3 Marks Question

Question 13 Marks

By what number should $(-3)^{-1} $ be multiplied so that the product blecomes $ 6^{-1}$?

Answer

Let the number be x
$\therefore(-3)^{-1}\times\text{x}=(6)^{-1}$
$\Rightarrow\frac{1}{-3}\times\text{x}=\frac{1}{6}$
$\Rightarrow\frac{1\times-1}{-3\times-1}\times\text{x}=\frac{1}{6}$
$\therefore\frac{-\text{x}}{3}=\frac{1}{6}$
On cross multiplying:
$-\text{x}\times6=1\times3$
$\Rightarrow-6\text{x}=3$
$\Rightarrow6\text{x}=-3$
$\therefore\text{x}=\frac{-3}{6}=\frac{-1}{2}$

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Question 23 Marks

Simplify: $\big(3^{-1}+6^{-1}\big)\div\Big(\frac{3}{4}\Big)^{-1}$

Answer

$\big(3^{-1}+6^{-1}\big)\div\Big(\frac{3}{4}\Big)^{-1}$ $=\Big(\frac{1}{3}+\frac{1}{6}\Big)\div\Big(\frac{4}{3}\Big)^1$ $=\bigg(\Big[\frac{1\times2}{3\times2}\Big]+\Big[\frac{1\times1}{6\times1}\Big]\bigg)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{2+1}{6}\Big)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{3}{6}\Big)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{1}{2}\Big)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{1}{2}\Big)\times\Big(\frac{3}{4}\Big)$ $=\Big(\frac{3}{8}\Big)$ $\therefore\big(3^{-1}+6^{-1}\big)\div\Big(\frac{3}{4}\Big)^{-1}=\frac{3}{8}$

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Question 33 Marks

Mass of earth is $\left(5.97 \times 10^{24}\right) kg$ and mass of moon is $\left(7.35 \times 10^{22}\right) kg$. What is the total mass of the two?

Answer

Mass of the earth $=\left(15.97 \times 10^{24}\right) kg$
and mass of the moon $=\left(7.35 \times 10^{22}\right) kg$
$=$ Total mass of the earth and moon
$=\left(5.97 \times 10^{24}\right)+\left(7.35 \times 10^{22}\right) kg$
$=\left(\frac{597}{100} \times 10^{24}+7.35 \times 10^{22}\right) kg$
$=\left(\frac{597}{10^{22}} \times 10^{24}+7.35 \times 10^{22}\right) kg$
$=\left(597 \times 10^{22}+7.35 \times 10^{22}\right) kg$
$=10^{22}(597+7.35) kg$
$=(604.35) 10^{22} kg$

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Question 43 Marks

If $5^{2 x+1} \div 25=125$, find the value of $x$.

Answer

$5^{2\text{x}+1}\div25=125$
$\Rightarrow5^{2\text{x}+1}\times\frac{1}{25}=125$
$\Rightarrow5^{2\text{x}+1}=125\times25$
$\Rightarrow5^{2\text{x}+1}=5^3\times5^2=5^{3+2}$
$\Rightarrow5^{2\text{x}+1}=5^5$
$\therefore2\text{x}+1=5$
$\Rightarrow2\text{x}=5-1=4$
$\Rightarrow2\text{x}=4$
$\Rightarrow\text{x}=\frac{4}{2}=2$
Hence, $\text{x}=2$

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Question 53 Marks

By what number should $(-6)^{-1}$ be multiplied so that the product becomes $9^{-1}$ ?

Answer

Let x be the required number which is multiplied
$\therefore(-6)^{-1}\times\text{x}=9^{-1}$
$\Rightarrow\Big(\frac{1}{-6}\Big)\times\text{x}=\frac{1}{9}$
$\text{x}=\frac{1}{9}\times\frac{-6}{1}$
$=\frac{-2}{3}$
Hence, required number $=\frac{-2}{3}$

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Question 63 Marks

Find the value of x for which $\Big(\frac{4}{9}\Big)^{4}\times\Big(\frac{4}{9}\Big)^{-7}=\Big(\frac{4}{9}\Big)^{2\text{x}-1}$

Answer

$\Big(\frac{4}{9}\Big)^{4}\times\Big(\frac{4}{9}\Big)^{-7}=\Big(\frac{4}{9}\Big)^{2\text{x}-1}$ $\Rightarrow\Big(\frac{4}{9}\Big)^{4-7}=\Big(\frac{4}{9}\Big)^{2\text{x}-1}$ $\Rightarrow\Big(\frac{4}{9}\Big)^{-3}=\Big(\frac{4}{9}\Big)^{2\text{x}-1}$ $\therefore 2\text{x}-1=-3$ $\Rightarrow2\text{x}=-3+1$ $\Rightarrow2\text{x}=-2$ $\Rightarrow\text{x}=\frac{-2}{2}$ $\Rightarrow\text{x}=-1$ Hence $\text{x}=-1$

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Question 73 Marks

Evaluate: $\Big(\frac{5}{9}\Big)^{-2}\times\Big(\frac{3}{5}\Big)^{-3}\times\Big(\frac{3}{5}\Big)^0$

Answer

$\Big(\frac{5}{9}\Big)^{-2}\times\Big(\frac{3}{5}\Big)^{-3}\times\Big(\frac{3}{5}\Big)^0$$=\Big(\frac{5}{9}\Big)^{-2}\times\Big(\frac{3}{5}\Big)^{-3+0}$
$=\Big(\frac{5}{9}\Big)^{-2}\times\Big(\frac{3}{5}\Big)^{-3}$
$=\Big(\frac{9}{5}\Big)^{2}\times\Big(\frac{5}{3}\Big)^{3}$
$=\frac{9^2}{5^2}\times\frac{5^3}{3^3}$
$=\frac{(3^2)^2}{5^2}\times\frac{5^3}{3^3}$
$=\frac{3^4}{5^2}\times\frac{5^3}{3^3}$
$=\Big(3^{(4-3)}\Big)\times\Big(5^{(3-2)}\Big)$
$=3\times5$
$=15$

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Question 83 Marks

Find the value of x for which $\Big(\frac{5}{3}\Big)^{-4}\times\Big(\frac{5}{3}\Big)^{-5}=\Big(\frac{5}{3}\Big)^{3\text{x}}$

Answer

$\Big(\frac{5}{3}\Big)^{-4}\times\Big(\frac{5}{3}\Big)^{-5}=\Big(\frac{5}{3}\Big)^{3\text{x}}$ $\Rightarrow\Big(\frac{5}{3}\Big)^{-4+(-5)}=\Big(\frac{5}{3}\Big)^{3\text{x}}$ $\Rightarrow\Big(\frac{5}{3}\Big)^{-9}=\Big(\frac{5}{3}\Big)^{3\text{x}}$ $\therefore 3\text{x}=-9$ $\Rightarrow\text{x}=\frac{-9}{3}$ $=-3$ Hence $\text{x}=-3$

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