Question 11 Mark
$\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)=\Big(\frac{-1}{4}\Big)^{11}$
Answer$\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{11}$ Solution: Let $\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)=\Big(\frac{-1}{4}\Big)^{11}$ $\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\frac{\Big(\frac{-1}{4}\Big)^{11}}{\Big(\frac{-1}{4}\Big)^{3}}$ $\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{\text{11-3}}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}}\text{m}>\text{n}\big]$ $\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{8}$ Since, base are equal do, ny equating the power the pawers, we get $\text{x}=8$ $\therefore\Big(-\frac{1}{4}\Big)^{3}\times\Big(-\frac{1}{4}\Big)^{8}=\Big(-\frac{1}{4}\Big)^{11}$
View full question & answer→Question 21 Mark
$\frac{-3^{100}}{5}=\frac{-3^{100}}{-5^{100}}$
Answer$\Big(\frac{-3}{5}\Big)=\Big(\frac{-1\times3}{5}\Big)^{100}$
$\big[\because-3=1\times3\big]$
$\frac{(-1)^{100}\times3^{100}}{5^{100}}$
$\big[\because(\text{a}\times\text{b}^{\text{m}})=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\big]$
$\frac{1\times3^{100}}{5^{100}}$
$\big[\because(-1)^{\text{n}}=1,\text{ if }\text{n}\text{ is even}\big]$
$\frac{3^{100}}{5^{100}}$
Now, taking $RHS,$ we have $\frac{-3^{100}}{-5^{100}}=\frac{3^{10}}{5^{100}}$
$[\because$ if both numerator and denominator have negative sign, then it is cacelled out$]$
$\therefore LHS = RHS$
Hence, $\frac{-3^{100}}{-5^{100}}=\frac{3^{10}}{5^{100}}$
View full question & answer→Question 31 Mark
Identify the greater number, in the following:
$2^9$ or $9^2$
AnswerWe have, $2^9$
$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=512$ and $9^2$
$=9 \times 9$
$=81$
So, $2^9>9^2$
View full question & answer→Question 41 Mark
Express the following in exponential form:
$3 \times 3 \times 3 \times a \times a \times a \times a$
AnswerWe know that,
$a \times a \times \ldots \times a(n$ times $)=a^n$
$3 \times 3 \times 3 \times a \times a \times a \times a$
$=3^3 \times a^4$
$=27 a^4$
View full question & answer→Question 51 Mark
Find the value of: $7^0$
Answer$\Big[\because\text{a}^{0}=1\Big] 7^0 = 1$
View full question & answer→Question 61 Mark
$x^m \times y^m=(x \times y)^{2 m}$, where $x$ and $y$ are non-zero rational numbers and $m$ is a positive integer.
AnswerIf $a$ and $b$ are rational numbers, then
$x^m \times y^m=(a b)^m$
$x^m \times y^m=(x y)^m=(x \times y)^m$
Hence, $x^{\mathrm{m}} \times \mathrm{y}^{\mathrm{m}} \neq\left(\mathrm{x} \times \mathrm{y}^{2 \mathrm{~m}}\right)$
View full question & answer→Question 71 Mark
Express the following in single exponential form:
$(-3)^3 \times(-10)^3$
AnswerWe have, $(-3)^3 \times(-10)^3$
$=[(-3) \times(-10)]^3\left[\because a^m \times b^m=(a \times b)^m\right]$
${[\because(-3) \times(-10)=30]}$
$=(30)^3$
View full question & answer→Question 81 Mark
$5^0 \times 25^0 \times 125^0=\left(5^0\right)^6$
AnswerTrue.
Solution:
Here, $5^0 \times 25^0 \times 125^0$
$=5^0 \times(5 \times 5)^0 \times(5 \times 5 \times 5)^0$
${[\therefore 25=5 \times 5 \text { and } 125=5 \times 5 \times 5]}$
$=5^0 \times 5^0 \times 5^0 \times 5^0 \times 5^0 \times 5^0\left[\therefore a^m \times b^m=a^m b^m\right]$
$=\left(5^0\right)^6$
Hence, $5^0 \times 25^0 \times 125^0$
$=\left(5^0\right)^6$
View full question & answer→Question 91 Mark
Find the value of:
$\left(2^0+3^0+4^0\right)\left(4^0-3^0-2^0\right)$
Answer$\left(2^0+3^0+4^0\right)\left(4^0-3^0-2^0\right)$
$=(1+1+1)(1+1+1)$
${\left[\because a^0=1\right]}$
$=(3)(-1)=-3$
View full question & answer→Question 101 Mark
$(-2)^{31} \times(-2)^{13}=(-2)$____.
Answer$(-2)^{31} \times(-2)^{31}=(-2)^{44}$
Solution:
Here, $(-2)^{31} \times(-2)^{31}=(-2)^{44}\left[\because \mathrm{a}^{\mathrm{m} \times \mathrm{n}}=\mathrm{a}^{\mathrm{m}+\mathrm{n}}\right]$
$=(-2)^{44}$
$\therefore(-2)^{31} \times(-2)^{13}=(-2)^{44}$
View full question & answer→Question 111 Mark
Find the value of:
$2 \times 3 \times 4 \div 2^0 \times 3^0 \times 4^0$
Answer$2 \times 3 \times 4 \div 2^0 \times 3^0 \times 4^0$
$=2 \times 3 \times 4+1 \times 1 \times 1$
${\left[\because \mathrm{a}^0=1\right]}$
$=\frac{2 \times 3 \times 4}{1 \times 1 \times 1}=2 \times 3 \times 4=24$
View full question & answer→Question 121 Mark
Find the value of:
$2^5$
AnswerWe knoe that, $a^n=a \times a \times a \times \ldots \times a\left(n\right.$ times)
$2^5=2 \times 2 \times 2 \times 2 \times 2=32$
View full question & answer→Question 131 Mark
One million $= 10^7$
AnswerOne million $= 10$ lakhs
$= 1000000 = 10^6$
Hence, $10^{6}\neq10^{7}$
View full question & answer→Question 141 Mark
$(10+10)^{10}=10^{10}+10^{10}$
AnswerFalse.
Soluton:
We know that, $(a \times b)^{\mathrm{m}}$
$=a^m \times b^m$
$\text { So, }(10 \times 10)^{10}$
$=10^{10} \times 10^{10}$
View full question & answer→Question 151 Mark
$10,000$ _____ $10^5$
Answer$10,000 \underline < 10^5$
Solution:
$\because 1000=10^4$
So, $10^4<10^5$
[ $\because$ base are same, so if power is greater then the number is greater]
View full question & answer→Question 161 Mark
$340900000 = 3.409 \times 10 $_____.
Answer$340900000=3.409 \times 10^8$
Solution:
For standard form, $340900000=3409 \times 10^5$
Also, $3409=3.409 \times 10^3$
$\text { So, } 3.409 \times 10^3 \times 10^5$
$=3.409 \times 10^8\left[\therefore a^m \times a^n=a^{m+n}\right]$
$340900000=3.409 \times 10^8$
View full question & answer→Question 171 Mark
$x^m \times x^n=x^m+n$, where $x$ is a non-zero rational number and $m, n$ are positive integers.
AnswerIf $x$ is a rational number and m and n are positive integers, then
$a^m \times a^n$
$=a^{m+n}$
View full question & answer→Question 181 Mark
Identify the greater number, in the following: $2^6$ or $6^2$
Answer$\text { We have, } 2^6$
$=2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=64 \text { and } 6^2=6 \times 6$
$=36 \text { So, } 2^6>6^2$
View full question & answer→Question 191 Mark
$\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)$
Answer$\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$Solution:
Here, $\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}$
$=\Big(\frac{6}{13}\Big)^{0}+\Big(\frac{6}{13}\Big)^{5\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}+\text{a}^{\text{mn}}$
$\Big(\frac{6}{13}^{10}+\Big(\frac{6}{13}\Big)^{10}=\Big(\frac{6}{13}\Big)^{10-10}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}},(\text{m}>\text{n})\big]$
$=\Big(\frac{6}{13}\Big)^{0}$
$\therefore\Big(\frac{6}{13}\Big)^{10}+\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$
View full question & answer→Question 201 Mark
$\Big(\frac{4}{3}\Big)^{5}\times\Big(\frac{5}{7}\Big)^{5}=\Big(\frac{4}{3}+\frac{5}{7}\Big)^{5}$
AnswerFalse. Solution: Here,$\Big(\frac{4}{3}\Big)^{5}\times\Big(\frac{5}{7}\Big)^{5}=\Big(\frac{4}{3}\times\frac{5}{7}\Big)^{5}$$\big[\because\text{a}^{\text{m}}\times\text{b}^{\text{m}}=(\text{ab})^{\text{m}}\big]$ and $\Bigg[\Big(\frac{4}{3}\Big)^{5}+\Big(\frac{5}{7}\Big)^{5}\Bigg]^{5}$ $\Big[=\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{c}}{\text{d}}\Big)=\frac{\text{a}}{\text{b}}\times\frac{\text{d}}{\text{c}}\Big]$ $=\Big(\frac{4}{3}\times\frac{5}{7}\Big)^{5}$ Hence, $\Big(\frac{4}{3}\times\frac{5}{7}\Big)^{5}\neq\Big(\frac{4}{3}\times\frac{7}{5}\Big)^{5}$
View full question & answer→Question 211 Mark
One hour $= 60^2$ seconds.
Answer$1h = 60$ min
$= 60 \times 60s$
$= 60^2s$
View full question & answer→Question 221 Mark
Express the following in usual form: $1.75 × 10^{-3}$
AnswerHere, $1.75\times10^{-3}=\frac{175}{100}\times\frac{1}{10}^{3}$
$=\frac{175}{100000}=0.00175$ $\Big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
View full question & answer→Question 231 Mark
Find the value of: $7^7 \div 7^7$
Answer$7^{7}\div7^{7}=\frac{7^{7}}{7^{7}}=7^{7-7}$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$=7^{0}=1$
View full question & answer→Question 241 Mark
$\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}=\Big(\frac{7}{3}\Big)^{10}$
Answer False.
Solution:
Here,$\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}=\Big(\frac{7}{3}\Big)^{2+5}$
$=\Big(\frac{7}{3}\Big)^{7}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\big]$
Here, $\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}\neq\Big(\frac{7}{3}\Big)^{10}$
View full question & answer→Question 251 Mark
$53700000=$________$\times 10^7$
AnswerGiven, $53700000$
For standard form, $53700000$
$=537 \times 10^5 \text { Also, } 537$
$=5.37 \times 10^2$
So, $5.37 \times 10^2 \times 10^5$
$=5.37 \times 10^7$
$53700000=5.37 \times 10^7$
View full question & answer→Question 261 Mark
$53700000=$______$\times 10^7$
AnswerGiven, $53700000$
For standard form, $53700000$
$=537 \times 10^5 \text { Also, } 537$
$=5.37 \times 10^2$
$\text { So, } 5.37 \times 10^2 \times 10^5$
$=5.37 \times 10^7$
$53700000=5.37 \times 10^7$
View full question & answer→Question 271 Mark
Express the following numbers in standard form:
$5,83,00,00,00,000$
Answer$5,83,00,00,00,000$
$=583000000000.00$
$=583 \times 10^9$
$=5.83 \times 10^2 \times 10^9$
$=5.83 \times 10^{11}$
View full question & answer→Question 281 Mark
$600060=6 \times 10^5+6 \times 10^2$
AnswerFalse.
Solution:
$\text { Take RHS }=6 \times 10^5+6 \times 10^2$
$=6 \times 100000+6 \times 100$
$=600000+600=600600\neq\text{LHS}$
Hence, $\text{RHS}\neq\text{LHS}$
View full question & answer→Question 291 Mark
$(-3)^4 = -12$
AnswerFalse.
Solution:
$(-3)^4=(-3) \times(-3) \times(-3) \times(-3)$
$=81\neq-12$
View full question & answer→Question 301 Mark
Express the following in single exponential form: $(-11)^2 \times(-2)^2$
AnswerWe have,$(-11)^2 \times(-2)^2$
$=[(-11) \times(-2)]^2\left[\because a^m \times b^m=(a \times b)^m\right]$
${[\because(-11) \times(-2)=22]}$
$=22^2$
View full question & answer→Question 311 Mark
$3^4 > 4^3$
Answer$\therefore3^{4}=3\times3\times3\times3=81$
and $4^3=4 \times 4 \times 4=64$
$81>64$
Hence, $3^4 > 4^3$
View full question & answer→Question 321 Mark
$4 \times 10^5+3 \times 10^4+2 \times 10^3+1 \times 10^0 = 432010$
AnswerTake $LHS$
$=4 \times 10^5+3 \times 10^4+2 \times 10^3+1 \times 10^0$
$=4 \times 100000+3 \times 10000+2 \times 1000+1 \times 1\left[\therefore a^0=1\right]$
$=400000+30000+2000+1$
$=432001\neq\text{RHS}$
Hence, $\text{LHS}\neq\text{RHS}$
View full question & answer→Question 331 Mark
Find the value of: $(-7)^{2 \times 7-6-8}$
Answer$(-7)^{2 \times 7-6-8}$
$=(-7)^{14-14}\left[\because a^0=1\right]$
$=(-7)^0$
$=1$
View full question & answer→Question 341 Mark
$(-3)^8 \div(-3)^5=(-3)$____.
Answer$(-3)^8 \div(-3)^5=(-3)^3$
Solution:
Here, $(-3)^8 \div(-3)^5=(-3)^{8-5}=(-3)^3\left[\because a^m+a^n=a^{m-n}, m>n\right]$
$\therefore(-3)^8 \div(-3)^5=(-3)^3$
View full question & answer→Question 351 Mark
$\Big(\frac{2}{5}\Big)^{3}\div\Big(\frac{5}{2}\Big)^{3}=1$
Answer False.
Solution:
Here, $\Big(\frac{2}{5}\Big)^{3}\div\Big(\frac{5}{2}\Big)^{3}$ $\big[\because=\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{c}}{\text{d}}\Big)=\frac{\text{a}}{\text{b}}\times\frac{\text{d}}{\text{c}}\big]$
$=\Big(\frac{2}{5}\Big)^{3}\times\Big(\frac{2}{5}\Big)^{3}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}\text{}^{\text{m+n}}\big]$
$\Big(\frac{2}{5}\Big)^{3+3}=\Big(\frac{5}{2}\Big)^{6}$
Here, $\Big(\frac{2}{5}\Big)^{3}\div\Big(\frac{5}{2}\Big)^{3}\neq1$
View full question & answer→Question 361 Mark
$6^3$ _____ $4^4$
Answer$6^3 \leq 4^4$
$\because 6^3=6 \times 6 \times 6=216$
$\text { and } 4^4=4 \times 4 \times 4 \times 4=256$
So, $216<256$
$\therefore 6^3<4^4$
View full question & answer→Question 371 Mark
Express the following in exponential form:
$s \times s \times t \times t \times s \times s \times t$
AnswerWe know that,
$a \times a \times \ldots \times a(n \text { times })=a^n$
$s \times s \times t \times t \times s \times s \times t$
$=s \times s \times s \times s \times t \times t \times t$
$=s^4 \times t^3$
View full question & answer→Question 381 Mark
$27500000=2.75 \times 10$_____.
AnswerGiven, $27500000$
For standard form, $27500000=275 \times 10^5$ Also, $275$
$=2.75 \times 10^2$
$\text { So, } 2.75 \times 10^2 \times 10^5$
$=2.75 \times 10^7\left[\therefore a^m \times a^n=a^{m+n}\right]$
$27500000=2.75 \times 10^7$
View full question & answer→Question 391 Mark
$2^3$ _____ $3^2$
Answer$2^3 \leq 3^2$
Solution:
$2^3=3^2$
$\because 2^3=2 \times 2 \times 2=8$
and $3^2=3 \times 3=9$
$\text { So, } 8<9$
$\therefore 2^3<3^2$
View full question & answer→Question 401 Mark
$432 = 2^4 × 3$ _____.
Answer$432 = 2^4 × 3^3$
Solution:
$\begin{array}{c|c} 4 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline & 1 \end{array}$
Firstly, we find out the factors of given expression.
$\text { So, } 432=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
$=2^4 \times 3^3$
$\therefore 432=2^4 \times 3^3$
View full question & answer→Question 411 Mark
$\Big(\frac{5}{8}\Big)^{9}\div\Big(\frac{5}{8}\Big)^{4}=\Big(\frac{5}{8}\Big)^{4}$
AnswerFalse. Solution: Here,$\Big(\frac{5}{8}\Big)^{9}\div\Big(\frac{5}{8}\Big)^{4}=\Big(\frac{5}{8}\Big)^{9-4}=\Big(\frac{5}{8}\Big)^{5}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{m}}=\text{a}^{\text{m-n}}\big]$ Hence, $\Big(\frac{5}{8}\Big)^{9}\div\Big(\frac{5}{8}\Big)^{4}\neq\Big(\frac{5}{8}\Big)^{4}$
View full question & answer→Question 421 Mark
$\Bigg[\Big(\frac{7}{11}\Big)^{3}\Bigg]^{4}=\Big(\frac{7}{11}\Big)$
Answer$\Bigg[\Big(\frac{7}{11}\Big)^{3}\Bigg]^{4}=\Big(\frac{7}{11}\Big)^{3\times4}$ Solution: Here, $\Bigg[\Big(\frac{7}{11}\Big)^{3}\Bigg]^{4}=\Big(\frac{7}{11}\Big)^{3\times4}=\Big(\frac{7}{11}\Big)^{12}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\big]$ $\therefore\bigg[\Big(\frac{7}{11}\Big)^{3}\bigg]=\Big(\frac{7}{11}\Big)^{12}$
View full question & answer→Question 431 Mark
$729 = 3$_____.
Answer$729 = 3^6$
Solution:
Here, firstly we find out the factors of given expression.
$\begin{array}{c|c} 3 & 729 \\ \hline 3 & 243 \\ \hline 3& 81\\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline &1 \end{array}$
So, $729=3 \times 3 \times 3 \times 3 \times 3 \times 3=3^6$
$\therefore 729=3^6$
View full question & answer→Question 441 Mark
$a^6 \times a^5 \times a^0=a$ _____.
Answer$a^6 \times a^5 \times a^0 = a^{11}$
Solution:
$\text { Since, } a^6 \times a^5 \times a^0$
$=a^{6+5+0}$
$=a^{11}$
$\therefore a^6 \times a^5 \times a^0$
$=a^{11}$
View full question & answer→Question 451 Mark
Find the value of:
$-(-4)^4$
Answer$-(-4)^4$
$=[(-4) \times(-4) \times(-4) \times(-4)]\left[\because(-1)^{\mathrm{n}}=-1, \mathrm{n} \text { is even }\right]$
$=-\left[(-1)^4 \times(4 \times 4 \times 4 \times 4)\right]$
$=(256)=-256$
View full question & answer→Question 461 Mark
Express the following numbers in standard form:
$8,19,00,000$
Answer$8,19,00,000$
$=81900000.00$
$=819 \times 10 \mathrm{~s}$
$=8.19 \times 10^2 \times 10^5$
$=8.19 \times 10^7$
View full question & answer→Question 471 Mark
$x^0 \times x^0=x^0 \div x^0$ is true for all non-zero values of $x$.
AnswerA number in standard form is written as $\mathrm{a} \times 10^k$, where $1 ≤ a ≤ 10$ and $k$ is any integer.
View full question & answer→Question 481 Mark
Express the following in exponential form: $a \times a \times b \times b \times b \times c \times c \times c \times c$
AnswerWe know that,
$\begin{aligned} & a \times a \times \ldots \times a(n \text { times })
=a^n \\ & a \times a \times b \times b \times b \times c \times c \times c \times c \\ &
=a^2 \times b^3 \times c^4\end{aligned}$
View full question & answer→Question 491 Mark
$876543=8 \times 10^5+7 \times 10^4+6 \times 10^3+5 \times 10^2+4 \times 10^1+3 \times 10^0$
Answer$\text { Take RHS }=8 \times 10^5+7 \times 10^4+6 \times 10^3+5 \times 10^2+4 \times 10^1+3 \times 10^0$
$=8 \times 100000+7 \times 10000+6 \times 1000+5 \times 100+4 \times 10+3 \times 1$
${\left[\therefore a^0=1\right]}$
$=800000+70000+6000+500+40+3$
$=876543$
$=\text { LHS }$
Hence, $RHS = LHS.$
View full question & answer→Question 501 Mark
Find the value of:
$\left(8^0-2^0\right) \times\left(8^0+2^0\right)$
Answer$\left(8^0-2^0\right) \times\left(8^0+2^0\right)$
$=(1-1) \times(1+1)$
${\left[\because a^0=1\right]}$
$=0 \times 2=0$
View full question & answer→