[5 marks sum]

Question 15 Marks

Solve $x$ and $y$ if $:(\sqrt{32 } )^x \div 2^{y+1}=1$ and $8^y-16^{4-x / 2}=0$

Answer

Consider the quation
$(\sqrt{32})^x \div 2^{y+1}=1$
$ \Rightarrow(\sqrt{2 \times 2 \times 2 \times 2 \times 2})^x \div 2^{y+1}=1$
$ \Rightarrow\left(\sqrt{2^5}\right)^x \div 2^{y+1}=1$
$ \Rightarrow\left[\left(2^5\right)^{\frac{1}{2}}\right]^x \div 2^{y+1}=x^0$
$ \Rightarrow 2^{5 \frac{x}{2}} \div 2^{y+1}=x^0$
$ \Rightarrow \frac{5 x}{2}-(y+1)=0$
$ \Rightarrow 5 \mathrm{x}-2(y+1)=0$
$\Rightarrow 5 x-2 y-2=0\ldots(1)$
Now consider the other equation
$8^y-16^{4-\frac{x}{2}}=0$
$ \Rightarrow\left(2^3\right)^y-\left(2^4\right)^{4-\frac{x}{2}}=0$
$ \Rightarrow 2^{3 y}-2^{4\left(4-\frac{x}{2}\right)}=0$
$ \Rightarrow 2^{3 y}=2^{4\left(4-\frac{x}{2}\right)}$
$ \Rightarrow 3 y=4\left(4-\frac{x}{2}\right)$
$ \Rightarrow 3 y=16-2 x$
$\Rightarrow 2 x+3 y=16\ldots(2)$
Thus, We have two equations,
$5 x-2 y=2\ldots(1)$
$2 x+3 y=16\ldots(2)$
Multiplying equation $(1)$ by $3$ and $(2)$ by $2 ,$ We have
$15 x-6 y=6\ldots .(3)$
$4 x+6 y=32\ldots .(4)$
Adding equation $(3)$ and $(4)$, We have
$19 x=38$
$ \Rightarrow x=2$
Substituting the value of $x$ in equation $(1),$ We have
$5(2)-2 y=2$
$ \Rightarrow 10-2 y=2$
$ \Rightarrow 2 y=10-2$
$ \Rightarrow 2 y=8$
$ \Rightarrow y=\frac{8}{2}$
$ \Rightarrow y=4$
Thus the values of $x$ and $y$ are : $x=2$ and $y=4$.

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Question 25 Marks

Find the values of $m$ and $n$ if :$4^{2 m}=(\sqrt[3]{16})^{-\frac{6}{n}}=(\sqrt{8})^2$

Answer

$4^{2 \mathrm{\sim m}}=(\sqrt[3]{16})^{-\frac{6}{n}}=(\sqrt{8})^2$
$\Rightarrow 4^{2 \mathrm{\sim m}}=(\sqrt{8})^2\ldots .(1)$
and
$(\sqrt[3]{16})^{-\frac{6}{n}}=(\sqrt{8})^2\ldots .(2)$
From $(1)$
$ 4^{2 \mathrm{\sim m}}=(\sqrt{8})^2$
$ \Rightarrow\left(2^2\right)^{2 \mathrm{\sim m}}=\left(\sqrt{2^3}\right)^2$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=\left[\left(2^3\right)^{\frac{1}{2}}\right]^2$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=\left[2^{3 \times \frac{1}{2}}\right]^2$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=2^{3 \times \frac{1}{2} \times 2}$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=2^3$
$ \Rightarrow 4 \mathrm{\sim m}=3$
$ \Rightarrow \mathrm{m}=\frac{3}{4}$
From $(2),$ We have
$(3 \sqrt{16})^{-\frac{6}{n}}=(\sqrt{8})^2$
$ \Rightarrow(\sqrt[3]{2 \times 2 \times 2 \times 2})^{-\frac{6}{n}}=(\sqrt{2 \times 2 \times 2})^2$
$ \Rightarrow\left(\sqrt[3]{2^4}\right)^{-\frac{6}{n}}=\left(\sqrt{2^3}\right)^2$
$ \Rightarrow\left[\left(2^4\right)^{\frac{1}{3}}\right]^{-\frac{6}{n}}=\left[\left(2^3\right)^{\frac{1}{2}}\right]^2$
$ \Rightarrow\left[2^{\frac{4}{3}}\right]^{-\frac{6}{n}}=\left[2^{\frac{3}{2}}\right]^2$
$ \Rightarrow 2^{\frac{4}{3} \times\left(-\frac{6}{n}\right)=2^{\frac{3}{2} \times 2}}$
$ \Rightarrow 2^{-\frac{8}{n}}=2^3$
$ \Rightarrow-\frac{8}{\mathrm{n}}=3$
$ \Rightarrow \mathrm{n}=-\frac{8}{3}$
Thus $\mathrm{m}=\frac{3}{4} \mathrm{n}=-\frac{8}{3}$

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Question 35 Marks

Solve : $8 \times 2^{2 x}+4 \times 2^{x+1}=1+2^x$

Answer

$8 \times 2^{2 x}+4 \times 2^{x+1}=1+2^x$
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1=1+2^x$
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1-1-2^x=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+2^x \times(8-1)-1=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+7\left(2^x\right)-1=0$
$\Rightarrow 8 y^2+7 y-1=0$
$\left[y=2^x\right]$
$ \Rightarrow 8 y^2+8 y-y-1=0$
$ \Rightarrow 8 y(y+1)-1(y+1)=0$
$ \Rightarrow(8 y-1)(y+1)=0$
$ \Rightarrow 8 y=1$ or $y=-1$
$ \Rightarrow y=\frac{1}{8}$ or $y=-1$
$ \Rightarrow 2^x=\frac{1}{8}$ or $2^x=-1$
$ \Rightarrow 2^x=\frac{1}{2^3}$ or $2^x=-1$
$ \Rightarrow 2^x=2^{-3}$ or $2^x=-1$
$ \Rightarrow x=-3$
$[\because 2^x=-1$ is not possible$]$

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Question 45 Marks

Simplify :$\frac{3 \times 27^{n+1}+9 \times 3^{3 n-1}}{8 \times 3^{3 n}-5 \times 27^n}$

Answer

$ \frac{3 \times 27^{n+1}+9 \times 3^{3 n-1}}{8 \times 3^{3 n}-5 \times 27^n}$
$ =\frac{3 \times(3 \times 3 \times 3)^{n+1}+3 \times 3 \times 3^{3 n-1}}{2 \times 2 \times 2 \times 3^{3 n}-5 \times(3 \times 3 \times 3)^n}$
$ =\frac{3 \times\left(3^3\right)^{n+1}+3^2 \times 3^{3 n-1}}{2^3 \times 3^{3 n}-5 \times\left(3^3\right)^n}$
$ =\frac{3 \times 3^{3 n+3}+3^{3 n+1}}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n+3+1}+3^{3 n+1}}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n+4}+3^{3 n+1}}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n} \times 3^4+3^{3 n} \times 3^1}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n}\left(3^4+3^1\right)}{\left(3^3\right)^n(8-5)}$
$ =\frac{3^{3 n}\left(3^4+3^1\right)}{3^{3 n} \times 3}$
$ =\frac{3 \times 3 \times 3 \times 3+3}{3}$
$ =\frac{81+3}{3}$
$ =\frac{84}{3}$
$ =28$

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Question 55 Marks

If $2160=2^a \cdot 3^b \cdot 5^c$, find $a, b$ and $c$. Hence calculate the value of $3^a \times 2^{-b} \times 5^{-c}$.

Answer

$2160=2^a \times 3^b \times 5^c$
$ \Rightarrow 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5=2^a \times 3^b \times 5^c$
$ \Rightarrow 2^4 \times 3^3 \times 5^1=2^a \times 3^b \times 5^c$
$ \Rightarrow 2^a \times 3^b \times 5^c=2^4 \times 3^3 \times 5^1$
Comparing powers of $2,3$ and $5$ on the both sides of equation, We have
$a=4 ; b=3$ and $c=1$
Hence,
Value of $3^a \times 2^{-b} \times 5^{-c}$
$=3^4 \times 2^{-3} \times 5^{-1}$
$ =3 \times 3 \times 3 \times 3 \times \frac{1}{2^3} \times \frac{1}{5}$
$=81 \times \frac{1}{2 \times 2 \times 2} \times \frac{1}{5}$
$=81 \times \frac{1}{8} \times \frac{1}{5}$
$=\frac{81}{40}$
$=2 \frac{1}{40}$

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Question 65 Marks

Simplify the following and express with positive index :$\left[1-\left\{1-(1-n)^{-1}\right\}^{-1}\right]^{-1}$

Answer

$ {\left[1-\left\{1-(1-n)^{-1}\right\}^{-1}\right]^{-1}}$
$ =\frac{1}{\left[1-\left\{1-(1-n)^{-1}\right\}^{-1}\right]^{+} 1}$
$ =\frac{1}{1-\frac{1}{1-(1-n)^{-1}}}$
$ =\frac{1}{1-\frac{1}{1-\frac{1}{1-\pi}}}$
$ =\frac{1}{1-\frac{\frac{1}{1(1-n)-1}}{1-n}}$
$ =\frac{1}{1-\frac{\frac{1}{-n}}{1-n}}$
$ =\frac{1}{1-\frac{1-n}{-n}}$
$ =\frac{1}{1+\frac{1-n}{n}}$
$ =\frac{1}{\frac{n+(1-n)}{n}}$
$ =\frac{\frac{1}{n+1-n}}{n}$
$ =\frac{n}{1}$
$ =\mathrm{n}$

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Question 75 Marks

Evaluate :$\left(\frac{27}{8}\right)^{\frac{2}{3}}-\left(\frac{1}{4}\right)^{-2}+5^0$

Answer

$ \left(\frac{27}{8}\right)^{\frac{2}{3}}-\left(\frac{1}{4}\right)^{-2}+5^0$
$ =\left(\frac{3 \times 3 \times 3}{2 \times 2 \times 2}\right)^{\frac{2}{3}}-\left(\frac{1 \times 1}{2 \times 2}\right)^{-2}+5^0$
$ =\left[\left(\frac{3}{2}\right)^3\right]^{\frac{2}{3}}-\left[\left(\frac{1}{2}\right)^2\right]^{-2}+1$
$ =\left(\frac{3}{2}\right)^{3 \times \frac{2}{3}}-\left(\frac{1}{2}\right)^{2 \times(-2)}+1$
$ =\left(\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^{-4}+1$
$ =\frac{3}{2} \times \frac{3}{2}-\frac{1}{\left(\frac{1}{2}\right)^4}+1$
$ =\frac{9}{4}-\frac{1}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}}+1$
$ =\frac{9}{4}-\frac{1}{\frac{1}{16}}+1$
$ =\frac{9}{4}-16+1$
$ =\frac{9-64+4}{4}$
$ =\frac{-51}{4}$

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Question 85 Marks

Evaluate :$\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times\left(\frac{49}{9}\right)^{\frac{3}{2}} \div\left(\frac{343}{216}\right)^{\frac{2}{3}}$

Answer

$ \left(\frac{16}{81}\right)^{-\frac{3}{4}} \times\left(\frac{49}{9}\right)^{\frac{3}{2}} \div\left(\frac{343}{216}\right)^{\frac{2}{3}}$
$ =\left(\frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3}\right)^{-\frac{3}{4}} \times\left(\frac{7 \times 7}{3 \times 3}\right)^{\frac{3}{2}} \div\left(\frac{7 \times 7 \times 7}{6 \times 6 \times 6}\right)^{\frac{2}{3}}$
$ =\left[\left(\frac{2}{3}\right)^4\right]^{-\frac{3}{4}} \times\left[\left(\frac{7}{3}\right)^2\right]^{\frac{3}{2}} \div\left[\left(\frac{7}{6}\right)^3\right]^{\frac{2}{3}}$
$ =\left(\frac{2}{3}\right)^{4 \times-\frac{3}{4}} \times\left(\frac{7}{3}\right)^{2 \times \frac{3}{2}} \div\left(\frac{7}{6}\right)^{3 \times \frac{2}{3}}$
$ =\left(\frac{2}{3}\right)^{-3} \times\left(\frac{7}{3}\right)^3 \div\left(\frac{7}{6}\right)^2$
$ =\frac{1}{\left(\frac{2}{3}\right)^3} \times\left(\frac{7}{3}\right)^3 \times \frac{1}{\left(\frac{7}{6}\right)^2}$
$ =\frac{1}{\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}} \times \frac{7}{3} \times \frac{7}{3} \times \frac{7}{3} \times \frac{1}{\frac{7}{6} \times \frac{7}{6}}$
$ =\frac{1 \times 3 \times 3 \times 3}{2 \times 2 \times 2} \times \frac{7}{3} \times \frac{7}{3} \times \frac{7}{3} \times \frac{1 \times 6 \times 6}{7 \times 7}$
$ =\frac{7 \times 3 \times 3}{2}$
$ =\frac{63}{2}$
$ =31.5$

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MATHEMATICS [5 marks sum]

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