
Answer$-38\times\frac{-7}{19}\times=\frac{-7}{19}\times(-38)$
View full question & answer→Question 831 Mark
Find the multiplicative inverse (i.e., reciprocal) of:
$\frac{2}{-5}$
Answer Multiplicative inverse of $\frac{2}{-5}=\frac{-5}{2}$
View full question & answer→Question 841 Mark
Find the multiplicative inverse (i.e., reciprocal) of: $\frac{-17}{12}$
AnswerMultiplicative inverse of $\frac{-17}{12}=\frac{-12}{17}$
View full question & answer→Question 851 Mark
Fill in the blanks with the correct symbol out of $ >, =$ and <. $-2\ ....\ \frac{-13}{5}$
Answer$-2\ \text{and}\ \frac{-13}{5}$ or $\frac{-2}{1}\ \text{and}\ \frac{-13}{5}$
$LCM$ of $1$ and $5 = 5$
$\therefore\frac{-2}{1}=\frac{-2\times5}{1\times5}=\frac{-10}{5}$
It is clear that between $\frac{-10}{5}$ and $\frac{-13}{5}$
$\frac{-10}{5}$ is greater than $\frac{-13}{5}$
$\therefore\frac{-10}{5}>\frac{-13}{5}$ or $-2>\frac{-13}{5}$
View full question & answer→Question 861 Mark
Fill in the blanks with the correct symbol out of $>, =$ and $<.$ $\frac{5}{-13}\ ....\ \frac{-35}{91}$
AnswerBetween $\frac{5}{-13}\ \text{and}\ \frac{-35}{91}$ or $\frac{-5}{13}\ \text{and}\ \frac{-35}{91}$
$LCM$ of $13$ and $91 = 91$
$\therefore\frac{-5}{13}=\frac{-5\times7}{13\times7}=\frac{-35}{91}$
$\therefore\ \frac{-35}{91}=\frac{-35}{91}\ \text{or}\ \frac{-5}{-13}=\frac{-35}{91}$
View full question & answer→Question 871 Mark
Which of the following statements are true and which are false$?$ $\frac{-3}{5}$ lies to the left of $0$ on the number line.
AnswerAs the number left of $0$ are negative.
View full question & answer→Question 881 Mark
Find the multiplicative inverse (i.e., reciprocal) of:
$(-7)^{-1}$
Answer$(-7)^{-1}=\Big(\frac{-1}{7}\Big)$
View full question & answer→Question 891 Mark
Fill in the blank: The reciprocal of a negative rational number is ______.
AnswerThe reciprocal of a negative rational number is negative.
View full question & answer→Question 901 Mark
Fill in the blanks with the correct symbol out of $>, = $ and $<.$
$\frac{-8}{9}\ ....\ \frac{-9}{10}$
AnswerBetween $\frac{-8}{9}\ \text{and}\ \frac{-9}{10}$
$LCM$ of $9$ and $10 = 90$
$\frac{-8}{9}=\frac{-8\times10}{9\times10}=\frac{-80}{90}$ and $\frac{-9}{10}=\frac{-9\times9}{10\times9}=\frac{-81}{90}$
It is clear that $\frac{-80}{90}>\frac{-81}{90}$
$\therefore\frac{-8}{9}>\frac{-9}{10}$
View full question & answer→Question 911 Mark
Is subtraction associative on rational numbers?
AnswerNo, subtraction is not associative.
View full question & answer→Question 921 Mark
Add the following rational numbers. $\frac{-8}{9}\ \text{and}\ \frac{11}{6}$
AnswerThe denominators of the given rational number are $9$ and $6.$
$LCM$ of $9, 6$ is $18$
Now,
$\frac{-8}{9}=\frac{-8\times2}{9\times2}$
$=\frac{-16}{18}$
and $\frac{11}{6}=\frac{11\times3}{6\times3}$
$=\frac{33}{18}$
$\therefore\frac{-8}{9}+\frac{11}{6}$
$=\frac{-16}{18}+\frac{33}{18}$
$=\frac{-16+33}{18}$
$=\frac{-17}{18}$
View full question & answer→Question 931 Mark
Name the property of multiplication illustrated of the following statement: $\frac{-7}{5}\times0=0$
Answer$\frac{-7}{5}\times0=0$ It is multiplicative property of $0.$
View full question & answer→Question 941 Mark
Name the property of multiplication shown by the following statement: $\frac{-12}{5}\times\frac{3}{4}=\frac{3}{4}\times\frac{-12}{5}$
AnswerCommutative law of multiplication.
View full question & answer→Question 951 Mark
Which of the following statement are true and which are false$?\ 0$ is a whole number but it is not a rational number.
AnswerAs $0$ is a whole number and set of whole number is a sub of rational numbers.
$\therefore$ $0$ is also a rational number.
View full question & answer→Question 961 Mark
Name the property of multiplication shown by the following statement: $\frac{-2}{3}\times0=0$
AnswerMultiplicative property of $0.$
View full question & answer→