Question 12 Marks
Simplify and write in exponential form.
(i) $(-2)^{-3} \times(-2)^{-4}$
(ii) $p^3 \times p^{-10}$
(iii) $3^2 \times 3^{-5} \times 3^6$
Answer(i) $(-2)^{-7}$ $\qquad$ (ii) $p^{-7}$ $\qquad$ (iii) $3^3$
View full question & answer→Question 22 Marks
Expand the following numbers using exponents.
(i) 1025.63 $\qquad$ (ii) 1256.249
Answer
$\begin{array}{l} \text { (i) } 1 \times 10^3+2 \times 10^1+5 \times 10^0+6 \times 10^{-1}+3 \times 10^{-2} \\ \text { (ii) } 1 \times 10^3+2 \times 10^2+5 \times 10^1+6 \times 10^0+2 \times 10^{-1} +4 \times 10^{-2}+9 \times 10^{-3}\end{array}$
View full question & answer→Question 32 Marks
Find the multiplicative inverse of the following
(i) $2^{-4}$ $\qquad$ (ii) $10^{-5}$ $\qquad$ (iii) $7^{-2}$
(iv) $5^{-3}$ $\qquad$ (v) $10^{-100}$
Answer(i) $2^4$ $\qquad$ (ii) $10^5$ $\qquad$ (iii) $7^2$ $\qquad$ (iv) $5^3$ $\qquad$ (v) $10^{100}$
View full question & answer→Question 42 Marks
Simplify
(i) $\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}(t \neq 0)$
(ii) $\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$
Answer(i) We have,
$
\begin{array}{l}
\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \\
=\frac{25 \times(5)^3 \times t^8}{10 \times t^4} \quad\left[\because a^{-m}=\frac{1}{a^m}\right] \\
=\frac{25 \times 125 \times t^{8-4}}{10} \quad\left[\because \frac{a^m}{a^n}=a^{m-n}\right] \\
=\frac{25 \times 125}{10} \times t^4 \\
=\frac{5 \times 125}{2} \times t^4 \\
=\frac{625}{2} t^4
\end{array}
$
(ii) Ans. $5^5$
View full question & answer→Question 52 Marks
Evaluate.
(i) $\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}$
(ii) $\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}$
Answer(i)
We have, $\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}$
$
\begin{array}{ll}
=\left\{\frac{(1)^{-1}}{(3)^{-1}}-\frac{(1)^{-1}}{(4)^{-1}}\right\}^{-1} \qquad {\left[\because\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\right]} \\
=\left\{\frac{3}{1}-\frac{4}{1}\right\}^{-1} \qquad {\left[\because a^{-m}=\frac{1}{a^m}\right]} \\
=(3-4)^{-1}=(-1)^{-1}=\frac{1}{(-1)^1}\qquad \left[\because a^{-m}=\frac{1}{a^m}\right] \\
=\frac{1}{-1}=-1
\end{array}
$
(ii) Ans. $\frac{512}{125}$
View full question & answer→Question 62 Marks
Find the value of $m$ for which $5^m+5^{-3}=5^5$.
View full question & answer→Question 72 Marks
Evaluate
(i) $\frac{8^{-1} \times 5^3}{2^{-4}}$
(ii) $\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}$
Answer(i) We have,
$
\begin{array}{l}
\frac{8^{-1} \times 5^3}{2^{-4}} \\
=\frac{2^4 \times 5^3}{8} \quad\left[\because a^{-m}=\frac{1}{a^m}\right] \\
=\frac{16 \times 125}{8}=2 \times 125=250
\end{array}
$
(ii) Ans. $\frac{1}{60}$
View full question & answer→Question 82 Marks
Evaluate
(i) $3^{-2}$ $\qquad$ (ii) $(-4)^{-2}$ $\qquad$ (iii) $\left(\frac{1}{2}\right)^{-5}$
Answer(i) $\frac{1}{9}$ $\qquad$ (ii) $\frac{1}{16}$ $\qquad$ (iii) 32
View full question & answer→Question 92 Marks
If $5^{3 x-1}÷ 25=125$, then find the value of $x$
AnswerWe have, $5^{3 x-1}÷25=125$
$
\begin{array}{l}
\Rightarrow \quad 5^{3 x-1} \times \frac{1}{25}=125 \Rightarrow \frac{5^{3 x}}{5} \times \frac{1}{5 \times 5}=5 \times 5 \times 5 \\
\Rightarrow \quad 5^{3 x} \times \frac{1}{5^3}=5^3 \Rightarrow 5^{3 x}=5^3 \times 5^3 \\
\Rightarrow \quad 5^{3 x}=5^{3+3}=5^6 \\
\Rightarrow \quad 3 x=6 \quad[\because \text { bases are same }] \\
\Rightarrow \quad x=6 \times \frac{1}{3}=2
\end{array}
$
View full question & answer→Question 102 Marks
Express the product of $3.2 \times 10^6$ and $4.1 \times 10^{-1}$ in the standard form.
AnswerWe have, $\left(3.2 \times 10^6\right) \times\left(4.1 \times 10^{-1}\right)$
$
\begin{array}{l}
=3.2 \times 10^6 \times 4.1 \times 10^{-1}=3.2 \times 4.1 \times 10^{6-1} \\
=13.12 \times 10^5=1.312 \times 10 \times 10^5=1.312 \times 10^6
\end{array}
$
View full question & answer→Question 112 Marks
Fill in the Blanks.
AnswerWe have, $144 \times 2^{-3}=\frac{144}{2^3}=\frac{12 \times 12}{2 \times 2 \times 2}=18$
Now, $\quad 18 \times 12^{-1}=\frac{18}{12}=\frac{3}{2}$
Again, $\quad \frac{3}{2} \times 3^{-2}=\frac{3}{2} \times \frac{1}{3^2}=\frac{1}{6}$
View full question & answer→Question 122 Marks
The cells of a bacteria double itself every hour. How many cells will there be after 12 h , if initially we start with 1 cell? Express the answer in powers.
AnswerThe cells of a bacteria double itself every hour i.e. the cells of a bacteria after $1 h=2$
$\therefore$ The cells of a bacteria after $2 h=2 \times 2=2^2$
The cells of a bacteria after $3 h=2 \times 2^2=2^3$
The cells of a bacteria after $12 h=2^{12}$
View full question & answer→Question 132 Marks
Convert 5 hectares in $m ^2$ and express it in standard form.
Answer
$\begin{array}{l}\because 1 \text { hectare }=10000 m^2 \\ \therefore 5 \text { hectares }=5 \times 10000 m^2=50000 m^2=5.0 \times 10^4 m^2\end{array}$
View full question & answer→Question 142 Marks
If $a=-1$ and $b=2$, then find the value of the following.
(i) $a^b+b^a$ $\qquad$ (ii) $a^b-b^a$ $\qquad$
(iii) $a^b \times b^a$ $\qquad$ (iv) $a^b \div b^a$
Answer(i) We have,
$\begin{aligned} a^b+b^a & =(-1)^2+(2)^{-1}=1+\frac{1}{2} \\ & =\frac{2+1}{2}=\frac{3}{2}\end{aligned}$
(ii) We have, $a^b-b^a=(-1)^2-(2)^{-1} \quad\left[\because a^{-n}=\frac{1}{a^n}\right]$
$
=1-\frac{1}{2}=\frac{2-1}{2}=\frac{1}{2}
$
(iii) We have, $a^b \times b^a=(-1)^2 \times(2)^{-1}=1 \times \frac{1}{2}=\frac{1}{2}$
$
\left[\because a^{-n}=\frac{1}{a^n}\right]
$
(iv) We have, $a$
$
\begin{aligned}
a^b \div b^a= & (-1)^2 \div(2)^{-1}=(-1)^2 \times \frac{1}{(2)^{-1}} \\
& =1 \times 2=2 \quad\left[\because \frac{1}{a^{-n}}=a^n\right]
\end{aligned}
$
View full question & answer→Question 152 Marks
If $\frac{5^m \times 5^3 \times 5^-2}{5^5}=5^{12}$. then find the value of $m$.
AnswerGiven, $\frac{5^m \times 5^3 \times 5^{-2}}{5^{-5}}=5^{12}$
$\Rightarrow \qquad \frac{5^{m+3-2}}{5^{-5}}=5^{12}$
$\Rightarrow \qquad 5^{m+1+5}=5^{12}$
$\Rightarrow \qquad(5)^{m+6}=(5)^{12}$
$\Rightarrow \qquad m+6=12 \qquad[\because$ bases are same $]$
$\Rightarrow \qquad m=12-6=6$
View full question & answer→Question 162 Marks
By what number should $(-15)^{-1}$ be divided, so that quotient may be equal to $(-15)^{-1}$ ?
AnswerLet $x$ be the required number.
Then, $\quad \frac{(-15)^{-1}}{x}=(-15)^{-1}$
$
\begin{array}{l}
\Rightarrow \quad \frac{1}{(-15) x}=\frac{1}{(-15)} \quad\left[\because(a)^{-1}=\frac{1}{a}\right] \\
\Rightarrow \quad \frac{1}{x}=1 \Rightarrow x=1
\end{array}
$
So, $(-15)^{-1}$ should be divided by 1 to get quotient $(-15)^{-1}$.
View full question & answer→Question 172 Marks
An electron's mass is approximately $9.1093826 \times 10^{-31} kg$. What is this mass in grams?
Answer
$
\begin{array}{l}
\text { An electron's mass }=91093826 \times 10^{-31} kg \\
\because \quad 1 kg=1000 g=10 \times 10 \times 10 g=10^3 g
\end{array}
$
$\therefore$ An electron's mass in grams
$
\begin{array}{l}
=9.1093826 \times 10^{-31} \times 10^3 g \\
=91093826 \times 10^{-31+3} g \\
=9.1093826 \times 10^{-28} g
\end{array}
$
View full question & answer→Question 182 Marks
Express $\frac{1.5 \times 10^6}{2.5 \times 10^4}$ in the standard form.
AnswerWe have, $\frac{1.5 \times 10^6}{2.5 \times 10^4}$
$
\begin{array}{l}
=\frac{1.5 \times 10^2 \times 10^4}{2.5 \times 10^4}=\frac{150}{2.5} \times 10^{4-4} \\
=\frac{150}{2.5} \times 10^0=60 \times 1=60 \\
=6.0 \times 10^1=6.0 \times 10
\end{array}
$
View full question & answer→Question 192 Marks
If $\frac{6^n}{6^{-2}}=6^3$, then find the value of $n$
Answer
$\begin{array}{l}\because \frac{6^n}{6^{-2}}=6^3 \\ \therefore 6^{n+2}=6^3 \Rightarrow n+2=3 \Rightarrow n=3-2 \\ \Rightarrow n=1 \quad \quad[\because \text { bases of two equal numbers are } \text { same, so exponents will be same }]\end{array}$
View full question & answer→Question 202 Marks
Express as a power of a rational number with negative exponent.
(i) $\left[\left(-\frac{3}{2}\right)^{-2}\right]^{-3}$
(ii) $\left(2^5+2^8\right) \times 2^{-7}$
Answer(i)
$
\begin{aligned}
{\left[\left(-\frac{3}{2}\right)^{-2}\right]^{-3} } & =\left(-\frac{3}{2}\right)^{-2 \times(-3)} =\left(-\frac{3}{2}\right)^6=\left(-\frac{2}{3}\right)^{-6}
\end{aligned}
$
(ii)
$
\begin{aligned}
\left(2^5+2^8\right) \times 2^{-7} & =\left(2^5 \times \frac{1}{2^8}\right) \times 2^{-7}=2^{5-8} \times 2^{-7} \\
& =2^{-3} \times 2^{-7}=2^{-3+(-7)}=2^{-3-7} \\
& =2^{-10}
\end{aligned}
$
View full question & answer→Question 212 Marks
Solve the following
(i) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{2}\right)^{-3}$
(ii) $\left(\frac{2}{3}\right)^{-2} \times\left(\frac{2}{3}\right)^5$
Answer(i)
$
\begin{aligned}
\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{2}\right)^{-3} & =\left(\frac{1}{2}\right)^{-2} \times\left(\frac{1}{2}\right)^3 =\left(\frac{1}{2}\right)^{-2+3}=\left(\frac{1}{2}\right)^1=\frac{1}{2}
\end{aligned}
$
(ii)
$
\begin{aligned}
\left(\frac{2}{3}\right)^{-2} \times\left(\frac{2}{3}\right)^5 & =\left(\frac{2}{3}\right)^{-2+5}=\left(\frac{2}{3}\right)^3=\frac{(2)^3}{(3)^3} \\
& =\frac{2 \times 2 \times 2}{3 \times 3 \times 3}=\frac{8}{27}
\end{aligned}
$
View full question & answer→Question 222 Marks
Find the value of $x^{-3}$, if $x=(100)^{1-4}+(100)^0$
Answer
$\begin{aligned} \because \quad x & =(100)^{1-4}+(100)^0=(100)^{-3} \quad\left[\because(100)^{0}=1\right] \\ \therefore x^{-3} & =\frac{1}{x^3}=\frac{1}{\left[(100)^{-3}\right]^3}=\frac{1}{100^{-9}} \quad\left[\because\left(a^m\right)^n=a^{m m}\right] \\ & =100^9\end{aligned}$
View full question & answer→Question 232 Marks
Write the value of $(13)^{-13}+(13)^{13}$.
Answer$(13)^{-13}+(13)^{13}=( r 3)^{-13} \times \frac{1}{(13)^{13}}=\frac{1}{(13)^{26}}$
View full question & answer→Question 242 Marks
Find the value of $\left[2^{-1} \times 3^{-1}\right]^{-1}$.
Answer
$\begin{array}{l}\because 2^{-1} \times 3^{-1}=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6} \\ \therefore\left[2^{-1} \times 3^{-1}\right]^{-1}=\left[\frac{1}{6}\right]^{-1}=6\end{array}$
View full question & answer→Question 252 Marks
Express 56 km in metres and then in standard form.
Answer
$\begin{array}{l}\because 1 km=1000 m \\ \therefore 56 km=56000 m=5.6 \times 10^4 m\end{array}$
View full question & answer→Question 262 Marks
Find the product of the cube of $(-2)$ and the square of $(+4)$.
Answer
$\begin{array}{l}\text { Cube of }(-2)=(-2)^3=-2 \times-2 \times-2=-8 \\ \text { Square of }(+4)=(4)^2=4 \times 4=16 \\ \therefore \quad \text { Product }=-8 \times 16=-128\end{array}$
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