Question 2011 Mark
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: -25 is not a rational number.
Reason : -25 can not be written in in the form of $\frac{\text{p}}{\text{q}}.$
Answer - Both assertion and reason are false.
View full question & answer→Question 2021 Mark
The value of $0.\overline{23}+0.\overline{22}$ is:
Answer- $0.\overline{45}$
Solution:
$0.\overline{23}=0.232323$
$0.\overline{22}=0.222222$
$0.\overline{23}+0.\overline{23}=0.45454545$
$=0.\overline{45}$
View full question & answer→Question 2031 Mark
Select the correct statement from the following.
Answer- $\frac{-2}{3}<\frac{-4}{5}$
Solution:
$\frac{-2}{3}<\frac{-4}{5}$
Taking LCM of 3 and 5,
LCM = 15,
So, $\frac{-2\times5}{3\times5},\frac{-4\times3}{5\times3}$
$\Rightarrow\frac{-10}{15},\frac{-12}{15}$
Now, since both the denominator is equal so, we compare its numerator,
and since, -10 < -12
So, $\frac{-10}{15}>\frac{-12}{15}$
thus, $\frac{-2}{3}>\frac{-4}{5}$
View full question & answer→Question 2041 Mark
Which of the following is a true statment?
Answer- Every real number is either rational or irrational.
Solution:
Consider, $\big(2+\sqrt{3}\big)$ and $\big(2-\sqrt{3}\big)$ which are two irrational numbers.
$\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4,$ which is a rational number.
Consider, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ which are two irrational numbers.
$\sqrt{3}\times\frac{1}{\sqrt{3}}=1,$ which is a rational number.
Every real number can either be a rational number or an irrational number.
Hence, the correct opion is (d).
View full question & answer→Question 2051 Mark
The value of $\Big(\frac{12^{\frac{1}{5}}}{27^{\frac{1}{5}}}\Big)^{\frac{5}{2}}.$
Answer- $\frac{2}{3}$
Solution:
$\Big(\frac{12^{\frac{1}{5}}}{27^{\frac{1}{5}}}\Big)^{\frac{5}{2}}.$
$\Rightarrow\Big(\frac{12}{27}\Big)^{\frac{1}{5}\times\frac{5}{2}}$
$\Rightarrow\Big(\frac{12}{27}\Big)^{\frac{1}{2}}$
$\Rightarrow\frac{2\sqrt{3}}{3\sqrt{3}}$
$\Rightarrow\frac{2}{3}$
View full question & answer→Question 2061 Mark
An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is :
Answer- $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
Solution :
An irrational number between a and b is given by $\sqrt{\text{ab}}$
So, an irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
Hence, the correct answer is option (c).
View full question & answer→Question 2071 Mark
If $\frac{\sqrt3-1}{\sqrt3+1}=\text{a}-\text{b}\sqrt3,$ then :
Answer- a = 2, b = 1
Solution :
$\frac{\sqrt3-1}{\sqrt3+1}$
Multiplying and dividing by the rationalisation factor of denominator, we get
$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}$
$=\frac{\big(\sqrt3-1\big)^2}{\big(\sqrt3\big)^2-1^2}$
$=\frac{3-2\sqrt3+1}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=\frac{2(2-\sqrt3)}{2}$
$=2-\sqrt3$
Comparing with $\text{a}-\text{b}\sqrt3,$ we get a = 2 and b = 1.
Hence, correct option is (a).
View full question & answer→Question 2081 Mark
Between any two rational numbers there :
Answer- are infinitely many rational numbers.
Solution :
Options (a), (b) and (d) are incorrect since between two rational numbers there are infinitely many rational and irrational numbers.
Hence, the correct opion is (c).
View full question & answer→Question 2091 Mark
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called :
Answer- An irrational number.
Solution :
An irrational number cannot be written in the form of $\frac{\text{p}}{\text{q}}.$
Irrational number can neither be expressed as terminating decimal nor as repeating decimal.
View full question & answer→Question 2101 Mark
If $\text{x}=\big(7+4\sqrt{3}\big)$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)=?$
Answer- 14
Solution:
$\text{x}=\big(7+4\sqrt{3}\big)$
$\therefore\frac{1}{\text{x}}=\frac{1}{\big(7+4\sqrt{3}\big)}$
$=\frac{1}{\big(7+4\sqrt{3}\big)}\times\frac{\big(7-4\sqrt{3}\big)}{\big(7-4\sqrt{3}\big)}$
$=\frac{\big(7-4\sqrt{3}\big)}{7^2-\big(4\sqrt{3}\big)^2}$
$=\frac{7-4\sqrt{3}}{49-48}$
$=7-4\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(7+4\sqrt{3}\big)+\big(7-4\sqrt{3}\big)=14$
Hence, the correct option is (b).
View full question & answer→Question 2111 Mark
Write the correct answer in each of the following:
Which of the following is irrational?
Answer- $0.4014001400014...$
Solution:
A number is irrational if and only of its decimal representation is non - terminating and nonrecurring.
- $0.14$ is a terminating decimal and therefore cannot be an irrational number.
- $0.14\overline{16}$ is a non - terminating and recurring decimal and therefore cannot be irrational.
- $0.\overline{1416}$ is a non - terminating and recurring decimal and therefore cannot be irrational.
- $0.4014001400014...$ is a non - terminating and non - recurring decimal and therefore is an irrational number.
View full question & answer→Question 2121 Mark
The decimal expansion that a rational number cannot have is:
Answer- 0.5030030003...
Solution:
The decimal expansion of a rational number is either terminating or non-terminating recurring.
The decimal expansion of 0.5030030003... is non-terminating, non-recurring, which is not a property of a rational number.
Hence, the correct opion is (d).
View full question & answer→Question 2131 Mark
The value of $\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$ is:
Answer- 1
Solution:
$\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{q}}{\text{p}}}.\sqrt{\frac{\text{r}}{\text{q}}}.\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=\sqrt{1}$
$=1$
Hence, the correct option is (c).
View full question & answer→Question 2141 Mark
The value of $(\frac{81}{16})^{\frac{-3}{4}}\times\{(\frac{25}{9})^{\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$ is:
Answer- 1
Solution:
$(\frac{81}{16})^{\frac{-3}{4}}\times\{(\frac{25}{9})^{\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{4\times\frac{-3}{4}}\times\{(\frac{5}{3})^{2\times\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{-3}\times\{(\frac{5}{3})^{-3}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{-3}\times(\frac{5}{3}\times\frac{2}{5})^{-3}$
$\Rightarrow(\frac{3}{2})^{-3}\times(\frac{2}{3})^{-3}$
$\Rightarrow(\frac{3}{2}\times\frac{2}{3})^{-3}$
$\Rightarrow(1)^{-3}=1$
View full question & answer→Question 2151 Mark
Which of the following numbers is irrational?
Answer- $\sqrt{8}$
Solution:
$\therefore\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}$
View full question & answer→Question 2161 Mark
The value of $x^{a-b} \times ^{b-c} x^{c-a}$ is:
Answer$x^{a-b} \times ^{b-c} x^{c-a}$
$\Rightarrow x^{a-b+c+c-a}$
$\Rightarrow x^0$
$= 1$
View full question & answer→Question 2171 Mark
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to :
Answer- $3+2\sqrt{2}$
Solution:
After rationalising
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{\sqrt{9}-\sqrt{8}}\times\frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}$
$=\frac{\sqrt{9}+\sqrt{8}}{(\sqrt{9})^2-(\sqrt{8})^2}$
$=\frac{\sqrt{3\times3}+\sqrt{2\times2\times2}}{9-8}$
$=\frac{3+2\sqrt{2}}{1}$
$=3+2\sqrt{2}$
View full question & answer→Question 2181 Mark
Write the correct answer in the following: Value of $(256)^{0.16} \times (256)^{0.09}$ is.
Answer$(256)^{0.16}\times(256)^{0.09}=(256)^{\frac{16}{100}}\times(256)^{\frac{9}{100}}$
$=(256)^{\frac{16}{100}+\frac{9}{100}}\ [\because\text{x}^\text{a}\cdot\text{x}^{\text{b}}=\text{x}^{\text{a+b}}]$
$(256)^{\frac{25}{100}}=(256)^{\frac{1}{4}}$
$=(4^4)^{\frac{1}{4}}=4\ [\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$
View full question & answer→Question 2191 Mark
If $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16,\ \frac{3^\text{p}}{2^\text{n}}=81$ and $\text{a}=2^{\frac{1}{10}},$ then $\frac{\text{a}^{2\text{m}+\text{n}-\text{p}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}=$
Answer- $2$
Solution:
Given: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16,\ \frac{3^\text{p}}{2^\text{n}}=81$ and $\text{a}=2^{\frac{1}{10}}$
To find: $\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}$
Find: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16$
By using rational components $\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}$ We get
$2^{\text{m}+\text{n}-\text{n}+\text{m}}=16$
$2^{2\text{m}}=2^4$
By equating rational exponents we get
$2\text{m}=4$
$\text{m}=\frac{4}{2}$
$\text{m}=2$
Now, $\frac{\text{a}^{2\text{m}+\text{n}-\text{p}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})}=(\text{a}^{2\text{m}+\text{n}-\text{p}}).(\text{a}^{\text{m}-2\text{n}+2\text{p})}$ we get
$=\text{a}^{2\text{mn}+\text{n}+\text{p}+\text{m}-2\text{n}+2\text{p}}$
$=\text{a}^{3\text{m}-\text{n}+\text{p}}$
Now putting value of $\text{a}=2^\frac{1}{10}$ we get,
$=2^{\frac{3\text{m}-\text{n}+\text{p}}{10}}$
$=2^{\frac{6-\text{n}+\text{p}}{10}}$
Also, $\frac{3^\text{p}}{3^\text{n}}=81$
$3^{\text{p}-\text{n}}=3^4$
On comparing LHS and RHS we get, p - n = 4.
Now,
$\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}=\text{a}^{3\text{m}-\text{n}+\text{p}}$
$=2^{\frac{6+(\text{p}-\text{n})}{10}}$
$=2^{\frac{6+4}{10}}$
$=2^{\frac{10}{10}}=2^1$
$=2$
So, option (a) is the correct answer.
View full question & answer→Question 2201 Mark
The value of $\sqrt{\text{p}^{-1}\text{q}}\times\sqrt{\text{q}^{-1}\text{r}}\times\sqrt{\text{r}^{-1}\text{p}}$ is:
Answer- 1
Solution:
$\sqrt{\text{p}^{-1}\text{q}}\times\sqrt{\text{q}^{-1}\text{r}}\times\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{p}}{\text{q}}}\times\sqrt{\frac{\text{r}}{\text{q}}}\times\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=1$
View full question & answer→Question 2211 Mark
The value of $(243)^{\frac{1}{5}}$ is:
Answer- 3
Solution:
$(243)^{\frac{1}{5}}$
$=(3^5)^{\frac{1}{5}}$
$=3$
View full question & answer→Question 2221 Mark
The simplest for of $0.\overline{32}$ is:
Answer- $\frac{29}{90}$
Solutions:
Let $\text{x}=0.\overline{32}$
Then, $\text{x}=0.3222 \ ...(\text{i})$
$\therefore10\text{x}=3.222 \ ...(\text{ii})$
and $100\text{x}=32.222 \ ...(\text{iii})$
On subtracting (ii) from (iii), we get
$90\text{x}=29$
$\Rightarrow\text{x}=\frac{29}{90}$
Hence, the correct option is (c).
View full question & answer→Question 2231 Mark
The value of $(x^{a-b})^{a+b} \times (x^{b-c})^{b+c} \times (x^{c-a})^{c+a}$ is:
Answer$(x^{a-b})^{a+b} \times (x^{b-c})^{b+c} \times (x^{c-a})^{c+a}$
$\Rightarrow\text{x}^{\text{a}^{2}-\text{b}^2}\times\text{x}^{\text{b}^2-\text{c}^2}\times\text{x}^{\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^0=1$
View full question & answer→Question 2241 Mark
$\sqrt[5]{6}\times\sqrt[5]{6}$ is equal to:
Answer- $\sqrt[5]{36}$
Solution:
$\sqrt[5]{6}=(6)^{\frac{1}{5}}$
So $\sqrt[5]{6}\times\sqrt[5]{6}=(6)^{\frac{1}{5}}\times(6)^{\frac{1}{5}}$
$=(6\times6)^{\frac{1}{5}}$
$(36)^{\frac{1}{5}}$
$=\sqrt[5]{36}$
Hence, correct option is (a).
View full question & answer→Question 2251 Mark
$\frac{1}{\sqrt9-\sqrt8}$ is equal to:
Answer- $3+2\sqrt2$
Solution:
$\frac{1}{\sqrt9-\sqrt8}$
$=\frac{1}{\sqrt9-\sqrt8}\times\frac{{\sqrt9+\sqrt8}}{{\sqrt9+\sqrt8}}$
$\frac{{\sqrt9+\sqrt8}}{\big(\sqrt9\big)^2-\big(\sqrt8\big)^2}$
$=\frac{{\sqrt9+\sqrt8}}{9-8}$
$={\sqrt9+\sqrt8}$
$=3+2\sqrt2$
Hence, correct option is (a).
View full question & answer→Question 2261 Mark
Write the correct answer in the following:
Which of the following is equal to x?
Answer- $(\sqrt{\text{x}^3})^{\frac{2}{3}}$
Solution:
- $\text{x}^{\frac{12}{7}}-\text{x}^{\frac{5}{7}}\neq\text{x}$
- $\sqrt[12]{\text{(x}^4})^{\frac{1}{3}}=\sqrt[12]{\text{x}^{4\times\frac{1}{3}}}=\Big(\text{x}^\frac{4}{3}\Big)^\frac{1}{12}=\text{x}^{\frac{4}{3}\times\frac{1}{12}}=\text{x}^\frac{1}{9}\neq\text{x}$
- $\Big((\text{x}^3)^\frac{1}{2}\Big)^\frac{2}{3}=\text{(x)}^{\frac{3}{2}\times\frac{2}{3}}=\text{x}^1=\text{x}$
- $\text{x}^{\frac{12}{7}}\times\text{x}^\frac{7}{12}=\text{x}^{\frac{12}{7}+\frac{7}{12}}=\text{x}^\frac{193}{84}\neq\text{x}$
Hence, (c) is the correct answer.
View full question & answer→Question 2271 Mark
If $\text{x}^{-2}=64,$ then $\text{x}^{\frac{1}{3}}+\text{x}^0=$
Answer- $\frac{3}{2}$
Solution:
We have to find the value of $\text{x}^{\frac{1}{3}}+\text{x}^0$ if $\text{x}^{-2}=64$
Consider,
$\text{x}^{-2}=2^6$
$\frac{1}{\text{x}^2}=2^6$
Multiply $\frac{1}{2}$ on both sides of powers we get
$\frac{1}{\text{x}^{2\times\frac{1}{2}}}=2^{6\times\frac{1}{6}}$
$\frac{1}{\text{x}}=2^3$
$\frac{1}{\text{x}}=\frac{8}{1}$
By taking reciprocal on both sides we get,
$\frac{1}{8}=\text{x}$
Substituting $\frac{1}{8}$ in $\text{x}^{\frac{1}{3}}+\text{x}^0$ we get
$=\Big(\frac{1}{8}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$
$=\Big(\frac{1}{2^2}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$
$=\frac{1}{2^{3\times\frac{1}{3}}}+1$
$=\frac{1}{2^1}+1$
$=\frac{1}{2}+1$
By taking least common multiply we get
$=\frac{1}{2}+\frac{1\times2}{1\times2}$
$=\frac{1}{2}+\frac{2}{2}$
$=\frac{1+2}{2}$
$=\frac{3}{2}$
Hence the correct choice is c.
View full question & answer→Question 2281 Mark
Which one of the following statements is true?
Answer- The sum of two irrational numbers may be a rational number or an irrational number.
Solution:
If two irrational numbers i.e. $\sqrt{2},\sqrt{5},2+\sqrt{3},2-\sqrt{3}$ etc. are added it is not necessary that sum comes out to be an irrational number always, or a rational nnumber always, or a rational number always...
Since $\sqrt{2}+\sqrt{5}=$ an irrational number
$2+\sqrt{\not\text{3}}+2-\sqrt{\not\text{3}}=4=$ a rational number
So we see that $\sqrt{2}$ and $\sqrt{5}$ are irrational numbers, and their sum is also irrational.
But $2+\sqrt{3}$ and $2-\sqrt{3}$ are also irrational numbers, and their sum is rational number '4'.
So sum of two irrational numbers can be either an irrational number or a rational number depending which numbers are being added.
So options (a) and (b) are totally wrong, because they are not 'always' true.
Option (c) is correcrt because sum can be either irrational or rational and option (c) is verifying this statement.
Option (d) - again it is not always true, if we add two irrational numbers like $2+\sqrt{3}$ and $2-\sqrt{3}.$
Sum is an integer = 4, but if we add $\sqrt{3}$ and $\sqrt{3},$ sum is $2\sqrt{3}$ which is not an integer but again an irrational number.
So option (d) is also incorrect.
Hence, correct option is (c).
View full question & answer→Question 2291 Mark
The simplest for of $1.\overline{6}$ is:
Answer- $\frac{5}{3}$
Solution:
Let $\text{x}=1.\overline{6}$
$\Rightarrow\text{x}=1.666 \ ...(\text{i})$
$\therefore10\text{x}=16.666 \ ...(\text{ii})$
On subtracting (i) from (ii), we get
$9\text{x}=15$
$\Rightarrow\text{x}=\frac{15}{9}=\frac{5}{3}$
Hence, the correct option is (c).
View full question & answer→Question 2301 Mark
$(1296)\frac{-1}{4}=$
Answer- $\frac{1}{6}$
Solution:
$(1296)\frac{-1}{4}=(6^4)^{\frac{-1}{4}}$
$=(6)^{-1}=\frac{1}{6}$
View full question & answer→Question 2311 Mark
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: Sum of two irrational numbers $2+\sqrt3$ is an irrational number.
Reason: Sum of a rational number and an irrational numbers is always an irrational number.
Answer - Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→Question 2321 Mark
$\frac{1}{(3+2\sqrt{2})}=\text{?}$
Answer- $(3-2\sqrt{2})$
Solution:
$\frac{1}{(3+2\sqrt{2})}$
$=\frac{3-2\sqrt{2}}{(3+2\sqrt{2})(3-2\sqrt{2})}$
$=(3-2\sqrt{2})$
View full question & answer→Question 2331 Mark
The product of two irrational number is:
Answer- sometimes rational and sometimes irrational.
Solution:
Consider, the irrational number, $\sqrt{3}.$
Product $=\sqrt{3}\times\sqrt{3}=3,$ which is a rational number.
Consider two irrational numbers, $\sqrt{2}$ and $\sqrt{3}.$
Product $=\sqrt{2}\times\sqrt{3}=\sqrt{6},$ which is an irrational number.
Hence, the product of two irrational numbers are sometimes rational and sometimes irrational.
Hence, the correct opion is (d).
View full question & answer→Question 2341 Mark
If $2^x = 4^y$ and $\frac{1}{2\text{x}}+\frac{1}{4\text{y}}+\tfrac{1}{4\text{z}}=4$ then the value of $x$ is:
Answer$2^x = 4^y = 8^z$
$\Rightarrow 2^x = 2^{2y} = 2^{3z}$
$\Rightarrow x = 2y = 3z$
$\Rightarrow\text{y}=\frac{\text{x}}{2}$ and
$\Rightarrow\text{z}=\frac{\text{x}}{3}$
So, $\frac{1}{2\text{x}}+\tfrac{1}{4\text{y}}+\frac{1}{4\text{z}}=4$
$\Rightarrow\frac{1}{2\text{x}}+\frac{2}{4\text{x}}+\frac{3}{4\text{x}}=4$
$\Rightarrow\frac{7}{4\text{x}}=4$
$\Rightarrow\text{x}=\frac{7}{16}$
View full question & answer→Question 2351 Mark
The product of the square root of x with the cube root of x is:
Answer- Sixth root of the fifth power of x.
Solution:
We have to find the product (say L) of the square root of x with the cube root of x is.
So,
$\text{L}=\sqrt[2]{\text{x}}\times\sqrt[3]{\text{x}}$
$=\text{x}^\frac{1}{2}\times\text{x}^\frac{1}{3}$
$=\text{x}^{\frac{1}{2}+\frac{1}{3}}$
$=\text{x}^{\frac{1\times3}{2\times3}+\frac{1\times2}{3\times2}}$
$=\text{x}^{\frac{3+2}{6}}=\text{x}^\frac{5}{6}$
The product of the square root of x with the cube root of x is $\text{x}^{\frac{5}{6}}$
Hence the correct alternative is b.
View full question & answer→Question 2361 Mark
The value of $\sqrt[4]{(64)^{-2}}$ is:
Answer- $\frac{1}{8}$
Solution:
$\sqrt[4]{(64)^{-2}}$
$\Rightarrow(64)^{\frac{-2}{4}}$
$\Rightarrow(64)^{\frac{-1}{2}}$ or $\frac{1}{\sqrt{64}}$
$\Rightarrow\frac{1}{8}$
View full question & answer→Question 2371 Mark
If $\text{x}=3+\sqrt{8},$ then the value of $(\text{x}^2+\frac{1}{\text{x}^2})$ is:
Answer- 34
Solution:
Given, $(\text{3}+\sqrt{8})$
$\frac{1}{\text{x}}=\frac{1}{(3+\sqrt{8})}=\frac{1}{(3+\sqrt{8})}\times\frac{(3-\sqrt{8})}{(3-\sqrt{8})}$
$=\frac{(3-\sqrt{8})}{(3^2-(\sqrt{8})^2)}=\frac{3+\sqrt{8}}{9-8}=(3-\sqrt{8})$
$(\text{x}+\frac{1}{\text{x}})=(3+\sqrt{8})+(3-\sqrt{8})=6$
$\Rightarrow(\text{x}+\frac{1}{\text{x}})^2=6^2=36$
$\Rightarrow(\text{x}^2+\frac{1}{\text{x}^2})+2\times\text{x}\times\frac{1}{\text{x}}=36$
$\Rightarrow(\text{x}^2+\frac{1}{\text{x}^2})+2=36$
$\Rightarrow(\text{x}+\frac{1}{\text{x}^2})=36-2=34$
View full question & answer→Question 2381 Mark
$16\sqrt{134}\div9\sqrt{52}$ is equal to :
Answer- $\frac{8}{9}$
Solution :
$16\sqrt{134}\div9\sqrt{52}$
$\frac{16\sqrt{13}}{9\sqrt{52}}=\frac{16}{9}\times\sqrt{\frac{13}{52}}=\frac{16}{9}\times\frac{1}{2}$
$=\frac{8}{9}$
View full question & answer→Question 2391 Mark
Which of the following is true statement ?
Answer- Every real number is either rational or irrational.
Solution :
Consider, $(2+\sqrt{3})$ and $(2-\sqrt{3})$ which are two irrational number.
$(2+\sqrt{3})+(2-\sqrt{3})=4,$ which is a rational number.
Consider, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ which is a rational number.
Every real number can either be a rational number or an irrational number.
View full question & answer→Question 2401 Mark
If $\text{x}=3+2\sqrt{2},$ than the value of $\text{x}+\frac{1}{\text{x}}$ is:
Answer- 6
Solution:
$\text{x}+\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}$
Put the value of x,
$\Rightarrow\frac{(3+2\sqrt{2})^2+1}{3+2\sqrt{2}}$
$\Rightarrow\frac{9+8+12\sqrt{2}+1}{3+2\sqrt{2}}$
$\Rightarrow\frac{18+12\sqrt{2}}{3+2\sqrt{2}}$
$\Rightarrow\frac{6(3+2\sqrt{2})}{3+2\sqrt{2}}$
$\Rightarrow6$
View full question & answer→Question 2411 Mark
Which of the following statement is true?
Answer- $\sqrt{\text{ab}}=\sqrt{\text{a}}\times\sqrt{\text{b}}$
Solution:
When we multiply two different root number then only number is multiplied not the root because we have both number power equal,
$\sqrt{\text{a}}\times\sqrt{\text{b}}=\sqrt{\text{ab}}$
or, $\text{a}\frac{1}{2}\times\text{b}\frac{1}{2}=(\text{ab})\frac{1}{2}$
View full question & answer→Question 2421 Mark
An irrational number between s and 6 is:
Answer
- $\sqrt{5\times6}$
Solution:
We know that, If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, Then $\sqrt{\text{ab}}$ is an irrational number lying between a and b,
Here also we have 5 and 6 two distinct rational numbers and 5 × 6 = 30 is not a perfect square,
So irrational number betweens and 6 $=\sqrt{5\times6}$
View full question & answer→Question 2431 Mark
The decimal form of $\frac{\text{1}}{999}$ is :
Answer- $0.\overline{001}$
Solution:
When we divide 1 by 999 result is 0.001001001001001.....
So, $0.\overline{001}=\frac{1}{999}$
View full question & answer→Question 2441 Mark
The value of $15\sqrt{15}\div3\sqrt{5}$ is:
Answer- $5\sqrt{3}$
Solution:
$\frac{15\sqrt{15}}{3\sqrt{5}}$
$=\frac{(3\times5)\sqrt{3}\times\sqrt{5}}{3\sqrt{3}}$
$=5\sqrt{3}$
View full question & answer→Question 2451 Mark
$\sqrt{8}+2\sqrt{32}-5\sqrt{2}$ is equal to:
Answer- $5\sqrt{2}$
Solution:
$\sqrt{8}+2\sqrt{32}-5\sqrt{2}$
$\Rightarrow2\sqrt{2}+2\times4\sqrt{2}-5\sqrt{2}$
$\Rightarrow10\sqrt{2}-5\sqrt{2}$
$\Rightarrow5\sqrt{2}$
View full question & answer→Question 2461 Mark
$2^{\frac{4}{3}}$ is same as.
Answer- $\sqrt[3]{2^4}$
Solution:
$\sqrt[3]{2^4}$
$=(2^4)^{\frac{1}{3}}$
$=2^{\frac{4}{3}}$
View full question & answer→Question 2471 Mark
The simplest form of $25^{\frac{1}{3}}\times5^{\frac{1}{3}}$ is:
Answer- 5
Solution:
$25^{\frac{1}{3}}\times5^{\frac{1}{3}}$
$=5^{\frac{2}{3}}\times5^{\frac{1}{3}}$
$=(5)^{\frac{2+1}{3}}\Leftrightarrow5$
View full question & answer→Question 2481 Mark
Which of the following is a rational number.
Answer- $\sqrt{225}$
Solution:
Because $\sqrt{225}$ is a square of 15, i.e., $\sqrt{225}=15,$ and it can be expressed in the pq from, it is a rational number.
View full question & answer→Question 2491 Mark
The simplest rationalisation factor of $\big(2\sqrt{2}-\sqrt{3}\big)$ is:
Answer- $2\sqrt{2}+\sqrt{3}$
Solution:
The simplest rationalisation factor of $\big(2\sqrt{2}-\sqrt{3}\big)$ is $2\sqrt{2}+\sqrt{3}$
Hence, the correct option is (b).
View full question & answer→Question 2501 Mark
If $\sqrt{2}=1.41$ then $\frac{1}{\sqrt{2}}=?$
Answer- 0.705
Solution:
$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}}{2}$
$=\frac{1.41}{2}$
$=0.705$
Hence, the correct option is (c).
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