Fundamental Concepts of Algebra

$x$ increased by 23 means 23 is added to $x$.

Hence, the required form is $\bm{x + 23}$.

$y$ decreased by 15 means 15 is subtracted from $y$.

Hence, the required form is $\bm{y - 15}$.

Thrice $y = 3y$.

Hence, the required form is $\bm{x - 3y}$.

$x$ taken away from 20 means $x$ is subtracted from 20.

Hence, the required form is $\bm{20 - x}$.

Four times $x = 4x$ and five times $y = 5y$.

Hence, the required form is $\bm{4x - 5y}$.

Quotient of $y$ by $10 = \frac{y}{10}$.

Hence, the required form is $\bm{x + \frac{y}{10}}$.

Sum of $a$ and $b = a + b$, and product of $a$ and $b = ab$.

Hence, the required form is $\bm{(a + b) - ab}$.

Quotient of $a$ by $3 = \frac{a}{3}$ and sum of $b$ and $3 = (b + 3)$.

Hence, the required form is $\bm{\frac{a}{3}(b + 3)}$.

$x$ exceeds 14 by 5 means $x - 14 = 5$.

Hence, the required form is $\bm{x - 14 = 5}$.

Twice $x = 2x$. Decreased by 9 gives $2x - 9$, and this equals 11.

Hence, the required form is $\bm{2x - 9 = 11}$.

45 decreased by $y = 45 - y$, and this equals 29.

Hence, the required form is $\bm{45 - y = 29}$.

Sum of $x$ and $4 = (x + 4)$. One third of it $= \frac{1}{3}(x + 4)$, which equals 10.

Hence, the required form is $\bm{\frac{1}{3}(x + 4) = 10}$.

Quotient of $a$ by 2 = $\frac{a}{2}$. 16 less than it = $\frac{a}{2} - 16$, which equals 8.

Hence, the required form is $\bm{\frac{a}{2} - 16 = 8}$.

Nine times $m = 9m$, which is greater than 18.

Hence, the required form is $\bm{9m \gt 18}$.

Twice $p = 2p$, and 7 is greater than it.

Hence, the required form is $\bm{7 \gt 2p}$.

Quotient of $a$ by $6 = \frac{a}{6}$, which is less than 1.

Hence, the required form is $\bm{\frac{a}{6} \lt 1}$.

Three times $x = 3x$ and twice $y = 2y$. Since $3x$ is 12 less than $2y$, we get $3x = 2y - 12$.

Hence, the required form is $\bm{3x = 2y - 12}$.

When the same factor is multiplied repeatedly, we write it using exponents (powers).

$a \times a \times a \times .....$ taken 10 times = $a^{10}$.

Hence, the exponential form is $\bm{a^{10}}$.

When the same factor is multiplied repeatedly, we write it using exponents (powers).

$r \times r \times r \times .....$ taken 6 times = $r^{6}$

Hence, the exponential form is $\bm{r^6}$.

When the same factor is multiplied repeatedly, we write it using exponents (powers).

$$\begin{aligned}\Rightarrow 2 \times m \times m \times n \times n \times n\\[1em] \Rightarrow 2 \times m^2 \times n^3\\[1em] \Rightarrow \bm{2m^2n^3}\end{aligned}$$

Hence, the exponential form is $\bm{2m^2n^3}$.

When the same factor is multiplied repeatedly, we write it using exponents (powers).

$$\begin{aligned}\Rightarrow 3 \times 5 \times p \times p \times p \times q \times q \times r\\[1em] \Rightarrow 15 \times p^3 \times q^2 \times r\\[1em] \Rightarrow \bm{15p^3q^2r}\end{aligned}$$

Hence, the exponential form is $\bm{15p^3q^2r}$.

Product form means expanding exponents into repeated multiplication.

Hence, the product form is $\bm{6 \times x \times x \times y \times y \times y}$.

Product form means expanding exponents into repeated multiplication.

Hence, the product form is $\bm{3 \times a \times a \times b \times b \times b \times c \times c \times c \times c}$.

Product form means expanding exponents into repeated multiplication.

Hence, the product form is $\bm{2 \times p \times p \times p \times q \times q \times q \times r \times r \times r}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{2 + 3x}$ are $\bm{2}$ and $\bm{3x}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{a + 2b - 3c}$ are $\bm{a}$, $\bm{2b}$ and $\bm{-3c}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{3 - 2x + 5x^2}$ are $\bm{3}$, $\bm{-2x}$ and $\bm{5x^2}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{1 - 8xyz}$ are $\bm{1}$ and $\bm{-8xyz}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{2p^2 - 3pq + q^2 - 1}$ are $\bm{2p^2}$, $\bm{-3pq}$, $\bm{q^2}$ and $\bm{-1}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{3x^2 - 5x + \frac{2}{x} - \frac{3}{x^2}}$ are $\bm{3x^2}$, $\bm{-5x}$, $\bm{\frac{2}{x}}$ and $\bm{-\frac{3}{x^2}}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{-5y^3 + 2y^2 + 8y - \frac{3}{4}}$ are $\bm{-5y^3}$, $\bm{2y^2}$, $\bm{8y}$ and $\bm{-\frac{3}{4}}$.

The parts of an expression separated by a + or − sign are called its terms.

Hence, the terms of $\bm{2xy + 3yz - 5zx + \frac{1}{9}}$ are $\bm{2xy}$, $\bm{3yz}$, $\bm{-5zx}$ and $\bm{\frac{1}{9}}$.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$5 + x$ has 2 terms, namely 5 and $x$. So, it is a binomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$5x$ has 1 term only. So, it is a monomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$\frac{5}{x}$ has 1 term only. So, it is a monomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$abc$ has 1 term only. So, it is a monomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$a + b + c$ has 3 terms, namely $a, b$ and $c$. So, it is a trinomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$-8x$ has 1 term only. So, it is a monomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$x^2 + x - \frac{1}{2}$ has 3 terms, namely $x^2, x \text{ and }-\frac{1}{2}$. So, it is a trinomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$3x + \frac{y}{2}$ has 2 terms, namely $3x \text{ and } \frac{y}{2}$. So, it is a binomial.

An expression having one term is a monomial, two terms is a binomial and three terms is a trinomial.

$3 - \frac{y}{2} + z$ has 3 terms, namely $3, -\frac{y}{2} \text{ and } z$. So, it is a trinomial.

Any factor of a monomial is called the coefficient of the remaining factor.

$-5ab^2 = a \times (-5b^2)$

Hence, the coefficient of $\bm{a}$ in $\bm{-5ab^2}$ is $\bm{-5b^2}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$-px = p \times (-x)$

Hence, the coefficient of $\bm{p}$ in $\bm{-px}$ is $\bm{-x}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$-x = x \times (-1)$

Hence, the coefficient of $\bm{x}$ in $\bm{-x}$ is $\bm{-1}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$\frac{1}{3}xy = x \times \frac{1}{3}y$

Hence, the coefficient of $\bm{x}$ in $\bm{\frac{1}{3}xy}$ is $\bm{\frac{1}{3}y}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$-3xyz = xz \times (-3y)$

Hence, the coefficient of $\bm{xz}$ in $\bm{-3xyz}$ is $\bm{-3y}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$abcp = p \times abc$

Hence, the coefficient of $\bm{p}$ in $\bm{abcp}$ is $\bm{abc}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$-6axy^2 = y^2 \times (-6ax)$

Hence, the coefficient of $\bm{y^2}$ in $\bm{-6axy^2}$ is $\bm{-6ax}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$abc = abc \times 1$

Hence, the coefficient of $\bm{abc}$ in $\bm{abc}$ is $\bm{1}$.

Any factor of a monomial is called the coefficient of the remaining factor.

$8abx^2yz = x^2y \times 8abz$

Hence, the coefficient of $\bm{x^2y}$ in $\bm{8abx^2yz}$ is $\bm{8abz}$.

Numerical coefficient is the number (along with the sign, + or −).

Literal coefficient is the variable part in the monomials.

Hence, in $\bm{9a^2b}$ the numerical coefficient is $\bm{9}$ and the literal coefficient is $\bm{a^2b}$.

Numerical coefficient is the number (along with the sign, + or −).

Literal coefficient is the variable part in the monomials.

Hence, in $\bm{-3xyz^2}$ the numerical coefficient is $\bm{-3}$ and the literal coefficient is $\bm{xyz^2}$.

Numerical coefficient is the number (along with the sign, + or −).

Literal coefficient is the variable part in the monomials.

Hence, in $\bm{\frac{3}{5}apq}$ the numerical coefficient is $\bm{\frac{3}{5}}$ and the literal coefficient is $\bm{apq}$.

Numerical coefficient is the number (along with the sign, + or −).

Literal coefficient is the variable part in the monomials.

Hence, in $\bm{-\frac{7}{2}pqr}$ the numerical coefficient is $\bm{-\frac{7}{2}}$ and the literal coefficient is $\bm{pqr}$.

Numerical coefficient is the number (along with the sign, + or −).

Literal coefficient is the variable part in the monomials.

Hence, in $\bm{-a^3b^2c}$ the numerical coefficient is $\bm{-1}$ and the literal coefficient is $\bm{a^3b^2c}$.

Numerical coefficient is the number (along with the sign, + or −).

Literal coefficient is the variable part in the monomials.

Hence, in $\bm{xy^2z^3}$ the numerical coefficient is $\bm{1}$ and the literal coefficient is $\bm{xy^2z^3}$.

In a monomial, each quantity of the given product is called a factor of the monomial.

$3a = 3 \times a$

Hence, the factors are $\bm{3}, \bm{a}$ and $\bm{3a}$.

In a monomial, each quantity of the given product is called a factor of the monomial.

$-x = (-1) \times x$

Hence, the factors are $\bm{-1}, \bm{x}$ and $\bm{-x}$.

In a monomial, each quantity of the given product is called a factor of the monomial.

$ab = a \times b$

Hence, the factors are $\bm{a}, \bm{b}$ and $\bm{ab}$.

In a monomial, each quantity of the given product is called a factor of the monomial.

$3pq^2 = 3 \times p \times q \times q$

Taking the factors one at a time, two at a time, three at a time and all four together, we get:

One at a time : $3, p, q$

Two at a time : $3p, 3q, pq, q^2$

Three at a time : $3pq, 3q^2, pq^2$

All four together : $3pq^2$

Hence, the factors are $\bm{3}, \bm{p}, \bm{q}, \bm{3p}, \bm{3q}, \bm{pq}, \bm{q^2}, \bm{3pq}, \bm{3q^2}, \bm{pq^2}$ and $\bm{3pq^2}$.

Terms having the same literal factors are like terms.

The terms containing $x$ are $5x, \frac{3}{5}x \text{ and } x$.

The terms containing $y$ are $-6y, -y, \frac{5}{7}y \text{ and }-\frac{1}{2}y$.

Hence, the groups of like terms are $\bm{\left(5x, \frac{3}{5}x, x\right)\text{ and }\left(-6y, -y, \frac{5}{7}y, -\frac{1}{2}y\right)}$.

Terms having the same literal factors are like terms.

The terms containing $xy^2$ are $xy^2, 5y^2x \text{ and } -\frac{3}{5}xy^2$.

The terms containing $x^2y$ are $-3x^2y \text{ and } \frac{2}{3}yx^2$.

Hence, the groups of like terms are

$\bm{\left(xy^2, 5y^2x, -\frac{3}{5}xy^2\right)\text{ and }\left(-3x^2y, \frac{2}{3}yx^2\right)}$.

Terms having the same literal factors are like terms.

The terms containing $p^2q$ are $2p^2q$ and $8qp^2$.

The terms containing $pq^2$ are $-3pq^2$ and $7q^2p$.

The terms containing $p^2q^2$ are $6p^2q^2$ and $-8q^2p^2$.

Hence, the groups of like terms are

$\bm{\left(2p^2q, 8qp^2\right), \left(-3pq^2, 7q^2p\right)\text{ and }\left(6p^2q^2, -8q^2p^2\right)}$.

(i) False. $8a$ is a single term (a monomial); 8 and $a$ are its factors, not its terms.

(ii) False. $6 + xy$ has 2 terms, so it is a binomial.

(iii) False. $3ab$ has only 1 term, so it is a monomial.

(iv) True. The literal factors $xy$ and $axy$ are different, so the terms are unlike.

(v) True. Both have the same literal factor $a^2b$, so they are like terms.

(vi) True. Since $yx = xy$, both have the same literal factor $xy$, so they are like terms.

(vii) False. 6 has no literal factor while $6p$ has the literal factor $p$, so they are unlike terms.

(viii) False. Both $x$ and $-x$ have the same literal factor $x$, so they are like terms.

(ix) True. Both have the same literal factor $yx$, so they are like terms.

(x) False. $-5xy = y \times (-5x)$, so the coefficient of $y$ is $-5x$.

An algebraic expression is a polynomial only when every power of the variable is a positive integer (or zero). No term of a polynomial can be of the form $\frac{1}{x}, \frac{1}{x^2}$, etc.

The degree of a polynomial is the highest power of the variable present in it.

(i) $4x^3 - 3x^2 + 5x - 2$

All powers of $x$ are positive integers, so it is a polynomial. The highest power of $x$ is 3.

Hence, it is a polynomial of degree 3.

(ii) $\frac{3}{5}x^2 - \frac{5}{4}x + \frac{2}{3}$

All powers of $x$ are positive integers, so it is a polynomial. The highest power of $x$ is 2.

Hence, it is a polynomial of degree 2.

(iii) $\frac{1}{4}x^4 + 2x^3 - x + 7$

All powers of $x$ are positive integers, so it is a polynomial. The highest power of $x$ is 4.

Hence, it is a polynomial of degree 4.

(iv) $x^3 + x^2 - x + \frac{1}{x} + 5$

The term $\frac{1}{x}$ does not have a positive integral power of $x$.

Hence, it is not a polynomial.

(v) $x^6 - 1$

All powers of $x$ are positive integers, so it is a polynomial. The highest power of $x$ is 6.

Hence, it is a polynomial of degree 6.

(vi) $x^2 + \frac{1}{x^2} - 1$

The term $\frac{1}{x^2}$ does not have a positive integral power of $x$.

Hence, it is not a polynomial.

(vii) $z - \frac{1}{z} + 3$

The term $\frac{1}{z}$ does not have a positive integral power of $z$.

Hence, it is not a polynomial.

(viii) $y^3 - 4y + 6$

All powers of $y$ are positive integers, so it is a polynomial. The highest power of $y$ is 3.

Hence, it is a polynomial of degree 3.

(ix) $z^2 - \frac{1}{8}$

All powers of $z$ are positive integers, so it is a polynomial. The highest power of $z$ is 2 and $\frac{1}{8}$ is a constant term.

Hence, it is a polynomial of degree 2.

The highest power of the variable is the degree of the polynomial.

In $2 + x + 3x^2$, the highest power of $x$ is 2.

Hence, the degree is 2.

The highest power of the variable is the degree of the polynomial.

In $x - 8$, the highest power of $x$ is 1.

Hence, the degree is 1.

The highest power of the variable is the degree of the polynomial.

In $6 - 5x + 2x^3$, the highest power of $x$ is 3.

Hence, the degree is 3.

The highest power of the variable is the degree of the polynomial.

In $4 + y^2$, the highest power of $y$ is 2.

Hence, the degree is 2.

The highest power of the variable is the degree of the polynomial.

In $y - y^3$, the highest power of $y$ is 3.

Hence, the degree is 3.

The highest power of the variable is the degree of the polynomial.

In $z - z^2 + 3z^5$, the highest power of $z$ is 5.

Hence, the degree is 5.

The highest power of the variable is the degree of the polynomial.

In $m^3 - 8$, the highest power of $m$ is 3.

Hence, the degree is 3.

The highest power of the variable is the degree of the polynomial.

In $3 + 2p + p^2 - 6p^4$, the highest power of $p$ is 4.

Hence, the degree is 4.

The highest power of the variable is the degree of the polynomial.

In $1 - 2t + t^2 - 3t^3 + t^4$, the highest power of $t$ is 4.

Hence, the degree is 4.

In a polynomial of two or more variables, every variable must have only positive integral powers. The degree is the highest sum of the powers of the variables in a term.

(i) $m^2 + n^2 - mn$

Degrees of the terms are 2, 2 and (1 + 1) = 2.

Hence, it is a polynomial of degree 2.

(ii) $x^2 + y^2 + z^2 - xy - yz - zx$

Degrees of the terms are 2, 2, 2, 2, 2 and 2.

Hence, it is a polynomial of degree 2.

(iii) $y^2 + z^2 - \frac{2}{y} + 3z$

The term $\frac{2}{y}$ does not have a positive integral power of $y$.

Hence, it is not a polynomial.

(iv) $x + \frac{1}{y} + \frac{2}{z}$

The terms $\frac{1}{y} \text{ and } \frac{2}{z}$ do not have positive integral powers.

Hence, it is not a polynomial.

(v) $a^2b + ab^2 - 3ab$

Degrees of the terms are (2 + 1), (1 + 2) and (1 + 1), i.e., 3, 3 and 2.

Hence, it is a polynomial of degree 3.

(vi) $x^2 + \frac{1}{x^2} + 4$

The term $\frac{1}{x^2}$ does not have a positive integral power of $x$.

Hence, it is not a polynomial.

(vii) $\frac{1}{2}p^2q^2 + \frac{3}{4}q^2r^2 + \frac{5}{6}p^2r^2$

Degrees of the terms are (2 + 2), (2 + 2) and (2 + 2), i.e., 4, 4 and 4.

Hence, it is a polynomial of degree 4.

(viii) $x + \frac{1}{x}$

The term $\frac{1}{x}$ does not have a positive integral power of $x$.

Hence, it is not a polynomial.

In $x + 2y - 3z$, degrees of the terms are 1, 1 and 1.

Hence, the degree is 1.

In $xy + yz + zx$, degrees of the terms are (1 + 1), (1 + 1) and (1 + 1), i.e., 2, 2 and 2.

Hence, the degree is 2.

In $x^2 + xyz$, degrees of the terms are 2 and (1 + 1 + 1) = 3.

Hence, the degree is 3.

In $m^2n + mn^2 + 5mn - 6$, degrees of the terms are (2 + 1), (1 + 2), (1 + 1) and 0, i.e., 3, 3, 2 and 0.

Hence, the degree is 3.

In $p^2q^2 + 2pq^2 - p^2q + 1$, degrees of the terms are (2 + 2), (1 + 2), (2 + 1) and 0, i.e., 4, 3, 3 and 0.

Hence, the degree is 4.

In $m^2n^3 + mn^2 + n^4$, degrees of the terms are (2 + 3), (1 + 2) and 4, i.e., 5, 3 and 4.

Hence, the degree is 5.

(i) True. In $2 - x^3$ the highest power of $x$ is 3, so its degree is 3.

(ii) False. All the powers of $x$ are positive integers, so $\frac{2}{3} + \frac{3}{5}x^2 + x^3$ is a polynomial of degree 3.

(iii) False. The term $\frac{1}{x}$ does not have a positive integral power of $x$, so it is not a polynomial.

(iv) True. The degrees of the terms are 2, 2 and 2, so the degree is 2.

(v) True. The degrees of the terms are 3, 3, 2 and 1, so the degree is 3.

(vi) False. The term $\frac{m}{n}$ does not have a positive integral power of $n$, so it is not a polynomial.

Substituting $x = 3$ and $y = 4$, we get:

$$\begin{aligned}\Rightarrow 2x + 3y\\[1em] \Rightarrow 2 \times 3 + 3 \times 4\\[1em] \Rightarrow 6 + 12\end{aligned}$$

Hence, $\bm{2x + 3y = 18}$.

Substituting $x = 5$ and $y = 3$, we get:

$$\begin{aligned}\Rightarrow 4x - 5y\\[1em] \Rightarrow 4 \times 5 - 5 \times 3\\[1em] \Rightarrow 20 - 15\end{aligned}$$

Hence, $\bm{4x - 5y = 5}$.

Substituting $a = 10, b = 7$ and $c = 5$, we get:

$$\begin{aligned}\Rightarrow 3a - 4b + 5c\\[1em] \Rightarrow 3 \times 10 - 4 \times 7 + 5 \times 5\\[1em] \Rightarrow 30 - 28 + 25\end{aligned}$$

Hence, $\bm{3a - 4b + 5c = 27}$.

Substituting $m = 15, n = -4$ and $p = 20$, we get:

$$\begin{aligned}\Rightarrow 8m - 5n - 3p\\[1em] \Rightarrow 8 \times 15 - 5 \times (-4) - 3 \times 20\\[1em] \Rightarrow 120 + 20 - 60\end{aligned}$$

Hence, $\bm{8m - 5n - 3p = 80}$.

Substituting $a = 2, b = 3$ and $c = 4$, we get:

$$\begin{aligned}\Rightarrow 5ab + 7bc - 3ca\\[1em] \Rightarrow 5 \times 2 \times 3 + 7 \times 3 \times 4 - 3 \times 4 \times 2\\[1em] \Rightarrow 30 + 84 - 24\end{aligned}$$

Hence, $\bm{5ab + 7bc - 3ca = 90}$.

Substituting $x = -3$ in the given expression, we get:

$$\begin{aligned}\Rightarrow 5 - 2x\\[1em] \Rightarrow 5 - 2 \times (-3)\\[1em] \Rightarrow 5 + 6\end{aligned}$$

Hence, $\bm{5 - 2x = 11}$.

Substituting $x = -3$ in the given expression, we get:

$$\begin{aligned}\Rightarrow x^2 + 2\\[1em] \Rightarrow (-3)^2 + 2\\[1em] \Rightarrow 9 + 2\end{aligned}$$

Hence, $\bm{x^2 + 2 = 11}$.

Substituting $x = -3$ in the given expression, we get:

$$\begin{aligned}\Rightarrow 6 - x^2\\[1em] \Rightarrow 6 - (-3)^2\\[1em] \Rightarrow 6 - 9\end{aligned}$$

Hence, $\bm{6 - x^2 = -3}$.

Substituting $x = 5$ in the given expression, we get:

$$\begin{aligned}\Rightarrow x^2 + x + 3\\[1em] \Rightarrow (5)^2 + 5 + 3\\[1em] \Rightarrow 25 + 5 + 3\end{aligned}$$

Hence, $\bm{x^2 + x + 3 = 33}$.

Substituting $x = 5$ in the given expression, we get:

$$\begin{aligned}\Rightarrow 5 + 4x - x^2\\[1em] \Rightarrow 5 + 4 \times 5 - (5)^2\\[1em] \Rightarrow 5 + 20 - 25\end{aligned}$$

Hence, $\bm{5 + 4x - x^2 = 0}$.

Substituting $x = 5$ in the given expression, we get:

$$\begin{aligned}\Rightarrow 9x^2 - 7x\\[1em] \Rightarrow 9 \times (5)^2 - 7 \times 5\\[1em] \Rightarrow 9 \times 25 - 35\\[1em] \Rightarrow 225 - 35\end{aligned}$$

Hence, $\bm{9x^2 - 7x = 190}$.

Substituting $x = 5$, we get:

$$\begin{aligned}\Rightarrow 6x^2 - 5x + 4\\[1em] \Rightarrow 6 \times (5)^2 - 5 \times 5 + 4\\[1em] \Rightarrow 150 - 25 + 4\end{aligned}$$

Hence, $\bm{6x^2 - 5x + 4 = 129}$.

Substituting $x = -4$, we get:

$$\begin{aligned}\Rightarrow 3x^2 - 4x - 2\\[1em] \Rightarrow 3 \times (-4)^2 - 4 \times (-4) - 2\\[1em] \Rightarrow 48 + 16 - 2\end{aligned}$$

Hence, $\bm{3x^2 - 4x - 2 = 62}$.

Substituting $x = 3$, we get:

$$\begin{aligned}\Rightarrow 2x^3 - 3x^2 + 5x + 6\\[1em] \Rightarrow 2 \times (3)^3 - 3 \times (3)^2 + 5 \times 3 + 6\\[1em] \Rightarrow 54 - 27 + 15 + 6\end{aligned}$$

Hence, $\bm{2x^3 - 3x^2 + 5x + 6 = 48}$.

Substituting $x = -2$, we get:

$$\begin{aligned}\Rightarrow x^4 - 2x^3 + x^2 - 7x - 10\\[1em] \Rightarrow (-2)^4 - 2 \times (-2)^3 + (-2)^2 - 7 \times (-2) - 10\\[1em] \Rightarrow 16 + 16 + 4 + 14 - 10\end{aligned}$$

Hence, $\bm{x^4 - 2x^3 + x^2 - 7x - 10 = 40}$.

Substituting $a = 4$ and $b = 5$, we get:

$$\begin{aligned}\Rightarrow ab - a^2\\[1em] \Rightarrow 4 \times 5 - (4)^2\\[1em] \Rightarrow 20 - 16\end{aligned}$$

Hence, $\bm{ab - a^2 = 4}$.

Substituting $a = 4$ and $b = 5$, we get:

$$\begin{aligned}\Rightarrow 6a^2 - 2ab\\[1em] \Rightarrow 6 \times (4)^2 - 2 \times 4 \times 5\\[1em] \Rightarrow 96 - 40\end{aligned}$$

Hence, $\bm{6a^2 - 2ab = 56}$.

Substituting $a = 4$ and $b = 5$, we get:

$$\begin{aligned}\Rightarrow 5a^2 - 3b^2\\[1em] \Rightarrow 5 \times (4)^2 - 3 \times (5)^2\\[1em] \Rightarrow 80 - 75\end{aligned}$$

Hence, $\bm{5a^2 - 3b^2 = 5}$.

Substituting $a = 4$ and $b = 5$, we get:

$$\begin{aligned}\Rightarrow 4b^2 - 3ab\\[1em] \Rightarrow 4 \times (5)^2 - 3 \times 4 \times 5\\[1em] \Rightarrow 100 - 60\end{aligned}$$

Hence, $\bm{4b^2 - 3ab = 40}$.

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$$\begin{aligned}\Rightarrow x - 2y + 3z\\[1em] \Rightarrow 4 - 2 \times 3 + 3 \times (-2)\\[1em] \Rightarrow 4 - 6 - 6\end{aligned}$$

Hence, $\bm{x - 2y + 3z = -8}$.

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$$\begin{aligned}\Rightarrow xy + yz + zx\\[1em] \Rightarrow 4 \times 3 + 3 \times (-2) + (-2) \times 4\\[1em] \Rightarrow 12 - 6 - 8\end{aligned}$$

Hence, $\bm{xy + yz + zx = -2}$.

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$$\begin{aligned}\Rightarrow -2xyz\\[1em] \Rightarrow -2 \times 4 \times 3 \times (-2)\\[1em] \Rightarrow 48\end{aligned}$$

Hence, $\bm{-2xyz = 48}$.

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$$\begin{aligned}\Rightarrow p^2 - 2qr\\[1em] \Rightarrow (3)^2 - 2 \times (-5) \times 4\\[1em] \Rightarrow 9 + 40\end{aligned}$$

Hence, $\bm{p^2 - 2qr = 49}$.

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$$\begin{aligned}\Rightarrow q^2 - 3pr\\[1em] \Rightarrow (-5)^2 - 3 \times 3 \times 4\\[1em] \Rightarrow 25 - 36\end{aligned}$$

Hence, $\bm{q^2 - 3pr = -11}$.

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$$\begin{aligned}\Rightarrow r^2 - pq\\[1em] \Rightarrow (4)^2 - 3 \times (-5)\\[1em] \Rightarrow 16 + 15\end{aligned}$$

Hence, $\bm{r^2 - pq = 31}$.

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$$\begin{aligned}\Rightarrow pq + qr + rp\\[1em] \Rightarrow 3 \times (-5) + (-5) \times 4 + 4 \times 3\\[1em] \Rightarrow -15 - 20 + 12\end{aligned}$$

Hence, $\bm{pq + qr + rp = -23}$.

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$$\begin{aligned}\Rightarrow 3pq + 10pr + 2qr\\[1em] \Rightarrow 3 \times 3 \times (-5) + 10 \times 3 \times 4 + 2 \times (-5) \times 4\\[1em] \Rightarrow -45 + 120 - 40\end{aligned}$$

Hence, $\bm{3pq + 10pr + 2qr = 35}$.

Substituting $p = 3, q = -5$ and $r = 4$, we get:

$$\begin{aligned}\Rightarrow p^2 + q^2 + r^2 - pq - qr - pr\\[1em] \Rightarrow (3)^2 + (-5)^2 + (4)^2 - 3 \times (-5) - (-5) \times 4 - 3 \times 4\\[1em] \Rightarrow 9 + 25 + 16 + 15 + 20 - 12\end{aligned}$$

Hence, $\bm{p^2 + q^2 + r^2 - pq - qr - pr = 73}$.

Substituting $a = 3$ and $b = -4$, we get:

$$\begin{aligned}\Rightarrow \frac{a + b}{a - b}\\[1em] \Rightarrow \frac{3 + (-4)}{3 - (-4)}\\[1em] \Rightarrow \frac{-1}{3 + 4}\\[1em] \Rightarrow -\frac{1}{7}\end{aligned}$$

Hence, the required value is $\bm{-\frac{1}{7}}$.

Substituting $a = 3$ and $b = -4$, we get:

$$\begin{aligned}\Rightarrow \frac{ab}{a^2 + b^2}\\[1em] \Rightarrow \frac{3 \times (-4)}{(3)^2 + (-4)^2}\\[1em] \Rightarrow \frac{-12}{9 + 16}\\[1em] \Rightarrow -\frac{12}{25}\end{aligned}$$

Hence, the required value is $\bm{-\frac{12}{25}}$.

Substituting $a = 3$ and $b = -4$, we get:

$$\begin{aligned}\Rightarrow \frac{a^2 + b^2}{b^2 - a^2}\\[1em] \Rightarrow \frac{(3)^2 + (-4)^2}{(-4)^2 - (3)^2}\\[1em] \Rightarrow \frac{9 + 16}{16 - 9}\\[1em] \Rightarrow \frac{25}{7}\end{aligned}$$

Hence, the required value is $\bm{\frac{25}{7}}$.

Substituting $a = 8, b = 2$ and $c = 3$, we get:

$$\begin{aligned}\Rightarrow \frac{16bc - 5ab}{abc - 3b^2}\\[1em] \Rightarrow \frac{16 \times 2 \times 3 - 5 \times 8 \times 2}{8 \times 2 \times 3 - 3 \times (2)^2}\\[1em] \Rightarrow \frac{96 - 80}{48 - 12}\\[1em] \Rightarrow \frac{16}{36}\\[1em] \Rightarrow \frac{4}{9}\end{aligned}$$

Hence, the required value is $\bm{\frac{4}{9}}$.

Substituting $x = 4, y = 3$ and $z = -2$, we get:

$$\begin{aligned}\Rightarrow \frac{x^2 + y^2 - z^2}{xy + yz - zx}\\[1em] \Rightarrow \frac{(4)^2 + (3)^2 - (-2)^2}{4 \times 3 + 3 \times (-2) - (-2) \times 4}\\[1em] \Rightarrow \frac{16 + 9 - 4}{12 - 6 + 8}\\[1em] \Rightarrow \frac{21}{14}\\[1em] \Rightarrow \frac{3}{2}\end{aligned}$$

Hence, the required value is $\bm{\frac{3}{2}}$.

(i) True

Reason :

L.H.S. $= x^2 + y^2 = (5)^2 + (12)^2 = 25 + 144 = 169$

R.H.S. $= z^2 = (13)^2 = 169$

Since L.H.S. = R.H.S., the statement is true.

(ii) False

Reason :

At $x = 0, 5x - 3 = 5 \times 0 - 3 = -3$, which is not 2.

(iii) False

Reason :

At $x = 3$ and $y = 2, x^y = 3^2 = 9$, which is not 8.

(iv) False.

Reason :

At $a = 3, a^2 = 9$ and $2a = 6$, which are not equal.

Adding,

$$\begin{aligned}\Rightarrow 9x + 7x\\[1em] \Rightarrow (9 + 7)x\\[1em] \Rightarrow \bm{16x}\end{aligned}$$

Hence, $\bm{9x + 7x = 16x}$

Adding,

$$\begin{aligned}\Rightarrow 13p + 6p + 2p\\[1em] \Rightarrow (13 + 6 + 2)p\\[1em] \Rightarrow \bm{21p}\end{aligned}$$

Hence, $\bm{13p + 6p + 2p = 21p}$

Adding,

$$\begin{aligned}\Rightarrow 8xy + 7xy + xy\\[1em] \Rightarrow (8 + 7 + 1)xy\\[1em] \Rightarrow \bm{16xy}\end{aligned}$$

Hence, $\bm{8xy + 7xy + xy = 16xy}$

Adding,

$$\begin{aligned}\Rightarrow 5y + (-13y)\\[1em] \Rightarrow (5 - 13)y\\[1em] \Rightarrow \bm{-8y}\end{aligned}$$

Hence, $\bm{5y + (-13y) = -8y}$

Adding,

$$\begin{aligned}\Rightarrow 4x^2 + (-9x^2)\\[1em] \Rightarrow (4 - 9)x^2\\[1em] \Rightarrow \bm{-5x^2}\end{aligned}$$

Hence, $\bm{4x^2 + (-9x^2) = -5x^2}$

Adding,

$$\begin{aligned}\Rightarrow 3a^2b + (-5a^2b)\\[1em] \Rightarrow (3 - 5)a^2b\\[1em] \Rightarrow \bm{-2a^2b}\end{aligned}$$

Hence, $\bm{3a^2b + (-5a^2b) = -2a^2b}$

Adding,

$$\begin{aligned}\Rightarrow \frac{1}{2}mn + 2mn\\[1em] \Rightarrow \left(\frac{1}{2} + 2\right)mn\\[1em] \Rightarrow \frac{1 + 4}{2}mn\\[1em] \Rightarrow \bm{\frac{5}{2}mn}\end{aligned}$$

Hence, $\bm{\frac{1}{2}mn + 2mn = \frac{5}{2}mn}$

Adding,

$$\begin{aligned}\Rightarrow \frac{1}{3}x^2y + 3x^2y\\[1em] \Rightarrow \left(\frac{1}{3} + 3\right)x^2y\\[1em] \Rightarrow \frac{1 + 9}{3}x^2y\\[1em] \Rightarrow \bm{\frac{10}{3}x^2y}\end{aligned}$$

Hence, $\bm{\frac{1}{3}x^2y + 3x^2y = \frac{10}{3}x^2y}$

Adding,

$$\begin{aligned}\Rightarrow 5ab^2 + (-ab^2)\\[1em] \Rightarrow (5 - 1)ab^2\\[1em] \Rightarrow \bm{4ab^2}\end{aligned}$$

Hence, $\bm{5ab^2 + (-ab^2) = 4ab^2}$

Adding,

$$\begin{aligned}\Rightarrow 5a + (-3a) + 4a\\[1em] \Rightarrow (5 - 3 + 4)a\\[1em] \Rightarrow \bm{6a}\end{aligned}$$

Hence, $\bm{5a + (-3a) + 4a = 6a}$

Adding,

$$\begin{aligned}\Rightarrow (-2b) + b + (-3b)\\[1em] \Rightarrow (-2 + 1 - 3)b\\[1em] \Rightarrow \bm{-4b}\end{aligned}$$

Hence, $\bm{(-2b) + b + (-3b) = -4b}$

Adding,

$$\begin{aligned}\Rightarrow 4ab + (-5ab) + (-3ab)\\[1em] \Rightarrow (4 - 5 - 3)ab\\[1em] \Rightarrow \bm{-4ab}\end{aligned}$$

Hence, $\bm{4ab + (-5ab) + (-3ab) = -4ab}$

Adding,

$$\begin{aligned}\Rightarrow 6xy^2 + (-8xy^2) + (-3xy^2)\\[1em] \Rightarrow (6 - 8 - 3)xy^2\\[1em] \Rightarrow \bm{-5xy^2}\end{aligned}$$

Hence, $\bm{6xy^2 + (-8xy^2) + (-3xy^2) = -5xy^2}$

Adding,

$$\begin{aligned}\Rightarrow 4m^2n + (-m^2n) + (-6m^2n)\\[1em] \Rightarrow (4 - 1 - 6)m^2n\\[1em] \Rightarrow \bm{-3m^2n}\end{aligned}$$

Hence, $\bm{4m^2n + (-m^2n) + (-6m^2n) = -3m^2n}$

Adding,

$$\begin{aligned}\Rightarrow 2xyz + (-5xyz) + (-xyz)\\[1em] \Rightarrow (2 - 5 - 1)xyz\\[1em] \Rightarrow \bm{-4xyz}\end{aligned}$$

Hence, $\bm{2xyz + (-5xyz) + (-xyz) = -4xyz}$

Adding,

$$\begin{aligned}\Rightarrow (-6x) + (-9x)\\[1em] \Rightarrow (-6 - 9)x\\[1em] \Rightarrow \bm{-15x}\end{aligned}$$

Hence, $\bm{(-6x) + (-9x) = -15x}$

Adding,

$$\begin{aligned}\Rightarrow (-3y) + (-y) + (-7y)\\[1em] \Rightarrow (-3 - 1 - 7)y\\[1em] \Rightarrow \bm{-11y}\end{aligned}$$

Hence, $\bm{(-3y) + (-y) + (-7y) = -11y}$

Adding,

$$\begin{aligned}\Rightarrow (-8xy) + (-4xy) + 6xy\\[1em] \Rightarrow (-8 - 4 + 6)xy\\[1em] \Rightarrow \bm{-6xy}\end{aligned}$$

Hence, $\bm{(-8xy) + (-4xy) + 6xy = -6xy}$

Adding,

$$\begin{aligned}\Rightarrow 7m^2n + (-2m^2n) + (-3m^2n)\\[1em] \Rightarrow (7 - 2 - 3)m^2n\\[1em] \Rightarrow \bm{2m^2n}\end{aligned}$$

Hence, $\bm{7m^2n + (-2m^2n) + (-3m^2n) = 2m^2n}$

Adding,

$$\begin{aligned}\Rightarrow (-2ab^2) + (-8ab^2) + 5ab^2\\[1em] \Rightarrow (-2 - 8 + 5)ab^2\\[1em] \Rightarrow \bm{-5ab^2}\end{aligned}$$

Hence, $\bm{(-2ab^2) + (-8ab^2) + 5ab^2 = -5ab^2}$

Adding,

$$\begin{aligned}\Rightarrow \frac{1}{2}yz + 2yz + (-yz)\\[1em] \Rightarrow \left(\frac{1}{2} + 2 - 1\right)yz\\[1em] \Rightarrow \frac{1 + 4 - 2}{2}yz\\[1em] \Rightarrow \bm{\frac{3}{2}yz}\end{aligned}$$

Hence, $\bm{\frac{1}{2}yz + 2yz + (-yz) = \frac{3}{2}yz}$

Adding,

$$\begin{aligned}\Rightarrow 5a^3 + (-3a^3) + \left(-\frac{1}{2}a^3\right)\\[1em] \Rightarrow \left(5 - 3 - \frac{1}{2}\right)a^3\\[1em] \Rightarrow \frac{10 - 6 - 1}{2}a^3\\[1em] \Rightarrow \bm{\frac{3}{2}a^3}\end{aligned}$$

Hence, $\bm{5a^3 + (-3a^3) + \left(-\frac{1}{2}a^3\right) = \frac{3}{2}a^3}$

Adding,

$$\begin{aligned}\Rightarrow \left(-\frac{1}{2}a^2\right) + \left(-\frac{1}{3}a^2\right) + 2a^2\\[1em] \Rightarrow \left(-\frac{1}{2} - \frac{1}{3} + 2\right)a^2\\[1em] \Rightarrow \frac{-3 - 2 + 12}{6}a^2\\[1em] \Rightarrow \bm{\frac{7}{6}a^2}\end{aligned}$$

Hence, $\bm{\left(-\frac{1}{2}a^2\right) + \left(-\frac{1}{3}a^2\right) + 2a^2 = \frac{7}{6}a^2}$

Adding columnwise, we get:

$$\begin{array}{rrrr} & 8a & - 3b & + 11c \\ & -6a & + 5b & + 7c \\ {+} & 2a & - 8b & + 9c \\ \hline & 4a & - 6b & + 27c \end{array}$$

Hence, the required sum is $\bm{4a - 6b + 27c}$.

Adding columnwise, we get:

$$\begin{array}{rrrr} & -6x & + 3y & + 4z \\ & 2x & - 4y & + z \\ {+} & -x & - 8y & + 5z \\ \hline & -5x & - 9y & + 10z \end{array}$$

Hence, the required sum is $\bm{-5x - 9y + 10z}$.

Adding columnwise, we get:

$$\begin{array}{rrrr} & 8x &- 5y &+ 3z \\ & -11x &+ 6y &- z \\ & & 7y &- 2z \\ {+} &x &- 2y& \\ \hline &-2x &+ 6y& \end{array}$$

Hence, the required sum is $\bm{-2x + 6y}$.

Adding columnwise, we get:

$$\begin{array}{rrrr} & 10x^2 & + 8x & - 7 \\ & -13x^2 & + 3x & + 5 \\ {+} & -5x^2 & - 6x & + 9 \\ \hline & -8x^2 & + 5x & + 7 \end{array}$$

Hence, the required sum is $\bm{-8x^2 + 5x + 7}$.

Adding columnwise, we get:

$$\begin{array}{rrrrr} & 5x^3 & - 8x^2 & + 6x & - 3 \\ & & x^2 & - 2x & + 15 \\ {+} & -2x^3 & & & + 16 \\ \hline & 3x^3 & - 7x^2 & + 4x & + 28 \end{array}$$

Hence, the required sum is $\bm{3x^3 - 7x^2 + 4x + 28}$.

Collecting the like terms, we get:

Coefficient of $x^2 = 3 - \frac{1}{4} - 2 = 1 - \frac{1}{4} = \frac{4 - 1}{4} = \frac{3}{4}$

Coefficient of $x = -\frac{1}{5} + \frac{1}{3} - \frac{1}{2} = \frac{-6 + 10 - 15}{30} = -\frac{11}{30}$

Constant term $= \frac{7}{3} - \frac{1}{6} + \frac{1}{15} = \frac{70 - 5 + 2}{30} = \frac{67}{30}$

Hence, the required sum is $\bm{\frac{3}{4}x^2 - \frac{11}{30}x + \frac{67}{30}}$.

Adding columnwise, we get:

$$\begin{array}{rrrr} & 2a^2 & - 3ab & + 4bc \\ & -a^2 & + 6ab & - 5bc \\ {+} & -8a^2 & + ab & + 13bc \\ \hline & -7a^2 & + 4ab & + 12bc \end{array}$$

Hence, the required sum is $\bm{-7a^2 + 4ab + 12bc}$.

Adding columnwise, we get:

$$\begin{array}{rrrr} & 6x^2 & + 2xy & - y^2 \\ & -x^2 & + xy & - 2y^2 \\ {+} & -8x^2 & + 3xy & + 5y^2 \\ \hline & -3x^2 & + 6xy & + 2y^2 \end{array}$$

Hence, the required sum is $\bm{-3x^2 + 6xy + 2y^2}$.

Adding columnwise, we get:

$$\begin{array}{rrrr} & 2x^2 & - 3y^2 & + 6z^2 \\ & -2x^2 & + y^2 & + z^2 \\ & -6x^2 & - 8y^2 & + z^2 \\ {+} & x^2 & - y^2 & \\ \hline & -5x^2 & - 11y^2 & + 8z^2 \end{array}$$

Hence, the required sum is $\bm{-5x^2 - 11y^2 + 8z^2}$.

Adding columnwise, we get:

$$\begin{array}{rrrr} & 6m^2n & + 3mn & - 2n^2 \\ & -m^2n & - mn & + n^2 \\ {+} & -4m^2n & + mn & - 3n^2 \\ \hline & m^2n & + 3mn & - 4n^2 \end{array}$$

Hence, the required sum is $\bm{m^2n + 3mn - 4n^2}$.

Given,

$2m^3 - 5m^2 + 3m - 6 - 5m^3 + 2m^2 - 8m + 2 + 6m + 5$

Re-arranging and collecting the like terms, we get:

$$\begin{aligned}\Rightarrow (2m^3 - 5m^3) + (-5m^2 + 2m^2) + (3m - 8m + 6m) + (-6 + 2 + 5)\\[1em] \Rightarrow (2 - 5)m^3 + (-5 + 2)m^2 + (3 - 8 + 6)m + 1\\[1em] \Rightarrow \bm{-3m^3 - 3m^2 + m + 1}\end{aligned}$$

Hence, the simplified expression is $\bm{-3m^3 - 3m^2 + m + 1}$.

Given,

$2x^4 - 5x^3 + 2x - 5 + 6x^3 - 3x + 7 - x^4 - 2x^3 + 2$

Re-arranging and collecting the like terms, we get:

$$\begin{aligned}\Rightarrow (2x^4 - x^4) + (-5x^3 + 6x^3 - 2x^3) + (2x - 3x) + (-5 + 7 + 2)\\[1em] \Rightarrow (2 - 1)x^4 + (-5 + 6 - 2)x^3 + (2 - 3)x + 4\\[1em] \Rightarrow \bm{x^4 - x^3 - x + 4}\end{aligned}$$

Hence, the simplified expression is $\bm{x^4 - x^3 - x + 4}$.

Given,

$3p^2 - 2pq + 5p - 4q + 5pq - 3p + 4p^2 + 3q$

Re-arranging and collecting the like terms, we get:

$$\begin{aligned}\Rightarrow (3p^2 + 4p^2) + (-2pq + 5pq) + (5p - 3p) + (-4q + 3q)\\[1em] \Rightarrow (3 + 4)p^2 + (-2 + 5)pq + (5 - 3)p + (-4 + 3)q\\[1em] \Rightarrow \bm{7p^2 + 3pq + 2p - q}\end{aligned}$$

Hence, the simplified expression is $\bm{7p^2 + 3pq + 2p - q}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow 15x - 8x\\[1em] \Rightarrow (15 - 8)x\\[1em] \Rightarrow 7x.\end{aligned}$$

Hence, the required difference is $\bm{7x}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow 7ab^2 - 3ab^2\\[1em] \Rightarrow (7 - 3)ab^2\\[1em] \Rightarrow 4ab^2.\end{aligned}$$

Hence, the required difference is $\bm{4ab^2}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow -b - 5b\\[1em] \Rightarrow (-1 - 5)b\\[1em] \Rightarrow -6b.\end{aligned}$$

Hence, the required difference is $\bm{-6b}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow 11x - (-4x)\\[1em] \Rightarrow 11x + 4x\\[1em] \Rightarrow (11 + 4)x\\[1em] \Rightarrow 15x.\end{aligned}$$

Hence, the required difference is $\bm{15x}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow x - (-x)\\[1em] \Rightarrow x + x\\[1em] \Rightarrow 2x.\end{aligned}$$

Hence, the required difference is $\bm{2x}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow -2ab - 2ab\\[1em] \Rightarrow (-2 - 2)ab\\[1em] \Rightarrow -4ab.\end{aligned}$$

Hence, the required difference is $\bm{-4ab}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow -7m^2n - (-3m^2n)\\[1em] \Rightarrow -7m^2n + 3m^2n\\[1em] \Rightarrow (-7 + 3)m^2n\\[1em] \Rightarrow -4m^2n.\end{aligned}$$

Hence, the required difference is $\bm{-4m^2n}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow -2b - \frac{1}{2}b\\[1em] \Rightarrow \left(-2 - \frac{1}{2}\right)b\\[1em] \Rightarrow \frac{-4 - 1}{2}b\\[1em] \Rightarrow -\frac{5}{2}b\end{aligned}$$

Hence, the required difference is $\bm{-\frac{5}{2}b}$.

To subtract, we change the sign of the term to be subtracted and then add.

$$\begin{aligned}\Rightarrow 3c - \left(-\frac{1}{3}c\right)\\[1em] \Rightarrow 3c + \frac{1}{3}c\\[1em] \Rightarrow \left(3 + \frac{1}{3}\right)c\\[1em] \Rightarrow \frac{9 + 1}{3}c\\[1em] \Rightarrow \frac{10}{3}c\end{aligned}$$

Hence, the required difference is $\bm{\frac{10}{3}c}$.

Required difference $= (3a - b + 2c) - (a + 2b - c)$

$$\begin{aligned}\Rightarrow 3a - b + 2c - a - 2b + c\\[1em] \Rightarrow (3a - a) + (-b - 2b) + (2c + c)\\[1em] \Rightarrow 2a - 3b + 3c.\end{aligned}$$

Hence, the required difference is $\bm{2a - 3b + 3c}$.

Required difference $= (5a - 3b - 4c) - (-3a + b + 6c)$

$$\begin{aligned}\Rightarrow 5a - 3b - 4c + 3a - b - 6c\\[1em] \Rightarrow (5a + 3a) + (-3b - b) + (-4c - 6c)\\[1em] \Rightarrow 8a - 4b - 10c.\end{aligned}$$

Hence, the required difference is $\bm{8a - 4b - 10c}$.

Required difference $= (-m + n - 2p) - (2m - 3n + p)$

$$\begin{aligned}\Rightarrow -m + n - 2p - 2m + 3n - p\\[1em] \Rightarrow (-m - 2m) + (n + 3n) + (-2p - p)\\[1em] \Rightarrow -3m + 4n - 3p.\end{aligned}$$

Hence, the required difference is $\bm{-3m + 4n - 3p}$.

Required difference $= (x - y - 3z) - (-6x - 9y + 15z)$

$$\begin{aligned}\Rightarrow x - y - 3z + 6x + 9y - 15z\\[1em] \Rightarrow (x + 6x) + (-y + 9y) + (-3z - 15z)\\[1em] \Rightarrow 7x + 8y - 18z.\end{aligned}$$

Hence, the required difference is $\bm{7x + 8y - 18z}$.

Required difference $= \left(\frac{1}{2}p - \frac{1}{3}q - \frac{3}{2}r\right) - \left(-\frac{1}{2}p + q - r\right)$

$$\begin{aligned}\Rightarrow \frac{1}{2}p - \frac{1}{3}q - \frac{3}{2}r + \frac{1}{2}p - q + r\\[1em] \Rightarrow \left(\frac{1}{2} + \frac{1}{2}\right)p + \left(-\frac{1}{3} - 1\right)q + \left(-\frac{3}{2} + 1\right)r\\[1em] \Rightarrow p - \frac{4}{3}q - \frac{1}{2}r\end{aligned}$$

Hence, the required difference is $\bm{p - \frac{4}{3}q - \frac{1}{2}r}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &+5x^2 &- x &- 3 \\ &+3x^2 &- 5x &+ 6 \\ &- &+ &- \\ \hline &+2x^2 &+ 4x &- 9 \end{array}$$

Hence, the required difference is $\bm{2x^2 + 4x - 9}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &+x^2 &+ x &- 5 \\ &+2x^2 &+ 3x &- 1 \\ &- &- &+ \\ \hline &-x^2 &- 2x &- 4 \end{array}$$

Hence, the required difference is $\bm{-x^2 - 2x - 4}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &+x^2 &+ x &- 1 \\ &+x^2 &- x &+ 1 \\ &- &+ &- \\ \hline & &+ 2x &- 2 \end{array}$$

Hence, the required difference is $\bm{2x - 2}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &-a^2 &+ a & \\ &+a^2 &+ a &- 1 \\ &- &- &+ \\ \hline &-2a^2 & &+ 1 \end{array}$$

Hence, the required difference is $\bm{-2a^2 + 1}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &-y^2 &+ 2y &+ 7 \\ &+6y^2 &- y &- 8 \\ &- &+ &+ \\ \hline &-7y^2 &+ 3y &+ 15 \end{array}$$

Hence, the required difference is $\bm{-7y^2 + 3y + 15}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &+p^2 &+ 2pq &- q^2 \\ &+p^2 &- 2pq &+ q^2 \\ &- &+ &- \\ \hline & &+ 4pq &- 2q^2 \end{array}$$

Hence, the required difference is $\bm{4pq - 2q^2}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &-2x^2 &+ 5xy &- 3y^2 \\ &+x^2 &- 2xy &+ y^2 \\ &- &+ &- \\ \hline &-3x^2 &+ 7xy &- 4y^2 \end{array}$$

Hence, the required difference is $\bm{-3x^2 + 7xy - 4y^2}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrrr} &+x^2 &- 2xy &+ 6y^2 &+ 8 \\ &-3x^2 &+ 5xy &- 2y^2 &- 3 \\ &+ &- &+ &+ \\ \hline &+4x^2 &- 7xy &+ 8y^2 &+ 11 \end{array}$$

Hence, the required difference is $\bm{4x^2 - 7xy + 8y^2 + 11}$.

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrrr} &-8x^3 &+ 6x^2 &- 4x &+ 5 \\ &+6x^3 &- 8x^2 &+ 4x &- 2 \\ &- &+ &- &+ \\ \hline &-14x^3 &+ 14x^2 &- 8x &+ 7 \end{array}$$

Hence, the required difference is $\bm{-14x^3 + 14x^2 - 8x + 7}$.

Required expression $= (a + 2b) - (b - a)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrr} &+a &+ 2b \\ &-a &+ b \\ &+ &- \\ \hline &+2a &+ b \end{array}$$

Hence, $\bm{a + 2b}$ is greater than $\bm{b - a}$ by $\bm{(2a + b)}$.

Required expression $= (a - 2b + 3c) - (2a + 3b - c)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &+a &- 2b &+ 3c \\ &+2a &+ 3b &- c \\ &- &- &+ \\ \hline &-a &- 5b &+ 4c \end{array}$$

Hence, $\bm{a - 2b + 3c}$ is greater than $\bm{2a + 3b - c}$ by $\bm{(-a - 5b + 4c)}$.

Required expression $= (3x^2 - 4x + 3) - (x^3 - x^2 + 2x - 1)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrrr} & &+ 3x^2 &- 4x &+ 3 \\ &+x^3 &- x^2 &+ 2x &- 1 \\ &- &+ &- &+ \\ \hline &-x^3 &+ 4x^2 &- 6x &+ 4 \end{array}$$

Hence, $\bm{3x^2 - 4x + 3}$ exceeds $\bm{x^3 - x^2 + 2x - 1}$ by $\bm{(-x^3 + 4x^2 - 6x + 4)}$.

Required expression $= (3r + 5) - (p - 2q + 3r)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrrr} & & &+ 3r &+ 5 \\ &+p &- 2q &+ 3r & \\ &- &+ &- & \\ \hline &-p &+ 2q & &+ 5 \end{array}$$

Hence, $\bm{p - 2q + 3r}$ is less than $\bm{3r + 5}$ by $\bm{(-p + 2q + 5)}$.

Required expression $= 3x - (5x + 1)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrr} &+3x & \\ &+5x &+ 1 \\ &- &- \\ \hline &-2x &- 1 \end{array}$$

Hence, $\bm{(-2x - 1)}$ must be subtracted from $\bm{3x}$ to get $\bm{5x + 1}$.

Required expression $= (2x + 3y) - (y - 3x)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrr} &+2x &+ 3y \\ &-3x &+ y \\ &+ &- \\ \hline &+5x &+ 2y \end{array}$$

Hence, $\bm{(5x + 2y)}$ must be subtracted from $\bm{2x + 3y}$ to get $\bm{y - 3x}$.

Required expression $= (x^2 + 2x - 3) - (2x^2 - 4)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} &+x^2 &+ 2x &- 3 \\ &+2x^2 & &- 4 \\ &- & &+ \\ \hline &-x^2 &+ 2x &+ 1 \end{array}$$

Hence, $\bm{(-x^2 + 2x + 1)}$ must be subtracted from $\bm{x^2 + 2x - 3}$ to get $\bm{2x^2 - 4}$.

Required expression $= (4x + 1) - 6x$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrr} &+4x &+ 1 \\ &+6x & \\ &- & \\ \hline &-2x &+ 1 \end{array}$$

Hence, $\bm{(-2x + 1)}$ must be added to $\bm{6x}$ to get $\bm{4x + 1}$.

Required expression $= (b - 2a) - (a + 2b)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrr} &-2a &+ b \\ &+a &+ 2b \\ &- &- \\ \hline &-3a &- b \end{array}$$

Hence, $\bm{(-3a - b)}$ must be added to $\bm{a + 2b}$ to get $\bm{b - 2a}$.

Required expression $= (3x + 2) - (5x^2 + 6x - 3)$

First we arrange the like terms columnwise. Change the sign of each term to be subtracted and then add.

$$\begin{array}{rrrr} & &+ 3x &+ 2 \\ &+5x^2 &+ 6x &- 3 \\ &- &- &+ \\ \hline &-5x^2 &- 3x &+ 5 \end{array}$$

Hence, $\bm{(-5x^2 - 3x + 5)}$ must be added to $\bm{5x^2 + 6x - 3}$ to get $\bm{3x + 2}$.

Adding the two expressions and then subtracting $5x^2 - 8x + 6$ from their sum columnwise, we get:

$$\begin{array}{rrrr} &+3x^2 &- 4x &+ 2 \\ {+} &-5x^2 &- 6x &+ 7 \\ \hline &-2x^2 &- 10x &+ 9 \\ &+5x^2 &- 8x &+ 6 \\ &- &+ &- \\ \hline &-7x^2 &- 2x &+ 3 \end{array}$$

Hence, the required difference is $\bm{-7x^2 - 2x + 3}$.

Adding the two expressions and then subtracting $3y^2 - 5y - 4$ from their sum columnwise, we get:

$$\begin{array}{rrrr} &-5y^2 &+ y &+ 1 \\ {+} &+4y^2 &+ 3y &- 5 \\ \hline &-y^2 &+ 4y &- 4 \\ &+3y^2 &- 5y &- 4 \\ &- &+ &+ \\ \hline &-4y^2 &+ 9y & \end{array}$$

Hence, the required difference is $\bm{-4y^2 + 9y}$.

$6x$ and $2y$ are unlike terms, so they cannot be added into a single term.

Sum of $6x$ and $2y = 6x + 2y$, which is not given in options 1, 2 and 3.

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 6 \times a \times a \times 3 \times b \times b \times b \times 2 \times c \times c \times c \times c\\[1em] \Rightarrow (6 \times 3 \times 2) \times (a \times a) \times (b \times b \times b) \times (c \times c \times c \times c)\\[1em] \Rightarrow 36 \times a^2 \times b^3 \times c^4\\[1em] \Rightarrow 36a^2b^3c^4\end{aligned}$$

Hence, option 2 is the correct option.

Substituting $x = 3$ and $y = -2$, we get:

$$\begin{aligned}\Rightarrow 7x - 9y + 2\\[1em] \Rightarrow 7 \times 3 - 9 \times (-2) + 2\\[1em] \Rightarrow 21 + 18 + 2\\[1em] \Rightarrow 41\end{aligned}$$

Hence, option 2 is the correct option.

Substituting $x = \frac{1}{2} \text{ and } y = \frac{1}{4}$, we get:

$$\begin{aligned}\Rightarrow 16x + 8y - 10\\[1em] \Rightarrow 16 \times \frac{1}{2} + 8 \times \frac{1}{4} - 10\\[1em] \Rightarrow 8 + 2 - 10\\[1em] \Rightarrow 0\end{aligned}$$

Hence, option 4 is the correct option.

solving,

$$\begin{aligned}\Rightarrow a + 2(b - c)\\[1em] \Rightarrow a + 2b - 2c\\[1em] \Rightarrow (a + 2b) - 2c\end{aligned}$$

Substituting $a + 2b = 5$ and $c = 3$, we get:

$$\begin{aligned}\Rightarrow 5 - 2 \times 3\\[1em] \Rightarrow 5 - 6\\[1em] \Rightarrow -1\end{aligned}$$

Hence, option 2 is the correct option.

Product of $x$ and $y = xy$.

9 taken away from it means 9 is subtracted from $xy$, i.e., $xy - 9$.

Hence, option 2 is the correct option.

$-\frac{7}{8}a^2b = b \times \left(\frac{-7}{8}a^2\right)$

So, the coefficient of $b$ is $\frac{-7}{8}a^2$.

Hence, option 3 is the correct option.

Degrees of the terms are (2 + 2), (1 + 2) and (1 + 1), i.e., 4, 3 and 2.

The highest of these is 4.

Hence, option 3 is the correct option.

solving,

$$\begin{aligned}\Rightarrow 6a + (-3b) + (-2a) + (-4b)\\[1em] \Rightarrow (6a - 2a) + (-3b - 4b)\\[1em] \Rightarrow (6 - 2)a + (-3 - 4)b\\[1em] \Rightarrow 4a - 7b\end{aligned}$$

Hence, option 1 is the correct option.

Let the variable be $x$.

Twice the variable added to $4 = 4 + 2x$.

3 subtracted from thrice the same variable $= 3x - 3$.

Adding the two expressions, we get $(4 + 2x) + (3x - 3)$.

Hence, option 2 is the correct option.

(i) $(2a^2 + 3b^2) - (3a^2 + 2b^2)$

$$\begin{aligned}\Rightarrow 2a^2 + 3b^2 - 3a^2 - 2b^2\\[1em] \Rightarrow (2a^2 - 3a^2) + (3b^2 - 2b^2)\\[1em] \Rightarrow {-a^2 + b^2}\end{aligned}$$

Hence, $\bm{(2a^2 + 3b^2) - (3a^2 + 2b^2) = -a^2 + b^2}$

(ii) $(5x - 6y) + (6x - 5y)$

$$\begin{aligned}\Rightarrow (5x + 6x) + (-6y - 5y)\\[1em] \Rightarrow {11x - 11y}\end{aligned}$$

Hence, $\bm{(5x - 6y) + (6x - 5y) = 11x - 11y}$

(iii) $(a^2 - 2ab + b^2) - (a^2 + 2ab - b^2)$

$$\begin{aligned}\Rightarrow a^2 - 2ab + b^2 - a^2 - 2ab + b^2\\[1em] \Rightarrow (a^2 - a^2) + (-2ab - 2ab) + (b^2 + b^2)\\[1em] \Rightarrow {-4ab + 2b^2}\end{aligned}$$

Hence, $\bm{(a^2 - 2ab + b^2) - (a^2 + 2ab - b^2) = -4ab + 2b^2}$

(iv) Required expression $= (a + 2b) - (b - a)$

$$\begin{aligned}\Rightarrow a + 2b - b + a\\[1em] \Rightarrow {2a + b}\end{aligned}$$

Hence, $\bm{(a + 2b) - (b - a) = 2a + b}$

(v) Required expression $= (3c + 5 - a) - (a - 2b + 3c)$

$$\begin{aligned}\Rightarrow 3c + 5 - a - a + 2b - 3c\\[1em] \Rightarrow (3c - 3c) + (-a - a) + 2b + 5\\[1em] \Rightarrow {-2a + 2b + 5}\end{aligned}$$

Hence, $\bm{(3c + 5 - a) - (a - 2b + 3c) = -2a + 2b + 5}$

(vi) Required expression $= (3x + 1) - (3x^2 + 5x - 2)$

$$\begin{aligned}\Rightarrow 3x + 1 - 3x^2 - 5x + 2\\[1em] \Rightarrow -3x^2 + (3x - 5x) + (1 + 2)\\[1em] \Rightarrow {-3x^2 - 2x + 3}\end{aligned}$$

Hence, $\bm{(3x + 1) - (3x^2 + 5x - 2) = -3x^2 - 2x + 3}$

(i) False (F).

Reason :

3 subtracted from −6 is −6 − 3 = −9, not −3.

(ii) False (F).

Reason :

$(a - b) - (a + b) = a - b - a - b = -2b$, not $2b$.

(iii) False (F).

Reason :

(−3) × (−6) = 18, since the product of two negative numbers is positive.

(iv) True (T).

Reason :

The highest power of $x$ in $x^2 + 5$ is 2, so its degree is 2.

(v) False (F).

Reason :

The degree of $x^2y^2$ is (2 + 2) = 4, so the polynomial is of degree 4.

(vi) False (F).

Reason :

A variable can take various values; it is a constant that has a fixed value.

A trinomial has 3 unlike terms. A monomial can be like at most one of these three terms, so on adding, at least two unlike terms will always remain. Hence, the sum can never be a monomial.

So, Assertion (A) is true.

While adding algebraic expressions, only like terms are added.

So, Reason (R) is true.

It is precisely because only like terms are added that the unlike terms of the trinomial remain in the sum. So, Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

In the expression, the term containing $a^2$ is $2a^2 = a^2 \times 2$. So, the coefficient of $a^2$ is 2.

So, Assertion (A) is true.

By definition, the term which does not contain any literal is called a constant term.

So, Reason (R) is true.

Reason (R) only defines a constant term and says nothing about how the coefficient of a term is obtained. So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

Substituting $m = 1, n = -1$ and $p = 2$, we get:

$$\begin{aligned}\Rightarrow m^2n^2 + n^2p^2 + p^2m^2\\[1em] \Rightarrow (1)^2(-1)^2 + (-1)^2(2)^2 + (2)^2(1)^2\\[1em] \Rightarrow 1 \times 1 + 1 \times 4 + 4 \times 1\\[1em] \Rightarrow 1 + 4 + 4\\[1em] \Rightarrow 9\end{aligned}$$

So, Assertion (A) is true.

In a polynomial of more than one variable, the values of all the variables can be substituted simultaneously.

So, Reason (R) is false.

Hence, option 3 is the correct option.

An equilateral triangle has 3 equal sides, each of length $x$ units.

Perimeter = sum of all sides $= x + x + x = 3x$ units.

Hence, option 4 is the correct option.

In $2n - 1$, the product of $n$ and 2 is $2n$, and 1 is subtracted from it.

So, $2n - 1$ means "1 subtracted from the product of $n$ and 2".

Hence, option 1 is the correct option.

Quotient of $x$ by $3 = \frac{x}{3}$.

Multiplying it by $y$, we get $\frac{x}{3} \times y = \frac{xy}{3}$.

Hence, option 4 is the correct option.

Substituting $p = -1$ and $q = -2$, we get:

$$\begin{aligned}\Rightarrow p - q + pq\\[1em] \Rightarrow (-1) - (-2) + (-1) \times (-2)\\[1em] \Rightarrow -1 + 2 + 2\\[1em] \Rightarrow 3\end{aligned}$$

Hence, option 4 is the correct option.

In the given polynomial:

Coefficient of $x^3 = -7$

Coefficient of $x^2 = 3$

Sum of the coefficients = −7 + 3 = −4

Hence, option 2 is the correct option.

$P - Q + R = (3a - 5b) - (4a + 7b) + (-2a + 9b)$

$$\begin{aligned}\Rightarrow 3a - 5b - 4a - 7b - 2a + 9b\\[1em] \Rightarrow (3a - 4a - 2a) + (-5b - 7b + 9b)\\[1em] \Rightarrow -3a - 3b\end{aligned}$$

Substituting $a = -1$ and $b = -2$, we get:

$$\begin{aligned}\Rightarrow -3 \times (-1) - 3 \times (-2)\\[1em] \Rightarrow 3 + 6\\[1em] \Rightarrow 9\end{aligned}$$

Hence, option 1 is the correct option.

Expanding the given expression, we get:

$$\begin{aligned}\Rightarrow 4pq^2 - 2q^2r + q^2p + q(3pr + qr) - 8\\[1em] \Rightarrow 4pq^2 - 2q^2r + pq^2 + 3pqr + q^2r - 8\end{aligned}$$

Collecting the like terms:

$$\begin{aligned}\Rightarrow (4pq^2 + pq^2) + (-2q^2r + q^2r) + 3pqr - 8\\[1em] \Rightarrow 5pq^2 - q^2r + 3pqr - 8\end{aligned}$$

This has 4 unlike terms. Hence, statement (i) is true.

An expression having exactly three terms is called a trinomial; an expression having two or more than two terms is called a multinomial. Hence, statement (ii) is false.

Hence, option 3 is the correct option.

Substituting $p = 2, q = 1, r = 3$ and $s = 4$ in each option:

Option 1: L.H.S. $= \frac{2 \times 1 + 3}{1} = \frac{5}{1} = 5$ and R.H.S. = 2 + 4 = 6. Since 5 ≠ 6, it is not correct.

Option 2: L.H.S. $= \frac{2 + 3}{1 + 3} = \frac{5}{4} \text{ and R.H.S. }= \frac{1}{3}. \text{ Since }\frac{5}{4} \neq \frac{1}{3}$, it is not correct.

Option 3: L.H.S. $= \frac{2 \times 1 + 2 \times 3}{2 \times 4} = \frac{2 + 6}{8} = \frac{8}{8} = 1 \text{ and R.H.S. }= \frac{1 + 3}{4} = \frac{4}{4} = 1$. Since L.H.S. = R.H.S., it is correct.

Option 4: L.H.S. $= \frac{2(1 + 3)}{2 + 4} = \frac{8}{6} = \frac{4}{3} \text{ and R.H.S. }= \frac{1 + 3}{4} = 1. \text{ Since }\frac{4}{3} \neq 1$, it is not correct.

Hence, option 3 is the correct option.

Number of boys in the class = $m$

4 times the number of boys = $4m$

Number of girls = 20 more than $4m = 4m + 20$

Hence, option 4 is the correct option.