Maths M.C.Q

4. 4

Solution:

$\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$

$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{3^4}{2^4}$

$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}+2\text{x}}=\Big(\frac{3}{2}\Big)^4$

$\Rightarrow\Big(\frac{3}{2}\Big)^{\text{x}}=\Big(\frac{3}{2}\Big)^4$

$\Rightarrow\text{x}=4$

Hence, the correct option is (d). 

2. 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).

3. $\sqrt{5\times6}$

Solution:

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

So, an irrational number between 5 and 6 is given by $\sqrt{5\times6}$

Hence, the correct answer is option (c).

3. 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).

2. $\frac{2}{9}$

Solution:

$\frac{2}{9}=0.2222222...=0.\overline{2}$

Hence, the correct option is (b).

1. $\frac{1}{8}$

Solution:

$\sqrt[4]{(64)^{-2}}=\sqrt[4]{(8^2)^{-2}}=8^{-4\times\frac{1}{4}}=8^{-1}=\frac{1}{8}$

Hence, the correct answer is option (a).

2. $\frac{6}{11}$

Solutions:

Let $\text{x}=0.\overline{54}$

Then, $\text{x}=54.5454 \ ...(\text{i})$

$\therefore100\text{x}=54.5454 \ ...(\text{ii})$

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

$99\text{x}=54$

$\Rightarrow\text{x}=\frac{54}{99}=\frac{6}{11}$

Hence, the correct option is (c).

2. $\frac{25}{7}$

Solution:

$\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$

$\Rightarrow\frac{5}{\sqrt{7}}=\text{p}\sqrt{7}$

$\Rightarrow\frac{25}{\sqrt{7}}=\text{p}\sqrt{7}$

$\Rightarrow\text{p}=\frac{25}{\sqrt{7}\times\sqrt{7}}=\frac{25}{7}$

Hence, the correct option is (b).

2. positive and rational.

Solution:

$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$

$=6+\sqrt{27}-3-\sqrt{3}-2\sqrt{3}$

$=6-3+3\sqrt{3}-\sqrt{3}-2\sqrt{3}$

$=3$

which is positive and rational number.

Hence, the correct answer is option (b).

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

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).

3. $\sqrt{3}+\sqrt{2}$

Solution:

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

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

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

Hence, the correct option is (c).

2. $\frac{3\text{y}}{4\text{x}^4}$

Solution:

$\Big(\frac{256\text{x}^{16}}{81\text{y}^4}\Big)^{-\frac{1}{4}}=\Big(\frac{81\text{y}^{4}}{256\text{x}^{16}}\Big)^{\frac{1}{4}}$

$=\Big(\frac{3^4\text{y}^{4}}{4^4(\text{x}^4)^4}\Big)^{\frac{1}{4}}=\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^{4\times\frac{1}{4}}=\frac{3\text{y}}{4\text{x}^4}$

Hence, the correct option is (b).

4. 5

Solution:

$(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}=5^{2\times\frac{1}{3}}\times5^\frac{1}{3}$

$=5^\frac{2}{3}\times5^\frac{1}{3}=5^{\frac{2}{3}+\frac{1}{3}}=5^{\frac{3}{3}}=5$

4. $0$

Solution:

Since, the sum and product of a rational and an irrational is always irrational.

So, $1+\sqrt{3}$ and $2\sqrt{3}$ are irrational numbers.

Also, $\pi$ is an irrational number.

And, 0 is an integer.

So, 0 is a rational number.

Hence, the correct option is (d).

1. 3

Solution:

$\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}=\Big[\big(9^2\big)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$

$=9^{2\times\frac{1}{2}\times\frac{1}{2}}=9^{\frac{1}{2}}=3^{2\times\frac{1}{2}}=3$

Hence, the correct option is (a).

2. $5\sqrt{6}$

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).

2. positive and rational.

Solution:

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

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

$=4-3=1,$ which is positive and rational number.

Hence, the correct answer is option (b).

4. $\sqrt{2}-1$

Solution:

$\sqrt{3-2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+(1)^2-2\times\sqrt{2}\times1}$

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

$=\sqrt{2}-1$

Hence, the correct option is (d).

1. 34

Solution:

$\text{x}=3+\sqrt{8}$

$\Rightarrow\text{x}^2=\big(3+\sqrt{8}\big)^2=9+8+6\sqrt{8}=17+6\sqrt{8}$

Now, $\frac{1}{\text{x}}=\frac{1}{3+\sqrt{8}}$

$=\frac{1}{3+\sqrt{8}}\times\frac{3-\sqrt{8}}{3-\sqrt{8}}$

$=\frac{3-\sqrt{8}}{3^2-\big(\sqrt{8}\big)^2}$

$=\frac{3-\sqrt{8}}{9-8}$

$=3-\sqrt{8}$

$\Rightarrow\Big(\frac{1}{\text{x}}\Big)^2=\big(3-\sqrt{8}\big)^2=9+8-6\sqrt{8}=17-6\sqrt{8}$

Then, $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=17+6\sqrt{8}+17-6\sqrt{8}=34$

Hence, the correct option is (a).

1. $2$

Solution:

$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$

$=\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{2^5}$

$=2^\frac{1}{3}\times2^\frac{1}{4}\times2^{5\times\frac{1}{12}}$

$=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$

$=2^{\frac{4+3+5}{12}}$

$=2^{\frac{12}{12}}$

$=2$

Hence, the correct option is (a).

4. $\pi$ is irrational and $\frac{22}{7}$ is rational.

Solution:

A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number.

So, $\pi=3.141592...$ is irrational.

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, $\frac{22}{7}$ is rational.

Hence, the correct option is (d).

2. $2\sqrt{2}+\sqrt{3}$

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).

3. $\frac{1}{5}$

Solution:

$(125)^{-\frac{1}{3}}=\frac{1}{(125)^{\frac{1}{3}}}$

$=\frac{1}{5^{3-\frac{1}{3}}}=\frac{1}{5}$

Hence, the correct answer is option (c).

2. $\frac{2}{9}$

Solution:

$\frac{4\sqrt{12}}{12\sqrt{27}}=\frac{4\sqrt{4\times3}}{12\sqrt{9\times3}}$

$=\frac{2\sqrt{3}}{3\times3\sqrt{3}}=\frac{2}{9}$

Hence, the correct answer is option (b).