Question 13 Marks
Find ten numbers between $\frac{1}{4}$ and $\frac{1}{2}.$
AnswerThe LCM of the denominators (2 and 4) is 4.
So, we ca write $\frac{1}{4}$ as it is.
Also, $\frac{1}{2}=\frac{1\times2}{2\times2}=\frac{2}{4}$
As the integers between the numerators 1 and 2 of both the fraction are not sufficient, we will multiply the fraction
Thus, $\frac{21}{40},\frac{22}{40},\frac{23}{40},\frac{24}{40},\frac{25}{40}..............\frac{38}{40}$ and $\frac{39}{40}$ are the fractions.
We can take any 10 of these.
View full question & answer→Question 23 Marks
$\Big(\frac{8}{5}\times\frac{-3}{2}\Big)+\Big(\frac{-3}{10}\times\frac{11}{16}\Big)$
Answer$\Big(\frac{8}{5}\times\frac{-3}{2}\Big)+\Big(\frac{-3}{10}\times\frac{11}{16}\Big)$
$=\frac{8\times(-30}{5\times2}+\frac{-3\times11}{10\times16}$
$=\frac{4\times(-3)}{5\times1}+\frac{-3\times11}{10\times16}$
$=\frac{-12}{5}+\frac{-33}{160}$
$=\frac{-384-33}{160}=\frac{-417}{160}$
View full question & answer→Question 33 Marks
Add the following number:
$\frac{-5}{16}$ and $\frac{-7}{24}$
Answer$\frac{-5}{16}$ and $\frac{-7}{24}$
The LCM of the denominators 16 and 26 is 48.
Now,
We will express $\frac{-15}{16}$ and $\frac{-7}{24}$ in the form in which it taken denominator as 48.
$\frac{-5}{16}\frac{-5\times3}{16\times3}=\frac{-15}{48}$
$\frac{7}{24}=\frac{7\times2}{24\times2}=\frac{14}{48}$
So,
$\frac{-15}{48}+\frac{14}{48}$
$=\frac{-15+14}{48}=\frac{-1}{48}$
View full question & answer→Question 43 Marks
$\Big(\frac{13}{5}\times\frac{8}{3}\Big)-\Big(\frac{-5}{2}\times\frac{11}{3}\Big)$
Answer$\Big(\frac{13}{5}\times\frac{8}{3}\Big)-\Big(\frac{-5}{2}\times\frac{11}{3}\Big)$
$=\frac{13\times8}{5\times3}-\frac{-5\times11}{2\times3}$
$=\frac{104}{15}-\frac{-55}{6}$
$=\frac{208+275}{30}$ (LCM of the 15, 6 = 30)
$=\frac{483}{30}=\frac{161}{10}$
View full question & answer→Question 53 Marks
Re-arrange suitably and find the sum in the following:
$\frac{3}{5}+\frac{7}{3}+\frac{9}{5}+\frac{-13}{15}+\frac{-7}{3}$
Answer$\frac{3}{5}+\frac{7}{3}+\frac{9}{5}+\frac{-13}{15}+\frac{-7}{7}$
$=\Big(\frac{3}{5}+\frac{9}{5}\Big)+\Big(\frac{7}{3}+\frac{-7}{3}\Big)+\frac{-13}{15}$
$=\frac{3+9}{5}+\frac{7-7}{3}+\frac{-13}{15}=\frac{12}{5}+0+\frac{-13}{15}$
$=\frac{12}{5}+\frac{-13}{15}$
$=\frac{36-13}{15}=\frac{23}{15}$
View full question & answer→Question 63 Marks
By what number should $\frac{-3}{4}$ be multiplied in order to produce $\frac{2}{3}$?
AnswerProduct $=\frac{2}{3}$
Let x be multiplied to $\frac{-3}{4}$ to get $\frac{2}{3}$
$\therefore\text{x}+\frac{-3}{4}=\frac{2}{3}$
$\Rightarrow\text{x}=\frac{2}{3}\div\Big(\frac{-3}{4}\Big)$
$\Rightarrow\text{x}=\frac{2}{3}\times\frac{4}{-3}=\frac{2\times4}{3\times(-3)}=\frac{8}{-9}$
$=\frac{8\times(-1)}{-9(-1)}=\frac{-8}{9}$
$\therefore$ Required number $=\frac{-8}{9}$
View full question & answer→Question 73 Marks
Simplify:
$\Big(\frac{3}{11}\times\frac{5}{6}\Big)-\Big(\frac{9}{12}\times\frac{4}{3}\Big)+\Big(\frac{5}{13}\times\frac{6}{15}\Big)$
Answer$\Big(\frac{3}{11}\times\frac{5}{6}\Big)-\Big(\frac{9}{12}\times\frac{4}{3}\Big)+\Big(\frac{5}{13}\times\frac{6}{15}\Big)$
$=\frac{3\times5}{11\times6}-\frac{9\times4}{12\times3}+\frac{5\times6}{13\times15}$
$=\frac{1\times5}{11\times2}-\frac{3\times1}{3}+\frac{1\times2}{13\times1}$
$=\frac{5}{22}-1+\frac{2}{13}$
$=\frac{65-286+44}{286}$ (LCM of 22, 13 = 286)
$=\frac{109-286}{286}=\frac{-177}{286}$
View full question & answer→Question 83 Marks
Using commutativity and associativity of addition of rational numbers, express the following as a rational number:
$\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}$
Answer$\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}$
$=\Big(\frac{2}{5}+\frac{4}{5}\Big)+\Big(\frac{8}{3}+\frac{-2}{3}\Big)+\frac{-11}{15}$
$=\frac{2+4}{5}+\frac{8-2}{3}+\frac{-11}{15}$
$=\frac{6}{5}+\frac{6}{3}+\frac{-11}{15}$
$=\frac{18+30-11}{15}=\frac{48-11}{15}=\frac{37}{15}$
View full question & answer→Question 93 Marks
Verify the property: x × (y + z) = x × y + x × z by taking:
$\text{x}=\frac{-3}{4},\text{y}=\frac{-5}{2},\text{z}=\frac{7}{6}$
Answer$\text{x}=\frac{-3}{4},\text{y}=\frac{-5}{2},\text{z}=\frac{7}{6}$
$\text{x}\times(\text{y}\times\text{z})=\frac{-3}{4}\times\bigg(\frac{-5}{2}+\frac{7}{6}\bigg)$
$=\frac{-3}{4}\times\frac{-15+7}{6}=\frac{-3}{4}\times\frac{-8}{6}=1$
$\text{x}\times\text{y}+\text{x}\times\text{z}=\frac{-3}{4}\times\frac{-5}{2}+\frac{-3}{4}\times\frac{7}{6}$
$=\frac{15}{8}+\frac{-7}{8}$
$=\frac{15-7}{8}$
$=1$
$\therefore \frac{-3}{4}\times\Big(\frac{-5}{2}+\frac{7}{6}\Big)$
$=\frac{-3}{4}\times\frac{-5}{2}+\frac{-3}{4}\times\frac{7}{6}$
Hence verified.
View full question & answer→Question 103 Marks
Using commutativity and associativity of addition of rational numbers, express the following as a rational number:
$\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}$
Answer$\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}$
$=\Big(\frac{3}{7}+\frac{-11}{7}\Big)+\Big(\frac{-4}{9}+\frac{7}{9}\Big)$
$=\frac{3-11}{7}+\frac{-4+7}{9}$
$=\frac{-8}{7}+\frac{3}{9}$ $\Big(\because\frac{3}{9}=\frac{3\div3}{9\div3}=\frac{1}{3}\Big)$
$=\frac{-24+7}{21}=\frac{-17}{21}$
View full question & answer→Question 113 Marks
Verify the property: x × (y + z) = x × y + x × z by taking:
$\text{x}=\frac{-12}{5},\text{y}=\frac{-15}{4},\text{z}=\frac{8}{3}$
Answer$\text{x}=\frac{-12}{5},\text{y}=\frac{-15}{4},\text{z}=\frac{8}{3}$
$\text{x}\times(\text{y}+\text{z})=\frac{-12}{5}\times\bigg(\frac{-15}{4}+\frac{8}{3}\bigg)$
$=\frac{-12}{5}\times\frac{-45+32}{12}=\frac{-12}{5}\times\frac{-13}{12}=\frac{13}{5}$
$\text{x}\times\text{y}+\text{x}\times\text{z}=\frac{-12}{5}\times\frac{-15}{4}+\frac{-12}{5}\times\frac{8}{3}$
$=\frac{9}{1}+\frac{-32}{5}$
$=\frac{45-32}{5}$
$=\frac{13}{5}$
$\therefore\frac{-12}{5}\times\bigg(\frac{-15}{4}+\frac{8}{3}\bigg)$
$=\frac{-12}{5}\times\frac{-15}{4}+\frac{-12}{5}\times\frac{8}{3}$
Hence verified.
View full question & answer→Question 123 Marks
Re-arrange suitably and find the sum in the following:
$\frac{2}{3}+\frac{-4}{5}+\frac{1}3{}+\frac{2}{5}$
Answer$\frac{2}{3}+\frac{-4}{5}+\frac{1}{3}+\frac{2}{5}$
$=\Big(\frac{2}{3}+\frac{1}{3}\Big)+\Big(\frac{-4}{5}+\frac{2}{5}\Big)$
$=\frac{2+1}{3}+\frac{-4+2}{5}$
$=\frac{3}{3}+\frac{-2}{5}$
$=1+\frac{-2}{5}=\frac{5-2}{5}=\frac{3}{5}$
View full question & answer→Question 133 Marks
Simplify:
$\frac{-8}{19}+\frac{-4}{57}$
Answer$\frac{-8}{19}+\frac{-4}{57}$
The LCM of the denominator of 19 and 57 is 57.
Now,
We will express $\frac{-8}{19}$ in the form in which it takes denominator as 57.
$\frac{-8}{19}=\frac{-8\times3}{19\times3}=\frac{-24}{57}$
So,
$\frac{-24}{57}+\frac{-4}{57}$
$=\frac{-24-4}{57}=\frac{-28}{57}$
View full question & answer→Question 143 Marks
Re-arrange suitably and find the sum in the following:
$\frac{11}{12}+\frac{-17}{3}+\frac{11}{2}+\frac{-25}{2}$
Answer$\frac{11}{12}+\frac{-17}{3}+\frac{11}{2}+\frac{-25}{2}$
$=\Big(\frac{11}{12}+\frac{-17}{3}\Big)+\Big(\frac{11}{2}+\frac{-25}{2}\Big)$
$=\frac{11-68}{12}+\frac{11-25}{2}$
$=\frac{-57}{12}+\frac{-14}{2}$
$=\frac{-52-84}{12}=\frac{-141}{12}$
View full question & answer→Question 153 Marks
Simplify:
$\Big(\frac{1}{4}\times\frac{2}{7}\Big)-\Big(\frac{5}{14}\times\frac{-2}{3}\Big)+\Big(\frac{3}{7}\times\frac{9}{2}\Big)$
Answer$\Big(\frac{1}{4}\times\frac{2}{7}\Big)-\Big(\frac{5}{14}\times\frac{-2}{3}\Big)+\Big(\frac{3}{7}\times\frac{9}{2}\Big)$
$=\frac{1\times2}{4\times7}-\frac{5\times(-2)}{14\times3}+\frac{3\times9}{7\times2}$
$=\frac{1\times1}{2\times7}-\frac{5\times(-1)}{7\times3}+\frac{3\times9}{7\times2}$
$=\frac{1}{14}-\frac{-5}{21}+\frac{27}{14}$
$=\frac{3+10+81}{42}$ (LCM of 14, 21 = 42)
$=\frac{94}{42}=\frac{94\div2}{42\div2}=\frac{47}{21}$
View full question & answer→Question 163 Marks
Using commutativity and associativity of addition of rational numbers, express the following as a rational number:
$\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}$
Answer$\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}$
$=\Big(\frac{4}{7}+\frac{-13}{7}\Big)+\Big(\frac{-8}{9}+\frac{17}{21}\Big)+0$
$=\frac{4-13}{7}+\frac{-56+51}{63}$
$=\frac{-9}{7}+\frac{-5}{63}$
$\frac{-81-5}{63}=\frac{-86}{63}$
View full question & answer→Question 173 Marks
Re-arrange suitably and find the sum in the following:
$\frac{4}{13}+\frac{-5}{8}+\frac{-8}{13}+\frac{9}{13}$
Answer$\frac{4}{13}+\frac{-5}{8}+\frac{-8}{13}+\frac{9}{13}$
$=\frac{4}{13}+\frac{-8}{13}+\frac{9}{13}+\frac{-5}{8}$
$=\frac{4-8+9}{13}+\frac{-5}{8}$
$=\frac{5}{13}+\frac{-5}{8}$
$=\frac{40-65}{104}$ (LCM of 13, 8 = 104)
$=\frac{-25}{104}$
View full question & answer→Question 183 Marks
Verify the property: x × (y + z) = x × y + x × z by taking:
$\text{x}=\frac{-8}{3},\text{y}=\frac{5}{6},\text{z}=\frac{-13}{12}$
Answer$\text{x}=\frac{-8}{3},\text{y}=\frac{5}{6},\text{z}=\frac{-13}{12}$
$\text{x}(\text{y}+\text{z})=\frac{-8}{3}\times\bigg(\frac{5}{6}+\frac{-13}{12}\bigg)$
$=\frac{-8}{3}\times\frac{10-13}{12}=\frac{-8}{2}\times\frac{-3}{12}=\frac{2}{3}$
$\text{x}\times\text{y}+\text{x}\times\text{z}=\frac{-8}{3}\times\frac{5}{6}+\frac{-8}{3}\times\frac{-13}{12}$
$=\frac{-20}{9}+\frac{29}{9}$
$=\frac{-20+26}{9}$
$=\frac{2}{3}$
$\therefore\frac{-8}{3}\times\bigg(\frac{5}{6}+\frac{-13}{12}\bigg)$
$=\frac{-8}{3}\times\frac{5}{6}+\frac{-8}{3}\times\frac{-13}{12}$
Hence verified.
View full question & answer→Question 193 Marks
Re-arrange suitably and find the sum in the following:
$\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}$
Answer$\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}$
$=\Big(\frac{-6}{7}+\frac{-15}{7}\Big)+\Big(\frac{-5}{6}+\frac{-4}{9}\Big)$
$=\frac{-6-15}{7}+\frac{-15-8}{18}$
$=\frac{-21}{7}+\frac{-23}{18}$
$=\frac{-378-161}{126}$
LCM of 7, 18 = 126
$=\frac{-539}{126}=\frac{-539\div7}{126\div7}=\frac{-77}{18}$
View full question & answer→Question 203 Marks
Simplify:
$\Big(\frac{3}{2}\times\frac{1}{6}\Big)+\Big(\frac{5}{3}\times\frac{7}{2}\Big)-\Big(\frac{13}{8}\times\frac{4}{3}\Big)$
Answer$\Big(\frac{3}{2}\times\frac{1}{6}\Big)+\Big(\frac{5}{3}\times\frac{7}{2}\Big)-\Big(\frac{13}{8}\times\frac{4}{3}\Big)$
$=\frac{3\times1}{2\times6}+\frac{5\times7}{3\times2}-\frac{13\times1}{2\times3}$
$=\frac{1\times1}{2\times2}+\frac{5\times7}{3\times2}-\frac{13\times1}{2\times3} $
$=\frac{1}{4}+\frac{5}{7}-\frac{13}{6}$
$=\frac{3+70-26}{12}$ (LCM of 4, 6 = 12)
$\frac{73-26}{12}=\frac{47}{12}$
View full question & answer→Question 213 Marks
Add the following rational number:
$\frac{7}{-18}$ and $\frac{8}{27}$
Answer$\frac{7}{-18}$ and $\frac{8}{27}$
$\frac{7}{-18}=\frac{-7}{18}$
The LCM of the denominator 18 and 27 is 54.
Now,
we will express $\frac{7}{-18}$ and $\frac{8}{27}$ in the form in which it taken denominator as 54.
$\frac{-7}{18}=\frac{-7\times3}{18\times3}=\frac{-21}{54}$
$\frac{8}{27}=\frac{8\times2}{27\times2}=\frac{16}{54}$
So,
$\frac{-21}{54}+\frac{16}{54}$
$=\frac{-21+16}{54}=\frac{-5}{54}$
View full question & answer→Question 223 Marks
Simplify:
$\Big(\frac{13}{9}\times\frac{-15}{2}\Big)+\Big(\frac{7}{3}\times\frac{8}{5}\Big)+\Big(\frac{3}{5}\times\frac{1}{2}\Big)$
Answer$\Big(\frac{13}{9}\times\frac{-15}{2}\Big)+\Big(\frac{7}{3}\times\frac{8}{5}\Big)+\Big(\frac{3}{5}\times\frac{1}{2}\Big)$
$=\frac{-13\times5}{6}+\frac{7\times8}{15}+\frac{3}{10}$
$=\frac{-65}{6}+\frac{56}{15}+\frac{3}{10}$
$=\frac{-65\times5+56\times2+3\times3}{30}$
$=\frac{-204}{30}=\frac{-34}{5}$
View full question & answer→Question 233 Marks
Simplify:
$\frac{8}{9}+\frac{-11}{6}$
Answer$\frac{8}{9}+\frac{-11}{6}$
The LCM of the denominator 9 and 6 is 18.
Now,
We will express $\frac{8}{9}$ and $\frac{-11}{6}$ in the form in which it taken denominator as 18.
$\frac{8}{9}=\frac{8\times2}{9\times2}=\frac{16}{18}$
$\frac{-11}{6}=\frac{-11\times3}{6\times3}=\frac{-33}{18}$
So,
$\frac{16}{18}+\frac{-33}{18}$
$=\frac{16-33}{18}=\frac{-17}{18}$
View full question & answer→Question 243 Marks
The cost of $2\frac{1}{3}\text{m}$ metres of cloth is Rs. $75\frac{1}{4}$ Find the cost of cloth per metre.
AnswerCost of $2\frac{1}{3}\text{m}$ or $\frac{7}{3}\text{m}$ of cloths = Rs. $75\frac{1}{4}$
$=\text{Rs.}\frac{301}{4}$
$\therefore$ Costof 1m cloth = Rs. $\frac{301}{4}\div\frac{7}{3}$
$=\text{Rs.}\frac{301}{4}\times\frac{3}{7} =\text{Rs.}\frac{43\times3}{4\times1}$
$=\text{Rs.}\frac{129}{4}=\text{Rs.}\ 32\frac{1}{4}$
$=\text{Rs.}\ 32.25$
View full question & answer→Question 253 Marks
Using commutativity and associativity of addition of rational numbers, express the following as a rational number:
$\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}$
Answer$\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}$$=\Big(\frac{2}{5}+\frac{-4}{5}\Big)+\Big(\frac{7}{3}+\frac{-1}{3}\Big)$
$=\Big(\frac{2-4}{5}\Big)+\Big(\frac{7-1}{3}\Big)$
$=\frac{-2}{5}+\frac{6}{3}=\frac{-2}{5}+\frac{2}{1}$
$=\frac{2}{1}-\frac{2}{5}$
$=\frac{10-2}{5}=\frac{8}{5}$
View full question & answer→Question 263 Marks
Re-arrange suitably and find the sum in the following:
$\frac{1}{8}+\frac{5}{12}+\frac{2}{7}+\frac{7}{12}+\frac{9}{7}+\frac{-5}{16}$
Answer$\frac{1}{8}+\frac{5}{12}+\frac{2}{7}+\frac{7}{12}+\frac{9}{7}+\frac{-5}{16}$
$=\Big(\frac{1}{8}+\frac{-5}{16}\Big)+\Big(\frac{5}{12}+\frac{7}{12}\Big)+\Big(\frac{2}{7}+\frac{9}{7}\Big)$
$=\frac{2-5}{16}+\frac{5+7}{12}+\frac{2+9}{7}$
$=\frac{-3}{16}+\frac{12}{12}+\frac{11}{7}$
$=\frac{-3}{16}+1+\frac{11}{7}$
$=\frac{-21+112+176}{112}=\frac{-21+228}{112}$
$=\frac{267}{112}$
View full question & answer→Question 273 Marks
Simplify:
$\Big(\frac{13}{9}\times\frac{-15}{2}\Big)+\Big(\frac{7}{3}\times\frac{8}{5}\Big)+\Big(\frac{3}{5}\times\frac{1}{2}\Big)$
Answer$\Big(\frac{13}{9}\times\frac{-15}{2}\Big)+\Big(\frac{7}{3}\times\frac{8}{5}\Big)+\Big(\frac{3}{5}\times\frac{1}{2}\Big)$
$=\frac{-13\times5}{6}+\frac{7\times8}{15}+\frac{3}{10}$
$=\frac{-65}{6}+\frac{56}{15}+\frac{3}{10}$
$=\frac{-65\times5+56\times2+3\times3}{30}$
$=\frac{-204}{30}=\frac{-34}{5}$
View full question & answer→Question 283 Marks
The cost of $7\frac{2}{3}$ metres of rope is Rs $12\frac{3}{4}$ Find its cost per metre.
AnswerCost of $7\frac{2}{3}\text{m}$ of rope = Rs. $12\frac{3}{4}$
or cost of $\frac{23}{3}\text{m}$ of rope = Rs. $\frac{51}{4}$
$\therefore$ Cost of 1m of rope = Rs. $\frac{51}{4}\div\frac{23}{3}$
$=\text{Rs.}\frac{51}{4}\times\frac{3}{23}$
$=\text{Rs.}\frac{153}{92}=\text{Rs.}1\frac{61}{92}$
View full question & answer→Question 293 Marks
Verify the property: x × (y × z) = (x × y) × z by taking:
$\text{x}=0,\text{y}=\frac{-3}{5},\text{z}\frac{-9}{4}$
Answer$\text{x}\times(\text{x}\times\text{z})=(\text{x}\times\text{y})\times\text{z}$
$\text{x}=0,\text{y}=\frac{-3}{5},\text{z}\frac{-9}{4}$
$\text{L.H.S.}=\text{x}\times(\text{y}\times\text{z})=0\times\Big(\frac{-3}{5}\times\frac{-9}{4}\Big)$
$=0\times\frac{(-3)\times(-9)}{5\times4}=0\times\frac{27}{20}=0$
$\text{R.H.S.}=(\text{x}\times\text{y})\times\text{z}\Big(0\times\frac{-3}{5}\Big)\times\frac{-9}{4}$
$=0\times\frac{-9}{4}=0$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 303 Marks
Simplify:
$\frac{-13}{8}+\frac{5}{36}$
Answer$\frac{-13}{8}+\frac{5}{36}$
The LCM of the denominator 8 and 36 is 72.
Now,
We will express $\frac{-13}{8}$ and $\frac{5}{36}$ in the form in which it takes danominator as 72.
$\frac{-13}{8}=\frac{-13\times9}{8\times9}=\frac{-117}{72}$
$\frac{5}{36}=\frac{5\times2}{36\times2}=\frac{10}{72}$
So,
$\frac{-117}{72}+\frac{10}{72}$
$=\frac{-117+10}{72}=\frac{-107}{72}$
View full question & answer→Question 313 Marks
Add and express the sum as a mixed fraction:
$\frac{-12}{5}$ and $\frac{43}{10}$
AnswerWe have:
$\frac{-12}{5}$ and $\frac{43}{10}$
The LCM of the denominator 5 and 10 is 10.
Now,
We will express $\frac{-12}{5}$ in the form in which it takes denominator as 10.
$\frac{-12}{5}=\frac{-12\times2}{5\times2}=\frac{-24}{10}$
So,
$\frac{-24}{10}+\frac{43}{10}$
$=\frac{-24+43}{10}=\frac{19}{10}$
View full question & answer→Question 323 Marks
Add the following rational number:
$\frac{-7}{27}$ and $\frac{11}{18}$
Answer$\frac{-7}{27}+\frac{11}{18}$
The LCM of the denominators 27 and 18 is 54.
Now,
We will express $\frac{-7}{27}$ and $\frac{11}{18}$ in the form in which it taken denominator as 54.
$\frac{-7}{27}=\frac{-7\times2}{27\times2}=\frac{-14}{54}$
$\frac{11}{28}=\frac{11\times3}{18\times3}=\frac{33}{54}$
So,
$\frac{-14}{54}+\frac{33}{54}$
$=\frac{-14+13}{54}=\frac{19}{54}$
View full question & answer→Question 333 Marks
Add and express the sum as a mixed fraction:
$\frac{101}{6}$ and $\frac{7}{8}$
AnswerWe have:
$\frac{101}{6}$ and $\frac{7}{8}$
The LCM o fthe denominator 6 and 8 is 24.
Now,
We will express $\frac{101}{6}$ and $\frac{7}{8}$ in the form in which it takes denominator as 24.
$\frac{101}{6}=\frac{101\times4}{6\times4}=\frac{404}{24}$
$\frac{7}{8}=\frac{7\times3}{8\times3}=\frac{21}{24}$
So,
$\frac{404}{24}+\frac{21}{24}$
$=\frac{404+21}{24}$
$=\frac{425}{24}=17\frac{17}{24}$
View full question & answer→Question 343 Marks
Verify the property: x × y = y × x by taking:
$\text{x}=\frac{-3}{5},\text{y}=\frac{-11}{13}$
Answer$\text{x}\times\text{y}=\text{y}\times\text{x}$ $\text{x}=\frac{-3}{5},\text{y}=\frac{-11}{13}$ $\text{L.H.S.}=\text{x}\times\text{y}=\frac{-3}{5}\times\frac{-11}{13}$ $=\frac{(-3)\times(-11)}{5\times13}=\frac{33}{65}$ $\text{R.H.S}.=\text{y}\times\text{x}=\frac{-11}{13}\times\frac{-3}{5}$ $=\frac{(-11)\times(-3)}{13\times5}=\frac{33}{65}$ $\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 353 Marks
$\Big(\frac{13}{7}+\frac{11}{26}\Big)-\Big(\frac{-4}{3}\times\frac{5}{6}\Big)$
Answer$\Big(\frac{13}{7}+\frac{11}{26}\Big)-\Big(\frac{-4}{3}\times\frac{5}{6}\Big)$
$=\frac{13\times11}{7\times26}-\frac{-4\times5}{3\times6}$
$=\frac{1\times11}{7\times2}-\frac{-4\times5}{3\times6}$
$=\frac{11}{14}-\frac{-10}{9}$
$=\frac{99+140}{126}$ (LCM of 14, 9 = 126)
$=\frac{239}{126}$
View full question & answer→Question 363 Marks
Simplify:
$\frac{5}{26}+\frac{11}{-39}$
Answer$\frac{11}{-39}=\frac{-11}{39}$
The LCM of the denominator 36 and 39 is 78.
Now,
We will express $\frac{-3}{4}$ and $\frac{-11}{39}$ in the form of which it takes denominator as 78.
$\frac{5}{26}=\frac{5\times3}{26\times3}=\frac{15}{78}$
$\frac{-11}{39}=\frac{-11\times2}{39\times2}=\frac{-22}{78}$
So,
$\frac{15}{78}+\frac{-22}{78}$
$=\frac{15-22}{78}=\frac{-7}{-78}$
View full question & answer→Question 373 Marks
Verify the property: x × y = y × x by taking:
$\text{x}=2,\text{y}=\frac{7}{-8}$
Answer$\text{x}\times\text{y}=\text{y}\times\text{x}$
$\text{x}=2,\text{y}=\frac{7}{-8}=\frac{7\times(-1)}{-8\times(-1)}=\frac{-7}{8}$
$\text{L.H.S.}=\text{x}\times\text{y}=2\times\frac{-7}{8}=\frac{2\times(-7)}{8}$
$=\frac{-7}{4}$
$\text{R.H.S.}=\text{x}\times\text{y}=\frac{-7}{8}\times2=\frac{-7}{4}\times1=\frac{-7}{4}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 383 Marks
Verify the property: x × (y + z) = x × y + x × z by taking:
$\text{x}=\frac{-3}{7},\text{y}=\frac{12}{13},\text{z}=\frac{-5}{6}$
AnswerWe have to verify that
$\text{x}\times(\text{y}\times\text{z})=\text{x}\times\text{y}+\text{x}\times\text{z}.$
$\text{x}=\frac{-3}{7},\text{y}=\frac{12}{13},\text{z}=\frac{-5}{6}$
$\text{x}\times(\text{y}+\text{z})=\frac{-3}{7}\times\bigg(\frac{12}{13}+\frac{-5}{6}\bigg)$
$=\frac{-3}{7}\times\frac{72-65}{78}=\frac{-3}{7}\times\frac{7}{78}=\frac{-1}{26}$
$\text{x}\times\text{y}+\text{x}\times\text{z}=\frac{-3}{7}\times\frac{12}{13}+\frac{-3}{7}\times\frac{-5}{6}$
$=\frac{-36}{91}+\frac{5}{14}$
$=\frac{-36\times2+5\times13}{182}=\frac{-72+65}{182}$
$=\frac{-1}{26}$
$\therefore\frac{-3}{7}\times\Big(\frac{12}{13}+\frac{-5}{6}\Big)$
$=\frac{-3}{7}\times\frac{12}{13}+\frac{-3}{7}\times\frac{-5}{6}$
Hence verified.
View full question & answer→Question 393 Marks
Add and express the sum as a mixed fraction:
$\frac{-31}{6}$ and $\frac{-27}{8}$
AnswerWe have:
$\frac{-31}{6}$ and $\frac{-27}{8}$
The LCM of the denominator 6 and 8 is 24.
Now,
We will express $\frac{-31}{6}$ and $\frac{-27}{8}$ in the form in which it takes denominator as 24.
$\frac{-31}{6}=\frac{-31\times4}{6\times4}=\frac{-124}{24}$
$\frac{-27}{8}=\frac{-27\times3}{8\times3}=\frac{-81}{24}$
$\frac{-124}{24}+\frac{-81}{24}$
$\frac{-124-18}{24}$
$=\frac{-205}{24}=-8\frac{13}{24}$
View full question & answer→Question 403 Marks
Express the following as a rational number of the form $\frac{\text{p}}{\text{q}}:$
$\frac{-8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}-3$
Answer$\frac{-8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}-3$
$=\frac{-64}{24}+\frac{-6}{24}+\frac{-44}{24}+\frac{9}{24}-\frac{72}{24}$
$=\frac{(-64)+(-6)+(-44)+9+(-72)}{24}$
$=\frac{-64-6-44+9-72}{24}$
$=\frac{-177}{24}$
$=\frac{-59}{8}$
View full question & answer→Question 413 Marks
Add and express the sum as a mixed fraction:
$\frac{24}{7}$ and $\frac{-11}{4}$
AnswerWe have:
$\frac{24}{7}$ and $\frac{-11}{4}$
The LCM of the denominator 7 and 4 is 28.
Now,
We will express $\frac{24}{7}$ and $\frac{-11}{4}$ in the form in which it takes denominator as 10.
$\frac{24}{7}=\frac{24\times4}{7\times4}=\frac{96}{28}$
$\frac{-11}{4}=\frac{-11\times7}{4\times7}=\frac{-77}{28}$
So,
$\frac{96}{28}+\frac{-77}{28}$
$=\frac{96-77}{28}=\frac{19}{28}$
View full question & answer→Question 423 Marks
Verify the property: x × y = y × x by taking:
$\text{x}=-\frac{1}{3},\text{y}=\frac{2}{7}$
Answer$\text{x}\times\text{y}=\text{y}\times\text{x}$
$\text{x}=-\frac{1}{3},\text{y}=\frac{2}{7}$
$\text{L.H.S.}=\text{x}\times\text{y}=-\frac{1}{3}\times\frac{2}{7}=\frac{-1\times2}{3\times7}=\frac{-2}{21}$
$\text{R.H.S.}=\text{y}\times\text{x}=\frac{2}{7}\times\frac{-1}{3}=\frac{2\times(-1)}{7\times3}$
$=\frac{-2}{21}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 433 Marks
Simplify:
$\frac{-16}{9}+\frac{-5}{12}$
Answer$\frac{-16}{9}+\frac{-5}{12}$
The LCM of the denominator 9 and 12 is 36.
Now,
We will express $\frac{-16}{9}$ and $\frac{-5}{12}$ in the form of in which it takes dinominator as 36.
$\frac{-16}{9}=\frac{-16\times4}{9\times4}=\frac{-64}{36}$
$\frac{-5}{12}=\frac{-5\times3}{12\times3}=\frac{-15}{36}$
So,
$\frac{-64}{36}+\frac{-15}{36}$
$=\frac{-64-15}{36}=\frac{-79}{36}$
View full question & answer→Question 443 Marks
$\Big(\frac{-4}{3}\times\frac{12}{-5}\Big)+\Big(\frac{3}{7}\times\frac{21}{15}\Big)$
Answer$\Big(\frac{-4}{3}\times\frac{12}{-5}\Big)+\Big(\frac{3}{7}\times\frac{21}{15}\Big)$
$=\frac{-4\times12}{3\times(-5)}+\frac{3\times21}{7\times15}$
$=\frac{-4\times4}{1\times(-5)}+\frac{1\times3}{1\times5}=\frac{-16}{-5}+\frac{3}{5}$
$=\frac{16}{5}+\frac{3}{5}=\frac{16+3}{5}=\frac{19}{5}$
View full question & answer→