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→Question 511 Mark
$4^0+5^0+6^0=(4+5+6)^0$
AnswerFalse.
Solution:
$\text { Here, } 4^0+5^0+6^0$
$=1+1+1$
$=3\left[\therefore a^0=1\right]$
$\text { and }(4+5+6)^0$
$=(15)^0$
$=1$
$\text { Hence, } 4^0+5^0+6^0$
$\neq(4+5+6)^0$
View full question & answer→Question 521 Mark
$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)$
Answer$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$ Solution: Here, $\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{16\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\big]$ $=\Big(\frac{-1}{4}\Big)^{32}$ $\therefore\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$
View full question & answer→Question 531 Mark
$1^0 \times 0^1=1$
AnswerFalse.
Solution:
$\therefore1^{0}\times0^{1}$
$=1\times0$
$=0\neq1$
View full question & answer→Question 541 Mark
$x^m+x^m=x^{2 m}$, where $x$ is a non-zero rational number and $m$ is a positive integer
Answer$a^m \times a^n=a^{m+n}$
$x^m \times x^m=x^{m+m}=x^{2 m}$
$\text { Also, } a^k+a^k=2 a^k$
$\text { So, } x^m+x^m=2 x^m$
View full question & answer→Question 551 Mark
$88880000000=$ ______ $\times 10^{10}$
AnswerGiven, $88880000000$
For standard form, $88880000000$
$=8888 \times 10^7 \text { Also, } 8888$
$=8.888 \times 10^3$
$\text { So, } 8.888 \times 10^3 \times 10^7$
$=8.888 \times 10^{10}$
$88880000000=8.888 \times 10^{10}$
View full question & answer→Question 561 Mark
Express the following numbers in standard form:
$24$ billion.
Answer$24 \text { billion }$
$=24,00,00,00,000$
$=24 \times 10^9$
$=2.4 \times 10^1 \times 10^9$
$=2.4 \times 10^{10}$
View full question & answer→Question 571 Mark
$4^9$ is greater than $16^3$
AnswerTrue.
Solution:
$\therefore 16^3=\left(4^2\right)^3$
${\left[\therefore 16=4 \times 4=4^2\right]}$
$=4^6$
Now, in $4^9$ and $4^6, 4^9>4^6$ as powers $9>6$
View full question & answer→Question 581 Mark
$a. 3^2 .........1^5$
Answer$\because 3^2=3 \times 3=9$
So, $9<15$
$\therefore 3^2<15$
View full question & answer→Question 591 Mark
Express the following in usual form:
$8.01 \times 10^7$
AnswerHere, $8.01\times10^{7}=\frac{801}{100}\times10000000$ $=80100000$
View full question & answer→Question 601 Mark
Identify the greater number, in the following:
$7.9 \times 104$ or $5.28 \times 105$
AnswerWe have, $7.9 \times 10^4$
$=7.9 \times 10000$
$=79000 \text { and } 5.28 \times 10^5$
$=5.28 \times 100000$
$=528000 \text { So, } 5.28 \times 10^5>7.9 \times 10^4$
View full question & answer→Question 611 Mark
In the standard form, a large number can be expressed as a decimal number between $0$ and $1,$ multiplied by a power of $10$
AnswerA number in standard form is written as $a \times 10^k$, where $1 ≤ a ≤ 10$ and $k$ is any integer.
View full question & answer→Question 621 Mark
$8 \times 106 + 2 \times 104 + 5 \times 102 + 9 \times 100 = 8020509$
AnswerTake $LHS$
$=8 \times 10^6+2 \times 10^4+5 \times 10^2+9 \times 10^0$
$=8 \times 1000000+2 \times 10000+5 \times 100+9 \times 1$
${\left[\therefore a^0=1\right]}$
$=8000000+2000+500+9$
$=8020509$
$=\text { RHS }$
Hence, $LHS = RHS.$
View full question & answer→Question 631 Mark
$7^4$ _____ $5^4$
Answer$7^4$ > $5^4$
Solution:
Obviously, $7^4$ > $5^4$
[$\because$ powers are same, so if base is greater then the number is greater]
View full question & answer→Question 641 Mark
Express the following in single exponential form:
$2^3 \times 3^3$
AnswerWe have, $2^3 \times 3^3=(2 \times 3)^3$
$=6^3\left[\because a^m b^m=(a \times b)^m\right]$
View full question & answer→Question 651 Mark
$1$ lakh $= 10 $_____.
AnswerHere, $1$ lakh
$=100000=10^5$
$\therefore 1$ lakh
$=10^5$
View full question & answer→Question 661 Mark
$1$ million $= 10$ _____.
AnswerHere, $1$ million
$=1000000=10^6$
$\therefore 1$ million $=10^6$
View full question & answer→Question 671 Mark
Find the value of:
$\left(-3^5\right)$
Answer$\left(-3^5\right)=\left(-1^5\right) \times 3^5$
$=1 \times 3 \times 3 \times 3 \times 3 \times 3$
$[\because(-1)=-1, \mathrm{n}$ is odd $]$
$=-243$
View full question & answer→Question 681 Mark
$\Big(\frac{11}{15}\Big)^{4}\times$ (_____)$^5$ = $\Big(\frac{11}{15}\Big)^{9}$
Answer$\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$ Solution: $\Rightarrow(\text{x})^{5}=\frac{\Big(\frac{11}{15}\Big)^{9}}{\Big(\frac{11}{15}\Big)^{4}}$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9}\Big(\frac{11}{15}\Big)^{-4}$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9-4}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{-n}}=\text{a}^{\text{m+(-n)}}=\text{a}^{\text{m-n}}\big]$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{5}$ Since, the powers are same $\therefore\text{x}=\frac{11}{15}$ Hence, $\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$
View full question & answer→Question 691 Mark
$x^m \div y^m=(x \div y)^m$, where $x$ and $y$ are non-zero rational numbers and $m$ is a positive integer.
AnswerIf $x$ and $y$ are rational numbers, then $\frac{\text{x}}{\text{y}^{\text{m}}}=\frac{\text{x}^{\text{m}}}{\text{y}^{\text{m}}}$ or $\text{x}^{\text{m}}\div\text{y}^{\text{m}}=(\text{x}\div\text{y})^{\text{m}}$
View full question & answer→Question 701 Mark
$4^2$ is greater than $2^4$
AnswerFalse.
Solution:
$4^2=4 \times 4=16\left[\therefore a^m=a \times a \times a \times \ldots \times a(m\right.$ times $\left.)\right]$
and $2^4=2 \times 2 \times 2 \times 2=16$
So, $4^2=2^4$
View full question & answer→Question 711 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 721 Mark
$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)$
Answer$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$
Solution:
Here, $\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{16\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\big]$
$=\Big(\frac{-1}{4}\Big)^{32}$
$\therefore\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$
View full question & answer→Question 731 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 741 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 751 Mark
$\Big(\frac{13}{14}\Big)^{5}\div(\ )^{2}=\Big(\frac{13}{14}\Big)^{3}$
Answer$\Big(\frac{13}{14}\Big)^{5}\div(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{3}$ Solution:
Let
$\Big(\frac{13}{14}\Big)^{5}\div(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{3}$ $\therefore(\text{x})^{2}=\frac{\Big(\frac{13}{14}^{5}\Big)}{\Big(\frac{13}{14}\Big)^{3}}$ $\therefore(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{5-3}$ $\big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\big]$ $\therefore(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{2}$ Since, powers are same. $\therefore\text{x}=\frac{13}{14}$ Hence, $\Big(\frac{13}{14}\Big)^{5}\div(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{3}$ View full question & answer→Question 761 Mark
$\Big(\frac{11}{15}\Big)^{4}\times$ (_____)5 = $\Big(\frac{11}{15}\Big)^{9}$
Answer$\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$
Solution:
$\Rightarrow(\text{x})^{5}=\frac{\Big(\frac{11}{15}\Big)^{9}}{\Big(\frac{11}{15}\Big)^{4}}$
$\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9}\Big(\frac{11}{15}\Big)^{-4}$
$\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9-4}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{-n}}=\text{a}^{\text{m+(-n)}}=\text{a}^{\text{m-n}}\big]$
$\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{5}$
Since, the powers are same
$\therefore\text{x}=\frac{11}{15}$
Hence, $\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$
View full question & answer→Question 771 Mark
Identify the greater number, in the following:
7.9 × 104 or 5.28 × 105
AnswerWe have, 7.9 x 104
= 7.9 × 10000
= 79000 and 5.28 × 105
= 5.28 × 100000
= 528000 So, 5.28 × 105 > 7.9 × 104
View full question & answer→Question 781 Mark
Identify the greater number, in the following:
29 or 92
AnswerWe have, 29
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 512 and 92
= 9 × 9
= 81
So, 29 > 92
View full question & answer→Question 791 Mark
Identify the greater number, in the following:
26 or 62
AnswerWe have, 26
= 2 × 2 × 2 × 2 × 2 × 2
= 64 and 62 = 6 × 6
= 36 So, 26 > 62
View full question & answer→Question 801 Mark
Find the value of:
(80 - 20) × (80 + 20)
Answer(80 - 20) × (80 + 20)
= (1 - 1) × (1 + 1)
$\Big[\because\text{a}^{0}=1\Big]$
= 0 × 2 =0
View full question & answer→Question 811 Mark
Find the value of:
77 ÷ 77
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 821 Mark
Find the value of:
(-7)2 × 7 - 6 - 8
Answer(-7)2 × 7 - 6 - 8
= (-7)14-14 $\Big[\because\text{a}^{0}=1\Big]$
= (-7)0
= 1
View full question & answer→Question 831 Mark
Answer$\Big[\because\text{a}^{0}=1\Big]$
70 = 1
View full question & answer→Question 841 Mark
Find the value of:
-(-4)4
Answer-(-4)4
= [(-4) ×(-4) × (-4) × (-4)] $\Big[\because(-1)^{\text{n}} =-1, \text{n is even}\Big]$
= -[(-1)4 × (4 × 4 × 4 × 4)]
= (256) = -256
View full question & answer→Question 851 Mark
Answer(-35) = (-15) × 35
= 1 × 3 × 3 × 3 × 3 × 3 $\Big[\because(-1)=-1, \text{n is odd}\Big]$
= -243
View full question & answer→Question 861 Mark
AnswerWe knoe that, an = a × a × a ×...×a (n times)
25 = 2 × 2 × 2 × 2 × 2 = 32
View full question & answer→Question 871 Mark
Find the value of:
(20 + 30 + 40) (40 - 30 - 20)
Answer(20 + 30 + 40) (40 - 30 - 20)
= (1 + 1 + 1) (1 + 1 + 1)
$\Big[\because\text{a}^{0}=1\Big]$
= (3) (-1) = -3
View full question & answer→Question 881 Mark
Find the value of:
2 × 3 × 4 ÷ 20 × 30 × 40
Answer2 × 3 × 4 ÷ 20 × 30 × 40
= 2 × 3 × 4 + 1 × 1 × 1
$\Big[\because\text{a}^{0}=1\Big]$
$=\frac{2\times3\times4}{1\times1\times1}=2\times3\times4=24$
View full question & answer→Question 891 Mark
Express the following numbers in standard form:
8,19,00,000
Answer8,19,00,000
= 81900000.00
= 819 × 10s
= 8.19 × 102 × 105
= 8.19 × 107
View full question & answer→Question 901 Mark
Express the following numbers in standard form:
5,83,00,00,00,000
Answer5,83,00,00,00,000
= 583000000000.00
= 583 × 109
= 5.83 × 102 × 109
= 5.83 × 1011
View full question & answer→Question 911 Mark
Express the following numbers in standard form:
24 billion.
Answer24 billion
= 24,00,00,00,000
= 24 × 109
= 2.4 × 101 × 109
= 2.4 × 1010
View full question & answer→Question 921 Mark
Express the following in usual form:
8.01 × 107
AnswerHere, $8.01\times10^{7}=\frac{801}{100}\times10000000$
$=80100000$
View full question & answer→Question 931 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 941 Mark
Express the following in single exponential form:
(-3)3 × (-10)3
AnswerWe have, (-3)3 × (-10)3
= [(-3) × (-10)]3 [$\because$ am × bm = (a × b)m]
[$\because$ (-3) × (-10) = 30]
= (30)3
View full question & answer→Question 951 Mark
Express the following in single exponential form:
(-11)2 × (-2)2
AnswerWe have, (-11)2 × (-2)2
= [(-11) × (-2)]2 [$\because$ am × bm = (a × b)m]
[$\because$ (-11) × (-2) = 22]
= 222
View full question & answer→Question 961 Mark
Express the following in exponential form:
s × s × t × t × s × s × t
AnswerWe know that,
a × a ×… × a (n times) = an
s × s × t × t × s × s × t
= s × s × s × s × t × t × t
= s4 × t3
View full question & answer→Question 971 Mark
Express the following in exponential form:
a × a × b × b × b × c × c × c × c
AnswerWe know that,
a × a ×… × a (n times) = an
a × a × b × b × b × c × c × c × c
= a2 × b3 × c4
View full question & answer→Question 981 Mark
Express the following in exponential form:
3 × 3 × 3 × a × a × a × a
AnswerWe know that,
a × a ×… × a (n times) = an
3 × 3 × 3 × a × a × a × a
=33 × a4
= 27 a4
View full question & answer→Question 991 Mark
Express the following in single exponential form:
23 × 33
AnswerWe have, 23 × 33 = (2 × 3)3
= 63 [$\because$ am bm = (a × b)m]
View full question & answer→