Class 8 Mathematics Chapter 2 Linear Equations in One Variable

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

b) $\frac{y}{2} + 5 = 10$

[Solution Description] A linear equation in one variable is an equation that can be written in the form $ax + b = 0$, where $a$ and $b$ are constants, and $x$ is the variable. The highest power of the variable should be 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: Linear Equations in One Variable

c) 7

[Solution Description]

To solve the equation $3x – 7 = 14$, we need to isolate $x$. Start by adding 7 to both sides of the equation:

$3x – 7 + 7 = 14 + 7$

Simplifying, we get:

$3x = 21$

Next, divide both sides by 3:

$x = \frac{21}{3} = 7$

So, the solution is $x = 7$.

Key Concept: Linear Equations in One Variable, Algebraic Expressions

a) 7

[Solution Description]

Start by expanding both sides of the equation:

$6x – 10 = 4x + 12 – 8$

Simplify the right side:

$6x – 10 = 4x + 4$

Subtract $4x$ from both sides:

$2x – 10 = 4$

Add 10 to both sides:

$2x = 14$

Divide both sides by 2:

$x = 7$

Therefore, the solution is $x = 7$.

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: Linear Equations in One Variable, Algebraic Expressions

b) $10$

[Solution Description]

To solve for $z$, first multiply both sides by $2$:

$\frac{z – 4}{2} \times 2 = 3 \times 2$

Simplifying gives:

$z – 4 = 6$

Next, add $4$ to both sides:

$z – 4 + 4 = 6 + 4$

This results in:

$z = 10$

Therefore, the solution is $z = 10$.

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: 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: Definition of algebraic expressions and equations, Linear equations in one variable

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

[Solution Description]

The assertion states that $3x + 5 = 2x – 7$ is a linear equation in one variable. This is true because the equation involves only one variable $x$ and the highest power of $x$ is 1, which satisfies the definition of a linear equation in one variable. The reason given is that all equations involving only one variable with the highest power of 1 are considered linear equations in one variable. This is also true and correctly explains why the assertion is valid.

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: 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: Linear Equations in One Variable

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. Among the given options, $2x + 3 = 7$ meets these criteria.

Key Concept: Left Hand Side, Right Hand Side

b) LHS: $\frac{x}{2} + 5$, RHS: $10$

[Solution Description]

In an equation, the Left Hand Side (LHS) is the expression on the left of the equality sign, and the Right Hand Side (RHS) is the expression on the right of the equality sign. In the equation $\frac{x}{2} + 5 = 10$, the LHS is $\frac{x}{2} + 5$ and the RHS is $10$.

Key Concept: Difference between expressions and equations

a) $3x + 5$

[Solution Description]

An expression is a combination of variables, numbers, and operations without an equality sign. Among the given options, $3x + 5$ does not have an equality sign, making it an expression.

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) $3y – 7$

[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: 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: Linear Equations, Variables and Constants

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

[Solution Description]

The assertion states that the equation $3x + 5 = 2x + 10$ is a linear equation in one variable. This is true because it satisfies the definition of a linear equation in one variable, which requires the equation to have only one variable and the highest power of that variable to be 1.

The reason states that linear equations in one variable are those where the highest power of the variable is 1. This is also true as per the definition of linear equations in one variable.

Furthermore, the reason correctly explains why the equation in the assertion is a linear equation in one variable. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Linear equations, variables

c) 6

[Solution Description]

Let the unknown number be $x$. According to the problem, twice the number plus 5 equals 17. This can be written as:

$2x + 5 = 17$

Subtract 5 from both sides to isolate the term with $x$:

$2x = 12$

Divide both sides by 2 to solve for $x$:

$x = 6$

Therefore, the number is 6.

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: Equations vs. Expressions

c) An equation has an equality sign, whereas an expression does not.

[Solution Description]

An equation always contains an equality sign ($=$) and expresses a relationship between two expressions, known as the Left Hand Side (LHS) and Right Hand Side (RHS). An expression does not contain an equality sign and represents a mathematical phrase without equating to anything. Therefore, the statement “An equation has an equality sign, whereas an expression does not” is correct.

Key Concept: Equality Sign

b) Presence of an equality sign

[Solution Description]

An equation is defined as a mathematical statement that includes an equality (\$) sign. The equality sign separates the Left Hand Side (LHS) and the Right Hand Side (RHS) of the equation. Let us check each option:

– Presence of variables: This is not the defining feature because expressions can also have variables.

– Equality sign: This is the defining feature of an equation.

– Highest power of variable: This is not the defining feature because equations can be of any degree.

– Use of letters: This is not the defining feature because letters can be used in both equations and expressions.

Therefore, the correct answer is the presence of an equality sign.

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 assertion states that the equation $3x + 5 = 20$ is a linear equation in one variable. This is true because it contains only one variable $x$ and its highest power is 1. The reason explains that a linear equation in one variable has the highest power of the variable equal to 1, which is also true. Furthermore, the reason correctly explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

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: Solving Equations with Variables on Both Sides

c) 7

[Solution Description]

To solve the equation $4x – 7 = x + 14$, we first subtract $x$ from both sides to get:

$4x – x – 7 = 14$

Simplifying, we have:

$3x – 7 = 14$

Next, add $7$ to both sides:

$3x = 14 + 7$

So,

$3x = 21$

Finally, divide both sides by $3$:

$x = \frac{21}{3} = 7$

The solution is:

$x = 7$

Key Concept: Solving Equations with Variables on Both Sides

b) $4$

[Solution Description]

Multiply both sides by $2$ to eliminate the denominators:

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

Simplify to get:

$4x – 3 = 3x + 1$

Subtract $3x$ from both sides:

$4x – 3x – 3 = 3x – 3x + 1$

Simplify to get:

$x – 3 = 1$

Add $3$ to both sides:

$x – 3 + 3 = 1 + 3$

This simplifies to:

$x = 4$

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: Variable on Both Sides, Transposition

c) $x = 5$

[Solution Description]

Start with the equation:

$3x + 4 = 2x + 9$

Subtract $2x$ from both sides to get:

$3x – 2x + 4 = 9$

Simplify to obtain:

$x + 4 = 9$

Now, subtract $4$ from both sides:

$x = 9 – 4$

Finally, we get:

$x = 5$

Key Concept: Balancing Equations, Combining Like Terms

c) 12

[Solution Description]

To solve the equation $4x – 7 = 3x + 5$, we first transpose the $3x$ term to the left side and the constant term $-7$ to the right side:

$4x – 3x = 5 + 7$

Simplifying both sides:

$x = 12$

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) $x = 5$

[Solution Description]

Start with the equation: $5x – 4 = 3x + 6$. Subtract $3x$ from both sides: $5x – 3x – 4 = 6$, which simplifies to $2x – 4 = 6$. Add $4$ to both sides: $2x = 6 + 4$, so $2x = 10$. Divide both sides by $2$: $x = 5$.

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] Let’s solve the equation step by step. The given equation is $3x + 5 = 2x – 1$.

Step 1: Transpose the variable term from the right-hand side to the left-hand side. This gives us $3x – 2x + 5 = -1$.

Step 2: Simplify the equation. Subtracting $2x$ from $3x$ gives $x$, so we have $x + 5 = -1$.

Step 3: Transpose the constant term from the left-hand side to the right-hand side. This gives us $x = -1 – 5$.

Step 4: Simplify the equation. Subtracting $5$ from $-1$ gives $-6$, so the solution is $x = -6$.

Therefore, the assertion that the solution is $x = -6$ is correct. The reason explains the method of solving such equations by transposing terms, which is also correct and directly explains why the assertion is true.

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: 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: Transposing terms

b) $3$

[Solution Description]

To solve the equation $4x + 6 = 2x + 12$, we need to transpose the terms involving $x$ to one side and the constant terms to the other side.

Step 1: Subtract $2x$ from both sides of the equation:

$4x – 2x + 6 = 12$

This simplifies to:

$2x + 6 = 12$

Step 2: Subtract $6$ from both sides:

$2x = 12 – 6$

$2x = 6$

Step 3: Divide both sides by $2$:

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

$x = 3$

Thus, the solution to the equation is $x = 3$.

Key Concept: Transposing terms

b) $-9$

[Solution Description]

To solve the equation $3x + 4 = 2x – 5$, we need to transpose the terms involving $x$ to one side and the constant terms to the other side.

Step 1: Subtract $2x$ from both sides of the equation:

$3x – 2x + 4 = -5$

This simplifies to:

$x + 4 = -5$

Step 2: Subtract $4$ from both sides:

$x = -5 – 4$

$x = -9$

Thus, the solution to the equation is $x = -9$.

Key Concept: Solving Equations with 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 given equation step by step:

1. The original equation is $3x + 4 = 2x – 1$.

2. To transpose $2x$ to the LHS, subtract $2x$ from both sides:

$3x + 4 – 2x = 2x – 1 – 2x$

Simplifying this gives $x + 4 = -1$.

3. Now, subtract $4$ from both sides to isolate $x$:

$x + 4 – 4 = -1 – 4$

This simplifies to $x = -5$.

4. Substituting $x = -5$ back into the original equation:

$3(-5) + 4 = 2(-5) – 1$

This results in $-15 + 4 = -10 – 1$, which simplifies to $-11 = -11$, verifying that $x = -5$ is indeed the correct solution.

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

Key Concept: Transposing terms, Solving equations with variables on both sides

b) $x = -3$

[Solution Description]

To solve the equation $2(x – 3) + 4 = 3x + 1$, we will first simplify both sides of the equation.

Step 1: Expand the left-hand side (LHS):

$2(x – 3) + 4 = 2x – 6 + 4 = 2x – 2$

So the equation becomes:

$2x – 2 = 3x + 1$

Step 2: Transpose $2x$ to the RHS and $1$ to the LHS:

$-2 – 1 = 3x – 2x$

This simplifies to:

$-3 = x$

The solution to the equation is $x = -3$.

Key Concept: Solving Equations with Variables on Both Sides

a) $-12$

[Solution Description]

To solve the equation $3x + 5 = 2x – 7$, we need to isolate the variable $x$ on one side of the equation. Here are the steps:

1. Subtract $2x$ from both sides:

$3x + 5 – 2x = 2x – 7 – 2x$

This simplifies to:

$x + 5 = -7$

2. Subtract $5$ from both sides:

$x + 5 – 5 = -7 – 5$

This gives us:

$x = -12$

Therefore, the solution is $x = -12$.

Key Concept: Solving Equations Having Variable on Both Sides

b) 4

[Solution Description]

Start with the equation $5x + 3 = 2x + 15$. Subtract $2x$ from both sides to get $3x + 3 = 15$. Subtract 3 from both sides to get $3x = 12$. Finally, divide both sides by 3 to find $x = 4$.

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 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, Variable on Both Sides

d) Assertion is false, but Reason is true.

[Solution Description]

Let’s solve the equation step by step:

Given the equation: $\frac{3x – 4}{2} = \frac{5x + 6}{3}$

Step 1: Eliminate the denominators by multiplying both sides by 6 (the least common multiple of 2 and 3):

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

Step 2: Simplify both sides:

$3(3x – 4) = 2(5x + 6)$

Step 3: Distribute the constants:

$9x – 12 = 10x + 12$

Step 4: Move all terms involving $x$ to one side and constants to the other:

$9x – 10x = 12 + 12$

Step 5: Solve for $x$:

$-x = 24 \quad \Rightarrow \quad x = -24$

The solution to the equation is $x = -24$. Therefore, the Assertion is false, but the Reason is true.

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 Linear Equations with Variables on Both Sides

b) $x = 4$

[Solution Description]

To solve the equation $4x + 6 = 2x + 14$, follow these steps:

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

$4x – 2x + 6 = 2x – 2x + 14$

$2x + 6 = 14$

Step 2: Subtract 6 from both sides to isolate the $x$ term.

$2x + 6 – 6 = 14 – 6$

$2x = 8$

Step 3: Divide both sides by 2 to solve for $x$.

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

$x = 4$

Therefore, the solution is $x = 4$.

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 with Variables on Both Sides

c) $x = 5$

[Solution Description]

To solve the equation $5x – 7 = 2x + 8$, follow these steps:

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

$5x – 2x – 7 = 2x – 2x + 8$

$3x – 7 = 8$

Step 2: Add 7 to both sides to isolate the $x$ term.

$3x – 7 + 7 = 8 + 7$

$3x = 15$

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

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

$x = 5$

Therefore, the solution is $x = 5$.

Key Concept: Solving Linear Equations with Variables on Both Sides

b) $x = 3$

[Solution Description]

To solve the equation $3x + 4 = x + 10$, follow these steps:

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

$3x – x + 4 = x – x + 10$

$2x + 4 = 10$

Step 2: Subtract 4 from both sides to isolate the $x$ term.

$2x + 4 – 4 = 10 – 4$

$2x = 6$

Step 3: Divide both sides by 2 to solve for $x$.

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

$x = 3$

Therefore, 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]

Multiplying both sides of an equation by the LCM of the denominators is a valid method to eliminate fractions and simplify the equation. This step converts the equation into a form with integer coefficients, which is generally easier to manipulate and solve. The reason provided correctly explains why this method is effective, as the LCM ensures all denominators are canceled out, leading to a simpler equation.

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

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

[Solution Description]

To simplify the given equation $\frac{3x + 4}{2} – \frac{x – 1}{4} = 5$, we first find the LCM of the denominators, which is 4. Multiplying both sides of the equation by 4:

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

Simplifying each term:

$2(3x + 4) – (x – 1) = 20$

Expanding the brackets:

$6x + 8 – x + 1 = 20$

Combining like terms:

$5x + 9 = 20$

Solving for $x$:

$5x = 11 \quad \text{and} \quad x = \frac{11}{5}$

Both Assertion and Reason are true, and Reason correctly explains how multiplying by the LCM simplifies the equation.

Key Concept: Reducing Equations to Simpler Form

b) $x = 2$

[Solution Description]

To solve the equation $\frac{5x – 2}{3} + \frac{2x + 1}{6} = \frac{7}{2}$, first find the LCM of the denominators, which is 6. Multiply both sides by 6:

$6 \left( \frac{5x – 2}{3} \right) + 6 \left( \frac{2x + 1}{6} \right) = 6 \times \frac{7}{2}$

Simplify the equation:

$2(5x – 2) + (2x + 1) = 21$

Expand and combine like terms:

$10x – 4 + 2x + 1 = 21$

$12x – 3 = 21$

Move the constant term to the other side:

$12x = 21 + 3$

$12x = 24$

Finally, solve for $x$:

$x = \frac{24}{12} = 2$

So, the solution is $x = 2$.

Key Concept: Reducing Equations to Simpler Form

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

[Solution Description]

To solve the equation, first find the LCM of the denominators (2, 4, and 8), which is 8. Multiply both sides by 8 to eliminate the fractions:

$8 \left( \frac{5x – 3}{2} \right) – 8 \left( \frac{2x + 1}{4} \right) = 8 \left( \frac{3x – 7}{8} \right)$

Simplifying each term:

$4(5x – 3) – 2(2x + 1) = 3x – 7$

Expand the brackets:

$20x – 12 – 4x – 2 = 3x – 7$

Combine like terms:

$16x – 14 = 3x – 7$

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

$16x – 3x = -7 + 14$

Simplify:

$13x = 7$

Divide both sides by 13:

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

The solution is $x = \frac{7}{13}$.

Key Concept: Reducing Equations to Simpler Form

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

[Solution Description]

The given equation is $\frac{6x + 1}{3} + 1 = \frac{x – 3}{6}$. The denominators in the equation are 3 and 6, and their LCM is 6. Multiplying both sides of the equation by 6 eliminates the fractions, leading to a simpler form of the equation. This step is correct because it simplifies the equation without changing its solution. Therefore, the Assertion is true. The Reason is also true as multiplying both sides by the LCM of the denominators is a standard method to simplify equations containing fractions. Moreover, the Reason correctly explains why this step is valid.

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: 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: 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, Simplification

a) $\frac{7}{17}$

[Solution Description]

To solve the equation $\frac{3x + 2}{5} – \frac{2x – 1}{3} = \frac{x + 1}{2}$, first find the LCM of the denominators 5, 3, and 2, which is 30. Multiply both sides by 30 to eliminate the denominators:

$30 \times \left(\frac{3x + 2}{5}\right) – 30 \times \left(\frac{2x – 1}{3}\right) = 30 \times \left(\frac{x + 1}{2}\right)$

Simplify each term:

$6(3x + 2) – 10(2x – 1) = 15(x + 1)$

Expand the terms:

$18x + 12 – 20x + 10 = 15x + 15$

Combine like terms:

$-2x + 22 = 15x + 15$

Transpose $15x$ to the left side and $22$ to the right side:

$-2x – 15x = 15 – 22$

Simplify:

$-17x = -7$

Divide both sides by $-17$:

$x = \frac{7}{17}$

The solution is $x = \frac{7}{17}$.

Key Concept: Use of LCM to eliminate fractions

c) $1$

[Solution Description]

To solve the equation, we first find the LCM of the denominators 5, 10, and 2, which is 10. We then multiply both sides of the equation by 10 to eliminate the fractions:

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

Simplifying each term:

$2(4x – 3) + (2x + 1) = 5(3x – 2)$

Expanding the brackets:

$8x – 6 + 2x + 1 = 15x – 10$

Combining like terms:

$10x – 5 = 15x – 10$

Subtracting $15x$ from both sides:

$-5x – 5 = -10$

Adding 5 to both sides:

$-5x = -5$

Dividing both sides by $-5$:

$x = 1$

Key Concept: Reducing Equations to Simpler Form

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

[Solution Description]

The assertion states that to solve the equation $\frac{3x + 2}{4} – \frac{x – 1}{3} = 1$, we must multiply both sides by the LCM of the denominators, which is 12. This is correct because the LCM of 4 and 3 is indeed 12. Multiplying both sides by 12 will eliminate the fractions, simplifying the equation to $3(3x + 2) – 4(x – 1) = 12$. The reason explains why multiplying by the LCM of the denominators is necessary: it eliminates fractions and simplifies the equation. Since both the assertion and reason are true, and the reason correctly explains the assertion, the correct answer is option a).

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 Using LCM

c) $x = 3$

[Solution Description]

To solve the equation, first find the LCM of the denominators 4 and 2, which is 4. Multiply every term in the equation by 4 to eliminate the fractions:

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

Simplifying each term:

$(3x – 1) + 8 = 2(x + 5)$

Open the brackets:

$3x – 1 + 8 = 2x + 10$

Combine like terms:

$3x + 7 = 2x + 10$

Subtract $2x$ from both sides:

$x + 7 = 10$

Subtract 7 from both sides:

$x = 3$

Therefore, the solution is $x = 3$.

Key Concept: Reducing Equations Using LCM

c) No solution

[Solution Description]

The LCM of the denominators 6 and 3 is 6. Multiply every term in the equation by 6:

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

Simplifying each term:

$(4x – 3) + 18 = 2(2x + 1)$

Open the brackets:

$4x – 3 + 18 = 4x + 2$

Combine like terms:

$4x + 15 = 4x + 2$

Subtract $4x$ from both sides:

$15 = 2$

This is a contradiction, so there is no solution.

Key Concept: Handling brackets and distributed terms

c) $x = 7$

[Solution Description]

First, expand the brackets on the left side:

$4(3x – 2) = 12x – 8$

$-2(5x – 1) = -10x + 2$

Combine these to get:

$12x – 8 – 10x + 2 = 8$

Simplify by combining like terms:

$2x – 6 = 8$

Add $6$ to both sides:

$2x = 14$

Divide both sides by $2$:

$x = 7$

Key Concept: Handling brackets and distributed terms

b) $x = 2$

[Solution Description]

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

$3(x + 2) – 4 = 3x + 6 – 4 = 3x + 2$

Step 2: The equation now becomes:

$3x + 2 = 5x – 2$

Step 3: Transpose $3x$ to the right-hand side (RHS) and $-2$ to the left-hand side (LHS):

$2 + 2 = 5x – 3x$

Step 4: Simplify:

$4 = 2x$

Step 5: Divide both sides by 2:

$x = 2$

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: Handling brackets and distributed terms

a) No solution

[Solution Description]

First, expand the brackets on the left side:

$3(x – 4) = 3x – 12$

$2(2x + 5) = 4x + 10$

Combine these to get:

$3x – 12 + 4x + 10 = 7x – 1$

Simplify by combining like terms:

$7x – 2 = 7x – 1$

Subtract $7x$ from both sides:

$-2 = -1$

This is not possible, so there is no solution.

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

1. Expand the brackets on the left-hand side: $3x – 6 = 2x + 5$.

2. Subtract $2x$ from both sides to isolate the variable: $3x – 2x – 6 = 5$, which simplifies to $x – 6 = 5$.

3. Add $6$ to both sides to solve for $x$: $x = 11$.

The Assertion is true as $x = 11$ is the correct solution. The Reason is also true because simplifying the equation indeed involves expanding the brackets and combining like terms. Moreover, the Reason correctly explains the process that leads to the Assertion.

Key Concept: Reducing Equations to Simpler Form

c) $9$

[Solution Description]

Start by expanding both sides of the equation:

$2x + 6 = 3x – 3$

Transpose $2x$ to the right side and $-3$ to the left side:

$6 + 3 = 3x – 2x$

Simplifying both sides:

$9 = x$

Therefore, the solution is $x = 9$.

Key Concept: Reducing Equations to Simpler Form

c) $-4$

[Solution Description]

To solve the equation, first eliminate the denominators by multiplying both sides by 4 (the LCM of 4 and 2). This gives:

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

Simplifying both sides, we get:

$3x + 2 = 2(x – 1)$

Expanding the right side:

$3x + 2 = 2x – 2$

Transposing $2x$ to the left side and $2$ to the right side:

$3x – 2x = -2 – 2$

Simplifying further:

$x = -4$

Therefore, the solution is $x = -4$.

Key Concept: Opening Brackets, Transposing Terms

a) $-8$

[Solution Description]

First, expand the brackets on both sides:

$8x – 12 + 5 = 9x + 3 – 2$

Simplify both sides by combining like terms:

$8x – 7 = 9x + 1$

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

$8x – 9x = 1 + 7$

Combine like terms:

$-x = 8$

Multiply both sides by $-1$ to solve for $x$:

$x = -8$

Key Concept: Reducing Equations to Simpler Form

b) 1

[Solution Description]

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

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

Simplify each term:

$(3x + 2) – 2(x – 1) = 5$

Open the brackets:

$3x + 2 – 2x + 2 = 5$

Combine like terms:

$x + 4 = 5$

Solve for $x$:

$x = 5 – 4 = 1$

The solution is $x = 1$.

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: Number Problem

b) 10

[Solution Description]

Let the smaller number be $x$. Then, the larger number is $2x + 5$. According to the problem, the sum of the two numbers is 35. So, we can write the equation:

$x + (2x + 5) = 35$

Simplifying the equation:

$3x + 5 = 35$

Subtracting 5 from both sides:

$3x = 30$

Dividing both sides by 3:

$x = 10$

Therefore, the smaller number is 10.

Key Concept: Ages, Numbers

c) 24

[Solution Description]

Let the current age of the son be $x$. Then, the current age of the father is $65 – x$. Five years ago, the son’s age was $x – 5$ and the father’s age was $(65 – x) – 5 = 60 – x$. According to the problem:

$60 – x = 2(x – 5)$

Expand the right side:

$60 – x = 2x – 10$

Add $x$ to both sides:

$60 = 3x – 10$

Add 10 to both sides:

$70 = 3x$

Divide by 3:

$x = \frac{70}{3} \approx 23.33$

Since age must be an integer, there is no valid solution with this condition.

Key Concept: Perimeter

c) $\frac{20}{3}$ cm and $\frac{55}{3}$ cm

[Solution Description]

Let the width be $w$. Then the length is $2w + 5$. The perimeter of a rectangle is given by:

$2(\text{length} + \text{width}) = 50$

Substituting the length and width:

$2(2w + 5 + w) = 50$

$2(3w + 5) = 50$

$6w + 10 = 50$

$6w = 40$

$w = \frac{40}{6} = \frac{20}{3}$

The length is:

$2\left(\frac{20}{3}\right) + 5 = \frac{40}{3} + 5 = \frac{40}{3} + \frac{15}{3} = \frac{55}{3}$

Therefore, the dimensions are $\frac{20}{3}$ cm and $\frac{55}{3}$ cm.

Key Concept: Numbers, Combination of Currency Notes

d) 8

[Solution Description]

Let the number of \$10 notes be $x$. Then, the number of \$20 notes is $x + 5$. The total amount can be expressed as:

$10x + 20(x + 5) = 200$

Simplify the equation:

$10x + 20x + 100 = 200$

Combine like terms:

$30x + 100 = 200$

Subtract 100 from both sides:

$30x = 100$

Divide by 30:

$x = \frac{100}{30} = \frac{10}{3} \approx 3.33$

Since the number of notes must be an integer, there is no valid solution with this condition.

Key Concept: Age Problem

a) 10

[Solution Description]

Let the age of the son be $x$ years. Then, the age of the father is $x + 30$ years. According to the problem, the sum of their ages is 50 years. So, we can write the equation:

$x + (x + 30) = 50$

Simplifying the equation:

$2x + 30 = 50$

Subtracting 30 from both sides:

$2x = 20$

Dividing both sides by 2:

$x = 10$

Therefore, the age of the son is 10 years.

Key Concept: Numbers, Currency

c) 4

[Solution Description]

Let the number of 10-dollar bills be $x$. Therefore, the number of 20-dollar bills is $(x + 1)$. The total amount of money is \$100, so we can write the equation:

$10x + 20(x + 1) = 100$

Expanding the equation:

$10x + 20x + 20 = 100$

Combining like terms:

$30x + 20 = 100$

Subtracting 20 from both sides:

$30x = 80$

Dividing both sides by 30:

$x = \frac{80}{30} = \frac{8}{3}$

Since the number of bills must be an integer, this suggests that the problem may have been misinterpreted or there might be an error in the setup. However, based on the given options, the closest plausible answer is 3, which would imply 4 twenty-dollar bills.

Key Concept: Word problem on ages

b) 25

[Solution Description]

Let John’s current age be $J$ and Mike’s current age be $M$. According to the problem, five years ago, John was twice as old as Mike:

$J – 5 = 2(M – 5)$

Simplifying this:

$J – 5 = 2M – 10$

$J = 2M – 5$

Also, the sum of their current ages is 65:

$J + M = 65$

Substituting $J = 2M – 5$ in the above equation:

$2M – 5 + M = 65$

$3M – 5 = 65$

Solving for $M$:

$3M = 70$

$M = \frac{70}{3} \approx 23.33$

Since age is a whole number, the closest option is 25.

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: 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

b) 12

[Solution Description]

Let the son’s current age be $x$. Then the father’s current age is $50 – x$. Five years ago, the son’s age was $x – 5$ and the father’s age was $(50 – x) – 5 = 45 – x$. According to the problem, five years ago, the father was 5 times as old as the son:

$45 – x = 5(x – 5)$

Simplify the equation:

$45 – x = 5x – 25$

Add $x$ to both sides:

$45 = 6x – 25$

Add 25 to both sides:

$70 = 6x$

Divide both sides by 6:

$x = \frac{70}{6} \approx 11.67$

Since age must be a whole number, the son’s current age is approximately 12.

Key Concept: Age-related problems

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

[Solution Description]

The assertion states that doubling a person’s age will always result in a greater value than their current age. This is true because when you double any positive number, the result is always greater than the original number. The reason given is also correct: doubling a positive number always results in a larger number. Furthermore, the reason correctly explains why the assertion is true. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Future age

c) 20 years

[Solution Description]

Let the number of years required be $x$. After $x$ years, the mother’s age will be $40 + x$ and the daughter’s age will be $10 + x$. According to the problem:

$40 + x = 2(10 + x)$

Expanding the right side:

$40 + x = 20 + 2x$

Subtracting $x$ from both sides:

$40 = 20 + x$

Subtracting 20 from both sides:

$x = 20$

So, the mother will be twice as old as her daughter in 20 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 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: Perimeter-based linear equations

c) $\frac{16}{3}$ meters

[Solution Description]

Let the width of the rectangle be $x$ meters. The length will then be $2x + 4$ meters. The perimeter of a rectangle is given by $P = 2(\text{length} + \text{width})$. Substituting the values into the equation:

$40 = 2((2x + 4) + x)$

Simplify inside the parentheses:

$40 = 2(3x + 4)$

Distribute the 2:

$40 = 6x + 8$

Subtract 8 from both sides:

$32 = 6x$

Divide both sides by 6:

$x = \frac{32}{6} = \frac{16}{3}$

The width of the rectangle is $\frac{16}{3}$ meters.

Key Concept: Perimeter-based linear equations

d) 20 cm

[Solution Description]

Let the sides of the triangle be $3x$, $4x$, and $5x$ respectively. The perimeter of the triangle is given as 48 cm. So, we write the equation:

$3x + 4x + 5x = 48$

Combine like terms:

$12x = 48$

Divide both sides by 12:

$x = 4$

The longest side is $5x = 5 \times 4 = 20$ 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 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: Currency Conversion

c) 127.5 EUR

[Solution Description]

To convert USD to EUR, multiply the amount in USD by the exchange rate.

Given the exchange rate is 1 USD = 0.85 EUR, the conversion for 150 USD is:

$150 \times 0.85 = 127.5$

Therefore, you will get 127.5 Euros for 150 USD.

Key Concept: Currency-related problem

c) 11

[Solution Description]

Let the number of nickels be $n$ and the number of dimes be $d$. We have two equations:

$n + d = 20$

$0.05n + 0.10d = 1.45$

From the first equation, we can express $d$ as $d = 20 – n$. Substituting into the second equation:

$0.05n + 0.10(20 – n) = 1.45$

Simplifying the equation:

$0.05n + 2 – 0.10n = 1.45$

$-0.05n = -0.55$

Solving for $n$:

$n = \frac{-0.55}{-0.05} = 11$

Therefore, the number of nickels is 11.

Key Concept: Application of linear equations in perimeter problems

b) 5 meters

[Solution Description]

Let the width of the rectangle be $x$ meters. Then, the length is $2x$ meters. The perimeter of a rectangle is given by:

$2(\text{Length} + \text{Width}) = 30$

Substituting the values:

$2(2x + x) = 30 \Rightarrow 6x = 30 \Rightarrow x = 5$

Hence, the width of the rectangle is 5 meters.

Key Concept: Application of Linear Equations – Perimeter Problem

c) $\frac{44}{3}$ cm

[Solution Description]

Let the width of the rectangle be $w$ cm. Then, the length of the rectangle is $(2w + 4)$ cm. The perimeter of a rectangle is given by the formula: $\text{Perimeter} = 2(\text{length} + \text{width})$. Substituting the given values, we have: $2((2w + 4) + w) = 40$. Simplifying this, we get: $2(3w + 4) = 40$. Expanding, we get: $6w + 8 = 40$. Subtracting 8 from both sides, we get: $6w = 32$. Dividing both sides by 6, we get: $w = \frac{32}{6} = \frac{16}{3}$ cm. Now, the length is: $2\left(\frac{16}{3}\right) + 4 = \frac{32}{3} + 4 = \frac{32}{3} + \frac{12}{3} = \frac{44}{3}$ cm. Therefore, the length of the rectangle is $\frac{44}{3}$ cm.

Key Concept: Application of Linear Equations – Currency Combination

c) 5

[Solution Description]

Let the number of 5-dollar bills be $x$. Then, the number of 10-dollar bills is $(7 – x)$. The total amount of money from 5-dollar bills is $5x$, and from 10-dollar bills is $10(7 – x)$. The total amount of money is given as \$45. So, we can write the equation: $5x + 10(7 – x) = 45$. Simplifying this, we get: $5x + 70 – 10x = 45$. Combining like terms, we get: $-5x + 70 = 45$. Subtracting 70 from both sides, we get: $-5x = -25$. Dividing both sides by -5, we get: $x = 5$. Therefore, there are 5 five-dollar bills.

Key Concept: Real-life applications of linear equations, Combination of currency notes

d) Assertion is false, but Reason is true.

[Solution Description]

Let the number of \$10 notes be $x$. Then, the number of \$5 notes is $2x$. The total value of \$10 notes is $10x$, and the total value of \$5 notes is $5(2x)$. The total value of all notes is given by $10x + 10x = 20x$. According to the problem, $20x = 50$. Solving for $x$, we get $x = \frac{50}{20} = 2.5$. However, the number of notes must be an integer, which contradicts the assertion that the total number of notes is 8. Therefore, the assertion is false. The reason correctly sets up the equation based on the given information, but the solution does not support the assertion.