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Maths 3 Mark Question

Real Numbers · Maharashtra Board STD 9 (MSBSHSE - Semi English Medium). 21 questions.

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3 Mark Question

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Rationalize the denominator : $\frac{12}{4 \sqrt{3}-\sqrt{2}}$
Answer
$\frac{12}{4 \sqrt{3}-\sqrt{2}}=\frac{12}{(4 \sqrt{3}-\sqrt{2})} \times \frac{(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3}+\sqrt{2})}$
...[Multiplying the numerator and
denominator by $(4 \sqrt{3}+\sqrt{2})]$
$= \frac{12(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3}-\sqrt{2})(4 \sqrt{3}+\sqrt{2})}$
$= \frac{12(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3})^2-(\sqrt{2})^2}$
$\ldots\left[\because(a+b)(a-b)=a^2-b^2\right]$
$= \frac{12(4 \sqrt{3}+\sqrt{2})}{(16 \times 3)-2}=\frac{12(4 \sqrt{3}+\sqrt{2})}{48-2}$
$\therefore \quad \frac{12}{4 \sqrt{3}-\sqrt{2}}=\left.\frac{6(4 \sqrt{3}+\sqrt{2})}{23}=\sqrt{2}\right)$
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Question 23 Marks
Rationalize the denominator : $\frac{1}{3 \sqrt{5}+2 \sqrt{2}}$
Answer
$\frac{1}{3 \sqrt{5}+2 \sqrt{2}}=\frac{1}{(3 \sqrt{5}+2 \sqrt{2})} \times \frac{(3 \sqrt{5}-2 \sqrt{2})}{(3 \sqrt{5}-2 \sqrt{2})}$
...[Multiplying the numerator and denominator by $(3 \sqrt{5}-2 \sqrt{2})$ ]
$
\begin{aligned}
& =\frac{1 \times(3 \sqrt{5}-2 \sqrt{2})}{(3 \sqrt{5}+2 \sqrt{2})(3 \sqrt{5}-2 \sqrt{2})} \\
& =\frac{3 \sqrt{5}-2 \sqrt{2}}{(3 \sqrt{5})^2-(2 \sqrt{2})^2} \\
& =\frac{3 \sqrt{5}-2 \sqrt{2}}{(9 \times 5)-(4 \times 2)} \\
& =\frac{3 \sqrt{5}-2 \sqrt{2}}{45-8} \\
\therefore \quad \frac{1}{3 \sqrt{5}+2 \sqrt{2}} & =\frac{3 \sqrt{5}-2 \sqrt{2}}{37}
\end{aligned}
$
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Question 33 Marks
Rationalize the denominator : $\frac{1}{\sqrt{3}-\sqrt{2}}$
Answer
$\frac{1}{\sqrt{3}-\sqrt{2}}=\frac{1}{(\sqrt{3}-\sqrt{2})} \times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}$
...[Multiplying the numerator and denominator by $(\sqrt{3}+\sqrt{2})]$
$
\begin{aligned}
& =\frac{1 \times(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} \\
= & \frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} \\
& \ldots\left[\because(a+b)(a-b)=a^2-b^2\right] \\
= & \frac{\sqrt{3}+\sqrt{2}}{3-2}=\frac{\sqrt{3}+\sqrt{2}}{1} \\
\therefore \quad \frac{1}{\sqrt{3}-\sqrt{2}}= & \sqrt{3}+\sqrt{2}
\end{aligned}
$
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Question 43 Marks
Simplify : $2 \sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}}$
Answer
$\begin{aligned} & 2 \sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}} \\ = & 2 \sqrt{16 \times 3}-\sqrt{25 \times 3}-\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ = & 2 \times 4 \sqrt{3}-5 \sqrt{3}-\frac{1}{3} \sqrt{3} \\ = & 8 \sqrt{3}-5 \sqrt{3}-\frac{1}{3} \sqrt{3} \\ = & \left(8-5-\frac{1}{3}\right) \sqrt{3} \\ = & \left(3-\frac{1}{3}\right) \sqrt{3} \\ = & \left(\frac{9-1}{3}\right) \sqrt{3} \\ = & \frac{8}{3} \sqrt{3} \\ \therefore \quad & 2 \sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}}=\frac{8}{3} \sqrt{3}\end{aligned}$
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Question 53 Marks
Simplify : $4 \sqrt{12}-\sqrt{75}-7 \sqrt{48}$
Answer
$\begin{aligned} & 4 \sqrt{12}-\sqrt{75}-7 \sqrt{48} \\ = & 4 \sqrt{4 \times 3}-\sqrt{25 \times 3}-7 \sqrt{16 \times 3} \\ = & 4 \times 2 \sqrt{3}-5 \sqrt{3}-7 \times 4 \sqrt{3} \\ = & 8 \sqrt{3}-5 \sqrt{3}-28 \sqrt{3} \\ = & (8-5-28) \sqrt{3} \\ = & (-25) \sqrt{3} \\ \therefore \quad & 4 \sqrt{12}-\sqrt{75}-7 \sqrt{48}=-25 \sqrt{3}\end{aligned}$
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Question 63 Marks
Simplify : $\sqrt{216}-5 \sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}}$
Answer
$\begin{aligned} & \sqrt{216}-5 \sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}} \\ = & \sqrt{36 \times 6}-5 \sqrt{6}+\sqrt{49 \times 6}-\frac{3}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} \\ = & 6 \sqrt{6}-5 \sqrt{6}+7 \sqrt{6}-\frac{3 \sqrt{6}}{6} \\ = & 6 \sqrt{6}-5 \sqrt{6}+7 \sqrt{6}-\frac{1}{2} \sqrt{6} \\ = & \left(6-5+7-\frac{1}{2}\right) \sqrt{6} \\ = & \left(8-\frac{1}{2}\right) \sqrt{6}=\left(\frac{16-1}{2}\right) \sqrt{6}=\frac{15}{2} \sqrt{6} \\ \therefore \quad & \sqrt{216}-5 \sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}}=\frac{15}{2} \sqrt{6}\end{aligned}$
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Question 73 Marks
Simplify : $5 \sqrt{3}+2 \sqrt{27}+\frac{1}{\sqrt{3}}$
Answer
$\begin{aligned} & 5 \sqrt{3}+2 \sqrt{27}+\frac{1}{\sqrt{3}} \\ = & 5 \sqrt{3}+2 \sqrt{9 \times 3}+\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ = & 5 \sqrt{3}+2 \times 3 \sqrt{3}+\frac{\sqrt{3}}{3} \\ = & 5 \sqrt{3}+6 \sqrt{3}+\frac{\sqrt{3}}{3} \\ = & \left(5+6+\frac{1}{3}\right) \sqrt{3} \\ = & \left(11+\frac{1}{3}\right) \sqrt{3}=\left(\frac{33+1}{3}\right) \sqrt{3}=\frac{34}{3} \sqrt{3} \\ \therefore \quad & 5 \sqrt{3}+2 \sqrt{27}+\frac{1}{\sqrt{3}}=\frac{34}{3} \sqrt{3}\end{aligned}$
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Question 83 Marks
Simplify : $\quad \frac{4}{7} \sqrt{147}+\frac{3}{8} \sqrt{192}-\frac{1}{5} \sqrt{75}$
Answer
$\quad \begin{aligned} & \frac{4}{7} \sqrt{147}+\frac{3}{8} \sqrt{192}-\frac{1}{5} \sqrt{75} \\ = & \frac{4}{7} \sqrt{49 \times 3}+\frac{3}{8} \sqrt{64 \times 3}-\frac{1}{5} \sqrt{25 \times 3} \\ = & \frac{4}{7} \times 7 \sqrt{3}+\frac{3}{8} \times 8 \sqrt{3}-\frac{1}{5} \times 5 \sqrt{3} \\ = & 4 \sqrt{3}+3 \sqrt{3}-\sqrt{3} \\ = & (4+3-1) \sqrt{3} \\ = & 6 \sqrt{3}\end{aligned}$
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Question 93 Marks
Write the following surds in simplest form. : $\quad \frac{3}{4} \sqrt{8}$
Answer
$\begin{aligned} \frac{3}{4} \sqrt{8} & =\frac{3}{4} \times \sqrt{4 \times 2} \\ & =\frac{3}{4} \times 2 \sqrt{2} \\ \therefore \quad \frac{3}{4} \sqrt{8} & =\frac{3}{2} \sqrt{2}\end{aligned}$
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Question 103 Marks
Rationalize the denominator $\frac{8}{3 \sqrt{2}+\sqrt{5}}$
Answer
The conjugate pair of $3 \sqrt{2}+\sqrt{5}$ is $3 \sqrt{2}-\sqrt{5}$
$
\begin{aligned}
\frac{8}{3 \sqrt{2}+\sqrt{5}} & =\frac{8}{3 \sqrt{2}+\sqrt{5}} \times \frac{3 \sqrt{2}-\sqrt{5}}{3 \sqrt{2}-\sqrt{5}} \\
& =\frac{8(3 \sqrt{2}-\sqrt{5})}{(3 \sqrt{2})^2-(\sqrt{5})^2} \\
& =\frac{8 \times 3 \sqrt{2}-8 \sqrt{5}}{9 \times 2-5}=\frac{24 \sqrt{2}-8 \sqrt{5}}{18-5}=\frac{24 \sqrt{2}-8 \sqrt{5}}{13}
\end{aligned}
$
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Question 113 Marks
Find the rationalizing factor of $\sqrt{27}$
Answer
$\sqrt{27}=\sqrt{9 \times 3}=3 \sqrt{3} \quad \therefore 3 \sqrt{3} \times \sqrt{3}=3 \times 3=9$ is a rational number.
$\therefore \sqrt{3}$ is the rationalizing factor of $\sqrt{27}$.
Note that, $\sqrt{27}=3 \sqrt{3}$ means $3 \sqrt{3} \times 3 \sqrt{3}=9 \times 3=27$.
Hence $3 \sqrt{3}$ is also a rationalizing factor of $\sqrt{27}$. In the same way $4 \sqrt{3}, 7 \sqrt{3}, \ldots$ are also the rationalizing factors of $\sqrt{27}$. Out of all these $\sqrt{3}$ is the simplest rationalizing factor.
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Question 123 Marks
Find the value.
i. | 15 – 2|
ii. | 4 – 9|
iii. | 7| x | -4|
Answer
i. |15 – 2| = |13| = 13
ii. |4 – 9| = |-5| = 5
iii. |7| x |- 4| = 7 x 4 = 28
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Question 133 Marks
Rationalize the denominator : $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
Answer
$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})} \times \frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})}$
...[Multiplying the numerator and denominator by $(\sqrt{5}-\sqrt{3})]$
$ =\frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5})^2-(\sqrt{3})^2}$
$\ldots\left[\because(a-b)(a+b)=a^2-b^2\right]$
$=\frac{(\sqrt{5})^2-2 \times \sqrt{5} \times \sqrt{3}+(\sqrt{3})^2}{5-3}$
$=\frac{5-2 \sqrt{15}+3}{2}$
$=\frac{8-2 \sqrt{15}}{2}$
$=\frac{2(4-\sqrt{15})}{2}=4-\sqrt{15}$
$\therefore \quad \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=4-\sqrt{15} $
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Question 143 Marks
Rationalize the denominator : $\frac{4}{7+4 \sqrt{3}}$
Answer
$ \frac{4}{7+4 \sqrt{3}}= \frac{4}{(7+4 \sqrt{3})} \times \frac{(7-4 \sqrt{3})}{(7-4 \sqrt{3})}$
$\ldots[\text { Multiplying the numerator and }$
$\text { denominator by }(7-4 \sqrt{3})]$
$= \frac{4(7-4 \sqrt{3})}{(7)^2-(4 \sqrt{3})^2}$
$= \frac{4(7-4 \sqrt{3})}{49-(16 \times 3)}$
$= \frac{4(7-4 \sqrt{3})}{49-48}=\frac{4(7-4 \sqrt{3})}{1}$
$\therefore \quad \frac{4}{7+4 \sqrt{3}}= 28-16 \sqrt{3}$
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Question 153 Marks
Rationalize the denominator : $\frac{3}{2 \sqrt{5}-3 \sqrt{2}}$
Answer
$\frac{3}{2 \sqrt{5}-3 \sqrt{2}}=\frac{3}{(2 \sqrt{5}-3 \sqrt{2})} \times \frac{(2 \sqrt{5}+3 \sqrt{2})}{(2 \sqrt{5}+3 \sqrt{2})}$
...[Multiplying the numerator and denominator by $(2 \sqrt{5}+3 \sqrt{2})]$
$\begin{aligned}
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{(2 \sqrt{5})^2-(3 \sqrt{2})^2} \\
& \ldots\left[\because(a-b)(a+b)=a^2-b^2\right] \\
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{(4 \times 5)-(9 \times 2)} \\
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{20-18} \\
\therefore \quad \frac{3}{2 \sqrt{5}-3 \sqrt{2}} & =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{2}
\end{aligned}$
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Question 163 Marks
Rationalize the denominator : $\frac{1}{\sqrt{7}+\sqrt{2}}$
Answer
$\quad \frac{1}{\sqrt{7}+\sqrt{2}}=\frac{1}{(\sqrt{7}+\sqrt{2})} \times \frac{(\sqrt{7}-\sqrt{2})}{(\sqrt{7}-\sqrt{2})}$
...[Multiplying the numerator and
denominator by $(\sqrt{7}-\sqrt{2})]$
$\begin{aligned}
= & \frac{\sqrt{7}-\sqrt{2}}{(\sqrt{7})^2-(\sqrt{2})^2} \\
& \ldots\left[\because(a-b)(a+b)=a^2-b^2\right] \\
= & \frac{\sqrt{7}-\sqrt{2}}{7-2} \\
\therefore \quad \frac{1}{\sqrt{7}+\sqrt{2}} & =\frac{\sqrt{7}-\sqrt{2}}{5}
\end{aligned}$
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Question 173 Marks
Write the following rational numbers in $\frac{p}{q}$ form. : $2 . \overline{514}$
Answer
$\text { Let } x =2 . \overline{514}$
$ x = 2.514514…$
Since, three numbers i.e. $5, 1$ and $4$ are repeating after the decimal point.
Thus, multiplying both sides by $1000,$
$1000x = 2514.514514…$
$1000 x =2514 . \overline{514} \text {...(ii) }$
$\text { Subtracting (i) from (ii), }$
$1000 x - x =2514 . \overline{514}-2 . \overline{514}$
$\therefore 999 x =2512$
$\therefore \quad x=\frac{2512}{999}$
$\therefore  2 . \overline{514}=\frac{2512}{999}$
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Question 183 Marks
Write the following rational numbers in $\frac{p}{q}$ form. : $15 . \overline{89}$
Answer
Let $x=15 . \overline{89} \ldots \ldots (i)$
$\therefore x = 15.8989…$
Since, two numbers i.e. $8$ and $9$ are repeating after the decimal point.
Thus, multiplying both sides by $100,$
$100x= 1589.8989…$
$\therefore 100 x =1589 . \overline{89} \ldots \text {.(ii) }$
$\text { Subtracting (i) from (ii), }$
$100 x - x =1589 . \overline{89}-15 . \overline{89}$
$\therefore 99 x =1574$
$\therefore \quad x=\frac{1574}{99}$
$\therefore \quad 15 . \overline{89}=\frac{1574}{99}$
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Question 193 Marks
Write the following rational numbers in $\frac{p}{q}$ form. : $\quad 3 . \overline{17}$
Answer
Letx $=3 . \overline{17}$...(i)
$\therefore x = 3.1717…$
Since, two numbers i.e. $1$ and $7$ are repeating after the decimal point.
Thus, multiplying both sides by $100,$
$100x = 317.1717…$
$\therefore 100x= 317.17 …(ii)$
Subtracting (i) from (ii),
$100 x - x =317 . \overline{17}-3 . \overline{17}$
$\therefore 99 x =314$
$\therefore \quad x=\frac{314}{99}$
$\therefore \quad 3 . \overline{17}=\frac{314}{99} $
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Question 203 Marks
Write the following rational numbers in $\frac{p}{q}$ form. : $\quad 0 . \overline{37}$
Answer
Let $x=0 . \overline{37}$
$\therefore x = 0.3737…$
Since, two numbers i.e. $3$ and $7$ are repeating after the decimal point.
Thus, multiplying both sides by $100,$
$100x = 37.3737……$
$\therefore 100 x =37 . \overline{37} \ldots \ldots \text {..(ii) }$
Subtracting (i) from (ii),
$100 x-x=37 . \overline{37}-0 . \overline{37}$
$\therefore 99x = 37$
$\therefore \quad x=\frac{37}{99}$
$\therefore \quad 0 . \overline{37}=\frac{37}{99}$
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Question 213 Marks
Write the following rational numbers in $\frac{p}{q}$ form. :$ 0 . \overline{6}$
Answer
Let $x=0.6$...(i)
$\therefore x = 0.666…$
Since, one number i.e. $6$ is repeating after the decimal point.
Thus, multiplying both sides by $10,$
$10x = 6.666…$
$\therefore 10 x 6.6 …(ii)$
Subtracting (i) from (ii),
$10x – x = 6.6 – 0.6$
$\therefore 9x = 6$
$ \therefore \quad x=\frac{6}{9}=\frac{3 \times 2}{3 \times 3}$
$\therefore \quad x=\frac{2}{3}$
$\therefore \quad 0 . \dot{6}=\frac{2}{3} $
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