5 Marks Questions

Question 15 Marks

Find the reciprocal of the rational number $\Big(\frac{1}{2}\Big)^{2}\div\Big(\frac{2}{3}\Big)^{3}$

Answer

Given, $\Big(\frac{1}{2}\Big)^{2}\div\Big(\frac{2}{3}\Big)^{3}$
$=\frac{\Big(\frac{1}{2}\Big)^{2}}{\Big(\frac{2}{3}\Big)^{3}}$
$\Big[\because\text{a+b}=\frac{\text{a}}{\text{b}}\Big]$
$=\frac{\frac{(1)^{2}}{(2)^{2}}}{\frac{(2)^{3}}{(3)^{3}}}=\frac{\Big(\frac{1}{4}\Big)}{\Big(\frac{8}{27}\Big)}$
$\Big[\because\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{n}}=\frac{\text{a}^{\text{n}}}{\text{b}^{\text{n}}}\Big]$
$=\frac{1}{4}\times\frac{27}{8}=\frac{27}{4\times8}=\frac{27}{32}$
$\big[\because1^{2}=1,2^{2}=4,2^{3}=8\text{ and }3^{3}=27\big]$
$\Big[\because\frac{\text{a}}{\text{b}}\div\frac{\text{c}}{\text{d}}=\frac{\text{a}}{\text{b}}\times\frac{\text{d}}{\text{c}}\Big]$
We know that, reciprocal of a rational number is obtained by interchaning number numerator and denominator. $\therefore$ Reciprocal of given number $=\frac{32}{27}$

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Question 25 Marks

If $\frac{p}{q}=\Big(\frac{3}{2}\Big)^{2}\div\Big(\frac{9}{4}\Big)^{0}$ find the value of $\Big(\frac{\text{p}}{\text{q}}\Big)^{3}$

Answer

We have, $\frac{\text{p}}{\text{q}}=\Big(\frac{3}{2}\Big)^{2}\div\Big(\frac{9}{4}\Big)^{0}$
$\Rightarrow\frac{\text{p}}{\text{q}}=\Big(\frac{3}{2}\Big)^{2}+\frac{1}{1}$$\Big[\because\text{a}^{0}=1\Big]$
$\Rightarrow\frac{\text{p}}{\text{q}}=\Big(\frac{3}{2}\Big)^{2}$
$\Bigg[\because\frac{\text{a}}{\text{b}}+\frac{\text{c}}{\text{d}}=\frac{\text{a}}{\text{b}}\times\frac{\text{d}}{\text{c}}\Bigg]$
$\Rightarrow\frac{\text{p}}{\text{q}}=\frac{3^{2}}{2^{2}}$
$\Bigg[\because\Big(\frac{\text{a}}{\text{b}}\Big)\text{n}=\frac{\text{a}^{\text{n}}}{\text{b}^{\text{n}}}\Bigg]$
$\Rightarrow\frac{\text{p}}{\text{q}}=\frac{9}{4}$ On taking cube both sides,
we get $\Big(\frac{\text{p}}{\text{q}}\Big)^{3}=\Big(\frac{9}{4}\Big)^{3}$
$\therefore\Big(\frac{\text{p}}{\text{q}}\Big)^{3}=\frac{9\times9\times9}{4\times4\times4}=\frac{729}{64}$

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Question 35 Marks

In our own planet Earth, $361,419,000$ square kilometre of area is covered with water and $148,647,000$ square kilometre of area is covered by land. Find the approximate ratio of area covered with water to area covered by land by converting these numbers into scientific notation.

Answer

$\text { Given, }$
$\text { Area covered by water }=361419000 km^2$
$\text { Area covered by land }=14864700 km^2$
$\text { Convertion of area into scientific notation, }$
$\text { Also, } 361419000=361419 \times 10^5$
$\text { So, } 361419 \times 10^5 \times 10^3=3.61419 \times 10^8$
$\therefore \text { Area covered by water }=3.61419 \times 10^8 km^2$
$\text { Similary, } 148647000=148647 \times 10^3$
$\text { Also, } 148647=1.48647 \times 10^5$
$\text { So, }=1.48647 \times 10^8$
$\therefore \text { Area covered by land }=1.48647 \times 10^8 km^2$
$\text { Let } 3.61419 \times 10^8=3.6 \times 10^8$
$\text { and } 1.48647 \times 10^8$
$=1.5 \times 10^8$
$\therefore \text { Ration of water to land }=\frac{3.6}{1.5}=12: 5$

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Question 45 Marks

Express the given information in Scientific notation (standard form) and then arrange them in ascending order of their size.

	Deserts of the world	Area (in sq km)
$1$	Kalahari, South Africa	$932,400$
$2$	Thar, India	$199,430$
$3$	Gibson, Australia	$155,400$
$4$	Great Victoria, Australia	$647,500$
$5$	Sahara, North Africa	$8,598,800$

Answer

1. Area of Kalahari, South Africa $=932,400$
$=932400.00\left[\therefore \text { standard form a } \times 10^{k}\right]$
$=9324 \times 10^2$
$=9.324 \times 10^3 \times 10^2$
$=9.324 \times 10^6$
2. Area of Thar, India $=199,430$
$=199430.00$
$=19943 \times 10^1$
$=1.9943 \times 10^4 \times 10^1$
$=1.9943 \times 10^5$
3. Area of Gibson, Australia $=155,400$
$=155400.00$
$=1554 \times 10^2$
$=1.554 \times 10^3 \times 10^2$
$=1.554 \times 10^5$
4. Area of Great Victoria, Australia $=647,500$
$=647500.00$$=6475 \times 10^2$
$=6.475 \times 10^3 \times 10^5$
$=6475 \times 10^5$
5. Area of Sahara, North-Africa $=8,598,800$
$=8598800.00$
$=85988 \times 10^2$
$=8.5988 \times 10^4 \times 10^2$
$=85988 \times 10^6$
Two numbers written in scientific notation can be compared.
The number with the larger power of $10$ is greater than the number with the smaller power of $10$.
If the powers of ten are the same, then the number with larger factor is the larger number.
Hence, required ascending order of the size will be
Gibson, Australia < Thar, India < Great Victoria, Australia < Kalahari, South Africa < Sahara, North-Africa.

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Question 55 Marks

Find the value of n, where n is an integer and $2^{\text{n-5}}\times6^{\text{2n-4}}=\frac{1}{12^{4}\times2}$

Answer

We have, $2^{\text{n-5}}\times6^{\text{2n-4}}=\frac{1}{12^{4}\times2}$
$\Rightarrow\frac{2^{\text{n}}}{2^{5}}\times\frac{6^{\text{2n}}}{6^{4}}=\frac{1}{12^{4}\times2}$ $\Big[\because\text{a}^{\text{m-n}}=\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}\Big]$
$\Rightarrow\frac{2^{\text{n}}}{2^{5}}\times\frac{6^{\text{2n}}}{6^{4}}=\frac{1}{(2\times6)^{4}\times2}$ $\Big[\because12=6\times2\Big]$
$\Rightarrow2^{\text{n}}\times(6^{2})^{\text{n}}=\frac{1}{(2\times6)^{4}\times2}$ [by cross-multiplicaton]
$\Rightarrow2^{\text{n}}\times(6^{2})^{\text{n}}=\frac{2^{5}\times6^{4}}{2\times6^{4}\times2}$ $\Big[\because\text{a}^{\text{mn}}=(\text{a}^{\text{m}})^{\text{n}}\text{ and }(\text{a}\times\text{b})^{\text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\Big]$
$\Rightarrow2^{\text{n}}\times36^{\text{n}}=\frac{2^{5}\times6^{4}}{2^{5}\times6^{4}}$
$\Rightarrow2^{\text{n}}\times36^{\text{n}}=1$ $\Big[\because\text{a}^{\text{m}}=\text{b}^{\text{m}}\Rightarrow(\text{ab})^{\text{m}}\Big]$
$\Rightarrow(2\times36)^{\text{n}}=1$ $\Big[\because\text{a}^{0}=1\Big]$
$\Rightarrow(72)^{\text{n}}=(72)^{0}=1$ $\Big[\because\text{a}^{\text{m}}=\text{a}^{\text{n}}\Rightarrow\text{m}=\text{n}\Big]$
$\therefore\text{n}=0$

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