Question 14 Marks
Arrange the following rational numbers in descending order:
$\frac{-5}{6},\frac{-7}{12},\frac{-13}{18},\frac{23}{-24}$
AnswerAmong $\frac{-5}{6},\frac{-7}{12},\frac{-13}{18},\frac{23}{-24}$ Making their denominator of $\frac{23}{-24}$ as positive, then $\frac{-5}{6},\frac{-7}{12},\frac{-13}{18},\frac{-23}{24}$ Now, LCM of 6, 12, 18, 24 = 72 $\therefore\frac{-5}{6}=\frac{-5\times12}{6\times12}=\frac{-60}{72}$ $\frac{-7}{12}=\frac{-7\times6}{12\times6}=\frac{-42}{72}$ $\frac{-13}{18}=\frac{-13\times4}{18\times4}=\frac{-52}{72}$ $\frac{-23}{24}=\frac{-23\times3}{24\times3}=\frac{-69}{72}$ Now, writing them in descending order, We get,$\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}$
Or $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}$ $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{23}{-24}$
View full question & answer→Question 24 Marks
Arrange the following rational numbers in descending order:
$-2,\frac{-13}{6},\frac{8}{-3},\frac{1}{3}$
AnswerAmong $-2,\frac{-13}{6},\frac{8}{-3},\frac{1}{3}$ Making their denominator of $\frac{8}{-3}$ as LCM of 1, 6, 3, 3 = 6 $\therefore\frac{-2}{1}=\frac{-2\times6}{1\times6}=\frac{-12}{6}$ $\frac{-13}{6}=\frac{-13\times1}{6\times1}=\frac{-13}{6}$ $\frac{-8}{3}=\frac{-8\times2}{3\times2}=\frac{-16}{6}$ $\frac{1}{3}=\frac{1\times2}{3\times2}=\frac{2}{6}$ Writing them in descending orders, We get,$\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}$
Or $\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}$ $\frac{1}{3}>-2>\frac{-13}{6}>\frac{8}{-3}$
View full question & answer→Question 34 Marks
Arrange the following rational numbers in ascending order:
$\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}$
AnswerAmong $\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}$ making their denominator positive, We get $\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}$ LCM of 4, 12, 16 and 24 = 48 $\therefore\frac{-3}{4}=\frac{-3\times12}{4\times12}=\frac{-36}{48}$ $\frac{-5}{12}=\frac{-5\times4}{12\times4}=\frac{-20}{48}$ $\frac{-7}{16}=\frac{-7\times3}{16\times3}=\frac{-21}{48}$ $\frac{-9}{24}=\frac{-9\times2}{24\times2}=\frac{-18}{48}$ Now, writing them in ascending orders,$\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{48}$
Or $\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}$
View full question & answer→Question 44 Marks
Find three rational numbers between 4 and 5.
AnswerFirst rational number between 4 and 5 $=\frac{1}{2}[4+5]$ $=\frac{1}{2}\times9$ $=\frac{9}{2}$ $\therefore4<\frac{9}{2}<5$Second rational number between 4 and $\frac{9}{2}$
$=\frac{1}{2}\Big[4+\frac{9}{2}\Big]$ $=\frac{1}{2}\Big[\frac{8+9}{2}\Big]$ $=\frac{1}{2}\times\frac{17}{2}$ $=\frac{17}{4}$and third rational number between $\frac{9}{2}$ and 5
$=\frac{1}{2}\Big[\frac{9}{2}+5\Big]$ $=\frac{1}{2}\Big[\frac{9+10}{2}\Big]$ $=\frac{1}{2}\times\frac{19}{2}$ $=\frac{19}{4}$ $\therefore4<\frac{17}{4}<\frac{9}{2}<\frac{19}{4}<5$ $\therefore$ Required three rational numbers and $\frac{17}{4},\frac{9}{2},\frac{19}{4}$
View full question & answer→Question 54 Marks
Find three rational numbers between $\frac{2}{3}$ and $\frac{3}{4}$.
AnswerFirst rational number between $\frac{2}{3}$ and $\frac{3}{4}$ $=\frac{1}{2}\Big[\frac{2}{3}+\frac{3}{4}\Big]$ $=\frac{1}{2}\Big[\frac{8+9}{12}\Big]$ $=\frac{1}{2}\times\frac{17}{12}$ $=\frac{17}{24}$ $\therefore\frac{2}{3}<\frac{17}{24}<\frac{3}{4}$Second rational number between $\frac{2}{3}$ and $\frac{17}{24}$
$=\frac{1}{2}\Big[\frac{2}{3}+\frac{17}{24}\Big]$ $=\frac{1}{2}\Big[\frac{16+17}{24}\Big]$ $=\frac{1}{2}\times\frac{33}{24}$ $=\frac{33}{48}$and third rational number between $\frac{17}{24}$ and $\frac{3}{4}$
$=\frac{1}{2}\Big[\frac{17}{24}+\frac{3}{4}\Big]$ $=\frac{1}{2}\Big[\frac{17+18}{24}\Big]$ $=\frac{1}{2}\times\frac{35}{24}$ $=\frac{35}{48}$ $\therefore\frac{2}{3}<\frac{33}{48}<\frac{17}{24}<\frac{35}{48}<\frac{3}{4}$ $\therefore$ Required three rational numbers $\frac{33}{48},\frac{17}{24},\frac{35}{48}$
View full question & answer→Question 64 Marks
Arrange the following rational numbers in ascending order:
$\frac{-4}{7},\frac{-9}{14},\frac{13}{-28},\frac{-23}{42}$
AnswerAmong $\frac{-4}{7},\frac{-9}{14},\frac{13}{-28},\frac{-23}{42}$ making their denominator of $\frac{13}{-28}$ positive, We get, $\frac{-4}{7},\frac{-9}{14},\frac{-13}{28},\frac{-23}{42}$ LCM of 7, 14, 28 and 42 = 84 $\therefore\frac{-4}{7}=\frac{-4\times12}{7\times12}=\frac{-48}{84}$ $\frac{-9}{14}=\frac{-9\times6}{14\times6}=\frac{-54}{84}$ $\frac{-13}{28}=\frac{-13\times3}{28\times3}=\frac{-39}{84}$ $\frac{-23}{42}=\frac{-23\times2}{42\times2}=\frac{-46}{84}$ Now writing them in ascending orders, We get,$\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84}$
Or $\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{-13}{28}$ $\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{13}{-28}$
View full question & answer→Question 74 Marks
Verify the following:
$\Big(\frac{-7}{11}+\frac{2}{-5}\Big)+\frac{-13}{22}=\frac{-7}{11}+\Big(\frac{2}{-5}+\frac{-13}{22}\Big)$
Answer$\text{L.H.S. }=\Big(\frac{-7}{11}+\frac{2}{-5}\Big)+\frac{-13}{22}$
$\Big(\frac{-7}{11}+\frac{-2}{5}\Big)+\frac{-13}{22}$
$=\frac{-35+(-22)}{55}+\frac{-13}{22}$
$=\frac{-35-22}{55}+\frac{-13}{22}$
$=\frac{-57}{55}+\frac{-13}{22}$
$=\frac{-114-65}{110}$
$=\frac{-179}{110}$
$\text{R.H.S. }=\frac{-7}{11}+\Big(\frac{2}{-5}+\frac{-13}{22}\Big)$
$=\frac{-7}{11}+\Big(\frac{-2}{5}+\frac{-13}{22}\Big)$
$=\frac{-7}{11}+\frac{-44+(-65)}{110}$
$=\frac{-7}{11}+\frac{-44-65}{110}$
$=\frac{-7}{11}+\frac{(-109)}{110}$
$=\frac{-70+(-109)}{110}$
$=\frac{-70-109}{110}$
$=\frac{-179}{110}$
$\text{L.H.S. }=\text{R.H.S. }$
$\therefore\Big(\frac{-7}{11}+\frac{2}{-5}\Big)+\frac{-13}{22}=\frac{-7}{11}+\Big(\frac{2}{-5}+\frac{-13}{22}\Big)$
View full question & answer→Question 84 Marks
Arrange the following rational numbers in ascending order:
$\frac{4}{-9},\frac{-5}{12},\frac{7}{-18},\frac{-2}{3}$
AnswerAmong $\frac{4}{-9},\frac{-5}{12},\frac{7}{-18},\frac{-2}{3}$ making their denominator positive, We get, $\frac{4}{-9},\frac{-5}{12},\frac{7}{-18},\frac{-2}{3}$ LCM of 9, 12, 18 and 3 = 36 $\therefore\frac{-4}{9}=\frac{4\times4}{9\times4}=\frac{-16}{36}$ $\frac{-5}{12}=\frac{-5\times3}{12\times3}=\frac{-15}{36}$ $\frac{-7}{18}=\frac{-7\times2}{18\times2}=\frac{-14}{36}$ $\frac{-2}{3}=\frac{-2\times12}{3\times12}=\frac{-24}{36}$ Now, writing them in ascending orders,$\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}$
Or $\frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}$ Or $\frac{-2}{3}<\frac{4}{-9}<\frac{-5}{12}<\frac{7}{-18}$
View full question & answer→Question 94 Marks
Arrange the following rational numbers in descending order:
$\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}$
AnswerAmong $\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}$ Making their denominator of $\frac{7}{-15}$ and $\frac{17}{-30}$, as positive then $\frac{-3}{10},\frac{-7}{15},\frac{-11}{20},\frac{-17}{30}$ Now, LCM of 10, 15, 20, 30 = 60 $\therefore\frac{-3}{10}=\frac{-3\times6}{10\times6}=\frac{-18}{60}$ $\frac{-7}{15}=\frac{-7\times4}{15\times4}=\frac{-28}{60}$ $\frac{-11}{20}=\frac{-11\times3}{20\times3}=\frac{-33}{60}$ $\frac{-17}{30}=\frac{-17\times2}{30\times2}=\frac{-34}{60}$ Now, writing them in descending orders, We get,$\frac{18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}$
Or $\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}$ $\frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}$
View full question & answer→Question 104 Marks
Amit earns Rs. 32000 per month. He spends $\frac{1}{4}$ of his income of food; $\frac{3}{10}$ of the remainder on house rent and $\frac{5}{21}$ of the remainder on the education of children. How much money is still left with him?
AnswerTotal amount eamed by Amit = Rs. 32000
Amount spent on food $=\frac{1}{4}$ of Rs. 32000
$=\frac{1}{4}\times\text{Rs.}\ 32000-\text{Rs.}\ 8000$
Balance = Rs. 32000 - Rs. 8000
= Rs. 24000
Amount spent on house rent $=\frac{3}{10}$ of Rs. 24000
$=\frac{3}{10}\times\text{Rs.}\ 24000$
$=3\times\text{Rs.}\ 2400$
$=\text{Rs.}\ 7200$
Amount left = Rs. 24000 - Rs. 37200
= Rs. 16800
Amount spent on education of children $=\frac{5}{21}$ of Rs. 16800
$=\frac{5}{21}\times\text{Rs.}\ 16800$
$=\text{Rs.}\ 5\times800$
$=\text{Rs.}\ 4000$
Amount left = Rs. 16800 - Rs. 4000
= Rs. 12800
View full question & answer→Question 114 Marks
Verify the following:
$\Big(\frac{3}{4}+\frac{-2}{5}\Big)+\frac{-7}{10}=\frac{3}{4}+\Big(\frac{-2}{5}+\frac{-7}{10}\Big)$
Answer$\text{L.H.S. }=\Big(\frac{3}{4}+\frac{-2}{5}\Big)+\frac{-7}{10}$
$=\frac{15+(-8)}{20}+\frac{-7}{10}$
$=\frac{15-8}{20}+\frac{-7}{10}$
$=\frac{7}{20}+\frac{-7}{10}$
$=\frac{7+(-14)}{20}$
$=\frac{7-14}{20}$
$=\frac{-7}{20}$
$\text{R.H.S. }=\frac{3}{4}+\Big(\frac{-2}{5}+\frac{-7}{10}\Big)$
$=\frac{3}{4}+\frac{-4+(-7)}{10}$
$=\frac{3}{4}+\frac{-4-7}{10}$
$=\frac{3}{4}+\frac{-11}{10}$
$=\frac{15+(-22)}{20}$
$=\frac{15-22}{20}$
$=\frac{-7}{20}$
$\text{L.H.S. }=\text{R.H.S. }$
$\therefore\Big(\frac{3}{4}+\frac{-2}{5}\Big)+\frac{-7}{10}=\frac{3}{4}+\Big(\frac{-2}{5}+\frac{-7}{10}\Big)$
View full question & answer→Question 124 Marks
Verify whether the given statement is true or false:
$\Big\{(-16)\div\frac{6}{5}\Big\}\div\frac{-9}{10}=(-16)\div\Big\{\frac{6}{5}\div\frac{-9}{10}\Big\}$
AnswerFalse.
Solution:
$\text{L.H.S.}=\Big\{(-16)\div\frac{6}{5}\Big\}\div\frac{-9}{10}$
$=\Big[(-16)\times\frac{5}{6}\Big]\times\frac{10}{-9}$
$=\frac{(-16)\times5\times10}{6\times(-9)}$
$=\frac{800}{54}$
$=\frac{400}{27}$
$\text{R.H.S.}=(-16)\div\Big\{\frac{6}{5}\div\frac{-9}{10}\Big\}$
$=(-16)\div\Big[\frac{6}{5}\times\frac{10}{-9}\Big]$
$=-16\div\Big(\frac{-60}{45}\Big)$
$=-16\times\Big(\frac{-45}{60}\Big)$
$=-16\times\Big(\frac{-3}{4}\Big)$
$=\frac{48}{4}$
$=12$
$\text{L.H.S}\neq\text{R.H.S}$
View full question & answer→Question 134 Marks
Verify the following:
$-1+\Big(\frac{-2}{3}+\frac{-3}{4}\Big)=\Big(-1+\frac{-2}{3}\Big)+\frac{-3}{4}$
Answer$\text{L.H.S. }=-1+\Big(\frac{-2}{3}+\frac{-3}{4}\Big)$
$=-1+\frac{-8+(-9)}{12}$
$=-1+\frac{-8-9}{12}$
$=\frac{-1}{1}+\frac{-17}{12}$
$=\frac{-12-17}{12}$
$=\frac{-29}{12}$
$\text{R.H.S. }=\Big(-1+\frac{-2}{3}\Big)+\frac{-3}{4}$
$=\frac{-3+(-2)}{3}+\frac{-3}{4}$
$=\frac{-3-2}{3}+\frac{-3}{4}$
$=\frac{-5}{3}+\frac{-3}{4}$
$=\frac{-20+(-9)}{12}$
$=\frac{-20-9}{12}$
$=\frac{-29}{12}$
$\text{L.H.S. }=\text{R.H.S. }$
$\therefore-1+\Big(\frac{-2}{3}+\frac{-3}{4}\Big)=\Big(-1+\frac{-2}{3}\Big)+\frac{-3}{4}$
View full question & answer→Question 144 Marks
Verify the following:
$\Big(\frac{-9}{5}\times\frac{-10}{3}\Big)\times\frac{21}{-4}=\frac{-9}{5}\times\Big(\frac{-10}{3}\times\frac{21}{-4}\Big)$
Answer$\text{L.H.S.}=\Big(\frac{-9}{5}\times\frac{-10}{3}\Big)\times\frac{21}{-4}$
$=\frac{-9\times(-10)}{5\times3}\times\frac{21}{-4}$
$=\frac{90}{15}\times\frac{21}{-4}$
$=\frac{90\times21}{15\times(-4)}$
$=\frac{1890}{-60}$
$=\frac{1890\div30}{-60\div30}$
$=\frac{63}{-2}=\frac{-63}{2}$
$\text{R.H.S.}=\frac{-9}{5}\times\Big(\frac{-10}{3}\times\frac{21}{-4}\Big)$
$=\frac{9}{5}\times\frac{-10\times21}{3\times(-4)}$
$=\frac{9}{5}\times\frac{-210}{-12}$
$=\frac{-9\times(-210)}{5\times(-12)}$
$=\frac{1890}{-60}$
$=\frac{1890\div30}{-60\div30}$
$=\frac{63}{-2}=\frac{-63}{2}$
$\text{L.H.S.}=\text{R.H.S.}$
$\therefore\Big(\frac{-9}{5}\times\frac{-10}{3}\Big)\times\frac{21}{-4}=\frac{-9}{5}\times\Big(\frac{-10}{3}\times\frac{21}{-4}\Big)$
View full question & answer→Question 154 Marks
Arrange the following rational numbers in descending order:
$\frac{-10}{11},\frac{-19}{22},\frac{-23}{33},\frac{-39}{44}$
AnswerAmong $\frac{-10}{11},\frac{-19}{22},\frac{-23}{33},\frac{-39}{44}$ Now, LCM of 11, 22, 33, 44 = 132 $\therefore\frac{-10}{11}=\frac{-10\times12}{11\times12}=\frac{-120}{132}$ $\frac{-19}{22}=\frac{-19\times6}{22\times6}=\frac{-114}{132}$ $\frac{-23}{33}=\frac{-23\times4}{33\times4}=\frac{-92}{132}$ $\frac{-39}{44}=\frac{-39\times3}{44\times3}=\frac{-117}{132}$ Now, writing them in descending order, We get,$\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}$
Or $\frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}$
View full question & answer→Question 164 Marks
Divide the sum of $\frac{13}{5}$ and $\frac{-12}{7}$ by the product of $\frac{-31}{7}$ and $\frac{1}{-2}$.
AnswerSum of $\frac{13}{5}$ and $\frac{-12}{7}$
$=\frac{13}{5}+\frac{-12}{7}$
$=\frac{91+(-60)}{35}$
$=\frac{91-60}{35}$
$=\frac{31}{35}$
Now, product of $\frac{-31}{7}$ and $\frac{1}{-2}$
$=\frac{-31}{7}\times\frac{1}{-2}$
$=\frac{-31}{-14}=\frac{31}{14}$
$\therefore\frac{-31}{-14}\div\frac{31}{14}$
$=\frac{31}{35}\times\frac{14}{31}$
$=\frac{14}{35}$
$=\frac{14\div7}{35\div7}$
$=\frac{2}{5}$
$\therefore$ Required number $=\frac{2}{5}$
View full question & answer→Question 174 Marks
On one day a rickshaw puller earned Rs. 160. Out of his earnings he spent $\text{Rs. }26\frac{3}{5}$ on tea and snacks, $\text{Rs. }50\frac{1}{2}$ on food and $\text{Rs. }16\frac{2}{5}$ on repairs of the rickshaw. How much did he save on that day?
AnswerTotal earnings $=\text{Rs. }160$
Spent on tea and snacks $=\text{Rs. }26\frac{3}{5}$
$=\text{Rs. }\frac{133}{5}$
Spent on food $=\text{Rs. }50\frac{1}{2}$
$=\text{Rs. }\frac{101}{2}$
Spent on repair of rickshaw $=\text{Rs. }16\frac{2}{5}$
$=\text{Rs. }\frac{82}{5}$
Total amount spent $=\text{Rs.}\ \frac{133}{5}+\frac{101}{2}+\frac{82}{5}$
$=\frac{266+505+164}{10}$
$=\text{Rs. }\frac{935}{10}$
$\therefore$ Savings $=\text{Rs. }160-\frac{935}{10}$
$=\frac{1600-935}{10}$
$=\frac{665}{10}$
$=\frac{665\div5}{10\div5}$
$=\frac{133}{2}$
$=\text{Rs. }66\frac{1}{2}$
View full question & answer→Question 184 Marks
Arrange the following rational numbers in ascending order:
$\frac{3}{-5},\frac{-7}{10},\frac{-11}{15},\frac{-13}{20}$
AnswerAmong $\frac{3}{-5},\frac{-7}{10},\frac{-11}{15},\frac{-13}{20}$ making their denominator of $\frac{3}{-5}$ positive, We get $\frac{-3}{5},\frac{-7}{10},\frac{-11}{15},\frac{-13}{20}$ LCM of 5, 10, 15 and 20 = 60 $\therefore\frac{-3}{5}=\frac{-3\times12}{5\times12}=\frac{-36}{60}$ $\frac{-7}{10}=\frac{-7\times6}{10\times6}=\frac{-42}{60}$ $\frac{-11}{15}=\frac{-11\times4}{15\times4}=\frac{-44}{60}$ $\frac{-13}{20}=\frac{-13\times3}{20\times3}=\frac{-39}{60}$ Now, writing them in ascending orders,$\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{60}<\frac{-36}{60}$
Or $\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{-3}{5}$ $\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{3}{-5}$
View full question & answer→Question 194 Marks
Verify the following:
$\frac{-13}{24}\times\Big(\frac{-12}{5}\times\frac{35}{36}\Big)=\Big(\frac{-13}{24}\times\frac{-12}{5}\Big)\times\frac{35}{36}$
Answer$\text{L.H.S.}=\frac{-13}{24}\times\Big(\frac{-12}{5}\times\frac{35}{36}\Big)$
$=\frac{-13}{24}\times\frac{-12\times35}{5\times36}$
$=\frac{-13}{24}\times\frac{-420}{180}$
$=\frac{-13\times(-420)}{24\times180}$
$=\frac{5460}{4320}$
$=\frac{5460\div60}{4320\div60}$
$=\frac{91}{72}$
$\text{R.H.S.}=\Big(\frac{-13}{24}\times\frac{-12}{5}\Big)\times\frac{35}{36}$
$=\frac{-13\times(-12)}{24\times5}\times\frac{35}{36}$
$=\frac{156\times35}{120\times36}$
$=\frac{5460}{4320}$
$=\frac{5460\div60}{4320\div60}$
$=\frac{91}{72}$
$\text{L.H.S.}=\text{R.H.S.}$
$\therefore\frac{-13}{24}\times\Big(\frac{-12}{5}\times\frac{35}{36}\Big)=\Big(\frac{-13}{24}\times\frac{-12}{5}\Big)\times\frac{35}{36}$
View full question & answer→Question 204 Marks
Verify whether the given statement is true or false:
$\Big(\frac{5}{9}\div\frac{1}{3}\Big)\div\frac{5}{2}=\frac{5}{9}\div\Big(\frac{1}{3}\div\frac{5}{2}\Big)$
AnswerFalse.
Solution:
$\text{L.H.S.}=\Big(\frac{5}{9}\div\frac{1}{3}\Big)\div\frac{5}{2}$
$=\Big(\frac{5}{9}\times\frac{3}{1}\Big)\times\frac{2}{5}$
$=\frac{5\times3\times2}{9\times1\times5}$
$=\frac{30}{45}$
$=\frac{2}{3}$
$\text{R.H.S.}=\frac{5}{9}\div\Big(\frac{1}{3}\div\frac{5}{2}\Big)$
$=\frac{5}{9}\div\Big(\frac{1}{3}\times\frac{2}{5}\Big)$
$=\frac{5}{9}\div\Big(\frac{2}{15}\Big)$
$=\frac{5}{9}\times\Big(\frac{15}{2}\Big)$
$=\frac{75}{18}$
$=\frac{25}{6}$
$\text{L.H.S}\neq\text{R.H.S}$
View full question & answer→Question 214 Marks
Verify whether the given statement is true or false:
$\Big(\frac{-3}{5}\div\frac{-12}{35}\Big)\div\frac{1}{14}=\frac{-3}{5}\div\Big(\frac{-12}{35}\div\frac{1}{14}\Big)$
AnswerFalse.
Solution:
$\text{L.H.S.}=\Big(\frac{-3}{5}\div\frac{-12}{35}\Big)\div\frac{1}{14}$
$=\Big(\frac{-3}{5}\times\frac{35}{-12}\Big)\times{14}$
$=\frac{(-3)\times35\times14}{5\times(-12)}$
$=\frac{1470}{60}$
$=\frac{49}{2}$
$\text{R.H.S.}=\frac{-3}{5}\div\Big(\frac{-12}{35}\div\frac{1}{14}\Big)$
$=\frac{-3}{5}\div\Big(\frac{-12}{35}\times\frac{14}{1}\Big)$
$=\frac{-3}{5}\div\Big(\frac{-12\times490}{35}\Big)$
$=\frac{-3}{5}\div\Big(\frac{-5880}{35}\Big)$
$=\frac{-3}{5}\times\frac{35}{-5880}$
$=\frac{-3\times35}{5\times-5880}$
$=\frac{105}{29400}$
$=\frac{3}{280}$
$\text{L.H.S}\neq\text{R.H.S}$
View full question & answer→