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MATHEMATICS [4 marks sum]

Indices · ICSE STD 9 (ICSE - English Medium). 17 questions.

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[4 marks sum]

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17 questions · timed · auto-graded

Question 14 Marks
Solve for $x:5^{x2} : 5^x= 25 : 1$
Answer
$5^{x 2}: 5^x=25: 1 $
$ \Rightarrow \frac{5 x^2}{5 x}=\frac{25}{1} $
$ \Rightarrow \frac{5 x^2}{5 x}=\frac{5^2}{1} $
$ \Rightarrow 5^{x 2}=5^2 \times 5^x $
$ \Rightarrow 5^{x 2}=5^{2+x} $
$ \Rightarrow x^2=2+x $
$ \Rightarrow x^2-x-2=0$
$\Rightarrow(x-2)(x+1)=0 $
$ \Rightarrow x-2=0$  or $x+1=0 $
$ \Rightarrow x=2$ or $x=-1 .$
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Question 24 Marks
Solve for $x:9 \times 3^x=(27)^{2 x-5}$
Answer
$9 \times 3^x=(27)^{2 x-5} $
$ \Rightarrow 3^2 \times 3^x=\left(3^3\right)^{2 x-5} $
$ \Rightarrow 3^2 \times 3^x=3^{3 x \times^{(2 x-5)}}$
$ \Rightarrow 3^{2+x}=3^{6 x-15}$
$ \Rightarrow 1=\frac{3^{6 x-15}}{3^{2+x}} $
$ \Rightarrow 1=3^{6 x-15-2-x} $
$ \Rightarrow 3^0=3^{5 x-17}$
$ \Rightarrow 5 \times 17=0$
$ \Rightarrow x=\frac{17}{5} .$
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Question 34 Marks
Simplify the following:$(81)^{\frac{3}{4}}-\left(\frac{1}{32}\right)^{-\frac{2}{5}}+8^{\frac{1}{3}} \cdot\left(\frac{1}{2}\right)^{-1} \cdot 2^0$
Answer
$(81)^{\frac{3}{4}}-\left(\frac{1}{32}\right)^{-\frac{2}{5}}+8^{\frac{1}{3}} \cdot\left(\frac{1}{2}\right)^{-1} \cdot 2^0$
$ =\left(3^4\right)^{\frac{3}{4}}-\left(\frac{1}{2^5}\right)^{\frac{-2}{6}}+\left(2^3\right)^{\frac{1}{3}} \cdot\left(\frac{1}{2}\right)^{-1} \times 1 \ldots \ldots($ Using  $a^0=1)$
$=3^{4 \times \frac{3}{4}}-\frac{1}{2^{5 \times\left(-\frac{2}{5}\right)}}+2^{3 \times \frac{1}{3}} \cdot(2)^1 \ldots \ldots($ Using $\left(a^{ m }\right)^n= a ^{ mn }) $
$ =3^3-\frac{1}{2^{-2}}+2^1 \cdot(2)^1 $
$ =3^3-2^2+2^{1+1} \ldots($ Using  $a ^{ m } \times a ^{ n }= a ^{ m + n }$  and  $a ^{- n }=\frac{1}{ a ^{ n }}) $
$ =3^3-2^2+2^2 $
$ =27 .$
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Question 44 Marks
Evaluate the following:$\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}$
Answer
$\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}} $
$ =\left(\frac{1}{2^2}\right)^{\frac{1}{2}}+(0.1)^{-1}-3^2 $
$ =\left(\frac{1}{2}\right)+(0.1)^{-1}-3^2 $
$=\frac{1}{2}+\frac{1}{0.1}-9$
$ =\frac{1}{2}+\frac{10}{1}-9$
$ =\frac{1}{2}+1$
$ =\frac{3}{2} .$
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Question 54 Marks
Evaluate the following:$(27)^{\frac{2}{3}} \times 8^{\frac{-1}{6}} \div 18^{\frac{-1}{2}}$
Answer
$(27)^{\frac{2}{3}} \times 8^{\frac{-1}{6}} \div 18^{\frac{-1}{2}} $
$ =3^{3 \times \frac{2}{3}} \times \frac{1}{2^{3 \times \frac{1}{6}}} \div\left(\frac{1}{18}\right)^{\frac{1}{2}}$
$ =\frac{3^2}{2^{\frac{1}{2}}} \times\left(2 \times 3^2\right)^{\frac{1}{2}}$
$ =\frac{3^2}{2^{\frac{1}{2}}} \times 2^{\frac{1}{2}} \times 3$
$=3^{2+1} $
$ =3^3 $
$=27 .$
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Question 64 Marks
Evaluate the following:$\left(\frac{8}{27}\right)^{\frac{-2}{3}}-\left(\frac{1}{3}\right)^{-2}-7^0$
Answer
$\left(\frac{8}{27}\right)^{\frac{2}{3}}-\left(\frac{1}{3}\right)^{-2}-7^0$
$=\left(\frac{27}{8}\right)^{\frac{2}{3}}-(3)^2-1 $
$ =\left(\frac{3}{2}\right)^{3 \times \frac{2}{3}}-9-1$
$ =\left(\frac{3}{2}\right)^2-10$
$=\frac{9}{4}-10 $
$ =\frac{9-40}{4} $
$=\frac{-31}{4} .$
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Question 74 Marks
Evaluate the following:$\frac{2^6 \times 5^{-4} \times 3^{-3} \times 4^2}{8^3 \times 15^{-3} \times 25^{-1}}$
Answer
$\frac{2^6 \times 5^{-4} \times 3^{-3} \times 4^2}{8^3 \times 15^{-3} \times 25^{-1}}$
$ =\frac{2^6 \times\left(2^2\right)^2 \times(3 \times 5)^3 \times\left(5^2\right)^1}{\left(2^3\right)^3 \times 5^4 \times 3^3}$
$ =\frac{2^{6+4} \times 3^3 \times 5^{3+2}}{2^9 \times 3^3 \times 5^4} $
$=2^{10-9} \times 5^{5-4} $
$ =2 \times 5$
$ =10 .$
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Question 84 Marks
Evaluate the following:$\frac{12^2 \times 75^{-2} \times 35 \times 400}{48^2 \times 15^{-3} \times 525}$
Answer
$\frac{12^2 \times 75^{-2} \times 35 \times 400}{48^2 \times 15^{-3} \times 525}$
$=\frac{\left(2^2 \times 3\right)^2 \times(7 \times 5) \times\left(2^4 \times 5^2\right) \times(3 \times 5)^3}{\left(2^4 \times 3^2\right) \times\left(3 \times 5^2 \times 7\right) \times\left(3 \times 5^2\right)^2} $
$ =\frac{2^4 \times 3^2 \times 7 \times 5 \times 2^4 \times 5^2 \times 3^3 \times 5^3}{2^8 \times 3^2 \times 3 \times 5^2 \times 7 \times 3^2 \times 5^4}$
$=\frac{2^{4+4} \times 3^{2+3} \times 5^{1+2+3} \times 7}{2^8 \times 3^{2+1+2} \times 5^{4+2} \times 7} $
$ =\frac{2^8 \times 3^5 \times 5^6 \times 7}{2^8 \times 3^5 \times 5^6 \times 7} $
$ =1 .$
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Question 94 Marks
Evaluate the following:$\frac{2^3 \times 3^5 \times 24^2}{12^2 \times 18^3 \times 27}$
Answer
$\frac{2^3 \times 3^5 \times 24^2}{12^2 \times 18^3 \times 27} $
$=\frac{2^3 \times 3^5 \times\left(2^3 \times 3\right)^2}{\left(2^2 \times 3^2\right)^2 \times\left(2 \times 3^2\right)^3 \times\left(3^3\right)} $
$ =\frac{2^3 \times 3^5 \times 2^6 \times 3^2}{2^4 \times 3^2 \times 2^3 \times 3^6 \times 3^3} $
$=\frac{2^9 \times 3^7}{2^7 \times 3^{11}} $
$=\frac{2^{9-7}}{3^{11-7}} $
$=\frac{2^2}{3^4} $
$=\frac{4}{81} .$
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Question 104 Marks
Prove the following:$\left(x^a\right)^{b-c} x\left(x^b\right)^{c-a} \times\left(x^c\right)^{a-b}=1$
Answer
$\text { L. H.S. }$
$=\left(x^a\right)^{b-c} \times\left(x^b\right)^{c-a} \times\left(x^c\right)^{a-b}$
$ =x^{a(b-c)} \times x^{b(c-a)} \times x^{c(a-b)} \ldots($ Using $\left(a^m\right)^n=a^{m n})$
$=x^{a b-a c} \times x^{b c-a b} \times x^{a c-b c}$
$=x^{a b-a c+b c-a b+a c-b c} \ldots($Using  $a^m \times a^n=a^{m+n})$
$ =x^a$
$ =1$
$ =\text { R.H.S }$
$=$ Hence proved.
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Question 114 Marks
Prove the following:$\sqrt{x^{-1} y} \cdot \sqrt{y^{-1} z} \cdot \sqrt{z^{-1} x}=1$
Answer
$\text { L.H.S. } $
$=\sqrt{x^{-1} y} \cdot \sqrt{y^{-1} z} \cdot \sqrt{z^{-1} x} $
$=\sqrt{\frac{y}{x}} \cdot \sqrt{\frac{z}{y}} \cdot \sqrt{\frac{x}{z}} $
$=\sqrt{\left(\frac{y}{x}\right)\left(\frac{z}{y}\right)\left(\frac{x}{z}\right)} $
$=\sqrt{x^{1-1} \cdot y^{1-1} \cdot z^{1-1}} $
$=\sqrt{x^* \cdot y \cdot z^*} $
$=\sqrt{1 \cdot 1 \cdot 1} $
$=1 $
$=\text { R.H.S. }$
Hence proved.
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Question 124 Marks
Evaluate the following:$7^{-4} \times(343)^{\frac{2}{3}} \div(49)^{-\frac{1}{2}}$
Answer
$7^{-4} \times(343)^{\frac{2}{3}} \div(49)^{-\frac{1}{2}} $
$=7^{-4} \times\left(7^3\right)^{\frac{2}{3}} \div\left(7^2\right)^{-\frac{1}{2}} $
$ =7^{-4} \times 7^{3 \times \frac{2}{3}} \div 7^{2 \times\left(-\frac{1}{2}\right)} $
$=7^{-4} \times 7^2 \div 7^{-1} $
$=7^{-4+2-(-1)} \quad \ldots($Using  $a^m \times a^n=a^{m+n}$ and $a^m \div a^n=a^{m-n})$
$=7^{-4+2+1}$
$=7^{-1} $
$=\frac{1}{7} \quad \cdots($ Using $a^{- m }=\frac{1}{a^m}).$
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Question 134 Marks
Evaluate the following:$9^4 \div 27^{-\frac{2}{3}}$
Answer
$9^4 \div 27^{-\frac{2}{3}} $
$=\left[(3)^2\right]^4 \div\left[(3)^3\right]^{-\frac{2}{3}}$
$ =(3)^{2 \times 4} \div(3)^{3 \times}\left(-\frac{2}{3}\right) \quad \ldots \ldots($ Using $\left(a^m\right)^n=a^{m n}) $
$=(3)^8 \div(3)-2$ 
$=(3)^{8-(-2)} \ldots . .($ Using  $a^m \div a^n=a^{m-n}) $
$ =(3)^{8+2} $
$ =3^{10}$
$ =(3)^{2 \times 5} $
$ =\left[(3)^2\right]^5$
$=[9]^5 $
$=59049 .$
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Question 144 Marks
If $x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0$, prove that $(x+y+z)^3=27\text{xyz}$
Answer
$x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0$
$\Rightarrow\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)+z^{\frac{1}{3}}=0$ cubing both sides, we get:   
$ \Rightarrow\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)^3+z+3\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right) z^{\frac{1}{3}}\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}\right)=0 $
$ \Rightarrow x+y+3 x^{\frac{1}{3}} y^{\frac{1}{3}}\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)+z+0=0 $
$ \Rightarrow x+y+3 x^{\frac{1}{3}} y^{\frac{1}{3}}\left(-z^{\frac{1}{3}}\right)+z=0 \quad \ldots($Using the given condition again$)$
$ \Rightarrow x + y + z =3 x^{\frac{1}{3}} y^{\frac{1}{3}} z^{\frac{1}{3}} $
$\Rightarrow( x + y + z )^3=27\text{xyz} .$
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Question 154 Marks
If $x=3^{\frac{2}{3}}+3^{\frac{1}{3}}$, prove that $x^3-9 x-12=0$
Answer
$x=3^{\frac{2}{3}}+3^{\frac{1}{3}}$
$\Rightarrow x^3=3^2+3+3 \times 3^{\frac{2}{3}} \times 3^{\frac{1}{3}}\left(3^{\frac{2}{3}}+3^{\frac{1}{3}}\right)$
$ \Rightarrow x^3=9+3+3 \times 3^{\frac{2}{3}+\frac{1}{3}}(x)$
$ \Rightarrow x^3=12+9 x$
$ \Rightarrow x^3-9 x-12=0 .$
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Question 164 Marks
If $a=2^{\frac{1}{3}}-2^{\frac{-1}{3}}$, prove that $2 a^3+6 a=3$
Answer
$a=2^{\frac{1}{3}}-2^{\frac{1}{3}} $
$ \Rightarrow a=2^{\frac{1}{3}}-\frac{1}{2^{\frac{1}{3}}} $
$ \Rightarrow a^3=\left(2^{\frac{1}{3}}-\frac{1}{2^{\frac{1}{3}}}\right)^3 $
$ =2-\frac{1}{2}-3\left(2^{\frac{1}{3}}-\frac{1}{2^{\frac{1}{3}}}\right) $
$ \Rightarrow a^3=\frac{4-1}{2}-3 a $
$\Rightarrow a^3=\frac{3}{2}-3 a $
$\Rightarrow 2 a^3+6 a=3 .$
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Question 174 Marks
Find the value of $k$ in each of the following:$\left(\frac{1}{3}\right)^{-4} \div 9^{\frac{-1}{3}}=3^k$
Answer
$\left(\frac{1}{3}\right)^{-4} \div 9^{\frac{-1}{3}}=3^k$
$ \Rightarrow\left(3^{-1}\right)^{-4} \div\left(3^2\right)^{\frac{-1}{2}}=3^k $
$\Rightarrow 3^4 \div 3^{\frac{-2}{3}}=3^k $
$ \Rightarrow 3^{4+\frac{2}{3}}=3^k$
$ \Rightarrow 3^{\frac{14}{3}}=3^k $
$ \Rightarrow k=\frac{14}{3} .$
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[4 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip