Chapter 10 Algebraic Expressions (Class 7 - Maths NCERT Exemplar Solutions)

Given:

Expression 1: $3x (3 - 2y)$

Expression 2: $2 (xy + x^2)$

Solution:

First, let us expand both expressions using the distributive property:

Expression 1: $3x \times 3 - 3x \times 2y = 9x - 6xy$

Expression 2: $2 \times xy + 2 \times x^2 = 2xy + 2x^2$

Like terms are terms that have the same variables with the same powers. Looking at the expanded terms:

From expression 1, we have: $9x$ and $-6xy$

From expression 2, we have: $2xy$ and $2x^2$

Comparing these, we find that $-6xy$ and $2xy$ are like terms because they both have the same algebraic factor $xy$.

Correct Option: (b)

Given:

Expression: $3x^2zy + 7xyz - 2z^2x$

To Find:

The coefficient of $xy$.

Solution:

To find the coefficient of $xy$, we identify the term that contains both $x$ and $y$.

In the term $7xyz$, the factors are $7, x, y,$ and $z$.

The coefficient of $xy$ is the product of the remaining factors in that term.

Remaining factors = $7 \times z = 7z$

Correct Option: (d)

Solution:

A term is a product of its factors. The given term is $-xy^2$.

We can break this down as follows:

1. The negative sign represents the numerical factor $-1$.

2. The variable factor $x$ is present once.

3. The variable factor $y^2$ means $y \times y$.

Therefore, the factors are: $-1 \times x \times y \times y$.

Correct Option: (d)

Solution:

According to the standard mathematical definition, an algebraic expression where the variables have only non-negative integer powers is known as a polynomial.

The blank should be filled with polynomial.

Solution:

In any algebraic term, the number that multiplies the variables is known as the numerical coefficient or simply the coefficient.

The blank should be filled with numerical coefficient (or coefficient).

Solution:

Terms that do not have the same literal (variable) factors are known as unlike terms.

The blank should be filled with unlike terms.

Solution:

Terms that consist of the same variables raised to the same corresponding powers are known as like terms.

The blank should be filled with like terms.

Solution:

By definition, an algebraic expression that consists of exactly two unlike terms is known as a binomial.

Example: $2x + 3y$ or $a^2 - 5$.

Therefore, the statement is True.

Solution:

A polynomial is a general category of algebraic expressions that can have one, two, three, or many terms. A monomial specifically refers to a polynomial with only one term.

For example, $x + y$ is a polynomial (specifically a binomial), but it is not a monomial.

Therefore, the statement is False.

Solution:

The word "variable" itself means something that can change. A variable is a symbol (like $x, y,$ or $z$) that can take on different numerical values depending on the context or equation. It is the "constant" whose value remains fixed.

Therefore, the statement is False.

Given:

Length of the rectangle = $x$

Breadth of the rectangle = $y$

To Find:

The algebraic expression for the perimeter.

Solution:

According to the problem, the perimeter is "twice the sum of length and breadth".

1. Sum of length and breadth = $(x + y)$

2. Twice the sum = $2 \times (x + y)$

So, the expression for the perimeter ($P$) is:

$P = 2(x + y)$

Given Expression:

$u^3 + 3u^2v + 3uv^2 + v^3$

Solution:

First, we identify the term that has the variable part $u^2$.

By observing the expression, the term containing $u^2$ is $3u^2v$.

Now, we find the coefficient of $u^2$. The coefficient is the factor that multiplies $u^2$ in that specific term.

In the term $3u^2v$, the factors are $3, u^2,$ and $v$.

Therefore, the coefficient of $u^2$ is $3v$.

Given Expression:

$7x^3 - 3x^2y + xy^2 + x^2y - y^3$

Solution:

To simplify, we group the like terms (terms with same variable factors) together.

The like terms in this expression are $-3x^2y$ and $+x^2y$.

Rearranging the expression:

$7x^3 + (-3x^2y + x^2y) + xy^2 - y^3$

Combining the coefficients of the like terms:

$7x^3 + (-3 + 1)x^2y + xy^2 - y^3$

$7x^3 - 2x^2y + xy^2 - y^3$

Therefore, the simplified expression is $7x^3 - 2x^2y + xy^2 - y^3$.

Step 1: Find the sum of the first two expressions

Let $S$ be the sum.

$S = (-3x^3y^2 + 2x^2y^3) + (-3x^2y^3 - 5y^4)$

$S = -3x^3y^2 + 2x^2y^3 - 3x^2y^3 - 5y^4$

Combining like terms ($2x^2y^3$ and $-3x^2y^3$):

$S = -3x^3y^2 - x^2y^3 - 5y^4$

Step 2: Subtract the sum from the third expression

We need to calculate: $(x^4 + x^3y^2 + x^2y^3 + y^4) - S$

$= (x^4 + x^3y^2 + x^2y^3 + y^4) - (-3x^3y^2 - x^2y^3 - 5y^4)$

Distributing the negative sign:

$= x^4 + x^3y^2 + x^2y^3 + y^4 + 3x^3y^2 + x^2y^3 + 5y^4$

Grouping and combining like terms:

$= x^4 + (x^3y^2 + 3x^3y^2) + (x^2y^3 + x^2y^3) + (y^4 + 5y^4)$

$= x^4 + 4x^3y^2 + 2x^2y^3 + 6y^4$

Therefore, the result is $x^4 + 4x^3y^2 + 2x^2y^3 + 6y^4$.

Given values: $a = 1, b = -2$

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

Substituting the values:

$= (1)^2 + (-2)^2 + 3(1)(-2)$

$= 1 + 4 - 6$

$= 5 - 6$

$= -1$

(ii) $a^3 + a^2b + ab^2 + b^3$

Substituting the values:

$= (1)^3 + (1)^2(-2) + (1)(-2)^2 + (-2)^3$

$= 1 + 1(-2) + 1(4) + (-8)$

$= 1 - 2 + 4 - 8$

$= 5 - 10$

$= -5$

Given:

The triangle is an equilateral triangle (all sides are equal).

Side length given in the figure = $2x + 3$

Perimeter of the triangle = $240$ cm

To Find:

The length of each side.

Solution:

We know that the perimeter of an equilateral triangle is given by:

$\text{Perimeter} = 3 \times \text{Side}$

Substituting the given values:

$240 = 3 \times (2x + 3)$

Dividing both sides by $3$:

$\frac{240}{3} = 2x + 3$

$80 = 2x + 3$

Subtracting $3$ from both sides:

$80 - 3 = 2x$

$77 = 2x$

$x = \frac{77}{2} = 38.5$

Now, we find the actual length of the side by substituting $x$ back into the expression:

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

$\text{Side} = 2(38.5) + 3$

$\text{Side} = 77 + 3 = 80$ cm

Alternate Solution:

Since it is an equilateral triangle, all three sides are equal.

$\text{Length of each side} = \frac{\text{Perimeter}}{3}$

$\text{Length of each side} = \frac{240}{3} = 80$ cm

Therefore, each side of the equilateral triangle is $80$ cm.

Solution:

In algebra, we classify expressions based on the number of terms they contain:

1. A monomial contains only one term.

2. A binomial contains exactly two unlike terms.

3. A trinomial contains exactly three unlike terms.

Since the question asks for an expression with three terms, it is called a trinomial.

Correct Option: (c)

Given Expression:

$3x^2y - 2y^2z - z^2x + 5$

Solution:

Terms are the parts of an algebraic expression that are separated by addition ($+$) or subtraction ($-$) signs. Let us identify the individual terms in the given expression:

1. $3x^2y$

2. $-2y^2z$

3. $-z^2x$

4. $5$

There are total 4 terms in the expression.

Correct Option: (c)

Given Expression:

$4x^2 - 3xy$

Solution:

Terms are the building blocks of an expression. In the expression $4x^2 - 3xy$, the terms are added or subtracted. Note that the sign ($+$ or $-$) is part of the term.

The first term is $4x^2$.

The second term is $-3xy$.

Therefore, the terms are $4x^2$ and $-3xy$.

Correct Option: (a)

Solution:

A term is formed by the product of its factors. The given term is $-5x^2y^2z$. Let us break it down into its irreducible factors:

1. The numerical factor is $-5$.

2. $x^2$ is $x \times x$.

3. $y^2$ is $y \times y$.

4. $z$ is $z$.

Multiplying these together: $-5 \times x \times x \times y \times y \times z$.

Correct Option: (c)

Given Term:

$-9xy^2z$

Solution:

The coefficient of a variable in a term is the product of all the other factors in that term. To find the coefficient of $x$, we look at what is multiplying $x$ in the given term.

Term = $(-9y^2z) \times x$

So, the coefficient of $x$ is $-9y^2z$.

Correct Option: (d)

Solution:

Like terms are terms that have exactly the same literal (variable) factors. This means the variables and their respective exponents must be identical.

Let's examine the options:

(a) $xy^2z$ and $x^2yz$: Variable parts are different. (Unlike)

(b) $xyz^2$ and $xyz^2$: Both have the same variable part $xyz^2$. (Like)

(c) $xyz$ and $x^2y^2z^2$: Exponents are different. (Unlike)

(d) $xyz^2$ and $x^2yz$: Variable parts are different. (Unlike)

Correct Option: (b)

Solution:

A binomial is an expression which, after simplification, contains exactly two unlike terms.

(a) $3xy^2 + 5y - x^2y$: It has 3 unlike terms (Trinomial).

(b) $x^2y - 5y - x^2y = (1-1)x^2y - 5y = -5y$: It simplifies to 1 term (Monomial).

(c) $xy + yz + zx$: It has 3 unlike terms (Trinomial).

(d) $3xy^2 + 5y - xy^2 = (3-1)xy^2 + 5y = 2xy^2 + 5y$: It simplifies to 2 unlike terms (Binomial).

Correct Option: (d)

To Find:

The sum of $(x^4 - xy + 2y^2)$ and $(-x^4 + xy + 2y^2)$.

Solution:

Sum $= (x^4 - xy + 2y^2) + (-x^4 + xy + 2y^2)$

Grouping like terms:

Sum $= (x^4 - x^4) + (-xy + xy) + (2y^2 + 2y^2)$

Sum $= 0 + 0 + 4y^2$

Sum $= 4y^2$

The resulting expression is $4y^2$. It has only one term, so it is a monomial. Since it contains only the variable $y$, it is a polynomial in $y$.

Correct Option: (a)

Solution:

1. "5 times of y" can be written as $5y$.

2. "Subtraction from x" means $x$ is the starting value from which something is taken away.

Combining these, the expression is $x - 5y$.

Correct Option: (c)

Solution:

The given expression is $-b - 0$.

Since subtracting zero does not change the value of a number, $-b - 0 = -b$.

Now let's check the options:

(a) $-1 \times b = -b$. (Matches)

(b) $1 - b - 0 = 1 - b$. (Incorrect)

(c) $0 - (-1) \times b = 0 - (-b) = +b$. (Incorrect)

(d) $-b - 0 - 1 = -b - 1$. (Incorrect)

Correct Option: (a)

Solution:

A square has four equal sides. The perimeter of a square is the sum of the lengths of its four sides.

$\text{Perimeter} = \text{Side} + \text{Side} + \text{Side} + \text{Side}$

$\text{Perimeter} = 4 \times \text{Side}$

Given the side length is $x$, the perimeter is $4x$.

Correct Option: (c)

Solution:

To find the number of pieces, we divide the total length of the cloth by the length of one scarf.

Total cloth $= y$ m

Length of one scarf $= \frac{1}{2}$ m

$\text{Number of scarfs} = y \div \frac{1}{2}$

$\text{Number of scarfs} = y \times \frac{2}{1} = 2y$

Correct Option: (a)

Solution:

Like terms are terms that have the same algebraic factors (variables and exponents).

The given expression $123x^2y - 138x^2y$ consists of terms with the algebraic factor $x^2y$.

Checking the options for the same algebraic factor $x^2y$:

(a) $10xy$ (Incorrect)

(b) $-15xy$ (Incorrect)

(c) $-15xy^2$ (Incorrect)

(d) $10x^2y$ (Matches)

Correct Option: (d)

Given:

Expression: $3x^2 - 5x + 3$

Value of $x = 1$

Solution:

Substitute $x = 1$ into the expression:

$= 3(1)^2 - 5(1) + 3$

$= 3(1) - 5 + 3$

$= 3 - 5 + 3$

$= 6 - 5 = 1$

Correct Option: (a)

Solution:

Consider a polygon with $n$ vertices. From one specific vertex, you cannot draw a diagonal to:

1. Itself (1 point)

2. The two adjacent vertices (2 points)

Since these 3 points must be excluded, the number of diagonals from a single vertex is $n - 3$.

Correct Option: (d)

Given:

Side of square $= 2x + 3$

Solution:

We know that the perimeter of a square $= 4 \times \text{side}$.

$\text{Perimeter} = 4 \times (2x + 3)$

Using the distributive property:

$\text{Perimeter} = 4 \times 2x + 4 \times 3$

$\text{Perimeter} = 8x + 12$

Correct Option: (d)

Solution:

Like terms are terms that have the same algebraic factors. When we add or subtract two like terms, we only operate on their numerical coefficients, while the variable part remains exactly the same.

For example: $3x + 5x = 8x$, where $3x, 5x,$ and $8x$ are all like terms.

Therefore, the blank should be filled with a like term.

Solution:

In the expression $\pi r^2$, the variable is $r$ (radius), which can change. The symbol $\pi$ (Pi) represents a fixed numerical value (approximately $3.14159...$ or $\frac{22}{7}$).

Therefore, the numerical constant is $\pi$.

Solution:

Let's look at the algebraic factors of both terms:

Term 1: $3a^2b \rightarrow$ Algebraic factors are $a, a, b$ (or $a^2b$).

Term 2: $-7ba^2 \rightarrow$ Algebraic factors are $b, a, a$ (or $ba^2$).

Since multiplication is commutative ($a^2b = ba^2$), the variable parts are identical.

Therefore, these are like terms.

Solution:

Let's compare the literal (variable) factors:

Term 1: $-5a^2b \rightarrow$ Power of $a$ is $2$, power of $b$ is $1$.

Term 2: $-5b^2a \rightarrow$ Power of $a$ is $1$, power of $b$ is $2$.

Since the powers of the corresponding variables are not the same, the variable parts are different.

Therefore, these are unlike terms.

Solution:

The expression $2\pi r$ represents the circumference of a circle. Here:

1. $2$ is a constant number.

2. $\pi$ is a mathematical constant.

3. $r$ stands for the radius, which can take different values for different circles.

Therefore, the algebraic variable is $r$.

Solution:

The word "mono" means "one". By definition, a monomial is an algebraic expression consisting of only one term.

Therefore, the number of terms is one.

Solution:

First, we expand the given expression:

$n(n + 1) + 6(n - 1) = (n \times n) + (n \times 1) + (6 \times n) - (6 \times 1)$

$= n^2 + n + 6n - 6$

In this expanded expression, the terms are $n^2, n, 6n,$ and $-6$. We look for terms with the same variable part.

The terms $n$ and $6n$ both have the variable $n$ raised to the power $1$.

Therefore, the like terms are $n$ and $6n$.

Solution:

The expression $13 + 90$ consists only of constants. When simplified, it equals $103$. Since it simplifies to a single numerical value, it can be classified as a monomial (specifically a numerical monomial or a constant expression).

Therefore, the blank should be filled with monomial (or constant).

Given:

Speed of the car = $55$ km/hrs

Time taken = $y$ hours

To Find:

Distance covered in $y$ hours.

Solution:

The relationship between speed, distance, and time is given by the formula:

Substituting the given values into the formula:

$\text{Distance} = 55 \times y$

$\text{Distance} = 55y$

The distance covered in $y$ hours is $55y$ km.

Solution:

In algebra, expressions are classified based on the number of terms they contain:

1. Monomial: An expression containing only one term.

2. Binomial: An expression containing exactly two terms.

3. Trinomial: An expression containing exactly three terms.

The given expression is $x + y + z$.

This expression has three distinct unlike terms: $x$, $y$, and $z$. Therefore, it is a trinomial.

Since it is a trinomial, it is neither a monomial nor a binomial.

To Find:

The coefficient of $y$ in the resulting expression after subtraction.

Solution:

Subtracting $(x^2y + y^2 + 3)$ from $(3x^2y + 2y^2 + 5)$:

$(3x^2y + 2y^2 + 5) - (x^2y + y^2 + 3)$

Expanding the brackets and changing signs of terms to be subtracted:

$3x^2y + 2y^2 + 5 - x^2y - y^2 - 3$

Grouping the like terms together:

$(3x^2y - x^2y) + (2y^2 - y^2) + (5 - 3)$

$2x^2y + y^2 + 2$

In the resulting expression $2x^2y + y^2 + 2$:

The term containing $y$ is $2x^2y$.

The coefficient of $y$ in the term $2x^2y$ is $2x^2$.

Solution:

We are given the expression $- a - b - c$.

We need to find the expression inside the bracket such that when a negative sign is applied, it results in $- b - c$.

Using the distributive property of signs:

$- a - b - c = - a - (b + c)$

When the negative sign outside the parenthesis is distributed, it changes the sign of every term inside the parenthesis.

$- (b + c) = - b - c$

Therefore, the blank should be filled with $b + c$.

Solution:

First, let's find the perimeter of the Rectangle (first figure):

Length ($l$) = $2x$

Breadth ($b$) = $y$

$\text{Perimeter of Rectangle} = 2(l + b)$

$\text{Perimeter} = 2(2x + y) = 4x + 2y$

Now, let's find the perimeter of the Parallelogram (second figure):

Base ($a$) = $x$

Adjacent side ($b$) = $y^2$

$\text{Perimeter of Parallelogram} = 2(a + b)$

$\text{Perimeter} = 2(x + y^2) = 2x + 2y^2$

Comparing the terms in both perimeters: $4x, 2y, 2x, 2y^2$.

Terms that have the same literal factor (variable and power) are like terms. Here, $4x$ and $2x$ are like terms.

Terms that have different literal factors are unlike terms. Here, $2y$ and $2y^2$ have the same variable but different powers.

Therefore, the unlike terms are $2y$ and $2y^2$.

Given:

Trinomial expression: $- 2x + 4y^2 + z$

Solution:

A binomial is an expression consisting of exactly two terms. To make the given trinomial a binomial, we must add a term that cancels out one of the existing three terms.

If we add the monomial $2x$ to the expression:

$(- 2x + 4y^2 + z) + 2x = (-2x + 2x) + 4y^2 + z$

$= 0 + 4y^2 + z = 4y^2 + z$

Since $4y^2 + z$ has only two terms, it is a binomial.

The monomial can be $2x$ (similarly, adding $-4y^2$ or $-z$ would also result in a binomial).

Solution:

Let the missing term be $P$.

$3x + 23x^2 + 6y^2 + 2x + y^2 + P = 5x + 7y^2$

First, simplify the terms on the left side by adding like terms:

$(3x + 2x) + 23x^2 + (6y^2 + y^2) + P = 5x + 7y^2$

$5x + 23x^2 + 7y^2 + P = 5x + 7y^2$

Now, subtract $(5x + 23x^2 + 7y^2)$ from both sides to find $P$:

$P = (5x + 7y^2) - (5x + 23x^2 + 7y^2)$

$P = 5x - 5x + 7y^2 - 7y^2 - 23x^2$

$P = -23x^2$

The blank should be filled with $-23x^2$.

Given:

Number of toffees Rohit has = $5xy$

Number of toffees Shantanu has = $20yx$

Solution:

In algebraic multiplication, the order of variables does not change the value (Commutative Property).

So, Shantanu has $20xy$ toffees.

To find how many more toffees Shantanu has compared to Rohit, we subtract:

$\text{Difference} = \text{Shantanu's toffees} - \text{Rohit's toffees}$

$\text{Difference} = 20xy - 5xy$

$\text{Difference} = 15xy$

Shantanu has $15xy$ more toffees.

Solution:

An algebraic expression is called a polynomial if the exponents (powers) of the variables involved are non-negative integers (i.e., $0, 1, 2, 3, \dots$).

The given expression is: $1 + \frac{x}{2} + x^3$

In this expression, the powers of the variable $x$ are $1$ (in $\frac{x}{2}$) and $3$ (in $x^3$). Since both $1$ and $3$ are positive integers, the expression satisfies the condition of being a polynomial.

Hence, the statement is True.

Solution:

Let's simplify the given expression by performing the subtraction:

$(3a - b + 3) - (a + b)$

$= 3a - b + 3 - a - b$

Grouping the like terms together:

$= (3a - a) + (-b - b) + 3$

$= 2a - 2b + 3$

A binomial is an expression that contains exactly two terms.

The simplified expression $2a - 2b + 3$ contains three unlike terms ($2a$, $-2b$, and $3$), which makes it a trinomial, not a binomial.

Hence, the statement is False.

Solution:

By definition, a polynomial is an algebraic expression consisting of one or more terms with non-negative integer powers of the variables.

Based on the number of terms, polynomials are classified as:

1. Monomial (1 term)

2. Binomial (2 terms)

3. Trinomial (3 terms)

Since a trinomial is simply a specific type of polynomial (one with exactly three terms), the statement is correct.

Hence, the statement is True.

Solution:

A polynomial with exactly three terms is called a trinomial.

The phrase "more than two terms" includes expressions with $3, 4, 5, \dots$ terms.

If a polynomial has 4 terms, it has "more than two terms," but it is called a quadrinomial (or generally a polynomial), not a trinomial.

Hence, the statement is False.

Solution:

In algebra, the word "Sum" represents the addition of two or more quantities.

To find the sum of two variables $x$ and $y$, we use the addition operator ($+$).

Therefore, the sum is written as $x + y$.

Hence, the statement is True.

Solution:

The sum of $2$ and $p$ refers to the addition of these two values, which is written as $2 + p$.

The expression $2p$ represents the product (multiplication) of $2$ and $p$.

Hence, the statement is False.

Solution:

The prefix "bi-" means two.

In algebra, a binomial is defined as an expression consisting of exactly two unlike terms.

An expression with "more than two terms" (such as 3 or 4 terms) would be a trinomial or a general polynomial, not a binomial.

Hence, the statement is False.

Solution:

The prefix "tri-" means three.

In algebra, a trinomial is defined as an algebraic expression containing exactly three unlike terms.

Hence, the statement is True.

Statement: In like terms, variables and their powers are the same.

Solution:

The given statement is True.

In algebra, Like Terms are defined as terms that contain the same variables (literal factors) raised to the same exponents (powers).

For example, in the expression $7x^2y$ and $-3x^2y$, both terms have the same variables $x$ and $y$ with the same powers $2$ and $1$ respectively. Therefore, they are like terms.

Statement: The expression $x + y + 5x$ is a trinomial.

Solution:

The given statement is False.

An algebraic expression is categorized as a monomial, binomial, or trinomial based on the number of terms it contains after it has been simplified by combining all like terms.

Given expression: $x + y + 5x$

The simplified expression $6x + y$ contains exactly two terms ($6x$ and $y$).

By definition, an expression with two terms is called a Binomial. Therefore, the original expression is not a trinomial.

Statement: $4p$ is the numerical coefficient of $q^2$ in $-4pq^2$.

Solution:

The given statement is False.

In the term $-4pq^2$, we look for the factors that are multiplied by $q^2$.

The coefficient of $q^2$ in the term $-4pq^2$ is $-4p$.

The numerical coefficient specifically refers to the constant or number part of the term. In the term $-4pq^2$, the numerical coefficient is $-4$.

Since the statement claims the coefficient is $4p$ (missing the negative sign) and identifies it as a "numerical coefficient" (despite containing the variable $p$), the statement is incorrect.

Statement: $5a$ and $5b$ are unlike terms.

Solution:

The given statement is True.

Terms are called unlike terms when they do not have the same algebraic (literal) factors. The numerical coefficients do not determine whether terms are "like" or "unlike".

In the terms $5a$ and $5b$:

1. The literal factor of $5a$ is $a$.

2. The literal factor of $5b$ is $b$.

Since the variables $a$ and $b$ are different, these are unlike terms.

Statement: Sum of $x^2 + x$ and $y + y^2$ is $2x^2 + 2y^2$.

Solution:

The given statement is False.

To find the sum, we add the two expressions together:

$\text{Sum} = (x^2 + x) + (y + y^2)$

$\text{Sum} = x^2 + x + y + y^2$

Since all four terms ($x^2$, $x$, $y$, and $y^2$) are unlike terms, they cannot be simplified further or combined into $2x^2$ or $2y^2$. Like terms must have the same variable raised to the same power to be added numerically.

Statement: Subtracting a term from a given expression is the same as adding its additive inverse to the given expression.

Solution:

The given statement is True.

In algebra, the operation of subtraction is equivalent to the addition of the additive inverse (or the negative) of the term being subtracted.

If we have an expression $E$ and we subtract a term $T$, the operation is:

Here, $-T$ is the additive inverse of $T$. For example, subtracting $5$ is the same as adding $-5$.

Statement: The total number of planets of Sun can be denoted by the variable $n$.

Solution:

The given statement is False.

A variable is a quantity that can take different values and does not have a fixed value. In mathematics, we use variables to represent quantities that change.

The number of planets in our solar system orbiting the Sun is a fixed, known value (which is 8). Because this value does not change, it is a constant, not a variable.

Statement: In like terms, the numerical coefficients should also be the same.

Solution:

The given statement is False.

For terms to be classified as like terms, only their algebraic (literal) factors—the variables and their respective powers—must be identical. Their numerical coefficients can be any real number and do not need to be the same.

For example, $4xy$ and $-9xy$ are like terms because they both have the same literal factor $xy$, even though their numerical coefficients ($4$ and $-9$) are different.

Statement: If we add a monomial and binomial, then answer can never be a monomial.

Solution:

The given statement is False.

To understand why, let us consider an example where the monomial is the additive inverse of one of the terms in the binomial.

Let the monomial be $-x$.

Let the binomial be $x + y$.

Now, let us find their sum:

$\text{Sum} = (-x) + (x + y)$

$\text{Sum} = -x + x + y$

In this case, the result is $y$, which is a monomial. Therefore, it is possible for the sum of a monomial and a binomial to be a monomial.

Statement: If we subtract a monomial from a binomial, then answer is atleast a binomial.

Solution:

The given statement is False.

Let us take an example where we subtract a monomial from a binomial containing a like term.

Let the binomial be $2x + 3y$.

Let the monomial be $2x$.

Subtracting the monomial from the binomial:

$\text{Result} = (2x + 3y) - (2x)$

$\text{Result} = 2x - 2x + 3y$

$\text{Result} = 3y$

Here, the result $3y$ is a monomial, not a binomial. Thus, the result can be a monomial.

Statement: When we subtract a monomial from a trinomial, then answer can be a polynomial.

Solution:

The given statement is True.

By definition, any algebraic expression consisting of one or more terms with non-negative integer exponents is a polynomial. This includes monomials, binomials, and trinomials.

When you subtract a monomial from a trinomial:

1. If there are no like terms, the result is an expression with four terms (a polynomial).

2. If there is a like term, the result is a binomial or a trinomial (both of which are polynomials).

Since the result will always be an algebraic expression of this nature, it will always be a polynomial.

Statement: When we add a monomial and a trinomial, then answer can be a monomial.

Solution:

The given statement is False.

A trinomial has three unlike terms. A monomial is a single term. When we add a monomial to a trinomial, the monomial can, at most, cancel out one of the terms of the trinomial (if it is the additive inverse of that term).

Let the trinomial be $a + b + c$.

Let the monomial be $-a$.

$\text{Sum} = (a + b + c) + (-a) = b + c$

The result $b + c$ is a binomial. Since a single monomial cannot eliminate two terms of a trinomial simultaneously, the result will always have at least two terms. Thus, it can never be a monomial.

(a) $x$ is multiplied by itself and then added to the product of $x$ and $y$.

Expression: $x \cdot x + x \cdot y = x^2 + xy$

Type: Binomial

(b) Three times of $p$ and two times of $q$ are multiplied and then subtracted from $r$.

Expression: $r - (3p \cdot 2q) = r - 6pq$

Type: Binomial

(c) Product of $p$, twice of $q$ and thrice of $r$.

Expression: $p \cdot 2q \cdot 3r = 6pqr$

Type: Monomial

(d) Sum of the products of $a$ and $b$, $b$ and $c$ and $c$ and $a$.

Expression: $ab + bc + ca$

Type: Trinomial

(e) Perimeter of an equilateral triangle of side $x$.

Expression: $x + x + x = 3x$

Type: Monomial

(f) Perimeter of a rectangle with length $p$ and breadth $q$.

Expression: $2(p + q) = 2p + 2q$

Type: Binomial

(g) Area of a triangle with base $m$ and height $n$.

Expression: $\frac{1}{2} \cdot m \cdot n = \frac{1}{2}mn$

Type: Monomial

(h) Area of a square with side $x$.

Expression: $x \cdot x = x^2$

Type: Monomial

(i) Cube of $s$ subtracted from cube of $t$.

Expression: $t^3 - s^3$

Type: Binomial

(j) Quotient of $x$ and $15$ multiplied by $x$.

Expression: $\left(\frac{x}{15}\right) \cdot x = \frac{x^2}{15}$

Type: Monomial

(k) The sum of square of $x$ and cube of $z$.

Expression: $x^2 + z^3$

Type: Binomial

(l) Two times $q$ subtracted from cube of $q$.

Expression: $q^3 - 2q$

Type: Binomial

Solution:

The coefficient of a variable in a term is the factor (numerical or literal) that multiplies that variable.

(i) In the expression $x^2 - x + 4$, the term containing $x^2$ is $1 \cdot x^2$.

Therefore, the coefficient of $x^2$ is $1$.

(ii) In the expression $x^3 - 2x^2 + 3x + 1$, the term containing $x^2$ is $-2x^2$.

Therefore, the coefficient of $x^2$ is $-2$.

(iii) In the expression $1 + 2x + 3x^2 + 4x^3$, the term containing $x^2$ is $3x^2$.

Therefore, the coefficient of $x^2$ is $3$.

(iv) In the expression $y + y^2x + y^3x^2 + y^4x^3$, the term containing $x^2$ is $y^3x^2$.

Therefore, the coefficient of $x^2$ is $y^3$.

Solution:

The numerical coefficient is the constant (number) part of an algebraic term.

(i) Given terms: $x^3y^2z$, $xy^2z^3$, $-3xy^2z^3$, $5x^3y^2z$, $-7x^2y^2z^2$

(ii) Given terms: $10xyz$, $-7xy^2z$, $-9xyz$, $2xy^2z$, $2x^2y^2z^2$

Solution:

(a) $3x^2yz^2 - 3xy^2z + x^2yz^2 + 7xy^2z$

Grouping like terms:

$= (3x^2yz^2 + x^2yz^2) + (-3xy^2z + 7xy^2z)$

$= 4x^2yz^2 + 4xy^2z$

The simplified expression has two terms. Therefore, it is a Binomial.

(b) $x^4 + 3x^3y + 3x^2y^2 - 3x^3y - 3xy^3 + y^4 - 3x^2y^2$

Grouping like terms:

$= x^4 + (3x^3y - 3x^3y) + (3x^2y^2 - 3x^2y^2) - 3xy^3 + y^4$

$= x^4 + 0 + 0 - 3xy^3 + y^4$

$= x^4 - 3xy^3 + y^4$

The simplified expression has three terms. Therefore, it is a Trinomial.

(c) $p^3q^2r + pq^2r^3 + 3p^2qr^2 - 9p^2qr^2$

Grouping like terms:

$= p^3q^2r + pq^2r^3 + (3 - 9)p^2qr^2$

$= p^3q^2r + pq^2r^3 - 6p^2qr^2$

The simplified expression has three terms. Therefore, it is a Trinomial.

(d) $2a + 2b + 2c - 2a - 2b - 2c - 2b + 2c + 2a$

Grouping like terms:

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

$= 2a - 2b + 2c$

The simplified expression has three terms. Therefore, it is a Trinomial.

(e) $50x^3 - 21x + 107 + 41x^3 - x + 1 - 93 + 71x - 31x^3$

Grouping like terms:

$= (50x^3 + 41x^3 - 31x^3) + (-21x - x + 71x) + (107 + 1 - 93)$

$= (91 - 31)x^3 + (-22 + 71)x + (108 - 93)$

$= 60x^3 + 49x + 15$

The simplified expression has three terms. Therefore, it is a Trinomial.

Solution:

(a) $(p^2 - 7pq - q^2) + (-3p^2 - 2pq + 7q^2)$

$\begin{array}{cc} & p^2 & -7pq & -q^2 \\ + & -3p^2 & -2pq & +7q^2 \\ \hline & -2p^2 & -9pq & +6q^2 \\ \hline \end{array}$

(b) $(x^3 - x^2y - xy^2 - y^3) + (x^3 - 2x^2y + 3xy^2 + 4y)$

Combining like terms:

$= (x^3 + x^3) + (-x^2y - 2x^2y) + (-xy^2 + 3xy^2) - y^3 + 4y$

$= 2x^3 - 3x^2y + 2xy^2 - y^3 + 4y$

(c) $(ab + bc + ca) + (-bc - ca - ab)$

$= ab - ab + bc - bc + ca - ca$

$= 0$

(d) $(p^2 - q + r) + (q^2 - r + p) + (r^2 - p + q)$

$= p^2 + q^2 + r^2 + (-q + q) + (r - r) + (p - p)$

$= p^2 + q^2 + r^2$

(e) $(x^3y^2 + x^2y^3 + 3y^4) + (x^4 + 3x^2y^3 + 4y^4)$

$= x^4 + x^3y^2 + (x^2y^3 + 3x^2y^3) + (3y^4 + 4y^4)$

$= x^4 + x^3y^2 + 4x^2y^3 + 7y^4$

(f) $(p^2qr + pq^2r + pqr^2) + (-3pq^2r - 2pqr^2)$

$= p^2qr + (pq^2r - 3pq^2r) + (pqr^2 - 2pqr^2)$

$= p^2qr - 2pq^2r - pqr^2$

(g) $(uv - vw) + (vw - wu) + (wu - uv)$

$= uv - uv - vw + vw - wu + wu$

$= 0$

(h) $(a^2 + 3ab - bc) + (b^2 + 3bc - ca) + (c^2 + 3ca - ab)$

$= a^2 + b^2 + c^2 + (3ab - ab) + (-bc + 3bc) + (-ca + 3ca)$

$= a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

(i) $(\frac{5}{8}p^4 + 2p^2 + \frac{5}{8}) + (\frac{1}{8} - 17p + \frac{9}{8}p^2) + (p^5 - p^3 + 7)$

Grouping like powers of $p$:

$= p^5 + \frac{5}{8}p^4 - p^3 + (2 + \frac{9}{8})p^2 - 17p + (\frac{5}{8} + \frac{1}{8} + 7)$

$= p^5 + \frac{5}{8}p^4 - p^3 + (\frac{16+9}{8})p^2 - 17p + (\frac{6}{8} + 7)$

$= p^5 + \frac{5}{8}p^4 - p^3 + \frac{25}{8}p^2 - 17p + (\frac{3}{4} + 7)$

$= p^5 + \frac{5}{8}p^4 - p^3 + \frac{25}{8}p^2 - 17p + \frac{31}{4}$

(j) Add: $(t - t^2 - t^3 - 14)$, $(15t^3 + 13 + 9t - 8t^2)$, $(12t^2 - 19 - 24t)$, $(4t - 9t^2 + 19t^3)$

Adding term by term:

Sum of $t^3$ terms: $-t^3 + 15t^3 + 19t^3 = 33t^3$

Sum of $t^2$ terms: $-t^2 - 8t^2 + 12t^2 - 9t^2 = -6t^2$

Sum of $t$ terms: $t + 9t - 24t + 4t = -10t$

Sum of constant terms: $-14 + 13 - 19 = -20$

Resulting Expression: $33t^3 - 6t^2 - 10t - 20$

(a) Subtract $-7p^2qr$ from $-3p^2qr$.

Difference $= (-3p^2qr) - (-7p^2qr)$

Difference $= -3p^2qr + 7p^2qr$

Difference $= (-3+7)p^2qr$

Difference $= 4p^2qr$

(b) Subtract $-a^2 - ab$ from $b^2 + ab$.

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

Difference $= b^2 + ab + a^2 + ab$

Difference $= a^2 + b^2 + ab + ab$

Difference $= a^2 + b^2 + 2ab$

(c) Subtract $-4x^2y - y^3$ from $x^3 + 3xy^2 - x^2y$.

Difference $= (x^3 + 3xy^2 - x^2y) - (-4x^2y - y^3)$

Difference $= x^3 + 3xy^2 - x^2y + 4x^2y + y^3$

Difference $= x^3 + y^3 + 3xy^2 - x^2y + 4x^2y$

Difference $= x^3 + y^3 + 3xy^2 + (-1+4)x^2y$

Difference $= x^3 + y^3 + 3xy^2 + 3x^2y$

(d) Subtract $x^4 + 3x^3y^3 + 5y^4$ from $2x^4 - x^3y^3 + 7y^4$.

Difference $= (2x^4 - x^3y^3 + 7y^4) - (x^4 + 3x^3y^3 + 5y^4)$

Difference $= 2x^4 - x^3y^3 + 7y^4 - x^4 - 3x^3y^3 - 5y^4$

Difference $= 2x^4 - x^4 - x^3y^3 - 3x^3y^3 + 7y^4 - 5y^4$

Difference $= (2-1)x^4 + (-1-3)x^3y^3 + (7-5)y^4$

Difference $= x^4 - 4x^3y^3 + 2y^4$

(e) Subtract $ab - bc - ca$ from $-ab + bc + ca$.

Difference $= (-ab + bc + ca) - (ab - bc - ca)$

Difference $= -ab + bc + ca - ab + bc + ca$

Difference $= -ab - ab + bc + bc + ca + ca$

Difference $= (-1-1)ab + (1+1)bc + (1+1)ca$

Difference $= -2ab + 2bc + 2ca$

(f) Subtract $-2a^2 - 2b^2$ from $-a^2 - b^2 + 2ab$.

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

Difference $= -a^2 - b^2 + 2ab + 2a^2 + 2b^2$

Difference $= -a^2 + 2a^2 - b^2 + 2b^2 + 2ab$

Difference $= (-1+2)a^2 + (-1+2)b^2 + 2ab$

Difference $= a^2 + b^2 + 2ab$

(g) Subtract $x^3y^2 + 3x^2y^2 - 7xy^3$ from $x^4 + y^4 + 3x^2y^2 - xy^3$.

Difference $= (x^4 + y^4 + 3x^2y^2 - xy^3) - (x^3y^2 + 3x^2y^2 - 7xy^3)$

Difference $= x^4 + y^4 + 3x^2y^2 - xy^3 - x^3y^2 - 3x^2y^2 + 7xy^3$

Difference $= x^4 + y^4 - x^3y^2 + (3x^2y^2 - 3x^2y^2) + (-xy^3 + 7xy^3)$

Difference $= x^4 + y^4 - x^3y^2 + 0 + (-1+7)xy^3$

Difference $= x^4 + y^4 - x^3y^2 + 6xy^3$

(h) Subtract $2(ab + bc + ca)$ from $-ab - bc - ca$.

$2(ab + bc + ca) = 2ab + 2bc + 2ca$

Difference $= (-ab - bc - ca) - (2ab + 2bc + 2ca)$

Difference $= -ab - bc - ca - 2ab - 2bc - 2ca$

Difference $= (-1-2)ab + (-1-2)bc + (-1-2)ca$

Difference $= -3ab - 3bc - 3ca$

Difference $= -3(ab + bc + ca)$

(i) Subtract $4.5x^5 - 3.4x^2 + 5.7$ from $5x^4 - 3.2x^2 - 7.3x$.

Difference $= (5x^4 - 3.2x^2 - 7.3x) - (4.5x^5 - 3.4x^2 + 5.7)$

Difference $= 5x^4 - 3.2x^2 - 7.3x - 4.5x^5 + 3.4x^2 - 5.7$

Difference $= -4.5x^5 + 5x^4 + (-3.2 + 3.4)x^2 - 7.3x - 5.7$

Difference $= -4.5x^5 + 5x^4 + 0.2x^2 - 7.3x - 5.7$

(j) Subtract $11 - 15y^2$ from $y^3 - 15y^2 - y - 11$.

Difference $= (y^3 - 15y^2 - y - 11) - (11 - 15y^2)$

Difference $= y^3 - 15y^2 - y - 11 - 11 + 15y^2$

Difference $= y^3 + (-15y^2 + 15y^2) - y + (-11 - 11)$

Difference $= y^3 + 0 - y - 22$

Difference $= y^3 - y - 22$

(a) To find what should be added to $x^3 + 3x^2y + 3xy^2 + y^3$ to get $x^3 + y^3$, we subtract the first expression from the second expression.

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

Required expression $= x^3 + y^3 - x^3 - 3x^2y - 3xy^2 - y^3$

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

Required expression $= 0 + 0 - 3x^2y - 3xy^2$

Required expression $= -3x^2y - 3xy^2$

(b) To find what should be added to $3pq + 5p^2q^2 + p^3$ to get $p^3 + 2p^2q^2 + 4pq$, we subtract the first expression from the second expression.

Required expression $= (p^3 + 2p^2q^2 + 4pq) - (3pq + 5p^2q^2 + p^3)$

Required expression $= p^3 + 2p^2q^2 + 4pq - 3pq - 5p^2q^2 - p^3$

Required expression $= (p^3 - p^3) + (2p^2q^2 - 5p^2q^2) + (4pq - 3pq)$

Required expression $= 0 + (2-5)p^2q^2 + (4-3)pq$

Required expression $= -3p^2q^2 + pq$

(a) Let the expression that should be subtracted be $E$.

According to the question:

$(2x^3 - 3x^2y + 2xy^2 + 3y^3) - E = x^3 - 2x^2y + 3xy^2 + 4y^3$

To find $E$, we rearrange the equation:

$E = (2x^3 - 3x^2y + 2xy^2 + 3y^3) - (x^3 - 2x^2y + 3xy^2 + 4y^3)$

$E = 2x^3 - 3x^2y + 2xy^2 + 3y^3 - x^3 + 2x^2y - 3xy^2 - 4y^3$

Combine like terms:

$E = (2x^3 - x^3) + (-3x^2y + 2x^2y) + (2xy^2 - 3xy^2) + (3y^3 - 4y^3)$

$E = (2-1)x^3 + (-3+2)x^2y + (2-3)xy^2 + (3-4)y^3$

$E = 1x^3 - 1x^2y - 1xy^2 - 1y^3$

$E = x^3 - x^2y - xy^2 - y^3$

Thus, $x^3 - x^2y - xy^2 - y^3$ should be subtracted.

(b) Let the expression that should be subtracted be $E$.

According to the question:

$(-7mn + 2m^2 + 3n^2) - E = m^2 + 2mn + n^2$

To find $E$, we rearrange the equation:

$E = (-7mn + 2m^2 + 3n^2) - (m^2 + 2mn + n^2)$

$E = -7mn + 2m^2 + 3n^2 - m^2 - 2mn - n^2$

Combine like terms:

$E = (2m^2 - m^2) + (3n^2 - n^2) + (-7mn - 2mn)$

$E = (2-1)m^2 + (3-1)n^2 + (-7-2)mn$

$E = 1m^2 + 2n^2 - 9mn$

$E = m^2 + 2n^2 - 9mn$

Thus, $m^2 + 2n^2 - 9mn$ should be subtracted.

To find how much $21a^3 - 17a^2$ is less than $89a^3 - 64a^2 + 6a + 16$, we need to subtract the first expression from the second expression.

Difference $= (89a^3 - 64a^2 + 6a + 16) - (21a^3 - 17a^2)$

Difference $= 89a^3 - 64a^2 + 6a + 16 - 21a^3 + 17a^2$

Group like terms:

Difference $= (89a^3 - 21a^3) + (-64a^2 + 17a^2) + 6a + 16$

Combine the coefficients:

Difference $= (89 - 21)a^3 + (-64 + 17)a^2 + 6a + 16$

Difference $= 68a^3 - 47a^2 + 6a + 16$

So, $21a^3 - 17a^2$ is $68a^3 - 47a^2 + 6a + 16$ less than $89a^3 - 64a^2 + 6a + 16$.

To find how much $y^4 - 12y^2 + y + 14$ is greater than $17y^3 + 34y^2 - 51y + 68$, we need to subtract the second expression from the first expression.

Difference $= (y^4 - 12y^2 + y + 14) - (17y^3 + 34y^2 - 51y + 68)$

Difference $= y^4 - 12y^2 + y + 14 - 17y^3 - 34y^2 + 51y - 68$

Group like terms, arranging in descending order of powers of $y$:

Difference $= y^4 - 17y^3 + (-12y^2 - 34y^2) + (y + 51y) + (14 - 68)$

Combine the coefficients of like terms:

Difference $= y^4 - 17y^3 + (-12 - 34)y^2 + (1 + 51)y + (14 - 68)$

Difference $= y^4 - 17y^3 - 46y^2 + 52y - 54$

So, $y^4 - 12y^2 + y + 14$ is $y^4 - 17y^3 - 46y^2 + 52y - 54$ greater than $17y^3 + 34y^2 - 51y + 68$.

To find how much $93p^2 - 55p + 4$ exceeds $13p^3 - 5p^2 + 17p - 90$, we need to subtract the second expression from the first expression.

Expression 1: $93p^2 - 55p + 4$

Expression 2: $13p^3 - 5p^2 + 17p - 90$

Difference $= (93p^2 - 55p + 4) - (13p^3 - 5p^2 + 17p - 90)$

Remove parentheses and change signs of the terms being subtracted:

Difference $= 93p^2 - 55p + 4 - 13p^3 + 5p^2 - 17p + 90$

Group like terms:

Difference $= -13p^3 + (93p^2 + 5p^2) + (-55p - 17p) + (4 + 90)$

Combine coefficients of like terms:

Difference $= -13p^3 + (93+5)p^2 + (-55-17)p + (4+90)$

Difference $= -13p^3 + 98p^2 - 72p + 94$

Thus, $93p^2 - 55p + 4$ exceeds $13p^3 - 5p^2 + 17p - 90$ by $98p^2 - 72p + 94 - 13p^3$.

We can also write the result as $-13p^3 + 98p^2 - 72p + 94$.

Let the expression that must be added be $E$.

According to the question, when $E$ is added to $99x^3 - 33x^2 - 13x - 41$, the sum is zero.

So, we have the equation:

$(99x^3 - 33x^2 - 13x - 41) + E = 0$

To find the expression $E$, we subtract $(99x^3 - 33x^2 - 13x - 41)$ from $0$:

$E = 0 - (99x^3 - 33x^2 - 13x - 41)$

$E = - (99x^3 - 33x^2 - 13x - 41)$

Distribute the negative sign to each term inside the parentheses:

$E = -99x^3 - (-33x^2) - (-13x) - (-41)$

$E = -99x^3 + 33x^2 + 13x + 41$

Thus, the expression that must be added is $-99x^3 + 33x^2 + 13x + 41$.

Unity means the number $1$.

To subtract the expression $9a^2 - 15a + 3$ from unity, we write:

Difference $= 1 - (9a^2 - 15a + 3)$

Remove the parentheses by distributing the negative sign to each term inside:

Difference $= 1 - 9a^2 - (-15a) - (+3)$

Difference $= 1 - 9a^2 + 15a - 3$

Combine the constant terms $1$ and $-3$:

Difference $= -9a^2 + 15a + (1 - 3)$

Difference $= -9a^2 + 15a - 2$

Thus, the result of subtracting $9a^2 - 15a + 3$ from unity is $-9a^2 + 15a - 2$.

We are given $a = -2$ and $b = 3$. We need to find the value of each expression by substituting these values.

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

Substitute $a=-2$ and $b=3$:

$(-2)^2 + 2(-2)(3) + (3)^2$

$= 4 + 2(-6) + 9$

$= 4 - 12 + 9$

$= -8 + 9$

$= 1$

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

Substitute $a=-2$ and $b=3$:

$(-2)^2 - 2(-2)(3) + (3)^2$

$= 4 - 2(-6) + 9$

$= 4 + 12 + 9$

$= 16 + 9$

$= 25$

(c) $a^3 + 3a^2b + 3ab^2 + b^3$

This is the expansion of $(a+b)^3$.

Substitute $a=-2$ and $b=3$:

$(-2+3)^3 = (1)^3 = 1$

Alternatively, substitute directly:

$(-2)^3 + 3(-2)^2(3) + 3(-2)(3)^2 + (3)^3$

$= -8 + 3(4)(3) + 3(-2)(9) + 27$

$= -8 + 36 - 54 + 27$

$= 63 - 62 = 1$

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

This is the expansion of $(a-b)^3$.

Substitute $a=-2$ and $b=3$:

$(-2-3)^3 = (-5)^3 = -125$

Alternatively, substitute directly:

$(-2)^3 - 3(-2)^2(3) + 3(-2)(3)^2 - (3)^3$

$= -8 - 3(4)(3) + 3(-2)(9) - 27$

$= -8 - 36 - 54 - 27$

$= -125$

(e) $\frac{a^2 + b^2}{3}$

Substitute $a=-2$ and $b=3$:

$\frac{(-2)^2 + (3)^2}{3} = \frac{4 + 9}{3} = \frac{13}{3}$

(f) $\frac{a^2 - b^2}{3}$

Substitute $a=-2$ and $b=3$:

$\frac{(-2)^2 - (3)^2}{3} = \frac{4 - 9}{3} = \frac{-5}{3}$

(g) $\frac{a}{b} + \frac{b}{a}$

Substitute $a=-2$ and $b=3$:

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

Find a common denominator:

$-\frac{2}{3} - \frac{3}{2} = -\frac{2 \times 2}{3 \times 2} - \frac{3 \times 3}{2 \times 3} = -\frac{4}{6} - \frac{9}{6} = \frac{-4-9}{6} = \frac{-13}{6}$

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

First, simplify the expression:

$a^2 + b^2 - ab - b^2 - a^2 = (a^2 - a^2) + (b^2 - b^2) - ab $$ = 0 + 0 - ab = -ab$

Now, substitute $a=-2$ and $b=3$ into the simplified expression:

$-(-2)(3) = -(-6) = 6$

We are given $m = 1$, $n = -1$, and $p = 2$. We will substitute these values into each expression.

(a) $m + n + p$

Substitute $m=1$, $n=-1$, $p=2$:

$1 + (-1) + 2$

$= 1 - 1 + 2$

$= 0 + 2$

$= 2$

(b) $m^2 + n^2 + p^2$

Substitute $m=1$, $n=-1$, $p=2$:

$(1)^2 + (-1)^2 + (2)^2$

$= 1 + 1 + 4$

$= 2 + 4$

$= 6$

(c) $m^3 + n^3 + p^3$

Substitute $m=1$, $n=-1$, $p=2$:

$(1)^3 + (-1)^3 + (2)^3$

$= 1 + (-1) + 8$

$= 1 - 1 + 8$

$= 0 + 8$

$= 8$

(d) $mn + np + pm$

Substitute $m=1$, $n=-1$, $p=2$:

$(1)(-1) + (-1)(2) + (2)(1)$

$= -1 + (-2) + 2$

$= -1 - 2 + 2$

$= -3 + 2$

$= -1$

(e) $m^3 + n^3 + p^3 - 3mnp$

From part (c), $m^3 + n^3 + p^3 = 8$.

Calculate $3mnp$:

$3(1)(-1)(2)$

$= 3(-2)$

$= -6$

Now substitute these values into the expression:

$(m^3 + n^3 + p^3) - 3mnp = 8 - (-6)$

$= 8 + 6$

$= 14$

Alternatively, substitute directly into the expression:

$(1)^3 + (-1)^3 + (2)^3 - 3(1)(-1)(2)$

$= 1 + (-1) + 8 - 3(-2)$

$= 1 - 1 + 8 + 6$

$= 0 + 8 + 6$

$= 14$

(f) $m^2n^2 + n^2p^2 + p^2m^2$

Substitute $m=1$, $n=-1$, $p=2$:

$(1)^2(-1)^2 + (-1)^2(2)^2 + (2)^2(1)^2$

$= (1)(1) + (1)(4) + (4)(1)$

$= 1 + 4 + 4$

$= 9$

Given the expressions:

$A = 3x^2 - 4x + 1$

$B = 5x^2 + 3x - 8$

$C = 4x^2 - 7x + 3$

(i) Find $(A + B) - C$

First, calculate $A + B$:

$A + B = (3x^2 - 4x + 1) + (5x^2 + 3x - 8)$

$A + B = 3x^2 + 5x^2 - 4x + 3x + 1 - 8$

$A + B = (3+5)x^2 + (-4+3)x + (1-8)$

$A + B = 8x^2 - x - 7$

Now, subtract $C$ from $A + B$:

$(A + B) - C = (8x^2 - x - 7) - (4x^2 - 7x + 3)$

$(A + B) - C = 8x^2 - x - 7 - 4x^2 + 7x - 3$

$(A + B) - C = 8x^2 - 4x^2 - x + 7x - 7 - 3$

$(A + B) - C = (8-4)x^2 + (-1+7)x + (-7-3)$

$(A + B) - C = 4x^2 + 6x - 10$

(ii) Find $B + C - A$

First, calculate $B + C$:

$B + C = (5x^2 + 3x - 8) + (4x^2 - 7x + 3)$

$B + C = 5x^2 + 4x^2 + 3x - 7x - 8 + 3$

$B + C = (5+4)x^2 + (3-7)x + (-8+3)$

$B + C = 9x^2 - 4x - 5$

Now, subtract $A$ from $B + C$:

$(B + C) - A = (9x^2 - 4x - 5) - (3x^2 - 4x + 1)$

$(B + C) - A = 9x^2 - 4x - 5 - 3x^2 + 4x - 1$

$(B + C) - A = 9x^2 - 3x^2 - 4x + 4x - 5 - 1$

$(B + C) - A = (9-3)x^2 + (-4+4)x + (-5-1)$

$(B + C) - A = 6x^2 + 0x - 6$

$(B + C) - A = 6x^2 - 6$

(iii) Find $A + B + C$

Add the three expressions:

$A + B + C = (3x^2 - 4x + 1) + (5x^2 + 3x - 8) + (4x^2 - 7x + 3)$

$A + B + C = 3x^2 + 5x^2 + 4x^2 - 4x + 3x - 7x + 1 - 8 + 3$

$A + B + C = (3+5+4)x^2 + (-4+3-7)x + (1-8+3)$

$A + B + C = (8+4)x^2 + (-1-7)x + (-7+3)$

$A + B + C = 12x^2 - 8x - 4$

Given the expressions:

$P = -(x - 2)$

$Q = -2(y + 1)$

$R = -x + 2y$

Simplify expressions P and Q:

$P = -x + 2$

$Q = -2y - 2$

$R = -x + 2y$

Calculate the sum $P + Q + R$:

$P + Q + R = (-x + 2) + (-2y - 2) + (-x + 2y)$

Group like terms:

$P + Q + R = (-x - x) + (-2y + 2y) + (2 - 2)$

Combine coefficients:

$P + Q + R = (-1 - 1)x + (-2 + 2)y + (2 - 2)$

$P + Q + R = -2x + 0y + 0$

$P + Q + R = -2x$

We are given that $P + Q + R = ax$.

Substitute the calculated sum into the given equation:

$-2x = ax$

To find the value of $a$, compare the coefficients of $x$ on both sides of the equation.

The coefficient of $x$ on the left side is $-2$.

The coefficient of $x$ on the right side is $a$.

Equating the coefficients:

Thus, the value of $a$ is $-2$.

First, find the sum of the three expressions: $x^2 - y^2 - 1$, $y^2 - x^2 - 1$, and $1 - x^2 - y^2$.

Sum $= (x^2 - y^2 - 1) + (y^2 - x^2 - 1) + (1 - x^2 - y^2)$

Sum $= x^2 - y^2 - 1 + y^2 - x^2 - 1 + 1 - x^2 - y^2$

Group the like terms:

Sum $= (x^2 - x^2 - x^2) + (-y^2 + y^2 - y^2) + (-1 - 1 + 1)$

Combine the coefficients:

Sum $= (1 - 1 - 1)x^2 + (-1 + 1 - 1)y^2 + (-1 - 1 + 1)$

Sum $= (-1)x^2 + (-1)y^2 + (-1)$

Sum $= -x^2 - y^2 - 1$

Next, identify the expression to be subtracted: $-(1 + y^2)$.

Simplify the expression to be subtracted:

$-(1 + y^2) = -1 - y^2$

Now, subtract the simplified expression from the sum obtained above.

Result $= (\text{Sum}) - (-1 - y^2)$

Result $= (-x^2 - y^2 - 1) - (-1 - y^2)$

Remove the parentheses and change the sign of each term being subtracted:

Result $= -x^2 - y^2 - 1 + 1 + y^2$

Group the like terms:

Result $= -x^2 + (-y^2 + y^2) + (-1 + 1)$

Combine the like terms:

Result $= -x^2 + 0 + 0$

Result $= -x^2$

The final result is $-x^2$.

First, find the sum of the expressions $12ab - 10b^2 - 18a^2$ and $9ab + 12b^2 + 14a^2$.

Sum 1 $= (12ab - 10b^2 - 18a^2) + (9ab + 12b^2 + 14a^2)$

Sum 1 $= 12ab + 9ab - 10b^2 + 12b^2 - 18a^2 + 14a^2$

Sum 1 $= (12+9)ab + (-10+12)b^2 + (-18+14)a^2$

Sum 1 $= 21ab + 2b^2 - 4a^2$

Rearranging in terms of descending powers of a (and b):

Sum 1 $= -4a^2 + 2b^2 + 21ab$

Next, find the sum of the expressions $ab + 2b^2$ and $3b^2 - a^2$.

Sum 2 $= (ab + 2b^2) + (3b^2 - a^2)$

Sum 2 $= ab + 2b^2 + 3b^2 - a^2$

Sum 2 $= -a^2 + (2+3)b^2 + ab$

Sum 2 $= -a^2 + 5b^2 + ab$

Now, subtract Sum 1 from Sum 2.

Result $= \text{Sum 2} - \text{Sum 1}$

Result $= (-a^2 + 5b^2 + ab) - (-4a^2 + 2b^2 + 21ab)$

Remove parentheses and change signs of the terms being subtracted:

Result $= -a^2 + 5b^2 + ab + 4a^2 - 2b^2 - 21ab$

Group the like terms:

Result $= (-a^2 + 4a^2) + (5b^2 - 2b^2) + (ab - 21ab)$

Combine the coefficients:

Result $= (-1+4)a^2 + (5-2)b^2 + (1-21)ab$

Result $= 3a^2 + 3b^2 - 20ab$

Given:

According to the corrected values provided:

To Find:

The algebraic expression represented by the combination:

Solution:

First, we find the sum of Triangle and Circle:

$\text{Sum} = (2x^2 + 3y) + (5x^2 + 3x)$

$\begin{array}{cc} & 2x^2 & + & 0x & + & 3y \\ + & 5x^2 & + & 3x & + & 0y \\ \hline & 7x^2 & + & 3x & + & 3y \\ \hline \end{array}$

Now, we subtract Square from this sum:

$\text{Result} = (7x^2 + 3x + 3y) - (8y^2 - 3x^2 + 2x + 3y)$

Subtracting by adding the additive inverse of Square:

$\begin{array}{cc} & 7x^2 & + & 3x & + & 3y & + & 0y^2 \\ - & (-3x^2 & + & 2x & + & 3y & + & 8y^2) \\ \hline & 10x^2 & + & x & + & 0y & - & 8y^2 \\ \hline \end{array}$

Alternate Solution:

We can solve it by grouping like terms directly:

$= (2x^2 + 3y) + (5x^2 + 3x) - (8y^2 - 3x^2 + 2x + 3y)$

$= 2x^2 + 3y + 5x^2 + 3x - 8y^2 + 3x^2 - 2x - 3y$

Grouping the $x^2$ terms, $x$ terms, $y^2$ terms, and $y$ terms:

$= (2x^2 + 5x^2 + 3x^2) + (3x - 2x) + (3y - 3y) - 8y^2$

$= 10x^2 + 1x + 0 - 8y^2$

$= 10x^2 + x - 8y^2$

Therefore, the expression represented by the symbols is $10x^2 + x - 8y^2$.

From the given chart, the amount of carbohydrates per unit (100g) is:

Potatoes: 22g

Rajma: 60g

Tomatoes: 4g

Apples: 14g

(a) y units of potatoes and 2 units of rajma

Amount of carbohydrates in y units of potatoes = $y \times 22\text{g} = 22y\text{g}$

Amount of carbohydrates in 2 units of rajma = $2 \times 60\text{g} = 120\text{g}$

Total amount of carbohydrates = (Carbs from potatoes) + (Carbs from rajma)

Total amount of carbohydrates = $22y + 120\text{g}$

The algebraic expression for the amount of carbohydrates is $\mathbf{22y + 120}$.

(b) 2x units tomatoes and y units apples

Amount of carbohydrates in 2x units of tomatoes = $2x \times 4\text{g} = 8x\text{g}$

Amount of carbohydrates in y units of apples = $y \times 14\text{g} = 14y\text{g}$

Total amount of carbohydrates = (Carbs from tomatoes) + (Carbs from apples)

Total amount of carbohydrates = $8x + 14y\text{g}$

The algebraic expression for the amount of carbohydrates is $\mathbf{8x + 14y}$.

Given:

Length of the rectangular plot = x

Breadth of the rectangular plot = y

Base of the triangular part sold = y

Height of the triangular part sold = z

To Find:

The area of the remaining part of the plot.

Solution:

The area of a rectangle is given by the formula: Area = length $\times$ breadth.

Area of the rectangular plot = $x \times y = xy$

The area of a triangle is given by the formula: Area = $\frac{1}{2} \times$ base $\times$ height.

Area of the triangular part sold = $\frac{1}{2} \times y \times z = \frac{1}{2}yz$

The area of the remaining part of the plot is the area of the rectangular plot minus the area of the triangular part sold.

Area of remaining part = (Area of rectangle) - (Area of triangle)

Area of remaining part = $xy - \frac{1}{2}yz$

The area of the remaining part of the plot is $\mathbf{xy - \frac{1}{2}yz}$.

Given:

Side of the square plot = m

Base of the triangular plot = m

Height of the triangular plot = m

To Find:

The total area of both plots.

Solution:

The area of a square is given by the formula: Area = side $\times$ side.

Area of the square plot = $m \times m = m^2$

The area of a triangle is given by the formula: Area = $\frac{1}{2} \times$ base $\times$ height.

Area of the triangular plot = $\frac{1}{2} \times m \times m = \frac{1}{2}m^2$

The total area of both plots is the sum of the area of the square plot and the area of the triangular plot.

Total Area = (Area of square plot) + (Area of triangular plot)

Total Area = $m^2 + \frac{1}{2}m^2$

Total Area = $(1 + \frac{1}{2})m^2$

Total Area = $(\frac{2}{2} + \frac{1}{2})m^2$

Total Area = $\frac{3}{2}m^2$

The total area of both plots is $\mathbf{\frac{3}{2}m^2}$.

Given:

Charge per km = $\textsf{₹}$ 8

Fixed charge = $\textsf{₹}$ 50

Distance covered = x km

To Find:

An algebraic expression for the total charge if the taxi is hired for x km.

Solution:

The total charge consists of two parts:

1. The fixed charge.

2. The variable charge which depends on the distance covered.

Fixed charge = $\textsf{₹}$ 50

Variable charge = (Charge per km) $\times$ (Number of km)

Variable charge = $\textsf{₹}$ 8 $\times$ x = $\textsf{₹}$ $8x$

Total charge = Fixed charge + Variable charge

Total charge = $\textsf{₹}$ 50 + $\textsf{₹}$ $8x$

Total charge = $\textsf{₹}$ $(50 + 8x)$

The algebraic expression for the total charge in $\textsf{₹}$ is $\mathbf{50 + 8x}$.

Given:

Hourly pay rate = $\textsf{₹}$ 50

Hours worked last week = 7 hours

Hours to work this week = x hours

To Find:

An algebraic expression for the total money paid to him for both the weeks.

Solution:

Money earned last week = Hourly pay $\times$ Hours worked last week

Money earned last week = $\textsf{₹ }50 \times 7$

Money earned last week = $\textsf{₹ }350$

Money earned this week = Hourly pay $\times$ Hours worked this week

Money earned this week = $\textsf{₹ }50 \times x$

Money earned this week = $\textsf{₹ }50x$

Total money paid for both weeks = (Money earned last week) + (Money earned this week)

Total money paid = $\textsf{₹ }350 + \textsf{₹ }50x$

Total money paid = $\textsf{₹ }(350 + 50x)$

The algebraic expression for the money paid to him for both the weeks is $\mathbf{350 + 50x}$.

Given:

Initially, Sonu collects 12 leaves.

Initially, Raj collects x leaves.

Sonu loses 3 leaves.

Raj collects an additional 2x leaves.

To Find:

An algebraic expression for the total number of leaves collected by both Sonu and Raj.

Solution:

First, let's find the number of leaves Sonu has after losing some.

Number of leaves Sonu has = (Initial leaves) - (Leaves lost)

Number of leaves Sonu has = $12 - 3 = 9$

Next, let's find the total number of leaves Raj has.

Number of leaves Raj has = (Initial leaves) + (Additional leaves collected)

Number of leaves Raj has = $x + 2x$

Number of leaves Raj has = $3x$

The total number of leaves collected by both of them is the sum of the leaves Sonu has and the leaves Raj has.

Total number of leaves = (Leaves Sonu has) + (Leaves Raj has)

Total number of leaves = $9 + 3x$

The algebraic expression for the total number of leaves collected by both of them is $\mathbf{9 + 3x}$.

Given:

From the figure and the question, we have:

To Find:

Total perimeter of both the playground and the lawn combined together.

Solution:

When the square lawn and the rectangular playground are combined as shown in the figure, the total perimeter is the sum of the lengths of all the outer boundaries of the combined shape.

The outer boundary consists of the following sides:

1. The top side of the square lawn = $x$

2. The left side of the square lawn = $x$

3. The left side (breadth) of the rectangular playground = $y$

4. The bottom side (length) of the rectangular playground = $x$

5. The right side (breadth) of the rectangular playground = $y$

6. The right side of the square lawn = $x$

Note: The side where the square and rectangle meet is an internal boundary and is not included in the perimeter.

Therefore, the Total Perimeter ($P$) is:

$P = \text{Top} + \text{Left Square} + \text{Left Rect} + \text{Bottom} + \text{Right Rect} $$ + \text{Right Square}$

$P = x + x + y + x + y + x$

Grouping the like terms together:

$P = (x + x + x + x) + (y + y)$

Alternate Method:

The combined figure forms a large rectangle where:

Perimeter of a rectangle is given by the formula:

$\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})$

Substituting the values:

$\text{Perimeter} = 2 \times (x + (x + y))$

$\text{Perimeter} = 2 \times (2x + y)$

Thus, the total perimeter of the combined figure is $4x + 2y$.

Given:

Rate of planting grass = $\textsf{₹}$ x per square metre

Base of the triangular lawn = y metres

Height of the triangular lawn = z metres

To Find:

The cost of planting the grass on the triangular lawn.

Solution:

First, we need to find the area of the triangular lawn.

The area of a triangle is given by the formula: Area = $\frac{1}{2} \times$ base $\times$ height.

Area of the triangular lawn = $\frac{1}{2} \times y \times z$ square metres

Area of the triangular lawn = $\frac{1}{2}yz$ square metres

Next, we need to find the total cost of planting the grass.

Cost of planting grass = Area $\times$ Rate per square metre

Cost of planting grass = $(\frac{1}{2}yz) \times x$

Cost of planting grass = $\frac{1}{2}xyz$

The cost is in Rupees ($\textsf{₹}$).

The cost of planting the grass on the triangular lawn is $\mathbf{\textsf{₹} \frac{1}{2}xyz}$.

Given:

From the given figure of the rectangle $ABCD$:

To Find:

Perimeter of the rectangle $ABCD$.

Solution:

The formula for the Perimeter of a rectangle is given by:

$\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})$

Substituting the given values in the formula:

$\text{Perimeter} = 2 \times [(5x - y) + 2(x + y)]$

First, let us simplify the expression inside the brackets:

$= 2 \times [5x - y + 2x + 2y]$

Grouping the like terms together:

$= 2 \times [(5x + 2x) + (2y - y)]$

$= 2 \times [7x + y]$

Now, multiplying by $2$:

$= 14x + 2y$

Thus, the perimeter of the given figure is $14x + 2y$ units.

Given:

Number of square flower beds = $5$

Side of each square flower bed ($s$) = $(x + 2)$ metres

Rate of fencing = $\textsf{₹} 50$ per $100$ metres

To Find:

Total cost of fencing the $5$ flower beds.

Solution:

First, we find the perimeter of one square flower bed:

$\text{Perimeter of 1 bed} = 4 \times \text{side}$

$\text{Perimeter of 1 bed} = 4 \times (x + 2)$

$\text{Perimeter of 1 bed} = 4x + 8$ metres

Now, we find the total length of fencing required for $5$ such beds:

$\text{Total Perimeter} = 5 \times (4x + 8)$

$\text{Total Perimeter} = 20x + 40$ metres

Calculation of Cost:

Given that the cost of fencing $100$ metres is $\textsf{₹} 50$.

Therefore, the cost of fencing $1$ metre is:

$\text{Cost per metre} = \textsf{₹} \frac{\cancel{50}^{1}}{\cancel{100}_{2}} = \textsf{₹} \frac{1}{2}$

Total Cost = $\text{Total Perimeter} \times \text{Cost per metre}$

$\text{Total Cost} = (20x + 40) \times \frac{1}{2}$

$\text{Total Cost} = \frac{20x}{2} + \frac{40}{2}$

$\text{Total Cost} = 10x + 20$

The total cost of fencing the $5$ flower beds is $\textsf{₹} (10x + 20)$.

Given:

Initial length of the wire = $(7x - 3)$ metres

Length of wire cut = $(3x - 4)$ metres

(a) How much wire is left?

The length of the wire left is the initial length minus the length cut for use.

Length of wire left = (Initial length) - (Length cut)

Length of wire left = $(7x - 3) - (3x - 4)$

Remove the parentheses. Remember to distribute the negative sign to both terms inside the second parenthesis:

Length of wire left = $7x - 3 - 3x + 4$

Combine the like terms:

Length of wire left = $(7x - 3x) + (-3 + 4)$

Length of wire left = $(7 - 3)x + (4 - 3)$

Length of wire left = $4x + 1$

The length of wire left is $\mathbf{(4x + 1)}$ metres.

(b) What is the length of each side of the triangle so formed?

The wire left is used to make an equilateral triangle. An equilateral triangle has three sides of equal length.

The perimeter of the equilateral triangle is equal to the length of the wire left.

Perimeter of equilateral triangle = $(4x + 1)$ metres

Let the length of each side of the equilateral triangle be 's'.

Perimeter of equilateral triangle = $3 \times s$

So, $3s = 4x + 1$

To find the length of each side, divide the perimeter by 3:

$s = \frac{4x + 1}{3}$

$s = \frac{4x}{3} + \frac{1}{3}$

The length of each side of the equilateral triangle is $\mathbf{\frac{4x + 1}{3}}$ metres or $\mathbf{(\frac{4}{3}x + \frac{1}{3})}$ metres.

Given:

Money given by mother = $\textsf{₹ }3xy^2$

Money given by father = $\textsf{₹ }5(xy^2 + 2)$

Money spent on birthday party = $\textsf{₹ }(10 - 3xy^2)$

To Find:

The amount of money left with Rohan.

Solution:

First, calculate the money given by the father:

Money from father = $5(xy^2 + 2) = 5 \times xy^2 + 5 \times 2 = 5xy^2 + 10$

So, money from father is $\textsf{₹ }(5xy^2 + 10)$.

Next, calculate the total money Rohan has before spending:

Total money = Money from mother + Money from father

Total money = $\textsf{₹ }3xy^2 + \textsf{₹ }(5xy^2 + 10)$

Total money = $\textsf{₹ }(3xy^2 + 5xy^2 + 10)$

Total money = $\textsf{₹ }(8xy^2 + 10)$

Now, calculate the money left with Rohan by subtracting the money spent from the total money:

Money left = Total money - Money spent

Money left = $\textsf{₹ }(8xy^2 + 10) - \textsf{₹ }(10 - 3xy^2)$

Money left = $8xy^2 + 10 - (10 - 3xy^2)$

Remove the parentheses, distributing the negative sign:

Money left = $8xy^2 + 10 - 10 + 3xy^2$

Combine the like terms:

Money left = $(8xy^2 + 3xy^2) + (10 - 10)$

Money left = $(8 + 3)xy^2 + 0$

Money left = $11xy^2$

The money left with Rohan is $\mathbf{\textsf{₹ }11xy^2}$.

(i) Solution for Triangle Pattern:

Given:

Length of one red stick = $r$

Length of one blue stick = $b$

In the given pattern, there are 6 upright triangles and 3 inverted triangles.

Total number of small triangles in the pattern = $6 + 3 = 9$

To Find:

Total length of sticks in the entire pattern.

Solution:

Each of the 9 triangles is formed using 2 red sticks as its slanting sides.

Total number of red sticks = $9 \times 2 = 18$

The blue sticks form the horizontal bases of the upright triangles. In the triangular arrangement:

Row 1 has 1 blue stick.

Row 2 has 2 blue sticks.

Row 3 has 3 blue sticks.

Total number of blue sticks = $1 + 2 + 3 = 6$

The expression for the total length of sticks is:

$\text{Total Length} = (\text{Number of red sticks} \times r) $$ + (\text{Number of blue sticks} \times b)$

$\text{Total Length} = 18r + 6b$

(ii) Solution for Joined Shapes Pattern:

Given:

Length of a green side = $g$

Length of a red side = $p$

The basic shape (kite-like) consists of 2 green sides and 2 red sides.

Total length of one shape = $2g + 2p$

To Find:

1. Expression for the pattern shown (consisting of 3 shapes).

2. Expression if 100 such shapes are joined together.

Solution:

The pattern shown in the full sequence consists of 3 such shapes.

Expression for 3 shapes = $3 \times (2g + 2p)$

Expression for 3 shapes = $6g + 6p$

Now, if 100 such shapes are joined together:

Expression for 100 shapes = $100 \times (2g + 2p)$

Expression for 100 shapes = $200g + 200p$

Taking 200 as common:

Expression for 100 shapes = $200(g + p)$

Therefore, the total length for the triangle pattern is $18r + 6b$. For the second part, the expression for the pattern is $6g + 6p$ and for 100 shapes, it is $200(g + p)$.

Given:

The sum of the first n natural numbers is given by the formula: $S_n = \frac{1}{2}n^2 + \frac{1}{2}n$

(i) The sum of first 5 natural numbers.

Here, we need to find the sum of the first 5 natural numbers, so n = 5.

Substitute $n=5$ into the formula:

$S_5 = \frac{1}{2}(5)^2 + \frac{1}{2}(5)$

$S_5 = \frac{1}{2}(25) + \frac{1}{2}(5)$

$S_5 = \frac{25}{2} + \frac{5}{2}$

$S_5 = \frac{25 + 5}{2}$

$S_5 = \frac{30}{2}$

$S_5 = 15$

The sum of the first 5 natural numbers is $\mathbf{15}$.

(ii) The sum of first 11 natural numbers.

Here, we need to find the sum of the first 11 natural numbers, so n = 11.

Substitute $n=11$ into the formula:

$S_{11} = \frac{1}{2}(11)^2 + \frac{1}{2}(11)$

$S_{11} = \frac{1}{2}(121) + \frac{1}{2}(11)$

$S_{11} = \frac{121}{2} + \frac{11}{2}$

$S_{11} = \frac{121 + 11}{2}$

$S_{11} = \frac{132}{2}$

$S_{11} = 66$

The sum of the first 11 natural numbers is $\mathbf{66}$.

(iii) The sum of natural numbers from 11 to 30.

To find the sum of natural numbers from 11 to 30, we subtract the sum of the first 10 natural numbers from the sum of the first 30 natural numbers.

Sum (11 to 30) = Sum (1 to 30) - Sum (1 to 10)

First, find the sum of the first 30 natural numbers (n = 30):

$S_{30} = \frac{1}{2}(30)^2 + \frac{1}{2}(30)$

$S_{30} = \frac{1}{2}(900) + \frac{1}{2}(30)$

$S_{30} = 450 + 15$

$S_{30} = 465$

Next, find the sum of the first 10 natural numbers (n = 10):

$S_{10} = \frac{1}{2}(10)^2 + \frac{1}{2}(10)$

$S_{10} = \frac{1}{2}(100) + \frac{1}{2}(10)$

$S_{10} = 50 + 5$

$S_{10} = 55$

Now, subtract the sum of the first 10 numbers from the sum of the first 30 numbers:

Sum (11 to 30) = $S_{30} - S_{10}$

Sum (11 to 30) = $465 - 55$

Sum (11 to 30) = $410$

The sum of natural numbers from 11 to 30 is $\mathbf{410}$.

Given:

The sum of squares of the first n natural numbers is given by the formula:

$S_n = \frac{1}{6}n(n + 1)(2n + 1)$ or $S_n = \frac{1}{6}(2n^3 + 3n^2 + n)$

To Find:

The sum of squares of the first 10 natural numbers.

Solution:

We need to find the sum of squares of the first 10 natural numbers, so we set $n = 10$ in the formula $S_n = \frac{1}{6}n(n + 1)(2n + 1)$.

Substitute $n = 10$ into the formula:

$S_{10} = \frac{1}{6}(10)(10 + 1)(2 \times 10 + 1)$

$S_{10} = \frac{1}{6}(10)(11)(20 + 1)$

$S_{10} = \frac{1}{6}(10)(11)(21)$

$S_{10} = \frac{10 \times 11 \times 21}{6}$

Perform the calculation by cancelling common factors:

$S_{10} = \frac{\cancel{10}^{5} \times 11 \times \cancel{21}^{7}}{\cancel{6}_{1}}$

$S_{10} = 5 \times 11 \times 7$

$S_{10} = 55 \times 7$

$S_{10} = 385$

The sum of squares of the first 10 natural numbers is $\mathbf{385}$.

Given:

The sum of the multiplication table of a natural number 'n' is given by the formula: Sum $= 55 \times n$

To Find:

The sum of the multiplication table of 7, 10, and 19.

Solution:

We use the given formula $Sum = 55 \times n$ for each part.

(a) Table of 7

Here, $n = 7$.

Sum of table of 7 $= 55 \times 7$

Sum $= 385$

The sum of the table of 7 is $\mathbf{385}$.

(b) Table of 10

Here, $n = 10$.

Sum of table of 10 $= 55 \times 10$

Sum $= 550$

The sum of the table of 10 is $\mathbf{550}$.

(c) Table of 19

Here, $n = 19$.

Sum of table of 19 $= 55 \times 19$

Sum $= 55 \times (20 - 1)$

Sum $= 55 \times 20 - 55 \times 1$

Sum $= 1100 - 55$

Sum $= 1045$

The sum of the table of 19 is $\mathbf{1045}$.

Given:

The rules for the symbols are provided as follows:

To Find:

(i) The value of $2 \times \text{Triangle}(6) + \text{Square}(3) - \text{Circle}(1)$

(ii) The value of $\frac{1}{2} \times \text{Square}(2) + \text{Circle}(8) - 3 \times \text{Triangle}(0)$

Solution for (i):

First, we evaluate each symbol individually based on the given rules:

Now, substitute these values back into the expression:

$= 2(15) + (11.5) - (-2)$

$= 30 + 11.5 + 2$

$= 43.5$

Solution for (ii):

Again, we evaluate each symbol individually:

Now, substitute these values into the required expression:

$= \frac{1}{2}(10) + (5) - 3(3)$

$= 5 + 5 - 9$

$= 10 - 9$

$= 1$

Final Answers:

The values are $43.5$ for (i) and $1$ for (ii).

Given:

The algebraic operations for the given symbols are defined as follows:

To Find:

(i) The value of $\text{Triangle}(10) - \text{Diamond}(4)$

(ii) The value of $2 \times \text{Diamond}(12) - \frac{3}{2} \times \text{Triangle}(1)$

Solution for (i):

First, we evaluate each symbol by substituting the given values into their respective expressions:

Step 1: Evaluate Triangle(10)

$\text{Triangle}(10) = \frac{3 \times \cancel{10}^5}{\cancel{4}_2} - 2$

$\text{Triangle}(10) = \frac{15}{2} - 2$

$\text{Triangle}(10) = 7.5 - 2 = 5.5$

Step 2: Evaluate Diamond(4)

Step 3: Calculate the final result for (i)

$\text{Value} = \text{Triangle}(10) - \text{Diamond}(4)$

$\text{Value} = 5.5 - 10$

$\text{Value} = -4.5$

Solution for (ii):

Now, we evaluate the expression $2 \times \text{Diamond}(12) - \frac{3}{2} \times \text{Triangle}(1)$:

Step 1: Evaluate Diamond(12)

Step 2: Evaluate Triangle(1)

$\text{Triangle}(1) = \frac{3}{4} - \frac{2}{1} = \frac{3 - 8}{4} = -\frac{5}{4}$

Step 3: Calculate the final result for (ii)

$\text{Value} = 2 \times (18) - \frac{3}{2} \times \left(-\frac{5}{4}\right)$

$\text{Value} = 36 + \left(\frac{3 \times 5}{2 \times 4}\right)$

$\text{Value} = 36 + \frac{15}{8}$

$\text{Value} = 36 + 1.875$

$\text{Value} = 37.875$

Final Answers:

The value for expression (i) is $-4.5$.

The value for expression (ii) is $37.875$.

The expression is $4b - 3$.

This expression involves multiplication (4 multiplied by b) and subtraction (3 is subtracted from the result).

We can translate this into words in several ways:

1. Three subtracted from four times b.

2. The product of 4 and b, reduced by 3.

3. 3 less than the product of 4 and b.

A common translation is: Three subtracted from four times b.

The expression is $8(m + 5)$.

This expression involves the sum of m and 5, and then the result is multiplied by 8.

We can translate this into words as:

1. 8 times the sum of m and 5.

2. The product of 8 and the sum of m and 5.

3. 8 multiplied by the quantity m plus 5.

A common and clear translation is: 8 times the sum of m and 5.

The expression is $\frac{7}{8 - x}$.

This expression is a fraction, which represents division. The numerator is 7 and the denominator is $(8 - x)$. The denominator involves subtraction.

First, translate the denominator $(8 - x)$: This can be expressed as "the difference between 8 and x", or "x subtracted from 8".

Now, translate the entire expression (7 divided by the denominator):

Using "the difference between 8 and x" for the denominator, the expression can be translated as:

7 divided by the difference between 8 and x.

Alternatively, we could say:

The quotient of 7 and the difference between 8 and x.

The expression is $7 \left( \frac{16}{w} \right)$.

This expression involves division (16 divided by w) inside parentheses, and then the result is multiplied by 7.

First, consider the part inside the parentheses: $\frac{16}{w}$.

This can be translated as "16 divided by w" or "the quotient of 16 and w".

Now, the entire expression means 7 multiplied by the result of the division.

Using "16 divided by w" for the inner part, the expression can be translated as:

7 times 16 divided by w.

Alternatively, using "the quotient of 16 and w":

7 times the quotient of 16 and w.

Or:

The product of 7 and the quotient of 16 and w.

(i) Solution:

To Find:

Two different algebraic expressions for the phrase "$\frac{1}{4}$ of the sum of $x$ and $7$".

Solution:

The phrase "sum of $x$ and $7$" is written as $(x + 7)$.

Taking "$\frac{1}{4}$ of" this sum can be represented in the following two ways:

1. Using multiplication: $\frac{1}{4}(x + 7)$

2. Using division: $\frac{x + 7}{4}$

Both expressions are mathematically identical and represent the given word phrase correctly.

(ii) Solution:

To Find:

The error made by the student in the expression $\frac{n}{3} - 5$ for the phrase "5 less than a number $n$ divided by 3".

Solution:

The phrase "5 less than a number $n$ divided by 3" is ambiguous and can be interpreted in two ways. However, in a mathematical context, it usually implies that the subtraction happens before the division or that the division applies to the entire quantity described first.

If the phrase intended the division to apply to the result of "5 less than $n$", the correct expression should be:

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

The Error:

The student wrote $\frac{n}{3} - 5$. This expression represents "the quotient of $n$ and $3$, decreased by $5$".

The student likely confused the order of operations. They divided the number $n$ by $3$ first and then subtracted $5$, whereas the wording "5 less than a number $n$ (pause) divided by 3" typically suggests that the quantity $(n - 5)$ is the one being divided by $3$.

(iii) Solution:

Given:

Weekly cost of toll tax = $\textsf{₹} 15$ per day.

Shashi used Addition.

Ravi used Multiplication.

Solution:

A week has 7 days. To find the total cost for the week:

Shashi's Method (Repeated Addition):

Shashi added the daily cost seven times:

$\text{Total Cost} = \textsf{₹} (15 + 15 + 15 + 15 + 15 + 15 + 15)$

$\text{Total Cost} = \textsf{₹} 105$

Ravi's Method (Multiplication):

Ravi multiplied the daily cost by the number of days:

$\text{Total Cost} = \textsf{₹} (15 \times 7)$

$\text{Total Cost} = \textsf{₹} 105$

Why they both got the correct answer:

This is possible because multiplication is defined as repeated addition. Multiplying a number by 7 is the same as adding that number to itself 7 times. Since both operations are mathematically equivalent, they both arrived at the same correct result of $\textsf{₹} 105$.

To Find:

An algebraic expression for the sum of 1 and twice a number $n$ and determine if the result is always odd when $n$ is an odd number.

Solution:

According to the statement, "twice a number $n$" is represented as $2n$.

The sum of 1 and $2n$ is written as:

Now, let us test if the result is always odd for any odd value of $n$.

Let $n$ be an odd number. Any odd number can be represented in the form $2k + 1$, where $k$ is an integer.

Substituting $n = 2k + 1$ in our expression:

$= 1 + 2(2k + 1)$

$= 1 + 4k + 2$

$= 4k + 3$

$= 4k + 2 + 1$

$= 2(2k + 1) + 1$

Since any number in the form $2(\text{integer}) + 1$ is always odd, the result will always be an odd number.

Alternate Explanation:

1. For any integer $n$, $2n$ is always an even number (since it is a multiple of 2).

2. We know that $\text{Even} + 1 = \text{Odd}$.

3. Therefore, $2n + 1$ will always be an odd number regardless of whether $n$ is odd or even.

Thus, the result is Yes, it will always be an odd number.

Given:

Expression = $11x$

Value of $x = -5$

Solution:

Let us find the value of the expression by substituting $x = -5$:

$= 11 \times (-5)$

$= -55$

Comparison:

We need to compare $-55$ and $11$.

On a number line, any negative number lies to the left of a positive number. Therefore, every negative number is smaller than every positive number.

Since $-55 < 11$, the value of $11x$ for $x = -5$ is less than 11.

Solution:

By converting the word phrases into algebraic expressions, we can match the columns as follows:

Given:

Age at 2 years = $24$ human years.

Rate of increase after 2 years = $4$ human years per actual year.

To Find:

1. The missing value in the expression $[24 + \_\_\_ (a - 2)]$.

2. The meaning of the variable '$a$'.

3. Completed table for ages 2 to 6.

Solution:

Since each year after age 2 adds 4 human years, the number of additional years must be multiplied by 4. Thus, the filled expression is $[24 + 4(a - 2)]$.

In this expression, '$a$' stands for the actual age of the cat or dog in years.

The term $(a - 2)$ represents the number of years passed after the initial 2 years.

Completed Table:

Solution:

Using the variables $x, y$ and $z$, the properties are expressed as follows:

(i) Commutative property of addition:

(ii) Commutative property of multiplication:

(iii) Associative property of addition:

(iv) Associative property of multiplication:

(v) Distributive property of multiplication over addition: