Question 11 Mark
The simplified form of $16^{\frac{-1}{4}}\times\sqrt[4]{16}$ is :
Answer- 1
Solution :
$16^{\frac{-1}{4}}\times\sqrt[4]{16}$
But, $16=2^4$
So,
$\Rightarrow16^{\frac{-1}{4}}\times\sqrt[4]{16}$
$\Rightarrow\{(2)^4\}^{\frac{-1}{4}}\times(2)^{4\times\frac{1}{4}}$
$\Rightarrow(2)^{4\times\frac{-1}{2}}\times2$
$\Rightarrow2^{-1}\times2$
$\Rightarrow\frac{2}{2}$
$\Rightarrow1$
View full question & answer→Question 21 Mark
Which of the following is an irrational number ?
Answer- 3.141141114...
Solution :
The decimal expansion of an irrational number is non-terminating recurring non-recurring.
Hence, 3.141141114... is an irrational number.
Hence, the correct opion is (d).
View full question & answer→Question 31 Mark
$\frac{1}{\big(3+2\sqrt{2}\big)}=?$
Answer- $\big(3-2\sqrt{2}\big)$
Solution:
$\frac{1}{\big(3+2\sqrt{2}\big)}=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=\frac{3-2\sqrt{2}}{9-8}$
$=\big(3-2\sqrt{2}\big)$
Hence, the correct option is (c).
View full question & answer→Question 41 Mark
The number obtained on rationalising the denominator $\frac{1}{\sqrt{7}-2}$ of is :
Answer- $\frac{\sqrt{7}+2}{3}$
Solution :
After rationalizing:
$\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times\frac{\sqrt{7}+2}{\sqrt{7}+2}$
$=\frac{\sqrt{7}+2}{(\sqrt{7})^2-(2)^2}$
$=\frac{\sqrt{7}+2}{7-4}$
$\frac{\sqrt{7}+2}{3}$
View full question & answer→Question 51 Mark
If $\text{x}=(7+4\sqrt{3})$ than $\Big(\text{x}+\frac{1}{\text{x}}\Big)=?$
Answerb. 14
Solution :
$\text{x}=(7+4\sqrt{3})$
$\frac{1}{\text{x}}=\frac{1}{7+4\sqrt{3}}=(7-4\sqrt{3})$
$\text{x}+\frac{1}{\text{x}}=(7+4\sqrt{3})+(7-4\sqrt{3})$
$=14$
View full question & answer→Question 61 Mark
Write the correct answer in the following :
Every rational number is
Answer
- A real number.
Solution:
Since, real numbers are the combination of rational and irrational numbers.
Hence, every rational number is a real number.
View full question & answer→Question 71 Mark
The value of $\frac{2^\circ+7^\circ}{5^\circ}$ is:
Answer- 2
Solution:
$\frac{2^\circ+7^\circ}{5^\circ}=\frac{1+1}{1}=\frac{2}{1}=2$
Hence, the correct answer is option (b).
View full question & answer→Question 81 Mark
The value of 1.9999...... in the from $\frac{\text{p}}{\text{q}},$ where 'p' and 'q' are integers and $\text{q}\not=0,$ is:
Answer- $2$
Solution:
1.9999 can be written as 2,
2 is taken as approx vlaue.
View full question & answer→Question 91 Mark
Write the correct answer in the following:
$2\sqrt{3}+\sqrt{3}$ is equal to
Answer- $3\sqrt{3}$
Solution:
Given $2\sqrt{3}+\sqrt{3}=(2+1)\sqrt{3}=3\sqrt{3}$
Hence, (c) is the correct answer.
View full question & answer→Question 101 Mark
When $15\sqrt{15}$ is divided by $3\sqrt{3},$ the quotient is :
Answer- $5\sqrt{5}$
Solution :
$\frac{15\sqrt{15}}{3\sqrt{3}}=\frac{5\sqrt{5\times3}}{\sqrt{3}}$
$\frac{5\sqrt{5}\times\sqrt{3}}{\sqrt{3}}=5\sqrt{5}$
Hence, the correct answer is option (c).
View full question & answer→Question 111 Mark
The simplest rationalising factor of $\sqrt[3]{500},$ is:
Answer- $\sqrt[3]{2}$
Solution:
$\sqrt[3]{500}=(500)^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}\\ \ =\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=(10^{\not3})^{\frac{1}{\not3}}.\frac{1}{2^{\frac{1}{3}}}\Rightarrow10.2^{\frac{-1}{3}}$
The simplest Rationalisation factor of $\sqrt[3]{500}$
After simplify it to $\Big(10.2^{\frac{-1}{3}}\Big)$ is $2^{\frac{1}{3}}$ or $\sqrt[3]{2}.$
Hence, correct option is (a).
View full question & answer→Question 121 Mark
A rational number between $\sqrt{2}$ and $\sqrt{3}$ is:
Answer- 1.5
Solution:
A rational number between two given number a and b is given by $\frac{\text{a}+\text{b}}{2},$
Hence, a rational number between two given number $\sqrt{2}=1.414$ and $\sqrt{3}=1.732.$
$=\frac{\sqrt{2}+\sqrt{3}}{2}=\frac{1.414+1.732}{2}=1.5$
View full question & answer→Question 131 Mark
Which one of the following is a correct statement?
Answer- Decimal expansion of an irrational number is non-terminating and non-repeating.
Solution:
Decimal Expansion of a Rational number is not only terminating,
It can be either terminating like $\frac{1}{2}=0.5$ or non-terminating Repeating like $\frac{1}{3}=0.3333333......$ So option (a) is not true alone.
Now we know that Non-Terminating numbers are of two types:
One is Non-Terminating Repeating and other is Non-Terminating Non-Repeating.
The Decimal expansion of a Rational number matches one of it's kind i.e Non-Terminating Repeating of Non-Terminating numbers.
So Rational number does not consist both the kinds of Non-Terminating numbers.
Hence, they are not Non-Terminating numbers.
An irrational number is always Non-Terminating in nature, but again not of both of it's kinds.
The decimal Expansion of an irrational number is Non-Terminating Non-Repeating in Nature.
So from all above points and theory we can conclude an Irrational number is Non-Terminating but Non-Repeating in nature
i.e. $\sqrt{2}=1.4142135623730...$
So, option (d) is correct.
View full question & answer→Question 141 Mark
$\sqrt{12}\times\sqrt{15}=$
Answer- $6\sqrt{5}$
Solution:
$\sqrt{12}=\sqrt{3\times2^2}=2\sqrt{3}$ and $\sqrt{15}=\sqrt{5}\times\sqrt{3}$
So, $\sqrt{12}\times\sqrt{15}=2\sqrt{3}\times\sqrt{3}\times\sqrt{5}$
$=2\times3\sqrt{5}=6\sqrt{5}$
View full question & answer→Question 151 Mark
$\Big(\frac{125}{216}\Big)^{\frac{-1}{3}}=$
Answer- $\frac{6}{5}$
Solution:
$\Big(\frac{125}{216}\Big)^{\frac{-1}{3}}$
$\Rightarrow\Big(\frac{5}{6}\Big)^{3\times\frac{-1}{3}}$
$\Rightarrow\Big(\frac{5}{6}\Big)^{-1}$
$\Rightarrow\frac{6}{5}$
View full question & answer→Question 161 Mark
$\sqrt{12}\times\sqrt{15}=$
Answer- $6\sqrt{5}$
Solution:
$\sqrt{12}=\sqrt{3\times2^2}=2\sqrt{3}$ and $\sqrt{15}=\sqrt{5}\times\sqrt{3}$
So, $\sqrt{12}\times\sqrt{15}=2\sqrt{3}\times\sqrt{3}\times\sqrt{5}$
$=2\times3\sqrt{5}=6\sqrt{6}$
View full question & answer→Question 171 Mark
$7.\bar{2}$ is Equal to:
Answer- $\frac{65}{9}$
Solution:
$7.\bar{2}=7+0.\bar{2}$
i.e., $7+\frac{2}{9}$
i.e., $\frac{63+2}{9}=\frac{65}{9}$
View full question & answer→Question 181 Mark
If $\sqrt{13-\text{a}\sqrt{10}}=\sqrt8+\sqrt5,$ then a =
Answer- -4
Solution:
$\sqrt{13-\text{a}\sqrt10}=\sqrt8+\sqrt5$
Squaring both sides, we get
$13-\text{a}\sqrt{10}=8+5+2\sqrt{40}$
$={13}-\text{a}\sqrt{10}-{13}=2\times2\sqrt{10}$
$=-\text{a}\sqrt{10}=4\sqrt{10}$
$\Rightarrow\text{a}=-4$
Hence, correct option is (c).
View full question & answer→Question 191 Mark
The simplest from of $0.\overline{32}$ is:
Answer- $\frac{29}{90}$
Solution:
x = 0.32222...
10x = 3.222... ----(i)
100x = 32.222... ----(ii)
Subtract eq. (i) from (ii)
90x = 29
$\text{x}=\frac{29}{90}$
View full question & answer→Question 201 Mark
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$
Answer- 4
Solution:
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{4\times8}+\sqrt{4\times12}}{\sqrt{8}+\sqrt{12}}$
$=\frac{2\sqrt{8}+2\sqrt{12}}{\sqrt{8}+\sqrt{12}}=\frac{2\big(\sqrt{8}+\sqrt{12}\big)}{\sqrt{8}+\sqrt{12}}=2$
Hence, the correct answer is option (b).
View full question & answer→Question 211 Mark
Answer- Either rational or irrational.
Solution:
Either you can make a fraction from two whole numbers (denominator $\not=$ 0), thus a rational number. Or you can't, thus irrational.
View full question & answer→Question 221 Mark
If $8=\text{t}^{\frac{2}{3}}+4\text{t}^{-\frac{1}{2}},$ What is the value of g when t = 64?
Answer- $\frac{33}{2}$
Solution:
Given $\text{t}=64,\ \text{g}=\text{t}^{\frac{2}{3}}+4\text{t}^{\frac{1}{2}}.$ We have to find the value of g
So,
$\text{g}=\text{t}^{\frac{2}{3}}+4\text{t}^{-\frac{1}{2}}$
$\text{g}=64^{\frac{2}{3}}+4\times64^{\frac{1}{2}}$
$\text{g}=(64)^{\frac{2}{3}}+4\times\frac{1}{64^{\frac{1}{2}}}$
$\text{g}=2^{6\times\frac{2}{3}}+4\times\frac{1}{2^{6\times\frac{1}{2}}}$
$\text{g}=2^{2\times2}+4\times\frac{1}{2^3}$
$\text{g}=2^4+4\times\frac{1}{8}$
$\text{g}=16+\frac{1}{2}$
$\text{g}=\frac{16\times2}{1\times2}+\frac{1}{2}$
$\text{g}=\frac{32}{2}+\frac{1}{2}$
$\text{g}=\frac{32+1}{2}$
$\text{g}=\frac{33}{2}$
The value of g is $\frac{33}{2}$
Hence the correct choice is b.
View full question & answer→Question 231 Mark
The value of $\frac{2}{\sqrt{5}-\sqrt{3}}$ is:
Answer- $\sqrt{5}+\sqrt{3}$
Solution:
$\frac{2}{\sqrt{5}-\sqrt{3}}$
Multiplying nu nominator and denominator by $\sqrt{5}+\sqrt{3},$
We get,
$\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$
$=\frac{2(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{5}+\sqrt{3}$
View full question & answer→Question 241 Mark
If $\frac{3^{5\text{x}}\times81^2\times6561}{3^{2\text{x}}}$ then x =
Answer- $-3$
Solution:
We have to find the value of x provided $\frac{3^{5\text{x}}\times81^2\times6561}{3^{2\text{x}}}=3^7$
So,
$\frac{3^{5\text{x}}\times81^2\times6561}{3^{2\text{x}}}=3^7$
By using law of rational exponents we get
$3^{5\text{x}+8+8-2\text{x}}=3^7$
By equating exponents we get
$5\text{x}+88-2\text{x}=7$
$3\text{x}+16=7$
$3\text{x}=7-16$
$3\text{x}=-9$
$\text{x}=\frac{-9}{3}$
$\text{x}=-3$
Hence the correct choice is b.
View full question & answer→Question 251 Mark
If $a = 2, b = 3,$ then the value of $(a^b + b^a)^{-1}$ is:
Answer$(a^b + b^a)^{-1}$
Put value of $a$ and $b$,
$(2^3 + 3^2)^{-1}$
$\Rightarrow (8 + 9)^{-1}$
$\Rightarrow (17)^{-1}$
$\Rightarrow\frac{1}{17}$
View full question & answer→Question 261 Mark
The simplest rationalisation factor of $(2\sqrt{2}-\sqrt{3})$ is:
View full question & answer→Question 271 Mark
If $64^{-\frac{1}{3}}\Big(64^{\frac{1}{3}}-64^\frac{2}{3}\Big)$ then $5\sqrt[\text{n}]{64}=$
Answer- $25$
Solution:
We have to find $5\sqrt[\text{n}]{64}$ provided $\sqrt{5^\text{n}}=125$
So,
$\sqrt{5^\text{n}}=125$
$5^{\text{n}\times\frac{1}{2}}=5^3$
$\frac{\text{n}}{2}=3$
$\text{n}=3\times2$
$\text{n}=6$
Substitute $\text{n}=6$ in $5^{\sqrt[\text{n}]{64}}$ to get
$5^{\sqrt[\text{n}]{64}}=5^{2^{6\times\frac{1}{6}}}$
$=5\times5$
$=25$
Hence the value of $5^{\sqrt[\text{n}]{64}}$ is 25
The correct choice is a.
View full question & answer→Question 281 Mark
The simplest rationalising factor of $2\sqrt5-\sqrt3,$ is:
Answer- $2\sqrt5+\sqrt3$
Solution:
Rationalising factor of any number of kind $\text{a}\sqrt{\text{a}}\pm\sqrt{\text{b}}$ is $\sqrt{\text{a}}\mp\sqrt{\text{b}}$
So. for given number $2\sqrt5-\sqrt3.$ Rationalising factor would be $2\sqrt5+\sqrt3.$
Hence, correct option is (b).
View full question & answer→Question 291 Mark
Write the correct answer in the following:
The product of any two irrational numbers is.
Answer- Sometimes rational, sometimes irrational.
Solution:
Sometimes rational, sometimes irrational The product of any two irrational numbers is sometimes rational and sometimes irrational.
Hence, (D) is the correct answer.
For example:
rational
$(2+\sqrt{3})(2-\sqrt{3})$
$(2)^2-(\sqrt{3})^2$
$4-3=1$
irrational
$(2+\sqrt{3})(1-\sqrt{3})$
$2(1-\sqrt{3})+\sqrt{3}(1-\sqrt{3})$
$2-2\sqrt{3}+\sqrt{3}-3$
$-1-\sqrt{3}$
View full question & answer→Question 301 Mark
The value of $\sqrt[3]{1000}$ is:
Answer- 10
Solution:
$(10)^3=1000$
So, $\sqrt[3]{1000}=1000^{\frac{1}{3}}=(10^3)^{\frac{1}{3}}$
View full question & answer→Question 311 Mark
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=$
Answer- 10
Solution:
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow\frac{(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
$\Rightarrow\frac{(3+2+2\sqrt{6})+3+2-2\sqrt{6}}{3-2}$
$\Rightarrow10$
View full question & answer→Question 321 Mark
If $\text{x}=\sqrt[3]{2+\sqrt3},$ then $\text{x}^3+\frac{1}{\text{x}^3}=$:
Answer- 4
Solution:
$\text{x}=\sqrt[3]{2+\sqrt3}+\big(2+\sqrt3\big)^{\frac{1}{3}}$
$\text{x}^3=\big\{\big({2+\sqrt3}\big)^{\frac{1}{\not3}}\big\}^{\not3}=\big(2+\sqrt3\big)$
$\Rightarrow\frac{1}{\text{x}^3}=\frac{1}{2+\sqrt3}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}\\ \ =\frac{2-\sqrt3}{4-3}=2-\sqrt3$
Now, $\text{x}^3+\frac{1}{\text{x}^3}=2+\sqrt3+2-\sqrt3=4$
Hence, correct option is (b).
View full question & answer→Question 331 Mark
Simplified value of $(16)^{-\frac{1}{4}}\times\sqrt[4]{16}$ is:
Answer- 1
Solution:
$(16)^{-\frac{1}{4}}\times\sqrt[4]{16}=(2^4)^{-\frac{1}{4}}\times(2^4)^\frac{1}{4}\\=2^{-1}\times2^1=\frac{1}{2}\times2=1$
Hence, the correct answer is option (a).
View full question & answer→Question 341 Mark
$\sqrt{10}\times\sqrt{15}=?$
Answer- $5\sqrt{6}$
Solution:
$\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt{3}=5\sqrt{6}$
Hence, the correct answer is option (b).
View full question & answer→Question 351 Mark
$9^3 + (-3)^3 - 6^3 =$ ?
Answer$9^3 + (-3)^3 - 6^3$
$= 729 - 27 - 216$
$= 729 - 243$
$= 486$
View full question & answer→Question 361 Mark
$2\sqrt{3}+\sqrt{3}$ is equal to :
Answerc. $3\sqrt{3}$
Solution :
$(\sqrt{\text{x}^3})^{\frac{2}{3}}$
$=(\text{x}\frac{3}{2})^{\frac{2}{3}}$
$=\text{x}$
View full question & answer→Question 371 Mark
Write the correct answer in the following:
$\sqrt[4]{\sqrt[3]{2^2}}$ equals.
Answer- $2^{\frac{1}{6}}$
Solution:
$\sqrt[4]{\sqrt[3]{2^2}}=\sqrt[4]{(2^2)^{\frac{1}{3}}}=\Big(2^{\frac{2}{3}}\Big)^{\frac{1}{4}}=2^{\frac{2}{3}\times\frac{1}{4}}=2^{\frac{1}{6}}$
Hence, (c) is the correct answer.
View full question & answer→Question 381 Mark
The value of $\{5(8^{\frac{1}{4}}+27^{\frac{1}{3}})^3\}^{\frac{1}{4}}$ is :
Answerc. 5
Solution :
$\{5(8^{\frac{1}{4}}+27^{\frac{1}{3}})^3\}^{\frac{1}{4}}$
$\Rightarrow\{5(2^{3\times\frac{1}{3}}+3^{3\times\frac{1}{3}})^3\}^{\frac{1}{4}}$
$\Rightarrow\{5(3+2)^3\}^{\frac{1}{4}}$
$\Rightarrow\{5\times5^3\}^{\frac{1}{4}}$
$\Rightarrow5^{4\times\frac{1}{4}}$
$\Rightarrow5$
View full question & answer→Question 391 Mark
The decimal representation of an irrational number is :
Answer- Neither terminating nor repeating.
Solution :
An irrational number can not be written in the form of $\frac{\text{p}}{\text{q}}.$
And decimal representation of it is neither terminating nor repeating.
View full question & answer→Question 401 Mark
The rationalisation factor of $\frac{1}{\big(2\sqrt{3}-\sqrt{5}\big)}$ is :
Answer- $\sqrt{12}+\sqrt{5}$
Solution:
The rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ is $2\sqrt{3}+\sqrt{5},$
i.e. $\sqrt{3\times4}+\sqrt{5}$
i.e. $\sqrt{12}+\sqrt{5}$
Hence, the correct option is (d).
View full question & answer→Question 411 Mark
If $\text{x}=7+4\sqrt3$ and xy = 1, then $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=$
Answer- 194
Solution :
$\text{x}=7+4\sqrt3,\ \text{xy}=1\Rightarrow\text{y}=\frac{1}{\text{x}}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}\\ \ =\frac{7-4\sqrt3}{(7)^2-\big(4\sqrt3\big)^2}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3$
Now, $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=\frac{\text{y}^2+\text{x}^2}{\text{x}^2\text{y}^2}=\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}$
$\text{x}^2=\big(7+4\sqrt3\big)^2=49+48+56\sqrt3=97+56\sqrt3$
$\text{y}^2=\big(7-4\sqrt3\big)^2=49+48-56\sqrt3=97-56\sqrt3$
$\therefore\text{x}^2+\text{y}^2=97+56\sqrt3+97-56\sqrt3=194$
$\text{xy} = 1$
$\therefore\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}=\frac{194}{(1)^2}=194$
Hence, correct option is (c).
View full question & answer→Question 421 Mark
An irrational number between 2 and 2.5 is :
Answer- $\sqrt{5}$
Solution :
$\sqrt{4}=2$ and $\sqrt{6.25}=2.5$
Option (a), (c) and (d): $\sqrt{11},\sqrt{22.5}$ and $\sqrt{12.5},$ all are greater than $\sqrt{6.25}$
⇒ Out of interval (2, 2.5)
Option (b): $\sqrt{4}<\sqrt{5}<\sqrt{6.25}\Rightarrow$ lies in the interval (2, 2.5)
Hence, option (b) is correct.
View full question & answer→Question 431 Mark
The decimal representation of a rational number is :
Answer- Either terminating or repeating.
Solution :
Rational numbers can be represented in decimal forms rather than representing infractions.
They can easily be represented as decimals by just dividing numerator 'p' by denominator 'q' $\Big($as rational numbers is in the form of $\frac{\text{p}
{\text{q}}\Big)$
A rational number can be expressed as a terminating or no terminating, recurring decimal.
For example:
- $\frac{5}{2}=2.5,$
$\frac{2}{8}=0.25,$
7 = 7.0, etc., are rational numbers which are terminating decimals.
- $\frac{5}{9}=0.555555555.......=0.5,$
$\frac{1}{6}=0.166666.....=0.16,$
$\frac{9}{11}=0.818181$
View full question & answer→Question 441 Mark
After simplification, $\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}$ is :
Answer- $13^{-\frac{2}{15}}$
Solution :
$\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}$
$=13^{\frac{1}{5}+\frac{1}{3}}$
$=13^{-\frac{2}{15}}$
View full question & answer→Question 451 Mark
If $(3^3)^2 = 9^x$ than $5^x = $?
Answer$(3^3)^2 = 9^x$
$(3^3)^3 = 9^x$
$9^3 = 9^x$
$\Rightarrow x = 3$
$\therefore 5^3 = 125$
View full question & answer→Question 461 Mark
Which of the following numbers is irrational?
Answer- $\sqrt{8}$
Solution :
The decimal expansion of $\sqrt{8}=2.82842712...,$ which is non-terminating, non-recurring.
Hence, it is an irrational number.
Hence, the correct opion is (c).
View full question & answer→Question 471 Mark
$(256)^{0.16} \times (256)^{0.09}$
AnswerWe have to find the value of $(256)^{0.16} \times (256)^{0.09}$. So,
By using law of rational exponents
$a^m \times a^n= a^{m+n}$ we get
$(265)^{0.16} \times (256)^{0.09} = (256)^{0.16} \times (256)^{0.09}$
$= (256)^{0.16+0.09}$
$=256^{0.25}$
$=256^{\frac{25}{100}}$
$(256)^{0.16}\times(256)^{0.09 }=2^{8\times\frac{25}{100}}$
$=2^{8\times\frac{25}{100}}$
$=2^{8\times\frac{1}{4}}$
$=2^{8\times\frac{1}{4}}=4$
The value of $(256)^{0.16} \times (256)^{0.09}$ is $4$
View full question & answer→Question 481 Mark
Answer- An irrational number.
Solution:
Some numbers cannot be written as a ratio of two integers they are called Irrational Numbers.
It is irrational because it cannot be written as a ratio (or fraction),
View full question & answer→Question 491 Mark
The value of $64^{-\frac{1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)$ is:
Answer- -3
Solution:
Find the value of $64^{\frac {1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)$
So,
$\Rightarrow64^{\frac {1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)=2^{6\times\frac{1}{3}}\Big(2^{6\times\frac{1}{3}}-2^{6\times\frac{2}{3}}\Big)$
$=2^{-2}(2^2-2^4)$
$=2^2(4-16)$
$\Rightarrow64^{\frac {1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)=\frac{1}{2^2}\times-12$
$=\frac{1}{4}\times-12$
$=-3$
Hence the correct statement is c.
View full question & answer→Question 501 Mark
If $\sqrt{2}=1.414$ than $\sqrt{\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}}=?$
Answer- 0.414
Solution:
$\sqrt{\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}}$
$=\sqrt{\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}\times\frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}}$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-(1)^2}}$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}$
$=\sqrt{(\sqrt{2}-1)^2}$
$=1.414-1$
$=0.414$
View full question & answer→