Question 13 Marks
Evaluate:
$\frac{(25)^{\frac{5}{2}}\times(729)^{\frac{1}{3}}}{(125)^{\frac{2}{3}}\times(27)^{\frac{2}{3}}\times8^{\frac{4}{3}}}$
Answer $\frac{(25)^{\frac{5}{2}}\times(729)^{\frac{1}{3}}}{(125)^{\frac{2}{3}}\times(27)^{\frac{2}{3}}\times8^{\frac{4}{3}}}$
$=\frac{(5^2)^{\frac{5}{2}}\times(9^3)^{\frac{1}{3}}}{(5^3)^{\frac{2}{3}}\times(3^3)^{\frac{2}{3}}\times(2^3)^{\frac{4}{3}}}$
$=\frac{5^{2\times\frac{5}{2}}\times9^{3\times\frac{1}{3}}}{5^{3\times\frac{2}{3}}\times3^{3\times\frac{2}{3}}\times2^{3\times\frac{4}{3}}}$
$=\frac{5^5\times9}{5^2\times3^2\times2^4}$
$=\frac{5^3}{2^4}$
$=\frac{125}{16}$
View full question & answer→Question 23 Marks
Find three rational numbers lying between $\frac{3}{5}$ and $\frac{7}{8}.$ How many rational numbers can be determined between these two numbers?
Answer $\text{x}=\frac{3}{5}$ and $\text{y}=\frac{7}{8}$
$\text{n}=3$
$\text{d}=\frac{(\text{y}-\text{x})}{\text{n}+1}=\frac{\frac{7}{8}-\frac{3}{5}}{3+1}=\frac{11}{40}\times\frac{1}{4}=\frac{11}{160}$
Rational numbers between $\text{x}=\frac{3}{5}$ and $\text{y}=\frac{7}{8}$ will be
$(\text{x}+\text{d}),(\text{x}+2\text{d}),...,(\text{x}+\text{n}\text{d})$
$\Rightarrow\Big(\frac{3}{5}+\frac{11}{160}\Big),\Big(\frac{3}{5}+2\times\frac{11}{160}\Big),\Big(\frac{3}{5}+3\times\frac{11}{160}\Big)$
$\Rightarrow\Big(\frac{107}{160}\Big),\Big(\frac{118}{160}\Big),\Big(\frac{129}{160}\Big)$
$\Rightarrow\Big(\frac{107}{160}\Big),\Big(\frac{59}{80}\Big),\Big(\frac{129}{160}\Big)$
There are infinitely many rational numbers between two given rational numbers.
View full question & answer→Question 33 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $1.3\overline{23}$
Answer$1.3\overline{23}$ Let $x = 1.32323 ...(i)$
we multiply $x$ by $10. 10x = 13.2323 ...(ii)$
Again multiplying by $100$ as there are $2$ repeating numbers after decimals we get $1000x = 1323.2323 ...(iii)$
Subtracting $(ii)$ from $(iii)$ we get $990x = 1310$
$\Rightarrow\text{x}=\frac{131}{99}$
View full question & answer→Question 43 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $0.\overline{53}$
Answer$0.\overline{53}$ Let $x = 0.5353 ...(i)$
Two digits are repeated so,we multiply $x$ by $100. 100x = 53.5353 ...(ii)$
Subtracting $(i)$ from $(ii)$ we get $99x = 53$ $\Rightarrow\text{x}=\frac{53}{99}$
View full question & answer→Question 53 Marks
Find six rational numbers between $2$ and $3$.
Answer$x = 2, y = 3$ and $n = 6$ $\text{d}=\frac{\text{y}-\text{x}}{\text{n}+1}=\frac{3-2}{6+1}=\frac{1}{7}$
Thus, the required numbers are $(\text{x}+\text{d}),(\text{x}+2\text{d}),(\text{x}+3\text{d}),...,(\text{x}+\text{n}\text{d})$
$=\Big(2+\frac{1}{7}\Big),\Big(2+2\times\frac{1}{7}\Big),\Big(2+3\times\frac{1}{7}\Big),$
$\Big(2+4\times\frac{1}{7}\Big),\Big(2+5\times\frac{1}{7}\Big),\Big(2+6\times\frac{1}{7}\Big)$
$=\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7}$
View full question & answer→Question 63 Marks
It being given that $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236$ and $\sqrt{10}=3.162,$ find the value of three places of decimals, of the following:
$\frac{2}{\sqrt{5}}$
Answer$\frac{2}{\sqrt{5}}$
$=\frac{2}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}$
$=\frac{2\sqrt{5}}{\big(\sqrt{5}\big)^2}$
$=\frac{2\sqrt{5}}{5}$
$=\frac{2\times2.236}{5}$
$=\frac{4.472}{5}$
$=0.8944$
$=0.894$ (three places of decimal)
View full question & answer→Question 73 Marks
It being given that $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236$ and $\sqrt{10}=3.162,$ find the value of three places of decimals, of the following: $\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}$
Answer$\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}$
$=\frac{\sqrt{5\times2}-\sqrt{5}}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{2\sqrt{5}-\sqrt{10}}{\big(\sqrt{2}\big)^2}$
$=\frac{2\sqrt{5}-\sqrt{10}}{2}$
$=\frac{2\times2.236-3.162}{2}$
$=\frac{4.472-3.162}{2}$
$=\frac{1.31}{2}$
$=0.655$ (three places of decimal)
View full question & answer→Question 83 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $2.\overline{93}$
Answer$2.\overline{93}$ Let $x = 2.9393 ...(i)$
Two digits are repeated so, we multiply $x$ by $100. 100x = 293.9393 ...(ii)$
Subtracting $(i)$ from $(ii)$ we get $99x = 291$
$\Rightarrow\text{x}=\frac{291}{99}=\frac{97}{33}$
View full question & answer→Question 93 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $0.\overline{2}$
Answer$0.\overline{2}$ Let $x = 0.222 ...(i)$
Only one digit is repeated so, we multiply $x$ by $10$ $10x = 2.222 ...(ii)$
Subtracting $(i)$ from $(ii)$ we get $9x = 2$
$\Rightarrow\text{x}=\frac{2}{9}$
View full question & answer→Question 103 Marks
Prove that: $\Big[8^{-\frac{2}{3}}\times2^{\frac{1}{2}}\times25^{-\frac{5}{4}}\Big]\div\Big[32^{-\frac{2}{5}}\times125^{-\frac{5}{6}}\Big]=\sqrt{2}$
Answer$\text{L.H.S.}=\Big[8^{-\frac{2}{3}}\times2^{\frac{1}{2}}\times25^{-\frac{5}{4}}\Big]\div\Big[32^{-\frac{2}{5}}\times125^{-\frac{5}{6}}\Big]$
$=\Big[2^{3\times\big(-\frac{2}{3}\big)}\times\sqrt{2}\times5^{2\times\big(-\frac{5}{4}\big)}\Big]\div\Big[2^{5\times\big(-\frac{2}{5}\big)}\times5^{3\times\big(-\frac{5}{6}\big)}\Big]$
$=\Big[2^{-2}\times\sqrt{2}\times5^{-\frac{5}{2}}\Big]\div\Big[2^{-2}\times5^{-\frac{5}{2}}\Big]$
$=\frac{2^{-2}\times\sqrt{2}\times5^{-\frac{5}{2}}}{2^{-2}\times5^{-\frac{5}{2}}}$
$=\sqrt{2}$
$=\text{R.H.S.}$
View full question & answer→Question 113 Marks
Prove that: $\Big(\frac{1}{\text{x}^{\text{a}-\text{b}}}\Big)^{\frac{1}{\text{a}-\text{c}}}\times\Big(\frac{1}{\text{x}^{\text{b}-\text{c}}}\Big)^{\frac{1}{\text{b}-\text{a}}}\times\Big(\frac{1}{\text{x}^{\text{c}-\text{a}}}\Big)^{\frac{1}{\text{c}-\text{b}}}=1$
Answer$\Big(\frac{1}{\text{x}^{\text{a}-\text{b}}}\Big)^{\frac{1}{\text{a}-\text{c}}}\times\Big(\frac{1}{\text{x}^{\text{b}-\text{c}}}\Big)^{\frac{1}{\text{b}-\text{a}}}\times\Big(\frac{1}{\text{x}^{\text{c}-\text{a}}}\Big)^{\frac{1}{\text{c}-\text{b}}}=1$
$=\big(\text{x}\big)^{\frac{1}{(\text{a}-\text{b})}\times\frac{1}{(\text{a}-\text{c})}}\times=\big(\text{x}\big)^{\frac{1}{(\text{b}-\text{c})}\times\frac{1}{(\text{b}-\text{a})}}\times\big(\text{x}\big)^{\frac{1}{(\text{c}-\text{a})}\times\frac{1}{(\text{c}-\text{b})}}$
$=\big(\text{x}\big)^{\frac{1}{(\text{a}-\text{b})(\text{a}-\text{c})}+\frac{1}{(\text{b}-\text{c})(\text{b}-\text{a})}+\frac{1}{(\text{c}-\text{a})(\text{c}-\text{b})}}$
$=\big(\text{x}\big)^{\frac{1}{(\text{a}-\text{b})(\text{c}-\text{a})}-\frac{1}{(\text{b}-\text{c})(\text{a}-\text{b})}-\frac{1}{(\text{c}-\text{a})(\text{b}-\text{c})}}$
$=\big(\text{x}\big)^{\frac{-(\text{b}-\text{c})-(\text{c}-\text{a})-(\text{a}-\text{b})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}}$
$=\big(\text{x}\big)^{\frac{-\text{b}+\text{c}-\text{c}+\text{a}-\text{a}+\text{b}}{(\text{a}-\text{b}(\text{b}-\text{c})(\text{c}-\text{a})}}$
$=\text{x}^0$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 123 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$
$18.\overline{48}$
Answer$18.\overline{48}$
Let $x = 18.4848 ...(i)$
Two digits are repeated so, we multiply $x$ by $100$.
$100x = 1848.4848 ...(ii)$
Subtracting $(i)$ from $(ii)$ we get
$99x = 1830$
$\Rightarrow\text{x}=\frac{1830}{99}=\frac{610}{33}$
View full question & answer→Question 133 Marks
If $\text{x}=3+2\sqrt{2},$ check whether $\text{x}+\frac{1}{\text{x}}$ is rational or irrational.
Answer$\text{x}=3+2\sqrt{2}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=3+2\sqrt{2}+\frac{1}{3+2\sqrt{2}}$
$=3+2\sqrt{2}+\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{9-8}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{1}$
$=3+2\sqrt{2}+3-2\sqrt{2}$
$=6$ Thus, the given number is rational.
View full question & answer→Question 143 Marks
Find five rational numbers between $\frac{3}{5}$ and $\frac{2}{3}.$
Answer$n = 5 n + 1 = 6$ $\text{x}=\frac{3}{5},\text{y}=\frac{2}{3}$
$\text{d}=\frac{\text{y}-\text{x}}{\text{n}+1}=\frac{\frac{2}{3}-\frac{3}{5}}{6}=\frac{10-9}{90}=\frac{1}{90}$
Thus, rational numbers between $\frac{3}{5}$ and $\frac{2}{3}$ will be
$(\text{x}+\text{d}),(\text{x}+2\text{d}),(\text{x}+3\text{d}),(\text{x}+4\text{d}),(\text{x}+5\text{d})$
$=\Big(\frac{3}{5}+\frac{1}{90}\Big),\Big(\frac{3}{5}+\frac{2}{90}\Big),\Big(\frac{3}{5}+\frac{3}{90}\Big),\\\Big(\frac{3}{5}+\frac{4}{90}\Big),\Big(\frac{3}{5}+\frac{5}{90}\Big)$
$=\Big(\frac{55}{90}\Big),\Big(\frac{56}{90}\Big),\Big(\frac{57}{90}\Big),\Big(\frac{58}{90}\Big),\Big(\frac{59}{90}\Big)$
$=\Big(\frac{11}{18}\Big),\Big(\frac{28}{45}\Big),\Big(\frac{19}{30}\Big),\Big(\frac{29}{46}\Big),\Big(\frac{59}{90}\Big)$
View full question & answer→Question 153 Marks
Evaluate: $\Big(\frac{81}{16}\Big)^{-\frac{3}{4}}\Big[\Big(\frac{25}{9}\Big)^{-\frac{3}{2}}\div\Big(\frac{5}{2}\Big)^{-3}\Big]$
Answer$\Big(\frac{81}{16}\Big)^{-\frac{3}{4}}\Big[\Big(\frac{25}{9}\Big)^{-\frac{3}{2}}\div\Big(\frac{5}{2}\Big)^{-3}\Big]$
$=\Big(\frac{16}{81}\Big)^{\frac{3}{4}}\Big[\Big(\frac{9}{25}\Big)^{\frac{3}{2}}\div\Big(\frac{2}{5}\Big)^{3}\Big]$
$=\Big(\frac{2^4}{3^4}\Big)^{\frac{3}{4}}\Big[\Big(\frac{3^2}{5^2}\Big)^{\frac{3}{2}}\div\frac{2^3}{5^3}\Big]$
$=\frac{2^{4\times\frac{3}{4}}}{3^{4\times\frac{3}{4}}}\Bigg[\frac{3^{2\times\frac{3}{2}}}{5^{2\times\frac{3}{2}}}\div\frac{8}{125}\Bigg]$
$=\frac{2^3}{3^3}\Big[\frac{3^3}{5^3}\times\frac{125}{8}\Big]$ $=\frac{8}{27}\Big[\frac{27}{125}\times\frac{125}{8}\Big]$
$=\frac{8}{27}\times\frac{27}{8}$
$=1$
View full question & answer→Question 163 Marks
Prove that: $\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$
Answer$\text{L.H.S.}=\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}$
$=\Big(\frac{125}{64}\Big)^{\frac{2}{3}}+\Big(\frac{625}{256}\Big)^{\frac{1}{4}}+\frac{\sqrt{5^2}}{\sqrt[3]{4^3}}$
$=\frac{5^{3\times\frac{2}{3}}}{4^{3\times\frac{2}{3}}}+\frac{5^{4\times\frac{1}{4}}}{4^{4\times\frac{1}{4}}}+\frac{5}{4}$
$=\frac{24}{16}+\frac{5}{4}+\frac{5}{4}$
$=\frac{25+20+20}{16}$
$=\frac{65}{16}$
$=\text{R.H.S.}$
View full question & answer→Question 173 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $0.\overline{235}$
Answer$0.\overline{235}$ Let $x = 0.235235 ...(i)$
Two digits are repeated so, we multiply $x$ by $1000. 1000x = 235.235235 ...(ii)$
Subtracting $(i)$ from $(ii)$ we get $999x = 235$
$\Rightarrow\text{x}=\frac{235}{999}$
View full question & answer→Question 183 Marks
If $x$ is a positive real number and exponents are rational numbers, simplify $\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{c}}\Big)^{\text{b+c-a}}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^{\text{c+a-b}}\times\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^{\text{a+b-c}}.$
Answer$\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{c}}\Big)^{\text{b+c-a}}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^{\text{c+a-b}}\times\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^{\text{a+b-c}}$
$=\bigg(\frac{\text{X}^{\text{b}^2+\text{bc}-\text{ab}}}{\text{X}^{\text{bc}+\text{c}^2-\text{ac}}}\bigg)\times\bigg(\frac{\text{X}^{\text{c}^2+\text{ac}-\text{bc}}}{\text{X}^{\text{ac}+\text{a}^2-\text{ab}}}\bigg)\times\bigg(\frac{\text{X}^{\text{a}^2+\text{ab}-\text{ac}}}{\text{X}^{\text{ab}+\text{b}^2-\text{bc}}}\bigg)$
$=\Big(\text{X}^{\text{b}^2+\text{bc}-\text{ab}-\text{bc}-\text{c}^2+\text{ac}}\Big)$
$\Big(\text{X}^{\text{c}^2+\text{ac}-\text{bc}-\text{ac}-\text{a}^2+\text{ab}}\Big)$
$\Big(\text{X}^{\text{a}^2+\text{ab}-\text{ac}-\text{ab}-\text{b}^2+\text{bc}}\Big)$
$=\Big(\text{X}^{\text{b}^2-\text{ab}-\text{c}^2+\text{ac}}\Big)\Big(\text{X}^{\text{c}^2-\text{bc}-\text{a}^2+\text{ab}}\Big)\Big(\text{X}^{\text{a}^2-\text{ac}-\text{b}^2+\text{bc}}\Big)$
$=\text{X}^{\text{b}^2-\text{ab}-\text{c}^2+\text{ac}+\text{c}^2-\text{bc}-\text{a}^2+\text{ab}+\text{a}^2-\text{ac}-\text{b}^2+\text{bc}}$
$=\text{X}^0$
$=1$
View full question & answer→Question 193 Marks
Simplify by rationalising the denominator: $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$
Answer$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$
$=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\times\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}$
$=\frac{2\sqrt{6}\times3\sqrt{5}+\big(2\sqrt{6}\big)^2-\sqrt{5}\times3\sqrt{5}-\sqrt{5}\times2\sqrt{6}}{\big(3\sqrt{5}\big)^2-\big(2\sqrt{6}\big)^2}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}$
$=\frac{4\sqrt{30}+9}{21}$
View full question & answer→Question 203 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $0.40\overline{7}$
Answer$0.40\overline{7}$ Let $x = 0.40777 ...(i)$
we multiply $x$ by $100. 100x = 40.7777 ...(ii)$
Again multiplying by $100$ as there are $1$ repeating numbers after decimals we get $1000x = 407.777 ...(iii)$
Subtracting $(ii)$ from $(iii)$ we get $900x = 367$
$\Rightarrow\text{x}=\frac{367}{900}$
View full question & answer→Question 213 Marks
Simplify: $\frac{2\sqrt{30}}{\sqrt{6}}-\frac{3\sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$
Answer$\frac{2\sqrt{30}}{\sqrt{6}}-\frac{3\sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$
$=\frac{2\sqrt{6\times5}}{\sqrt{6}}-\frac{3\sqrt{28\times5}}{\sqrt{28}}+\frac{\sqrt{11\times5}}{\sqrt{11\times9}}$
$=\frac{2\sqrt{6}\times\sqrt{5}}{\sqrt{6}}-\frac{3\sqrt{28}\times\sqrt{5}}{\sqrt{28}}+\frac{\sqrt{11}\times\sqrt{5}}{3\sqrt{11}}$
$=2\sqrt{5}-3\sqrt{5}+\frac{\sqrt{5}}{3}$
$=-\sqrt{5}+\frac{\sqrt{5}}{3}$
$=\frac{-3\sqrt{5}+\sqrt{5}}{3}$
$=\frac{-2\sqrt{5}}{3}$
View full question & answer→Question 223 Marks
Prove that: $\frac{(\text{x}^{\text{a}+\text{b}})^2(\text{x}^{\text{b}+\text{c}})^2(\text{x}^{\text{c}+\text{a}})^2}{(\text{x}^\text{a}\text{x}^\text{b}\text{x}^\text{c})}=1$
Answer$\text{L.H.S}=\frac{(\text{x}^{\text{a}+\text{b}})^2(\text{x}^{\text{b}+\text{c}})^2(\text{x}^{\text{c}+\text{a}})^2}{(\text{x}^\text{a}\text{x}^\text{b}\text{x}^\text{c})}=1$
$=\frac{(\text{X}^{2\text{a}+2\text{b}})(\text{X}^{2\text{b}+2\text{c}})(\text{X}^{2\text{c}+2\text{a}})}{\text{X}^{4\text{a}}\text{X}^{4\text{b}}\text{X}^{4\text{c}}}$
$=\frac{\text{X}^{2\text{a}+2\text{b}+2\text{b}+2\text{c}+2\text{c}+2\text{a}}}{\text{X}^{4\text{a}+4\text{b}+4\text{c}}}$
$=\frac{\text{X}^{4\text{a}+4\text{b}+4\text{c}}}{\text{X}^{4\text{a}+4\text{b}+4\text{c}}}$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 233 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $32.12\overline{35}$
Answer$32.12\overline{35}$ Let $x = 32.123535 ...(i)$
we multiply $x$ by $100. 100x = 3212.3535 ...(ii)$
Again multiplying by $100$ as there are $2$ repeating numbers after decimals we get $10000x = 321235.35 ...(iii)$
Subtracting $(ii)$ from $(iii)$ we get $9900x = 318023$
$\Rightarrow\text{x}=\frac{318023}{9900}$
View full question & answer→Question 243 Marks
If $\text{x}=\sqrt{13}+2\sqrt{3},$ find the value of $\text{x}-\frac{1}{\text{x}}.$
Answer$\text{x}=\sqrt{13}+2\sqrt{3}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt{13}+2\sqrt{3}}$
$=\frac{1}{\sqrt{13}+2\sqrt{3}}\times\frac{\sqrt{13}-2\sqrt{3}}{\sqrt{13}-2\sqrt{3}}$
$=\frac{\sqrt{13}-2\sqrt{3}}{\big(\sqrt{13}\big)^2-\big(2\sqrt{3}\big)^2}$
$=\frac{\sqrt{13}-2\sqrt{3}}{13-12}$
$=\sqrt{13}-2\sqrt{3}$
$\therefore \ \text{x}-\frac{1}{\text{x}}=\sqrt{13}+2\sqrt{3}-\sqrt{13}+2\sqrt{3}=4\sqrt{3}$
View full question & answer→Question 253 Marks
Prove that: $\bigg[7\Big\{(81)^\frac{1}{4}+(256)^\frac{1}{4}\Big\}^\frac{1}{4}\bigg]^4=16807$
Answer$\text{L.H.S}=\bigg[7\Big\{(81)^\frac{1}{4}+(256)^\frac{1}{4}\Big\}^\frac{1}{4}\bigg]^4=16807$
$=\bigg[7\Big\{3^{4\times\frac{1}{4}}+4^{4\times\frac{1}{4}}\Big\}^\frac{1}{4}\bigg]^4$
$=\bigg[7\big\{3+4\big\}^\frac{1}{4}\bigg]^4$
$=\Big[7\big\{7\big\}^\frac{1}{4}\Big]^4$
$=7^4\times7^{\frac{1}{4}\times4}$
$=7^4\times7$
$=7^5$
$=16807$
$=\text{R.H.S.}$
View full question & answer→Question 263 Marks
Evaluate: $\big[(16)^{\frac{1}{2}}\big]^{\frac{1}{2}}$
Answer$\big[(16)^{\frac{1}{2}}\big]^{\frac{1}{2}}$ $=\Big[(4^2)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$ $=\Big[4^{\frac{1}{2}}\Big]$ $=2^{2\times\frac{1}{2}}$ $=2$
View full question & answer→Question 273 Marks
Write the following in ascending order of magnitude. $\sqrt[6]{6},\ \sqrt[3]{7},\ \sqrt[4]{8}.$
Answer$\sqrt[6]{6}=6^\frac{1}{6}$
$\sqrt[3]{7}=7^\frac{1}{3}$
$\sqrt[4]{8}=8^\frac{1}{4}$ $L.C.M$ of $6, 3$ and $4 = 12$ Thus,
we have $\Rightarrow \sqrt[6]{6}=6^\frac{1}{6}=6^{\frac{1}{6}\times\frac{2}{2}}$
$=6^\frac{2}{12}=(6^2)^\frac{1}{12}=(36)^\frac{1}{12}$
$\Rightarrow \sqrt[3]{7}=7^\frac{1}{3}=7^{\frac{1}{3}\times\frac{4}{4}}$
$=7^\frac{4}{12}=(7^4)^\frac{1}{12}=(2401)^\frac{1}{12}$
$\Rightarrow \sqrt[4]{8}=8^\frac{1}{4}=8^{\frac{1}{4}\times\frac{3}{3}}$
$=8^\frac{3}{12}=(8^3)^\frac{1}{12}=(512)^\frac{1}{12}$ Since $36<512<2401,$
$\Rightarrow(36)^\frac{1}{12}<(512)^\frac{1}{12}<(2401)^\frac{1}{12}$
$\Rightarrow\sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7}$
View full question & answer→Question 283 Marks
Simplify: $\big(3+\sqrt{3}\big)\big(2+\sqrt{2}\big)^2$
Answer$\big(3+\sqrt{3}\big)\big(2+\sqrt{2}\big)^2$
$=\big(3+\sqrt{3}\big)\Big[(2)^2+2\times2\times\sqrt{2}+\big(\sqrt{2}\big)^2\Big]$
$=\big(3+\sqrt{3}\big)\big[4+4\sqrt{2}+2\big]$
$=\big(3+\sqrt{3}\big)\big(6+4\sqrt{2}\big)$
$=3\times6+3\times4\sqrt{2}+6\sqrt{3}+4\sqrt{2}\times\sqrt{3}$
$=18+12\sqrt{2}+6\sqrt{3}+4\sqrt{6}$
View full question & answer→Question 293 Marks
It being given that $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236$ and $\sqrt{10}=3.162,$ find the value of three places of decimals, of the following:
$\frac{2-\sqrt{3}}{\sqrt{3}}$
Answer $\frac{2-\sqrt{3}}{\sqrt{3}}$
$=\frac{2-\sqrt{3}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$=\frac{\sqrt{3}\big(2-\sqrt{3}\big)}{\big(\sqrt{3}\big)^2}$
$=\frac{2\sqrt{3}-3}{3}$
$=\frac{3.464-3}{3}$
$=\frac{0.464}{3}$
$=0.1546$
$=0.155$ (three places of decimal)
View full question & answer→Question 303 Marks
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following:
$\frac{1}{4\sqrt{3}-3\sqrt{5}}$
Answer$\frac{1}{4\sqrt{3}-3\sqrt{5}}$
$=\frac{1}{4\sqrt{3}-3\sqrt{5}}\times\frac{4\sqrt{3}+3\sqrt{5}}{4\sqrt{3}+3\sqrt{5}}$
$=\frac{4\sqrt{3}+3\sqrt{5}}{\big(4\sqrt{3}\big)^2-\big(3\sqrt{5}\big)^2}$
$=\frac{4\times1.732+3\times2.236}{16\times3-9\times5}$
$=\frac{6.928+6.708}{48-45}$
$=\frac{13.636}{3}$
$=4.545$
View full question & answer→Question 313 Marks
Simplify: $\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}$
Answer$\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}$
$=\frac{1}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}$
$\times\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}\times\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}$
$=\frac{\sqrt{3}+\sqrt{2}}{\big(\sqrt{3}\big)^2-\big(\sqrt{2}\big)^2}-\frac{2\big(\sqrt{5}+\sqrt{3}\big)}{\big(\sqrt{5}\big)^2-\big(\sqrt{3}\big)^2}-\frac{3\big(\sqrt{2}+\sqrt{5}\big)}{\big(\sqrt{2}\big)^2-\big(\sqrt{5}\big)^2}$
$=\frac{\sqrt{3}+\sqrt{2}}{3-2}-\frac{2\big(\sqrt{5}+\sqrt{3}\big)}{5-3}-\frac{3\big(\sqrt{2}+\sqrt{5}\big)}{2-5}$
$=\frac{\sqrt{3}+\sqrt{2}}{1}-\frac{2\big(\sqrt{5}+\sqrt{3}\big)}{2}-\frac{3\big(\sqrt{2}+\sqrt{5}\big)}{-3}$
$=\big(\sqrt{3}+\sqrt{2}\big)-\big(\sqrt{5}+\sqrt{3}\big)+\big(\sqrt{2}+\sqrt{5}\big)$
$=\sqrt{3}+\sqrt{2}-\sqrt{5}-\sqrt{3}+\sqrt{2}+\sqrt{5}$
$=2\sqrt{2}$
View full question & answer→Question 323 Marks
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following: $\frac{1+2\sqrt{3}}{2-\sqrt{3}}$
Answer$\frac{1+2\sqrt{3}}{2-\sqrt{3}}$
$=\frac{1+2\sqrt{3}}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}$
$=\frac{2+\sqrt{3}+2\sqrt{3}\times2+2\sqrt{3}\times\sqrt{3}}{(2)^2-\big(\sqrt{3}\big)^2}$
$=\frac{2+\sqrt{3}+4\sqrt{3}+6}{4-3}$
$=\frac{8+5\sqrt{3}}{1}$
$=8+5\times1.732$
$=8+8.660$
$=16.660$
View full question & answer→Question 333 Marks
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following: $\frac{6}{\sqrt{5}+\sqrt{3}}$
Answer$\frac{6}{\sqrt{5}+\sqrt{3}}$$=\frac{6}{\sqrt{5}+\sqrt{3}}\times\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
$=\frac{6\big(\sqrt{5}-\sqrt{3}\big)}{\big(\sqrt{5}\big)^2-\big(\sqrt{3}\big)^2}$
$=\frac{6\big(\sqrt{5}-\sqrt{3}\big)}{5-3}$
$=\frac{6\big(\sqrt{5}-\sqrt{3}\big)}{2}$
$=3\big(\sqrt{5}-\sqrt{3}\big)$
$=3(2.236-1.732)$
$=3\times0.504$
$=1.512$
View full question & answer→Question 343 Marks
Evaluate:
$\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\Big(\frac{256}{625}\Big)^{-\frac{1}{4}}+\Big(\frac{3}{7}\Big)^0$
Answer $\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\Big(\frac{256}{625}\Big)^{-\frac{1}{4}}+\Big(\frac{3}{7}\Big)^0$
$\Big(\frac{125}{64}\Big)^{\frac{2}{3}}+\Big(\frac{625}{256}\Big)^{\frac{1}{4}}+1$
$=\Big(\frac{5^3}{4^3}\Big)^{\frac{2}{3}}+\Big(\frac{5^4}{4^4}\Big)^{\frac{1}{4}}+1$
$=\frac{5^{3\times\frac{2}{3}}}{4^{3\times\frac{2}{3}}}+\frac{5^{4\times\frac{1}{4}}}{4^{4\times\frac{1}{4}}}+1$
$=\frac{5^2}{4^2}+\frac{5}{4}+1$
$=\frac{25}{16}+\frac{5}{4}+1$
$=\frac{25+20+16}{16}$
$=\frac{61}{16}$
View full question & answer→Question 353 Marks
Insert $16$ rational numbers between $2.1$ and $2.2$.
AnswerLet: $x = 2.1, y = 2.2$ and $n = 16$
We know: $\text{d}=\frac{\text{y}-\text{x}}{\text{n}+1}=\frac{2.2-2.1}{16+1}=\frac{0.1}{17}=\frac{1}{170}=0.005\text{(approx.)}$
So, $16$ rational numbers between $2.1$ and $2.2$ are: $(x + d), (x + 2d),...(x + 16d) = [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)] = 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175$ and $2.18$.
View full question & answer→Question 363 Marks
Evaluate: $\Big[5\Big(8^{\frac{1}{3}}+27^{\frac{1}{3}}\Big)^3\Big]^{\frac{1}{4}}$
Answer$\Big[5\Big(8^{\frac{1}{3}}+27^{\frac{1}{3}}\Big)^3\Big]^{\frac{1}{4}}$
$=\Big[5\Big(2^{3\times\frac{1}{3}}+3^{3\times\frac{1}{3}}\Big)^3\Big]^{\frac{1}{4}}$
$=\big[5(2+3)^3\big]^{\frac{1}{4}}$
$=\big[5(5)^3\big]^{\frac{1}{4}}$
$=\big[5^4\big]^{\frac{1}{4}}$
$=5$
View full question & answer→Question 373 Marks
Evaluate: $\frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}$
Answer$\frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}$
$=\frac{4}{(6^3)^{-\frac{2}{3}}}+\frac{1}{(4^4)^{-\frac{3}{4}}}+\frac{2}{(3^5)^{-\frac{1}{5}}}$
$=\frac{4}{6^{3\times\big(-\frac{2}{3}\big)}}+\frac{1}{4^{4\times\big(-\frac{3}{4}\big)}}+\frac{2}{3^{5\times\big(-\frac{1}{5}\big)}}$
$=\frac{4}{6^{-2}}+\frac{1}{4^{-3}}+\frac{2}{3^{-1}}$
$=4\times6^2+1\times4^3+2\times3$
$=4\times36+64+6$
$=144+70$
$=214$
View full question & answer→Question 383 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $0.00\overline{32}$
Answer$0.00\overline{32}$ Let $x = 0.003232 ...(i)$
we multiply $x$ by $100. 100x = 0.3232 ...(ii)$
Again multiplying by $100$ as there are $2$ repeating numbers after decimals we get $10000x = 32.3232 ...(iii)$
Subtracting $(ii)$ from $(iii)$ we get $9900x = 32$
$\Rightarrow\text{x}=\frac{32}{9900}=\frac{8}{2475}$
View full question & answer→Question 393 Marks
If $\text{x}=2-\sqrt{3},$ find the value of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^3.$
Answer$\text{x}=2-\sqrt{3}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2-\sqrt{3}-\frac{1}{2-\sqrt{3}}$
$=2-\sqrt{3}-\frac{1}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}$
$=2-\sqrt{3}-\frac{2+\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=2-\sqrt{3}-\frac{2+\sqrt{3}}{4-3}$
$=2-\sqrt{3}-2-\sqrt{3}$
$=-2\sqrt{3}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\big(-2\sqrt{3}\big)^3=-24\sqrt{3}$
View full question & answer→Question 403 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$ $0.3\overline{178}$
Answer$0.3\overline{178}$ Let $x = 0.3178178 ...(i)$
we multiply $x$ by $10. 10x = 3.178178 ...(ii)$
Again multiplying by $1000$ as there are $3$ repeating numbers after decimals we get $10000x = 3178.178178 ...(iii)$
Subtracting $(ii)$ from $(iii)$ we get $9990x = 3175$
$\Rightarrow\text{x}=\frac{3175}{9990}=\frac{635}{1998}$
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