2 Marks Questions

Question 512 Marks

Express the following in exponential form: $\frac{-1296}{14641}$

Answer

Given, $\frac{-1296}{14641}$
Since, $(6) × (6) × (6) × (6) = 1296 = (6)^4$ and $11 × 11 × 11 × 11 = 14641 = (11)^4$
Exponential form of $\frac{-1296}{14641}=-\frac{(6)^4}{(11)^4}=-\big(\frac{6}{11}\big)^4$

View full question & answer→

Question 522 Marks

A sugar factory has annual sales of $3$ billion $720$ million kilograms of sugar. Express this number in the standard form.

Answer

Annual sales of a sugar factory $= 3$ billion $720$ million kilograms $= 3720000\ kg$
Standard form of $3720000 = 372 \times 10 \times 10 \times 10 \times 10$
$ =372 \times 10^4 \mathrm{~kg} $
$ =3.72 \times 10^4 \times 10^2 $
$ =3.72 \times 10^6 \mathrm{~kg}\left[\because a^{\mathrm{m}} \times a^{\mathrm{n}}=(a)^{\mathrm{m}+\mathrm{n}}\right]$

View full question & answer→

Question 532 Marks

For hook-up, determine whether there is a single repeater machine that will do the same work. If so, describe or draw it.

Answer

Using law of exponents, $(a^m× a^n= a^{m+n})$ $[\because$ a is non-zero integer$]$
Hook-up machine work $= 12^2× 12^3= 12^5$
So, $(x12^5)$ machine can do the same work.
Diagram of single $x12^5$ machine.

View full question & answer→

Question 542 Marks

The left column of the chart lists the lengths of input chains of gold.Repeater machines are listed across the top. The other entries are the outputs you get when you send the input chain from that row through the repeater machine from that column. Copy and complete the chart.

Input Length	Repeater Machine
	$x2^3$
	$40$		$125$
$2$
		$162$

Answer

In the given table, the left column of the chart lists is the length of input chains of gold. Thus, the output we get when we send the input chain from the row through the repeater machine are detailed in the following table.

Input Length	Repeater Machine
	$x3$	$x12$	$x9$
$13.3$	$40$	$160$	$125$
$2$	$6$	$24$	$18$
$13.5$	$41$	$162$	$121$

View full question & answer→

Question 552 Marks

Express the following in exponential form: $\frac{400}{3969}$

Answer

Given, $\frac{400}{3969}$
Since, $20 × 20 = 400 = (20)^2$ and $63 × 63 = 3969 = (63)^2$
Exponential form of $\frac{-400}{3969}=\frac{(20)^2}{(63)^2}=-\big(\frac{20}{63}\big)^2$

View full question & answer→

Question 562 Marks

Express the following in exponential form: $\frac{-125}{343}$

Answer

Given, $\frac{-125}{343}$
Since, $(-5) × (-5) × (-5) = -125 = (-5)^3$ and $7 × 7 × 7 =343 = (-7)^3$
Exponential form of $\frac{-125}{343}=\frac{(-5)^3}{(7)^3}=\big(\frac{-5}{7}\big)^3$

View full question & answer→

Question 572 Marks

Shrinking Machine: In a shrinking machine, a piece of stick is compressed to reduce its length. If $9\ cm$ long sandwich is put into the shrinking machine below, how many cm long will it be when it emerges?

Answer

According to the question, in a shrinking machine, a piece of stick is compressed to reduce its length. If $9\ cm$ long sandwich is put into the shrinking machine, then the length of sandwich will be $9\times\frac{1}{3^{-1}}=9\times3=27\text{cm}.$

View full question & answer→

Question 582 Marks

The left column of the chart lists the lengths of input pieces of ribbon. Stretching machines are listed across the top. The other entries are the outputs for sending the input ribbon from that row through the machine from that column. Copy and complete the chart.

Input Length	Machine
	$x2$
	$1$	$5$
$3$				$15$
	$14$		$7$

Answer

In the given table, the left column of chart list is the length of input piece of ribbon. Thus, the outputs for sending the input ribbon are given in the following table.

Input Length	Machine
	$x2$	$x10$	$x5$
$5$	$1$	$5$	$2.5$
$3$	$6$	$30$	$15$
$7$	$14$	$70$	$35$

View full question & answer→

Question 592 Marks

Ajay had a $1\ cm$ piece of gum. He put it through repeater machine given below and it came out $\frac{1}{100,000}$ cm long. What is the missing value?

Answer

Since, Ajay put a $1\ cm$ pieace of gum and came out $\frac{1}{100000}$. So, it is shrinking machine.
Hence, it is a $\Big(\text{x}\frac{1^1}{10}\Big)^5$ type shrinking machine. Thus, missing value is $5$.

View full question & answer→

Question 602 Marks

Find the product of the cube of $(-2)$ and the square of $(+4)$.

Answer

$\because$ Cube of $(-2) = (-2)^3$
and square of $(+4) = (+4)^2$
$\therefore$ The product $= (-2)^3\times (4)^2$
$= (-8) \times 16$
$= -128$

View full question & answer→

Question 612 Marks

By what number should $(–8)^{-3}$ be multiplied so that that the product may be equal to $(–6)^{-3}?$

Answer

Let $x$ be the number multiplied with $(-8)^{-3}= (-6)^{-3}$
$\Rightarrow \ \text{x}\times(-8)^{-3}=(-6)^{-3}$
$\Rightarrow \ \text{x}=\frac{(-6)^{-3}}{(-8)^{-3}}=\frac{(-8)^{-3}}{(-6)^{-3}}=\frac{512}{216}=\frac{64}{27}$

View full question & answer→

Question 622 Marks

Find the multiplicative inverse of $(-7)^{-2}÷ (90)^{-1}$.

Answer

a is called multiplicative inverse of b, if $a \times b = 1$
We have, $(-7)^{-2}÷ (90)^{-1}$
$=\frac{1}{(7)^2}+\frac{1}{(90)^1}$
$=\frac{1}{49}\div\frac{1}{90}$
$=\frac{1}{49}\times\frac{90}{1}$
$=\frac{90}{49}$
$\Big[\because$ $(-a)^m= a^m$ if $m$ is an even number and $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}\Big]$
Put $\text{b}=\frac{90}{49}$
$\therefore$ $\text{a}\times\frac{90}{49}=1$
$\Rightarrow\text{a}\times\frac{90}{49}$

View full question & answer→

Question 632 Marks

Express $\frac{1.5\times10^6}{2.5\times10^{-4}}$ in the standard form.

Answer

Given,
$\frac{1.5\times10^6}{2.5\times10^{-4}}=\frac{15}{25}\times10^{6+4}$ [$\because$ $a^m÷ a^n= (a)^{m-n}$]
$=\frac{3}{5}\times10^{10}$
$=0.6\times10^{10}$
$=0.6\times10^{10}\times10^{-1}$
$=6\times10^9$ [$\because$ $a^m× a^n= (a)^{m-n}$]

View full question & answer→

Question 642 Marks

Mass of Mars is $6.42 \times 10^{29} \mathrm{~kg}$ and mass of the Sun is $1.99 \times 10^{30} \mathrm{~kg}$. What is the total mass?

Answer

Mass of Mars $=6.42 \times 10^{29} \mathrm{~kg}$
Mass of the Sun $=1.99 \times 10^{30} \mathrm{~kg}$
Total mass of Mars and Sun together $=6.42 \times 10^{29}+1.99 \times 10^{30}$
$=6.42 \times 10^{29}+19.9 \times 10^{29}=26.32 \times 10^{29} \mathrm{~kg}$

View full question & answer→

Question 652 Marks

Express the following in standard form: A Helium atom has a diameter of $0.000000022\ cm$.

Answer

A helium atom has a diameter of $0.000000022\ cm$.
Standard form of $0.000000022\ cm$
$ =0.22 \times 10^{-7} $
$ =2.2 \times 10^{-7} \times 10^{-1} $
$ =2.2 \times 10^{-8} \mathrm{~cm}\left[\because a^{\mathrm{m}} \times a^{\mathrm{n}}=(a)^{\mathrm{m}+\mathrm{n}}\right] $

View full question & answer→

Question 662 Marks

Fill in the blanks:

Answer

$144\times2^{-3}=144\times\frac{1}{8}=18$
$18\times12^{-1}=18\times\frac{1}{12}=\frac{3}{2}$
$\frac{3}{2}\times3^{-2}=\frac{3}{2}\times\frac{1}{3^{2}}=\frac{1}{2\times3}=\frac{1}{6}$

View full question & answer→

Question 672 Marks

The diameter of the Sun is $1.4 \times 10^9m$ and the diameter of the Earth is $1.2756 \times 10^7m$. Compare their diameters by division.

Answer

Diameter of the sun $= 1.4 \times 10^9m$
Diameter of the Earth $= 1.2756 \times 10^7m$
For comparison, we have to change both diameter in same powers of $10$ i.e. $1.2756 \times 10^7m = 0.012756 \times 10^9$
Hence, if we divide diameter of sun by diameter of Earth,we get
$\frac{1.4 \times 109\text{m}}{0.012756\times10^9}=110$
So, diameter of sun is $110$ times the diameter of Earth.

View full question & answer→

Question 682 Marks

The paper clip below has the indicated length. What is the length in standard form. Length of the paper clip $= 0.05m$.
In standard form, $0.05m$ $=0.5 \times 10^{-1}=5.0 \times 10^{-2} \mathrm{~m}$.
Hence, the length of the paper clip in standard form is $5.0 \times 10^{-2} \mathrm{~m}$.

Answer

Length of the paper clip $= 0.05m$.
In standard form, $0.05\ m$ $=0.5 \times 10^{-1}=5.0 \times 10^{-2} \mathrm{~m}$.
Hence, the length of the paper clip in standard form is $5.0 \times 10^{-2} \mathrm{~m}$.

View full question & answer→

MATHS 2 Marks Questions