Maths M.C.Q

2. $-\frac{5}{3}$

Solution:

$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$

$=\frac{5^{\text{n}+1}(5-6)}{5^{\text{n}}(13-2\times5)}$

$=\frac{5^{\text{n}}\times5\times(-1)}{5^{\text{n}}(13-10)}$

$=-\frac{5}{3}$

Hence, the correct option is (b).

3. $\sqrt{\frac{1}{7}\times\frac{2}{7}}$

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).

4. $\sqrt{12}+\sqrt{5}$

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).

4. $\frac{\sqrt{5}-\sqrt{2}}{3}$

Solution:

$\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}\\=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$

Hence, the correct option is (d).

3. 4

Solution:

$\text{x}=2+\sqrt{3}$

$\therefore\frac{1}{\text{x}}=\frac{1}{2+\sqrt{3}}$

$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$

$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$

$=\frac{2-\sqrt{3}}{4-3}$

$=2-\sqrt{3}$

$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4$

Hence, the correct option is (c).

3. $5\sqrt{5}$

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).

4. $\sqrt[3]{2}$

Solution:

$\sqrt[3]{500}=500^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}$ $=\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=\frac{10^{3\times\frac{1}{3}}}{2^{\frac{1}{3}}}=\frac{10}{\sqrt[3]{2}}$

Thus, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}.$

Hence, the correct option is (d).

3. $\sqrt{8}$

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).

2. 6

Solution:

$\frac{17}{7}=2.\overline{428571}$

Hence, the correct opion is (b).

4. $\sqrt[4]{2}$

Solution:

$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times2}=2^{\frac{1}{2}}=\sqrt{2},$ which is irrational.

$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times4}=2^1=2,$ which is rational.

Hence, the correct option is (d).

3. $\frac{5}{6}$ and $\frac{7}{6}$

Solution:

We have,

$\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{5}{3}=\frac{5\times2}{3\times2}=\frac{10}{6}$

And, $\frac{1}{2}=\frac{1\times3}{2\times3}=\frac{3}{6}$ and $\frac{2}{1}=\frac{2\times6}{1\times6}=\frac{12}{6}$

Also, $\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{4}{3}=\frac{4\times2}{3\times2}=\frac{8}{6}$

Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\Big(\frac{1}{2}\Big)<\frac{4}{6}\Big(=\frac{2}{3}\Big)<\frac{5}{6}<\frac{7}{6}\\<\frac{8}{6}\Big(=\frac{4}{3}\Big)<\frac{10}{6}\Big(=\frac{5}{3}\Big)<\frac{12}{6}\Big(=\frac{2}{1}\Big)$

So, the two rational numbers between $\frac{2}{3}$ and $\frac{5}{3}$ are $\frac{5}{6}$ and $\frac{7}{6}.$

Hence, the correct opion is (c).

2. 4

Solution:

$\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)$

$\big(\sqrt{11}\big)^2-\big(\sqrt{7}\big)^2$

$11-7=4$

Hence, the correct answer is option (b).

4. nonterminating, nonrecurring.

Solution:

The decimal expansion of $\sqrt{2}=1.41421356...,$ which is non-terminating, nonrecurring.

Hence, the correct opion is (d).

3. an irrational number.

Solution:

$\pi=3.14159265359...,$ which is non-terminating non-recurring.

Hence, it is an irrational number.

Hence, the correct opion is (c).

3. $\big(3-2\sqrt{2}\big)$

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).

2. 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).

1. $10$

Solution:

$\sqrt{20}\times\sqrt{5}=\sqrt{4\times5}\times\sqrt{5}$

$=2\sqrt{5}\times\sqrt{5}=2\times5=10$

Hence, the correct answer is option (a).

3. $\sqrt{225}$

Solution:

The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.

$\sqrt{2},$ $\sqrt{23}$ and 0.1010010001... are irrational numbers, since they cannot be expressed in the form $\frac{\text{p}}{\text{q}}.$

$\sqrt{225}=15=\frac{15}{1}$ is a rational number.

Hence, the correct opion is (c).

2.  $\frac{7}{9}$

Solution:

Let $\text{x}=0.\overline{3}$

i.e., $\text{x}=0.3333 \ ...(\text{i})​​$

$\Rightarrow10\text{x}=3.3333 \ ...(\text{ii})​​$

On subtracting (i) and (ii), we get

$9\text{x}=3$

$\Rightarrow\text{x}=\frac{3}{9}$

Let $\text{y}=0.\overline{4}$

i.e., $\text{y}=0.4444 \ ...(\text{i})​​$

$\Rightarrow10\text{y}=4.4444 \ ...(\text{ii})​​$

On subtracting (i) and (ii), we get

$9\text{y}=4$

$\Rightarrow\text{y}=\frac{4}{9}$

$\therefore0.\overline{3}+0.\overline{4}=\text{x}+\text{y}=\frac{\text{3}}{\text{9}}+\frac{\text{4}}{\text{9}}=\frac{\text{7}}{\text{9}}$

Hence, the correct answer is option (b). 

4. $6^{\frac{1}{4}}$

Solution:

An irrational number between a and b is given by $\sqrt{\text{ab}}$

An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is

$\sqrt{\sqrt{2}+\sqrt{3}}=\big(\sqrt{6}\big)^{\frac{1}{2}}$

$=6^{\frac{1}{2}\times\frac{1}{2}}$

$=6^{\frac{1}{4}}$

Hence, the correct answer is option (d).

1. $\sqrt{23}$

Solution:

The decimal expansion of $\sqrt{23}=4.79583152...,$ which is non-terminating, nonrecurring.

Hence, it is an irrational number.

Hence, the correct opion is (a).

2. either terminating or repeating.

Solution:

The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.

Decimal representation of a rational number is either terminating or a repeating decimal, since every decimal of this form can be expressed in the form $\frac{\text{p}}{\text{q}}.$

Hence, the correct opion is (b).

4. $13^\frac{-2}{15}$

Solution:

$\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}=13^{\frac{1}{5}-\frac{1}{3}}=13^{\frac{3-5}{15}}=13^{\frac{-2}{15}}$

Hence, the correct answer is option (d).

2. $(56)^\frac{1}{2}$

Solution:

$(7)^{\frac{1}{2}}\times(8)^{\frac{1}{2}}=(7\times8)^{\frac{1}{2}}=(56)^{\frac{1}{2}}$

Hence, the correct answer is option (b).

3. 1.6

Solution:

$\sqrt{2}=1. 414213562...$ and $\sqrt{3}=1. 7320508075...$

Hence, 1.6 lies between $\sqrt{2}$ and $\sqrt{3}$

Hence, the correct option is (c).

4. None of these.

Solution:

Let $\text{x}=0.12\overline{3}$

Then, $\text{x}=0.12333 \ ...(\text{i})$

$\therefore100\text{x}=12.333... \ (\text{ii})$

and $1000\text{x}=123.333... \ (\text{iii})$

On subtracting (ii) from (iii), we get

$900\text{x}=111$

$\Rightarrow\text{x}=\frac{111}{900}=\frac{37}{300}$

3. 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).

1. 3

Solution:

$(243)^{\frac{1}{5}}=(3^5)^{\frac{1}{5}}=3^{5\times\frac{1}{5}}=3$

Hence, the correct answer is option (a).

3.  $\frac{31}{11}$

Solution:

Let $\text{x}=2.\overline{45}$

i.e., $\text{x}=2.4545 \ ...(\text{i})​​$

$\Rightarrow100\text{x}=245.4545 \ ...(\text{ii})​​$

On subtracting (i) and (ii), we get

$99\text{x}=243$

$\Rightarrow\text{x}=\frac{243}{99}$

Let $\text{y}=0.\overline{36}$

i.e., $\text{y}=0.3636 \ ...(\text{iii})​​$

$\Rightarrow100\text{y}=36.3636 \ ...(\text{iv})​​$

On subtracting (iii) and (iv), we get

$99\text{y}=36$

$\Rightarrow\text{y}=\frac{36}{99}$

$\therefore2.\overline{45}+0.\overline{36}=\text{x}+\text{y}=\frac{\text{243}}{\text{99}}+\frac{\text{36}}{\text{99}}=\frac{\text{279}}{\text{99}}=\frac{\text{31}}{\text{11}}$

Hence, the correct answer is option (c).

2. 0.378

Solution:

$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$

$=\frac{\sqrt{7}}{7}$

$=\frac{1}{7}\times\sqrt{7}$

$=\frac{1}{7}\times2.646$

$=0.378$

Hence, the correct option is (b).

1. 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).

2. 2

Solution:

$\frac{2^\circ+7^\circ}{5^\circ}=\frac{1+1}{1}=\frac{2}{1}=2$

Hence, the correct answer is option (b).

2. $\frac{3}{10}$

Solution:

Given two rational numbers are negative and $\frac{3}{10}$ is a positive rational number.

 So, it does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$

Hence, the correct opion is (b).

4. 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).

4. 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).

3. $2^{\frac{1}{6}}$

Solution:

$\sqrt[4]{\sqrt[3]{2^2}}=\sqrt[4]{\sqrt[3]{4}}=\sqrt[4]{4^{\frac{1}{3}}}=4^{\frac{1}{3}\times\frac{1}{4}}=4^{\frac{1}{12}}=2^{2\times\frac{1}{12}}=2^{\frac{1}{6}}$

Hence, the correct answer is option (c).

4. $\frac{21}{57}$

Solution:

$\frac{7}{17}=\frac{7\times3}{17\times3}=\frac{21}{57}$

Hence, the correct opion is (d).

4. a real number.

Solution:

A number whose square is non-negative, is called a real number.

The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.

So, a rational number is a real number since p and q which form a rational number are integers.

Hence, the correct opion is (d).

3. 0.414

Solution:

$\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}\times\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}-1\big)}}$

$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{\big(\sqrt{2}\big)^2-(1)^2}}$

$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{1}}$

$=\sqrt{\big(\sqrt{2}-1\big)^2}$

$=\sqrt{2}-1$

$=1.414-1$

$=0.414$

Hence, the correct option is (c).

3. $\frac{29}{90}$

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).