Chapter 1 Number Systems (Additional Questions)

A rational number has a terminating decimal expansion if and only if the prime factors of its denominator (in the simplest form of the fraction) are only 2 and 5.

---

Let's examine each option:

(A) $\frac{10}{3}$. The denominator is $3$. The prime factor is $3$. Since $3$ is not $2$ or $5$, the decimal expansion is non-terminating repeating.

(B) $\frac{7}{11}$. The denominator is $11$. The prime factor is $11$. Since $11$ is not $2$ or $5$, the decimal expansion is non-terminating repeating.

(C) $\frac{15}{1600}$. First, simplify the fraction:

$15 = 3 \times 5$

$1600 = 16 \times 100 = 2^4 \times 10^2 = 2^4 \times (2 \times 5)^2 = 2^4 \times 2^2 \times 5^2 = 2^{4+2} \times 5^2 = 2^6 \times 5^2$

So, $\frac{15}{1600} = \frac{3 \times 5}{2^6 \times 5^2} = \frac{3}{2^6 \times 5^1}$.

The denominator of the simplified fraction is $2^6 \times 5^1$. The prime factors are $2$ and $5$. Since the prime factors are only $2$ and $5$, the decimal expansion is terminating.

(D) $\frac{29}{343}$. The denominator is $343$. We find the prime factors of $343$:

$343 = 7 \times 49 = 7 \times 7 \times 7 = 7^3$.

The prime factor is $7$. Since $7$ is not $2$ or $5$, the decimal expansion is non-terminating repeating.

---

Based on the analysis, only the fraction $\frac{15}{1600}$ has a terminating decimal expansion.

---

The correct option is (C).

The point P is at a distance of $\sqrt{17}$ units from the origin on the number line. This means the number represented by P is $\sqrt{17}$ (assuming it's on the positive side, but the interval determination is the same for $-\sqrt{17}$).

---

To find the interval in which $\sqrt{17}$ lies, we should compare $17$ with the perfect squares of integers near it.

The perfect square less than $17$ is $4^2 = 16$.

The perfect square greater than $17$ is $5^2 = 25$.

---

We have the inequality:

$16 < 17 < 25$

---

Taking the square root of all parts of the inequality (since all numbers are positive):

$\sqrt{16} < \sqrt{17} < \sqrt{25}$

$4 < \sqrt{17} < 5$

---

This inequality shows that the value of $\sqrt{17}$ is greater than $4$ but less than $5$. Therefore, the point P, representing $\sqrt{17}$, lies between $4$ and $5$ on the number line.

---

The correct option is (A).

We need to find the value of $\sqrt{10} \times \sqrt{15}$.

---

Using the property of square roots that $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$ for non-negative numbers $a$ and $b$, we have:

$\sqrt{10} \times \sqrt{15} = \sqrt{10 \times 15}$

$\sqrt{10 \times 15} = \sqrt{150}$

---

Now, we can simplify $\sqrt{150}$ by finding its prime factors or finding perfect square factors.

Let's find the prime factors of $150$:

$150 = 10 \times 15 = (2 \times 5) \times (3 \times 5) = 2 \times 3 \times 5^2$.

So, $\sqrt{150} = \sqrt{2 \times 3 \times 5^2}$.

---

Using the property $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ and $\sqrt{c^2} = c$:

$\sqrt{2 \times 3 \times 5^2} = \sqrt{2 \times 3} \times \sqrt{5^2} = \sqrt{6} \times 5 = 5\sqrt{6}$.

---

Comparing this result with the given options:

(A) $\sqrt{25} = 5$

(B) $5\sqrt{6}$

(C) $6\sqrt{5}$

(D) $\sqrt{150}$ (which simplifies to $5\sqrt{6}$)

---

Both options (B) and (D) represent the same value. However, option (B) is the simplified form. Option (D) is also a correct representation of the value.

Often, questions ask for the value in the simplest radical form. The simplest form is $5\sqrt{6}$.

---

The correct option, usually implying the simplest form, is (B).

A number is rational if its decimal expansion is either terminating or non-terminating repeating. An irrational number has a non-terminating non-repeating decimal expansion.

---

Let's examine each option:

(A) $0.3796$ is a terminating decimal. Therefore, it is a rational number.

---

(B) $7.478478\dots$ can be written as $7.\overline{478}$. This is a non-terminating repeating decimal. Therefore, it is a rational number.

---

(C) $1.10110111011110\dots$ is a decimal expansion where the pattern of digits after the decimal point does not repeat in a fixed cycle. The number of $1$s between consecutive $0$s increases ($1$ one, $2$ ones, $3$ ones, $4$ ones, and so on). This is a non-terminating non-repeating decimal. Therefore, it is an irrational number.

---

(D) $4.\overline{123}$ is a non-terminating repeating decimal with the repeating block $123$. Therefore, it is a rational number.

---

Based on the analysis, the number with a non-terminating non-repeating decimal expansion is $1.10110111011110\dots$.

---

The correct option is (C).

A rational number in its simplest form, $\frac{p}{q}$, has a terminating decimal expansion if the prime factors of the denominator $q$ are only $2$ and $5$. Otherwise, it has a non-terminating recurring (repeating) decimal expansion.

---

The given fraction is $\frac{5}{6}$.

First, check if the fraction is in its simplest form. The numerator is $5$ and the denominator is $6$. The prime factors of $5$ is $5$. The prime factors of $6$ are $2$ and $3$. There are no common factors other than $1$, so the fraction $\frac{5}{6}$ is in its simplest form.

---

Next, examine the prime factors of the denominator, which is $6$.

The prime factorization of $6$ is $2 \times 3$.

---

Since the prime factors of the denominator include $3$ (which is not $2$ or $5$), the decimal expansion of $\frac{5}{6}$ will be non-terminating recurring.

---

We can verify this by performing the division:

$\frac{5}{6} = 0.8333\dots = 0.8\overline{3}$.

This is a non-terminating repeating decimal.

---

The correct option is (B).

We need to simplify the expression $(\sqrt{5}+\sqrt{2})^2$.

---

We can use the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$.

In this case, let $a = \sqrt{5}$ and $b = \sqrt{2}$.

---

Substitute these values into the identity:

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

---

Now, evaluate each term:

$(\sqrt{5})^2 = 5$

$(\sqrt{2})^2 = 2$

$2(\sqrt{5})(\sqrt{2}) = 2\sqrt{5 \times 2} = 2\sqrt{10}$

---

Substitute these back into the expanded expression:

$(\sqrt{5}+\sqrt{2})^2 = 5 + 2\sqrt{10} + 2$

---

Combine the constant terms:

$5 + 2 + 2\sqrt{10} = 7 + 2\sqrt{10}$

---

The simplified form of $(\sqrt{5}+\sqrt{2})^2$ is $7 + 2\sqrt{10}$.

---

Comparing this result with the given options, we find that it matches option (A).

---

The correct option is (A).

To rationalize the denominator of $\frac{1}{\sqrt{7}-2}$, we multiply both the numerator and the denominator by the conjugate of the denominator.

---

The denominator is $\sqrt{7}-2$. Its conjugate is $\sqrt{7}+2$.

---

Multiply the fraction by $\frac{\sqrt{7}+2}{\sqrt{7}+2}$:

$\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2} = \frac{1 \times (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}$

---

In the numerator, $1 \times (\sqrt{7}+2) = \sqrt{7}+2$.

In the denominator, we use the difference of squares formula: $(a-b)(a+b) = a^2 - b^2$.

Here, $a = \sqrt{7}$ and $b = 2$.

$(\sqrt{7}-2)(\sqrt{7}+2) = (\sqrt{7})^2 - (2)^2$

$(\sqrt{7})^2 = 7$

$(2)^2 = 4$

So, the denominator is $7 - 4 = 3$.

---

Putting the numerator and denominator together, we get:

$\frac{\sqrt{7}+2}{3}$

---

Comparing this result with the given options, we find that it matches option (A).

---

The correct option is (A).

We need to find the value of $(125)^{\frac{1}{3}}$.

---

The expression $a^{\frac{1}{n}}$ is equivalent to $\sqrt[n]{a}$, which represents the $n$-th root of $a$.

In this case, we have $(125)^{\frac{1}{3}}$, which means we need to find the cube root of 125.

So, $(125)^{\frac{1}{3}} = \sqrt[3]{125}$.

---

We are looking for a number that, when multiplied by itself three times, gives 125.

Let's consider the integer 5:

$5 \times 5 = 25$

$25 \times 5 = 125$

---

Since $5 \times 5 \times 5 = 5^3 = 125$, the cube root of 125 is 5.

Therefore, $(125)^{\frac{1}{3}} = 5$.

---

Comparing this result with the given options:

(A) 5

(B) 25

(C) $\frac{1}{5}$

(D) $\sqrt[3]{25}$

---

The calculated value matches option (A).

---

The correct option is (A).

We are asked to find the expression equal to $x^m \cdot x^n$.

---

This is a fundamental property of exponents, known as the Product Rule.

---

The rule states that when multiplying two exponential expressions with the same base, we keep the base and add the exponents.

Mathematically, this rule is expressed as:

$x^m \cdot x^n = x^{m+n}$

---

Comparing this result with the given options:

(A) $x^{m-n}$ (This is the rule for division: $\frac{x^m}{x^n} = x^{m-n}$)

(B) $x^{m+n}$ (This matches the Product Rule)

(C) $x^{mn}$ (This is the rule for a power of a power: $(x^m)^n = x^{mn}$)

(D) $x^{\frac{m}{n}}$ (This is not a standard exponent rule in this form)

---

Therefore, the expression $x^m \cdot x^n$ is equal to $x^{m+n}$.

---

The correct option is (B).

We want to convert the non-terminating repeating decimal $0.\overline{6}$ into the form $\frac{p}{q}$.

---

Let $x = 0.\overline{6}$.

Since there is only one repeating digit (6), we multiply equation (i) by $10^1 = 10$.

---

Now, subtract equation (i) from equation (ii).

Equation (ii) - Equation (i):

$10x - x = 6.6666\dots - 0.6666\dots$

$9x = 6$

---

Now, solve for $x$ by dividing both sides by 9.

$x = \frac{6}{9}$

---

Simplify the fraction $\frac{6}{9}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 3.

$\frac{\cancel{6}^2}{\cancel{9}_3} = \frac{2}{3}$

So, $x = \frac{2}{3}$.

---

Thus, $0.\overline{6}$ is equal to $\frac{2}{3}$.

---

Comparing this result with the given options, we find that it matches option (C).

---

The correct option is (C).

Let's examine each statement regarding the relationships between different sets of numbers:

---

We recall the definitions of the number sets:

Natural Numbers ($\mathbb{N}$): $\{1, 2, 3, 4, \dots\}$

Whole Numbers ($\mathbb{W}$): $\{0, 1, 2, 3, 4, \dots\}$

Integers ($\mathbb{Z}$): $\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$

Rational Numbers ($\mathbb{Q}$): Numbers that can be expressed in the form $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$. These include terminating or non-terminating repeating decimals.

Real Numbers ($\mathbb{R}$): All rational and irrational numbers.

---

Now, let's evaluate each statement:

(A) Every whole number is a natural number.

Whole numbers include $0$, whereas natural numbers start from $1$. The number $0$ is a whole number, but it is not a natural number. Therefore, this statement is false.

---

(B) Every integer is a rational number.

An integer $z$ can be written as $\frac{z}{1}$. Since $z$ is an integer and $1$ is a non-zero integer, any integer can be expressed in the form $\frac{p}{q}$. Therefore, every integer is a rational number. This statement is true.

---

(C) Every rational number is a real number.

The set of real numbers consists of all rational and irrational numbers. Thus, all rational numbers are included within the set of real numbers. Therefore, this statement is true.

---

(D) Every natural number is a whole number.

Natural numbers are $\{1, 2, 3, \dots\}$. Whole numbers are $\{0, 1, 2, 3, \dots\}$. Every number in the set of natural numbers is also present in the set of whole numbers. Therefore, every natural number is a whole number. This statement is true.

---

The statement that is false is (A).

---

The correct option is (A).

We need to find a rational number that lies between $\frac{1}{4}$ and $\frac{1}{2}$.

---

One way to find a rational number between two given rational numbers, say $a$ and $b$, is to calculate their average, which is $\frac{a+b}{2}$. This average will always lie between $a$ and $b$ (assuming $a \neq b$).

---

Let $a = \frac{1}{4}$ and $b = \frac{1}{2}$.

Their average is $\frac{\frac{1}{4} + \frac{1}{2}}{2}$.

---

First, find the sum of the fractions in the numerator:

$\frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{1 \times 2}{2 \times 2} = \frac{1}{4} + \frac{2}{4} = \frac{1+2}{4} = \frac{3}{4}$.

---

Now, divide the sum by 2:

$\frac{\frac{3}{4}}{2} = \frac{3}{4} \times \frac{1}{2} = \frac{3 \times 1}{4 \times 2} = \frac{3}{8}$.

---

So, $\frac{3}{8}$ is a rational number between $\frac{1}{4}$ and $\frac{1}{2}$.

---

Let's check if $\frac{3}{8}$ is indeed between $\frac{1}{4}$ and $\frac{1}{2}$. We can compare the fractions by finding a common denominator, which is 8.

$\frac{1}{4} = \frac{1 \times 2}{4 \times 2} = \frac{2}{8}$

$\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}$

The numbers are $\frac{2}{8}$, $\frac{3}{8}$, and $\frac{4}{8}$.

We can see that $\frac{2}{8} < \frac{3}{8} < \frac{4}{8}$, which means $\frac{1}{4} < \frac{3}{8} < \frac{1}{2}$.

---

Comparing $\frac{3}{8}$ with the given options:

(A) $\frac{1}{8}$ (This is less than $\frac{1}{4}$)

(B) $\frac{3}{8}$ (This is the number we found)

(C) $\frac{3}{4}$ (This is greater than $\frac{1}{2}$)

(D) $\frac{1}{3}$ (To compare $\frac{1}{3}$ with $\frac{1}{4}$ and $\frac{1}{2}$, find a common denominator, e.g., 12: $\frac{1}{4}=\frac{3}{12}$, $\frac{1}{3}=\frac{4}{12}$, $\frac{1}{2}=\frac{6}{12}$. Since $\frac{3}{12} < \frac{4}{12} < \frac{6}{12}$, $\frac{1}{3}$ is also between $\frac{1}{4}$ and $\frac{1}{2}$. However, the question asks for *a* rational number, and $\frac{3}{8}$ is one of the options.)

---

Both $\frac{3}{8}$ and $\frac{1}{3}$ are rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$. Since $\frac{3}{8}$ is explicitly given as an option, it is the correct answer.

---

The correct option is (B).

We need to determine the type of number $2-\sqrt{5}$.

---

Let's analyze the components of the expression $2-\sqrt{5}$:

The number $2$ is an integer, a whole number, and a rational number (since it can be written as $\frac{2}{1}$).

---

The number $\sqrt{5}$ is the square root of $5$. Since $5$ is not a perfect square (i.e., there is no integer $n$ such that $n^2 = 5$), $\sqrt{5}$ is an irrational number.

---

Now consider the operation between the two numbers: subtraction ($2 - \sqrt{5}$).

A key property of real numbers is that the sum or difference of a rational number and an irrational number is always an irrational number.

---

In this case, $2$ is a rational number and $\sqrt{5}$ is an irrational number. Their difference is $2-\sqrt{5}$.

Applying the property mentioned above, the result $2-\sqrt{5}$ must be an irrational number.

---

Therefore, the number $2-\sqrt{5}$ is an irrational number.

---

Comparing this conclusion with the given options:

(A) A rational number (False)

(B) An integer (False, as it is irrational and not an integer)

(C) An irrational number (True)

(D) A whole number (False, as it is irrational and not a whole number)

---

The correct option is (C).

We need to simplify each expression and determine if the result is a rational number.

---

Recall that a rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. An irrational number cannot be expressed in this form; its decimal expansion is non-terminating and non-repeating.

---

Let's evaluate each option:

(A) $(1+\sqrt{3}) - (1-\sqrt{3})$

$(1+\sqrt{3}) - (1-\sqrt{3}) = 1 + \sqrt{3} - 1 + \sqrt{3}$

Combine like terms: $(1 - 1) + (\sqrt{3} + \sqrt{3}) = 0 + 2\sqrt{3} = 2\sqrt{3}$.

Since $\sqrt{3}$ is an irrational number, and the product of a non-zero rational number (2) and an irrational number is irrational, $2\sqrt{3}$ is an irrational number.

---

(B) $(2+\sqrt{5})(2-\sqrt{5})$

This expression is in the form $(a+b)(a-b)$, which simplifies to $a^2 - b^2$.

Here, $a=2$ and $b=\sqrt{5}$.

$(2+\sqrt{5})(2-\sqrt{5}) = (2)^2 - (\sqrt{5})^2 = 4 - 5 = -1$.

$-1$ is an integer, and every integer can be written as a fraction with denominator 1 (e.g., $\frac{-1}{1}$). Therefore, $-1$ is a rational number.

---

(C) $2\sqrt{7} / \sqrt{2}$

We can simplify this by rationalizing the denominator or by combining the square roots:

$\frac{2\sqrt{7}}{\sqrt{2}} = 2 \times \frac{\sqrt{7}}{\sqrt{2}} = 2 \sqrt{\frac{7}{2}} = 2\sqrt{3.5}$.

Alternatively, multiply numerator and denominator by $\sqrt{2}$:

$\frac{2\sqrt{7}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{14}}{(\sqrt{2})^2} = \frac{2\sqrt{14}}{2} = \sqrt{14}$.

Since $14$ is not a perfect square, $\sqrt{14}$ is an irrational number.

---

(D) $\sqrt{12} / \sqrt{3}$

We can simplify this by combining the square roots:

$\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4}$.

$\sqrt{4} = 2$.

$2$ is an integer, and every integer is a rational number.

---

Based on the simplification, options (B) and (D) both simplify to rational numbers ($-1$ and $2$ respectively).

Assuming this is a single-choice question and there might be an error in the options provided, both (B) and (D) are valid results that are rational numbers based on the mathematical evaluation.

In typical exam questions of this format, only one option simplifies to a rational number. However, based *strictly* on the input provided and standard mathematical simplification:

Option (A) simplifies to an irrational number.

Option (B) simplifies to a rational number ($-1$).

Option (C) simplifies to an irrational number ($\sqrt{14}$).

Option (D) simplifies to a rational number ($2$).

---

Both (B) and (D) satisfy the condition of simplifying to a rational number. However, given the format of the options, it's possible only one intended answer exists. If forced to choose one based on common question patterns, option (B) involves a direct application of a common rationalization/difference of squares technique often tested.

---

Assuming the question intends for only one correct option, and given the choices, there appears to be an error in the question's options as provided. However, if we must select from the given options which simplifies to a rational number, both (B) and (D) do.

Let's highlight the results:

(A) Result: $2\sqrt{3}$ (Irrational)

(B) Result: $-1$ (Rational)

(C) Result: $\sqrt{14}$ (Irrational)

(D) Result: $2$ (Rational)

---

Both (B) and (D) simplify to rational numbers.

We need to find the value of $32^{\frac{2}{5}}$.

---

We can use the property of exponents that $a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m = (\sqrt[n]{a})^m$.

In this expression, $a = 32$, $m = 2$, and $n = 5$.

So, $32^{\frac{2}{5}} = (\sqrt[5]{32})^2$.

---

First, let's find the fifth root of 32, i.e., $\sqrt[5]{32}$. This means finding a number that, when multiplied by itself 5 times, equals 32.

Let's test small integer values:

$1^5 = 1 \times 1 \times 1 \times 1 \times 1 = 1$

$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 4 \times 2 \times 2 \times 2 = 8 \times 2 \times 2 = 16 \times 2 = 32$

So, $\sqrt[5]{32} = 2$.

---

Now, we need to calculate the square of this result:

$(\sqrt[5]{32})^2 = (2)^2 = 2 \times 2 = 4$.

---

Alternatively, we could use the property $a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}}$.

$32^{\frac{2}{5}} = (32^2)^{\frac{1}{5}} = (1024)^{\frac{1}{5}} = \sqrt[5]{1024}$.

Since $4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 16 \times 4 \times 4 \times 4 = 64 \times 4 \times 4 = 256 \times 4 = 1024$, we have $\sqrt[5]{1024} = 4$.

---

Both methods give the same result: $32^{\frac{2}{5}} = 4$.

---

Comparing this result with the given options, we find that it matches option (B).

---

The correct option is (B).

A number whose decimal expansion is non-terminating and non-recurring is an irrational number. A rational number has a decimal expansion that is either terminating or non-terminating repeating (recurring).

---

Let's examine the decimal expansion of each given option:

(A) $0.101010\dots$

This decimal expansion can be written as $0.\overline{10}$. The block of digits '$10$' repeats infinitely. This is a non-terminating repeating decimal. Therefore, it represents a rational number.

---

(B) $0.10110111011110\dots$

This decimal expansion shows a pattern where the number of $1$s between consecutive $0$s increases ($1$ one, then $2$ ones, then $3$ ones, etc.). There is no repeating block of digits. This is a non-terminating and non-recurring decimal. Therefore, it represents an irrational number.

---

(C) $0.\overline{1}$

This decimal expansion means $0.111111\dots$. The digit '$1$' repeats infinitely. This is a non-terminating repeating decimal. Therefore, it represents a rational number (equal to $\frac{1}{9}$).

---

(D) $0.125$

This decimal expansion terminates after three decimal places. This is a terminating decimal. Therefore, it represents a rational number (equal to $\frac{125}{1000} = \frac{1}{8}$).

---

Based on the analysis, the number with a non-terminating and non-recurring decimal expansion is $0.10110111011110\dots$.

---

The correct option is (B).

We need to determine the nature of the product of a non-zero rational number and an irrational number.

---

Let $r$ be a non-zero rational number and $i$ be an irrational number.

We want to find the type of number represented by the product $r \times i$.

---

Let's assume, for the sake of contradiction, that the product $r \times i$ is a rational number. Let this rational number be $q$.

Since $r$ is a non-zero rational number, its multiplicative inverse (reciprocal) $\frac{1}{r}$ exists and is also a rational number.

---

Multiply both sides of the equation $r \times i = q$ by $\frac{1}{r}$:

$\frac{1}{r} \times (r \times i) = \frac{1}{r} \times q$

Using the associative property of multiplication:

$(\frac{1}{r} \times r) \times i = \frac{q}{r}$

$1 \times i = \frac{q}{r}$

---

Since $q$ is a rational number and $r$ is a non-zero rational number, the quotient $\frac{q}{r}$ is also a rational number (the set of rational numbers is closed under division by non-zero elements).

So, the equation $i = \frac{q}{r}$ implies that $i$ is a rational number.

---

However, our initial assumption was that $i$ is an irrational number. This contradicts the conclusion that $i$ is a rational number.

Therefore, our initial assumption (that the product $r \times i$ is a rational number) must be false.

---

Hence, the product of a non-zero rational number and an irrational number must always be an irrational number.

---

The correct option is (B).

We are asked to determine the number of rational numbers between 0 and 1.

---

A key property of rational numbers is that they are dense on the number line. This means that between any two distinct rational numbers, there exists infinitely many other rational numbers.

---

Let $a$ and $b$ be two distinct rational numbers such that $a < b$. For example, take $a=0$ and $b=1$. Both 0 and 1 are rational numbers.

---

Consider the average of $a$ and $b$: $\frac{a+b}{2}$. Since $a$ and $b$ are rational, $\frac{a+b}{2}$ is also a rational number (as the set of rational numbers is closed under addition and division by a non-zero rational). Also, if $a < b$, then $a < \frac{a+b}{2} < b$.

---

Let's apply this to the interval between 0 and 1.

Between 0 and 1, we can find a rational number, for example, $\frac{0+1}{2} = \frac{1}{2}$.

Now consider the interval $(0, \frac{1}{2})$. Between 0 and $\frac{1}{2}$, we can find another rational number, for example, $\frac{0+1/2}{2} = \frac{1/2}{2} = \frac{1}{4}$.

Consider the interval $(\frac{1}{2}, 1)$. Between $\frac{1}{2}$ and 1, we can find another rational number, for example, $\frac{1/2+1}{2} = \frac{3/2}{2} = \frac{3}{4}$.

---

We can continue this process indefinitely. Between any two rational numbers we find, we can always find another rational number. This demonstrates that there is no end to finding new rational numbers in the interval (0, 1).

---

Therefore, there are infinitely many rational numbers between 0 and 1.

---

The term "countably many" refers to whether the elements of a set can be put into a one-to-one correspondence with the set of natural numbers. The set of all rational numbers is countably infinite. The interval (0, 1) contains an infinite subset of rational numbers, and this subset is also countably infinite. However, option (D) says "Countably many but finite", which is incorrect because the number is infinite.

---

The correct option is (C).

We need to find the value of the expression $\frac{1}{2+\sqrt{3}} + \frac{1}{2-\sqrt{3}}$.

---

To add these fractions, we can find a common denominator, which is the product of the individual denominators $(2+\sqrt{3})(2-\sqrt{3})$.

Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$, the common denominator is:

$(2+\sqrt{3})(2-\sqrt{3}) = (2)^2 - (\sqrt{3})^2 = 4 - 3 = 1$.

---

Now, rewrite each fraction with the common denominator 1:

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

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

---

Now, add the simplified expressions:

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

Combine the terms:

$2 - \sqrt{3} + 2 + \sqrt{3} = (2+2) + (-\sqrt{3}+\sqrt{3}) = 4 + 0 = 4$.

---

Alternatively, we could add the fractions directly using the common denominator:

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

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

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

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

$= \frac{4 + 0}{1} = \frac{4}{1} = 4$.

---

The value of the expression is 4.

---

Comparing this result with the given options, we find that it matches option (D).

---

The correct option is (D).

To find the number of digits in the repeating block of the decimal expansion of $\frac{1}{11}$, we perform the division of 1 by 11.

---

Dividing 1 by 11:

$1 \div 11 = 0.090909\dots$

---

We can see the digits '09' repeat indefinitely after the decimal point.

The decimal expansion is $0.\overline{09}$.

---

The repeating block of digits is $09$.

The digits in the repeating block are $0$ and $9$.

---

The number of digits in the repeating block is 2.

---

The correct option is (B).

We need to check each option to see which one correctly applies the laws of exponents.

---

Let's recall the basic exponent laws:

• Product Rule: $a^m \times a^n = a^{m+n}$
• Quotient Rule: $a^m / a^n = a^{m-n}$ ($a \neq 0$)
• Power of a Power Rule: $(a^m)^n = a^{mn}$
• Zero Exponent Rule: $a^0 = 1$ ($a \neq 0$)

---

Evaluate each option:

(A) $2^3 \times 2^4 = 4^7$

Using the Product Rule, $2^3 \times 2^4 = 2^{3+4} = 2^7$.

The statement $2^7 = 4^7$ is incorrect, as the bases are different. ($4^7 = (2^2)^7 = 2^{14}$).

---

(B) $(3^2)^3 = 3^5$

Using the Power of a Power Rule, $(3^2)^3 = 3^{2 \times 3} = 3^6$.

The statement $3^6 = 3^5$ is incorrect.

---

(C) $5^0 = 0$

Using the Zero Exponent Rule, $5^0 = 1$ for any non-zero base.

The statement $5^0 = 0$ is incorrect.

---

(D) $7^5 / 7^3 = 7^2$

Using the Quotient Rule, $7^5 / 7^3 = 7^{5-3} = 7^2$.

The statement $7^5 / 7^3 = 7^2$ is correct as per the exponent law.

---

Based on the evaluation, option (D) shows the correct use of exponent laws.

---

The correct option is (D).

The standard geometric method to represent $\sqrt{x}$ on the number line (where $x > 0$) involves constructing a semicircle. The steps typically are:

---

1. On a number line, mark a point O as the origin (representing 0).

2. Mark a point A such that OA = $x$ units (so A represents $x$). In this question, $x = 5.3$, so OA = 5.3 units.

3. Extend the line segment OA to a point B such that AB = 1 unit. So B represents $x+1$.

4. The total length of the segment OB is OA + AB = $x + 1$. This segment OB is the diameter of the semicircle.

5. Find the midpoint M of OB. Draw a semicircle with center M and radius MO or MB.

6. Draw a line perpendicular to OB passing through point A. This perpendicular intersects the semicircle at a point P.

7. The length of the segment AP is $\sqrt{x}$.

---

In this specific case, we want to represent $\sqrt{5.3}$, so $x = 5.3$.

According to the method, the diameter of the semicircle is $x+1$ units.

Diameter = $5.3 + 1$ units.

Diameter = $6.3$ units.

---

Comparing this value with the given options:

(A) 5.3 units (Incorrect, this is $x$)

(B) 6.3 units (Correct, this is $x+1$)

(C) $\sqrt{5.3}$ units (Incorrect, this is the length AP)

(D) $5.3+1$ units (This expression evaluates to 6.3, also correct, but option (B) provides the evaluated value)

---

Both options (B) and (D) represent the correct length of the diameter. However, option (B) provides the simplified numerical value, which is typically expected when asked for the "value". Option (D) shows the calculation based on the construction method ($x+1$). Assuming the question asks for the resulting value, option (B) is the most direct answer.

---

The correct option is (B).

We are given that the price of an item is $\textsf{₹} \sqrt{400}$. We need to determine if this number is rational or irrational.

---

First, let's calculate the value of $\sqrt{400}$.

We need to find the square root of $400$. This is the number that, when multiplied by itself, gives $400$.

We know that $20 \times 20 = 400$.

So, $\sqrt{400} = 20$.

---

The price of the item is $\textsf{₹} 20$.

---

Now, we need to classify the number $20$.

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. An irrational number cannot be expressed in this form; its decimal expansion is non-terminating and non-repeating.

The number $20$ can be expressed as $\frac{20}{1}$. Here, $p=20$ and $q=1$, both are integers, and $q \neq 0$.

Since $20$ can be written in the form $\frac{p}{q}$, it is a rational number.

---

Let's consider the options:

(A) A rational number (Correct, as 20 is rational)

(B) An irrational number (Incorrect, as 20 is rational)

(C) Neither rational nor irrational (Incorrect, as 20 is a real number and all real numbers are either rational or irrational)

(D) A complex number (Incorrect, a complex number is of the form $a+bi$, where $i=\sqrt{-1}$. $20$ is a real number, and real numbers are a subset of complex numbers, but the primary classification among the given options based on rationality is required).

---

The price, $\textsf{₹} 20$, is a rational number.

---

The correct option is (A).

We need to simplify the expression $(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3})$.

---

This expression is in the form $(a+b)(a-b)$. We can use the algebraic identity:

$(a+b)(a-b) = a^2 - b^2$

---

In this expression, we have $a = 3\sqrt{2}$ and $b = 2\sqrt{3}$.

---

Now, we calculate $a^2$ and $b^2$:

$a^2 = (3\sqrt{2})^2 = 3^2 \times (\sqrt{2})^2 = 9 \times 2 = 18$

$b^2 = (2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12$

---

Substitute these values into the identity $a^2 - b^2$:

$(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3}) = (3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12$

---

Finally, perform the subtraction:

$18 - 12 = 6$

---

The simplified value of the expression is 6.

---

Comparing this result with the given options, we find that it matches option (A).

---

The correct option is (A).

We need to find which of the given options, when multiplied by $\sqrt{8}$, results in a rational number.

---

First, let's simplify $\sqrt{8}$.

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.

---

Now, let's multiply $2\sqrt{2}$ by each of the options:

(A) Multiply by $\sqrt{2}$:

$(2\sqrt{2}) \times \sqrt{2} = 2 \times (\sqrt{2} \times \sqrt{2}) = 2 \times 2 = 4$.

$4$ is an integer, and every integer is a rational number. So, the product is rational.

---

(B) Multiply by $\sqrt{3}$:

$(2\sqrt{2}) \times \sqrt{3} = 2 \times (\sqrt{2} \times \sqrt{3}) = 2\sqrt{6}$.

Since $6$ is not a perfect square, $\sqrt{6}$ is an irrational number. The product of a non-zero rational number ($2$) and an irrational number ($\sqrt{6}$) is irrational.

---

(C) Multiply by $\sqrt{5}$:

$(2\sqrt{2}) \times \sqrt{5} = 2 \times (\sqrt{2} \times \sqrt{5}) = 2\sqrt{10}$.

Since $10$ is not a perfect square, $\sqrt{10}$ is an irrational number. The product of a non-zero rational number ($2$) and an irrational number ($\sqrt{10}$) is irrational.

---

(D) Multiply by $\sqrt{6}$:

$(2\sqrt{2}) \times \sqrt{6} = 2 \times (\sqrt{2} \times \sqrt{6}) = 2\sqrt{12}$.

We can simplify $\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$.

So, $2\sqrt{12} = 2 \times (2\sqrt{3}) = 4\sqrt{3}$.

Since $3$ is not a perfect square, $\sqrt{3}$ is an irrational number. The product of a non-zero rational number ($4$) and an irrational number ($\sqrt{3}$) is irrational.

---

Only when $\sqrt{8}$ is multiplied by $\sqrt{2}$ do we get a rational number (which is 4).

---

The correct option is (A).

We need to find the value of $(64)^{-\frac{1}{2}}$.

---

We can use the properties of exponents:

1. Negative exponent rule: $a^{-n} = \frac{1}{a^n}$ (for $a \neq 0$)

2. Fractional exponent rule: $a^{\frac{1}{n}} = \sqrt[n]{a}$

---

Apply the negative exponent rule first:

$(64)^{-\frac{1}{2}} = \frac{1}{(64)^{\frac{1}{2}}}$.

---

Now, apply the fractional exponent rule to the denominator $(64)^{\frac{1}{2}}$:

$(64)^{\frac{1}{2}} = \sqrt[2]{64} = \sqrt{64}$.

---

Find the square root of 64. We are looking for a non-negative number that, when multiplied by itself, equals 64.

We know that $8 \times 8 = 64$. So, $\sqrt{64} = 8$.

---

Substitute this value back into the expression:

$\frac{1}{(64)^{\frac{1}{2}}} = \frac{1}{8}$.

---

Alternatively, we could address the fractional exponent first: $(64)^{\frac{1}{2}} = \sqrt{64} = 8$. Then apply the negative exponent: $8^{-1} = \frac{1}{8}$.

---

Both methods yield the same result: $\frac{1}{8}$.

---

Comparing this result with the given options, we find that it matches option (B).

---

The correct option is (B).

We need to identify the statement that represents a correct property or law of exponents or roots.

---

Let's examine each option:

(A) $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$

This is generally false. For example, let $a=9$ and $b=16$.

$\sqrt{9+16} = \sqrt{25} = 5$.

$\sqrt{9} + \sqrt{16} = 3 + 4 = 7$.

Since $5 \neq 7$, the statement is incorrect.

---

(B) $(a^{\frac{1}{m}})^{\frac{1}{n}} = a^{\frac{1}{m+n}}$

Using the Power of a Power rule $(x^p)^q = x^{pq}$, we have:

$(a^{\frac{1}{m}})^{\frac{1}{n}} = a^{\frac{1}{m} \times \frac{1}{n}} = a^{\frac{1}{mn}}$.

The statement claims the result is $a^{\frac{1}{m+n}}$. This is generally incorrect, as $\frac{1}{mn} \neq \frac{1}{m+n}$ for most values of $m$ and $n$.

---

(C) $a^m \cdot b^m = (ab)^m$

This is a correct property of exponents, stating that when multiplying powers with the same exponent, we can multiply the bases and keep the exponent. This is sometimes called the Power of a Product rule in reverse, or the Product Rule for Same Exponents.

For example, $2^3 \times 5^3 = (2 \times 2 \times 2) \times (5 \times 5 \times 5) = (2 \times 5) \times (2 \times 5) \times (2 \times 5) = (2 \times 5)^3 = 10^3 = 1000$.

Also, $2^3 = 8$ and $5^3 = 125$. $8 \times 125 = 1000$. The statement holds true.

---

(D) $a^m / a^n = a^{\frac{m}{n}}$

Using the Quotient Rule for exponents, $a^m / a^n = a^{m-n}$ (for $a \neq 0$).

The statement claims the result is $a^{\frac{m}{n}}$. This is generally incorrect, as $m-n \neq \frac{m}{n}$ for most values of $m$ and $n$.

---

Based on the evaluation of each statement, only option (C) is a correct property of exponents.

---

The correct option is (C).

We need to classify each number given in Column A and match it with the appropriate description in Column B.

---

Let's analyze each number:

(i) $\sqrt{9}$

$\sqrt{9} = 3$. The number 3 is a counting number, a whole number, an integer, and a rational number.

Comparing with options in Column B:

(a) Irrational Number - False, 3 is rational.

(b) Whole Number but not Natural Number - False, 3 is a Natural Number.

(c) Rational Number (Terminating or Recurring) - True, 3 can be written as 3.0 (terminating) or 3/1.

(d) Natural Number - True, 3 is a Natural Number.

Among the provided options in Column B, 'Natural Number' (d) is the most specific correct classification listed for 3.

So, (i) matches with (d).

---

(ii) $0.333\dots$

This is a non-terminating repeating decimal, which is a characteristic of rational numbers. It can be written as $0.\overline{3}$ or $\frac{1}{3}$.

Comparing with options in Column B:

(a) Irrational Number - False, it's a repeating decimal.

(b) Whole Number but not Natural Number - False.

(c) Rational Number (Terminating or Recurring) - True, it is a recurring decimal and thus rational.

(d) Natural Number - False.

So, (ii) matches with (c).

---

(iii) $\sqrt{7}$

Since 7 is not a perfect square, its square root $\sqrt{7}$ is an irrational number. Its decimal expansion is non-terminating and non-repeating.

Comparing with options in Column B:

(a) Irrational Number - True.

(b) Whole Number but not Natural Number - False.

(c) Rational Number (Terminating or Recurring) - False.

(d) Natural Number - False.

So, (iii) matches with (a).

---

(iv) 0

The number 0 is a whole number and an integer. It is not a natural number.

Comparing with options in Column B:

(a) Irrational Number - False.

(b) Whole Number but not Natural Number - True, 0 is in the set of whole numbers $\{0, 1, 2, \dots\}$ but not in the set of natural numbers $\{1, 2, 3, \dots\}$.

(c) Rational Number (Terminating or Recurring) - True, 0 can be written as 0.0 (terminating) or 0/1, so it is rational. However, option (b) is a more specific classification given as an option.

(d) Natural Number - False.

Among the provided options in Column B, 'Whole Number but not Natural Number' (b) is the most specific correct classification listed for 0.

So, (iv) matches with (b).

---

Based on the analysis, the correct matching is:

(i) - (d)

(ii) - (c)

(iii) - (a)

(iv) - (b)

---

Let's check which option corresponds to this matching.

(A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d) - Incorrect.

(B) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b) - Correct.

(C) (i)-(d), (ii)-(a), (iii)-(c), (iv)-(b) - Incorrect.

(D) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b) - Incorrect.

---

The correct option is (B).

Let's analyze the Assertion and the Reason separately.

---

Assertion (A): $\sqrt{2}$ is an irrational number.

An irrational number is a number that cannot be expressed as a simple fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is not zero. The decimal expansion of an irrational number is non-terminating and non-repeating.

The number $\sqrt{2}$ is the square root of 2. Since 2 is not a perfect square, $\sqrt{2}$ cannot be expressed as a fraction of two integers. Its decimal expansion is $1.41421356\dots$, which is non-terminating and non-repeating.

Therefore, the Assertion (A) is true.

---

Reason (R): A number is irrational if its decimal expansion is non-terminating recurring.

A decimal expansion that is non-terminating but recurring (repeating) represents a rational number. For example, $0.333\dots = 0.\overline{3} = \frac{1}{3}$, which is a rational number. Similarly, $0.142857142857\dots = 0.\overline{142857} = \frac{1}{7}$, which is a rational number.

An irrational number has a decimal expansion that is non-terminating non-recurring.

Therefore, the Reason (R) is false.

---

Based on the analysis, Assertion (A) is true, and Reason (R) is false.

The correct option is the one that states A is true and R is false.

---

The correct option is (C).

Given:

The size of the bacterial colony on day 'n' is $\sqrt{n}$ units.

---

To Find:

The day 'n' when the colony size, $\sqrt{n}$, is a rational number, among the given options.

---

Solution:

A number is rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. A number of the form $\sqrt{n}$, where $n$ is a positive integer, is rational if and only if $n$ is a perfect square (i.e., $n = k^2$ for some integer $k$).

---

We need to check which of the given day numbers ($n$) results in $\sqrt{n}$ being a rational number.

---

Let's evaluate $\sqrt{n}$ for each option:

(A) Day 7: $n=7$. Colony size is $\sqrt{7}$. Since 7 is not a perfect square, $\sqrt{7}$ is an irrational number.

(B) Day 8: $n=8$. Colony size is $\sqrt{8}$. $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$. Since 8 is not a perfect square, $\sqrt{8}$ is an irrational number.

(C) Day 9: $n=9$. Colony size is $\sqrt{9}$. $\sqrt{9} = 3$. Since $9 = 3^2$, 9 is a perfect square. The number 3 can be written as $\frac{3}{1}$, which is in the form $\frac{p}{q}$. Therefore, 3 is a rational number.

(D) Day 10: $n=10$. Colony size is $\sqrt{10}$. Since 10 is not a perfect square, $\sqrt{10}$ is an irrational number.

---

The colony size is a rational number only on Day 9, where the size is $\sqrt{9}=3$.

---

The correct option is (C).

We need to identify the number that is NOT a rational number among the given options. A number is rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Numbers that cannot be expressed in this form are irrational.

---

Let's analyze each option:

(A) $- \frac{3}{4}$

This number is already in the form $\frac{p}{q}$, where $p = -3$ and $q = 4$. Both $-3$ and $4$ are integers, and $q=4 \neq 0$. Thus, $- \frac{3}{4}$ is a rational number.

---

(B) $\sqrt{4}$

The value of $\sqrt{4}$ is $2$. The number $2$ can be expressed in the form $\frac{p}{q}$ as $\frac{2}{1}$. Here, $p=2$ and $q=1$, both are integers, and $q=1 \neq 0$. Thus, $\sqrt{4}$ is a rational number.

---

(C) $0$

The number $0$ can be expressed in the form $\frac{p}{q}$ as $\frac{0}{1}$. Here, $p=0$ and $q=1$, both are integers, and $q=1 \neq 0$. Thus, $0$ is a rational number.

---

(D) $\pi$

The number $\pi$ is a mathematical constant representing the ratio of a circle's circumference to its diameter. Its value is approximately $3.1415926535\dots$. The decimal expansion of $\pi$ is non-terminating and non-repeating. It has been proven that $\pi$ cannot be expressed as a fraction of two integers.

Therefore, $\pi$ is an irrational number.

---

The question asks for the number that is NOT a rational number. Based on our analysis, $\pi$ is an irrational number, while the other options are rational numbers.

---

The correct option is (D).

We are asked to complete the identity $(a-\sqrt{b})(a+\sqrt{b}) = \dots$

---

This expression is in the form of the algebraic identity known as the difference of squares.

The identity states that $(x-y)(x+y) = x^2 - y^2$.

---

In the given expression $(a-\sqrt{b})(a+\sqrt{b})$, we can let $x = a$ and $y = \sqrt{b}$.

---

Applying the difference of squares identity with $x=a$ and $y=\sqrt{b}$:

$(a-\sqrt{b})(a+\sqrt{b}) = (a)^2 - (\sqrt{b})^2$

---

Now, simplify the terms:

$(a)^2 = a^2$

$(\sqrt{b})^2 = b$ (assuming $b \geq 0$ for $\sqrt{b}$ to be a real number)

---

Substituting these back into the expression:

$a^2 - b$

---

So, the identity is $(a-\sqrt{b})(a+\sqrt{b}) = a^2 - b$.

---

Comparing this result with the given options:

(A) $a^2 + b$

(B) $a^2 - b$

(C) $a - \sqrt{b}$

(D) $a + \sqrt{b}$

---

The result matches option (B).

---

The correct option is (B).

Given:

Initial investment = $\textsf{₹} 1000$.

The total amount after some time involves $\sqrt{2}$.

---

To Determine:

Whether the total amount is a rational or an irrational number.

---

Solution:

The initial investment of $\textsf{₹} 1000$ is an integer, and thus a rational number (it can be written as $\frac{1000}{1}$).

The number $\sqrt{2}$ is an irrational number because 2 is not a perfect square, and its decimal expansion is non-terminating and non-repeating ($1.41421356\dots$).

---

The problem states that the total amount involves $\sqrt{2}$. This suggests that $\sqrt{2}$ is part of the calculation for the total amount.

---

Recall the properties of operations involving rational and irrational numbers:

1. The sum or difference of a rational number and an irrational number is always irrational.

2. The product or quotient of a non-zero rational number and an irrational number is always irrational.

---

Since the initial investment is a non-zero rational number ($\textsf{₹} 1000$), and the total amount involves $\sqrt{2}$ (an irrational number), the calculation leading to the total amount is likely to be one of the operations described above, or a combination where the irrational component remains.

For example, if the total amount is calculated as $1000 + k\sqrt{2}$ or $1000 \times k\sqrt{2}$, where $k$ is a rational number, the result will typically be irrational.

---

Without the specific formula for the total amount, we rely on the implication of the phrase "involving $\sqrt{2}$". In the context of classifying numbers, this phrase implies that $\sqrt{2}$ contributes to the irrationality of the final amount.

If the total amount is, for instance, $\textsf{₹} (1000 + \sqrt{2})$, it's a rational number plus an irrational number, which is irrational.

If the total amount is $\textsf{₹} (1000 \times \sqrt{2})$, it's a rational number multiplied by an irrational number, which is irrational.

The only way the total amount might be rational is if the $\sqrt{2}$ term cancelled out or was part of an expression like $\sqrt{2} \times \sqrt{2} = 2$, but the phrasing suggests a more direct involvement of $\sqrt{2}$ in the final numerical value.

---

Based on the standard interpretation of such problems, the total amount involving $\sqrt{2}$ is expected to be an irrational number.

---

Comparing with the options:

(A) A rational number (Unlikely)

(B) An irrational number (Likely)

(C) An integer (Unlikely, as it would be rational)

(D) Cannot be determined (While the exact value isn't determined, the type of number can be inferred from the involvement of $\sqrt{2}$)

---

The most likely classification for a number involving $\sqrt{2}$ in such a context is an irrational number.

---

The correct option is (B).

We need to simplify $(16)^{\frac{3}{4}}$ using exponent laws and identify the correct simplification process among the options.

---

We have the expression $16^{\frac{3}{4}}$. We can use the property $a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m$ or $a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}}$. It's often easier to calculate the $n$-th root first, especially if the base is a perfect $n$-th power.

---

Let's express the base 16 as a power of a smaller number.

$16 = 2 \times 8 = 2 \times 2 \times 4 = 2 \times 2 \times 2 \times 2 = 2^4$.

---

Substitute $16 = 2^4$ into the expression:

$(16)^{\frac{3}{4}} = (2^4)^{\frac{3}{4}}$.

---

Now, apply the Power of a Power rule: $(a^m)^n = a^{mn}$.

$(2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}}$.

Multiply the exponents: $4 \times \frac{3}{4} = \cancel{4} \times \frac{3}{\cancel{4}} = 3$.

---

So, $(16)^{\frac{3}{4}} = 2^3$.

Calculate $2^3$: $2^3 = 2 \times 2 \times 2 = 8$.

---

Let's examine the given options to see which one follows a correct simplification path and reaches the correct result.

(A) $16 \times \frac{3}{4} = 12$

This is incorrect. Exponents are not simply multiplied by the base.

---

(B) $(16^3)^{\frac{1}{4}}$

This shows a correct application of the rule $a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}}$. However, it doesn't complete the simplification to a numerical value.

---

(C) $(2^4)^{\frac{3}{4}} = 2^3 = 8$

This option correctly expresses 16 as $2^4$, applies the Power of a Power rule $(2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3$, and calculates $2^3 = 8$. This shows a complete and correct simplification process.

---

(D) $(16^{\frac{1}{4}})^3 = 4^3 = 64$

This option correctly applies the rule $a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m$, where $(16^{\frac{1}{4}})^3$. It calculates $16^{\frac{1}{4}} = \sqrt[4]{16}$. The fourth root of 16 is 2, not 4 (since $2^4=16$, not $4^4$). So, $16^{\frac{1}{4}} = 2$. The next step should be $(2)^3$, not $4^3$. The calculation $4^3 = 64$ is correct, but the value of $(16^{\frac{1}{4}})^3$ is $(2)^3 = 8$. So, this option shows an incorrect intermediate step ($16^{\frac{1}{4}}=4$) and arrives at the wrong final result.

---

Option (C) follows a correct application of exponent laws and arrives at the correct value.

---

The correct option is (C).

We want to convert the repeating decimal $0.1\overline{23}$ into the form $\frac{p}{q}$.

Let $x$ be the given number:

---

The repeating part is '23' and the non-repeating part is '1'.

To convert such a decimal to a fraction, we use a method involving multiplying by powers of 10 to align the repeating part.

---

First, multiply equation (i) by a power of 10 such that the non-repeating part is just before the decimal point. There is one non-repeating digit ('1'), so we multiply by $10^1 = 10$.

---

Next, multiply equation (i) by a power of 10 such that the decimal point is just after the first occurrence of the repeating block. The repeating block is '23' (two digits). So, we need to move the decimal point past the '1' and the first '23', which is $1 + 2 = 3$ places. We multiply by $10^3 = 1000$.

---

Subtract equation (ii) from equation (iii) to eliminate the repeating part:

$1000x - 10x = 123.232323\dots - 1.232323\dots$

$990x = 122$

---

Solve for $x$:

$x = \frac{122}{990}$

This fraction can be simplified by dividing the numerator and denominator by 2: $\frac{61}{495}$.

---

Now let's check the given options against the steps we performed:

(A) Let $x = 0.12323\dots$. Multiply by 10: $10x = 1.2323\dots$. This matches step (ii).

(B) Let $x = 0.12323\dots$. Multiply by 100: $100x = 12.32323\dots$. This is not one of the crucial multiplication steps for the standard subtraction method.

(C) Let $x = 0.12323\dots$. Multiply by 1000: $1000x = 123.2323\dots$. This matches step (iii).

(D) Both (A) and (C). Since both step (A) (multiplying by 10) and step (C) (multiplying by 1000) are correct steps used in the conversion process, this option is correct.

---

The correct option is (D).

We need to classify each given number as rational or irrational.

---

Recall the definitions:

A rational number can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Its decimal expansion is either terminating or non-terminating repeating.

An irrational number cannot be expressed in the form $\frac{p}{q}$. Its decimal expansion is non-terminating and non-repeating.

---

Let's examine each number:

1. $\sqrt{2}$: This is the square root of a non-perfect square (2). Therefore, $\sqrt{2}$ is an irrational number. Its decimal expansion is non-terminating and non-repeating (e.g., $1.41421356\dots$).

---

2. $\frac{22}{7}$: This number is in the form $\frac{p}{q}$, where $p=22$ and $q=7$ are integers and $q \neq 0$. Therefore, $\frac{22}{7}$ is a rational number. Its decimal expansion is non-terminating repeating ($3.\overline{142857}$). Note that $\frac{22}{7}$ is a rational approximation of $\pi$, but it is not $\pi$ itself.

---

3. $3.14159$: This is a terminating decimal. Any terminating decimal can be written as a fraction with an integer numerator and a power of 10 as the denominator (e.g., $3.14159 = \frac{314159}{100000}$). Therefore, $3.14159$ is a rational number.

---

4. $\pi$: This is a fundamental mathematical constant. It is a well-known irrational number. Its decimal expansion is non-terminating and non-repeating (e.g., $3.1415926535\dots$).

---

Based on the classifications:

• $\sqrt{2}$ is Irrational.
• $\frac{22}{7}$ is Rational.
• $3.14159$ is Rational.
• $\pi$ is Irrational.

---

The irrational numbers in the list are $\sqrt{2}$ and $\pi$.

---

Comparing this finding with the given options:

(A) $\sqrt{2}$ and $\frac{22}{7}$ - Incorrect ( $\frac{22}{7}$ is rational)

(B) $\sqrt{2}$ and $\pi$ - Correct

(C) $\frac{22}{7}$ and $3.14159$ - Incorrect (Both are rational)

(D) $\pi$ and $3.14159$ - Incorrect ($3.14159$ is rational)

---

The correct option is (B).

The concept of successive magnification is a process used to visualize and locate the position of real numbers, particularly those with non-terminating decimal expansions, on the number line.

---

It involves taking a section of the number line, magnifying it (like looking through a magnifying glass), and repeatedly doing so on smaller and smaller intervals based on the digits of the decimal expansion.

---

This method is used to show that every real number has a unique point on the number line, and every point on the number line represents a unique real number. Real numbers include both rational and irrational numbers.

---

Let's consider the options:

(A) Converting decimals to fractions is an algebraic process, not related to locating numbers on the number line by zooming.

(B) Locating irrational numbers (which have non-terminating non-repeating decimal expansions) on the number line is precisely what successive magnification helps visualize. The method can also be used for rational numbers with non-terminating repeating decimal expansions.

(C) Simplifying expressions with exponents involves applying exponent rules, which is unrelated to the number line visualization technique.

(D) Rationalizing denominators is an algebraic technique used to eliminate radicals from the denominator of a fraction, unrelated to number line representation.

---

Therefore, the concept of successive magnification is used to locate real numbers, including irrational numbers, on the number line.

---

The correct option is (B).

We need to simplify the expression $\frac{\sqrt[4]{81}}{\sqrt[3]{8}}$.

---

Let's evaluate the numerator $\sqrt[4]{81}$ and the denominator $\sqrt[3]{8}$ separately.

---

Numerator: $\sqrt[4]{81}$. This is the fourth root of 81. We need to find a number that, when multiplied by itself four times, equals 81.

Consider the number 3: $3 \times 3 \times 3 \times 3 = 9 \times 3 \times 3 = 27 \times 3 = 81$.

So, $\sqrt[4]{81} = 3$.

---

Denominator: $\sqrt[3]{8}$. This is the cube root of 8. We need to find a number that, when multiplied by itself three times, equals 8.

Consider the number 2: $2 \times 2 \times 2 = 4 \times 2 = 8$.

So, $\sqrt[3]{8} = 2$.

---

Now, substitute the values back into the fraction:

$\frac{\sqrt[4]{81}}{\sqrt[3]{8}} = \frac{3}{2}$.

---

The simplified value of the expression is $\frac{3}{2}$.

---

Comparing this result with the given options, we find that it matches option (A).

---

The correct option is (A).

We need to find the value of $(\frac{27}{8})^{\frac{2}{3}}$.

---

We can use the exponent property $(a^m)^n = a^{mn}$ and $(\frac{a}{b})^n = \frac{a^n}{b^n}$.

Also, $a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m = (\sqrt[n]{a})^m$.

---

First, let's find the cube root of the fraction $\frac{27}{8}$, which is $(\frac{27}{8})^{\frac{1}{3}}$.

$(\frac{27}{8})^{\frac{1}{3}} = \frac{27^{\frac{1}{3}}}{8^{\frac{1}{3}}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}}$.

---

We know that $3^3 = 27$, so $\sqrt[3]{27} = 3$.

We know that $2^3 = 8$, so $\sqrt[3]{8} = 2$.

Thus, $(\frac{27}{8})^{\frac{1}{3}} = \frac{3}{2}$.

---

Now, we need to square this result:

$(\frac{27}{8})^{\frac{2}{3}} = \left( (\frac{27}{8})^{\frac{1}{3}} \right)^2 = (\frac{3}{2})^2$.

---

Square the fraction:

$(\frac{3}{2})^2 = \frac{3^2}{2^2} = \frac{9}{4}$.

---

Alternatively, we could first square the fraction and then take the cube root:

$(\frac{27}{8})^{\frac{2}{3}} = \sqrt[3]{(\frac{27}{8})^2} = \sqrt[3]{\frac{27^2}{8^2}} = \sqrt[3]{\frac{729}{64}}$.

Since $9^3 = 729$ and $4^3 = 64$, $\sqrt[3]{\frac{729}{64}} = \frac{\sqrt[3]{729}}{\sqrt[3]{64}} = \frac{9}{4}$.

---

Both methods yield the same result: $\frac{9}{4}$.

---

Comparing this result with the given options, we find that it matches option (B).

---

The correct option is (B).

A rational number is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is not equal to zero ($q \neq 0$).

---

Here are five examples of rational numbers:

1. $\frac{1}{2}$ (Here $p=1$ and $q=2$, both are integers and $q \neq 0$)

2. $5$ (This can be written as $\frac{5}{1}$, where $p=5$ and $q=1$)

3. $-\frac{3}{4}$ (Here $p=-3$ and $q=4$)

4. $0$ (This can be written as $\frac{0}{1}$, where $p=0$ and $q=1$)

5. $2.5$ (This terminating decimal can be written as $\frac{25}{10}$ or simplified to $\frac{5}{2}$)

An irrational number is any real number that cannot be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is not equal to zero ($q \neq 0$). In decimal form, irrational numbers are non-terminating and non-repeating.

---

Here are five examples of irrational numbers:

1. $\sqrt{2}$ (The square root of 2; its decimal expansion is non-terminating and non-repeating)

2. $\sqrt{3}$ (The square root of 3; its decimal expansion is non-terminating and non-repeating)

3. $\pi$ (Pi; the ratio of a circle's circumference to its diameter, approximately 3.14159..., its decimal expansion is non-terminating and non-repeating)

4. $e$ (Euler's number, approximately 2.71828..., the base of the natural logarithm; its decimal expansion is non-terminating and non-repeating)

5. $0.10110111011110...$ (A number with a decimal expansion that follows a pattern but does not repeat periodically)

The key difference between rational and irrational numbers lies in the nature of their decimal expansions.

---

Rational Numbers:

The decimal expansion of a rational number is either terminating or non-terminating repeating.

Examples:

• Terminating decimal: $\frac{1}{2} = 0.5$
• Terminating decimal: $\frac{3}{4} = 0.75$
• Non-terminating repeating decimal: $\frac{1}{3} = 0.333... = 0.\overline{3}$
• Non-terminating repeating decimal: $\frac{2}{7} = 0.285714285714... = 0.\overline{285714}$

---

Irrational Numbers:

The decimal expansion of an irrational number is non-terminating and non-repeating. This means the decimal digits go on infinitely without any block of digits repeating periodically.

Examples:

• $\sqrt{2} \approx 1.41421356...$
• $\pi \approx 3.14159265...$
• $e \approx 2.718281828...$
• $0.01001000100001...$ (A number where the pattern is clear but doesn't repeat periodically)

(a) For $\frac{5}{8}$:

To convert the fraction $\frac{5}{8}$ to a decimal, we divide the numerator (5) by the denominator (8).

$5 \div 8 = 0.625$

The decimal expansion is $0.625$. Since the decimal expansion ends after a finite number of digits, it is a terminating decimal.

---

(b) For $\frac{4}{11}$:

To convert the fraction $\frac{4}{11}$ to a decimal, we divide the numerator (4) by the denominator (11).

$4 \div 11 = 0.363636...$

The decimal expansion is $0.3636...$, which can be written as $0.\overline{36}$. Since the decimal expansion repeats a block of digits infinitely, it is a non-terminating repeating decimal.

Let $x = 0.\overline{6}$.

Since only one digit is repeating, we multiply equation (i) by 10.

Now, subtract equation (i) from equation (ii):

Equation (ii) - Equation (i):

$10x - x = 6.666... - 0.666...$

$9x = 6$

Solve for $x$:

$x = \frac{6}{9}$

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:

$x = \frac{\cancel{6}^2}{\cancel{9}_3} = \frac{2}{3}$

Thus, $0.\overline{6}$ can be expressed as $\frac{2}{3}$ in the form $\frac{p}{q}$, where $p=2$ and $q=3$ are integers and $q \neq 0$.

Let $x = 0.4\overline{7}$.

Since one digit (4) is not repeating after the decimal point, multiply equation (i) by 10:

Since one digit (7) is repeating, multiply equation (ii) by 10:

Now, subtract equation (ii) from equation (iii) to eliminate the repeating part:

Equation (iii) - Equation (ii):

$100x - 10x = 47.777... - 4.777...$

$90x = 43$

Solve for $x$:

$x = \frac{43}{90}$

The fraction $\frac{43}{90}$ is already in its simplest form as 43 is a prime number and 90 is not divisible by 43.

Thus, $0.4\overline{7}$ can be expressed as $\frac{43}{90}$ in the form $\frac{p}{q}$, where $p=43$ and $q=90$ are integers and $q \neq 0$.

Real numbers are the union of the set of rational numbers and the set of irrational numbers. They include all the numbers that can be represented on the number line.

---

The set of real numbers, denoted by $\mathbb{R}$, is formed by combining all rational numbers and all irrational numbers.

Mathematically, we can say:

$\mathbb{R} = \text{Rational Numbers} \cup \text{Irrational Numbers}$

Rational numbers are numbers that can be expressed as a fraction $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$ (e.g., $5, -3/4, 0.25$). Their decimal expansions are either terminating or non-terminating repeating.

Irrational numbers are numbers that cannot be expressed as a fraction $\frac{p}{q}$ (e.g., $\sqrt{2}, \pi, e$). Their decimal expansions are non-terminating and non-repeating.

Every real number is either rational or irrational, and no number can be both rational and irrational. This means the set of rational numbers and the set of irrational numbers are disjoint.

Therefore, the real number line is completely filled by the rational and irrational numbers together.

To represent the rational number $\frac{9}{5}$ on the number line, we first convert it to a mixed number or a decimal to understand its position.

$\frac{9}{5} = 1 \frac{4}{5}$ or $1.8$ in decimal form.

This tells us that the number $\frac{9}{5}$ lies between the integers 1 and 2.

---

Steps to represent $\frac{9}{5}$ on the number line:

1. Draw a number line and mark the integers 0, 1, 2, etc.

2. Since the number is $1 \frac{4}{5}$, we know it is between 1 and 2.

3. Look at the fractional part, which is $\frac{4}{5}$. The denominator is 5. This means we need to divide the segment between 1 and 2 into 5 equal parts.

4. Each of these parts represents a length of $\frac{1}{5}$.

5. Starting from 1, move 4 parts towards the right (towards 2). The point where you land after moving 4 parts from 1 represents $\frac{9}{5}$.

Let's denote the point corresponding to 1 as A and the point corresponding to 2 as B. Divide the segment AB into 5 equal parts. The points will represent $1\frac{1}{5}, 1\frac{2}{5}, 1\frac{3}{5}, 1\frac{4}{5}$. The fourth point from 1 is $\frac{9}{5}$.

---

A visual representation would look like this (conceptually, as drawing is not directly possible here):

      <----|----|----|----|----|----|----|----|----|----|----|---->
          -1   0    1              9/5  2
                    ^    ^    ^    ^    ^    ^
                    |    |    |    |    |    |
                    1   1+1/5 1+2/5 1+3/5 1+4/5 2
                    5/5  6/5  7/5  8/5  9/5  10/5
          

The point marked at $1\frac{4}{5}$ or $\frac{9}{5}$ is the required representation.

To represent the rational number $\frac{-7}{4}$ on the number line, we first convert it to a mixed number or a decimal to understand its position.

$\frac{-7}{4} = -\frac{7}{4} = -1 \frac{3}{4}$ or $-1.75$ in decimal form.

This tells us that the number $\frac{-7}{4}$ lies between the integers -2 and -1.

---

Steps to represent $\frac{-7}{4}$ on the number line:

1. Draw a number line and mark the integers ..., -2, -1, 0, 1, ...

2. Since the number is $-1 \frac{3}{4}$, we know it is between -2 and -1.

3. Look at the fractional part, which is $\frac{3}{4}$. The denominator is 4. This means we need to divide the segment between -2 and -1 into 4 equal parts.

4. Each of these parts represents a length of $\frac{1}{4}$.

5. Starting from -1, move 3 parts towards the left (towards -2). The point where you land after moving 3 parts from -1 represents $-1 \frac{3}{4}$, which is $\frac{-7}{4}$.

Let's denote the point corresponding to -1 as A and the point corresponding to -2 as B. Divide the segment AB into 4 equal parts. The points will represent $-1\frac{1}{4}, -1\frac{2}{4}, -1\frac{3}{4}$. The third point from -1 towards -2 is $\frac{-7}{4}$.

---

A visual representation would look like this (conceptually):

      <----|----|----|----|----|----|----|----|----|----|---->
          -2       -7/4 -1    0    1
           ^    ^    ^    ^
           |    |    |    |
          -2 -1-3/4 -1-2/4 -1-1/4 -1
          -8/4 -7/4 -6/4 -5/4 -4/4
          

The point marked at $-1\frac{3}{4}$ or $\frac{-7}{4}$ is the required representation.

To add the rational numbers $\frac{3}{8}$ and $\frac{-5}{12}$, we need to find a common denominator.

---

The denominators are 8 and 12. The least common multiple (LCM) of 8 and 12 is 24.

Let's find the LCM of 8 and 12 using prime factorization:

$\begin{array}{c|cc} 2 & 8 \;, & 12 \\ \hline 2 & 4 \; , & 6 \\ \hline 2 & 2 \; , & 3 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$

LCM $(8, 12) = 2 \times 2 \times 2 \times 3 = 24$.

---

Now, we convert both fractions to have a denominator of 24:

For $\frac{3}{8}$:

$\frac{3}{8} = \frac{3 \times 3}{8 \times 3} = \frac{9}{24}$

For $\frac{-5}{12}$:

$\frac{-5}{12} = \frac{-5 \times 2}{12 \times 2} = \frac{-10}{24}$

---

Now, add the fractions with the common denominator:

$\frac{3}{8} + \frac{-5}{12} = \frac{9}{24} + \frac{-10}{24}$

$= \frac{9 + (-10)}{24}$

$= \frac{9 - 10}{24}$

$= \frac{-1}{24}$

---

The sum of $\frac{3}{8}$ and $\frac{-5}{12}$ is $\frac{-1}{24}$.

We need to subtract $\frac{-4}{7}$ from $\frac{3}{14}$.

This means we need to calculate the difference $\frac{3}{14} - \left(\frac{-4}{7}\right)$.

---

Subtracting a negative number is equivalent to adding the corresponding positive number. So, the expression becomes:

$\frac{3}{14} - \left(\frac{-4}{7}\right) = \frac{3}{14} + \frac{4}{7}$

---

To add fractions, they must have a common denominator.

The denominators are 14 and 7.

The least common multiple (LCM) of 14 and 7 is 14.

We need to convert the fraction $\frac{4}{7}$ so that it has a denominator of 14.

To do this, we multiply both the numerator and the denominator of $\frac{4}{7}$ by 2:

$\frac{4}{7} = \frac{4 \times 2}{7 \times 2} = \frac{8}{14}$

---

Now we can rewrite the addition with the common denominator:

$\frac{3}{14} + \frac{8}{14}$

Now that the denominators are the same, we add the numerators and keep the common denominator:

$\frac{3 + 8}{14}$

---

Performing the addition in the numerator:

$\frac{3 + 8}{14} = \frac{11}{14}$

---

Thus, subtracting $\frac{-4}{7}$ from $\frac{3}{14}$ gives $\frac{11}{14}$.

We are asked to multiply the fraction $\frac{-9}{16}$ by the fraction $\frac{4}{-3}$.

---

To multiply fractions, we multiply the numerators together and multiply the denominators together.

So, the product is given by:

$\frac{-9}{16} \times \frac{4}{-3} = \frac{(-9) \times 4}{16 \times (-3)}$

---

Now, we perform the multiplication in the numerator and the denominator:

$(-9) \times 4 = -36$

$16 \times (-3) = -48$

So the product is $\frac{-36}{-48}$.

---

Next, we simplify the fraction $\frac{-36}{-48}$.

Since both the numerator and the denominator are negative, the fraction is positive:

$\frac{-36}{-48} = \frac{36}{48}$

To simplify this fraction, we find the greatest common divisor (GCD) of 36 and 48. The GCD of 36 and 48 is 12.

We divide both the numerator and the denominator by 12:

$\frac{36}{48} = \frac{\cancel{36}^{3}}{\cancel{48}_{4}} = \frac{3}{4}$

---

Therefore, the result of multiplying $\frac{-9}{16}$ by $\frac{4}{-3}$ is $\frac{3}{4}$.

---

Alternate Method: Simplifying before Multiplication

We can simplify the expression by canceling common factors between the numerators and denominators before performing the multiplication.

Consider the expression:

$\frac{-9}{16} \times \frac{4}{-3}$

We can cancel -9 in the numerator and -3 in the denominator by dividing both by -3:

$\cancel{-9} \div (-3) = 3$

$\cancel{-3} \div (-3) = 1$

We can cancel 4 in the numerator and 16 in the denominator by dividing both by 4:

$\cancel{4} \div 4 = 1$

$\cancel{16} \div 4 = 4$

Applying these cancellations, the expression becomes:

$\frac{\cancel{-9}^{3}}{\cancel{16}_{4}} \times \frac{\cancel{4}^{1}}{\cancel{-3}_{1}}$

Now, multiply the simplified fractions:

$\frac{3}{4} \times \frac{1}{1} = \frac{3 \times 1}{4 \times 1} = \frac{3}{4}$

---

Both methods yield the same result, $\frac{3}{4}$.

We are asked to divide the fraction $\frac{15}{28}$ by the fraction $\frac{-5}{7}$.

---

To divide a fraction by another fraction, we multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of $\frac{-5}{7}$ is obtained by swapping its numerator and denominator, which is $\frac{7}{-5}$.

---

So, the division becomes a multiplication:

$\frac{15}{28} \div \frac{-5}{7} = \frac{15}{28} \times \frac{7}{-5}$

---

Now, we multiply the numerators and the denominators. We can also simplify by canceling common factors before multiplying.

We can cancel 15 in the numerator and -5 in the denominator by dividing both by 5:

$\cancel{15} \div 5 = 3$

$\cancel{-5} \div 5 = -1$

We can cancel 7 in the numerator and 28 in the denominator by dividing both by 7:

$\cancel{7} \div 7 = 1$

$\cancel{28} \div 7 = 4$

Applying these cancellations, the expression becomes:

$\frac{\cancel{15}^{3}}{\cancel{28}_{4}} \times \frac{\cancel{7}^{1}}{\cancel{-5}_{-1}}$

---

Now, multiply the simplified fractions:

$\frac{3}{4} \times \frac{1}{-1} = \frac{3 \times 1}{4 \times (-1)} = \frac{3}{-4}$

The fraction $\frac{3}{-4}$ can be written as $-\frac{3}{4}$.

---

Therefore, the result of dividing $\frac{15}{28}$ by $\frac{-5}{7}$ is $-\frac{3}{4}$.

Given:

The expression $\sqrt{5} + 2\sqrt{5} - 3\sqrt{5}$.

---

To Simplify:

The given expression.

---

Solution:

The given expression is $\sqrt{5} + 2\sqrt{5} - 3\sqrt{5}$.

We observe that all terms in the expression contain the same radical, $\sqrt{5}$. These are called like terms.

Just as we can combine terms like $x + 2x - 3x$ by combining their coefficients, we can combine terms with the same radical by combining their coefficients.

The coefficients of the terms are:

• For $\sqrt{5}$, the coefficient is 1 (since $\sqrt{5} = 1\sqrt{5}$).
• For $2\sqrt{5}$, the coefficient is 2.
• For $-3\sqrt{5}$, the coefficient is -3.

---

To simplify the expression, we add and subtract the coefficients of the like terms and keep the common radical:

$(1 + 2 - 3)\sqrt{5}$

---

Now, we calculate the sum and difference of the coefficients:

$1 + 2 = 3$

$3 - 3 = 0$

So, the combined coefficient is 0.

---

Substitute the combined coefficient back into the expression:

$0 \times \sqrt{5}$

Any number multiplied by 0 is 0.

$0 \times \sqrt{5} = 0$

---

Thus, the simplified form of the expression $\sqrt{5} + 2\sqrt{5} - 3\sqrt{5}$ is 0.

Given:

The expression $\sqrt{12} + \sqrt{27}$.

---

To Simplify:

The given expression.

---

Solution:

To simplify the sum of radicals $\sqrt{12} + \sqrt{27}$, we first need to simplify each radical individually by finding perfect square factors within the radicand.

---

Simplifying $\sqrt{12}$:

We look for the largest perfect square factor of 12. The factors of 12 are 1, 2, 3, 4, 6, 12. The perfect square factors are 1 and 4. The largest perfect square factor is 4.

We can write 12 as the product of 4 and 3:

$12 = 4 \times 3$

Now, we use the property of radicals $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$:

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3}$

Since $\sqrt{4} = 2$, we get:

$\sqrt{12} = 2\sqrt{3}$

---

Simplifying $\sqrt{27}$:

We look for the largest perfect square factor of 27. The factors of 27 are 1, 3, 9, 27. The perfect square factors are 1 and 9. The largest perfect square factor is 9.

We can write 27 as the product of 9 and 3:

$27 = 9 \times 3$

Now, we use the property of radicals $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$:

$\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3}$

Since $\sqrt{9} = 3$, we get:

$\sqrt{27} = 3\sqrt{3}$

---

Now we substitute the simplified radicals back into the original expression:

$\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3}$

We can now combine these terms because they are like terms (they both have $\sqrt{3}$ as the radical part). We add their coefficients:

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

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

---

Thus, the simplified form of the expression $\sqrt{12} + \sqrt{27}$ is $5\sqrt{3}$.