Question 512 Marks
Find two rational and two irrational number between $0.5$ and $0.55.$
AnswerThe two rational numbers between $0.5$ and $0.55$ are: $0.51$ and $0.52$
The two irrational numbers between $0.5$ and $0.55$ are: $0.505005000...$ and $0.5101100111000...$
Disclaimer: There are infinite number of rational and irrational numbers between $0.5$ and $0.55.$
View full question & answer→Question 522 Marks
Multiply: $6\sqrt{15}$ by $4\sqrt{3}$
Answer$6\sqrt{15}$ by $4\sqrt{3}$
$6\sqrt{15}\times4\sqrt{3}=6\times4\times\sqrt{15}\times\sqrt{3}$
$=24\times\sqrt{15\times3}$
$=24\times\sqrt{3\times5\times3}$
$=24\times3\sqrt{5}=72\sqrt{5}$
View full question & answer→Question 532 Marks
If $a = 2, b = 3$, find the values of:
$\big(\text{a}^{\text{b}}+\text{b}^{\text{a}}\big)^{-1}$
AnswerGiven, $a = 2$ and $b = 3$
$\therefore\big(\text{a}^{\text{b}}+\text{b}^{\text{a}}\big)^{-1}=\frac{1}{\text{a}^{\text{b}}+\text{b}^{\text{a}}}$
$=\frac{1}{2^3+3^2}$
$=\frac{1}{8+9}$
$=\frac{1}{17}$
View full question & answer→Question 542 Marks
Add: $\big(2\sqrt{3}-5\sqrt{2}\big)$ and $\big(\sqrt{3}+2\sqrt{2}\big)$
AnswerWe have: $=\big(2\sqrt{3}-5\sqrt{2}\big)+\big(\sqrt{3}+2\sqrt{2}\big)$
$=\big(2\sqrt{3}+\sqrt{3}\big)+\big(-5\sqrt{2}+2\sqrt{2}\big)$
$=(2+1)\sqrt{3}+(-5+2)\sqrt{2}$
$=3\sqrt{3}-3\sqrt{2}$
View full question & answer→Question 552 Marks
Examine whether the following numbers are rational or irrational.
$\sqrt{8}\times\sqrt{2}$
Answer As $\sqrt{8}\times\sqrt{2}$
$=\sqrt{8\times2}$
$=\sqrt{16}$
$=4,$ which is an integer
Hence, $\sqrt{8}\times\sqrt{2}$ is rational.
View full question & answer→Question 562 Marks
Rationalise the denominator of the following: $\frac{1+\sqrt{2}}{2-\sqrt{2}}$
Answer$\frac{1+\sqrt{2}}{2-\sqrt{2}}$ $=\frac{1+\sqrt{2}}{2-\sqrt{2}}\times\frac{2+\sqrt{2}}{2+\sqrt{2}}$ $=\frac{\big(1+\sqrt{2}\big)\big(2+\sqrt{2}\big)}{(2)^2-\big(\sqrt{2}\big)^2}$ $=\frac{1\times2+\sqrt{2}+2\sqrt{2}+\big(\sqrt{2}\big)^2}{4-2}$ $=\frac{2+3\sqrt{2}+2}{2}$ $=\frac{4+3\sqrt{2}}{2}$
View full question & answer→Question 572 Marks
Find the value of x in the following:
$5^{x-3} \times 3^{2 x-8}=225$
Answer$5^{x-3} \times 3^{2 x-8}=225$
$\Rightarrow 5^{x-3} \times 3^{2 x-8}=5^2 \times 3^2$
$\Rightarrow x-3=2 \text { and } 2 x-8=2$
$\Rightarrow x=5 \text { and } 2 x=10$
$\Rightarrow x=5$
View full question & answer→Question 582 Marks
Simplify: $3\sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$
Answer$3\sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$
$3\sqrt{9\times5}-\sqrt{25\times5}+\sqrt{100\times2}-\sqrt{25\times2}$
$=3\times3\sqrt{5}-5\sqrt{5}+10\sqrt{2}-5\sqrt{2}$
$=9\sqrt{5}-5\sqrt{5}+10\sqrt{2}-5\sqrt{2}$
$=4\sqrt{5}+5\sqrt{2}$
View full question & answer→Question 592 Marks
Write the following in decimal form and say what kind of decimal expansion has. $\frac{7}{25}$
Answer$\frac{7}{25}=0.28$ By actual division, we have:
It is a terminating decimal expansion. View full question & answer→Question 602 Marks
Solve for $\text{x}\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\frac{32}{3125}.$
Answer $\text{x}\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\frac{32}{3125}$
$\Rightarrow\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\frac{2^5}{5^5}$
$\Rightarrow\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\Big(\frac{2}{5}\Big)^5$
$\Rightarrow2\text{x}-2=5$
$\Rightarrow2\text{x}=7$
$\Rightarrow\text{x}=\frac{7}{2}$
View full question & answer→Question 612 Marks
Multiply: $3\sqrt{8}$ by $3\sqrt{2}$
Answer$3\sqrt{8}$ by $3\sqrt{2}$ $3\sqrt{8}\times3\sqrt{2}=3\times3\times\sqrt{8}\times\sqrt{2}$ $=9\times\sqrt{8\times2}$ $=9\times\sqrt{2\times2\times2\times2}$ $=(9\times2\times2)=36$
View full question & answer→Question 622 Marks
Simplify: $\big(-3+\sqrt{5}\big)\big(-3-\sqrt{5}\big)$
Answer$\big(-3+\sqrt{5}\big)\big(-3-\sqrt{5}\big)$ $=(-3)^2-\big(\sqrt{5}\big)^2$ $=9-5$ $=4$
View full question & answer→Question 632 Marks
Give an example of two irrational numbers whose sum as well as product is rational.
AnswerLet the two irrational numbers be $2+\sqrt{3}$ and $2-\sqrt{3}.$ Sum of these irrational numbers $=\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4,$ which is rational Product of these irrational numbers $=\big(2+\sqrt{3}\big)\big(2-\sqrt{3}\big)=2^2-\big(\sqrt{3}\big)^2=4-3=1,$ which is rational
View full question & answer→Question 642 Marks
Examine whether the following numbers are rational or irrational.
$\sqrt[3]{5}\times\sqrt[3]{25}$
Answer $\sqrt[3]{5}\times\sqrt[3]{25}$
$=\sqrt[3]{5\times25}$
$=\sqrt[3]{125}$
$=5,$ which is an integer
Hence, $\sqrt[3]{5}\times\sqrt[3]{25}$ is rational.
View full question & answer→Question 652 Marks
Simplify: $\big(3-\sqrt{11}\big)\big(3+\sqrt{11}\big)$
Answer$\big(3-\sqrt{11}\big)\big(3+\sqrt{11}\big)$ $=(3)^2-\big(\sqrt{11}\big)^2$ $=9-11$ $=-2$
View full question & answer→Question 662 Marks
Multiply: $16\sqrt{6}$ by $4\sqrt{2}$
Answer$16\sqrt{6}$ by $4\sqrt{2}$ $16\sqrt{6}\div4\sqrt{2}=\frac{16\sqrt{6}}{4\sqrt{2}}=\frac{4\sqrt{6}}{\sqrt{2}}=\frac{4\sqrt{6}\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}$ $=\frac{4\sqrt{6\times2}}{2}=\frac{4\sqrt{2\times3\times2}}{2}$ $=\frac{4\times2\sqrt{3}}{2}=4\sqrt{3}$
View full question & answer→Question 672 Marks
Examine whether the following numbers are rational or irrational. $\sqrt{7}-2$
AnswerLet us assume, to the contrary, that $\sqrt{7}-2$ is rational. Then, $\sqrt{7}-2=\frac{\text{p}}{\text{q}},$ where p and q are coprime and $\text{q}\neq0.$ $\Rightarrow\sqrt{7}=\frac{\text{p}}{\text{q}}+2$ $\Rightarrow\sqrt{7}=\frac{\text{p}+2\text{q}}{\text{q}}$ Since, p and q are are integers. $\Rightarrow\frac{\text{p}+2\text{q}}{\text{q}}$ is rational. So, $\sqrt{7}$ is also rational. But this contradicts the fact that $\sqrt{7}$ is irrational. This contradiction has arisen because of our incorrect assumption that $\sqrt{7}-2$ is rational. Hence, $\sqrt{7}-2$ is irrational.
View full question & answer→Question 682 Marks
Prove that:
$\sqrt{\text{x}^{-1}\text{y}}\times\sqrt{\text{y}^{-1}\text{z}}\times\sqrt{\text{z}^{-1}\text{x}}=1.$
Answer $\text{L.H.S}=\sqrt{\text{x}^{-1}\text{y}}\times\sqrt{\text{y}^{-1}\text{z}}\times\sqrt{\text{z}^{-1}\text{x}}=1.$
$=\sqrt{\frac{\text{y}}{\text{x}}}\times\sqrt{\frac{\text{z}}{\text{y}}}\times\sqrt{\frac{\text{x}}{\text{z}}}$
$=\sqrt{\frac{\text{y}}{\text{x}}\times\frac{\text{z}}{\text{y}}\times\frac{\text{x}}{\text{z}}}$
$=\sqrt1$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 692 Marks
Is the product of a rational and an irrational number always irrational? Give an example.
AnswerYes, the product of a rational and an irrational number is always an irrational number. Example: $2$ is a rational number and $\sqrt{3}$ is an irrational number. Now, $2\times\sqrt{3}=2\sqrt{3},$ which is an irrational number.
View full question & answer→Question 702 Marks
Insert a rational and an irrational number between $2$ and $2.5$
AnswerAs, few rational numbers between $2$ and $2.5$ are: $2.1, 2.2, 2.3, 2.4, ...$
And,
Since, $2=\sqrt{4}$ and $2.5=\sqrt{6.25}$
So, irrational number between $2$ ans $2.5$ are: $\sqrt{4.1},\sqrt{4.2}, \ ...,\sqrt{5} \ ...$
Hence, a rational and an irrational number can be $2.1$ and $\sqrt{5},$ respectively.
Disclaimer: There are infinite rational and irrational numbers between any two rational numbers.
View full question & answer→Question 712 Marks
Multiply: $12\sqrt{15}$ by $4\sqrt{3}$
Answer$12\sqrt{15}$ by $4\sqrt{3}$ $12\sqrt{15}\div4\sqrt{3}=\frac{12\sqrt{15}}{4\sqrt{3}}=\frac{3\sqrt{15}}{\sqrt{3}}=\frac{3\sqrt{15}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$ $=\frac{3\sqrt{15\times3}}{3}=\sqrt{3\times5\times3}=3\sqrt{5}$
View full question & answer→Question 722 Marks
Find the value of $(1296)^{0.17}\times(1296)^{0.08}.$
Answer$(1296)^{0.17}\times(1296)^{0.08}$ $=(6^4)^{\frac{17}{100}}\times(6^4)^{\frac{8}{100}}$ $=6^{4\times\frac{17}{100}}\times6^{4\times\frac{8}{100}}$ $=6^{\frac{17}{25}}\times6^{\frac{8}{25}}$ $=6^{\frac{17}{25}+\frac{8}{25}}$ $=6^{\frac{25}{25}}$ $=6$
View full question & answer→Question 732 Marks
Find the value of x in the following: $\sqrt[3]{3\text{x}-2}=4$
Answer$\sqrt[3]{3\text{x}-2}=4$$\Rightarrow(3\text{x}-2)^\frac{1}{3}=4$
$\Rightarrow\bigg[(3\text{x}-2)^\frac{1}{3}\bigg]=4^3$
$\Rightarrow3\text{x}-2=64$
$\Rightarrow3\text{x}=66$
$\Rightarrow\text{x}=22$
View full question & answer→Question 742 Marks
Rationalise the denominator of the following: $\frac{1}{2+\sqrt{3}}$
AnswerIf a and b are integers, then $\big(\text{a}+\sqrt{\text{b}}\big)$ and $\big(\text{a}-\sqrt{\text{b}}\big)$ are rationalising factor of each other, as $\big(\text{a}+\sqrt{\text{b}}\big)\big(\text{a}-\sqrt{\text{b}}\big)=\big(\text{a}^2-\text{b}\big),$ which is rational. Therefore, we have, $=\frac{1}{\big(2+\sqrt{3}\big)}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $=\frac{2-\sqrt{3}}{(2)^2-\big(\sqrt{3}\big)^2}=\frac{2-\sqrt{3}}{4-3}$ $=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}$
View full question & answer→Question 752 Marks
Simplify: $\big(5+\sqrt{7}\big)\big(2+\sqrt{5}\big)$
Answer$\big(5+\sqrt{7}\big)\big(2+\sqrt{5}\big)$ $=5\times2+5\times\sqrt{5}+2\times\sqrt{7}+\sqrt{5}\times\sqrt{7}$ $=10+5\sqrt{5}+2\sqrt{7}+\sqrt{35}$
View full question & answer→Question 762 Marks
Simplify: $\Big(\frac{81}{49}\Big)^{-\frac{3}{2}}$
Answer$\Big(\frac{81}{49}\Big)^{-\frac{3}{2}}$ $=\Big(\frac{49}{81}\Big)^{\frac{3}{2}}$ $=\Big(\frac{7^2}{9^2}\Big)^{\frac{3}{2}}$ $=\frac{7^{2\times\frac{3}{2}}}{9^{2\times\frac{3}{2}}}$ $=\frac{7^3}{9^3}$ $=\frac{343}{729}$
View full question & answer→Question 772 Marks
Write the following in decimal form and say what kind of decimal expansion has. $\frac{261}{400}$
Answer$\frac{261}{400}=0.6525$
It is a terminating decimal expansion. View full question & answer→Question 782 Marks
Find a rational number between $\frac{1}{9}$ and $\frac{2}{9}$
Answer $\frac{1}{9}$ and $\frac{2}{9}$ A rational number lying between $\frac{1}{9}$ and $\frac{2}{9}$ will be
$\frac{1}{2}\Big(\frac{1}{9}+\frac{2}{9}\Big)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$
View full question & answer→Question 792 Marks
Evaluate $\frac{2^{\text{n}}+2^{\text{n}-1}}{2^{\text{n}+1}-2^{\text{n}}}.$
Answer$\frac{2^{\text{n}}+2^{\text{n}-1}}{2^{\text{n}+1}-2^{\text{n}}}$ $=\frac{2^{\text{n}-1}(2+1)}{2^{\text{n}}(2-1)}$ $=\frac{2^{\text{n}-1}\times3}{2^{\text{n}}\times1}$ $=\frac{3}{2^{\text{n}-\text{n}+1}}$ $=\frac{3}{2}$
View full question & answer→Question 802 Marks
Rationalise the denominator of the following: $\frac{1}{5+3\sqrt{2}}$
AnswerIf a and b are integers and x is a natural number, then $\big(\text{a}+\text{b}\sqrt{\text{x}}\big)$ and $\big(\text{a}-\text{b}\sqrt{\text{x}}\big)$ are rationalising factor of each other, as $\big(\text{a}+\text{b}\sqrt{\text{x}}\big)\big(\text{a}-\text{b}\sqrt{\text{x}}\big)=\big(\text{a}^2-\text{b}^2\text{x}\big),$ which is rational. Therefore, we have, $=\frac{1}{\big(5+3\sqrt{2}\big)}=\frac{1}{5+3\sqrt{2}}\times\frac{5-3\sqrt{2}}{5-3\sqrt{2}}$ $=\frac{5-3\sqrt{2}}{(5)^2-\big(3\sqrt{2}\big)^2}=\frac{5-3\sqrt{2}}{25-18}=\Big(\frac{5-3\sqrt{2}}{7}\Big)$
View full question & answer→Question 812 Marks
Rationalise the denominator of the following: $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
Answer$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$ $=\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$ $=\frac{\big(3-2\sqrt{2}\big)^2}{(3)^2-\big(2\sqrt{2}\big)^2}$ $=\frac{(3)^2-2\times3+2\sqrt{2}+\big(2\sqrt{2}\big)^2}{9-4\times2}$ $=\frac{9-12\sqrt{2}+8}{9-8}$ $=\frac{17-12\sqrt{2}}{1}$ $=17-12\sqrt{2}$
View full question & answer→Question 822 Marks
Find a rational number between $-\frac{3}{4}$ and $-\frac{2}{5}$
Answer$-\frac{3}{4}$ and $-\frac{2}{5}$ Let: $\text{x}=-\frac{3}{4}$ and $\text{y}=-\frac{2}{5}$ Rational number lying between x and y. $\frac{1}{2}(\text{x}+\text{y})=\frac{1}{2}\Big(-\frac{3}{4}-\frac{2}{5}\Big)$ $\frac{1}{2}\Big(\frac{-15-8}{20}\Big)=-\frac{23}{40}$
View full question & answer→Question 832 Marks
Add: $=\big(2\sqrt{2}+5\sqrt{3}-7\sqrt{5}\big)$ and $\big(3\sqrt{3}-\sqrt{2}+\sqrt{5}\big)$
AnswerWe have, $=\big(2\sqrt{2}+5\sqrt{3}-7\sqrt{5}\big)+\big(3\sqrt{3}-\sqrt{2}+\sqrt{5}\big)$ $=\big(2\sqrt{2}-\sqrt{2}\big)+\big(5\sqrt{3}+3\sqrt{3}\big)+\big(-7\sqrt{5}+\sqrt{5}\big)$ $=(2-1)\sqrt{2}+(5+3)\sqrt{3}+(-7+1)\sqrt{5}$ $=\sqrt{2}+8\sqrt{3}-6\sqrt{5}.$
View full question & answer→Question 842 Marks
Examine whether the following number are rational or irrational:
$\big(5-\sqrt{5}\big)\big(5+\sqrt{5}\big)$
Answer $\big(5-\sqrt{5}\big)\big(5+\sqrt{5}\big)$
$=(5)^2-\big(\sqrt{5}\big)^2$
$=25-5$
$=20$
Thus, the given number is rational.
View full question & answer→Question 852 Marks
Find a rational number between $\frac{3}{8}$ and $\frac{2}{5}$
Answer $\frac{3}{8}$ and $\frac{2}{5}$
Let:
$\text{x}=\frac{3}{8}$ and $\text{y}=\frac{2}{5}$
Rational number lying between x and y.
$\frac{1}{2}(\text{x}+\text{y})=\frac{1}{2}\Big(\frac{3}{8}+\frac{2}{5}\Big)$
$=\frac{1}{2}\Big(\frac{15+16}{40}\Big)=\frac{31}{80}$
View full question & answer→Question 862 Marks
Find four rational numbers lying between $\frac{3}{7}$ and $\frac{5}{7}.$
Answern = 4 n + 1 = 4 + 1 = 5 $\frac{3}{7}=\frac{3}{7}\times\frac{5}{5}=\frac{15}{35}$
$\frac{5}{7}=\frac{5}{7}\times\frac{5}{5}=\frac{25}{35}$
Thus, rational numbers between $\frac{3}{7}$ and $\frac{5}{7}$ are $\frac{16}{35},\frac{17}{35},\frac{18}{35},\frac{19}{35}.$
View full question & answer→Question 872 Marks
Evaluate: $\frac{2^0+7^0}{5^0}$
Answer$\frac{2^0+7^0}{5^0}=\frac{1+1}{1}\frac{2}{1}=2$
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