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Maths 2 Marks Questions

Number Systems · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 18 questions.

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2 Marks Questions

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

Question 12 Marks
Simplify the following: $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$
Answer
$\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}=\frac{\sqrt{4\times5}}{8}+\frac{\sqrt{9\times6}}{9}$ $\frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}=\frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}$ $=\sqrt{6}\Big(\frac{1}{4}+\frac{1}{3}\Big)$ $=\sqrt{6}\Big(\frac{3+4}{12}\Big)=\frac{7\sqrt{6}}{12}$
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Question 22 Marks
Simplify: $\frac{9^{\frac{1}{3}}\times27^{-\frac{1}2{}}}{3^\frac{1}{6}\times3^{-\frac{2}{3}}}$
Answer
$\frac{9^{\frac{1}{3}}\times27^{-\frac{1}2{}}}{3^\frac{1}{6}\times3^{-\frac{2}{3}}}=\frac{{3^3}^{\frac{1}{3}}\times{3^3}^{-\frac{1}2{}}}{3^\frac{1}{6}\times3^{-\frac{2}{3}}}$ $[\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$ $=\frac{{3}^{\frac{2}{3}}\times{3}^{-\frac{3}{2}}}{3^\frac{1}{6}\times3^{-\frac{2}{3}}}=\frac{3^{\frac{2}{3}-\frac{3}{2}}}{3^{\frac{1}{6}-\frac{2}{3}}}$ $[\because\text{a}^\text{m}\times\text{a}^\text{n}=\text{a}^\text{m+n}]$ $\frac{3^{\frac{4-9}{6}}}{3^{\frac{1-4}{6}}}=\frac{3^{-\frac{5}{6}}}{3^{-\frac{3}{6}}}=3^{-\frac{5}{6}+\frac{3}{6}}=3^{-\frac{2}{6}}=3^{-\frac{1}{3}}$ $\Big[\because\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}\Big]$
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Question 32 Marks
Simplify: $\frac{8^{\frac{1}{3}}\times16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$
Answer
$\frac{8^{\frac{1}{3}}\times16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}=\frac{(2^3)^{\frac{1}{3}}\times(2^4)^{\frac{1}{3}}}{(2^5)^{-\frac{1}{3}}}=\frac{2^{3\times\frac{1}{3}}\times2^{4\times\frac{1}{3}}}{2^{5\times-\frac{1}{3}}}$ $[\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$ $\frac{2^{\frac{3}{3}+\frac{4}{3}}}{2^{-\frac{5}{3}}}=\frac{2^{\frac{7}{3}}}{2^{-\frac{5}{3}}}=2^{\frac{7}{3}+\frac{5}{3}}=2^{\frac{12}{3}}=2^4=16$ $\Big[\because\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}\Big]$
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Question 42 Marks
Simplify:
$\frac{3}{5}^4\frac{8}{5}^{-12}\frac{32}{5}^6$
Answer
$\frac{3}{5}^4\frac{8}{5}^{-12}\frac{32}{5}^6=\frac{3^4}{5^4}\times\Big(\frac{5}{2^3}\Big)^{12}\times\Big(\frac{2^5}{5}\Big)^6\Big(\because\text{a}^{-1}=\frac{1}{\text{a}}\Big)$
$=\frac{3^4}{5^4}\times\frac{5^{12}}{2^{36}}\times\frac{2^{30}}{5^6}$ $[\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$
$=\frac{3^4\times5^{12-4-6}}{2^{36-30}}$ $[\because\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^\text{m}-\text{n}]$
$=\frac{3^4}{2^6}\times5^2=\frac{81\times25}{64}=\frac{2025}{64}$
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Question 52 Marks
Simplify: $(256)^{-\Big(4^{-\frac{3}{2}}\Big)}$
Answer
$(256)^{-\Big(4^{-\frac{3}{2}}\Big)}=(256)^{-(4)^{-\frac{3}{2}}}=(256)^{-(2^2)^{-\frac{3}{2}}}$ $=(256)^{-\Big(2^{2\times-\frac{3}{2}}\Big)}=(256)^{-(2^{-3})}$ $\Big[\because\ \text{(b)}^{(\text{a}^\text{m})^\text{n}}=\text{b}^{\text{a}^\text{mn}}\Big]$ $=(2^8)^{-\Big(\frac{1}{2^3}\Big)}=(2^8)^{-\frac{1}{8}}$ $=2^{8\times-\frac{1}{8}}=2^{-1}=\frac{1}{2}$
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Question 62 Marks
Simplify the following: $3\sqrt{3}+2\sqrt{27}+\frac{7}{3}$
Answer
$3\sqrt{3}+2\sqrt{27}+\frac{7}{3}$ $=3\sqrt{3}+2\sqrt{3\times3\times3}+\frac{7}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$ $=3\sqrt{3}+6\sqrt{3}+\frac{7\sqrt{3}}{3}$ $=(3+6+\frac{7}{3})\sqrt{3}$ $=\frac{34}{3}\sqrt{3}$
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Question 72 Marks
Simplify: $64^{-\frac{1}{3}}64^{\frac{1}{3}}-64^{\frac{2}{3}}$
Answer
$64^{-\frac{1}{3}}\Big[64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big]=(4^3)^{-\frac{1}{3}}\Big[(4^3)^{\frac{1}{3}}-(4^3)^{\frac{2}{3}}\Big]$ $4^{3\times-\frac{1}{3}}\Big(4^{3\times\frac{1}{3}}-4^{3\times\frac{1}{3}}\Big)=4^{-1}(4-4^2)$ $[\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$ $=\frac{1}{4}(4-16)=-\frac{12}{4}=-3$
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Question 82 Marks
Rationalise the denominator of the following: $\frac{3+\sqrt{2}}{4\sqrt{2}}$
Answer
Let $\text{E}=\frac{3+\sqrt{2}}{4\sqrt{2}}$ For rationalising the denominator, multiplying numerator and denominator by $\sqrt{2}$ $\text{E}=\frac{3+\sqrt{2}}{4\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}+(\sqrt{2})^2}{4(\sqrt{2})^2}$ $=\frac{3\sqrt{2}+2}{4\times2}=\frac{3\sqrt{2}+2}{8}$
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Question 92 Marks
Rationalise the denominator of the following: $\frac{2}{3\sqrt{3}}$
Answer
Let $\text{E}=\frac{2}{3\sqrt{3}}$ For rationalising the denominator, multiplying numerator and denominator by $\sqrt{3}$ $\text{E}=\frac{2}{3\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$ $=\frac{2\sqrt{3}}{3\times3}\times\frac{2\sqrt{3}}{9}$
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Question 102 Marks
Express the following in the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0$: $5.\overline2$
Answer
Let $x =5 . \overline{2}=5.222 \ldots (i)$ on multiplying both sides of Eq. $(i)$ by $10$ , we get $10 x=52.222 \ldots (ii)$ on subtraking Eq. $(i)$ from Eq. $(ii),$
we get $10 x-x=(52.222 \ldots)-(5.222 \ldots)$
$\Rightarrow 9 x=47 \therefore x=\frac{47}{9}$
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Question 122 Marks
Express the following in the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0: 0.888...$
Answer
Let $x = 0.888... (i)$ on multiplying both sides of Eq. $(i)$
by $10$, we get $10x = 0.888... (ii)$
on subtraking Eq. $(i)$ from Eq. $(ii),$
we get $10x - x = (8.88) - (0.888)$
$\Rightarrow 9x = 8 \therefore\text{x}=\frac{8}{9}$
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Question 132 Marks
If $\text{a}=2+\sqrt{3},$ then find the value of $\text{a}-\frac{1}{\text{a}}.$
Answer
We have $\text{a}=2+\sqrt{3},$
$\therefore\ \frac{1}{\text{a}}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}$
$=\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}$
$\therefore\ \text{a}-\frac{1}{\text{a}}=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}$
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Question 142 Marks
Simplify the following: $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$
Answer
$\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$ $=\frac{3}{\sqrt{4\times2}}+\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\Big(\frac{3}{2}+1\Big)$ $=\frac{5}{2\sqrt{2}}=\frac{5}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{4}$
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Question 152 Marks
Simplify the following: $\sqrt[4]{81}–8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$
Answer
$\sqrt[4]{81}–8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$ $=\sqrt[4]{3}^4–8\sqrt[3]{6^3}+15\sqrt[5]{2^5}+\sqrt{225}$ $=3-(8\times6)+(15\times2)+15$ $=3-48+30+15=0$
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Question 162 Marks
Simplify the following: $\sqrt[4]{12}\times\sqrt[7]{6}$
Answer
$\sqrt[4]{12}\times\sqrt[7]{6}$ $=\sqrt[4]{2\times2\times3}\times7\sqrt{2\times3}$ $=8\sqrt{3}\times7\sqrt{2}\times\sqrt{3}$ $=24\times7\sqrt{2}=168\sqrt{2}$
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Question 172 Marks
Express the following in the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0$: $0.\overline{001}$
Answer
Let $x =0 . \overline{001}=0.001001$ $(i)$ on multiplying both sides of Eq. $(i)$ by $1000$ , we get $1000 x =001.001 \ldots$ $(ii)$ on subtraking Eq. $(i)$ from Eq. $(ii)$, we get $1000 x-x=001.001-(0.01001 \ldots) \Rightarrow 999 x=001 \therefore x=\frac{1}{999}$
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Question 182 Marks
Rationalise the denominator of the following: $\frac{\sqrt{40}}{\sqrt{3}}$
Answer
Let $\text{E}=\frac{\sqrt{40}}{\sqrt{3}}$ For rationalising the denominator, multiplying numerator and denominator by $\sqrt{3}$ $\text{E}=\frac{\sqrt{40}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{40\times3}}{(\sqrt{3})^2}=\frac{\sqrt{120}}{3 }$ $=\frac{\sqrt{2\times2\times2\times5\times3}}{3}=\frac{2}{3}\sqrt{30}$
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