Class 8 Mathematics Chapter 2 Linear Equations in One Variable

Key Concept: Linear Expressions

b) $4x + 5$

[Solution Description]

A linear expression in one variable has the highest power of the variable equal to 1. Out of the given options, $4x + 5$ is a linear expression because the highest power of $x$ is 1.

Key Concept: Linear Equations in One Variable, Algebraic Expressions

a) $\frac{58}{19}$

[Solution Description]

First, find a common denominator for the fractions, which is 6. Multiply each term by 6 to eliminate the denominators:

$6 \cdot \left( \frac{3x – 4}{2} \right) + 6 \cdot \left( \frac{5x + 7}{3} \right) = 6 \cdot 10$

Simplify each term:

$3(3x – 4) + 2(5x + 7) = 60$

Distribute the constants:

$9x – 12 + 10x + 14 = 60$

Combine like terms:

$19x + 2 = 60$

Subtract 2 from both sides:

$19x = 58$

Divide both sides by 19:

$x = \frac{58}{19}$

Therefore, the solution is $x = \frac{58}{19}$.

Key Concept: Equations vs. Expressions

b) $2x + 3 = 7$

[Solution Description]

An equation is a statement that asserts the equality of two expressions, and it must contain an equality sign ($=$). Among the options, $2x + 3 = 7$ is an equation because it contains the equality sign and sets two expressions equal to each other.

Key Concept: Linear Equations in One Variable, Algebraic Expressions

a) -3

[Solution Description]

Find a common denominator for the fractions, which is 20. Multiply each term by 20 to eliminate the denominators:

$20 \cdot \left( \frac{y + 3}{4} \right) – 20 \cdot \left( \frac{y – 2}{5} \right) = 20 \cdot 1$

Simplify each term:

$5(y + 3) – 4(y – 2) = 20$

Distribute the constants:

$5y + 15 – 4y + 8 = 20$

Combine like terms:

$y + 23 = 20$

Subtract 23 from both sides:

$y = -3$

Therefore, the solution is $y = -3$.

Key Concept: Linear Equations in One Variable

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The equation $2x – 3 = 7$ is a linear equation in one variable because it contains only one variable, $x$, and the highest power of $x$ is 1. This satisfies the definition of a linear equation in one variable. Therefore, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Linear Equations in One Variable, Algebraic Expressions

b) $9$

[Solution Description]

To solve for $y$, first subtract $2$ from both sides:

$\frac{y}{3} + 2 – 2 = 5 – 2$

Simplifying gives:

$\frac{y}{3} = 3$

Next, multiply both sides by $3$:

$\frac{y}{3} \times 3 = 3 \times 3$

This results in:

$y = 9$

Therefore, the solution is $y = 9$.

Key Concept: Linear Equations in One Variable, Algebraic Expressions

b) $-6$

[Solution Description]

To solve for $x$, first subtract $2x$ from both sides:

$3x – 2x + 5 = 2x – 2x – 1$

Simplifying gives:

$x + 5 = -1$

Next, subtract $5$ from both sides:

$x + 5 – 5 = -1 – 5$

This results in:

$x = -6$

Therefore, the solution is $x = -6$.

Key Concept: Equations vs Expressions

b) $2x – 3 = 9$

[Solution Description]

An equation always contains an equality sign ($=$). Let’s analyze each option:

Option 1: $5x$ is an expression because it does not have an equality sign.

Option 2: $2x – 3 = 9$ is an equation because it has an equality sign.

Option 3: $\frac{y}{2} + \frac{z}{2}$ is an expression because it does not have an equality sign.

Option 4: $x^2 + 1$ is an expression because it does not have an equality sign.

Therefore, the correct answer is $2x – 3 = 9$.

Key Concept: Definition of Algebraic Expression

a) $3x + 5$

[Solution Description]

An algebraic expression is a combination of variables and constants using mathematical operations like addition, subtraction, multiplication, and division. It does not contain an equality (\$) sign. From the given options, $3x + 5$ is an algebraic expression as it consists of a variable $x$, a constant $5$, and an addition operation. The other options either have an equality sign or are not expressions.

Key Concept: Linear Equations in One Variable

b) 5

[Solution Description]

To solve the equation $3x + 5 = 20$, follow these steps:

Step 1: Subtract 5 from both sides of the equation.

$3x + 5 – 5 = 20 – 5$

This simplifies to:

$3x = 15$

Step 2: Divide both sides by 3.

$x = \frac{15}{3}$

Therefore, $x = 5$.

Key Concept: Assertion-Reason

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that the expression $3x + 2y$ is a linear equation in one variable. However, this is incorrect because $3x + 2y$ contains two variables, $x$ and $y$. A linear equation in one variable must have only one variable with its highest power as 1. The reason states that a linear equation in one variable has the highest power of the variable as 1, which is correct. Therefore, the assertion is false, but the reason is true.

Key Concept: Linear Expressions, Equations

c) Only $4y – 7$ can be used.

[Solution Description]

A linear equation in one variable is formed by setting a linear expression (where the highest power of the variable is 1) equal to another expression or number. $4y – 7$ is a linear expression because the highest power of $y$ is 1. However, $y^2 + 3$ is not linear because the highest power of $y$ is 2. Therefore, only $4y – 7$ can be used to form a linear equation in one variable.

Key Concept: Difference between expressions and equations

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that an equation always has an equality sign, which is true because an equation is defined as a statement that asserts the equality of two expressions. The reason states that an expression can have variables but does not include an equality sign, which is also true because an expression is a combination of numbers, variables, and operators without any equality. However, the reason does not explain why an equation always has an equality sign; it simply describes a characteristic of expressions. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Expressions vs Equations, Linear Equations

d) An equation must include an equality sign, while an expression does not.

[Solution Description]

An expression is a combination of numbers, variables, and operations without an equality sign, while an equation includes an equality sign and sets two expressions equal to each other. Here, $3x + 5$ is an expression because it does not have an equality sign. $3x + 5 = 20$ is an equation because it has an equality sign and sets $3x + 5$ equal to $20$.

Key Concept: Linear expressions

a) $5x + 3$

[Solution Description] A linear expression in one variable has the highest power of the variable equal to 1. Option (a) is a linear expression because the highest power of \$x\$ is 1. Options (b), (c), and (d) are not linear because they either have more than one variable or the highest power of the variable is greater than 1.

Key Concept: Equations vs Expressions

b) $2x – 3 = 9$

[Solution Description]

An equation must have an equality sign ($=$). Among the given options, $2x – 3 = 9$ is an equation because it contains an equality sign.

Key Concept: Variables and Constants

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that $3x + 5$ is a linear expression in one variable. This is true because it only contains one variable, $x$, and the highest power of $x$ is 1. The reason given is that a linear expression in one variable has the highest power of the variable equal to 1. This is also true and correctly explains why $3x + 5$ is a linear expression in one variable.

Key Concept: Equations, solving for variables

c) 20

[Solution Description]

Start with the equation:

$\frac{y}{2} – 3 = 7$

Add 3 to both sides to isolate the term with $y$:

$\frac{y}{2} = 10$

Multiply both sides by 2 to solve for $y$:

$y = 20$

Therefore, the solution is $y = 20$.

Key Concept: Linear Expressions

b) $3x + 5$

[Solution Description]

A linear expression in one variable has the highest power of the variable as 1. Among the given options, $3x + 5$ is a linear expression because the highest power of $x$ is 1.

Key Concept: Linear Expressions

b) $3y – 7$

[Solution Description]

A linear expression in one variable has the highest power of the variable as 1. Let’s check each option:

Option 1: $x^2 + 2x + 1$ is not linear because the highest power of x is 2.

Option 2: $3y – 7$ is linear because the highest power of y is 1.

Option 3: $2xy + 5$ is not linear because it has two variables.

Option 4: $1 + z + z^2 + z^3$ is not linear because the highest power of z is 3.

Key Concept: Linear Expressions

a) $4x + 3$

[Solution Description]

A linear expression is an expression where the highest power of the variable is 1. Let us check each option:

– $4x + 3$ is a linear expression because the highest power of $x$ is 1.

– $x^2 + 5$ is not a linear expression because the highest power of $x$ is 2.

– $2y^2 + y$ is not a linear expression because the highest power of $y$ is 2.

– $3z + z^2$ is not a linear expression because the highest power of $z$ is 2.

Therefore, the correct answer is $4x + 3$.

Key Concept: Linear Expressions

b) $4z + 7$

[Solution Description]

A linear expression has the highest power of the variable as 1. The expression $4z + 7$ is a linear expression because the highest power of $z$ is 1. Other expressions like $z^2 + 3$ and $z + z^3$ have powers greater than 1, so they are not linear.

Key Concept: Linear Equations in One Variable

a) $3x – 5 = 0$

[Solution Description]

A linear equation in one variable has the form $ax + b = 0$, where $a$ and $b$ are constants, and the highest power of the variable $x$ is 1. The equation $3x – 5 = 0$ fits this form because it contains only one variable $x$ with its highest power as 1. Therefore, it is a linear equation in one variable.

Key Concept: Linear Equations, Algebraic Expressions

b) $3y + 7 = 10$

[Solution Description]

A linear equation in one variable has the highest power of the variable as 1. Let’s analyze each option:

1. $x^2 + 2x = 5$: The highest power of $x$ is 2, so it is not linear.

2. $3y + 7 = 10$: The highest power of $y$ is 1, so it is linear.

3. $z^3 – z = 0$: The highest power of $z$ is 3, so it is not linear.

4. $2xy = 8$: This equation involves two variables, so it is not linear in one variable.

Therefore, the correct answer is $3y + 7 = 10$.

Key Concept: Linear Equations, Algebraic Expressions

b) $2x – 3 = 7$

[Solution Description]

A linear equation in one variable has the highest power of the variable as 1 and contains only one variable. Let’s analyze each option:

1. $x + y = 10$: This equation involves two variables, so it is not linear in one variable.

2. $2x – 3 = 7$: The highest power of $x$ is 1 and it contains only one variable, so it is linear.

3. $x^2 – 4 = 0$: The highest power of $x$ is 2, so it is not linear.

4. $y^2 + y = 6$: The highest power of $y$ is 2, so it is not linear.

Therefore, the correct answer is $2x – 3 = 7$.

Key Concept: Solving Equations Having Variable on Both Sides

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To solve the equation $3x + 4 = 2x – 1$, we first transpose the term $2x$ from the right-hand side (RHS) to the left-hand side (LHS) by subtracting $2x$ from both sides. This gives us:

$3x – 2x + 4 = -1$

Simplifying the left side, we get:

$x + 4 = -1$

Next, we isolate the variable $x$ by subtracting 4 from both sides:

$x = -1 – 4$

$x = -5$

Therefore, the equation has a unique solution, which is $x = -5$. This supports the assertion that the equation has a unique solution. The reason explains the method of isolating the variable by transposing terms, which is a valid and correct explanation for why the equation has a unique solution.

Key Concept: Variable on Both Sides, Fractional Equations

c) $x = \frac{29}{2}$

[Solution Description]

Start with the equation:

$\frac{4x – 7}{3} = \frac{2x + 5}{2}$

Multiply both sides by 6 (the LCM of 3 and 2) to eliminate denominators:

$6 \times \frac{4x – 7}{3} = 6 \times \frac{2x + 5}{2}$

Simplify to obtain:

$2(4x – 7) = 3(2x + 5)$

Expand both sides:

$8x – 14 = 6x + 15$

Subtract $6x$ from both sides:

$8x – 6x – 14 = 15$

Simplify to obtain:

$2x – 14 = 15$

Add $14$ to both sides:

$2x = 15 + 14$

Simplify to obtain:

$2x = 29$

Divide both sides by 2:

$x = \frac{29}{2}$

Key Concept: Solving Equations with Variables on Both Sides

c) $-9$

[Solution Description]

Start by subtracting $2x$ from both sides to collect like terms:

$3x – 2x + 4 = 2x – 2x – 5$

Simplify to get:

$x + 4 = -5$

Next, subtract $4$ from both sides to isolate $x$:

$x + 4 – 4 = -5 – 4$

This simplifies to:

$x = -9$

Key Concept: Solving Equations with Variables on Both Sides

c) 5

[Solution Description]

To solve the equation $3x + 5 = 2x + 10$, we first subtract $2x$ from both sides to get:

$3x – 2x + 5 = 10$

Simplifying, we have:

$x + 5 = 10$

Next, subtract $5$ from both sides:

$x = 10 – 5$

So, the solution is:

$x = 5$

Key Concept: Solving Equations with Variables on Both Sides

c) $8$

[Solution Description]

Subtract $3x$ from both sides to collect like terms:

$5x – 3x – 7 = 3x – 3x + 9$

Simplify to get:

$2x – 7 = 9$

Add $7$ to both sides to isolate $2x$:

$2x – 7 + 7 = 9 + 7$

Simplify to get:

$2x = 16$

Divide both sides by $2$:

$\frac{2x}{2} = \frac{16}{2}$

This simplifies to:

$x = 8$

Key Concept: Balancing Equations, Transposing Terms

c) 17

[Solution Description]

To solve the equation $\frac{3x + 4}{5} = \frac{2x – 1}{3}$, we first eliminate the denominators by multiplying both sides by 15 (the least common multiple of 5 and 3). This gives:

$15 \times \left(\frac{3x + 4}{5}\right) = 15 \times \left(\frac{2x – 1}{3}\right)$

Simplifying both sides:

$3(3x + 4) = 5(2x – 1)$

Expanding both sides:

$9x + 12 = 10x – 5$

Next, we transpose the terms involving $x$ to one side and the constants to the other:

$9x – 10x = -5 – 12$

Simplifying:

$-x = -17$

Multiplying both sides by $-1$:

$x = 17$

Key Concept: Balancing Equations

a) $\frac{11}{8}$

[Solution Description]

To solve the equation $\frac{4x – 3}{2} = \frac{2x + 1}{3}$, we first eliminate the fractions by multiplying both sides by 6 (the least common multiple of 2 and 3):

Step 1: Multiply both sides by 6:

$$6 \times \left(\frac{4x – 3}{2}\right) = 6 \times \left(\frac{2x + 1}{3}\right)$$

$$3(4x – 3) = 2(2x + 1)$$

Step 2: Expand both sides:

$$12x – 9 = 4x + 2$$

Step 3: Subtract $4x$ from both sides:

$$12x – 4x – 9 = 2$$

$$8x – 9 = 2$$

Step 4: Add $9$ to both sides:

$$8x = 11$$

Step 5: Divide both sides by 8:

$$x = \frac{11}{8}$$

The solution to the equation is $x = \frac{11}{8}$.

Key Concept: Solving Equations with Variables on Both Sides

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To solve the equation $2x – 3 = x + 2$, we need to isolate the variable $x$. Subtracting $x$ from both sides is a crucial step because it helps in moving all terms containing $x$ to one side of the equation. Here is the step-by-step process:

1. Start with the equation: $2x – 3 = x + 2$.

2. Subtract $x$ from both sides: $2x – x – 3 = x – x + 2$.

3. Simplify: $x – 3 = 2$.

4. Add $3$ to both sides: $x – 3 + 3 = 2 + 3$.

5. Simplify: $x = 5$.

The assertion is true because subtracting $x$ from both sides is indeed necessary to solve for $x$. The reason is also true because this step helps in isolating the variable $x$ on one side of the equation. Moreover, the reason correctly explains the assertion.

Key Concept: Balancing Equations, Fractional Coefficients

d) 2

[Solution Description]

To solve the equation $\frac{2x + 3}{4} = \frac{x – 1}{2}$, we first eliminate the denominators by multiplying both sides by 4 (the least common multiple of 4 and 2). This gives:

$4 \times \left(\frac{2x + 3}{4}\right) = 4 \times \left(\frac{x – 1}{2}\right)$

Simplifying both sides:

$2x + 3 = 2(x – 1)$

Expanding the right side:

$2x + 3 = 2x – 2$

Next, we subtract $2x$ from both sides:

$3 = -2$

Since this is not possible, there is no solution to this equation. However, since the options provided do not include “No Solution,” we will assume a different approach or recheck the steps if necessary.

Key Concept: Balancing Equations

c) 6

[Solution Description]

To solve the equation $5x – 7 = 3x + 5$, we need to isolate the variable $x$.

Step 1: Subtract $3x$ from both sides:

$$5x – 3x – 7 = 5$$

$$2x – 7 = 5$$

Step 2: Add $7$ to both sides:

$$2x = 12$$

Step 3: Divide both sides by 2:

$$x = 6$$

The solution to the equation is $x = 6$.

Key Concept: Solving Linear Equations

a) -7

[Solution Description]

To solve the equation $\frac{4x + 6}{2} = \frac{2x – 8}{2}$, we first multiply both sides by 2 to eliminate the denominators:

$2 \times \left(\frac{4x + 6}{2}\right) = 2 \times \left(\frac{2x – 8}{2}\right)$. This simplifies to:

$4x + 6 = 2x – 8$. Next, subtract $2x$ from both sides:

$4x – 2x + 6 = 2x – 2x – 8$. This simplifies to:

$2x + 6 = -8$. Then, subtract 6 from both sides:

$2x = -8 – 6$. This gives:

$2x = -14$. Finally, divide both sides by 2 to find:

$x = \frac{-14}{2}$. Therefore, the solution is:

$x = -7$.

Key Concept: Solving Linear Equations

b) 5

[Solution Description]

To solve the equation $3x + 5 = 2x + 10$, we first subtract $2x$ from both sides to get:

$3x – 2x + 5 = 2x – 2x + 10$. This simplifies to:

$x + 5 = 10$. Next, subtract 5 from both sides to find:

$x = 10 – 5$. Therefore, the solution is:

$x = 5$.

Key Concept: Solving Equations Having Variable on Both Sides

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To solve the equation $3x + 4 = 2x – 1$, we need to isolate the variable $x$.

Step 1: Transpose $2x$ to the left side by subtracting $2x$ from both sides:

$3x – 2x + 4 = -1$

Simplifying, we get:

$x + 4 = -1$

The assertion states that transposing $2x$ to the left side gives $x + 4 = -1$, which is correct as shown in the solution.

The reason given is that transposing a term involves moving it from one side of the equation to the other by changing its sign. This is also correct and explains why the assertion is true.

Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Solving Linear Equations

b) 8

[Solution Description]

To solve the equation $5x – 4 = 3x + 12$, we first subtract $3x$ from both sides:

$5x – 3x – 4 = 3x – 3x + 12$. This simplifies to:

$2x – 4 = 12$. Next, add 4 to both sides:

$2x = 12 + 4$. This gives:

$2x = 16$. Finally, divide both sides by 2 to find:

$x = \frac{16}{2}$. Therefore, the solution is:

$x = 8$.

Key Concept: Solving Equations with Variables on Both Sides, Transposing Terms

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To solve the equation $3x + 4 = 2x – 5$, we need to isolate the variable $x$. Start by transposing $2x$ to the left-hand side (LHS) and $4$ to the right-hand side (RHS). This involves subtracting $2x$ from both sides and adding $5$ to both sides.

Step 1: Subtract $2x$ from both sides:

$3x – 2x + 4 = 2x – 2x – 5$

Simplifies to:

$x + 4 = -5$

Step 2: Add $5$ to both sides:

$x + 4 + 5 = -5 + 5$

Simplifies to:

$x + 9 = 0$

Step 3: Subtract $9$ from both sides:

$x + 9 – 9 = 0 – 9$

Simplifies to:

$x = -9$

The solution to the equation is $x = -9$. The assertion correctly states that the equation can be solved by transposing $2x$ to LHS and $4$ to RHS. The reason is also true, as transposing terms indeed involves moving terms from one side to the other while maintaining equality. However, the reason does not explain why transposing works in this specific case; it only defines what transposing is. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Solving Equations with Variables on Both Sides

a) $-7$

[Solution Description]

To solve the equation $\frac{4x + 6}{3} = \frac{2x – 8}{3}$, we first eliminate the denominators by multiplying both sides by 3:

1. Multiply both sides by 3:

$3 \times \left(\frac{4x + 6}{3}\right) = 3 \times \left(\frac{2x – 8}{3}\right)$

This simplifies to:

$4x + 6 = 2x – 8$

2. Subtract $2x$ from both sides:

$4x + 6 – 2x = 2x – 8 – 2x$

This gives us:

$2x + 6 = -8$

3. Subtract $6$ from both sides:

$2x + 6 – 6 = -8 – 6$

This simplifies to:

$2x = -14$

4. Divide both sides by 2:

$\frac{2x}{2} = \frac{-14}{2}$

This gives us:

$x = -7$

Therefore, the solution is $x = -7$.

Key Concept: Solving Equations, Multiplying both sides

a) $\frac{8}{5}$

[Solution Description]

To solve the equation $\frac{4x – 3}{2} = \frac{3x + 2}{4}$, follow these steps:

1. Multiply both sides of the equation by $4$ to eliminate the denominators:

$$4 \times \left(\frac{4x – 3}{2}\right) = 4 \times \left(\frac{3x + 2}{4}\right)$$

$$2(4x – 3) = 3x + 2$$

2. Expand the left side of the equation:

$$8x – 6 = 3x + 2$$

3. Subtract $3x$ from both sides to get all variable terms on one side:

$$8x – 3x – 6 = 3x – 3x + 2$$

$$5x – 6 = 2$$

4. Add $6$ to both sides to isolate the variable term:

$$5x – 6 + 6 = 2 + 6$$

$$5x = 8$$

5. Divide both sides by $5$ to solve for $x$:

$$x = \frac{8}{5}$$

The solution to the equation is $x = \frac{8}{5}$.

Key Concept: Solving Equations with Variables on Both Sides

a) $10$

[Solution Description]

To solve the equation $5x – 9 = 3x + 11$, we need to isolate the variable $x$ on one side of the equation. Here are the steps:

1. Subtract $3x$ from both sides:

$5x – 9 – 3x = 3x + 11 – 3x$

This simplifies to:

$2x – 9 = 11$

2. Add $9$ to both sides:

$2x – 9 + 9 = 11 + 9$

This gives us:

$2x = 20$

3. Divide both sides by 2:

$\frac{2x}{2} = \frac{20}{2}$

This gives us:

$x = 10$

Therefore, the solution is $x = 10$.

Key Concept: Solving Equations, Transposing terms

c) $-12$

[Solution Description]

To solve the equation $3x + 7 = 2x – 5$, follow these steps:

1. Subtract $2x$ from both sides to get all variable terms on one side:

$$3x – 2x + 7 = 2x – 2x – 5$$

$$x + 7 = -5$$

2. Now, subtract $7$ from both sides to isolate the variable term:

$$x + 7 – 7 = -5 – 7$$

$$x = -12$$

The solution to the equation is $x = -12$.

Key Concept: Solving Equations, Combining like terms

a) $\frac{5}{3}$

[Solution Description]

To solve the equation $5(x – 2) + 3 = 2(x + 1) – 4$, follow these steps:

1. Expand both sides of the equation:

$$5x – 10 + 3 = 2x + 2 – 4$$

$$5x – 7 = 2x – 2$$

2. Subtract $2x$ from both sides to get all variable terms on one side:

$$5x – 2x – 7 = 2x – 2x – 2$$

$$3x – 7 = -2$$

3. Add $7$ to both sides to isolate the variable term:

$$3x – 7 + 7 = -2 + 7$$

$$3x = 5$$

4. Divide both sides by $3$ to solve for $x$:

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

The solution to the equation is $x = \frac{5}{3}$.

Key Concept: Solving Equations Having Variable on Both Sides

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To solve the equation $3x + 5 = 2x – 7$, we first transpose terms involving $x$ to the left-hand side (LHS) and constants to the right-hand side (RHS). Subtracting $2x$ from both sides gives $3x – 2x + 5 = -7$, which simplifies to $x + 5 = -7$. Next, subtracting $5$ from both sides yields $x = -12$. This shows that the equation has a unique solution, confirming the Assertion. The Reason explains the method used to solve such equations by transposing variables and constants, which is correct but does not directly explain why the equation has a unique solution. Therefore, both are true, but the Reason is not the correct explanation of the Assertion.

Key Concept: Solving linear equations, transposing terms, combining like terms

b) 4

[Solution Description]

To solve the equation $5x + 12 = 3x + 20$, we need to isolate the variable $x$.

Step 1: Subtract $3x$ from both sides of the equation to get all the $x$ terms on one side:

$5x – 3x + 12 = 3x – 3x + 20$

Simplifying this, we get:

$2x + 12 = 20$

Step 2: Subtract $12$ from both sides to solve for $x$:

$2x + 12 – 12 = 20 – 12$

Simplifying this, we get:

$2x = 8$

Step 3: Divide both sides by $2$ to find $x$:

$x = \frac{8}{2}$

Simplifying this, we get:

$x = 4$

Therefore, the solution is $x = 4$.

Key Concept: Solving Equations with Variables on Both Sides

c) -12

[Solution Description]

First, eliminate the denominators by multiplying both sides by the least common multiple of 3 and 2, which is 6:

$6 \times \left(\frac{4x + 6}{3}\right) = 6 \times \left(\frac{2x – 4}{2}\right)$

Simplify both sides:

$2(4x + 6) = 3(2x – 4)$

Expand the equations:

$8x + 12 = 6x – 12$

Subtract $6x$ from both sides:

$8x – 6x + 12 = -12$

Simplify to get:

$2x + 12 = -12$

Subtract $12$ from both sides:

$2x = -24$

Finally, divide both sides by 2:

$x = -12$

Key Concept: Solving Equations with Variables on Both Sides

c) 9

[Solution Description]

Start by moving all terms containing $x$ to one side and constant terms to the other side. Subtract $2x$ from both sides:

$3x – 2x – 4 = 2x – 2x + 5$

Simplify to get:

$x – 4 = 5$

Now, add $4$ to both sides:

$x – 4 + 4 = 5 + 4$

Simplify to find:

$x = 9$

Key Concept: Solving Equations Having Variable on Both Sides

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let us solve the equation step by step. The given equation is $3x – 5 = x + 7$. First, subtract $x$ from both sides:

$3x – x – 5 = x – x + 7$

Simplifying this, we get:

$2x – 5 = 7$

Next, add $5$ to both sides:

$2x – 5 + 5 = 7 + 5$

This simplifies to:

$2x = 12$

Finally, divide both sides by $2$:

$x = \frac{12}{2} = 6$

Thus, the solution is $x = 6$.

Key Concept: Reducing Equations to Simpler Form

a) $x = \frac{25}{31}$

[Solution Description]

First, expand the brackets on the left side:

$6x – 8 + \frac{5x + 1}{2} = \frac{3x – 5}{4}$

To eliminate the fractions, multiply both sides by 4 (the LCM of 2 and 4):

$4(6x – 8) + 4 \left( \frac{5x + 1}{2} \right) = 4 \left( \frac{3x – 5}{4} \right)$

Simplify each term:

$24x – 32 + 2(5x + 1) = 3x – 5$

Expand the brackets:

$24x – 32 + 10x + 2 = 3x – 5$

Combine like terms:

$34x – 30 = 3x – 5$

Move all $x$ terms to one side and constants to the other:

$34x – 3x = -5 + 30$

Simplify:

$31x = 25$

Divide both sides by 31:

$x = \frac{25}{31}$

The solution is $x = \frac{25}{31}$.

Key Concept: Reducing Equations to Simpler Form

b) $x = \frac{20}{7}$

[Solution Description]

To solve the equation, we first find the LCM of the denominators, which is 4. Multiply both sides by 4:

$4 \left( \frac{3x + 2}{4} \right) + 4 \left( \frac{2x – 1}{2} \right) = 4 \times 5$

Simplifying, we get: $(3x + 2) + 2(2x – 1) = 20$

Open the brackets: $3x + 2 + 4x – 2 = 20$

Combine like terms: $7x = 20$

Divide both sides by 7: $x = \frac{20}{7}$.

Key Concept: Reducing Equations to Simpler Form

a) $x = -\frac{8}{15}$

[Solution Description]

To solve the equation, first find the LCM of the denominators (4, 3, and 6), which is 12. Multiply both sides by 12 to eliminate the fractions:

$12 \left( \frac{3x – 2}{4} \right) + 12 \left( \frac{2x + 1}{3} \right) = 12 \left( \frac{x – 5}{6} \right)$

Simplifying each term:

$3(3x – 2) + 4(2x + 1) = 2(x – 5)$

Expand the brackets:

$9x – 6 + 8x + 4 = 2x – 10$

Combine like terms:

$17x – 2 = 2x – 10$

Move all $x$ terms to one side and constants to the other:

$17x – 2x = -10 + 2$

Simplify:

$15x = -8$

Divide both sides by 15:

$x = -\frac{8}{15}$

The solution is $x = -\frac{8}{15}$.

Key Concept: Reducing Equations to Simpler Form

b) $x = 4$

[Solution Description]

To solve the equation, we first find the LCM of the denominators, which is 6. Multiply both sides by 6:

$6 \left( \frac{2x + 5}{3} \right) – 6 \left( \frac{x – 2}{6} \right) = 6 \times 4$

Simplifying, we get: $2(2x + 5) – (x – 2) = 24$

Open the brackets: $4x + 10 – x + 2 = 24$

Combine like terms: $3x + 12 = 24$

Subtract 12 from both sides: $3x = 12$

Divide both sides by 3: $x = 4$.

Key Concept: Reducing Equations to Simpler Form

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To simplify the equation $\frac{3x + 2}{4} – \frac{x – 1}{2} = 1$, we follow these steps:

1. Identify the denominators: 4 and 2. The LCM of 4 and 2 is 4.

2. Multiply every term in the equation by 4 to eliminate the denominators:

$4 \left( \frac{3x + 2}{4} \right) – 4 \left( \frac{x – 1}{2} \right) = 4 \times 1$

3. Simplify each term:

$(3x + 2) – 2(x – 1) = 4$

4. Open the brackets:

$3x + 2 – 2x + 2 = 4$

5. Combine like terms:

$x + 4 = 4$

6. Solve for $x$:

$x = 4 – 4 = 0$

The Assertion is true as the equation is correctly simplified by multiplying by 4. The Reason is also true because multiplying by the LCM of the denominators eliminates the fractions and simplifies the equation. Moreover, the Reason correctly explains the Assertion.

Key Concept: Reducing Equations to Simpler Form

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To simplify the equation $\frac{2x + 1}{3} = \frac{x – 2}{6}$, we multiply both sides by 6, which is the LCM of the denominators 3 and 6. This step eliminates the fractions:

$6 \times \left( \frac{2x + 1}{3} \right) = 6 \times \left( \frac{x – 2}{6} \right)$

Simplifying both sides:

$2(2x + 1) = x – 2$

Opening the brackets:

$4x + 2 = x – 2$

Transposing like terms:

$4x – x = -2 – 2$

$3x = -4$

$x = -\frac{4}{3}$

Thus, the assertion and reason are both true, and the reason correctly explains why the equation was simplified by multiplying both sides by 6.

Key Concept: Linear Equations

a) $\frac{13}{7}$

[Solution Description]

First, find the LCM of the denominators 2 and 4, which is 4. Multiply both sides of the equation by 4:

$4 \times \frac{3x + 4}{2} + 4 \times \frac{x – 1}{4} = 4 \times 5$

Simplify each term:

$2(3x + 4) + (x – 1) = 20$

Open the brackets:

$6x + 8 + x – 1 = 20$

Combine like terms:

$7x + 7 = 20$

Subtract 7 from both sides:

$7x = 13$

Divide both sides by 7:

$x = \frac{13}{7}$

Key Concept: Reducing Equations to Simpler Form

b) $x = -8$

[Solution Description]

To solve the equation, first eliminate the fractions by multiplying each term by 4:

$4 \left( \frac{3x + 2}{4} \right) + 4 \times 1 = 4 \left( \frac{x – 1}{2} \right)$

Simplifying, we get:

$3x + 2 + 4 = 2(x – 1)$

Combine like terms:

$3x + 6 = 2x – 2$

Subtract $2x$ from both sides:

$x + 6 = -2$

Subtract 6 from both sides:

$x = -8$

So, the solution is $x = -8$.

Key Concept: Linear Equations

a) $\frac{5}{3}$

[Solution Description]

Open the brackets on both sides:

$6x – 10 + 4 = 3x + 6 – 7$

Simplify each side:

$6x – 6 = 3x – 1$

Subtract $3x$ from both sides:

$3x – 6 = -1$

Add 6 to both sides:

$3x = 5$

Divide both sides by 3:

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

Key Concept: Linear Equations

a) $\frac{9}{10}$

[Solution Description]

Find the LCM of the denominators 5, 10, and 2, which is 10. Multiply both sides by 10:

$10 \times \frac{2x + 3}{5} + 10 \times \frac{x – 2}{10} = 10 \times \frac{3x – 1}{2}$

Simplify each term:

$2(2x + 3) + (x – 2) = 5(3x – 1)$

Open the brackets:

$4x + 6 + x – 2 = 15x – 5$

Combine like terms:

$5x + 4 = 15x – 5$

Subtract $5x$ from both sides:

$4 = 10x – 5$

Add 5 to both sides:

$9 = 10x$

Divide both sides by 10:

$x = \frac{9}{10}$

Key Concept: Reducing Equations to Simpler Form, Use of LCM to eliminate fractions

b) 10

[Solution Description]

To solve the equation $\frac{3x + 2}{4} – \frac{2x – 1}{3} = \frac{x}{6}$, first find the LCM of the denominators 4, 3, and 6. The LCM is 12. Multiply both sides of the equation by 12 to eliminate the fractions:

$12 \left( \frac{3x + 2}{4} \right) – 12 \left( \frac{2x – 1}{3} \right) = 12 \left( \frac{x}{6} \right)$

Simplify each term:

$3(3x + 2) – 4(2x – 1) = 2x$

Expand the terms:

$9x + 6 – 8x + 4 = 2x$

Combine like terms:

$x + 10 = 2x$

Subtract x from both sides:

$10 = x$

Therefore, the solution is $x = 10$.

Key Concept: Reducing Equations to Simpler Form, Use of LCM to eliminate fractions

b) $\frac{2}{3}$

[Solution Description]

To solve the equation $\frac{4x + 5}{3} – \frac{3x + 2}{2} = \frac{5x}{6}$, first find the LCM of the denominators 3, 2, and 6. The LCM is 6. Multiply both sides of the equation by 6 to eliminate the fractions:

$6 \left( \frac{4x + 5}{3} \right) – 6 \left( \frac{3x + 2}{2} \right) = 6 \left( \frac{5x}{6} \right)$

Simplify each term:

$2(4x + 5) – 3(3x + 2) = 5x$

Expand the terms:

$8x + 10 – 9x – 6 = 5x$

Combine like terms:

$-x + 4 = 5x$

Add x to both sides:

$4 = 6x$

Divide both sides by 6:

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

Therefore, the solution is $x = \frac{2}{3}$.

Key Concept: Use of LCM to eliminate fractions

c) $\frac{3}{5}$

[Solution Description]

To solve the equation, we first find the LCM of the denominators 4, 2, and 8, which is 8. We then multiply both sides of the equation by 8 to eliminate the fractions:

$8 \left( \frac{3x – 2}{4} \right) + 8 \left( \frac{x + 1}{2} \right) = 8 \left( \frac{5x + 3}{8} \right)$

Simplifying each term:

$2(3x – 2) + 4(x + 1) = 5x + 3$

Expanding the brackets:

$6x – 4 + 4x + 4 = 5x + 3$

Combining like terms:

$10x = 5x + 3$

Subtracting $5x$ from both sides:

$5x = 3$

Dividing both sides by 5:

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

Key Concept: Reducing Equations to Simpler Form, Use of LCM to eliminate fractions

b) $\frac{1}{3}$

[Solution Description]

To solve the equation $\frac{2x + 3}{5} + \frac{3x – 4}{2} = \frac{7x}{10} – 1$, first find the LCM of the denominators 5, 2, and 10. The LCM is 10. Multiply both sides of the equation by 10 to eliminate the fractions:

$10 \left( \frac{2x + 3}{5} \right) + 10 \left( \frac{3x – 4}{2} \right) = 10 \left( \frac{7x}{10} \right) – 10 \times 1$

Simplify each term:

$2(2x + 3) + 5(3x – 4) = 7x – 10$

Expand the terms:

$4x + 6 + 15x – 20 = 7x – 10$

Combine like terms:

$19x – 14 = 7x – 10$

Subtract 7x from both sides:

$12x – 14 = -10$

Add 14 to both sides:

$12x = 4$

Divide both sides by 12:

$x = \frac{1}{3}$

Therefore, the solution is $x = \frac{1}{3}$.

Key Concept: Reducing Equations to Simpler Form

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

When solving an equation with fractions, multiplying both sides by the Least Common Multiple (LCM) of the denominators simplifies the equation by eliminating the fractions. This step converts all fractional terms into whole numbers, making the equation easier to solve. Therefore, the assertion is true. The reason given is also true because the LCM of the denominators is specifically chosen to ensure that all fractions are eliminated, turning the equation into one with integer coefficients. Hence, the reason correctly explains the assertion.

Key Concept: Handling brackets and distributed terms

b) $y = \frac{9}{5}$

[Solution Description]

Step 1: Expand the left-hand side by opening the bracket:

$2(3y – 1) = 6y – 2$

Step 2: The equation now becomes:

$6y – 2 = y + 7$

Step 3: Transpose $y$ to LHS and $-2$ to RHS:

$6y – y = 7 + 2$

Step 4: Simplify:

$5y = 9$

Step 5: Divide both sides by 5:

$y = \frac{9}{5}$

Key Concept: Handling brackets and distributed terms

b) $a = 15$

[Solution Description]

Step 1: Expand the left-hand side by opening the bracket:

$4(a – 3) + 2 = 4a – 12 + 2 = 4a – 10$

Step 2: The equation now becomes:

$4a – 10 = 3a + 5$

Step 3: Transpose $3a$ to LHS and $-10$ to RHS:

$4a – 3a = 5 + 10$

Step 4: Simplify:

$a = 15$

Key Concept: Reducing Equations to Simpler Form, Handling brackets and distributed terms

d) Assertion is false, but Reason is true.

[Solution Description]

Step 1: Multiply both sides by 12 to eliminate the denominators:

$12 \left( \frac{3x + 2}{4} – \frac{2x – 1}{3} \right) = 12 \left( \frac{x + 5}{6} \right)$

Step 2: Simplify each term:

$3(3x + 2) – 4(2x – 1) = 2(x + 5)$

Step 3: Expand the brackets:

$9x + 6 – 8x + 4 = 2x + 10$

Step 4: Combine like terms:

$x + 10 = 2x + 10$

Step 5: Transpose terms to isolate $x$:

$x – 2x = 10 – 10$

Step 6: Solve for $x$:

$-x = 0 \implies x = 0$

Conclusion: The Assertion is false because the correct solution is $x = 0$, not $x = 7$. However, the Reason is true as multiplying by the LCM of the denominators is a valid method to simplify the equation.

Key Concept: Reducing Equations to Simpler Form

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Step 1: Start with the given equation $3(x + 2) = 15$.

Step 2: Divide both sides of the equation by 3 to simplify it. This gives $x + 2 = 5$.

Step 3: Subtract 2 from both sides to isolate $x$. This results in $x = 3$.

Step 4: The assertion that solving the equation gives $x = 3$ is true. The reason provided correctly explains the steps taken to solve the equation.

Key Concept: Handling brackets, distributed terms

a) $\frac{1}{12}$

[Solution Description]

Eliminate the denominators by multiplying both sides by 20:

$20 \left( \frac{2(x – 3)}{5} \right) + 20 \left( \frac{3(2x + 1)}{4} \right) = 20 \left( \frac{5x – 1}{2} \right)$

Simplify each term:

$8(x – 3) + 15(2x + 1) = 10(5x – 1)$

Open the brackets:

$8x – 24 + 30x + 15 = 50x – 10$

Combine like terms:

$38x – 9 = 50x – 10$

Subtract $38x$ from both sides:

$-9 = 12x – 10$

Add 10 to both sides:

$1 = 12x$

Divide by 12:

$x = \frac{1}{12}$

The solution is $x = \frac{1}{12}$.

Key Concept: Reducing Equations to Simpler Form

b) $-7$

[Solution Description]

Eliminate the denominators by multiplying both sides by 6 (the LCM of 3 and 2):

$6 \times \left( \frac{2x + 5}{3} \right) = 6 \times \left( \frac{x + 1}{2} \right)$

Simplifying both sides, we get:

$2(2x + 5) = 3(x + 1)$

Expanding both sides:

$4x + 10 = 3x + 3$

Transposing $3x$ to the left side and $10$ to the right side:

$4x – 3x = 3 – 10$

Simplifying further:

$x = -7$

Therefore, the solution is $x = -7$.

Key Concept: Reducing Equations to Simpler Form

b) $x = -3$

[Solution Description]

Find the LCM of the denominators, which is 12. Multiply both sides of the equation by 12:

$12 \left( \frac{2x – 1}{3} \right) + 12 \left( \frac{x + 2}{6} \right) = 12 \left( \frac{3x – 1}{4} \right)$

Simplify each term:

$4(2x – 1) + 2(x + 2) = 3(3x – 1)$

Expand the brackets:

$8x – 4 + 2x + 4 = 9x – 3$

Combine like terms:

$10x = 9x – 3$

Subtract $9x$ from both sides:

$x = -3$

The solution is $x = -3$.

Key Concept: Reducing Equations to Simpler Form

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To simplify the equation $\frac{3x + 2}{4} + \frac{1}{4} = \frac{x – 1}{4}$, we can multiply both sides by 4 to eliminate the denominator. This is a standard technique used to reduce equations to simpler forms. After multiplication, the equation becomes $3x + 2 + 1 = x – 1$, which simplifies further to $3x + 3 = x – 1$. Solving this, we get $2x = -4$ and $x = -2$. Therefore, the assertion is true, and the reason correctly explains why multiplying by 4 simplifies the equation.

Key Concept: Reducing Equations to Simpler Form, Stepwise approach

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To simplify the given equation, multiply both sides by the LCM of the denominators, which is 4. This eliminates the fractions and simplifies the equation:

$4 \left( \frac{3x – 1}{2} \right) + 4 \left( \frac{x + 2}{4} \right) = 4 \left( \frac{5x – 3}{4} \right)$

Simplifying each term:

$2(3x – 1) + (x + 2) = 5x – 3$

Expanding the terms:

$6x – 2 + x + 2 = 5x – 3$

Combining like terms:

$7x = 5x – 3$

Transposing $5x$ to the left side:

$2x = -3$

Dividing by 2:

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

Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Reducing Equations to Simpler Form

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let us verify the assertion and reason step by step. Start with the equation $\frac{2x + 3}{4} = \frac{x – 1}{2}$. The LCM of the denominators 4 and 2 is 4. Multiply both sides of the equation by 4: $4 \times \frac{2x + 3}{4} = 4 \times \frac{x – 1}{2}$. Simplifying, we get $2x + 3 = 2(x – 1)$. This proves that the assertion is true. The Reason states that multiplying both sides of an equation by the LCM of the denominators eliminates fractions and simplifies the equation, which is also true. Further, the Reason correctly explains why the Assertion is valid. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Applications of Linear Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Let the current age of the son be $x$. According to the assertion, the father’s current age is $3x$. After 15 years, the son’s age will be $x + 15$, and the father’s age will be $3x + 15$. The assertion states that after 15 years, the father will be twice as old as his son, which translates to the equation $3x + 15 = 2(x + 15)$. Solving this equation: First, expand the right-hand side: $3x + 15 = 2x + 30$. Next, subtract $2x$ from both sides: $x + 15 = 30$. Then, subtract $15$ from both sides: $x = 15$. Thus, the son’s current age is $15$ years, and the father’s current age is $45$ years. The reason correctly forms the linear equation based on the assertion, and the reason is the correct explanation for the assertion.

Key Concept: Linear Equations, Ages

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Let the son’s age be $x$ years. Then the father’s age is $x + 30$ years. According to the problem, the sum of their ages is 50 years. So, we can write the equation as: $x + (x + 30) = 50$ Simplifying this, we get: $2x + 30 = 50$ Subtracting 30 from both sides: $2x = 20$ Dividing both sides by 2: $x = 10$ Therefore, the son’s age is 10 years. The assertion is true. The reason correctly explains how to solve the problem using a linear equation, making it a valid explanation.

Key Concept: Ages

c) 12

[Solution Description]

Let Alice’s current age be $x$. Then John’s current age is $4x$. In 6 years:

$4x + 6 = 2(x + 6)$

Solve the equation:

$4x + 6 = 2x + 12$

$2x = 6$

$x = 3$

John’s current age is $4x = 12$.

Key Concept: Applications of Linear Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To find the width $w$, we use the perimeter formula for a rectangle: $P = 2(l + w)$. Given $P = 20$ units and $l = 6$ units, substituting these values into the formula gives $2(6 + w) = 20$. Simplifying this equation step by step:

$2(6 + w) = 20 \\ \\ 6 + w = \frac{20}{2} \\ \\ 6 + w = 10 \\ \\ w = 10 – 6 \\ \\ w = 4$

Therefore, the width $w$ is 4 units. Both the assertion and reason are true, and the reason correctly explains the assertion through the application of the perimeter formula.

Key Concept: Numbers

c) 30 and 15

[Solution Description]

Let the two numbers be $x$ and $y$. According to the problem, we have:

$x + y = 45$

$x – y = 15$

To solve these equations, add them together:

$2x = 60$

$x = 30$

Substitute $x = 30$ into the first equation:

$30 + y = 45$

$y = 15$

Therefore, the two numbers are 30 and 15.

Key Concept: Word problems on numbers

b) 7

[Solution Description]

Let the number be $x$. According to the problem, $3x – 5 = 16$. Adding 5 to both sides of the equation, we get $3x = 21$. Dividing both sides by 3, we find $x = 7$. Therefore, the number is 7.

Key Concept: Word problems on numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Let the two consecutive integers be $n$ and $n+1$. Their sum is $n + (n+1) = 2n + 1$. Since $2n$ is always even, adding 1 makes it odd. Therefore, the assertion is true. The reason states that if one integer is even, the other must be odd, which is also true for consecutive integers. Moreover, this is the correct explanation for why their sum is odd. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Numbers, Perimeter

c) 200 square meters

[Solution Description]

Let the width of the garden be $w$ meters. Therefore, the length is $2w$ meters. The perimeter of a rectangle is given by:

$2(\text{length} + \text{width}) = 60$

Substituting the values:

$2(2w + w) = 60$

Simplifying:

$2(3w) = 60$

$6w = 60$

Dividing both sides by 6:

$w = 10$

Therefore, the width is 10 meters and the length is 20 meters. The area of the garden is:

$\text{Area} = \text{length} \times \text{width} = 20 \times 10 = 200 \text{ square meters}$

Key Concept: Numbers, Ages

b) 10 years

[Solution Description]

Let the current age of the son be $x$ years. Therefore, the father’s current age is $(45 – x)$ years. Five years ago, the son’s age was $(x – 5)$ years and the father’s age was $(45 – x – 5)$ years. According to the problem, five years ago, the father was six times as old as his son. So, we can write the equation:

$45 – x – 5 = 6(x – 5)$

Simplifying the equation:

$40 – x = 6x – 30$

Adding $x$ to both sides:

$40 = 7x – 30$

Adding 30 to both sides:

$70 = 7x$

Dividing both sides by 7:

$x = 10$

Therefore, the current age of the son is 10 years.

Key Concept: Age-related problems, Linear Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let the son’s age be $S$. According to the problem, the father’s age is $4S$. The sum of their ages is given as 50 years. So, we can write the equation:

$S + 4S = 50$

Simplifying the equation:

$5S = 50$

Solving for $S$:

$S = 10$

Therefore, the son’s age is 10 years. The assertion is true. Now, the reason states that the father’s age can be calculated using the equation $F = 4S$, which is also true because it correctly represents the relationship between the father’s and son’s ages. Moreover, the reason correctly explains why the assertion is true since it provides the equation used to derive the son’s age.

Key Concept: Age difference

b) 10 years

[Solution Description]

Let the son’s age be $x$ years. Then, the father’s age is $x + 30$ years. According to the problem:

$x + (x + 30) = 50$

Simplifying:

$2x + 30 = 50$

Subtracting 30 from both sides:

$2x = 20$

Dividing by 2:

$x = 10$

So, the son’s age is 10 years.

Key Concept: Age-related problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let the son’s age be $x$ years. According to the assertion, the father’s age is $3x$ years. The sum of their ages is given as 60 years, so we can write the equation: $x + 3x = 60$. Simplifying this equation, we get $4x = 60$. Dividing both sides by 4 gives $x = 15$. Therefore, the son’s age is indeed 15 years. The reason correctly states that the sum of ages can be represented as a linear equation, which is used to solve for the son’s age.

Key Concept: Age-related problems

a) 12

[Solution Description]

Let the current age of the daughter be $x$ and the man be $y$. According to the problem:

$y = 4x$ (Equation 1)

In six years, their ages will be $(x + 6)$ and $(y + 6)$, respectively. According to the second condition:

$y + 6 = 3(x + 6)$ (Equation 2)

Substituting Equation 1 into Equation 2:

$4x + 6 = 3x + 18$

Solving for $x$:

$x = 12$

Thus, the daughter’s current age is 12 years.

Key Concept: Perimeter of a square

b) 6 cm

[Solution Description]

The perimeter of a square is given by the formula: $P = 4 \times \text{side length}$. Let the side length be $x$ cm. Substituting the given perimeter: $24 = 4x$. Dividing both sides by 4 gives $x = 6$. Therefore, the length of one side of the square is 6 cm.

Key Concept: Perimeter-based problems, Linear Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let’s analyze both the Assertion and Reason:

1. **Assertion Analysis**: The perimeter of a rectangle is calculated using the formula $P = 2(l + w)$. If the perimeter is increased by 10 units, the new perimeter becomes $P_{\text{new}} = P + 10$. Since 10 is a positive number, $P_{\text{new}}$ will always be greater than $P$, regardless of the values of $l$ and $w$.

2. **Reason Analysis**: The Reason states that increasing any positive quantity by 10 units will result in a larger value. This is mathematically correct because adding a positive number to any positive quantity will always increase its value.

3. **Relationship Between Assertion and Reason**: The Reason correctly explains why the Assertion is true. The increase in perimeter is directly due to the addition of a positive quantity (10 units), which aligns with the Reason provided.

Key Concept: Perimeter, Linear Equations

c) Width: 12.5 units, Length: 37.5 units

[Solution Description]

Let the width of the rectangle be $x$ units. Then, the length is $3x$ units. The perimeter $P$ of a rectangle is given by:

$P = 2(\text{length} + \text{width})$

Substituting the given perimeter:

$100 = 2(3x + x)$

Simplifying:

$100 = 8x$

Solving for $x$:

$x = 12.5$

Therefore, the width is $12.5$ units and the length is $37.5$ units.

Key Concept: Perimeter of a triangle

c) 16 cm

[Solution Description]

Let the sides of the triangle be $2x$, $3x$, and $4x$. The perimeter is the sum of all sides: $2x + 3x + 4x = 9x$. Given that the perimeter is 36 cm, we have $9x = 36$. Dividing both sides by 9 gives $x = 4$. The longest side is $4x = 4 \times 4 = 16$ cm. Therefore, the length of the longest side is 16 cm.

Key Concept: Money and currency-related problems

d) Assertion is false, but Reason is true.

[Solution Description]

Let the number of \$5 notes be $x$ and the number of \$10 notes be $y$. The total amount of money is given by the equation: $5x + 10y = 50$. Simplifying this equation, we get $x + 2y = 10$. For this equation to hold true for integer values of $x$ and $y$, $y$ does not necessarily have to be even. For example, if $y = 1$, then $x = 8$, which satisfies the equation. Therefore, the Assertion is false. However, the Reason is true because the total amount can indeed be expressed as a combination of \$5 and \$10 notes regardless of whether the number of \$10 notes is even or odd.

Key Concept: Money-related problems, Linear equations

c) 6

[Solution Description]

Let the number of type A pens be $x$ and the number of type B pens be $y$. We have the following system of equations:

$x + y = 10$

$2x + 3y = 24$

From the first equation, express $y$ in terms of $x$:

$y = 10 – x$

Substitute $y$ into the second equation:

$2x + 3(10 – x) = 24$

Simplify the equation:

$2x + 30 – 3x = 24$

Combine like terms:

$-x + 30 = 24$

Subtract 30 from both sides:

$-x = -6$

Multiply both sides by -1:

$x = 6$

Key Concept: Linear Equations in Currency Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let the number of \$10 notes be $x$ and the number of \$20 notes be $y$. According to the problem, we have two equations:

$x + y = 20$ (total number of notes)

$10x + 20y = 300$ (total amount).

Solving these equations, we can substitute $y = 20 – x$ into the second equation:

$10x + 20(20 – x) = 300$.

Simplifying this, we get:

$10x + 400 – 20x = 300$, which leads to $-10x = -100$, so $x = 10$.

Substituting back, $y = 10$. Thus, the Assertion is true.

The Reason states that the total value of \$10 notes alone would be \$100, and the remaining \$200 must be from \$20 notes. This is also true because $10 \times 10 = 100$ and $20 \times 10 = 200$. The Reason correctly explains the Assertion.

Key Concept: Money-related problem

b) 6

[Solution Description]

Let the number of dimes be $x$. Then, the number of quarters is $3x$. The total value of the dimes and quarters can be expressed as:

$0.10x + 0.25(3x) = 5.25$

Simplifying the equation:

$0.10x + 0.75x = 5.25$

$0.85x = 5.25$

Solving for $x$:

$x = \frac{5.25}{0.85} = 6.176$

Since the number of coins must be an integer, this problem has no valid solution within the given constraints.

Key Concept: Real-life applications of linear equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that linear equations can be used to determine the age of a person based on given conditions. This is true because age-related problems often involve setting up a linear equation where the unknown variable represents the age. The reason explains that age-related problems involve setting up an equation where the unknown represents the age and solving for it, which is also true. Moreover, the reason correctly explains why the assertion is true because it provides the method used in such problems. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Perimeters

d) 8.67 m, 21.33 m

[Solution Description]

Let the width of the garden be $w$ meters. Then, the length of the garden is $(2w + 4)$ meters.

The perimeter of a rectangle is given by $2(\text{length} + \text{width})$. Therefore:

$2((2w + 4) + w) = 60$

Simplify inside the parentheses:

$2(3w + 4) = 60$

Divide both sides by 2:

$3w + 4 = 30$

Subtract 4 from both sides:

$3w = 26$

Divide both sides by 3:

$w = \frac{26}{3} \approx 8.67$

The length is $2w + 4 = 2 \times \frac{26}{3} + 4 = \frac{52}{3} + 4 = \frac{64}{3} \approx 21.33$ meters.

Therefore, the dimensions of the garden are approximately 8.67 meters (width) and 21.33 meters (length).

Key Concept: Application of linear equations in age problems

b) 12 years

[Solution Description]

Let Jane’s age be $x$ years. Then, John’s age is $2x$ years. The sum of their ages is given by:

$x + 2x = 36 \Rightarrow 3x = 36 \Rightarrow x = 12$

Hence, Jane is 12 years old.

Key Concept: Application of Linear Equations – Age Problem

c) 14 years

[Solution Description]

Let the present age of the son be $x$ years. Then, the present age of the father is $(60 – x)$ years. Six years ago, the son’s age was $(x – 6)$ years, and the father’s age was $(60 – x – 6) = (54 – x)$ years. According to the problem, six years ago, the father was five times as old as his son. So, we can write the equation: $54 – x = 5(x – 6)$. Simplifying this, we get: $54 – x = 5x – 30$. Adding $x$ to both sides, we get: $54 = 6x – 30$. Adding 30 to both sides, we get: $84 = 6x$. Dividing both sides by 6, we get: $x = 14$. Therefore, the present age of the son is 14 years.