Chapter 4 Simple Equations (Additional Questions)

The correct option is (C).

A linear equation in one variable is an equation that can be written in the form $ax + b = 0$, where $x$ is the variable, and $a$ and $b$ are constants with $a \neq 0$. It involves only one variable raised to the power of 1.

Let's examine the given options:

(A) $2x + 3y = 5$: This equation contains two variables, $x$ and $y$. Therefore, it is a linear equation in two variables.

(B) $x^2 + 2 = 6$: This equation contains only one variable, $x$, but the highest power of $x$ is 2. Therefore, it is a quadratic equation, not a linear equation.

(C) $5m - 1 = 9$: This equation contains only one variable, $m$, and the highest power of $m$ is 1. This can be rewritten in the form $5m - 1 - 9 = 0$, or $5m - 10 = 0$. This matches the form $ax + b = 0$ with $a=5$ and $b=-10$. Therefore, it is a simple linear equation in one variable.

(D) $a > 7$: This is an inequality, not an equation.

Thus, the only simple linear equation in one variable among the given options is $5m - 1 = 9$.

The correct option is (B).

In an algebraic equation, a variable is a symbol, typically a letter, that represents an unknown value that can change or vary. Constants, on the other hand, are fixed values.

In the given equation, $3p + 7 = 16$, we have the following components:

$\bullet$ $3$: This is a coefficient, a constant value multiplying the variable.

$\bullet$ $p$: This is the symbol representing the unknown value we are trying to find. It is the quantity that can vary.

$\bullet$ $7$: This is a constant term.

$\bullet$ $16$: This is also a constant term.

Therefore, the variable in the equation $3p + 7 = 16$ is $p$.

The correct option is (B).

To find the solution to the equation $x + 5 = 12$, we need to isolate the variable $x$ on one side of the equation.

The given equation is:

$x + 5 = 12$

To get $x$ by itself, we need to eliminate the $+5$ on the left side. We can do this by performing the inverse operation, which is subtraction. We subtract 5 from both sides of the equation to maintain equality.

Subtract 5 from both sides:

$x + 5 - 5 = 12 - 5$

Simplify both sides:

$x = 7$

So, the solution to the equation $x + 5 = 12$ is $x = 7$.

Let's check the other options:

(A) If $x=5$, then $5 + 5 = 10 \neq 12$. Incorrect.

(C) If $x=12$, then $12 + 5 = 17 \neq 12$. Incorrect.

(D) If $x=17$, then $17 + 5 = 22 \neq 12$. Incorrect.

Only $x=7$ satisfies the equation.

The correct option is (C).

To find the value of $y$ that satisfies the equation $y - 8 = 3$, we need to solve the equation for $y$.

The given equation is:

$y - 8 = 3$

To isolate $y$ on the left side of the equation, we need to eliminate the $-8$. We can do this by adding 8 to both sides of the equation.

Add 8 to both sides:

$y - 8 + 8 = 3 + 8$

Simplify both sides:

$y = 11$

So, the value of $y$ that satisfies the equation $y - 8 = 3$ is $11$.

Let's check the other options:

(A) If $y=5$, then $5 - 8 = -3 \neq 3$. Incorrect.

(B) If $y=8$, then $8 - 8 = 0 \neq 3$. Incorrect.

(D) If $y=3$, then $3 - 8 = -5 \neq 3$. Incorrect.

Only $y=11$ makes the equation true.

The correct option is (B).

To find the value of $k$ in the equation $4k = 28$, we need to isolate the variable $k$ on one side of the equation.

The given equation is:

$4k = 28$

This equation means "4 times $k$ equals 28". To find the value of $k$, we need to perform the inverse operation of multiplication, which is division. We divide both sides of the equation by 4.

Divide both sides by 4:

$\frac{4k}{4} = \frac{28}{4}$

Simplify both sides:

$\cancel{\frac{4}{4}}k = 7$

$k = 7$

So, the value of $k$ is 7.

Let's check the other options:

(A) If $k=4$, then $4 \times 4 = 16 \neq 28$. Incorrect.

(C) If $k=24$, then $4 \times 24 = 96 \neq 28$. Incorrect.

(D) If $k=32$, then $4 \times 32 = 128 \neq 28$. Incorrect.

Only $k=7$ makes the equation true.

The correct option is (A).

To find the value of $z$ in the equation $\frac{z}{6} = 5$, we need to isolate the variable $z$ on one side of the equation.

The given equation is:

$\frac{z}{6} = 5$

This equation means "$z$ divided by 6 equals 5". To find the value of $z$, we need to perform the inverse operation of division, which is multiplication. We multiply both sides of the equation by 6.

Multiply both sides by 6:

$\frac{z}{6} \times 6 = 5 \times 6$

Simplify both sides:

$z = 30$

So, the value of $z$ is 30.

Let's check the other options:

(B) If $z=11$, then $\frac{11}{6} \neq 5$. Incorrect.

(C) If $z=6$, then $\frac{6}{6} = 1 \neq 5$. Incorrect.

(D) If $z=5$, then $\frac{5}{6} \neq 5$. Incorrect.

Only $z=30$ makes the equation true.

The correct option is (B).

The rule of transposition is a method used to solve equations by moving terms from one side of the equality sign to the other. When a term is transposed to the other side, its operation is reversed.

Specifically:

$\bullet$ If a term is being added on one side, it becomes subtraction on the other side.

$\bullet$ If a term is being subtracted on one side, it becomes addition on the other side.

$\bullet$ If a term is being multiplied on one side, it becomes division on the other side.

$\bullet$ If a term is being divided on one side, it becomes multiplication on the other side.

For example, in the equation $x + 5 = 12$, the term $+5$ is being added on the left side. When transposed to the right side, it becomes $-5$:

$x = 12 - 5$

$x = 7$

Therefore, if a term is being added on one side, it moves to the other side as subtraction.

The correct option is (B).

The given equation is $2x - 3 = 7$.

To solve for the variable $x$ using transposition, the general strategy is to first isolate the term containing the variable (the $2x$ term in this case) on one side of the equation, and then isolate the variable itself.

In the equation $2x - 3 = 7$, the term $-3$ is a constant term being subtracted from the variable term $2x$. To isolate the $2x$ term on the Left Hand Side (LHS), we need to move the constant term $-3$ to the Right Hand Side (RHS).

According to the rule of transposition, a term being subtracted on one side of the equation moves to the other side as addition.

So, transposing $-3$ from the LHS to the RHS, it becomes $+3$.

The equation becomes:

$2x = 7 + 3$

$2x = 10$

This is the equation after the first step of transposition. The next step would be to transpose the coefficient 2, which is multiplying $x$, to the RHS as division.

Therefore, the first step in solving the equation $2x - 3 = 7$ using transposition is to transpose $-3$ to the RHS.

The correct option is (A).

We are asked to solve the equation $5m + 1 = 16$ using transposition.

The given equation is:

$5m + 1 = 16$

To solve for $m$, we first transpose the constant term $+1$ from the Left Hand Side (LHS) to the Right Hand Side (RHS). When a term is transposed, its sign changes. Since $+1$ is being added on the LHS, it will be subtracted on the RHS.

Transpose $+1$ to the RHS:

$5m = 16 - 1$

Simplify the RHS:

$5m = 15$

Now, the coefficient $5$ is multiplying the variable $m$ on the LHS. To isolate $m$, we transpose $5$ to the RHS. Since $5$ is multiplying on the LHS, it will divide on the RHS.

Transpose $5$ to the RHS:

$m = \frac{15}{5}$

Perform the division:

$m = 3$

Thus, the solution to the equation $5m + 1 = 16$ is $m=3$.

The correct option is (B).

To translate the statement "5 added to a number is 18" into an equation, we first need to represent the unknown "a number" with a variable.

Let the unknown number be represented by the variable $x$.

The phrase "5 added to a number" means we are adding 5 to the number. This can be written as $x + 5$ or $5 + x$. Addition is commutative, so the order doesn't matter here.

The word "is" in a word problem often translates to an equality sign ($=$).

The number "18" is the result of the addition.

Combining these parts, we get the equation:

The number ($x$) plus 5 ($\text{added to}$) equals ($\text{is}$) 18.

$x + 5 = 18$

Comparing this equation with the given options:

(A) $5 + 18 = x$: This translates to "The sum of 5 and 18 is a number". Incorrect.

(B) $x + 5 = 18$: This translates to "A number plus 5 is 18" or "5 added to a number is 18". Correct.

(C) $5x = 18$: This translates to "5 times a number is 18" or "The product of 5 and a number is 18". Incorrect.

(D) $x - 5 = 18$: This translates to "A number minus 5 is 18" or "5 subtracted from a number is 18". Incorrect.

Therefore, the correct equation is $x + 5 = 18$.

The correct option is (B).

Let the unknown number be represented by the variable $n$.

The statement "twice a number" can be written as $2n$.

The statement "7 is subtracted from twice a number" can be written as $2n - 7$.

The statement "the result is 11" means that $2n - 7$ is equal to 11.

So, we can translate the given word problem into the following linear equation:

$2n - 7 = 11$

Now, we solve this equation for $n$. We can use the method of transposition.

First, transpose the constant term $-7$ from the LHS to the RHS. Since it is subtraction on the LHS, it becomes addition on the RHS.

$2n = 11 + 7$

$2n = 18$

Next, the coefficient $2$ is multiplying $n$ on the LHS. Transpose $2$ to the RHS. Since it is multiplication on the LHS, it becomes division on the RHS.

$n = \frac{18}{2}$

Performing the division:

$n = 9$

Thus, the number is 9.

To verify, substitute $n=9$ back into the original equation: $2(9) - 7 = 18 - 7 = 11$. This is true.

The correct option is (C).

Let the unknown number be represented by the variable $x$.

The phrase "Three-fourths of a number" translates to $\frac{3}{4}$ times the number, which is $\frac{3}{4}x$.

The word "is" translates to an equality sign ($=$).

The number given is 15.

So, the statement "Three-fourths of a number is 15" can be written as the equation:

$\frac{3}{4}x = 15$

To solve for $x$, we need to isolate $x$ on one side of the equation. The term $\frac{3}{4}$ is multiplying $x$. To remove this coefficient, we multiply both sides of the equation by the reciprocal of $\frac{3}{4}$, which is $\frac{4}{3}$.

Multiply both sides by $\frac{4}{3}$:

$\frac{4}{3} \times \frac{3}{4}x = 15 \times \frac{4}{3}$

Simplify both sides:

$\cancel{\frac{4}{3}} \times \cancel{\frac{3}{4}}x = \frac{\cancel{15}^{5} \times 4}{\cancel{3}_{1}}$

$x = 5 \times 4$

$x = 20$

Thus, the number is 20.

To check the answer, calculate three-fourths of 20: $\frac{3}{4} \times 20 = \frac{3 \times \cancel{20}^{5}}{\cancel{4}_{1}} = 3 \times 5 = 15$. This matches the given information.

The correct option is (C).

In an algebraic expression or equation, the coefficient of a variable is the numerical factor that multiplies the variable.

The given equation is $7 - 2p = 1$.

We can rewrite the left side of the equation to clearly see the terms:

$7 + (-2p) = 1$

The terms on the left side are $7$ and $-2p$.

The variable in this equation is $p$. The term containing the variable is $-2p$.

In the term $-2p$, the number that is multiplying the variable $p$ is $-2$.

Therefore, the coefficient of $p$ is $-2$.

Based on the structure of the question and options, we will determine the value that satisfies the equation first, and then identify which option is not that value.

We are given the equation $\frac{x}{2} + 3 = 7$. To find the solution, we need to isolate the variable $x$.

Subtract 3 from both sides of the equation:

$\frac{x}{2} + 3 - 3 = 7 - 3$

Simplify:

$\frac{x}{2} = 4$

Multiply both sides by 2:

$\frac{x}{2} \times 2 = 4 \times 2$

Simplify:

$x = 8$

The solution to the equation $\frac{x}{2} + 3 = 7$ is $x=8$.

The question asks which of the given options is NOT a solution. This means we are looking for a value among the options that is not equal to 8.

Let's check each option:

(A) $x=8$: This is the solution we found. $\frac{8}{2} + 3 = 4 + 3 = 7$. This satisfies the equation.

(B) $x=4$: Let's substitute $x=4$ into the equation: $\frac{4}{2} + 3 = 2 + 3 = 5$. Since $5 \neq 7$, $x=4$ is NOT a solution.

(C) $x=6$: Let's substitute $x=6$ into the equation: $\frac{6}{2} + 3 = 3 + 3 = 6$. Since $6 \neq 7$, $x=6$ is NOT a solution.

(D) $x=10$: Let's substitute $x=10$ into the equation: $\frac{10}{2} + 3 = 5 + 3 = 8$. Since $8 \neq 7$, $x=10$ is NOT a solution.

Since the question asks for a value that is NOT a solution, options (B), (C), and (D) are all technically correct as they are not equal to the solution $x=8$. However, in a standard multiple-choice question, there is usually only one correct answer among the options. Assuming there might be a specific intended answer among the non-solution options provided, we can state that option (B) is one such value that is not a solution.

The correct option is (B).

The correct option is (A).

We need to solve each equation to find its solution and then match it with the given options.

(i) $x + 4 = 9$

To solve for $x$, subtract 4 from both sides:

$x + 4 - 4 = 9 - 4$

$x = 5$

So, (i) matches with (c) $x=5$.

(ii) $3y = 21$

To solve for $y$, divide both sides by 3:

$\frac{3y}{3} = \frac{21}{3}$

$y = 7$

So, (ii) matches with (a) $y=7$.

(iii) $m - 2 = 6$

To solve for $m$, add 2 to both sides:

$m - 2 + 2 = 6 + 2$

$m = 8$

So, (iii) matches with (d) $m=8$.

(iv) $\frac{p}{5} = 3$

To solve for $p$, multiply both sides by 5:

$\frac{p}{5} \times 5 = 3 \times 5$

$p = 15$

So, (iv) matches with (b) $p=15$.

The correct matching is: (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b).

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

The correct option is (B).

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): To solve the equation $x - 10 = 4$, we should add 10 to both sides.

The equation is $x - 10 = 4$. To isolate $x$, we need to cancel out the $-10$ on the left side. The inverse operation of subtracting 10 is adding 10. If we add 10 to both sides, we get $x - 10 + 10 = 4 + 10$, which simplifies to $x = 14$. This is the correct step to solve the equation.

Thus, Assertion (A) is true.

Reason (R): Subtracting the same number from both sides of an equation maintains the equality.

This statement describes one of the properties of equality. If we have an equation $a = b$, then subtracting the same number $c$ from both sides gives $a - c = b - c$, and the equality holds. This property is correct.

Thus, Reason (R) is true.

Now, let's check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) describes the step of adding 10 to both sides to solve the equation $x - 10 = 4$. The principle used here is that adding the same number to both sides of an equation maintains equality ($a = b \implies a + c = b + c$). Reason (R) describes the property of subtracting the same number from both sides. While both are true properties of equality, Reason (R) does not explain why adding 10 is the correct step in Assertion (A). The correct explanation for A would be the property of adding the same number to both sides.

Therefore, both A and R are true statements, but R is not the correct explanation of A.

The correct option is (A).

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The equation $2x = 8$ is equivalent to the equation $x + 1 = 5$.

Two equations are equivalent if they have the same solution set.

Let's solve the first equation, $2x = 8$.

Divide both sides by 2:

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

$x = 4$

The solution to the first equation is $x=4$.

Let's solve the second equation, $x + 1 = 5$.

Subtract 1 from both sides:

$x + 1 - 1 = 5 - 1$

$x = 4$

The solution to the second equation is also $x=4$.

Since both equations have the same solution ($x=4$), they are equivalent.

Thus, Assertion (A) is true.

Reason (R): Equivalent equations have the same solution.

This statement is the definition of equivalent equations. By definition, equivalent equations are equations that have exactly the same solution(s).

Thus, Reason (R) is true.

Now let's consider if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) claims that two specific equations are equivalent. Reason (R) provides the fundamental property that defines equivalent equations – having the same solution. To verify Assertion (A), we would indeed check if the two equations have the same solution, which directly uses the concept stated in Reason (R).

Therefore, Reason (R) correctly explains why Assertion (A) is true.

The correct option is (B).

Let $n$ be the number of ice creams Rakesh bought.

The cost of each ice cream is $\textsf{₹ }15$.

The total cost of $n$ ice creams before applying the discount is the number of ice creams multiplied by the cost per ice cream, which is $15 \times n = 15n$.

Rakesh has a discount coupon of $\textsf{₹ }10$. A discount reduces the total cost. So, the amount paid after the discount is the cost before the discount minus the discount amount.

Amount paid = (Cost before discount) - (Discount amount)

Amount paid = $15n - 10$

The problem states that the total amount he paid is $\textsf{₹ }80$.

So, we can set the expression for the amount paid equal to 80:

$15n - 10 = 80$

This equation represents the given situation.

Let's briefly look at why other options are incorrect:

(A) $15n + 10 = 80$: This would imply a $\textsf{₹ }10$ charge is added to the cost, not a discount.

(C) $10n + 15 = 80$: This suggests the ice cream costs $\textsf{₹ }10$ and there is an additional $\textsf{₹ }15$ charge.

(D) $10n - 15 = 80$: This suggests the ice cream costs $\textsf{₹ }10$ and there is a $\textsf{₹ }15$ discount.

The equation that correctly represents the situation is $15n - 10 = 80$.

The correct option is (C).

From the case study in Question 18, we established the equation representing the situation where Rakesh bought $n$ ice creams, each costing $\textsf{₹ }15$, with a $\textsf{₹ }10$ discount, and paid a total of $\textsf{₹ }80$.

The equation is:

$15n - 10 = 80$

To find the number of ice creams Rakesh bought, we need to solve this equation for $n$.

First, we transpose the constant term $-10$ from the Left Hand Side (LHS) to the Right Hand Side (RHS). When a term is transposed, its sign changes. So, $-10$ becomes $+10$ on the RHS.

$15n = 80 + 10$

Simplify the RHS:

$15n = 90$

Now, the coefficient $15$ is multiplying $n$ on the LHS. To isolate $n$, we transpose $15$ to the RHS. Since $15$ is multiplying on the LHS, it becomes division on the RHS.

$n = \frac{90}{15}$

Perform the division:

$n = 6$

Therefore, Rakesh bought 6 ice creams.

We can check this answer: If Rakesh bought 6 ice creams, the cost before discount is $15 \times 6 = \textsf{₹ }90$. After applying the $\textsf{₹ }10$ discount, the amount paid is $90 - 10 = \textsf{₹ }80$, which matches the information given in the case study.

The correct option is (A).

We are asked to solve the equation $3x + 2 = 14$.

To solve for $x$, we need to isolate the variable term ($3x$) first, and then isolate the variable ($x$).

The given equation is:

$3x + 2 = 14$

Subtract 2 from both sides of the equation to eliminate the constant term on the left side:

$3x + 2 - 2 = 14 - 2$

Simplify both sides:

$3x = 12$

Now, the variable $x$ is multiplied by 3. To isolate $x$, divide both sides of the equation by 3:

$\frac{3x}{3} = \frac{12}{3}$

Simplify both sides:

$x = 4$

The solution to the equation $3x + 2 = 14$ is $x=4$.

Let's check the answer by substituting $x=4$ back into the original equation:

$3(4) + 2 = 12 + 2 = 14$

The left side equals the right side, so the solution is correct.

The correct option is (B).

We are asked to solve the equation $5(a - 2) = 15$.

There are two common methods to solve this equation:

Method 1: Using Division First

The entire expression $(a - 2)$ is multiplied by 5. We can divide both sides of the equation by 5 to isolate the term $(a - 2)$.

$5(a - 2) = 15$

Divide both sides by 5:

$\frac{5(a - 2)}{5} = \frac{15}{5}$

Simplify both sides:

$a - 2 = 3$

Now, add 2 to both sides to isolate $a$:

$a - 2 + 2 = 3 + 2$

$a = 5$

Method 2: Using the Distributive Property First

We can first distribute the 5 to the terms inside the parentheses on the left side.

$5(a - 2) = 15$

Apply the distributive property ($5 \times a$ and $5 \times -2$):

$5a - 10 = 15$

Now, add 10 to both sides to isolate the term with the variable:

$5a - 10 + 10 = 15 + 10$

$5a = 25$

Finally, divide both sides by 5 to isolate $a$:

$\frac{5a}{5} = \frac{25}{5}$

$a = 5$

Both methods give the same solution: $a=5$.

To check the answer, substitute $a=5$ back into the original equation: $5(5 - 2) = 5(3) = 15$. This is true.

The correct option is (B).

To determine which equation has $x=4$ as its solution, we need to substitute $x=4$ into each equation and check if the equation holds true.

(A) $x + 2 = 7$

Substitute $x=4$: $4 + 2 = 6$.

Is $6 = 7$? No.

So, $x=4$ is not the solution for this equation.

(B) $2x - 3 = 5$

Substitute $x=4$: $2(4) - 3 = 8 - 3 = 5$.

Is $5 = 5$? Yes.

So, $x=4$ is the solution for this equation.

(C) $\frac{x}{2} + 1 = 4$

Substitute $x=4$: $\frac{4}{2} + 1 = 2 + 1 = 3$.

Is $3 = 4$? No.

So, $x=4$ is not the solution for this equation.

(D) $x - 1 = 2$

Substitute $x=4$: $4 - 1 = 3$.

Is $3 = 2$? No.

So, $x=4$ is not the solution for this equation.

Only equation (B) is satisfied when $x=4$. Therefore, the equation that has $x=4$ as its solution is $2x - 3 = 5$.

The correct option is (A).

Let the smallest of the three consecutive integers be represented by the variable $n$.

Since the integers are consecutive, the next integer is $n+1$, and the third consecutive integer is $(n+1)+1 = n+2$.

The three consecutive integers are $n$, $n+1$, and $n+2$.

The problem states that the sum of these three integers is 33.

So, we can write the equation:

$n + (n + 1) + (n + 2) = 33$

Now, we need to solve this equation for $n$. Combine the terms with $n$ and the constant terms on the left side:

$(n + n + n) + (1 + 2) = 33$

$3n + 3 = 33$

Now, solve this linear equation for $n$. Subtract 3 from both sides:

$3n + 3 - 3 = 33 - 3$

$3n = 30$

Divide both sides by 3:

$\frac{3n}{3} = \frac{30}{3}$

$n = 10$

The variable $n$ represents the smallest integer. So, the smallest integer is 10.

The three consecutive integers are 10, $10+1=11$, and $10+2=12$.

Let's check their sum: $10 + 11 + 12 = 33$. This is correct.

The question asks for the smallest integer, which is $n=10$.

Let's check the options:

(A) 10: If the smallest is 10, the integers are 10, 11, 12. Sum is 33. Correct.

(B) 11: If the smallest is 11, the integers are 11, 12, 13. Sum is 36. Incorrect.

(C) 12: If the smallest is 12, the integers are 12, 13, 14. Sum is 39. Incorrect.

(D) 9: If the smallest is 9, the integers are 9, 10, 11. Sum is 30. Incorrect.

The correct option is (B).

We are given the equation:

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

We are asked to multiply both sides of this equation by 3.

Multiply the Left Hand Side (LHS) by 3:

$\frac{x}{3} \times 3$

Multiply the Right Hand Side (RHS) by 3:

$6 \times 3$

Now, set the results equal to each other to form the new equation:

$\frac{x}{3} \times 3 = 6 \times 3$

Simplify the LHS: The 3 in the numerator and the 3 in the denominator cancel out.

$x = 6 \times 3$

Simplify the RHS:

$x = 18$

The result of multiplying both sides of the equation $\frac{x}{3} = 6$ by 3 is $x = 18$.

The correct option is (C).

An equation is a mathematical statement that asserts that two expressions are equal. It is denoted by the equality sign ($=$).

A simple equation typically involves a variable and constants, related by arithmetic operations, and the entire expression on one side of the equality sign is set to be equal to the expression on the other side.

For example, in the equation $x + 5 = 12$, the expression on the left side is $x + 5$, and the expression on the right side is $12$. The equality sign indicates that the value of the expression $x + 5$ is the same as the value of the expression $12$ for the specific value of $x$ that is the solution to the equation.

Therefore, in a simple equation, the expression on the left side must be equal to the expression on the right side.

The correct option is (B).

Let the unknown number be represented by the variable $n$.

The phrase "A number increased by 10" means we are adding 10 to the number. This can be written as $n + 10$.

The word "is" translates to an equality sign ($=$).

The number given is 25.

So, the statement "A number increased by 10 is 25" can be written as the equation:

$n + 10 = 25$

To solve for $n$, we need to isolate $n$ on one side of the equation. Subtract 10 from both sides:

$n + 10 - 10 = 25 - 10$

$n = 15$

The number is 15.

To verify, substitute $n=15$ back into the original statement: "15 increased by 10 is $15 + 10 = 25$". This is correct.

The correct option is (B).

The perimeter of a polygon is the total length of its boundary. A square has four sides of equal length.

Let the side length of the square be $s$ cm.

The perimeter of the square is the sum of the lengths of its four sides:

Perimeter = $s + s + s + s$

Perimeter = $4s$

We are given that the perimeter of the square is 36 cm.

So, we can set the expression for the perimeter equal to 36:

$4s = 36$

This equation represents the given situation.

Let's look at the other options:

(A) $s + 4 = 36$: This suggests that adding 4 to the side length gives the perimeter, which is incorrect for a square.

(C) $s^2 = 36$: This represents the area of the square, not the perimeter.

(D) $2s + 2(4) = 36$: This formula ($2l + 2w$) is for the perimeter of a rectangle. If the width were fixed at 4, this would represent a rectangle with length $s$ and width 4.

Therefore, the equation that represents the perimeter of a square with side length $s$ and perimeter 36 cm is $4s = 36$.

The correct option is (A).

We are asked to solve the equation $7 - x = 2$.

To solve for $x$, we need to isolate the variable on one side of the equation.

The given equation is:

$7 - x = 2$

One way to solve this is to isolate the term with $x$ first. Subtract 7 from both sides of the equation:

$7 - x - 7 = 2 - 7$

Simplify both sides:

$-x = -5$

Now, to find $x$, we multiply or divide both sides by $-1$:

$-x \times (-1) = -5 \times (-1)$

$x = 5$

Alternatively, we can move the variable term to the other side to make it positive. Add $x$ to both sides:

$7 - x + x = 2 + x$

$7 = 2 + x$

Now, subtract 2 from both sides to isolate $x$:

$7 - 2 = 2 + x - 2$

$5 = x$

Both methods give the same solution: $x=5$.

Let's check the answer by substituting $x=5$ back into the original equation:

$7 - 5 = 2$

The left side equals the right side, so the solution is correct.

The correct option is (D).

The given equation is $3x = 12$. This means "3 times $x$ equals 12".

To solve for the variable $x$, we need to isolate it on one side of the equation.

The operation being performed on $x$ is multiplication by 3.

To undo multiplication, we perform the inverse operation, which is division.

We must perform the same operation on both sides of the equation to maintain equality.

Therefore, we should divide both sides of the equation by 3.

Dividing both sides by 3 gives:

$\frac{3x}{3} = \frac{12}{3}$

This simplifies to:

$x = 4$

Let's consider why the other options are incorrect:

(A) Adding 3 to both sides: $3x + 3 = 12 + 3 \implies 3x + 3 = 15$. This does not isolate $x$.

(B) Subtracting 3 from both sides: $3x - 3 = 12 - 3 \implies 3x - 3 = 9$. This does not isolate $x$.

(C) Multiplying both sides by 3: $3x \times 3 = 12 \times 3 \implies 9x = 36$. This changes the coefficient of $x$ but does not isolate $x$.

The correct step to solve $3x = 12$ is to divide both sides by 3.

The correct option is (A).

Transposition is a method used to solve linear equations by moving terms from one side of the equality sign to the other.

When a term is moved (transposed) from one side of an equation to the other, the operation associated with that term is reversed, which results in the change of its sign (for additive or subtractive terms) or the operation for multiplicative or divisive terms.

Specifically:

$\bullet$ An additive term ($+a$) on one side becomes a subtractive term ($-a$) on the other side.

$\bullet$ A subtractive term ($-a$) on one side becomes an additive term ($+a$) on the other side.

For example, in the equation $10 = x + 3$, if we transpose the $+3$ from the RHS to the LHS, it becomes $-3$: $10 - 3 = x$, which simplifies to $7 = x$. The sign of the term 3 changed from positive to negative.

In the equation $10 = x - 3$, if we transpose the $-3$ from the RHS to the LHS, it becomes $+3$: $10 + 3 = x$, which simplifies to $13 = x$. The sign of the term 3 changed from negative to positive.

While the operation also reverses for multiplication and division (multiplication becomes division, and division becomes multiplication), the question specifically asks about the "sign". For terms being added or subtracted, the sign *changes* when transposed.

Therefore, if you transpose a term from the RHS to the LHS, its sign changes.

The correct option is (C).

Let the unknown number be represented by the variable $x$.

According to the problem statement:

"you multiply it by 5" means $5x$.

"and subtract 7" means $5x - 7$.

"you get 13" means the result is equal to 13.

So, we can write the equation that represents this situation:

$5x - 7 = 13$

Now, we need to solve this equation for $x$.

To isolate the term with the variable ($5x$), add 7 to both sides of the equation:

$5x - 7 + 7 = 13 + 7$

Simplify both sides:

$5x = 20$

Now, the variable $x$ is multiplied by 5. To isolate $x$, divide both sides of the equation by 5:

$\frac{5x}{5} = \frac{20}{5}$

Simplify both sides:

$x = 4$

The number is 4.

To check the answer, substitute $x=4$ back into the original statement: "multiply 4 by 5" gives $5 \times 4 = 20$. "subtract 7" gives $20 - 7 = 13$. This matches the result given in the problem.

The correct option is (A).

To determine which equations have the same solution, we need to solve each equation for the variable $x$.

Equation 1: $x + 6 = 10$

Subtract 6 from both sides of the equation:

$x + 6 - 6 = 10 - 6$

$x = 4$

The solution for Equation 1 is $x = 4$.

Equation 2: $2x = 8$

Divide both sides of the equation by 2:

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

$x = 4$

The solution for Equation 2 is $x = 4$.

Equation 3: $x - 1 = 2$

Add 1 to both sides of the equation:

$x - 1 + 1 = 2 + 1$

$x = 3$

The solution for Equation 3 is $x = 3$.

Comparing the solutions:

Equation 1: $x = 4$

Equation 2: $x = 4$

Equation 3: $x = 3$

Equations 1 and 2 have the same solution ($x=4$), while Equation 3 has a different solution ($x=3$).

Therefore, Equations 1 and 2 have the same solution.

The correct option is (A).

Let the cost of the notebook be $\textsf{₹ }x$.

Let the cost of the pen be $\textsf{₹ }p$. We are given that $p = \textsf{₹ }30$.

The statement is "The cost of a pen is $\textsf{₹ }10$ less than the cost of a notebook".

This can be translated as:

(Cost of pen) = (Cost of notebook) - 10

Substitute the given values:

$30 = x - 10$

We can rewrite this equation with the variable term on the left side:

$x - 10 = 30$

This equation represents the relationship described in the problem.

Let's examine the other options:

(B) $10 - x = 30$: This would mean the cost of the notebook is $\textsf{₹ }10$ less than something equal to 30, which contradicts the statement.

(C) $x + 10 = 30$: This would mean the cost of the pen is $\textsf{₹ }10$ more than the cost of the notebook.

(D) $10x = 30$: This would mean 10 times the cost of the notebook is $\textsf{₹ }30$.

The equation that represents the relationship is $x - 10 = 30$.

The correct option is (B).

We are given the equation $2(x+3) = 10$.

The question asks for the value of the expression $x+3$.

In the given equation, the term $(x+3)$ is being multiplied by 2 on the left side.

To find the value of $(x+3)$, we can isolate this expression by performing the inverse operation of multiplying by 2, which is dividing by 2.

We must divide both sides of the equation by 2 to maintain the equality.

Divide both sides by 2:

$\frac{2(x+3)}{2} = \frac{10}{2}$

Simplify both sides:

$\cancel{\frac{2}{2}}(x+3) = 5$

$x+3 = 5$

Thus, the value of $x+3$ is 5.

We can also find the value of $x$ first and then calculate $x+3$. From $x+3 = 5$, subtract 3 from both sides: $x = 5 - 3 = 2$. Then $x+3 = 2+3 = 5$. This confirms the result.

The correct option is (C).

We are asked to find the value of $y$ that satisfies the equation $\frac{y}{4} - 1 = 2$.

The given equation is:

$\frac{y}{4} - 1 = 2$

To isolate the term containing the variable, $\frac{y}{4}$, we first add 1 to both sides of the equation:

$\frac{y}{4} - 1 + 1 = 2 + 1$

Simplify both sides:

$\frac{y}{4} = 3$

Now, to isolate $y$, which is being divided by 4, we multiply both sides of the equation by 4:

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

Simplify both sides:

$y = 12$

The value of $y$ that satisfies the equation is 12.

Let's check the answer by substituting $y=12$ back into the original equation:

$\frac{12}{4} - 1 = 3 - 1 = 2$

The left side equals the right side, so the solution is correct.

The correct option is (B).

An equilateral triangle is a triangle with all three sides of equal length.

Let the side length of the equilateral triangle be $s$ cm.

The perimeter of a triangle is the sum of the lengths of its three sides.

For an equilateral triangle with side length $s$, the perimeter is:

Perimeter = $s + s + s$

Perimeter = $3s$

We are given that the perimeter of the equilateral triangle is 18 cm.

So, we can set the expression for the perimeter equal to 18:

$3s = 18$

This equation represents the given situation.

Let's look at the other options:

(A) $3+s=18$: This suggests that adding 3 to the side length gives the perimeter, which is incorrect.

(C) $s/3=18$: This suggests that dividing the side length by 3 gives the perimeter, which is incorrect.

(D) $s^3=18$: This does not represent the perimeter of a triangle.

The equation that represents the perimeter of an equilateral triangle with side length $s$ and perimeter 18 cm is $3s = 18$.

The correct option is (C).

When solving an equation, we perform operations on both sides to isolate the variable while maintaining the equality of the equation. The fundamental properties of equality allow us to perform certain operations on both sides without changing the solution set.

Let's examine each option:

(A) Adding the same number to both sides: This is a valid step. If $a = b$, then $a + c = b + c$. This property is frequently used to move constant terms across the equality sign.

(B) Subtracting the same number from both sides: This is also a valid step. If $a = b$, then $a - c = b - c$. This is the inverse operation of adding and is also used to move constant terms.

(C) Multiplying both sides by zero: While multiplying both sides by zero results in a true statement ($0 = 0$), it is generally not a correct or useful step in the process of finding the specific solution(s) for the variable in an equation. Multiplying by zero can eliminate the variable entirely (if it appears with a coefficient) and reduces the equation to a trivial identity ($0=0$), which is true for any value of the variable. This prevents us from determining the unique value of the variable that satisfies the original equation (unless the original equation was $0=0$ or similar).

(D) Dividing both sides by the same non-zero number: This is a valid and essential step for isolating a variable when it is multiplied by a non-zero coefficient. If $a = b$ and $c \neq 0$, then $\frac{a}{c} = \frac{b}{c}$. It is crucial that the number is non-zero because division by zero is undefined.

Therefore, multiplying both sides of an equation by zero is NOT a correct step in solving for the variable, as it typically obscures or eliminates the solution.

A linear equation in one variable is an equation that can be written in the form $ax + b = 0$, where $x$ is the only variable, and $a$ and $b$ are real numbers with $a \neq 0$. The highest power of the variable in such an equation is always 1.

For example, $2x + 5 = 0$ is a linear equation in one variable ($x$), where $a=2$ and $b=5$.

In the equation $3x - 5 = 10$:

The variable is $x$.

The Left Hand Side (LHS) is $3x - 5$.

The Right Hand Side (RHS) is $10$.

To check if $x = 4$ is a solution to the equation $2x + 3 = 11$, we substitute $x = 4$ into the Left Hand Side (LHS) of the equation.

LHS $= 2x + 3$

Substitute $x = 4$:

LHS $= 2(4) + 3$

Evaluate the expression:

LHS $= 8 + 3$

LHS $= 11$

The Right Hand Side (RHS) of the equation is $11$.

Since LHS $=$ RHS ($11 = 11$), the value $x = 4$ satisfies the equation.

Therefore, $x = 4$ is a solution to the equation $2x + 3 = 11$.

The statement "The sum of a number $y$ and $7$ is $15$" can be written as an equation as follows:

The sum of a number $y$ and $7$ is represented by $y + 7$.

The word "is" translates to an equals sign ($=$).

Therefore, the equation is:

$y + 7 = 15$

The statement "Four times a number $p$ is $28$" can be written as an equation as follows:

"Four times a number $p$" is represented by multiplying the number $p$ by $4$, which gives $4p$.

"is $28$" translates to an equals sign ($=$) followed by the number $28$.

Therefore, the equation is:

$4p = 28$

The equation $a - 5 = 12$ can be written as a statement in words.

The term $a - 5$ represents "5 subtracted from a number $a$" or "a number $a$ decreased by 5" or "the difference between a number $a$ and 5".

The symbol $=$ means "is" or "equals".

The number on the right side is $12$.

Combining these, the statement is:

"5 subtracted from a number $a$ is 12."

Alternatively, it can be stated as:

"A number $a$ decreased by 5 is 12."

The equation $\frac{m}{3} = 9$ can be written as a statement in words.

The term $\frac{m}{3}$ represents "a number $m$ divided by 3" or "one-third of a number $m$".

The symbol $=$ means "is" or "equals".

The number on the right side is $9$.

Combining these, the statement is:

"A number $m$ divided by 3 is 9."

Alternatively, it can be stated as:

"One-third of a number $m$ is 9."

Given equation:

$x + 8 = 20$

To isolate $x$ on the Left Hand Side (LHS), we need to remove the '+ 8'. We can do this by subtracting 8 from both sides of the equation. This is the balancing method, as we keep the equation balanced by performing the same operation on both sides.

Subtract 8 from both sides:

$(x + 8) - 8 = 20 - 8$

Simplify both sides:

$x + (8 - 8) = 12$

$x + 0 = 12$

$x = 12$

The solution to the equation $x + 8 = 20$ is $x = 12$.

Given equation:

$y - 6 = 5$

To isolate $y$ on the Left Hand Side (LHS), we need to remove the '- 6'. We can do this by adding 6 to both sides of the equation. This is the balancing method, as we keep the equation balanced by performing the same operation on both sides.

Add 6 to both sides:

$(y - 6) + 6 = 5 + 6$

Simplify both sides:

$y + (-6 + 6) = 11$

$y + 0 = 11$

$y = 11$

The solution to the equation $y - 6 = 5$ is $y = 11$.

Given equation:

$4a = 24$

To isolate $a$ on the Left Hand Side (LHS), we need to remove the multiplication by 4. We can do this by dividing both sides of the equation by 4. This is the balancing method, as we keep the equation balanced by performing the same operation on both sides.

Divide both sides by 4:

$\frac{4a}{4} = \frac{24}{4}$

Simplify both sides:

$\frac{\cancel{4}a}{\cancel{4}} = \frac{\cancel{24}^{6}}{\cancel{4}_{1}}$

$a = 6$

The solution to the equation $4a = 24$ is $a = 6$.

Given equation:

$\frac{b}{5} = 7$

To isolate $b$ on the Left Hand Side (LHS), we need to remove the division by 5. We can do this by multiplying both sides of the equation by 5. This is the balancing method, as we keep the equation balanced by performing the same operation on both sides.

Multiply both sides by 5:

$\left(\frac{b}{5}\right) \times 5 = 7 \times 5$

Simplify both sides:

$\frac{b}{\cancel{5}} \times \cancel{5} = 35$

$b = 35$

The solution to the equation $\frac{b}{5} = 7$ is $b = 35$.

The rule of transposition is a convenient way to think about moving terms from one side of an equation to the other while solving it. It is essentially a shortcut for the balancing method.

For addition and subtraction, the rule states that if a term is being added on one side of the equation, it can be moved to the other side by changing its sign to subtraction. Conversely, if a term is being subtracted on one side, it can be moved to the other side by changing its sign to addition.

In simple terms:

+ term on LHS $\to$ - term on RHS

- term on LHS $\to$ + term on RHS

Example:

Let's solve the equation $x + 5 = 10$ using the rule of transposition.

We want to isolate $x$. The term $+5$ is being added to $x$ on the LHS.

According to the rule of transposition, we can move the $+5$ to the RHS by changing its sign to $-5$.

$x + 5 = 10$

$x = 10 - 5$

Simplify the RHS:

$x = 5$

Similarly, let's solve the equation $y - 3 = 7$ using the rule of transposition.

We want to isolate $y$. The term $-3$ is being subtracted from $y$ on the LHS.

According to the rule of transposition, we can move the $-3$ to the RHS by changing its sign to $+3$.

$y - 3 = 7$

$y = 7 + 3$

Simplify the RHS:

$y = 10$

The rule of transposition provides a simplified way to move terms involving multiplication or division from one side of an equation to the other.

For multiplication and division, the rule states that if a term is multiplying the variable on one side of the equation, it can be moved to the other side by changing the operation to division. Conversely, if a term is dividing the variable on one side, it can be moved to the other side by changing the operation to multiplication.

In simple terms:

Multiply on LHS $\to$ Divide on RHS

Divide on LHS $\to$ Multiply on RHS

Example (Multiplication):

Let's solve the equation $3x = 12$ using the rule of transposition.

We want to isolate $x$. The number $3$ is multiplying $x$ on the LHS.

According to the rule of transposition, we can move the $3$ to the RHS by changing the operation from multiplication to division.

$3x = 12$

$x = \frac{12}{3}$

Simplify the RHS:

$x = 4$

Example (Division):

Let's solve the equation $\frac{y}{4} = 5$ using the rule of transposition.

We want to isolate $y$. The number $4$ is dividing $y$ on the LHS.

According to the rule of transposition, we can move the $4$ to the RHS by changing the operation from division to multiplication.

$\frac{y}{4} = 5$

$y = 5 \times 4$

Simplify the RHS:

$y = 20$

Given equation:

$z + 10 = -3$

We want to solve for $z$. The term $+10$ is being added to $z$ on the Left Hand Side (LHS) of the equation.

Using the rule of transposition for addition and subtraction, we can move the term $+10$ from the LHS to the Right Hand Side (RHS) by changing its sign from positive to negative.

Move $+10$ to the RHS:

$z = -3 - 10$

Perform the subtraction on the RHS:

$z = -13$

The solution to the equation $z + 10 = -3$ is $z = -13$.

Given equation:

$t - (-4) = 10$

First, simplify the equation by dealing with the double negative. Subtracting a negative number is the same as adding the positive number.

$t + 4 = 10$

We want to solve for $t$. The term $+4$ is being added to $t$ on the Left Hand Side (LHS) of the equation.

Using the rule of transposition for addition and subtraction, we can move the term $+4$ from the LHS to the Right Hand Side (RHS) by changing its sign from positive to negative.

Move $+4$ to the RHS:

$t = 10 - 4$

Perform the subtraction on the RHS:

$t = 6$

The solution to the equation $t - (-4) = 10$ is $t = 6$.

Given equation:

$7m = -49$

We want to solve for $m$. The number $7$ is multiplying $m$ on the Left Hand Side (LHS) of the equation.

Using the rule of transposition for multiplication and division, we can move the multiplier $7$ from the LHS to the Right Hand Side (RHS) by changing the operation from multiplication to division.

Move $7$ to the RHS as a divisor:

$m = \frac{-49}{7}$

Perform the division on the RHS:

$m = -7$

The solution to the equation $7m = -49$ is $m = -7$.

Given equation:

$\frac{n}{-2} = 6$

We want to solve for $n$. The variable $n$ is being divided by $-2$ on the Left Hand Side (LHS) of the equation.

Using the rule of transposition for multiplication and division, we can move the divisor $-2$ from the LHS to the Right Hand Side (RHS) by changing the operation from division to multiplication.

Move $-2$ to the RHS as a multiplier:

$n = 6 \times (-2)$

Perform the multiplication on the RHS:

$n = -12$

The solution to the equation $\frac{n}{-2} = 6$ is $n = -12$.

Let the unknown number be represented by the variable $x$.

The phrase "a number increased by $12$" means we add $12$ to the number. This can be written as $x + 12$.

The word "is" in the statement represents the equals sign $(=)$.

The number on the right side of the equation is $30$.

Combining these parts, we get the equation:

$x + 12 = 30$

Let the cost of one pen be $\textsf{₹}x$.

The cost of $5$ pens will be $5$ times the cost of one pen.

Cost of $5$ pens $= 5 \times (\text{cost of one pen})$

Cost of $5$ pens $= 5 \times \textsf{₹}x$

Cost of $5$ pens $= \textsf{₹}5x$

According to the problem, the cost of $5$ pens is given as $\textsf{₹}75$.

Therefore, we can set the expression for the cost of $5$ pens equal to the given total cost:

$\textsf{₹}5x = \textsf{₹}75$

Dropping the unit symbol $\textsf{₹}$, the equation is:

$5x = 75$

Given equation:

$2x + 1 = 9$

We want to isolate the term containing the variable $x$ on one side of the equation. We can start by moving the constant term $1$ from the Left Hand Side (LHS) to the Right Hand Side (RHS). The term $1$ is being added on the LHS, so when we move it to the RHS, its sign changes from positive to negative (using the rule of transposition).

Transpose $1$ to the RHS:

$2x = 9 - 1$

Simplify the RHS:

$2x = 8$

Now, the variable $x$ is being multiplied by $2$ on the LHS. To isolate $x$, we need to move the multiplier $2$ from the LHS to the RHS. When a term is multiplying on one side, it divides on the other side (using the rule of transposition).

Transpose $2$ to the RHS as a divisor:

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

Perform the division:

$x = 4$

The solution to the equation $2x + 1 = 9$ is $x = 4$.

Given equation:

$3y - 4 = 11$

To solve for $y$, we first isolate the term containing $y$. The term $-4$ is being subtracted on the Left Hand Side (LHS). Using the rule of transposition, we move it to the Right Hand Side (RHS) by changing its sign to positive.

Transpose $-4$ to the RHS:

$3y = 11 + 4$

Simplify the RHS:

$3y = 15$

Now, the variable $y$ is being multiplied by $3$ on the LHS. To isolate $y$, we move the multiplier $3$ from the LHS to the RHS. Using the rule of transposition, a multiplier on one side becomes a divisor on the other side.

Transpose $3$ to the RHS as a divisor:

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

Perform the division:

$y = 5$

The solution to the equation $3y - 4 = 11$ is $y = 5$.

Given equation:

$5p + 2 = 17$

To find the value of $p$, we need to isolate $p$ on one side of the equation. First, transpose the constant term $+2$ from the Left Hand Side (LHS) to the Right Hand Side (RHS). When a term is moved across the equals sign, its sign changes.

Transpose $+2$ to the RHS:

$5p = 17 - 2$

Simplify the RHS:

$5p = 15$

Now, the variable $p$ is being multiplied by $5$ on the LHS. To isolate $p$, transpose the multiplier $5$ from the LHS to the RHS. A multiplier on one side becomes a divisor on the other side.

Transpose $5$ to the RHS as a divisor:

$p = \frac{15}{5}$

Perform the division:

$p = 3$

The value of $p$ in the equation $5p + 2 = 17$ is $3$.

Given equation:

$\frac{x}{2} - 3 = 4$

To find the value of $x$, we first isolate the term containing $x$. The constant term $-3$ is being subtracted on the Left Hand Side (LHS). Using the rule of transposition, we move $-3$ to the Right Hand Side (RHS) by changing its sign from negative to positive.

Transpose $-3$ to the RHS:

$\frac{x}{2} = 4 + 3$

Simplify the RHS:

$\frac{x}{2} = 7$

Now, the variable $x$ is being divided by $2$ on the LHS. To isolate $x$, we move the divisor $2$ from the LHS to the RHS. Using the rule of transposition, a divisor on one side becomes a multiplier on the other side.

Transpose $2$ to the RHS as a multiplier:

$x = 7 \times 2$

Perform the multiplication:

$x = 14$

The value of $x$ in the equation $\frac{x}{2} - 3 = 4$ is $14$.

Given equation: $3x - 8 = 7$

Given value of $x$: $x = 5$

To check if $x=5$ satisfies the equation, we substitute $x=5$ into the Left Hand Side (LHS) of the equation and see if it equals the Right Hand Side (RHS).

LHS $= 3x - 8$

Substitute $x = 5$ into the LHS:

LHS $= 3(5) - 8$

Perform the multiplication and subtraction:

LHS $= 15 - 8$

LHS $= 7$

The Right Hand Side (RHS) of the given equation is $7$.

Since LHS $=$ RHS ($7 = 7$), the value $x=5$ makes the equation true.

Therefore, $x=5$ satisfies the equation $3x - 8 = 7$.

We are asked to translate the given statement into a mathematical equation.

Let the unknown number be represented by the variable $k$, as stated in the problem.

The phrase "three times a number $k$" means we multiply the number $k$ by $3$. This can be written as $3k$.

The phrase "Seven subtracted from three times a number $k$" means we subtract $7$ from the expression "$3k$". This is written as $3k - 7$.

The word "is" in the statement represents the equals sign $(=)$.

The number on the right side of the equation is $11$.

Combining these parts, we get the equation:

$3k - 7 = 11$

Explanation of the Balancing Method:

The balancing method is a technique used to solve equations by keeping both sides of the equation balanced, much like a weighing scale. The fundamental principle is that if you perform an operation on one side of the equation, you must perform the exact same operation on the other side to maintain equality.

The goal is to isolate the variable (get the variable by itself) on one side of the equation. To do this, we systematically eliminate terms from the side containing the variable by applying inverse operations:

- To remove a term that is added, subtract it from both sides.

- To remove a term that is subtracted, add it to both sides.

- To remove a coefficient that is multiplying the variable, divide both sides by that coefficient.

- To remove a divisor that is dividing the variable, multiply both sides by that divisor.

By applying these operations symmetrically to both sides, we ensure that the equation remains true at each step until the variable is isolated, revealing its value.

Solving the equation $3x + 5 = 14$ using the balancing method:

Given equation:

$3x + 5 = 14$

Our aim is to isolate the variable $x$. The terms on the LHS are $3x$ and $+5$. We first remove the constant term, $+5$. To do this, we perform the inverse operation of adding 5, which is subtracting 5.

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

$(3x + 5) - 5 = 14 - 5$

Simplify both sides:

$3x + (5 - 5) = 9$

$3x + 0 = 9$

$3x = 9$

Now, the term containing the variable is $3x$. This means $3$ is multiplying $x$. To isolate $x$, we perform the inverse operation of multiplying by 3, which is dividing by 3.

Step 2: Divide both sides of the equation by 3.

$\frac{3x}{3} = \frac{9}{3}$

Simplify both sides:

$\frac{\cancel{3}x}{\cancel{3}} = \frac{\cancel{9}^{3}}{\cancel{3}_{1}}$

$x = 3$

The variable $x$ is now isolated on the LHS, and the solution is on the RHS.

Thus, the solution to the equation $3x + 5 = 14$ is $x = 3$.

Explanation of the Rule of Transposition:

The rule of transposition is a convenient method for solving linear equations. It is derived from the balancing method but provides a quicker way to move terms from one side of the equation to the other. The rule states that any term on one side of an equation can be moved to the other side by changing its sign (for addition/subtraction) or changing its operation (for multiplication/division).

Specifically:

1. A term being added on one side becomes a term being subtracted when moved to the other side.

Example: If $x + a = b$, then $x = b - a$.

2. A term being subtracted on one side becomes a term being added when moved to the other side.

Example: If $x - a = b$, then $x = b + a$.

3. A term that is multiplying the variable on one side becomes a divisor when moved to the other side.

Example: If $ax = b$ ($a \neq 0$), then $x = \frac{b}{a}$.

4. A term that is dividing the variable on one side becomes a multiplier when moved to the other side.

Example: If $\frac{x}{a} = b$ ($a \neq 0$), then $x = b \times a$.

The rule of transposition helps in solving linear equations by allowing us to efficiently isolate the variable step-by-step, moving all terms without the variable to one side and the term with the variable to the other, thereby making the solution process faster than explicitly writing down operations on both sides as in the balancing method.

Solving the equation $5y - 7 = 23$ using the rule of transposition:

Given equation:

$5y - 7 = 23$

We want to isolate $y$. First, we move the constant term $-7$ from the Left Hand Side (LHS) to the Right Hand Side (RHS). According to the rule of transposition, a term being subtracted becomes added on the other side.

Transpose $-7$ to the RHS:

$5y = 23 + 7$

Simplify the RHS:

$5y = 30$

Now, the variable $y$ is being multiplied by $5$ on the LHS. To isolate $y$, we move the multiplier $5$ from the LHS to the RHS. According to the rule of transposition, a term multiplying on one side becomes a divisor on the other side.

Transpose $5$ to the RHS as a divisor:

$y = \frac{30}{5}$

Perform the division:

$y = 6$

The solution to the equation $5y - 7 = 23$ is $y = 6$.

Verification of the solution:

To verify the solution $y = 6$, we substitute this value back into the original equation and check if the LHS equals the RHS.

Original equation:

$5y - 7 = 23$

Substitute $y = 6$ into the LHS:

LHS $= 5(6) - 7$

Perform the operations:

LHS $= 30 - 7$

LHS $= 23$

Compare LHS with RHS:

RHS $= 23$

Since LHS $=$ RHS ($23 = 23$), the solution $y = 6$ is correct.

Solution and Explanation:

Given equation:

$2(x - 3) = 10$

Our goal is to isolate the variable $x$ on one side of the equation. We achieve this by performing inverse operations to eliminate the terms around $x$.

Step 1: Remove the multiplier outside the parenthesis.

The expression $(x - 3)$ is multiplied by $2$. To isolate $(x - 3)$, we perform the inverse operation of multiplication, which is division. We divide both sides of the equation by $2$ to maintain equality.

Simplifying both sides:

$\frac{\cancel{2}(x - 3)}{\cancel{2}} = \frac{\cancel{10}^{5}}{\cancel{2}_{1}}$

$x - 3 = 5$

Step 2: Isolate the variable $x$.

Now we have the equation $x - 3 = 5$. The term $-3$ is being subtracted from $x$. To isolate $x$, we perform the inverse operation of subtracting $3$, which is adding $3$. We add $3$ to both sides of the equation (or use the rule of transposition by moving $-3$ to the RHS and changing its sign to $+3$).

Simplifying both sides:

$x = 8$

The variable $x$ is now isolated, and we have found its value.

The solution to the equation $2(x - 3) = 10$ is $x = 8$.

Let the first consecutive integer be represented by the variable $x$.

Since the integers are consecutive, the second integer will be one more than the first, which is $x + 1$.

The third consecutive integer will be one more than the second, which is $(x + 1) + 1 = x + 2$.

The problem states that the sum of these three consecutive integers is $54$. We can write this as an equation:

First integer + Second integer + Third integer = 54

$x + (x + 1) + (x + 2) = 54$

Now, we solve this linear equation for $x$:

$x + x + 1 + x + 2 = 54$

Combine like terms on the Left Hand Side (LHS):

$(x + x + x) + (1 + 2) = 54$

$3x + 3 = 54$

To isolate the term with $x$, transpose the constant term $+3$ from the LHS to the Right Hand Side (RHS). When a term is moved across the equals sign, its sign changes.

Simplify the RHS:

$3x = 51$

Now, the variable $x$ is being multiplied by $3$. To isolate $x$, transpose the multiplier $3$ from the LHS to the RHS. A multiplier on one side becomes a divisor on the other side.

Perform the division:

$x = 17$

So, the first integer is $x = 17$.

The second integer is $x + 1 = 17 + 1 = 18$.

The third integer is $x + 2 = 17 + 2 = 19$.

The three consecutive integers are $17$, $18$, and $19$.

Check the sum: $17 + 18 + 19 = 54$. The sum is correct.

Given:

Breadth of the rectangle = $b$ cm

Length of the rectangle is 5 cm more than its breadth.

Perimeter of the rectangle = 50 cm

To Find: The length and breadth of the rectangle.

Let the breadth of the rectangle be $b$ cm.

According to the problem, the length is 5 cm more than the breadth.

Length = Breadth + 5

Length = $b + 5$ cm

The formula for the perimeter of a rectangle is:

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

Substitute the given values and expressions into the perimeter formula:

$50 = 2 \times ((b + 5) + b)$

Now, we solve this equation for $b$. First, simplify the expression inside the parenthesis:

$50 = 2 \times (b + 5 + b)$

$50 = 2 \times (2b + 5)$

Apply the distributive property (or divide both sides by 2):

$25 = 2b + 5$

To isolate the term with $b$, transpose the constant term $+5$ from the RHS to the LHS. When a term is moved across the equals sign, its sign changes.

Simplify the LHS:

$20 = 2b$

Now, the variable $b$ is being multiplied by $2$. To isolate $b$, transpose the multiplier $2$ from the RHS to the LHS. A multiplier on one side becomes a divisor on the other side.

Perform the division:

$10 = b$

So, the breadth of the rectangle is $b = 10$ cm.

Now, calculate the length using the expression Length = $b + 5$:

Length = $10 + 5$

Length = $15$ cm

Thus, the breadth of the rectangle is 10 cm and the length is 15 cm.

Verification:

Perimeter = $2 \times (\text{Length} + \text{Breadth}) = 2 \times (15 + 10) = 2 \times 25 = 50$ cm, which matches the given perimeter.

Given:

Raju's age = $r$ years

Raju's father's age is 5 years more than three times Raju's age.

Raju's father's age = 44 years

To Find: Raju's age ($r$).

Let Raju's age be $r$ years.

According to the problem, three times Raju's age is $3 \times r = 3r$ years.

Raju's father's age is 5 years more than three times Raju's age. So, Raju's father's age can be expressed as $3r + 5$ years.

We are given that Raju's father is 44 years old. Therefore, we can set up the equation:

$3r + 5 = 44$

Now, we solve this linear equation for $r$ using the rule of transposition.

First, transpose the constant term $+5$ from the Left Hand Side (LHS) to the Right Hand Side (RHS). When a term is moved across the equals sign, its sign changes from positive to negative.

Simplify the RHS:

$3r = 39$

Now, the variable $r$ is being multiplied by $3$. To isolate $r$, transpose the multiplier $3$ from the LHS to the RHS. A term multiplying on one side becomes a divisor on the other side.

Perform the division:

$r = 13$

So, Raju's age is 13 years.

Verification:

Three times Raju's age $= 3 \times 13 = 39$ years.

5 years more than three times Raju's age $= 39 + 5 = 44$ years.

This matches the given age of Raju's father.

Thus, Raju's age is 13 years.

Solving the equation $\frac{x}{3} + \frac{x}{2} = 15$:

Given equation:

$\frac{x}{3} + \frac{x}{2} = 15$

Step 1: Combining the terms with the variable.

The terms involving the variable $x$ are on the Left Hand Side (LHS) as fractions with different denominators ($3$ and $2$). To combine these terms, we need to find a common denominator for the fractions. The least common multiple (LCM) of $3$ and $2$ is $6$.

Rewrite each fraction with the common denominator $6$:

For the first term, $\frac{x}{3}$, multiply the numerator and the denominator by $2$ to get the denominator $6$:

$\frac{x}{3} = \frac{x \times 2}{3 \times 2} = \frac{2x}{6}$

For the second term, $\frac{x}{2}$, multiply the numerator and the denominator by $3$ to get the denominator $6$:

$\frac{x}{2} = \frac{x \times 3}{2 \times 3} = \frac{3x}{6}$

Now substitute these equivalent fractions back into the equation:

$\frac{2x}{6} + \frac{3x}{6} = 15$

Since the fractions now have the same denominator, we can combine the numerators over the common denominator:

$\frac{2x + 3x}{6} = 15$

Combine the terms in the numerator:

$\frac{5x}{6} = 15$

We have now successfully combined the terms with the variable $x$ into a single term.

Step 2: Isolating the variable $x$.

Our equation is now $\frac{5x}{6} = 15$. To isolate $x$, we need to remove the operations being performed on it. The term $\frac{5x}{6}$ means $5x$ is being divided by $6$.

First, we remove the division by $6$. We do this by multiplying both sides of the equation by $6$. This is the inverse operation of division.

Simplify both sides:

$\frac{5x}{\cancel{6}} \times \cancel{6} = 90$

$5x = 90$

Now, the variable $x$ is being multiplied by $5$. To isolate $x$, we perform the inverse operation of multiplication, which is division. We divide both sides of the equation by $5$ (or use the rule of transposition by moving the multiplier $5$ to the RHS as a divisor).

Simplify both sides:

$\frac{\cancel{5}x}{\cancel{5}} = \frac{\cancel{90}^{18}}{\cancel{5}_{1}}$

$x = 18$

The variable $x$ is now isolated, and we have found its value.

The solution to the equation $\frac{x}{3} + \frac{x}{2} = 15$ is $x = 18$.

Given equation:

$4(m + 2) - 5(m - 1) = 7$

To solve this equation, we first need to simplify the Left Hand Side (LHS) by opening the brackets. We will use the distributive property: $a(b+c) = ab + ac$ and $a(b-c) = ab - ac$. We must be particularly careful with the second term, where we are distributing $-5$.

Step 1: Open the brackets.

Distribute $4$ into the first bracket $(m + 2)$: $4 \times m + 4 \times 2 = 4m + 8$.

Distribute $-5$ into the second bracket $(m - 1)$: $-5 \times m + (-5) \times (-1) = -5m + 5$. Note that multiplying a negative number by a negative number gives a positive number.

Rewrite the equation after opening the brackets:

$(4m + 8) - (5m - 5) = 7$ (Incorrect intermediate step - the sign change needs care)

$4m + 8 - 5(m - 1) = 7$ (Correct step before distributing -5)

$4m + 8 - 5m + 5 = 7$ (Correct step after distributing -5)

Step 2: Combine like terms on the LHS.

Group the terms containing $m$ and the constant terms:

$(4m - 5m) + (8 + 5) = 7$

Perform the addition/subtraction of like terms:

$(4 - 5)m + (8 + 5) = 7$

$-m + 13 = 7$

Step 3: Transpose the constant term to the RHS.

The constant term $+13$ is on the LHS. To isolate the term with $m$, we move $+13$ to the Right Hand Side (RHS). When a term is moved across the equals sign, its sign changes from positive to negative (using the rule of transposition).

Step 4: Simplify the RHS.

$-m = -6$

Step 5: Solve for $m$.

The equation is $-m = -6$, which is the same as $-1 \times m = -6$. The variable $m$ is being multiplied by $-1$. To isolate $m$, we divide both sides by $-1$ (or multiply both sides by $-1$, which is equivalent, or use the rule of transposition for multiplication).

$m = 6$

The solution to the equation $4(m + 2) - 5(m - 1) = 7$ is $m = 6$.

Given:

Let the cost of a notebook be $\textsf{₹}n$.

The cost of a book is $\textsf{₹}10$ more than the cost of a notebook.

Total cost of $2$ books and $3$ notebooks is $\textsf{₹}110$.

To Find: The cost of each book and each notebook.

According to the problem, the cost of a notebook is $\textsf{₹}n$.

The cost of a book is $\textsf{₹}10$ more than the cost of a notebook.

Cost of a book = (Cost of a notebook) + $\textsf{₹}10$

Cost of a book = $\textsf{₹}n + \textsf{₹}10 = \textsf{₹}(n + 10)$

The total cost of $2$ books is $2 \times (\text{Cost of a book}) = 2 \times \textsf{₹}(n + 10) = \textsf{₹}2(n + 10)$.

The total cost of $3$ notebooks is $3 \times (\text{Cost of a notebook}) = 3 \times \textsf{₹}n = \textsf{₹}3n$.

The total cost of $2$ books and $3$ notebooks is the sum of their individual costs, which is given as $\textsf{₹}110$. We can form the equation:

Total cost of 2 books + Total cost of 3 notebooks = $\textsf{₹}110$

$\textsf{₹}2(n + 10) + \textsf{₹}3n = \textsf{₹}110$

Dropping the unit symbol $\textsf{₹}$ for calculations, the equation is:

$2(n + 10) + 3n = 110$

Now, we solve this linear equation for $n$. First, open the bracket using the distributive property:

$2 \times n + 2 \times 10 + 3n = 110$

$2n + 20 + 3n = 110$

Combine the like terms on the Left Hand Side (LHS):

$(2n + 3n) + 20 = 110$

$5n + 20 = 110$

Transpose the constant term $+20$ from the LHS to the Right Hand Side (RHS). Its sign changes from positive to negative.

Simplify the RHS:

$5n = 90$

Transpose the multiplier $5$ from the LHS to the RHS. A multiplier on one side becomes a divisor on the other side.

Perform the division:

$n = 18$

So, the cost of a notebook is $\textsf{₹}18$.

Now, find the cost of a book using the expression Cost of a book = $\textsf{₹}(n + 10)$.

Cost of a book = $\textsf{₹}(18 + 10)$

Cost of a book = $\textsf{₹}28$

Thus, the cost of a notebook is $\textsf{₹}18$ and the cost of a book is $\textsf{₹}28$.

Verification:

Cost of 2 books = $2 \times \textsf{₹}28 = \textsf{₹}56$.

Cost of 3 notebooks = $3 \times \textsf{₹}18 = \textsf{₹}54$.

Total cost = $\textsf{₹}56 + \textsf{₹}54 = \textsf{₹}110$, which matches the given total cost.

Given equation:

$7x - (2x + 5) = 3(x - 2) + 10$

To solve this equation, we first simplify both sides by removing the brackets and combining like terms.

Simplify the Left Hand Side (LHS):

LHS $= 7x - (2x + 5)$

To remove the bracket preceded by a minus sign, we change the sign of each term inside the bracket.

LHS $= 7x - 2x - 5$

Combine the terms with $x$:

LHS $= (7 - 2)x - 5$

LHS $= 5x - 5$

Simplify the Right Hand Side (RHS):

RHS $= 3(x - 2) + 10$

Distribute $3$ into the bracket:

RHS $= 3 \times x - 3 \times 2 + 10$

RHS $= 3x - 6 + 10$

Combine the constant terms:

RHS $= 3x + (-6 + 10)$

RHS $= 3x + 4$

Now, rewrite the original equation with the simplified LHS and RHS:

$5x - 5 = 3x + 4$

To solve for $x$, we need to collect all terms containing $x$ on one side and all constant terms on the other side. Let's move terms with $x$ to the LHS and constant terms to the RHS using transposition.

Transpose $3x$ from the RHS to the LHS. Its sign changes from positive to negative.

Transpose $-5$ from the LHS to the RHS. Its sign changes from negative to positive.

Combine like terms on both sides:

$(5 - 3)x = 4 + 5$

$2x = 9$

Now, the variable $x$ is being multiplied by $2$. To isolate $x$, transpose the multiplier $2$ from the LHS to the RHS. A multiplier on one side becomes a divisor on the other side.

The value can be left as a fraction or converted to a decimal:

$x = 4.5$

The solution to the equation $7x - (2x + 5) = 3(x - 2) + 10$ is $x = \frac{9}{2}$ or $x = 4.5$.

Given:

Total sum distributed = $\textsf{₹}500$

Relationship between shares: B gets $\textsf{₹}50$ more than A, and C gets $\textsf{₹}50$ more than B.

Let A's share be $\textsf{₹}x$.

To Find: The share of A, B, and C.

Let A's share be $\textsf{₹}x$.

According to the problem, B gets $\textsf{₹}50$ more than A.

B's share = A's share + $\textsf{₹}50$

B's share = $\textsf{₹}x + \textsf{₹}50 = \textsf{₹}(x + 50)$

C gets $\textsf{₹}50$ more than B.

C's share = B's share + $\textsf{₹}50$

C's share = $\textsf{₹}(x + 50) + \textsf{₹}50 = \textsf{₹}(x + 100)$

The total sum distributed among A, B, and C is $\textsf{₹}500$. So, the sum of their shares equals the total sum.

A's share + B's share + C's share = $\textsf{₹}500$

$\textsf{₹}x + \textsf{₹}(x + 50) + \textsf{₹}(x + 100) = \textsf{₹}500$

Dropping the unit symbol $\textsf{₹}$ for calculations, we get the equation:

$x + (x + 50) + (x + 100) = 500$

Now, we solve this linear equation for $x$. First, combine the like terms on the Left Hand Side (LHS).

$(x + x + x) + (50 + 100) = 500$

$3x + 150 = 500$

To isolate the term with $x$, transpose the constant term $+150$ from the LHS to the Right Hand Side (RHS). When a term is moved across the equals sign, its sign changes from positive to negative (using the rule of transposition).

Simplify the RHS:

$3x = 350$

Now, the variable $x$ is being multiplied by $3$. To isolate $x$, transpose the multiplier $3$ from the LHS to the RHS. A term multiplying on one side becomes a divisor on the other side (using the rule of transposition).

So, A's share is $\textsf{₹}\frac{350}{3}$.

Now, calculate the shares of B and C:

B's share = $\textsf{₹}(x + 50) = \textsf{₹}\left(\frac{350}{3} + 50\right)$

To add the numbers, find a common denominator (which is 3):

$\frac{350}{3} + 50 = \frac{350}{3} + \frac{50 \times 3}{1 \times 3} = \frac{350}{3} + \frac{150}{3} = \frac{350 + 150}{3} = \frac{500}{3}$

B's share = $\textsf{₹}\frac{500}{3}$

C's share = $\textsf{₹}(x + 100) = \textsf{₹}\left(\frac{350}{3} + 100\right)$

To add the numbers, find a common denominator (which is 3):

$\frac{350}{3} + 100 = \frac{350}{3} + \frac{100 \times 3}{1 \times 3} = \frac{350}{3} + \frac{300}{3} = \frac{350 + 300}{3} = \frac{650}{3}$

C's share = $\textsf{₹}\frac{650}{3}$

The shares can also be expressed as mixed numbers or decimals:

A's share = $\frac{350}{3} = 116 \frac{2}{3} \approx \textsf{₹}116.67$

B's share = $\frac{500}{3} = 166 \frac{2}{3} \approx \textsf{₹}166.67$

C's share = $\frac{650}{3} = 216 \frac{2}{3} \approx \textsf{₹}216.67$

Thus, the share of A is $\textsf{₹}\frac{350}{3}$, the share of B is $\textsf{₹}\frac{500}{3}$, and the share of C is $\textsf{₹}\frac{650}{3}$.

Verification:

Total share = $\frac{350}{3} + \frac{500}{3} + \frac{650}{3} = \frac{350 + 500 + 650}{3} = \frac{1500}{3} = 500$. This matches the total sum $\textsf{₹}500$.

B's share - A's share = $\frac{500}{3} - \frac{350}{3} = \frac{150}{3} = 50$. B gets $\textsf{₹}50$ more than A.

C's share - B's share = $\frac{650}{3} - \frac{500}{3} = \frac{150}{3} = 50$. C gets $\textsf{₹}50$ more than B.

The conditions are satisfied.

Given:

The numerator of the original fraction is $2$ less than the denominator.

$1$ is added to both the numerator and the denominator, resulting in a new fraction $\frac{2}{3}$.

Let the denominator of the original fraction be $d$.

To Find: The original fraction.

Let the denominator of the original fraction be $d$.

Since the numerator is $2$ less than the denominator, the numerator is $d - 2$.

The original fraction is $\frac{\text{Numerator}}{\text{Denominator}} = \frac{d - 2}{d}$.

When $1$ is added to both the numerator and the denominator:

New Numerator = (Original Numerator) + $1 = (d - 2) + 1 = d - 1$

New Denominator = (Original Denominator) + $1 = d + 1$

The new fraction is $\frac{d - 1}{d + 1}$.

According to the problem, the new fraction is $\frac{2}{3}$. So, we can set up the equation:

$\frac{d - 1}{d + 1} = \frac{2}{3}$

Now, we solve this equation for $d$ using cross-multiplication:

Open the brackets on both sides using the distributive property:

Collect the terms containing $d$ on one side (LHS) and the constant terms on the other side (RHS) using transposition:

Combine like terms:

$d = 5$

So, the denominator of the original fraction is $d = 5$.

The numerator of the original fraction is $d - 2 = 5 - 2 = 3$.

Therefore, the original fraction is $\frac{3}{5}$.

Verification:

Original fraction: $\frac{3}{5}$. Numerator (3) is 2 less than the denominator (5).

Add 1 to numerator: $3 + 1 = 4$

Add 1 to denominator: $5 + 1 = 6$}

New fraction: $\frac{4}{6} = \frac{2}{3}$. This matches the given new fraction.

The original fraction is $\frac{3}{5}$.