Question 15 Marks
Arrange the following rational numbers in ascending order:
$\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}$
Answer$\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}$
$LCM$ of $5, 10, 15, 30 = 30$
$\begin{array}{c|c}2&5,10,15,30\\\hline5&5,5,15,15\ \ \ \\\hline3&1,1,3,3\ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \end{array}$
$\therefore LCM = 2 \times 3 \times 5 = 30$
$\therefore\frac{2}{5}=\frac{2\times6}{5\times6}=\frac{12}{30}$
$\frac{7}{10}=\frac{7\times3}{10\times3}=\frac{21}{30}$
$\frac{8}{15}=\frac{8\times2}{15\times2}=\frac{16}{30}$
$\frac{13}{30}=\frac{13}{30}$
Now, writing in ascending order we get:
$\frac{12}{30}<\frac{13}{30}<\frac{16}{30}<\frac{21}{30}$
$\text{or }\frac{2}{5}<\frac{13}{30}<\frac{8}{15}<\frac{7}{10}$
View full question & answer→Question 25 Marks
Simplify: $\frac{-11}{39}+\frac{5}{26}+\frac{2}{1}$
Answer$\frac{-11}{39}+\frac{5}{26}+\frac{2}{1}$
$LCM$ of $39$ and $26 = 78$
$\begin{array}{c|c}13&39,26\\\hline&3,2\end{array}$
$\therefore\text{ LCM}=13\times3\times2=78$
$\frac{-11}{39}=\frac{-11\times2}{39\times2}=\frac{-22}{78}$
$\frac{5}{26}=\frac{5\times3}{26\times3}=\frac{15}{78}$
$\frac{2}{1}=\frac{2\times78}{1\times78}=\frac{156}{78}$
$\therefore\frac{-11}{39}+\frac{5}{26}+\frac{2}{1}$
$=\frac{-22}{78}+\frac{15}{78}+\frac{156}{78}$
$=\frac{-22+15+156}{78}$
$=\frac{-22+171}{78}$
$=\frac{149}{78}$
View full question & answer→Question 35 Marks
Evaluate: $\frac{-16}{9}+\frac{-5}{12}+\frac{7}{18}$
Answer$\frac{-16}{9}+\frac{-5}{12}+\frac{7}{18}$
$LCM$ of $9,12,18 = 36$
$\begin{array}{c|c}2&9,12,18\\\hline3 &9,6,9\\\hline3&3,2,3\\\hline&1,2,1\end{array}$ $\therefore\text{LCM}=2\times3\times3\times2=36$
$\frac{-16}{9}=\frac{-16\times4}{9\times4}=\frac{-64}{36}$
$\frac{-5}{12}=\frac{-5\times3}{12\times3}=\frac{-15}{36}$
$\frac{7}{18}=\frac{7\times2}{18\times2}=\frac{14}{36}$
$\therefore \ \frac{-16}{9}+\frac{-5}{12}+\frac{7}{18}$
$=\frac{-64}{36}+\frac{-15}{36}+\frac{14}{36}$
$=\frac{-64+(-15)+14}{36}$
$=\frac{-64-15+14}{36}=\frac{-79+14}{36}$
$=\frac{-65}{36}$
View full question & answer→Question 45 Marks
Simplify: $-1+\frac{7}{9}+\frac{11}{12}$
Answer$-1+\frac{7}{9}+\frac{11}{12}$
$\frac{7}{-9}=\frac{7\times(-1)}{-9\times(-1)}=\frac{-7}{9}$
$LCM$ of $9$ and $12 = 36$
$\begin{array}{c|c}3&9,12\\\hline&3,4\end{array}$
$\text{LCM}=3\times3\times4=36$
$\frac{-1}{1}=\frac{-1\times36}{1\times36}=\frac{-36}{36}$
$\frac{-7}{9}=\frac{-7\times4}{9\times4}=\frac{-28}{36}$
$\frac{11}{12}=\frac{11\times3}{12\times3}=\frac{33}{36}$
$\therefore\frac{-1}{1}+\frac{-7}{9}+\frac{11}{12}$
$=\frac{-36}{36}+\frac{-28}{36}+\frac{33}{36}$
$=\frac{-36+(-28)+33}{36}$
$=\frac{-36-28+33}{36}$
$=\frac{-64+33}{36}$
$=\frac{-31}{36}$
View full question & answer→Question 55 Marks
Arrange the following rational numbers in ascending order: $\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}$
Answer$\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}$
Firstly, making positive denominators of $\frac{5}{-12}$ and $\frac{9}{-24}$
$\frac{5}{-12}=\frac{5\times(-1)}{-12\times(-1)}=\frac{-5}{12}$
$\frac{9}{-24}=\frac{9\times(-1)}{-24\times(-1)}=\frac{-9}{24}$
$LCM$ of $4, 12, 16$ and $24 = 48$
$\begin{array}{c|c}2&4,12,16,24\\\hline2&2,6,8,12\ \ \ \ \\\hline2&1,3,4,6\ \ \ \ \ \ \\\hline3&1,3,2,3\ \ \ \ \ \ \\\hline&1,1,2,1\ \ \ \ \ \end{array}$
$\therefore LCM = 2 \times 2 \times 2 \times 3 \times 2 = 48$
$\frac{-3}{4}=\frac{-3\times12}{4\times12}=\frac{-36}{48}$
$\frac{-5}{12}=\frac{-5\times4}{12\times4}=\frac{-20}{48}$
$\frac{-7}{16}=\frac{-7\times3}{16\times3}=\frac{-21}{48}$
$\frac{-9}{24}=\frac{-9\times2}{24\times2}=\frac{-18}{48}$
Now, writing in ascending order we get: $\frac{-36}{48}<\frac{21}{48}<\frac{-20}{48}<\frac{-18}{48}$
$\text{or }\frac{-3}{4}<\frac{-7}{16}<\frac{5}{-12}<\frac{9}{-24}$
View full question & answer→Question 65 Marks
Evaluate: $\frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}$
Answer$\frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}$
$\frac{11}{-12}=\frac{11\times(-1)}{-12\times(-1)}=\frac{-11}{12}$
$\frac{3}{8}=\frac{3\times(-1)}{-8\times(-1)}=\frac{-3}{8}$
Now $LCM of 12,8,4 = 24$
$\begin{array}{c|c}2&12,8,4\\\hline2 &6,4,2\\\hline&3,2,1\end{array}$
$\text{LCM}=2\times2\times2\times3=24$
$\therefore\frac{-11}{12}=\frac{-11\times2}{12\times2}=\frac{-22}{24}$
$\frac{-3}{8}=\frac{-3\times3}{8\times3}=\frac{-9}{24}$
$\frac{1}{4}=\frac{1\times6}{4\times6}=\frac{6}{24}$
$\therefore\frac{-11}{12}+\frac{-3}{8}+\frac{1}{4}$
$=\frac{-22}{24}+\frac{-9}{24}+\frac{6}{24}$
$=\frac{-22+(-9)+6}{24}$
$=\frac{-22-9+6}{24}=\frac{-31+6}{24}$
$=\frac{-25}{24}$
View full question & answer→Question 75 Marks
Arrange the following rational numbers in ascending order: $\frac{2}{3},\frac{3}{4},\frac{5}{-6},\frac{-7}{12}$
Answer$\frac{2}{3},\frac{3}{4},\frac{5}{-6},\frac{-7}{12}$ Firstly, making positive denominators of $\frac{5}{-6}$
$\frac{5}{-6}=\frac{5\times(-1)}{-6\times(-1)}=\frac{-5}{6}$
$LCM of 3, 4, 6, 12 = 12$
$\begin{array}{c|c}2&3,4,6,12\\\hline2&3,2,3,6\ \ \\\hline3&3,1,3,3\ \ \ \\\hline&1,1,1\ \ \ \ \ \ \ \end{array}$
$\therefore LCM = 2 \times 2 \times 3 = 12$
$\therefore\frac{2}{3}=\frac{2\times4}{3\times4}=\frac{8}{12}$
$\frac{3}{4}=\frac{3\times3}{4\times3}=\frac{9}{12}$
$\frac{-5}{6}=\frac{-5\times2}{6\times2}=\frac{-10}{12}$
$\frac{-7}{12}=\frac{-7}{12}$ Now, writing in ascending order
we get: $\frac{-10}{12}<\frac{-7}{12}<\frac{8}{12}<\frac{9}{12}$
$\text{or }\frac{-5}{6}<\frac{-7}{12}<\frac{2}{3}<\frac{3}{4}$
View full question & answer→Question 85 Marks
Arrange the following rational numbers in descending order: $\frac{-4}{9},\frac{5}{-12},\frac{-7}{18},\frac{2}{-3}$
Answer$\frac{-4}{9},\frac{5}{-12},\frac{-7}{18},\frac{2}{-3}$Making positive denominators of $\frac{5}{-12}$
$\frac{2}{-3}$
$\frac{5}{-12}=\frac{5\times(-1)}{-12\times(-1)}=\frac{-5}{12}$
$\frac{2}{-3}=\frac{2\times(-1)}{-3\times(-1)}=\frac{-2}{3}$
Now LCM of 9, 12, 18, 3 = 36
$\begin{array}{c|c}2&9,12,18,13\\\hline3&9,6,9,3\ \ \ \ \ \ \\\hline3&3,2,3,1\ \ \ \ \ \ \\\hline&1,2,1,1\ \ \ \ \ \ \end{array}$
$\therefore LCM = 2 \times 3 \times 3 \times 2 = 36$
$\therefore\frac{-4}{9}=\frac{-4\times4}{9\times4}=\frac{-16}{36}$
$\frac{-5}{12}=\frac{-5\times3}{12\times3}=\frac{-15}{36}$
$\frac{-7}{18}=\frac{-7\times2}{18\times2}=\frac{-14}{36}$
$\frac{-2}{3}=\frac{-2\times12}{3\times12}=\frac{-24}{36}$
Writing in descending order we get:
$\frac{-14}{36}>\frac{75}{36}>\frac{-16}{36}>\frac{-24}{36}$
$\frac{-7}{18}>\frac{11}{-15}>\frac{-4}{9}>\frac{-2}{3}$
View full question & answer→Question 95 Marks
Evaluate: $-3+\frac{1}{8}+\frac{-2}{5}$
Answer$-3+\frac{1}{8}+\frac{-2}{5}$
$LCM$ of $8, 5 = 40$
$\frac{-3}{1}=\frac{-3\times40}{1\times40}=\frac{-120}{40}$
$\frac{1}{8}=\frac{1\times5}{8\times5}=\frac{5}{40}$
$\frac{-2}{5}=\frac{-2\times8}{5\times8}=\frac{-16}{40}$
$\therefore \ \frac{-3}{1}+\frac{1}{8}+\frac{-2}{5}$
$=\frac{-120}{40}+\frac{5}{40}+\frac{-16}{40}$
$=\frac{-120+5+(-16)}{40}$
$=\frac{-120+5+(-16)}{40}=\frac{-136+5}{40}$
$=\frac{-131}{40}$
View full question & answer→Question 105 Marks
Arrange the following rational numbers in descending order: $\frac{17}{-30},\frac{11}{-15},\frac{-7}{10},\frac{10}{5}$
Answer$\frac{17}{-30},\frac{11}{-15},\frac{-7}{10},\frac{10}{5}$
Making positive denominators of $\frac{17}{-30}$ and $\frac{11}{-15}$
$\frac{17}{-30}=\frac{17\times(-1)}{-30\times(-1)}=\frac{-17}{30}$
$\frac{11}{-15}=\frac{1\times(-1)}{-15\times(-1)}=\frac{-11}{15}$
Now $LCM$ of $30, 15, 5 = 30$
$\begin{array}{c|c}5&30,15,10,5\ \ \ \\\hline2&6,3,2,1 \ \ \ \ \ \ \ \ \\\hline3&3,3,1,1\ \ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \end{array}$
$\therefore LCM = 5 \times 2 \times 3 = 30$
$\therefore\frac{-17}{30}=\frac{-17}{30}\frac{-11}{15}=\frac{-11\times2}{15\times2}=\frac{-22}{30}$
$\frac{-7}{10}=\frac{-7\times3}{10\times3}=\frac{-21}{30}$
$\frac{3}{5}=\frac{3\times6}{5\times6}=\frac{18}{30}$
Now writing in descending order we get:
$\frac{18}{30}>\frac{-17}{30}>\frac{-21}{30}>\frac{-22}{30}$
$\text{or }\frac{3}{5}>\frac{17}{-30}>\frac{-7}{10}>\frac{11}{-15}$
View full question & answer→Question 115 Marks
Arrange the following rational numbers in descending order: $-2,\frac{-13}{6},\frac{8}{-3},\frac{1}{3}$
Answer$-2,\frac{-13}{6},\frac{8}{-3},\frac{1}{3}$Making positive denominators of $\frac{8}{-3}$
$\frac{8}{-3}=\frac{8\times(-1)}{-3\times(-1)}=\frac{-8}{3}$
Now $LCM$ of $6, 3, 3 = 6$
$\therefore\frac{-2}{1}=\frac{-2\times6}{1\times6}=\frac{-12}{6}$
$\frac{-13}{6}=\frac{-13}{6}\frac{-8}{3}=\frac{-8\times2}{3\times2}=\frac{-16}{6}$
$\frac{1}{3}=\frac{1\times2}{3\times2}=\frac{2}{6}$
Now, writing in descending order we get:
$\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}$
$\text{or }\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}$
View full question & answer→Question 125 Marks
Arrange the following rational numbers in ascending order: $\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}$
Answer$\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}$
Firstly, making positive denominators of $\frac{7}{-15}$ and $\frac{17}{-30}$
$\frac{7}{-15}=\frac{7\times(-1)}{-15\times(-1)}=\frac{-7}{15}$
$\frac{17}{-30}=\frac{17\times(-1)}{-30\times(-1)}=\frac{-17}{30}$
$LCM$ of $10, 15, 20$ and $30 = 60$
$\begin{array}{c|c}2&10,15,20,30\ \ \ \\\hline3&5,15,10,15\ \ \ \ \\\hline5&5,15,10,5\ \ \ \ \ \ \\\hline2&1,1,2,1\ \ \ \ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \ \ \ \end{array}$
$\therefore LCM = 2 \times 3 \times 5 \times 2 = 60$
$\frac{-3}{10}=\frac{-3\times6}{10\times6}=\frac{-18}{60}$
$\frac{-7}{15}=\frac{-7\times4}{15\times4}=\frac{-28}{60}$
$\frac{-11}{20}=\frac{-11\times3}{20\times3}=\frac{-33}{60}$
$\frac{-17}{30}=\frac{-17\times2}{30\times2}=\frac{-34}{60}$ .
Now, writing in ascending order we get:
$\frac{-34}{60}<\frac{-33}{60}<\frac{-28}{60}<\frac{-18}{60}$
$\text{or }\frac{-17}{30}<\frac{-11}{20}<\frac{-7}{15}<\frac{-3}{10}$
View full question & answer→Question 135 Marks
Evaluate: $\frac{-13}{8}+\frac{5}{16}+\frac{-1}{4}$
Answer$\frac{-13}{8}+\frac{5}{16}+\frac{-1}{4}$
$LCM$ of $8, 16, 4 = 16$
$\frac{-13}{8}=\frac{-13\times2}{8\times2}=\frac{-26}{16}$
$\frac{5}{6}=\frac{5}{6}$
$\frac{-1}{4}=\frac{-1\times4}{4\times4}=\frac{-4}{16}$
$\therefore\frac{-13}{8}+\frac{5}{6}+\frac{(-1)}{4}$
$=\frac{-26}{16}+\frac{5}{16}+\frac{-4}{16}$
$=\frac{-26+5+(-4)}{16}$
$=\frac{-26+5-4}{16}=\frac{-30+5}{16}$
$=\frac{-25}{16}$
View full question & answer→Question 145 Marks
Simplify: $\frac{-7}{10}+\frac{13}{-15}+\frac{27}{20}$
Answer$\frac{-7}{10}+\frac{13}{-15}+\frac{27}{20}$
$\frac{13}{-15}=\frac{13\times(-1)}{-15\times(-1)}=\frac{-13}{15}$
Now $LCM$ of $10, 15, 20 = 60$
$\begin{array}{c|c}2&10,15,20\\\hline5&5,15,10\\\hline&1,3,2\end{array}$
$\therefore\text{LCM}=2\times5\times3\times2=60$
$\frac{-7}{10}=\frac{-7\times6}{10\times6}=\frac{-42}{60}$
$\frac{-13}{15}=\frac{-13\times4}{15\times4}=\frac{-52}{60}$
$\frac{27}{20}=\frac{27\times3}{20\times3}=\frac{81}{60}$
$\therefore\frac{-7}{10}+\frac{-13}{15}+\frac{27}{20}$
$=\frac{-42}{60}+\frac{-52}{60}+\frac{81}{60}$
$=\frac{-42+(-52)+81}{60}$
$=\frac{-42-52+81}{60}=\frac{-94+81}{60}$
$=\frac{-13}{60}$
View full question & answer→Question 155 Marks
Subtract the sum of $\frac{-36}{11}$ and $\frac{49}{22}$ from the sum of $\frac{33}{8}$ and $\frac{-19}{4}$
Answer$\Big(\frac{33}{8}+\frac{-19}{4}\Big)-\Big(\frac{-36}{11}+\frac{49}{22}\Big)$
$=\Big(\frac{33}{8}-\frac{19}{4}\Big)-\Big(\frac{-36}{11}+\frac{49}{22}\Big)$
$=\Big(\frac{33-38}{8}\Big)-\Big(\frac{-72+49}{22}\Big)$
$=\Big(\frac{-5}{8}\Big)-\Big(\frac{-23}{22}\Big)$
$=\frac{-5}{8}+\frac{23}{22}$
$LCM$ of $8$ and $22 = 88$
$\therefore\frac{-5}{8}=\frac{-5\times11}{8\times11}=\frac{-55}{88}$
$\frac{23}{22}=\frac{23\times4}{22\times}=\frac{92}{88}$
$\therefore\frac{-55}{88}+\frac{92}{88}$
$=\frac{-55+92}{88}=\frac{37}{88}$
View full question & answer→Question 165 Marks
Arrange the following rational numbers in descending order: $\frac{-2}{5},\frac{7}{-10},\frac{-11}{15},\frac{19}{-30}$
Answer$\frac{-2}{5},\frac{7}{-10},\frac{-11}{15},\frac{19}{-30}$
Making positive denominators of $\frac{7}{-10}$ and $\frac{19}{-30}$
$\frac{-7}{10}=\frac{7\times(-1)}{-10\times(-1)}=\frac{-7}{10}$
$\frac{19}{-30}=\frac{19\times(-1)}{-30\times(-1)}=\frac{-19}{30}$
Now $LCM$ of $5, 10, 15$ and $30 = 30$
$\begin{array}{c|c}2&5,10,15,30\ \\\hline3&5,5,15,15\ \ \ \\\hline5&5,5,5,5\ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \end{array}$
$\therefore LCM = 2 \times 3 \times 5 = 30$
$\therefore\frac{-7}{10}=\frac{-7\times3}{10\times3}=\frac{-12}{30}$
$\frac{-7}{10}=\frac{-7\times3}{10\times3}=\frac{-21}{30}$
$\frac{-11}{15}=\frac{-11\times2}{15\times2}=\frac{-22}{30}$
$\frac{-19}{30}=\frac{-19}{30}$ Now, writing in ascending order
we get: $\frac{-12}{30}>\frac{-19}{30}>\frac{-21}{30}>\frac{-22}{30}$
$\text{or }\frac{-2}{5}>\frac{19}{-30}>\frac{7}{-10}>\frac{-11}{15}$
View full question & answer→