2 Marks Questions - Maths STD 7 Questions - Vidyadip

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22 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Life Science Application: The major components of human blood are red blood cells, white blood cells, platelets and plasma. A typical red blood cell has a diameter of approximately $7 \times 10^{-6}$ metres. A typical platelet has a diameter of approximately $2.33 \times 10^{-6}$ metre. Which has a greater diameter, a red blood cell or a platelet?

Answer

Given, diameter of red blood cell
$=7 \times 10^{-6} \mathrm{~m}$ and diameter of platelet
$=2.33 \times 10^{-6} \mathrm{~m}$
We know that, 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 the larger factor is the larger number. Therefore, red blood cell has a greater diameter than a platelet.

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Question 22 Marks

Express following as a product of powers of their prime factors: $2025$

Answer

$\begin{array}{c|c} 3 &2250 \\ \hline 3 &675 \\ \hline 3 &225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$
Using prime factorisation of $2025$,
we have $2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5$
$\therefore 2025 =3^4 \times 5^2$

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Question 32 Marks

Express the following numbers using exponential notations:$1024$

Answer

For solving these type of question, we use prime factorisation method. Given, $1024$
$\begin{array}{c|c}2 & 1024 \\ \hline2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$
Using prime factorisatin of $1024$,
we have $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}$

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Question 42 Marks

What’s the error?
A student said that $\frac{3^{5}}{9^{5}}$ is the same as $\frac{1}{3}$ What mistake has the student made?

Answer

We have $\frac{3^{5}}{9^{5}}=\frac{3^{5}}{(3^{2})^{5}}$
$=\frac{3^{5}}{3^{10}}=\frac{1}{3^{10-5}}=\frac{1}{3^{5}}$ [$\because 9 = 3 \times 3 = 3^2$]
So, $\frac{1}{3}$ is not same as $\frac{1}{3^{5}}$
$\big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
Student has multiplied the base by its exponent.
This is the error.

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Question 52 Marks

Blubber: makes up $27$ per cent of a blue whale’s body weight. Deepak found the average weight of blue whales and used it to calculate the average weight of their blubber. He wrote the amount as $=2^2 \times 3^2 \times 5 \times 17 \mathrm{~kg}$
. Evaluate this amount.

Answer

Weight calculated by Deepak
$=2^2 \times 3^2 \times 5 \times 17 \mathrm{~kg}$
$=2 \times 2 \times 3 \times 3 \times 5 \times 17$
$=4 \times 9 \times 5 \times 17$
$=36 \times 5 \times 17$
$=180 \times 17$
$=3060 \mathrm{~kg} \text { Hence, weight calculated by Deepak was } 3060 \mathrm{~kg}$

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Question 62 Marks

Express the following numbers using exponential notations:$1029$

Answer

For solving these type of question, we use prime factorisation method. Given, $1029$
$\begin{array}{c|c} 3 &1029 \\ \hline 7 & 343\\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$
Using prime factorisatin of $1029,$
we have$1029 = 3 \times 7 \times 7 \times 7$
$=3 \times 7^3$

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Question 72 Marks

Simplify and express the following in exponential form:
$\left(5^{15} \div 5^{10}\right) \times 5^5$

Answer

We have, $(5^{15} \div 5^{10}) \times 5^{5}\Big(\frac{5^{15}}{5^{10}}\Big)\times5^{5}$$=5^{15-10}\times5^{5}$
$5^{5}=5^{15-10}\times5^{5}$ $\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$$ $=5^{5}\times5^{5}=5^{5+5}=5^{10}$ $\Big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}={\text{a}^{\text{m+n}}}\Big]$

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Question 82 Marks

Express the following numbers in standard form:
$76,47,000$

Answer

We have, $76,47,000$
$=7647000,00$
$\mathrm{A}$ number in standard form is written as a $\times 10^{\mathrm{k}}$, where a is the terminating decimal such that $1<\mathrm{a}<10$ and k is any integer.
So, $7647000=7647 \times 10^3=7.647 \times 10^3 \times 10^3$
$=7.647 \times 10^6$
Similarly,

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Question 92 Marks

Express the following in single exponential form:
$\left(-5^5\right) \times(-5)$

Answer

$\text { We have, }\left(5^5\right) \times(-5)$
$=(5)^{5+1} \times(-5)^6$
$=(-1 \times 5)^6\left[\because a^m \times a^n=a^{m+n}\right]$
$=(-1)^6 \times(5)^6\left[\because(a \times b)^m=a^m \times b^m\right]$
$=1 \times 5^6\left[\because(-1)^n=1, \text { if } n \text { is even }\right]$
$=5^6$

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Question 102 Marks

Simplify and express the following in exponential form: $\Bigg[\Big(\frac{3}{7}\Big)^{4}\times\Big(\frac{5}{7}\Big)^{5}\Bigg]\div\Big(\frac{3}{7}\Big)^{7}$

Answer

We have, $\Bigg[\Big(\frac{3}{7}\Big)^{4}\times\Big(\frac{5}{7}\Big)^{5}\Bigg]\div\Big(\frac{3}{7}\Big)^{7}$
$=\Big(\frac{3}{7}\Big)^{4+5}+\Big(\frac{3}{7}\Big)^{7}$
$\Big[\because \text{a}^{\text{m}}\times \text{a}^{\text{n}}=\text{a}^{\text{m-n}}\Big]$
$=\Big(\frac{3}{7}\Big)^{9}+\Big(\frac{3}{7}\Big)^{7}=\frac{\Big(\frac{3}{7}\Big)^{9}}{\Big(\frac{3}{7}\Big)^{7}}$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$=\Big(\frac{3}{7}\Big)^{9-7}=\Big(\frac{3}{7}\Big)^{2}$

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Question 112 Marks

Express following as a product of powers of their prime factors: $800$

Answer

$\begin{array}{c|c} 2 & 8000 \\ \hline 2 & 400 \\ \hline 2 &200 \\ \hline 2 &100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$
Using prime factorisation of $800,$
we have $800 = 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5$
$\therefore$ 800 = $2^5 \times 5^2$

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Question 122 Marks

Simplify and express the following in exponential form: $\Bigg[\Big(\frac{3}{5}\Big)^{3}\times\Big(\frac{3}{5}\Big)^{8}\Bigg]\div\Bigg[\Big(\frac{3}{5}\Big)^{2}\times\Big(\frac{3}{5}\Big)^{4}\Bigg]$

Answer

We have, $\Bigg[\Big(\frac{3}{5}\Big)^{3}\times\Big(\frac{3}{5}\Big)^{8}\Bigg]\div\Bigg[\Big(\frac{3}{5}\Big)^{2}\times\Big(\frac{3}{5}\Big)^{4}\Bigg]$
$=\Big(\frac{3}{5}\Big)^{3+8}\times\Big(\frac{3}{5}\Big)^{2+4}$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$=\Big(\frac{3}{5}\Big)^{1}+\Big(\frac{3}{5}\Big)^{6}$
$\Big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}={\text{a}^{\text{m+n}}}\Big]$
$=\frac{\Big(\frac{3}{5}\Big)^{11}}{\Big(\frac{6}{5}\Big)^{6}}=\Big(\frac{3}{5}\Big)^{11-6}=\Big(\frac{3}{5}\Big)^{5}$

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Question 132 Marks

Express the following numbers using exponential notations:$\frac{144}{875}$

Answer

For solving these type of question, we use prime factorisation method.
Given, $\frac{144}{875}$
$\begin{array}{c|c} 2 &144 \\ \hline2 &72 \\ \hline2 &36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$
$\begin{array}{c|c} 5 & 875 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$
Using prime factorisatin of $144$ and $875$, we have
$\frac{144}{875}=\frac{2\times2\times2\times2\times3\times3}{5\times5\times5\times7}=\frac{2^{4}\times3^{2}}{5^{3}\times7^{1}}$

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Question 142 Marks

Express following as a product of powers of their prime factors: $9000$

Answer

$\begin{array}{c|c} 2 &9000 \\ \hline 2 &4500 \\ \hline 2 &2250 \\ \hline 3 &1125 \\ \hline 3 &375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$
Using prime factorisation of $9000,$
we have $9000 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5$
$\therefore$ $9000$ $=2^2 \times 3^2 \times 5^3$

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Question 152 Marks

Simplify and express the following in exponential form: $\Bigg[\Big(\frac{7}{11}\Big)^{5}\div\Big(\frac{7}{11}\Big)^{2}\Bigg]\times\Big(\frac{7}{11}\Big)^{2}$

Answer

We have, $\Bigg[\Big(\frac{7}{11}\Big)^{5}\div\Big(\frac{7}{11}\Big)^{2}\Bigg]\times\Big(\frac{7}{11}\Big)^{2}$
$=\Bigg[\frac{\Big(\frac{7}{11}\Big)^{5}}{\Big(\frac{7}{11}\Big)^{2}}\Bigg]\times\Big(\frac{7}{11}\Big)^{2}=\Big(\frac{7}{11}\Big)^{5-2}\times\Big(\frac{7}{11}\Big)^{2}$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$=\Big(\frac{7}{11}\Big)^{3}\times\Big(\frac{7}{11}\Big)^{2}$
$\Big[\because \text{a}^{\text{m}}\times \text{a}^{\text{n}}=\text{a}^{\text{m-n}}\Big]$
$=\Big(\frac{7}{11}\Big)^{3+2}=\Big(\frac{7}{11}\Big)^{5}$

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Question 162 Marks

Express the following in single exponential form:
$5^2 \times 7^2$

Answer

We have, $5^2 \times 7^2=(5 \times 7)^2$
$=35^2\left[\because a^m \times b^m=(a \times b)^m\right]$

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Question 172 Marks

Express the following in single exponential form:
$2^4 \times 4^2$

Answer

We have, $2^4 \times 4^2=2^4 \times\left(2^2\right)^2$
$=2^4 \times 2^4\left[\because 4=2^2\right]$
$=2^{4+4}\left[\because\left(a^m\right)^n=a^{m n}\right]$
$=2^8$

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Question 182 Marks

Simplify and express the following in exponential form: $\Big(\frac{\text{a}^{6}}{\text{a}^{4}}\Big)\times\text{a}^{2}\times\text{a}^{0}$

Answer

We have, $\Big(\frac{\text{a}^{6}}{\text{a}^{4}}\Big)\times\text{a}^{2}\times\text{a}^{0}$
$=(\text{a}^{6-4}\times\text{a}^{5}\times1)$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}} \text{ and }\text{a}^{0}=1\Big]$
$\text{a}^{2}\times\text{a}^{5}=\text{a}^{2+5}$
$\Big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}={\text{a}^{\text{m+n}}}\Big]$
$=\text{a}^{7}$

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Question 192 Marks

A googol is the number $1$ followed by $100$ zeroes.
$a.$ How is a googol written as a power?
$b.$ How is a googol times a googol written as a power?

Answer

$1.$ Googal $=\frac{{\underbrace{100}\dots}0=1\times10^{100}}{\text{100 times}}$ [as there are $100$ zeroes after $1]$
$2.$ Googal times googal means multiply googal by googal

$\therefore$ Requrired number $=$ googal $ \times$ googol $=10^{100} \times 100^{100}$
$[\because 1$ googol $=10^{100}]$
$ =10^{100}+100^{100}\left[\because a^m \times a^n=a^{m+n}\right]$
$=100^{200}$

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Question 202 Marks

Life Science: Bacteria can divide in every $20$ minutes. So $1$ bacterium can multiply to $2$ in $20$ minutes. $4$ in $40$ minutes, and so on. How many bacteria will there be in $6$ hours? Write your answer using exponents, and then evaluate.

Most bacteria reproduce by a type of simple cell division known as binary fission. Each species reproduce best at a specific temperature and moisture level.

Answer

We know that, $1$h $= 60$min $6$h $= 60 \times 6$min $= 360$min
Given, a bacteria doubles itself in every $20$min.
Number of times it will double itself $= 360$min/ $20$min $= 18$
$\therefore$ Bacteria will there in $6$h $= 2 \times 2 \times 2 \times …\times 2 (18$ times) = $2^{18}$

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Question 212 Marks

The speed of light in vaccum is $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$. Sunlight takes about $8$ minutes to reach the earth. Express distance of Sun from Earth in standard form.

Answer

It is given that,
Speed of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Time taken by light to reach the Earth $=8 \mathrm{~min}=8 \times 60 \mathrm{~s}=480 \mathrm{~s}[\therefore 1 \mathrm{~min}=60 \mathrm{~s}]$
We know that,
$\text { Distance }=\text { Speed } \times \text { Time }=3 \times 10^8 \times 480=1440 \times 10^8$
$=1.440 \times 10^3 \times 10^8$
$=1.44 \times 10^{11}\left[\therefore 10^3 \times 10^8=10^{11}\right]$
Hence, the distance of Sun from the Earth is $1.44 \times 10^{11} \mathrm{~m}$.

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Question 222 Marks

Simplify and express the following in exponential form: $\left(3^7 \div 3^5\right)^4$

Answer

We have, $(3^{7} \div 3^{5})^{4}=\Big(\frac{3^{7}}{3^{5}}\Big)^{4}=(3^{7} \div 3^{5})^{4}$
$=(3^{2} \div 3^{4})^{2\times4}$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$=3^{8}$
$\Big[\because (\text{a}^{\text{m}})^{\text{n}}={\text{mn}}\Big]$

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