3 Marks Question - Maths STD 7 Questions - Vidyadip

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Question 13 Marks

How many times of $30$ must be added together to get a sum equal to $30^7$?

Answer

Let $n$ be the number of times that $30$ must be added together to get a sun equal to $30^7$
therefore, we can write that $\frac{\underbrace{30+30+\dots+30}=30^{7}}{\text{n times}}$
$\Rightarrow30\times\text{n}=30^{7}$
$\Big[\frac{{\underbrace{\because\text{a}+\text{a}+\dots+\text{a}\times}\text{n}}}{\text{n times}}\Big]$
$\Rightarrow\frac{30\times\text{n}}{30}=\frac{30^{7}}{30}[$dividing both sides by $30 ]$
$\Rightarrow\text{n}=30^{7-1}$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$\therefore\text{n}=30^{6}$
Hence, if $30$ is added $30^6$ times, then we get $30^7$

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Question 23 Marks

Evaluate: $\frac{3^{4}\times12^{3}\times36}{2^{5}\times6^{3}}$

Answer

We have, $\frac{3^{4}\times12^{3}\times36}{2^{5}\times6^{3}}$$=\frac{3^{4}\times(2^{2}\times3)^{3}\times(2^{2}\times3^{2})}{2^{5}\times(2\times3)^{3}}$ $\Big[\because 12=2\times2\times3\text{ and }36=2\times2\times3\times3\Big]$
$=\frac{3^{4}\times2^{6}\times3^{3}\times2^{2}\times3^{2}}{2^{5}\times2^{3}\times3^{3}}$ $\Big[\because(\text{a}\times\text{b})^{ \text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\Big]$
$=\frac{(3^{4}\times3^{2}\times3^{3})\times(2 ^{6}\times2^{2})}{(2^{5}\times2^{3})\times3^{3}}$
$\frac{3^{4+2+3}\times2^{6+2}}{2^{5+3}\times3^{3}}$ $\Big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\Big]$
$\frac{3^{9}\times2^{8}}{3^{3}\times2^{8}}=3^{9-3}\times2^{8-8}$ $\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$=3^{6}\times2^{0}=3^{6}\times1=729$

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Question 33 Marks

Evaluate: $\frac{7^{8}\text{a}^{10}\text{b}^{7}\text{c}^{12}}{7^{6}\text{a}^{8}\text{b}^{4}\text{c}^{12}}$

Answer

We have, $\frac{7^{8}\text{a}^{10}\text{b}^{7}\text{c}^{12}}{7^{6}\text{a}^{8}\text{b}^{4}\text{c}^{12}}$ $=\Big(\frac{7^{8}}{7^{6}}\Big)\times\Big(\frac{\text{a}^{10}}{\text{a}^{8}}\Big)\times\Big(\frac{\text{b}^{7}}{\text{b}^{8}}\Big)\times\Big(\frac{\text{c}^{12}}{\text{c}^{12}}\Big)$ $7^{8-6}\times\text{a}^{10-8}\times\text{b}^{7-4}\times\text{c}^{12-12}$ $\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$ $$ $7^{2}\times\text{a}^{2}\times\text{b}^{3}\times\text{c}^{0}=49\text{ a}^{2}\text{b}^{3}$ $\Big[\because\text{c}^{0}=1\Big]$

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Question 43 Marks

Write the number of seconds in scientific notation.

Unit		Value in Seconds
$1$	$1$ Minute	$60$
$2$	$1$ Hour	$3,600$
$3$	$1$ Day	$86,400$
$4$	$1$ Month	$2,600,000$
$5$	$1$ Year	$32,000,000$
$6$	$10$ Years	$3,20,000,000$

Answer

$\text { 1. } 1 \mathrm{~min}=60 \mathrm{~s}$
$=60 \times 10^1 \mathrm{~s}$
$=6 \times 101 \mathrm{~s}$
$\text { 2. } 1 \mathrm{~h}=3,600 \mathrm{~s}$
$=36 \times 10^2 \mathrm{~s}$
$=3.6 \times 10 \times 10^2 \mathrm{~s}$
$=3.6 \times 10^3 \mathrm{~s}$
$\text { 3. } 1 \text { day }=86,400 \mathrm{~s}$
$=8.64 \times 10^2 \mathrm{~s}$
$=8.64 \times 10^2 \times 10^2 \mathrm{~s}$
$=8.64 \times 10^4 \mathrm{~s}$
$=8.6 \times 10^4 \mathrm{~s}$
$\text { 4. } 1 \text { month }=2,600,000 \mathrm{~s}$
$=26 \times 10^5 \mathrm{~s}$
$=2.6 \times 10 \times 10^5 \mathrm{~s}$
$=2.6 \times 10^6 \mathrm{~s}$
$\text { 5. }1 \text { year }=32,000,000 \mathrm{~s}$
$=32 \times 10^6 \mathrm{~s}$
$=32 \times 10^6 \mathrm{~s}$
$=3.2 \times 10 \times 10^6 \mathrm{~s}$
$=32 \times 10^7 \mathrm{~s}$
$\text { 6. } 10 \text { year }=320,000,000 \mathrm{~s}$
$=32 \times 10^7 \mathrm{~S}$
$=32 \times 10 \times 10^7 \mathrm{~s}$
$=3.2 \times 10^8 \mathrm{~s} \text {. }$

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Question 53 Marks

Evaluate: $\frac{6^{4}\times12^{3}\times36}{2^{5}\times36}$

Answer

We have, $\frac{6^{4}\times12^{3}\times36}{2^{5}\times36}$ $=\frac{(2\times3)^{4}\times(3^{2})^{}2\times(5^{2})^{3}}{3^{2}\times(2^{2})^{2}\times3^{6}\times5^{6}}$ $\Big[\because(\text{a}\times\text{b})^{\text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\Big]$ $=2^{4-4}\times3^{4+4-2-6}\times5^{6-6}$ $\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$ $$ $=2^{0}\times3^{0}\times5^{0}$ $=1\times1\times1=1$

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Question 63 Marks

Find m so that $\Big(\frac{2}{9}\Big)^{3}\times\Big(\frac{2}{9}\Big)^{6}=\Big(\frac{2}{9}\Big)^{\text{2m-1}}$

Answer

We have, $\Big(\frac{2}{9}\Big)^{3}\times\Big(\frac{2}{9}\Big)^{6}=\Big(\frac{2}{9}\Big)^{\text{2m-1}}$
$\Rightarrow\Big(\frac{2}{9}\Big)^{3+6}=\Big(\frac{2}{9}\Big)^{\text{2m-1}} [ \because a^m \times a^n=a^{m+n}]$
$\Rightarrow\Big(\frac{2}{9}\Big)^{9}=\Big(\frac{2}{9}\Big)^{\text{2m-1}}$
$\Rightarrow9=2\text{m}-1$ [$\because$
$a^m=a^n \Rightarrow m=n]$
$\Rightarrow9+1=2\text{m} [$transposing $(-1)$ to $LHS]$
$\Rightarrow10=2\text{m}[$diving both sides by $2]$
$\Rightarrow\frac{10}{2}=\frac{2\text{m}}{2}$
$\Rightarrow5=\text{m}$
Hence, $m = 5$

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Question 73 Marks

By what number should $(-4)^5$ be divided so that the quotient may be equal to $(-4)^3$ ?

Answer

In order to find the number, which should divide $(-4)^5$ to get the quotient $(-4)^3$,
we will. divide $(-4)^5$ by $(-4)^3$
Hence, required number $=$ $(-4)^{5-3}
=(-4)^2$

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Question 83 Marks

Evaluate:
$\Big(\frac{6\times10}{2^{2}\times5^{3}}\Big)^{2}\times\frac{25}{27}$

Answer

We have, $\Big(\frac{6\times10}{2^{2}\times5^{3}}\Big)^{2}\times\frac{25}{27}$
$=\Big(\frac{2\times3\times2\times5}{2^{2}\times5^{3}}\Big)^{2}\times\frac{5^{2}}{3^{2}}$ $\Big[\because 6=2\times3\text{ and }10=2\times5\Big]$
$\Big(\frac{2^{2}\times3\times5}{2^{2}\times5^{3}}\Big)^{2}\times\frac{5^{2}}{5^{3}}$ $\Big[\because(\text{a}\times\text{b})^{ \text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\Big]$
$$$=\Big(\frac{3}{5^{2}}\Big)^{2}\times\frac{5^{2}}{3^{3}}$
$=\frac{3^{2}}{5^{4}}\times\frac{5^{2}}{3^{3}}$$\Big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\Big]$
$=\frac{1}{5^{2}\times3}=\frac{1}{25\times3}=\frac{1}{75}$

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Question 93 Marks

If $2^{n+2}-2^{n+1}+2^n=c \times 2^n$, find the value of $c$.

Answer

$\text { We have, } 2^{n+2}+2^{n+1}+2 n=c \times 2^n$
$\Rightarrow 2^n 2^2+2^n 2^1+2^n$
$=c \times 2^n\left[\therefore a^{m+n}=a^m \times a^n\right]$
$\Rightarrow 2^n\left[2^2-2^1+1\right]$
$=c \times 2^n\left[\text { taking common } 2^n \text { in LHS }\right]$
$\Rightarrow 2^n[4-2+1]=c \times 2^n$
$\Rightarrow 3 \times 2^n=c \times 2^n 3 \times 2^n \times 2^{-n}$
$=c \times 3^n \times c^{-n}\left[\text { multiplying both sides by } 2^{-n}\right]$
$\Rightarrow 3 \times 2^{n-1}=c \times 2^{n-n}\left[\therefore a^{m+n}=a^m \times a^n\right]$
$\Rightarrow 3 \times 2^0=c \times 2^0$
$\Rightarrow 3 \times 1=c \times 1\left[\therefore a^0=1\right]$
$\therefore 3=c$

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Question 103 Marks

Evaluate:
$\frac{15^{4}\times18^{3}}{3^{3}\times5^{2}\times12^{2}}$

Answer

We have, $\frac{15^{4}\times18^{3}}{3^{3}\times5^{2}\times12^{2}}=\frac{(3\times5)^{4}\times(2\times3^{2})^{3}}{3^{3}\times5^{2}(2^{2}\times3)^{2}}$
$=\frac{3^{4}\times5^{4}\times2^{3}\times3^{6}}{3^{3}\times5^{2}\times2^{4}\times3^{2}}$ $\Big[\because 18=2\times3\times3\text{ and }12=2\times2\times3\Big]$
$=\frac{3^{4+6-3-2}\times5^{4-2}}{2^{4-3}}$ $\Big[\because(\text{a}\times\text{b})^{ \text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\Big]$
$=\frac{3^{5}\times5^{2}}{2}=\frac{243\times25}{2}=\frac{6075}{2}$ $\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\text{ and }\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\Big]$

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Question 113 Marks

Express the given information in Scientific notation and then arrange them in descending order of their size.

Name of the Planet		Mass (in kg)
$1$	Mercury	$330000000000000000000000$
$2$	Venus	$4870000000000000000000000$
$3$	Earth	$5980000000000000000000000$
$4$	Mars	$642000000000000000000000$
$5$	Jupiter	$1900000000000000000000000000$
$6$	Saturn	$569000000000000000000000000$
$7$	Uranus	$86900000000000000000000000$
$8$	Neptune	$102000000000000000000000000$
$9$	Pluto	$13100000000000000000000$

Answer

A number is written in standard form as a $\times 10^k$, where a is terminating decimal and $k$ is an integer.

Name of the Planet		Mass (in kg)
$1$	Mercury	$3.3 \times 10^{23}$
$2$	Venus	$4.87 \times 40^{24}$
$3$	Earth	$5.98 \times 10^{24}$
$4$	Mars	$6.42 \times 10^{27}$
$5$	Jupiter	$1.9 \times 10^{27}$
$6$	Saturn	$5.69 \times 10^{26}$
$7$	Uranus	$8.69 \times 10^{25}$
$8$	Neptune	$1.02 \times 10^{26}$
$9$	Pluto	$1.31 \times 10^{22}$

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, the required descending order of the size will be
Jupiter > Saturn > Neptune > Uranus > Earth > Venus > Mars > Mercury > Pluto.

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Question 123 Marks

Geometry Application: The number of diagonals of an n-sided figure is $\frac{1}{2}(\text{n}^{2}-3\text{n})$ Use the formula to find the number of diagonals for a $6-$sided figure (hexagon).

Answer

Given, a plygon has n sides, then number of diagonals is $\frac{1}{2}(\text{n}^{2}-3\text{n})$
In hexagon, there are six sides. therefore for calcutating number of diagonals in haxagon,
put $n = 6$ in the above formula.
$\therefore$ Number of diagonals $=\frac{1}{2}(\text{n}^{2}-3\text{n})=\frac{1}{2}(6^{2}-3\times6)$
$=\frac{1}{2}(6\times6-3\times6)$
$=\frac{1}{2}(36-18)=\frac{1}{2}(18)=9$
Hence, a hexagon has 9 diagonals.

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Question 133 Marks

Evaluate: $\frac{5^{4}\times7^{4}\times2^{7}}{8\times49\times5^{3}}$

Answer

We have, $\frac{5^{4}\times7^{4}\times2^{7}}{8\times49\times5^{3}}$ $\frac{5^{4}\times7^{4}\times2^{7}}{2^{3}\times7^{2}\times5^{3}}$ $\Big[\because8=2^{3}\text{ and }49=7^{2}\Big]$ $=\Big(\frac{5^{4}}{5^{3}}\Big)\times\Big(\frac{2^{7}}{2^{3}}\Big)\times\Big(\frac{7^{4}}{7^{2}}\Big)$ $\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$ $=5^{4}\times2^{7-3}\times7^{7-2}$ $=5\times2^{4}\times7$ $=5\times16\times49=3920$

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Question 143 Marks

A light year is the distance that light can travel in one year. $1$ light year $= 9,460,000,000,000\ km.$
$a.$ Express one light year in scientific notation.
$b.$ The average distance between Earth and Sun is $1.496 \times 108\ km.$ Is the distance between Earth and the Sun greater than, less than or equal to one light year?

Answer

$a.$ Given, $1$ light year$ = 9,460,000,000,000\ km$
For standard from $=946\times10^{10}\ \text{km}$
$=\frac{946}{100}\times10^{10}\times100\ \text{km}$
$=9.46\times10^{12}\ \text{km}$
$b.$ the average distance between Earth and sun $= 1.496 \times 10^8 \mathrm{~km}$
$\therefore$ Distance between Earth and sun $=\frac{1.496}{10000}\times10^{8}\times10^{4}$
$=0.0001496\times10^{12}\ \text{km}$
Since, $9.46>0.0001496$
So, the distance between Earth and sun less than one light year.

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Question 153 Marks

Evaluate: $\frac{125\times5^{2}\times\text{a}^{7}}{10^{3}\times\text{a}^{4}}$

Answer

We have, $\frac{125\times5^{2}\times\text{a}^{7}}{10^{3}\times\text{a}^{4}}$ $\frac{5^{3}\times5^{2}\times\text{a}^{7}}{(2\times5)^{3}\times\text{a}^{4}}$ $\Big[\because 125=5^{3}\Big]$ $\frac{5^{2+3}\times\text{a}^{7}}{2^{3}\times5^{3}\times\text{a}^{4}}$ $\Big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\text{ and }(\text{a}\times\text{b})^{ \text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\Big]$ $\frac{5^{5}\times\text{a}^{7}}{2^{3}\times5^{3}\times\text{a}^{4}}$ $=\Big(\frac{5^{5}}{5^{3}}\Big)\times\Big(\frac{\text{a}^{7}}{\text{a}^{4}}\Big)\times\Big(\frac{1}{2^{3}}\Big)$ $\frac{5^{5-3}\times\text{a}^{7-4}}{2^{3}}$ $\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$ $\frac{5^{2}\times\text{a}^{3}}{2^{3}}=\frac{25\text{a}^{3}}{8}$

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Question 163 Marks

Arrange in ascending order:
$2^{2+3},\left(2^2\right)^3, 2 \times 2^2, \frac{3^5}{3^2} 3^2 \times 3^0, 2^3 \times 5^2$

Answer

In descending order, the numbers are arranged from largest to smallest.
We have, $22+3=25$
$=2 \times 2 \times 2 \times 2 \times 2=32$
$\left(2^2\right)^3=2^6$
$=2 \times 2 \times 2 \times 2 \times 2 \times 2=64$
$2 \times 2^2=2^{1+2}$
$=2^3=8$
$\frac{3^5}{3^2}=3^{5-2}$
$=3^3=27$
$3^2 \times 3^0=3^{2+0}$
$=3^2=9$
$2^3 \times 5^2=2 \times 2 \times 2 \times 5 \times 5$
$=8 \times 25=200$
Thus, the required descending order will be.
$\left(2^2 \times 5^2\right)>\left(2^2\right)^3>2^{2+3}>\frac{3^5}{3^2}>\left(3^2 \times 3^0\right)>\left(2 \times 2^2\right)$

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Question 173 Marks

Arrange in ascending order:
$2^5, 3^3, 2^3 \times 2,\left(3^3\right)^2, 3^5, 4^0, 2^3 \times 3^1$

Answer

In ascending order, the numbers are arranged from smallest to largest.
We have, $2^5=2 \times 2 \times 2 \times 2 \times 2=32$
$3^3=3 \times 3 \times 3=27$
$2^3 \times 2=2 \times 2 \times 2 \times 2=16$
$\left(3^3\right)^2=3^3 \times 2\left[\therefore\left(a^m\right)^n=a^{m n}\right]$
$3^6=3 \times 3 \times 3 \times 3 \times 3 \times 3=729$
$3^5=3 \times 3 \times 3 \times 3 \times 3=243$
$4^0=1\left[\therefore a^0=1\right]$
$\text { and } 2^3 \times 3^1$
$=2 \times 2 \times 2 \times 3$
$=24$ Thus, the required ascending order will be.
$4^0<2^3 \times 2<23 \times 31<33<25<35<\left(3^3\right)^2$

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