Chapter 2 Linear Equations in One Variable (Additional Questions)

Solution:

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

Let's examine each option:

(A) $2x + 3y = 5$

This equation contains two variables, $x$ and $y$. Thus, it is a linear equation in two variables, not one variable.

(B) $x^2 + 5 = 9$

This equation contains the term $x^2$. The highest power of the variable $x$ is 2. Thus, this is a quadratic equation, not a linear equation.

(C) $4x - 7 = 13$

This equation contains only one variable, $x$. The highest power of the variable $x$ is 1. We can rewrite the equation as $4x - 7 - 13 = 0$, which simplifies to $4x - 20 = 0$. This is in the form $ax + b = 0$, where $a=4$ and $b=-20$. Thus, this is a linear equation in one variable.

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

This equation can be rewritten by multiplying all terms by $x$ (assuming $x \neq 0$):

$x(x) + x\left(\frac{1}{x}\right) = 2x$

$x^2 + 1 = 2x$

Rearranging the terms gives:

$x^2 - 2x + 1 = 0$

This equation contains the term $x^2$. The highest power of the variable $x$ is 2. Thus, this is a quadratic equation, not a linear equation.

Based on the analysis, only option (C) satisfies the definition of a linear equation in one variable.

The correct option is (C).

Solution:

The given linear equation is:

$3x - 5 = 10$

To solve for $x$, we need to isolate the term containing $x$. First, we add 5 to both sides of the equation to eliminate the constant term on the left side.

$3x - 5 + 5 = 10 + 5$

$3x = 15$

Next, we divide both sides of the equation by the coefficient of $x$, which is 3, to find the value of $x$.

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

$x = 5$

Thus, the solution to the equation $3x - 5 = 10$ is $x = 5$.

Let's verify the solution by substituting $x = 5$ back into the original equation:

$3(5) - 5 = 15 - 5 = 10$

Since the left side equals the right side ($10 = 10$), the solution $x=5$ is correct.

Comparing the solution with the given options, we find that the correct option is (B).

The correct option is (B).

Solution:

The given equation is:

$2(y + 3) = 14$

To find the root of the equation, we need to solve for the variable $y$. We can start by dividing both sides of the equation by 2.

$\frac{2(y + 3)}{2} = \frac{14}{2}$

$y + 3 = 7$

Next, to isolate $y$, we subtract 3 from both sides of the equation.

$y + 3 - 3 = 7 - 3$

$y = 4$

Thus, the root of the equation $2(y + 3) = 14$ is $y = 4$.

Let's verify the solution by substituting $y = 4$ back into the original equation:

$2(4 + 3) = 2(7) = 14$

Since the left side equals the right side ($14 = 14$), the solution $y=4$ is correct.

Comparing the root with the given options, we find that the correct option is (A).

The correct option is (A).

Solution:

The given equation is:

$\frac{z}{3} + 1 = \frac{5}{3}$

To solve for $z$, we first isolate the term containing $z$. Subtract 1 from both sides of the equation:

$\frac{z}{3} + 1 - 1 = \frac{5}{3} - 1$

$\frac{z}{3} = \frac{5}{3} - \frac{3}{3}$

$\frac{z}{3} = \frac{5 - 3}{3}$

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

Now, to find the value of $z$, multiply both sides of the equation by 3:

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

$z = 2$

Thus, the solution to the equation is $z = 2$.

Let's verify the solution by substituting $z = 2$ back into the original equation:

$\frac{2}{3} + 1 = \frac{2}{3} + \frac{3}{3} = \frac{2 + 3}{3} = \frac{5}{3}$

Since the left side equals the right side ($\frac{5}{3} = \frac{5}{3}$), the solution $z=2$ is correct.

Comparing the solution with the given options, we find that the correct option is (A).

The correct option is (A).

Solution:

The given linear equation is:

$5m - 6 = 3m + 4$

To solve for $m$, we need to isolate the terms containing $m$ on one side of the equation and the constant terms on the other side.

First, subtract $3m$ from both sides of the equation:

$5m - 6 - 3m = 3m + 4 - 3m$

This simplifies to:

$2m - 6 = 4$

Next, add 6 to both sides of the equation to move the constant term to the right side:

$2m - 6 + 6 = 4 + 6$

This simplifies to:

$2m = 10$

Finally, divide both sides of the equation by the coefficient of $m$, which is 2, to find the value of $m$:

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

$m = 5$

Thus, the value of $m$ that satisfies the equation is 5.

Let's verify the solution by substituting $m=5$ back into the original equation:

Left side: $5(5) - 6 = 25 - 6 = 19$

Right side: $3(5) + 4 = 15 + 4 = 19$

Since the left side equals the right side ($19 = 19$), the solution $m=5$ is correct.

Comparing the solution with the given options, we find that the correct option is (B).

The correct option is (B).

Solution:

Let's represent the current ages of Priya and Rahul using variables.

Let Priya's current age be $p$ years.

According to the problem, Rahul is 5 years older than his sister Priya. So, Rahul's current age is $(p + 5)$ years.

The sum of their current ages is given as 35 years.

So, we can write the equation:

Priya's age + Rahul's age = 35

$p + (p + 5) = 35$

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

Combine like terms on the left side:

$p + p + 5 = 35$

$2p + 5 = 35$

Subtract 5 from both sides of the equation to isolate the term with $p$:

$2p + 5 - 5 = 35 - 5$

$2p = 30$

Divide both sides by 2 to find the value of $p$:

$\frac{2p}{2} = \frac{30}{2}$

$p = 15$

So, Priya's current age is 15 years.

Rahul's age is $p + 5 = 15 + 5 = 20$ years.

Let's verify the sum of their ages: $15 + 20 = 35$ years, which matches the given information.

The question asks for Priya's age, which is $p$. We found $p = 15$.

Comparing our answer with the given options, we find that the correct option is (A).

The correct option is (A) $15$ years.

Solution:

Let the width of the rectangle be $w$ cm.

According to the problem, the length of the rectangle is $5 \text{ cm}$ more than its width.

So, the length $l = (w + 5)$ cm.

The perimeter of a rectangle is given by the formula $P = 2 \times (\text{length} + \text{width})$.

We are given that the perimeter $P = 50 \text{ cm}$.

Substituting the expressions for length and width into the perimeter formula, we get the equation:

$2 \times (l + w) = 50$

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

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

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

Divide both sides of the equation by 2:

$\frac{2(2w + 5)}{2} = \frac{50}{2}$

$2w + 5 = 25$

Subtract 5 from both sides of the equation:

$2w + 5 - 5 = 25 - 5$

$2w = 20$

Divide both sides by 2 to find the value of $w$:

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

$w = 10$

So, the width of the rectangle is $10 \text{ cm}$.

To verify the answer, if the width is $10 \text{ cm}$, the length is $10 + 5 = 15 \text{ cm}$.

The perimeter is $2 \times (15 + 10) = 2 \times 25 = 50 \text{ cm}$, which matches the given information.

Comparing the calculated width with the given options, we find that the correct option is (A).

The correct option is (A) $10 \text{ cm}$.

Solution:

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

According to the problem statement, "A number increased by 12 equals 37". We can translate this sentence into a linear equation:

Number + 12 = 37

$x + 12 = 37$

Now, we need to solve this equation for $x$. To isolate $x$, we subtract 12 from both sides of the equation:

$x + 12 - 12 = 37 - 12$

$x = 25$

Thus, the unknown number is 25.

Let's verify the answer: If the number is 25, then 25 increased by 12 is $25 + 12 = 37$, which matches the condition given in the problem.

Comparing the calculated number with the given options, we find that the correct option is (B).

The correct option is (B) $25$.

Solution:

Let the smaller of the two numbers be $x$.

According to the problem, one number is twice the other. Since $x$ is the smaller number, the larger number must be $2x$.

The difference between the two numbers is given as 10.

The difference is calculated as the larger number minus the smaller number.

So, we can write the equation:

Larger number - Smaller number = 10

$2x - x = 10$

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

Combine the terms on the left side:

$(2-1)x = 10$

$x = 10$

The value of $x$ is 10. Since we defined $x$ as the smaller number, the smaller number is 10.

The larger number is $2x = 2 \times 10 = 20$.

Let's verify the conditions: The two numbers are 10 and 20. Their difference is $20 - 10 = 10$, which is correct. One number (20) is twice the other (10), which is also correct.

The question asks for the smaller number, which is $x$. We found $x = 10$.

Comparing our answer with the given options, we find that the correct option is (A).

The correct option is (A) $10$.

Solution:

The given linear equation is:

$\frac{2x}{3} + 5 = x - 2$

To solve for $x$, we want to gather the terms containing $x$ on one side of the equation and the constant terms on the other side.

Subtract $\frac{2x}{3}$ from both sides of the equation:

$5 = x - 2 - \frac{2x}{3}$

To combine the terms on the right side involving $x$, we find a common denominator, which is 3:

$5 = \frac{3x}{3} - \frac{2x}{3} - 2$

$5 = \frac{3x - 2x}{3} - 2$

$5 = \frac{x}{3} - 2$

Now, add 2 to both sides of the equation to isolate the term with $x$:

$5 + 2 = \frac{x}{3} - 2 + 2$

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

To find the value of $x$, multiply both sides of the equation by 3:

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

$21 = x$

So, the solution is $x = 21$.

Let's verify the solution by substituting $x = 21$ back into the original equation:

Left side: $\frac{2(21)}{3} + 5 = \frac{42}{3} + 5 = 14 + 5 = 19$

Right side: $21 - 2 = 19$

Since the left side equals the right side ($19 = 19$), the solution $x = 21$ is correct.

Comparing the solution with the given options, we find that the correct option is (A).

The correct option is (A) $x = 21$.

Solution:

To determine which equation has $x = \frac{1}{2}$ as a solution, we need to substitute $x = \frac{1}{2}$ into each equation and check if the left side (LHS) equals the right side (RHS).

(A) Consider the equation $4x + 1 = 3$.

Substitute $x = \frac{1}{2}$ into the equation:

LHS = $4\left(\frac{1}{2}\right) + 1$

LHS = $\frac{4}{2} + 1$

LHS = $2 + 1$

LHS = $3$

RHS = $3$

Since LHS = RHS, $x = \frac{1}{2}$ is a solution to the equation $4x + 1 = 3$.

(B) Consider the equation $2x - 3 = 1$.

Substitute $x = \frac{1}{2}$ into the equation:

LHS = $2\left(\frac{1}{2}\right) - 3$

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

LHS = $1 - 3$

LHS = $-2$

RHS = $1$

Since LHS $\neq$ RHS, $x = \frac{1}{2}$ is not a solution to the equation $2x - 3 = 1$.

(C) Consider the equation $6x + 2 = 4$.

Substitute $x = \frac{1}{2}$ into the equation:

LHS = $6\left(\frac{1}{2}\right) + 2$

LHS = $\frac{6}{2} + 2$

LHS = $3 + 2$

LHS = $5$

RHS = $4$

Since LHS $\neq$ RHS, $x = \frac{1}{2}$ is not a solution to the equation $6x + 2 = 4$.

(D) Consider the equation $8x - 3 = 1$.

Substitute $x = \frac{1}{2}$ into the equation:

LHS = $8\left(\frac{1}{2}\right) - 3$

LHS = $\frac{8}{2} - 3$

LHS = $4 - 3$

LHS = $1$

RHS = $1$

Since LHS = RHS, $x = \frac{1}{2}$ is a solution to the equation $8x - 3 = 1$.

Both option (A) and option (D) have $x = \frac{1}{2}$ as a solution. Assuming this is a standard multiple-choice question with a single correct answer, there might be an error in the question or options provided. However, based on our calculations, both (A) and (D) are correct.

Following the typical format of providing one correct option, we choose (A).

The correct option is (A) $4x + 1 = 3$.

Solution:

Let the number of $\textsf{₹} 10$ notes be $n_{10}$.

Let the number of $\textsf{₹} 20$ notes be $n_{20}$.

We are given that the total number of notes is 30. So, we can write the equation:

We are also given that the total value of the notes is $\textsf{₹} 500$. The value from $\textsf{₹} 10$ notes is $10 \times n_{10}$, and the value from $\textsf{₹} 20$ notes is $20 \times n_{20}$.

So, we can write the equation for the total value:

We now have a system of two linear equations with two variables. We can solve this system using substitution or elimination.

From equation (i), we can express $n_{20}$ in terms of $n_{10}$:

$n_{20} = 30 - n_{10}$

Substitute this expression for $n_{20}$ into equation (ii):

$10 n_{10} + 20 (30 - n_{10}) = 500$

Now, solve this equation for $n_{10}$. First, distribute the 20:

$10 n_{10} + 20 \times 30 - 20 \times n_{10} = 500$

$10 n_{10} + 600 - 20 n_{10} = 500$

Combine the terms containing $n_{10}$:

$(10 - 20) n_{10} + 600 = 500$

$-10 n_{10} + 600 = 500$

Subtract 600 from both sides of the equation:

$-10 n_{10} = 500 - 600$

$-10 n_{10} = -100$

Divide both sides by -10:

$\frac{-10 n_{10}}{-10} = \frac{-100}{-10}$

$n_{10} = 10$

The number of $\textsf{₹} 10$ notes is 10.

We can also find the number of $\textsf{₹} 20$ notes: $n_{20} = 30 - n_{10} = 30 - 10 = 20$.

Verification: Total notes = $10 + 20 = 30$. Total value = $10 \times 10 + 20 \times 20 = 100 + 400 = 500$. This matches the given information.

The question asks for the number of $\textsf{₹} 10$ notes, which is $n_{10}$. We found $n_{10} = 10$.

Comparing our answer with the given options, we find that the correct option is (A).

The correct option is (A) $10$.

Solution:

The given equation is:

$x - 5 = 8$

To solve for $x$, we add 5 to both sides of the equation:

$(x - 5) + 5 = 8 + 5$

$x + (-5 + 5) = 13$

$x + 0 = 13$

$x = 13$

The action of adding the same number (which is 5 in this case) to both sides of an equation while maintaining the equality is justified by a fundamental property of equality.

Let's consider the given options:

(A) Associative Property: This property deals with the grouping of numbers in addition or multiplication. For example, $(a+b)+c = a+(b+c)$. This is not what is being applied when adding 5 to both sides of the equation.

(B) Commutative Property: This property deals with the order of numbers in addition or multiplication. For example, $a+b = b+a$. This is not directly related to adding the same quantity to both sides of an equation.

(C) Additive Inverse Property: This property states that for any number $a$, there exists an additive inverse $-a$ such that $a + (-a) = 0$. When we add 5 to the left side of the equation ($x - 5 + 5$), we are using the additive inverse property for the terms $-5$ and $+5$, as $-5 + 5 = 0$. However, the property that allows us to add 5 to *both sides* of the equation is different; it's the property that ensures the equation remains balanced.

(D) Addition Property of Equality: This property states that if $a = b$, then $a + c = b + c$ for any number $c$. In our equation $x - 5 = 8$, we are adding $c = 5$ to both sides $(x - 5) + 5 = 8 + 5$. This property guarantees that the resulting equation is equivalent to the original one and has the same solution.

Therefore, the property used to solve the equation by adding 5 to both sides is the Addition Property of Equality.

The correct option is (D) Addition Property of Equality.

Solution:

The given equation is:

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

To solve for $x$, we can cross-multiply the terms:

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

Apply the distributive property on both sides of the equation:

$3x + 3 \times 1 = 2x - 2 \times 2$

$3x + 3 = 2x - 4$

Now, we need to collect the terms involving $x$ on one side of the equation and the constant terms on the other side. Subtract $2x$ from both sides:

$3x + 3 - 2x = 2x - 4 - 2x$

$x + 3 = -4$

Next, subtract 3 from both sides of the equation to isolate $x$:

$x + 3 - 3 = -4 - 3$

$x = -7$

Thus, the value of $x$ that satisfies the equation is $-7$.

Let's verify the solution by substituting $x = -7$ back into the original equation:

Left side (LHS) = $\frac{x+1}{2} = \frac{-7+1}{2} = \frac{-6}{2} = -3$

Right side (RHS) = $\frac{x-2}{3} = \frac{-7-2}{3} = \frac{-9}{3} = -3$

Since LHS = RHS ($-3 = -3$), the solution $x = -7$ is correct.

Comparing the solution with the given options, we find that the correct option is (A).

The correct option is (A) $-7$.

Solution:

Let the three consecutive integers be represented by variables.

Since they are consecutive, they differ by 1.

Let the smallest integer be $x$.

Then the next integer is $x + 1$.

And the largest integer is $(x + 1) + 1 = x + 2$.

We are given that the sum of these three consecutive integers is 51.

So, we can write the equation:

Smallest integer + Next integer + Largest integer = 51

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

Combine the like terms on the left side of the equation:

$(x + x + x) + (1 + 2) = 51$

$3x + 3 = 51$

Subtract 3 from both sides of the equation to isolate the term containing $x$:

$3x + 3 - 3 = 51 - 3$

$3x = 48$

Divide both sides by 3 to find the value of $x$:

$\frac{3x}{3} = \frac{48}{3}$

$x = 16$

The value of $x$ is 16. This is the smallest of the three consecutive integers.

The three integers are:

Smallest: $x = 16$

Middle: $x + 1 = 16 + 1 = 17$

Largest: $x + 2 = 16 + 2 = 18$

We can verify the sum: $16 + 17 + 18 = 33 + 18 = 51$. This matches the given condition.

The question asks for the largest integer, which is $x + 2$.

Largest integer $= 18$.

Comparing our answer with the given options, we find that the correct option is (C).

The correct option is (C) $18$.

Solution:

Let the rational number be $x$.

According to the problem statement, when the number is multiplied by $\frac{5}{2}$, the product is $\frac{5}{2}x$.

When $\frac{2}{3}$ is added to this product, the result is $-\frac{7}{12}$.

This can be written as the equation:

To solve for $x$, we first isolate the term containing $x$. Subtract $\frac{2}{3}$ from both sides of the equation:

$\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3}$

To subtract the fractions on the right side, we need a common denominator. The least common multiple of 12 and 3 is 12.

We rewrite $\frac{2}{3}$ with a denominator of 12:

$\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

Now, substitute this back into the equation:

$\frac{5}{2}x = -\frac{7}{12} - \frac{8}{12}$

Combine the fractions on the right side:

$\frac{5}{2}x = \frac{-7 - 8}{12}$

$\frac{5}{2}x = -\frac{15}{12}$

We can simplify the fraction on the right side by dividing both the numerator and the denominator by their greatest common divisor, which is 3:

$-\frac{15}{12} = -\frac{\cancel{15}^5}{\cancel{12}_4} = -\frac{5}{4}$

So, the equation becomes:

$\frac{5}{2}x = -\frac{5}{4}$

To solve for $x$, multiply both sides of the equation by the reciprocal of $\frac{5}{2}$, which is $\frac{2}{5}$:

$x = -\frac{5}{4} \times \frac{2}{5}$

Multiply the numerators and the denominators:

$x = -\frac{\cancel{5}^1 \times \cancel{2}^1}{\cancel{4}_2 \times \cancel{5}_1}$

$x = -\frac{1 \times 1}{2 \times 1}$

$x = -\frac{1}{2}$

The rational number is $-\frac{1}{2}$.

Let's verify the solution by substituting $x = -\frac{1}{2}$ back into the original equation:

LHS = $\frac{5}{2}\left(-\frac{1}{2}\right) + \frac{2}{3}$

LHS = $-\frac{5}{4} + \frac{2}{3}$

Find a common denominator (12) for $-\frac{5}{4}$ and $\frac{2}{3}$:

$-\frac{5}{4} = -\frac{5 \times 3}{4 \times 3} = -\frac{15}{12}$

$\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

LHS = $-\frac{15}{12} + \frac{8}{12}$

LHS = $\frac{-15 + 8}{12}$

LHS = $-\frac{7}{12}$

RHS = $-\frac{7}{12}$

Since LHS = RHS, the solution $x = -\frac{1}{2}$ is correct.

Comparing our answer with the given options, we find that the correct option is (A).

The correct option is (A) $-\frac{1}{2}$.

Solution:

Let the present age of Sharma ji's son be $s$ years.

Let the present age of Sharma ji be $j$ years.

According to the first condition, the present age of Sharma ji's son is half of Sharma ji's age:

$s = \frac{1}{2} j$

This implies $j = 2s$.

Now, let's consider their ages ten years ago.

Sharma ji's son's age ten years ago was $s - 10$ years.

Sharma ji's age ten years ago was $j - 10$ years.

According to the second condition, ten years ago, Sharma ji's age was three times his son's age:

$j - 10 = 3(s - 10)$

We have two equations:

1. $j = 2s$

2. $j - 10 = 3(s - 10)$

Substitute the expression for $j$ from equation (1) into equation (2):

Now, solve equation (iii) for $s$. Distribute the 3 on the right side:

$2s - 10 = 3s - 30$

Subtract $2s$ from both sides of the equation:

$2s - 10 - 2s = 3s - 30 - 2s$

$-10 = s - 30$

Add 30 to both sides of the equation:

$-10 + 30 = s - 30 + 30$

$20 = s$

The value of $s$ is 20. This represents the present age of Sharma ji's son.

To verify, the son's present age is 20 years. Sharma ji's present age is $j = 2s = 2 \times 20 = 40$ years.

Ten years ago, the son was $20 - 10 = 10$ years old, and Sharma ji was $40 - 10 = 30$ years old. $30 = 3 \times 10$, which is correct.

The present age of Sharma ji's son is 20 years.

Comparing our answer with the given options, we find that the correct option is (B).

The correct option is (B) $20$ years.

Solution:

Let the two parts into which 184 is divided be $x$ and $y$.

The sum of the two parts is 184.

According to the second condition, one-third of one part may exceed one-seventh of the other part by 8.

Let's assume one-third of the first part ($x$) exceeds one-seventh of the second part ($y$) by 8. We can write this as an equation:

We now have a system of two linear equations with two variables $x$ and $y$. We can use the substitution method to solve this system.

From equation (i), express $y$ in terms of $x$:

Substitute this expression for $y$ into equation (ii):

$\frac{x}{3} - \frac{184 - x}{7} = 8$

To eliminate the denominators, multiply every term by the least common multiple (LCM) of 3 and 7, which is 21.

$21 \times \frac{x}{3} - 21 \times \frac{184 - x}{7} = 21 \times 8$

$7x - 3(184 - x) = 168$

Apply the distributive property on the left side:

$7x - (3 \times 184) - (3 \times -x) = 168$

$7x - 552 + 3x = 168$

Combine the terms containing $x$:

$(7x + 3x) - 552 = 168$

$10x - 552 = 168$

Add 552 to both sides of the equation to isolate the term with $x$:

$10x = 168 + 552$

$10x = 720$

Divide both sides by 10 to find the value of $x$:

$x = \frac{720}{10}$

$x = 72$

Now that we have the value of $x$, substitute it back into equation (iii) to find the value of $y$:

$y = 184 - x$

$y = 184 - 72$

$y = 112$

The two parts are 72 and 112.

Let's verify the second condition with these parts:

One-third of $x = \frac{72}{3} = 24$

One-seventh of $y = \frac{112}{7} = 16$

The difference is $24 - 16 = 8$. This matches the condition that one-third of one part exceeds one-seventh of the other by 8.

The question asks for the smaller part. Comparing the two parts, 72 and 112, the smaller part is 72.

Comparing our answer with the given options, we find that the correct option is (C).

The correct option is (C) $72$.

Solution:

Let the numerator of the rational number be $n$.

According to the first condition, the denominator is greater than the numerator by 8.

So, the denominator is $n + 8$.

The original rational number is $\frac{\text{Numerator}}{\text{Denominator}} = \frac{n}{n+8}$.

Now, consider the changes described in the problem.

The numerator is increased by 17. The new numerator is $n + 17$.

The denominator is decreased by 1. The new denominator is $(n + 8) - 1 = n + 7$.

The new rational number obtained is $\frac{\text{New Numerator}}{\text{New Denominator}} = \frac{n+17}{n+7}$.

According to the problem, this new number is equal to $\frac{3}{2}$.

So, we can set up the equation:

Note that for the original number to be defined, $n+8 \neq 0$, i.e., $n \neq -8$. Also, for the new number to be defined, $n+7 \neq 0$, i.e., $n \neq -7$.

To solve equation (i) for $n$, we can cross-multiply:

$2(n+17) = 3(n+7)$

Apply the distributive property on both sides of the equation:

$2n + 2 \times 17 = 3n + 3 \times 7$

$2n + 34 = 3n + 21$

To isolate the term containing $n$, subtract $2n$ from both sides:

$2n + 34 - 2n = 3n + 21 - 2n$

$34 = n + 21$

Subtract 21 from both sides to find the value of $n$:

$34 - 21 = n + 21 - 21$

$13 = n$

So, the numerator $n = 13$.

Now, find the denominator of the original rational number using the relation $n+8$:

Denominator $= n + 8 = 13 + 8 = 21$.

The original rational number is $\frac{n}{n+8} = \frac{13}{21}$.

Let's verify the conditions with the rational number $\frac{13}{21}$.

The denominator (21) is greater than the numerator (13) by $21 - 13 = 8$. This condition is satisfied.

Now, increase the numerator by 17: $13 + 17 = 30$.

Decrease the denominator by 1: $21 - 1 = 20$.

The new number is $\frac{30}{20}$. Simplify this fraction:

$\frac{30}{20} = \frac{\cancel{30}^3}{\cancel{20}_2} = \frac{3}{2}$.

This matches the given condition that the new number is $\frac{3}{2}$.

The rational number is $\frac{13}{21}$.

Comparing our answer with the given options, we find that the correct option is (A).

The correct option is (A) $\frac{13}{21}$.

Solution:

A linear equation in one variable is an equation that can be written in the standard form $ax + b = 0$, where $x$ is the variable, $a$ and $b$ are constant real numbers, and the coefficient $a$ is not equal to zero ($a \neq 0$).

Let's examine each statement based on this definition:

(A) It has only one solution.

For a linear equation in one variable in the form $ax + b = 0$ with $a \neq 0$, we can solve for $x$ by subtracting $b$ from both sides and then dividing by $a$:

$ax = -b$

$x = -\frac{b}{a}$

Since $a \neq 0$, the value $-\frac{b}{a}$ is a unique real number. Therefore, a linear equation in one variable always has exactly one solution.

This statement is TRUE.

(B) The power of the variable is exactly 1.

In the standard form $ax + b = 0$, the variable is $x$. The power of $x$ is $1$ (since $x = x^1$). If the highest power of the variable were 0, the term $ax$ would become $a \times x^0 = a \times 1 = a$, resulting in $a+b=0$, which is a constant equation. If the highest power were greater than 1 (e.g., $x^2$, $x^3$), the equation would be quadratic, cubic, or of a higher degree, not linear.

This statement is TRUE.

(C) It can be written in the form $ax+b=0$, where $a \neq 0$.

This is the precise definition of a linear equation in one variable that guarantees a unique solution. The condition $a \neq 0$ distinguishes it from cases like $0x+b=0$ (which has infinitely many solutions if $b=0$ or no solution if $b \neq 0$).

This statement is TRUE.

Since statements (A), (B), and (C) are all true characteristics of a linear equation in one variable, the option stating "All of the above" is the correct choice.

The correct option is (D) All of the above.

Solution:

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

Assertion (A): The equation $7x - 9 = 12$ is a linear equation in one variable.

The given equation is $7x - 9 = 12$. We can rewrite this equation by moving all terms to one side to match the general form of a linear equation:

$7x - 9 - 12 = 0$

$7x - 21 = 0$

This equation involves only one variable, $x$. The highest power of the variable $x$ is 1. This fits the definition of a linear equation in one variable.

So, Assertion (A) is TRUE.

Reason (R): Any equation of the form $ax+b=0$, where $a, b$ are real numbers and $a \neq 0$, is a linear equation in the variable $x$.

This statement provides the standard definition of a linear equation in one variable. The condition $a \neq 0$ is crucial because if $a=0$, the term $ax$ becomes 0, and the equation reduces to $b=0$, which is either always true (if $b=0$) or always false (if $b \neq 0$), and does not represent a linear equation with a unique solution in $x$. The form $ax+b=0$ with $a \neq 0$ guarantees that the variable $x$ exists (with power 1) and has a unique solution $x = -\frac{b}{a}$.

So, Reason (R) is TRUE.

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

The equation in Assertion (A) is $7x - 9 = 12$, which was rewritten as $7x - 21 = 0$. This equation is in the form $ax+b=0$, where $a = 7$ and $b = -21$. Both $a$ and $b$ are real numbers, and $a = 7 \neq 0$. According to the definition given in Reason (R), this equation fits the criteria for a linear equation in the variable $x$. Therefore, Reason (R) correctly explains why the equation in Assertion (A) is a linear equation in one variable.

Both Assertion (A) and Reason (R) are true, and Reason (R) provides the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Solution:

Let the original speed of the train be $s$ km/hr.

Let the original time taken to cover the distance be $t$ hours.

Let the distance covered be $d$ km.

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

Distance = Speed $\times$ Time

Scenario 1: If the speed were $6 \text{ km/hr}$ more, the speed would be $(s+6)$ km/hr. It would take 4 hours less, so the time would be $(t-4)$ hours. The distance remains the same.

Using the distance formula:

$d = (s+6)(t-4)$

Substitute $d = st$ from equation (i):

$st = (s+6)(t-4)$

Expand the right side:

$st = st - 4s + 6t - 24$

Subtract $st$ from both sides:

$0 = -4s + 6t - 24$

Rearrange the terms to form a linear equation in $s$ and $t$:

$4s - 6t = -24$

Divide the entire equation by 2:

Scenario 2: If the speed were $6 \text{ km/hr}$ less, the speed would be $(s-6)$ km/hr. It would take 6 hours more, so the time would be $(t+6)$ hours. The distance remains the same.

Using the distance formula:

$d = (s-6)(t+6)$

Substitute $d = st$ from equation (i):

$st = (s-6)(t+6)$

Expand the right side:

$st = st + 6s - 6t - 36$

Subtract $st$ from both sides:

$0 = 6s - 6t - 36$

Rearrange the terms to form a linear equation in $s$ and $t$:

$6s - 6t = 36$

Divide the entire equation by 6:

We now have a system of two linear equations with two variables $s$ and $t$:

$2s - 3t = -12$ (from ii)

$s - t = 6$ (from iii)

We can solve this system using the substitution method. From equation (iii), we can express $s$ in terms of $t$:

Substitute the expression for $s$ from equation (iv) into equation (ii):

$2(t + 6) - 3t = -12$

Apply the distributive property:

$2t + 12 - 3t = -12$

Combine like terms:

$(2t - 3t) + 12 = -12$

$-t + 12 = -12$

Subtract 12 from both sides to isolate the term with $t$:

$-t = -12 - 12$

$-t = -24$

Multiply by -1:

$t = 24$

The original time taken is 24 hours.

Now substitute the value of $t$ into equation (iv) to find the value of $s$:

$s = t + 6$

$s = 24 + 6$

$s = 30$

The original speed is 30 km/hr.

Finally, we can find the distance covered by the train using equation (i):

$d = s \times t$

$d = 30 \text{ km/hr} \times 24 \text{ hours}$

$d = 720 \text{ km}$

The distance covered by the train is 720 km.

Let's quickly verify the scenarios:

Original: Speed = 30 km/hr, Time = 24 hr, Distance = $30 \times 24 = 720$ km.

Scenario 1: Speed = $30+6 = 36$ km/hr, Time = $24-4 = 20$ hr, Distance = $36 \times 20 = 720$ km.

Scenario 2: Speed = $30-6 = 24$ km/hr, Time = $24+6 = 30$ hr, Distance = $24 \times 30 = 720$ km.

All conditions are satisfied.

Comparing our calculated distance with the given options, we find that the correct option is (C).

The correct option is (C) $720 \text{ km}$.

Solution:

The given equation is:

$0.25(4x - 3) = 0.05(10x - 9)$

First, distribute the decimal values on both sides of the equation:

$0.25 \times 4x - 0.25 \times 3 = 0.05 \times 10x - 0.05 \times 9$

$1.00x - 0.75 = 0.50x - 0.45$

This simplifies to:

$x - 0.75 = 0.5x - 0.45$

Next, collect the terms involving $x$ on one side of the equation and the constant terms on the other side.

Subtract $0.5x$ from both sides of the equation:

$x - 0.75 - 0.5x = 0.5x - 0.45 - 0.5x$

Combine like terms on the left side:

$(1 - 0.5)x - 0.75 = -0.45$

$0.5x - 0.75 = -0.45$

Now, add $0.75$ to both sides of the equation to isolate the term containing $x$:

$0.5x - 0.75 + 0.75 = -0.45 + 0.75$

$0.5x = 0.30$

Finally, divide both sides of the equation by the coefficient of $x$, which is $0.5$, to find the value of $x$:

$x = \frac{0.30}{0.5}$

$x = \frac{0.3}{0.5}$

To perform the division with decimals, we can multiply the numerator and denominator by 10 to remove the decimals:

$x = \frac{0.3 \times 10}{0.5 \times 10} = \frac{3}{5}$

Convert the fraction to a decimal:

$x = 0.6$

Thus, the solution to the equation is $x = 0.6$.

Let's verify the solution by substituting $x = 0.6$ back into the original equation:

LHS = $0.25(4(0.6) - 3)$

LHS = $0.25(2.4 - 3)$

LHS = $0.25(-0.6)$

LHS = $-0.15$

RHS = $0.05(10(0.6) - 9)$

RHS = $0.05(6 - 9)$

RHS = $0.05(-3)$

RHS = $-0.15$

Since LHS = RHS ($-0.15 = -0.15$), the solution $x = 0.6$ is correct.

Comparing the solution with the given options, we find that the correct option is (B).

The correct option is (B) $x = 0.6$.

Solution:

Let the units digit of the two-digit number be $u$.

Let the tens digit of the two-digit number be $t$.

The value of the original two-digit number is $10 \times \text{tens digit} + \text{units digit}$.

Original number $= 10t + u$.

According to the first condition, the sum of the digits is 9.

If the digits are reversed, the new number has the units digit as the tens digit and the tens digit as the units digit.

The value of the new number is $10 \times \text{new tens digit} + \text{new units digit}$.

New number $= 10u + t$.

According to the second condition, the new number is 27 more than the original number.

New number = Original number + 27

Now, simplify equation (ii) and solve the system of equations (i) and (ii).

Rearrange equation (ii):

$10u + t - 10t - u = 27$

Combine like terms:

$(10u - u) + (t - 10t) = 27$

$9u - 9t = 27$

Divide the entire equation by 9:

We have a system of two linear equations with two variables $t$ and $u$:

$t + u = 9$ (from i)

$u - t = 3$ (from iii)

We can solve this system by adding the two equations:

$(t + u) + (u - t) = 9 + 3$

$t + u + u - t = 12$

$(t - t) + (u + u) = 12$

$0 + 2u = 12$

$2u = 12$

Divide by 2 to find the value of $u$:

$u = \frac{12}{2}$

$u = 6$

The units digit is 6.

Substitute the value of $u$ into equation (i) to find the value of $t$:

$t + u = 9$

$t + 6 = 9$

Subtract 6 from both sides:

$t = 9 - 6$

$t = 3$

The tens digit is 3.

The original two-digit number is $10t + u = 10(3) + 6 = 30 + 6 = 36$.

Let's verify the conditions with the number 36.

Sum of digits: $3 + 6 = 9$. This is correct.

Reversed number: 63.

Difference between reversed number and original number: $63 - 36 = 27$. This is correct.

The original number is 36.

Comparing our answer with the given options, we find that the correct option is (A).

The correct option is (A) $36$.

Solution:

A linear equation in one variable is an equation that can be written in the form $ax + b = 0$, where $x$ is the variable, $a$ and $b$ are real numbers, and the coefficient $a$ is non-zero ($a \neq 0$). The key characteristic is that the highest power of the variable must be exactly 1.

Let's examine each option:

(A) $\frac{x-1}{2} = \frac{x+1}{3}$

This equation involves the variable $x$. We can simplify it by cross-multiplying:

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

$3x - 3 = 2x + 2$

$3x - 2x = 2 + 3$

$x = 5$

This equation can be written as $x - 5 = 0$, which is in the form $ax+b=0$ with $a=1$ and $b=-5$. The highest power of $x$ is 1, and $a \neq 0$. Thus, this is a linear equation in one variable.

(B) $y^2 - y = 0$

This equation involves the variable $y$. The terms are $y^2$ and $-y$. The highest power of the variable $y$ in this equation is 2. Since the highest power of the variable is 2, this is a quadratic equation, not a linear equation.

(C) $3(t+5) = t - 1$

This equation involves the variable $t$. We can simplify it:

$3t + 15 = t - 1$

$3t - t = -1 - 15$

$2t = -16$

This equation can be written as $2t + 16 = 0$, which is in the form $at+b=0$ with $a=2$ and $b=16$. The highest power of $t$ is 1, and $a \neq 0$. Thus, this is a linear equation in one variable.

(D) $5p + 8 = p$

This equation involves the variable $p$. We can simplify it:

$5p - p + 8 = 0$

$4p + 8 = 0$

This equation is in the form $ap+b=0$ with $a=4$ and $b=8$. The highest power of $p$ is 1, and $a \neq 0$. Thus, this is a linear equation in one variable.

Comparing the options, only option (B) has the variable raised to a power other than 1 (specifically, power 2). Therefore, option (B) is NOT a linear equation in one variable.

The correct option is (B) $y^2 - y = 0$.

Solution:

We need to solve each equation and match the solution with the options provided.

(i) Solve $2x + 7 = 15$:

$2x = 15 - 7$

$2x = 8$

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

$x = 4$

(ii) Solve $5(x - 2) = 15$:

$5x - 10 = 15$

$5x = 15 + 10$

$5x = 25$

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

$x = 5$

(iii) Solve $\frac{x}{4} - 1 = 2$:

$\frac{x}{4} = 2 + 1$

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

$x = 3 \times 4$

$x = 12$

(iv) Solve $3x - 8 = 5x - 12$:

Collect terms with $x$ on one side and constants on the other.

$3x - 5x = -12 + 8$

$-2x = -4$

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

$x = 2$

Our calculated solutions are:

(i) $x = 4$

(ii) $x = 5$

(iii) $x = 12$

(iv) $x = 2$

The provided solution options are:

(a) $x = 4$

(b) $x = 5$

(c) $x = 12$

(d) $x = 4$

Matching the calculated solutions with the provided options:

(i) $x = 4$ matches (a) $x=4$ and (d) $x=4$.

(ii) $x = 5$ matches (b) $x=5$.

(iii) $x = 12$ matches (c) $x=12$.

(iv) $x = 2$ does not match any of the options (a), (b), (c), or (d).

Let's examine the given multiple-choice matching options:

(A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d)

(B) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(a)

(C) (i)-(d), (ii)-(b), (iii)-(c), (iv)-(a)

(D) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(a) and (i)-(d), (iv)-(a)

Based on our calculations, the first three pairings are consistent with options (A), (B), and (C): (i) matches (a) or (d) ($x=4$), (ii) matches (b) ($x=5$), and (iii) matches (c) ($x=12$).

However, for equation (iv), our calculated solution is $x=2$, which is not among the options (a), (b), (c), (d). All provided multiple-choice options (A, B, C, D) suggest that the solution for (iv) is $x=4$ (mapping to either (a) or (d)). This indicates a likely error in the problem statement, either in equation (iv) or the provided solution options (a)-(d).

Assuming the first three matches are correct and following the pattern suggested by the provided options, the intended answer likely corresponds to the pairings where (i) maps to $x=4$, (ii) maps to $x=5$, (iii) maps to $x=12$, and (iv) is incorrectly mapped to $x=4$ in the given options.

Considering the provided options (A), (B), (C), (D), options (A), (B), and (C) have the same first three pairings consistent with our calculations. Options (B) and (C) are identical in their first part. Option (D) seems to be a compound or erroneous listing but also implies the same primary matching as (B) and (C).

Since (i)-(a), (ii)-(b), (iii)-(c) are correctly matched by our calculations, and options (B) and (C) (and the first part of D) propose (iv)-(a) ($x=4$), let's select one of them, acknowledging the error in (iv).

Based on our calculations:

(i) $\rightarrow$ (a) or (d) ($x=4$)

(ii) $\rightarrow$ (b) ($x=5$)

(iii) $\rightarrow$ (c) ($x=12$)

(iv) $\rightarrow$ None ($x=2$)

However, if we are forced to choose from the given options (A), (B), (C), (D), and assuming there is an error in the question/options for part (iv), options (A), (B), and (C) are consistent for the first three parts. Option (B) is (i)-(a), (ii)-(b), (iii)-(c), (iv)-(a).

Assuming option (B) is the intended answer key despite the discrepancy in (iv), we provide the matching based on our calculations for the first three parts and the likely intended match for the fourth part as per the options.

(i) corresponds to $x=4$, which is option (a).

(ii) corresponds to $x=5$, which is option (b).

(iii) corresponds to $x=12$, which is option (c).

(iv) corresponds to $x=2$ by calculation, but is matched with (a) $x=4$ in option (B).

The correct option, assuming there is a typo in the question or answer options for part (iv), and following the pattern of options (B) and (C), is likely intended to be (B) or (C).

Choosing option (B): (i)-(a), (ii)-(b), (iii)-(c), (iv)-(a).

The final answer is $\boxed{B}$.

Solution:

Let Shyam's age be $s$ years.

According to the problem, Ram is 3 years older than Shyam.

So, Ram's age is $(s + 3)$ years.

According to the problem, Mohan is 2 years younger than Shyam.

So, Mohan's age is $(s - 2)$ years.

The sum of their ages is 58 years.

Sum of ages = Shyam's age + Ram's age + Mohan's age

$s + (s + 3) + (s - 2) = 58$

Let's compare this equation with the given options:

(A) $s + (s-3) + (s+2) = 58$: This represents Shyam's age, (Shyam's age - 3), and (Shyam's age + 2). This does not match the conditions given in the problem.

(B) $s + (s+3) + (s-2) = 58$: This represents Shyam's age, (Shyam's age + 3), and (Shyam's age - 2). This matches the ages we derived for Shyam, Ram, and Mohan respectively.

(C) $s + 3s + 2s = 58$: This represents Shyam's age, 3 times Shyam's age, and 2 times Shyam's age. This does not match the conditions given in the problem.

(D) $s + (s+3) + (2-s) = 58$: This represents Shyam's age, (Shyam's age + 3), and (2 minus Shyam's age). This does not match the conditions given in the problem.

The equation that correctly represents the sum of their ages based on the given information is $s + (s + 3) + (s - 2) = 58$. This matches option (B).

The correct option is (B) $s + (s+3) + (s-2) = 58$.

Given:

Ram is 3 years older than Shyam.

Mohan is 2 years younger than Shyam.

Sum of their ages = 58 years.

To Find:

The age of Shyam.

Solution:

Let the present age of Shyam be $s$ years.

According to the problem, Ram is 3 years older than Shyam.

Ram's age $= (s + 3)$ years.

Mohan is 2 years younger than Shyam.

Mohan's age $= (s - 2)$ years.

The sum of their ages is 58 years.

Shyam's age + Ram's age + Mohan's age = 58

Now, solve the equation for $s$:

$s + s + 3 + s - 2 = 58$

Combine like terms:

$(s + s + s) + (3 - 2) = 58$

$3s + 1 = 58$

Subtract 1 from both sides:

$3s = 58 - 1$

$3s = 57$

Divide both sides by 3:

$s = \frac{57}{3}$

$s = 19$

So, the present age of Shyam is 19 years.

To verify:

Shyam's age = 19 years.

Ram's age = $19 + 3 = 22$ years.

Mohan's age = $19 - 2 = 17$ years.

Sum of ages = $19 + 22 + 17 = 41 + 17 = 58$ years. This matches the given condition.

Comparing our answer with the given options, we find that the correct option is (A).

The correct option is (A) $19$ years.

Solution:

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

The square of the number is $x^2$.

The number itself is $x$.

The problem states that "The difference between the square of a number and the number itself is 20".

The phrase "the difference between A and B" is usually interpreted as A - B.

In this case, A is "the square of a number" ($x^2$) and B is "the number itself" ($x$).

So, the difference between the square of the number and the number itself is $x^2 - x$.

This difference is equal to 20.

Therefore, the equation that represents this statement is:

$x^2 - x = 20$

Let's compare this equation with the given options:

(A) $x - x^2 = 20$: This represents the difference between the number itself and its square.

(B) $x^2 - x = 20$: This represents the difference between the square of the number and the number itself, which matches our derivation.

(C) $x^2 + x = 20$: This represents the sum of the square of the number and the number itself.

(D) $x - x^2 + 20 = 0$: This is equivalent to $x - x^2 = -20$, which is not what is stated in the problem.

The equation that correctly represents the statement is $x^2 - x = 20$. This is a quadratic equation, not a linear equation, but the question asks which equation represents the statement.

Comparing our derived equation with the options, we find that option (B) is the correct representation.

The correct option is (B) $x^2 - x = 20$.

Solution:

In any algebraic equation, the value or values of the variable(s) that make the equation true are called the solution(s) of the equation.

For a linear equation in one variable, there is typically one such value.

Let's consider the options:

(A) Coefficient: A coefficient is the numerical factor of a term containing a variable (e.g., in $3x$, 3 is the coefficient). It is not the value of the variable that satisfies the equation.

(B) Constant: A constant is a term without a variable (e.g., in $3x + 5$, 5 is the constant). It is not the value of the variable that satisfies the equation.

(C) Root or Solution: The terms "root" and "solution" are used to describe the value(s) of the variable(s) that satisfy an equation. For a polynomial equation (like a linear equation), the solution(s) are also often called roots.

(D) Term: A term is a single mathematical expression, such as $3x$, $5$, or $x$. It is not the value of the variable.

The value of the variable that makes a linear equation true is called its root or solution.

Completing the sentence:

The value of the variable that satisfies a linear equation is called its Root or Solution.

The correct option is (C) Root or Solution.

Given:

The ratio of two numbers is $3:5$.

If 8 is added to each number, the ratio becomes $2:3$.

To Find:

The original numbers.

Solution:

Let the two original numbers be $3k$ and $5k$, where $k$ is a non-zero constant. Since the numbers are in the ratio $3:5$, we can represent them in this form.

When 8 is added to each number, the new numbers are obtained:

First new number $= 3k + 8$

Second new number $= 5k + 8$

According to the problem, the ratio of these new numbers is $2:3$. We can set up the equation based on this information:

To solve for $k$, we cross-multiply the terms in the equation:

$3 \times (3k + 8) = 2 \times (5k + 8)$

Apply the distributive property on both sides of the equation:

$3 \times 3k + 3 \times 8 = 2 \times 5k + 2 \times 8$

$9k + 24 = 10k + 16$

Now, we need to collect the terms containing $k$ on one side of the equation and the constant terms on the other side. Subtract $9k$ from both sides:

$9k + 24 - 9k = 10k + 16 - 9k$

$24 = k + 16$

Subtract 16 from both sides to find the value of $k$:

$24 - 16 = k + 16 - 16$

$8 = k$

So, the value of $k$ is 8.

Now that we have the value of $k$, we can find the original numbers using our initial representations $3k$ and $5k$.

The first original number is $3k = 3 \times 8 = 24$.

The second original number is $5k = 5 \times 8 = 40$.

The original numbers are 24 and 40.

Let's verify the conditions with the numbers 24 and 40:

The ratio of the original numbers is $\frac{24}{40}$. We can simplify this fraction by dividing the numerator and the denominator by their greatest common divisor, which is 8:

$\frac{24}{40} = \frac{\cancel{24}^3}{\cancel{40}_5} = \frac{3}{5}$. The original ratio condition is satisfied.

Now, add 8 to each number:

New first number $= 24 + 8 = 32$

New second number $= 40 + 8 = 48$

The ratio of the new numbers is $\frac{32}{48}$. We can simplify this fraction by dividing the numerator and the denominator by their greatest common divisor, which is 16:

$\frac{32}{48} = \frac{\cancel{32}^2}{\cancel{48}_3} = \frac{2}{3}$. The condition about the new ratio is also satisfied.

The original numbers are 24 and 40.

Comparing our calculated numbers with the given options, we find that the correct option is (A).

The correct option is (A) 24, 40.

Given:

1. The original length ($l$) of a rectangle is three times its width ($w$).

2. The new length ($l'$) is the original length decreased by $3 \text{ m}$.

3. The new width ($w'$) is the original width increased by $1 \text{ m}$.

4. The area of the new rectangle is equal to the area of the original rectangle.

To Find:

The original width ($w$).

Solution:

Let the original width of the rectangle be $w$ meters.

According to the given information, the original length is $l = 3w$ meters.

The area of the original rectangle is given by the formula $A = \text{length} \times \text{width}$.

When the length is decreased by $3 \text{ m}$, the new length is $l' = l - 3 = 3w - 3$ meters.

When the width is increased by $1 \text{ m}$, the new width is $w' = w + 1$ meters.

The area of the new rectangle is $A' = l' \times w'$.

$A' = (3w - 3)(w + 1)$

Expanding the expression for $A'$:

$A' = 3w(w + 1) - 3(w + 1)$

$A' = 3w^2 + 3w - 3w - 3$

According to the problem, the area of the new rectangle is equal to the area of the original rectangle.

So, $A' = A$.

Substituting the expressions from (1) and (2):

$3w^2 - 3 = 3w^2$

Subtracting $3w^2$ from both sides of the equation, we get:

$-3 = 0$

This is a contradiction, which indicates that there is no value of $w$ that satisfies the given conditions when the original length is exactly three times the width.

Analysis and Assuming a Typo:

Since the problem is presented as an objective question with specific options, it is highly probable that there is a typo in the statement. A common error in such problems involves the initial relationship between the length and the width.

Let's assume that the original length was intended to be four times its width, instead of three times, as this assumption often leads to a solvable equation that matches one of the options in problems of this type.

Let's proceed with the assumption that the original length $l = 4w$.

The original area $A = l \times w = (4w) \times w = 4w^2$.

The new length $l' = l - 3 = 4w - 3$ meters.

The new width $w' = w + 1$ meters.

The area of the new rectangle is $A' = l' \times w' = (4w - 3)(w + 1)$.

Expanding the expression for $A'$:

$A' = 4w(w + 1) - 3(w + 1)$

$A' = 4w^2 + 4w - 3w - 3$

$A' = 4w^2 + w - 3$.

Setting the new area equal to the original area ($A' = A$):

$4w^2 + w - 3 = 4w^2$

Subtracting $4w^2$ from both sides:

Solving for $w$:

$w = 3$

Thus, assuming the original length was four times the width, the original width is $3$ meters.

Verification (based on the assumed typo $l=4w$):

If the original width $w = 3 \text{ m}$, then the original length $l = 4w = 4 \times 3 = 12 \text{ m}$.

The original area $A = 12 \times 3 = 36 \text{ m}^2$.

The new length $l' = l - 3 = 12 - 3 = 9 \text{ m}$.

The new width $w' = w + 1 = 3 + 1 = 4 \text{ m}$.

The new area $A' = l' \times w' = 9 \times 4 = 36 \text{ m}^2$.

Since the new area ($36 \text{ m}^2$) is equal to the original area ($36 \text{ m}^2$), the value $w=3$ is consistent with the problem's conditions under the assumption that the original length was four times the width.

Based on the probable intended question (assuming the typo), the original width is $3 \text{ m}$.

Comparing this to the given options, the value $3 \text{ m}$ matches option (C).

The final answer is $\boxed{\text{3 m}}$.

Solution:

We are asked to solve the linear equation:

To eliminate the denominators, we find the Least Common Multiple (LCM) of the denominators 4 and 3. The LCM of 4 and 3 is 12.

Multiply every term on both sides of the equation by 12:

$12 \left( \frac{3t-2}{4} \right) - 12 \left( \frac{2t+3}{3} \right) = 12 \left( \frac{2}{3} \right) - 12(t)$

Simplify each term:

$\cancel{12}^3 \left( \frac{3t-2}{\cancel{4}_1} \right) - \cancel{12}^4 \left( \frac{2t+3}{\cancel{3}_1} \right) = \cancel{12}^4 \left( \frac{2}{\cancel{3}_1} \right) - 12t$

$3(3t - 2) - 4(2t + 3) = 4(2) - 12t$

Distribute the numbers into the parentheses:

$9t - 6 - (8t + 12) = 8 - 12t$

Remove the parenthesis on the left side, remembering to change the signs of the terms inside because of the negative sign in front:

$9t - 6 - 8t - 12 = 8 - 12t$

Combine like terms on the left side:

$(9t - 8t) + (-6 - 12) = 8 - 12t$

$t - 18 = 8 - 12t$

Now, we want to isolate the terms containing $t$ on one side and the constant terms on the other side.

Add $12t$ to both sides of the equation:

$t - 18 + 12t = 8 - 12t + 12t$

Combine like terms:

$(t + 12t) - 18 = 8$

$13t - 18 = 8$

Add 18 to both sides of the equation:

$13t - 18 + 18 = 8 + 18$

$13t = 26$

Finally, divide both sides by 13 to solve for $t$:

$t = 2$

Verification:

Substitute $t=2$ into the original equation (1):

Left Hand Side (LHS): $\frac{3(2)-2}{4} - \frac{2(2)+3}{3}$

LHS = $\frac{6-2}{4} - \frac{4+3}{3}$

LHS = $\frac{4}{4} - \frac{7}{3}$

LHS = $1 - \frac{7}{3}$

LHS = $\frac{3}{3} - \frac{7}{3}$

LHS = $\frac{3-7}{3} = -\frac{4}{3}$

Right Hand Side (RHS): $\frac{2}{3} - t$

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

RHS = $\frac{2}{3} - \frac{6}{3}$

RHS = $\frac{2-6}{3} = -\frac{4}{3}$

Since LHS = RHS, the solution $t=2$ is correct.

Comparing the result with the given options, $t=2$ matches option (A).

The final answer is $\boxed{t = 2}$.

Solution:

Let the three consecutive multiples of 8 be $8n$, $8(n+1)$, and $8(n+2)$, where $n$ is an integer.

According to the problem, the sum of these three multiples is 888.

So, we can write the equation:

Expand the terms inside the parentheses:

$8n + 8n + 8 + 8n + 16 = 888$

Combine the terms involving $n$ and the constant terms:

$(8n + 8n + 8n) + (8 + 16) = 888$

$24n + 24 = 888$

To isolate the term with $n$, subtract 24 from both sides of the equation:

$24n + 24 - 24 = 888 - 24$

$24n = 864$

Now, divide both sides by 24 to find the value of $n$:

$n = 36$

Now that we have the value of $n$, we can find the three consecutive multiples of 8:

First multiple = $8n = 8 \times 36 = 288$

Second multiple = $8(n+1) = 8(36+1) = 8(37) = 296$

Third multiple = $8(n+2) = 8(36+2) = 8(38) = 304$

The three consecutive multiples of 8 are 288, 296, and 304.

Verification:

Let's check if the sum of these multiples is 888:

$288 + 296 + 304 = 584 + 304 = 888$

The sum is indeed 888, so our answer is correct.

Comparing the result with the given options, the multiples 288, 296, and 304 match option (A).

The final answer is $\boxed{\text{288, 296, 304}}$.

Given:

1. The bag contains $\textsf{₹} 1$, $\textsf{₹} 2$, and $\textsf{₹} 5$ coins.

2. The number of $\textsf{₹} 2$ coins is 3 times the number of $\textsf{₹} 5$ coins.

3. The total value of the coins is $\textsf{₹} 150$.

4. The total number of coins is 60.

To Find:

The number of each type of coin ($\textsf{₹} 1$, $\textsf{₹} 2$, and $\textsf{₹} 5$).

Solution:

Let $n_1$ be the number of $\textsf{₹} 1$ coins.

Let $n_2$ be the number of $\textsf{₹} 2$ coins.

Let $n_5$ be the number of $\textsf{₹} 5$ coins.

From the given information, we can set up the following equations:

1. The number of $\textsf{₹} 2$ coins is 3 times the number of $\textsf{₹} 5$ coins:

2. The total number of coins is 60:

3. The total value of the coins is $\textsf{₹} 150$. The value of the $\textsf{₹} 1$ coins is $1 \times n_1 = n_1$, the value of the $\textsf{₹} 2$ coins is $2 \times n_2 = 2n_2$, and the value of the $\textsf{₹} 5$ coins is $5 \times n_5 = 5n_5$.

We now have a system of three linear equations with three variables:

(i) $n_2 = 3n_5$

(ii) $n_1 + n_2 + n_5 = 60$

(iii) $n_1 + 2n_2 + 5n_5 = 150$

Substitute equation (i) into equation (ii):

$n_1 + (3n_5) + n_5 = 60$

Substitute equation (i) into equation (iii):

$n_1 + 2(3n_5) + 5n_5 = 150$

$n_1 + 6n_5 + 5n_5 = 150$

Now we have a system of two equations with two variables, $n_1$ and $n_5$. Subtract equation (iv) from equation (v):

$(n_1 + 11n_5) - (n_1 + 4n_5) = 150 - 60$

$n_1 + 11n_5 - n_1 - 4n_5 = 90$

$7n_5 = 90$

Solving for $n_5$:

$n_5 = \frac{90}{7}$

Since the number of coins must be a whole number (integer), the result $n_5 = \frac{90}{7}$ indicates that there is no set of integer coin counts that satisfies all three conditions simultaneously as stated in the problem. This suggests there might be a typo in the problem statement.

Checking the Options:

Let's examine the given options to see if any of them satisfy the conditions, assuming one of the conditions in the problem statement might be misprinted.

Option (A): $n_1=30, n_2=20, n_5=10$

- Condition (i): $n_2 = 3n_5 \implies 20 = 3 \times 10 \implies 20 = 30$ (False)

Option (B): $n_1=10, n_2=30, n_5=20$

- Condition (i): $n_2 = 3n_5 \implies 30 = 3 \times 20 \implies 30 = 60$ (False)

Option (C): $n_1=20, n_2=30, n_5=10$

- Condition (i): $n_2 = 3n_5 \implies 30 = 3 \times 10 \implies 30 = 30$ (True)

- Condition (ii): $n_1 + n_2 + n_5 = 60 \implies 20 + 30 + 10 = 60 \implies 60 = 60$ (True)

- Condition (iii): $n_1 + 2n_2 + 5n_5 = 150 \implies 20 + 2(30) + 5(10) = 150 \implies 20 + 60 + 50 = 150 \implies 130 = 150$ (False)

Option (D): $n_1=20, n_2=10, n_5=30$

- Condition (i): $n_2 = 3n_5 \implies 10 = 3 \times 30 \implies 10 = 90$ (False)

Based on the verification, option (C) satisfies two out of the three conditions (the ratio of $\textsf{₹} 2$ to $\textsf{₹} 5$ coins and the total number of coins). It only fails the condition regarding the total value, yielding $\textsf{₹} 130$ instead of $\textsf{₹} 150$. Given the nature of multiple-choice questions, it is highly probable that the intended total value in the question was $\textsf{₹} 130$ instead of $\textsf{₹} 150$, and option (C) is the correct answer based on the likely intended problem.

Assuming that option (C) represents the intended solution set despite the inconsistency with the stated total value of $\textsf{₹} 150$, the number of coins are:

Number of $\textsf{₹} 1$ coins = 20

Number of $\textsf{₹} 2$ coins = 30

Number of $\textsf{₹} 5$ coins = 10

The final answer based on the provided options and the likelihood of a typo in the question is $\boxed{\text{₹} 1 \text{ coins: 20, ₹} 2 \text{ coins: 30, ₹} 5 \text{ coins: 10}}$.

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

Example:

$2x + 5 = 11$

Identification:

In the equation $2x + 5 = 11$:

Variable: The variable is $x$. It is the unknown quantity whose value we want to find.

Constants: The constants are the numerical values, which are $2$, $5$, and $11$.

LHS (Left Hand Side): The expression on the left side of the equals sign is $2x + 5$.

RHS (Right Hand Side): The expression on the right side of the equals sign is $11$.

No.

Justification:

A linear equation in one variable is defined as an equation that involves only one variable and the highest power of that variable is $1$.

The given equation is $x^2 + 2x - 3 = 0$.

In this equation, there is only one variable, which is $x$.

However, the highest power of the variable $x$ in this equation is $2$ (due to the term $x^2$).

Since the highest power of the variable is $2$, and not $1$, the equation $x^2 + 2x - 3 = 0$ is not a linear equation in one variable. It is a quadratic equation in one variable.

To check if $x = 5$ is a solution to the equation $3x - 7 = 8$, we substitute $x=5$ into the Left Hand Side (LHS) of the equation.

Given equation:

$3x - 7 = 8$

Substitute $x = 5$ into the LHS:

LHS $= 3x - 7$

LHS $= 3(5) - 7$

LHS $= 15 - 7$

LHS $= 8$

Now, we compare the calculated LHS with the RHS of the given equation.

RHS $= 8$

Since LHS $=$ RHS ($8 = 8$), the equation is satisfied when $x = 5$.

Therefore, $x = 5$ is a solution to the equation $3x - 7 = 8$.

To check if $y = -2$ is a solution to the equation $5y + 6 = y - 2$, we substitute $y = -2$ into both the Left Hand Side (LHS) and the Right Hand Side (RHS) of the equation.

Given equation:

$5y + 6 = y - 2$

Substitute $y = -2$ into the LHS:

LHS $= 5y + 6$

LHS $= 5(-2) + 6$

LHS $= -10 + 6$

LHS $= -4$

Substitute $y = -2$ into the RHS:

RHS $= y - 2$

RHS $= (-2) - 2$

RHS $= -4$

Now, we compare the calculated LHS and RHS.

LHS $= -4$

RHS $= -4$

Since LHS $=$ RHS ($-4 = -4$), the equation is satisfied when $y = -2$.

Therefore, $y = -2$ is a solution to the equation $5y + 6 = y - 2$.

Let the number be represented by the variable $p$.

The phrase "five times a number $p$" can be written mathematically as $5 \times p$, or simply $5p$.

The phrase "Seven subtracted from five times a number $p$" means we take $5p$ and subtract $7$. This can be written as $5p - 7$.

The word "is" indicates equality, and the number after "is" is $18$.

Combining these parts, the statement "Seven subtracted from five times a number $p$ is $18$" translates to the equation:

$5p - 7 = 18$

This is a linear equation in one variable $p$, as the variable $p$ appears with a power of $1$.

Let the number be represented by the variable $k$.

The phrase "one-third of a number $k$" can be written mathematically as $\frac{1}{3} \times k$, or simply $\frac{k}{3}$.

The phrase "When 6 is added to one-third of a number $k$" means we take $\frac{k}{3}$ and add $6$. This can be written as $\frac{k}{3} + 6$.

The phrase "the result is 10" indicates that the expression on the left side is equal to $10$.

Combining these parts, the statement "When 6 is added to one-third of a number $k$, the result is 10" translates to the equation:

$\frac{k}{3} + 6 = 10$

This is a linear equation in one variable $k$, as the variable $k$ appears with a power of $1$.

We are asked to solve the equation $x + 9 = -5$ using the balancing method.

The goal of the balancing method is to isolate the variable ($x$ in this case) on one side of the equation by performing the same operation on both sides.

Given equation:

$x + 9 = -5$

To eliminate the $+9$ on the left side, we subtract $9$ from both sides of the equation.

Subtract $9$ from LHS: $(x + 9) - 9$

Subtract $9$ from RHS: $-5 - 9$

Performing the operation on both sides:

$x + 9 - 9 = -5 - 9$

Simplify both sides:

On the LHS, $+9 - 9 = 0$, so we are left with $x$.

On the RHS, $-5 - 9 = -14$.

The equation becomes:

$x = -14$

Thus, the solution to the equation $x + 9 = -5$ is $x = -14$.

We are asked to solve the equation $y - 12 = 8$ using the balancing method.

The goal of the balancing method is to isolate the variable ($y$ in this case) on one side of the equation by performing the same operation on both sides.

Given equation:

$y - 12 = 8$

To eliminate the $-12$ on the left side, we add $12$ to both sides of the equation.

Add $12$ to LHS: $(y - 12) + 12$

Add $12$ to RHS: $8 + 12$

Performing the operation on both sides:

$y - 12 + 12 = 8 + 12$

Simplify both sides:

On the LHS, $-12 + 12 = 0$, so we are left with $y$.

On the RHS, $8 + 12 = 20$.

The equation becomes:

$y = 20$

Thus, the solution to the equation $y - 12 = 8$ is $y = 20$.

We are asked to solve the equation $6z = 42$ using the balancing method.

The goal is to isolate the variable $z$ on one side of the equation.

Given equation:

$6z = 42$

The variable $z$ is multiplied by $6$. To isolate $z$, we perform the inverse operation, which is division by $6$. We must do this on both sides of the equation to maintain the balance.

Divide both sides by $6$:

$\frac{6z}{6} = \frac{42}{6}$

Simplify both sides:

On the LHS, $\frac{6z}{6}$ simplifies to $z$.

On the RHS, $\frac{42}{6}$ simplifies to $7$.

The equation becomes:

$z = 7$

Thus, the solution to the equation $6z = 42$ is $z = 7$.

We are asked to solve the equation $\frac{m}{5} = 15$ using the balancing method.

The goal is to isolate the variable $m$ on one side of the equation.

Given equation:

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

The variable $m$ is being divided by $5$. To isolate $m$, we perform the inverse operation, which is multiplication by $5$. We must do this on both sides of the equation to maintain the balance.

Multiply both sides by $5$:

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

Simplify both sides:

On the LHS, $\frac{m}{5} \times 5$ simplifies to $m$.

On the RHS, $15 \times 5 = 75$.

The equation becomes:

$m = 75$

Thus, the solution to the equation $\frac{m}{5} = 15$ is $m = 75$.

We are asked to solve the equation $2x + 7 = 17$ using the rule of transposition.

Transposition involves moving terms from one side of the equation to the other while changing their signs.

Given equation:

$2x + 7 = 17$

Transpose the constant term $+7$ from the LHS to the RHS. When $+7$ is moved to the RHS, it becomes $-7$.

$2x = 17 - 7$

Simplify the RHS:

$2x = 10$

Now, the term $2x$ means $2 \times x$. The coefficient $2$ is multiplying the variable $x$ on the LHS. To isolate $x$, we transpose the coefficient $2$ from the LHS to the RHS. When $2$ is moved to the RHS, it will divide the term on the RHS.

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

Simplify the RHS:

$x = 5$

Thus, the solution to the equation $2x + 7 = 17$ is $x = 5$.

We are asked to solve the equation $4a - 3 = 9$ using the rule of transposition.

Transposition involves moving terms from one side of the equation to the other while changing their signs or inverse operations.

Given equation:

$4a - 3 = 9$

Transpose the constant term $-3$ from the LHS to the RHS. When $-3$ is moved to the RHS, it becomes $+3$.

$4a = 9 + 3$

Simplify the RHS:

$4a = 12$

Now, the term $4a$ means $4 \times a$. The coefficient $4$ is multiplying the variable $a$ on the LHS. To isolate $a$, we transpose the coefficient $4$ from the LHS to the RHS. When $4$ is moved to the RHS, it will divide the term on the RHS.

$a = \frac{12}{4}$

Simplify the RHS:

$a = 3$

Thus, the solution to the equation $4a - 3 = 9$ is $a = 3$.

We are asked to solve the equation $\frac{p}{3} - 4 = 1$ using the rule of transposition.

Transposition involves moving terms from one side of the equation to the other while changing their signs or inverse operations.

Given equation:

$\frac{p}{3} - 4 = 1$

Transpose the constant term $-4$ from the LHS to the RHS. When $-4$ is moved to the RHS, it becomes $+4$.

$\frac{p}{3} = 1 + 4$

Simplify the RHS:

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

Now, the variable $p$ is being divided by $3$ on the LHS. To isolate $p$, we transpose the divisor $3$ from the LHS to the RHS. When $3$ is moved to the RHS, it will multiply the term on the RHS.

$p = 5 \times 3$

Simplify the RHS:

$p = 15$

Thus, the solution to the equation $\frac{p}{3} - 4 = 1$ is $p = 15$.

We need to solve the equation $5 - 2q = 11$.

Given equation:

$5 - 2q = 11$

Transpose the constant term $5$ from the LHS to the RHS. When $5$ is moved to the RHS, it changes its sign from positive to negative.

$-2q = 11 - 5$

Simplify the RHS:

$11 - 5 = 6$

So, the equation becomes:

$-2q = 6$

Now, the term $-2q$ means $-2 \times q$. The coefficient $-2$ is multiplying the variable $q$ on the LHS. To isolate $q$, we transpose the coefficient $-2$ from the LHS to the RHS. When $-2$ is moved to the RHS, it will divide the term on the RHS.

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

Simplify the RHS by performing the division:

$q = -3$

Thus, the solution to the equation $5 - 2q = 11$ is $q = -3$.

Framing the Equation:

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

The statement "The sum of a number and $15$" translates to $x + 15$.

The statement "is $40$" indicates that the sum is equal to $40$.

Therefore, the linear equation representing the given statement is:

$x + 15 = 40$

Finding the Number (Solving the Equation):

We need to solve the equation $x + 15 = 40$ for $x$.

Given equation:

$x + 15 = 40$

To isolate $x$, we transpose the constant term $+15$ from the LHS to the RHS. When $+15$ is moved to the RHS, it becomes $-15$.

$x = 40 - 15$

Simplify the RHS by performing the subtraction:

$40 - 15 = 25$

So, the value of $x$ is:

$x = 25$

Thus, the unknown number is $25$.

Framing the Equation:

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

The phrase "Three times a number" translates to $3 \times n$, or simply $3n$.

The phrase "decreased by $8$" means we subtract $8$ from $3n$. This translates to $3n - 8$.

The phrase "is $16$" indicates that the expression on the left side is equal to $16$.

Therefore, the linear equation representing the given statement is:

$3n - 8 = 16$

Finding the Number (Solving the Equation):

We need to solve the equation $3n - 8 = 16$ for $n$.

Given equation:

$3n - 8 = 16$

Transpose the constant term $-8$ from the LHS to the RHS. When $-8$ is moved to the RHS, it becomes $+8$.

$3n = 16 + 8$

Simplify the RHS:

$16 + 8 = 24$

So, the equation becomes:

$3n = 24$

Now, the term $3n$ means $3 \times n$. The coefficient $3$ is multiplying the variable $n$ on the LHS. To isolate $n$, we transpose the coefficient $3$ from the LHS to the RHS. When $3$ is moved to the RHS, it will divide the term on the RHS.

$n = \frac{24}{3}$

Simplify the RHS by performing the division:

$n = 8$

Thus, the unknown number is $8$.

Let the width of the rectangle be $w$ cm (as given).

The length of the rectangle is $4$ cm more than its width. So, the length ($l$) can be expressed in terms of $w$ as:

$l = w + 4$ cm

The formula for the perimeter of a rectangle is:

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

We are given that the perimeter is $28$ cm. Substituting the expressions for length and width, and the given perimeter into the formula:

$28 = 2 \times ((w + 4) + w)$

Simplify the expression inside the parentheses:

$28 = 2 \times (w + 4 + w)$

$28 = 2 \times (2w + 4)$

Distribute the $2$ on the RHS:

$28 = 4w + 8$

This is the linear equation for the perimeter of the rectangle in terms of the width $w$.

So, the equation is:

$4w + 8 = 28$

We need to solve the equation $1.5x = 4.5$.

Given equation:

$1.5x = 4.5$

The variable $x$ is being multiplied by $1.5$. To isolate $x$, we need to divide both sides of the equation by $1.5$.

Divide both sides by $1.5$:

$\frac{1.5x}{1.5} = \frac{4.5}{1.5}$

Simplify both sides:

On the LHS, $\frac{1.5x}{1.5}$ simplifies to $x$.

On the RHS, we perform the division $\frac{4.5}{1.5}$. This is equivalent to $\frac{45}{15}$.

$\frac{45}{15} = 3$

The equation becomes:

$x = 3$

Thus, the solution to the equation $1.5x = 4.5$ is $x = 3$.

We need to solve the equation $x + \frac{3}{5} = \frac{7}{5}$ for the variable $x$.

Given equation:

$x + \frac{3}{5} = \frac{7}{5}$

To isolate $x$ on the left side of the equation, we need to eliminate the term $+\frac{3}{5}$. We can do this by transposing $+\frac{3}{5}$ from the LHS to the RHS. When a term is transposed to the other side of the equals sign, its sign changes.

$x = \frac{7}{5} - \frac{3}{5}$

Now, we perform the subtraction on the right side. Since the fractions have the same denominator ($5$), we can subtract the numerators directly.

$x = \frac{7 - 3}{5}$

Calculate the numerator:

$7 - 3 = 4$

So, the equation becomes:

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

Thus, the solution to the equation $x + \frac{3}{5} = \frac{7}{5}$ is $x = \frac{4}{5}$.

Let the number of books Rahul buys be represented by the variable $n$.

The cost of one book is given as $\textsf{₹}50$.

If Rahul buys $n$ books, the total cost will be the number of books multiplied by the cost per book.

Total cost $= (\text{Cost per book}) \times (\text{Number of books})$

Total cost $= \textsf{₹}50 \times n$

Total cost $= 50n$

We are given that the total cost is $\textsf{₹}300$.

Therefore, we can set the expression for the total cost equal to $300$:

$50n = 300$

This is the linear equation representing the given scenario.

The equation is:

$50n = 300$

We need to solve the equation $\frac{2}{3}y = 12$ for the variable $y$.

Given equation:

$\frac{2}{3}y = 12$

To isolate the variable $y$, we need to eliminate the coefficient $\frac{2}{3}$ which is multiplying $y$ on the left side. We can do this by multiplying both sides of the equation by the reciprocal of $\frac{2}{3}$, which is $\frac{3}{2}$.

Multiply both sides by $\frac{3}{2}$:

$\left(\frac{3}{2}\right) \times \left(\frac{2}{3}y\right) = 12 \times \left(\frac{3}{2}\right)$

Simplify both sides:

On the LHS, $\frac{3}{2} \times \frac{2}{3} = \frac{6}{6} = 1$. So, $\left(\frac{3}{2}\right) \times \left(\frac{2}{3}y\right) = 1y = y$.

On the RHS, we calculate $12 \times \frac{3}{2}$. We can simplify this by dividing $12$ by $2$ first:

$12 \times \frac{3}{2} = \cancel{12}^{6} \times \frac{3}{\cancel{2}_{1}} = 6 \times 3 = 18$

The equation becomes:

$y = 18$

Thus, the solution to the equation $\frac{2}{3}y = 12$ is $y = 18$.

Let the price of one movie ticket be $\textsf{₹}x$ (as given).

The cost of $4$ tickets will be the number of tickets multiplied by the price per ticket.

Cost of $4$ tickets $= 4 \times \textsf{₹}x = \textsf{₹}4x$.

The cost of $10$ tickets will be the number of tickets multiplied by the price per ticket.

Cost of $10$ tickets $= 10 \times \textsf{₹}x = \textsf{₹}10x$.

The problem states that "The cost of 4 tickets is $\textsf{₹}600$ less than the cost of 10 tickets". We can write this relationship as:

Cost of $4$ tickets $=$ Cost of $10$ tickets $- \textsf{₹}600$

Substitute the expressions for the costs in terms of $x$ into this relationship:

$4x = 10x - 600$

This is the required linear equation.

First, we need to find the value of $x$ from the given equation.

Given equation:

$x - 7 = 10$

To isolate $x$ on the left side, we transpose the constant term $-7$ from the LHS to the RHS. When $-7$ is moved to the RHS, it becomes $+7$.

$x = 10 + 7$

Perform the addition on the RHS:

$x = 17$

Now that we have found the value of $x$, we need to find the value of the expression $x + 2$.

Substitute the value of $x=17$ into the expression $x + 2$:

$x + 2 = 17 + 2$

Perform the addition:

$17 + 2 = 19$

Therefore, the value of $x + 2$ is $19$.

We are asked to solve the equation $3x - 5 = x + 7$ using the rule of transposition and show each step clearly.

The goal is to isolate the variable term ($x$) on one side of the equation and the constant terms on the other side.

Given equation:

$3x - 5 = x + 7$

Step 1: Transpose the variable term $x$ from the Right Hand Side (RHS) to the Left Hand Side (LHS). When $x$ (which is $+x$) moves from RHS to LHS, its sign changes to $-x$.

$3x - x - 5 = 7$

Step 2: Simplify the LHS by combining the like terms ($3x$ and $-x$).

$(3 - 1)x - 5 = 7$

$2x - 5 = 7$

Step 3: Transpose the constant term $-5$ from the LHS to the RHS. When $-5$ moves from LHS to RHS, its sign changes to $+5$.

$2x = 7 + 5$

Step 4: Simplify the RHS by performing the addition.

$2x = 12$

Step 5: The term $2x$ means $2 \times x$. The coefficient $2$ is multiplying the variable $x$ on the LHS. To isolate $x$, we transpose the coefficient $2$ from the LHS to the RHS. When a multiplicative coefficient moves to the other side, it becomes a divisor.

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

Step 6: Simplify the RHS by performing the division.

$x = 6$

Thus, the solution to the equation $3x - 5 = x + 7$ is $x = 6$.

We need to solve the equation $5(p + 3) = 2(p + 6)$ for the variable $p$.

Given equation:

$5(p + 3) = 2(p + 6)$

Step 1: Handle the brackets (Expand both sides).

We use the distributive property, which states that $a(b+c) = ab + ac$. We apply this to both sides of the equation.

Expand the left side: $5 \times (p + 3) = 5 \times p + 5 \times 3 = 5p + 15$.

Expand the right side: $2 \times (p + 6) = 2 \times p + 2 \times 6 = 2p + 12$.

The equation now becomes:

$5p + 15 = 2p + 12$

Step 2: Collect like terms.

We want to bring all the terms containing the variable ($p$) to one side of the equation (usually the left side) and all the constant terms to the other side (usually the right side). This is done using transposition, where a term moved across the equals sign changes its operation (addition becomes subtraction, multiplication becomes division, and vice versa).

Transpose the term $2p$ from the RHS to the LHS. Since it is $+2p$ on the RHS, it becomes $-2p$ on the LHS.

$5p - 2p + 15 = 12$

Now, transpose the constant term $15$ from the LHS to the RHS. Since it is $+15$ on the LHS, it becomes $-15$ on the RHS.

$5p - 2p = 12 - 15$

Step 3: Simplify both sides.

Combine the like terms on both sides of the equation.

On the LHS: $5p - 2p = 3p$.

On the RHS: $12 - 15 = -3$.

The equation simplifies to:

$3p = -3$

Step 4: Solve for the variable $p$.

The variable $p$ is currently multiplied by $3$. To isolate $p$, we need to perform the inverse operation of multiplication by $3$, which is division by $3$. We transpose the coefficient $3$ from the LHS to the RHS, where it will divide the term on the RHS.

$p = \frac{-3}{3}$

Perform the division on the RHS:

$p = -1$

Thus, the solution to the equation $5(p + 3) = 2(p + 6)$ is $p = -1$.

We need to solve the equation $\frac{x}{3} + \frac{x}{4} = \frac{7}{12}$ for the variable $x$. This equation involves fractions.

Step 1: Find a common denominator.

To clear the fractions from the equation, we find the Least Common Multiple (LCM) of the denominators of all the terms in the equation. The denominators are $3$, $4$, and $12$.

The multiples of $3$ are $3, 6, 9, 12, 15, ...$

The multiples of $4$ are $4, 8, 12, 16, ...$

The multiples of $12$ are $12, 24, ...$

The least common multiple of $3$, $4$, and $12$ is $12$.

Step 2: Clear the fractions by multiplying by the common denominator.

We multiply every term on both sides of the equation by the LCM, which is $12$.

$12 \times \left(\frac{x}{3}\right) + 12 \times \left(\frac{x}{4}\right) = 12 \times \left(\frac{7}{12}\right)$

Step 3: Simplify each term.

Perform the multiplication and cancellation for each term:

$12 \times \frac{x}{3} = \frac{\cancel{12}^{4} \times x}{\cancel{3}_{1}} = 4x$

$12 \times \frac{x}{4} = \frac{\cancel{12}^{3} \times x}{\cancel{4}_{1}} = 3x$

$12 \times \frac{7}{12} = \frac{\cancel{12}^{1} \times 7}{\cancel{12}_{1}} = 7$

The equation without fractions is:

$4x + 3x = 7$

Step 4: Combine like terms.

Combine the terms involving $x$ on the left side of the equation.

$(4 + 3)x = 7$

$7x = 7$

Step 5: Solve for the variable $x$.

The variable $x$ is multiplied by $7$. To isolate $x$, we divide both sides of the equation by $7$.

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

Simplify both sides:

$x = 1$

Thus, the solution to the equation $\frac{x}{3} + \frac{x}{4} = \frac{7}{12}$ is $x = 1$.

Defining the Variable:

Let the first of the three consecutive integers be represented by the variable $x$.

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

The third integer will be one more than the second (or two more than the first), i.e., $(x + 1) + 1 = x + 2$.

So, the three consecutive integers are $x$, $x+1$, and $x+2$.

Framing the Equation:

The problem states that the sum of these three consecutive integers is $84$. Therefore, we can write the equation as:

$x + (x + 1) + (x + 2) = 84$

Solving the Equation:

We need to solve the equation $x + (x + 1) + (x + 2) = 84$ for $x$.

Step 1: Combine the like terms on the left side of the equation.

Combine the $x$ terms: $x + x + x = 3x$.

Combine the constant terms: $1 + 2 = 3$.

The equation becomes:

$3x + 3 = 84$

Step 2: Transpose the constant term $+3$ from the LHS to the RHS. When $+3$ is moved to the RHS, it becomes $-3$.

$3x = 84 - 3$

Step 3: Simplify the RHS.

$84 - 3 = 81$

The equation is now:

$3x = 81$

Step 4: The variable $x$ is multiplied by $3$. To isolate $x$, transpose the coefficient $3$ from the LHS to the RHS. When $3$ is moved to the RHS, it will divide the term on the RHS.

$x = \frac{81}{3}$

Step 5: Simplify the RHS by performing the division.

$x = 27$

Finding the Integers:

Now that we have found the value of $x$, we can find the three consecutive integers:

First integer $= x = 27$

Second integer $= x + 1 = 27 + 1 = 28$

Third integer $= x + 2 = 27 + 2 = 29$

Thus, the three consecutive integers are $27$, $28$, and $29$.

We can check the sum: $27 + 28 + 29 = 84$, which matches the given condition.

Defining the Variable and Dimensions:

Let the breadth of the rectangle be represented by the variable $b$ cm.

We are given that the length is $10$ cm more than its breadth.

So, the length ($l$) of the rectangle can be expressed as $l = b + 10$ cm.

The perimeter of the rectangle is given as $60$ cm.

Framing the Equation:

The formula for the perimeter of a rectangle is:

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

Substitute the given perimeter and the expressions for length and breadth into the formula:

$60 = 2 \times ((b + 10) + b)$

Simplify the expression inside the parentheses:

$60 = 2 \times (b + 10 + b)$

$60 = 2 \times (2b + 10)$

Distribute the $2$ on the RHS:

$60 = 4b + 20$

This is the linear equation representing the given problem.

So, the equation is:

$4b + 20 = 60$

Solving the Equation:

We need to solve the equation $4b + 20 = 60$ for $b$.

Given equation:

$4b + 20 = 60$

Transpose the constant term $+20$ from the LHS to the RHS. When $+20$ moves, it becomes $-20$.

$4b = 60 - 20$

Simplify the RHS:

$4b = 40$

The variable $b$ is multiplied by $4$. To isolate $b$, transpose the coefficient $4$ from the LHS to the RHS. When $4$ moves, it divides the term on the RHS.

$b = \frac{40}{4}$

Simplify the RHS:

$b = 10$

Finding the Dimensions:

Now that we have found the value of the breadth $b$, we can find the dimensions:

Breadth $= b = 10$ cm

Length $= b + 10 = 10 + 10 = 20$ cm

Thus, the dimensions of the rectangle are: Length $= 20$ cm and Breadth $= 10$ cm.

Defining the Variable and Ages:

Let the present age of B be $x$ years.

The present age of A is twice the present age of B.

So, the present age of A is $2x$ years.

Now consider their ages five years ago:

B's age five years ago was $x - 5$ years.

A's age five years ago was $2x - 5$ years.

Framing the Equation:

We are given that five years ago, A's age was three times B's age.

We can write this relationship as:

A's age 5 years ago $= 3 \times$ B's age 5 years ago

$(2x - 5) = 3 \times (x - 5)$

Solving the Equation:

We need to solve the equation $2x - 5 = 3(x - 5)$ for $x$.

Given equation:

$2x - 5 = 3(x - 5)$

Step 1: Expand the right side by distributing the $3$ into the bracket.

$2x - 5 = 3x - 15$

Step 2: Transpose the variable term $3x$ from the RHS to the LHS. It becomes $-3x$.

$2x - 3x - 5 = -15$

Step 3: Transpose the constant term $-5$ from the LHS to the RHS. It becomes $+5$.

$2x - 3x = -15 + 5$

Step 4: Combine like terms on both sides.

$(2 - 3)x = -10$

$-x = -10$

Step 5: Multiply or divide both sides by $-1$ to solve for $x$.

$\frac{-x}{-1} = \frac{-10}{-1}$

$x = 10$

Finding the Present Ages:

Now that we have found the value of $x$, we can find the present ages:

Present age of B $= x = 10$ years.

Present age of A $= 2x = 2 \times 10 = 20$ years.

Thus, the present age of A is $20$ years and the present age of B is $10$ years.

We need to solve the equation $\frac{5x - 2}{3} - \frac{7x - 3}{5} = \frac{1}{2}$.

Step 1: Find the Least Common Multiple (LCM) of the denominators.

The denominators in the equation are $3$, $5$, and $2$.

The LCM of $3$, $5$, and $2$ is $2 \times 3 \times 5 = 30$.

Step 2: Multiply each term by the LCM to clear the fractions.

Multiply both sides of the equation by $30$:

$30 \times \left(\frac{5x - 2}{3}\right) - 30 \times \left(\frac{7x - 3}{5}\right) = 30 \times \left(\frac{1}{2}\right)$

Step 3: Simplify each term.

For the first term on the LHS: $\cancel{30}^{10} \times \frac{(5x - 2)}{\cancel{3}_{1}} = 10(5x - 2)$

For the second term on the LHS: $\cancel{30}^{6} \times \frac{(7x - 3)}{\cancel{5}_{1}} = 6(7x - 3)$

For the term on the RHS: $\cancel{30}^{15} \times \frac{1}{\cancel{2}_{1}} = 15$

The equation becomes:

$10(5x - 2) - 6(7x - 3) = 15$

Step 4: Expand the brackets using the distributive property.

Left side:

$10(5x - 2) = 10 \times 5x - 10 \times 2 = 50x - 20$

$-6(7x - 3) = -6 \times 7x -6 \times (-3) = -42x + 18$

The equation is now:

$50x - 20 - 42x + 18 = 15$

Step 5: Collect like terms.

Combine the $x$ terms and the constant terms on the LHS:

$(50x - 42x) + (-20 + 18) = 15$

$8x - 2 = 15$

Step 6: Isolate the variable term.

Transpose the constant term $-2$ from the LHS to the RHS. When $-2$ moves, it becomes $+2$.

$8x = 15 + 2$

$8x = 17$

Step 7: Solve for $x$.

The variable $x$ is multiplied by $8$. Transpose the coefficient $8$ from the LHS to the RHS. When $8$ moves, it divides the term on the RHS.

$x = \frac{17}{8}$

Thus, the solution to the equation $\frac{5x - 2}{3} - \frac{7x - 3}{5} = \frac{1}{2}$ is $x = \frac{17}{8}$.

Defining the Variable and Shares:

Let the share of B be represented by the variable $\textsf{₹}x$.

We are given that A gets twice as much as B.

So, the share of A is $2 \times (\textsf{₹}x) = \textsf{₹}2x$.

We are given that C gets $\textsf{₹}100$ more than B.

So, the share of C is $\textsf{₹}x + \textsf{₹}100 = \textsf{₹}(x + 100)$.

The total sum to be distributed is $\textsf{₹}1200$.

Framing the Equation:

The sum of the shares of A, B, and C must be equal to the total amount distributed.

Share of A $+$ Share of B $+$ Share of C $=$ Total Amount

$2x + x + (x + 100) = 1200$

This is the linear equation representing the problem.

Solving the Equation:

We need to solve the equation $2x + x + (x + 100) = 1200$ for $x$.

Given equation:

$2x + x + x + 100 = 1200$

Step 1: Combine the like terms on the left side.

Combine the $x$ terms: $2x + x + x = 4x$.

The equation becomes:

$4x + 100 = 1200$

Step 2: Transpose the constant term $+100$ from the LHS to the RHS. It becomes $-100$.

$4x = 1200 - 100$

Step 3: Simplify the RHS.

$1200 - 100 = 1100$

The equation is now:

$4x = 1100$

Step 4: The variable $x$ is multiplied by $4$. To isolate $x$, transpose the coefficient $4$ from the LHS to the RHS. When $4$ moves, it divides the term on the RHS.

$x = \frac{1100}{4}$

Step 5: Simplify the RHS by performing the division.

$x = 275$

Finding the Shares:

Now that we have found the value of $x$, we can find the share of each person:

Share of B $= x = \textsf{₹}275$

Share of A $= 2x = 2 \times 275 = \textsf{₹}550$

Share of C $= x + 100 = 275 + 100 = \textsf{₹}375$

Thus, the shares are: A gets $\textsf{₹}550$, B gets $\textsf{₹}275$, and C gets $\textsf{₹}375$.

We can check the sum: $550 + 275 + 375 = 1200$, which matches the total amount.

Defining the Variable:

Let the numerator of the rational number be represented by the variable $x$.

We are given that the denominator is $5$ more than its numerator.

So, the denominator of the rational number is $x + 5$.

The original rational number is $\frac{\text{Numerator}}{\text{Denominator}} = \frac{x}{x+5}$.

Analysis of the problem statement as written:

The numerator is increased by $3$. The new numerator is $x + 3$.

The denominator is decreased by $2$. The new denominator is $(x+5) - 2 = x+3$.

The new rational number, as stated, becomes $\frac{x+3}{x+3}$.

The equation given is $\frac{x+3}{x+3} = \frac{4}{3}$.

For any value of $x$ where $x+3 \neq 0$ (i.e., $x \neq -3$), the fraction $\frac{x+3}{x+3}$ simplifies to $1$.

The equation would then be $1 = \frac{4}{3}$, which is a false statement. This indicates that there might be a typo in the problem statement.

If $x=-3$, the original number would be $\frac{-3}{-3+5} = \frac{-3}{2}$. The new numerator is $-3+3=0$ and the new denominator is $(-3+5)-2 = 2-2=0$. The new fraction $\frac{0}{0}$ is undefined.

Assuming a likely typo in the problem statement:

It is common in such problems that both the numerator and denominator expressions after the change are not identical. Let's assume that the phrase "denominator is decreased by $2$" was intended to be "denominator is increased by $2$".

Let's proceed with the assumption that the numerator is increased by $3$ and the denominator is increased by $2$.

New numerator = Original numerator $+ 3 = x + 3$.

New denominator = Original denominator $+ 2 = (x+5) + 2 = x+7$.

The new rational number under this assumption is $\frac{x+3}{x+7}$.

We are given that this new number is $\frac{4}{3}$.

Framing the Equation (with assumed typo correction):

The equation is:

$\frac{x+3}{x+7} = \frac{4}{3}$

Solving the Equation:

To solve for $x$, we can cross-multiply:

$3 \times (x+3) = 4 \times (x+7)$

Expand both sides using the distributive property:

$3x + 9 = 4x + 28$

Collect like terms by transposing:

Transpose $4x$ from RHS to LHS (becomes $-4x$).

Transpose $9$ from LHS to RHS (becomes $-9$).

$3x - 4x = 28 - 9$

Simplify both sides:

$-x = 19$

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

$x = -19$

Finding the Original Rational Number:

The original numerator is $x = -19$.

The original denominator is $x + 5 = -19 + 5 = -14$.

The original rational number is $\frac{-19}{-14} = \frac{19}{14}$.

Verification (with assumed typo correction):

Original number: $\frac{19}{14}$. Numerator $= 19$, Denominator $= 14$. Denominator is $14-19 = -5$ more than numerator. This does not match the initial condition (denominator is 5 more than numerator). Let's re-check the variable definition.

Correction and Re-solving:

Let the numerator be $x$. Denominator is $x+5$. Original fraction $\frac{x}{x+5}$. This part is correct.

Let's re-check the algebra for $\frac{x+3}{x+7} = \frac{4}{3}$.

$3(x+3) = 4(x+7)$

$3x + 9 = 4x + 28$

$9 - 28 = 4x - 3x$

$-19 = x$. The algebra is correct.

Original numerator $= -19$. Original denominator $= -19+5 = -14$. Original fraction $= \frac{-19}{-14}$.

Let's check the original condition: Denominator is 5 more than numerator. Is $-14$ five more than $-19$? Yes, $-19 + 5 = -14$. The original condition is satisfied by $x=-19$.

Now let's re-check the assumed change: Numerator increased by 3. New numerator $=-19+3=-16$. Denominator increased by 2. New denominator $=-14+2=-12$. New fraction $=\frac{-16}{-12} = \frac{\cancel{-16}^{4}}{\cancel{-12}_{3}} = \frac{4}{3}$. This matches the given new number.

Therefore, assuming the typo was "denominator increased by 2" instead of "decreased by 2", the original number is $\frac{-19}{-14}$.

Given the high likelihood of a typo in the original question statement as presented, the answer based on the corrected assumption (denominator increased by 2) is provided.

The original rational number is $\frac{-19}{-14}$ or $\frac{19}{14}$.

We need to solve the equation $0.3(6 - 2x) = 0.4(x + 1) + 0.5$ for the variable $x$.

Given equation:

$0.3(6 - 2x) = 0.4(x + 1) + 0.5$

Step 1: Expand the brackets on both sides.

Apply the distributive property ($a(b-c) = ab - ac$ and $a(b+c) = ab + ac$).

Left side: $0.3 \times 6 - 0.3 \times 2x = 1.8 - 0.6x$

Right side: $0.4 \times x + 0.4 \times 1 + 0.5 = 0.4x + 0.4 + 0.5$

The equation becomes:

$1.8 - 0.6x = 0.4x + 0.4 + 0.5$

Step 2: Simplify the right side by combining constant terms.

$0.4 + 0.5 = 0.9$

The equation is now:

$1.8 - 0.6x = 0.4x + 0.9$

Step 3: Collect variable terms on one side and constant terms on the other.

Transpose the term $-0.6x$ from the LHS to the RHS. When it moves, it becomes $+0.6x$.

$1.8 = 0.4x + 0.6x + 0.9$

Transpose the constant term $0.9$ from the RHS to the LHS. When it moves, it becomes $-0.9$.

$1.8 - 0.9 = 0.4x + 0.6x$

Step 4: Simplify both sides.

Left side: $1.8 - 0.9 = 0.9$

Right side: $0.4x + 0.6x = (0.4 + 0.6)x = 1.0x = x$

The equation simplifies to:

$0.9 = x$

Thus, the solution to the equation $0.3(6 - 2x) = 0.4(x + 1) + 0.5$ is $x = 0.9$.

Alternate Solution (Multiplying by 10):

We can clear the decimals by multiplying every term on both sides of the equation by 10.

$10 \times [0.3(6 - 2x)] = 10 \times [0.4(x + 1) + 0.5]$

$10 \times 0.3 \times (6 - 2x) = 10 \times 0.4 \times (x + 1) + 10 \times 0.5$

$3(6 - 2x) = 4(x + 1) + 5$

Now, expand the brackets:

$18 - 6x = 4x + 4 + 5$

$18 - 6x = 4x + 9$

Collect like terms:

$18 - 9 = 4x + 6x$

$9 = 10x$

Solve for $x$:

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

$x = 0.9$

Both methods give the same result.

Defining the Variable:

Let the original speed of the train be $x$ km/hr.

Known Information:

Distance to be covered $= 400$ km.

Increased speed $= (x + 10)$ km/hr.

Time saved with increased speed $= 2$ hours.

Relationship: Time $= \frac{\text{Distance}}{\text{Speed}}$

Framing the Equation:

Calculate the original time taken ($t_1$) at speed $x$:

$t_1 = \frac{400}{x}$ hours.

Calculate the new time taken ($t_2$) at speed $(x + 10)$:

$t_2 = \frac{400}{x+10}$ hours.

We are given that the new time is 2 hours less than the original time.

This can be written as:

$t_2 = t_1 - 2$

Substitute the expressions for $t_1$ and $t_2$ into this equation:

$\frac{400}{x+10} = \frac{400}{x} - 2$

This is the linear equation (which will lead to a quadratic equation upon simplification) representing the problem.

Solving the Equation:

We need to solve the equation $\frac{400}{x+10} = \frac{400}{x} - 2$ for $x$.

Given equation:

$\frac{400}{x+10} = \frac{400}{x} - 2$

Step 1: Find a common denominator for the terms on the RHS.

$\frac{400}{x} - 2 = \frac{400}{x} - \frac{2x}{x} = \frac{400 - 2x}{x}$

The equation is now:

$\frac{400}{x+10} = \frac{400 - 2x}{x}$

Step 2: Cross-multiply to eliminate the denominators.

$400 \times x = (400 - 2x) \times (x + 10)$

$400x = 400x + 4000 - 2x^2 - 20x$

Step 3: Move all terms to one side to form a standard quadratic equation ($ax^2 + bx + c = 0$).

Add $2x^2$ and $20x$ to both sides, and subtract $400x$ from both sides.

$2x^2 + 20x - 4000 + 400x - 400x = 0$

$2x^2 + 20x - 4000 = 0$

Step 4: Divide the entire equation by $2$ to simplify the coefficients.

$\frac{2x^2}{2} + \frac{20x}{2} - \frac{4000}{2} = \frac{0}{2}$

$x^2 + 10x - 2000 = 0$

Step 5: Solve the quadratic equation. We can solve this by factoring.

We need two numbers that multiply to $-2000$ and add up to $10$. These numbers are $50$ and $-40$.

So, we can factor the quadratic expression as:

$(x + 50)(x - 40) = 0$

Step 6: Set each factor equal to zero and solve for $x$.

$x + 50 = 0 \implies x = -50$

$x - 40 = 0 \implies x = 40$

Step 7: Interpret the solution in the context of the problem.

The variable $x$ represents the speed of the train. Speed cannot be negative. Therefore, the solution $x = -50$ is not physically possible.

The only valid solution is $x = 40$.

Thus, the original speed of the train is $40$ km/hr.

Defining the Variable:

Let the unit digit of the two-digit number be $x$.

Defining the Tens Digit (Considering Cases):

We are given that the digits differ by $3$. This means the absolute difference between the tens digit and the unit digit is $3$.

So, $|\text{Tens digit} - x| = 3$.

This implies that the Tens digit $- x = 3$ or the Tens digit $- x = -3$.

Therefore, the Tens digit is either $x + 3$ or $x - 3$. We need to consider both cases.

Case 1: Tens digit is $x + 3$

Unit digit $= x$

Tens digit $= x + 3$

The original two-digit number is formed by (Tens digit $\times 10$) $+$ (Unit digit).

Original number $= 10(x + 3) + x$

Original number $= 10x + 30 + x$

Original number $= 11x + 30$

When the digits are interchanged:

New unit digit $= x + 3$

New tens digit $= x$

The number obtained by interchanging the digits is:

Interchanged number $= 10x + (x + 3)$

Interchanged number $= 10x + x + 3$

Interchanged number $= 11x + 3$

We are given that the sum of the original number and the interchanged number is $121$.

Framing the Equation (Case 1):

(Original number) $+$ (Interchanged number) $= 121$

$(11x + 30) + (11x + 3) = 121$

Solving the Equation (Case 1):

Combine like terms on the LHS:

$11x + 11x + 30 + 3 = 121$

$22x + 33 = 121$

Transpose the constant term $33$ to the RHS:

$22x = 121 - 33$

$22x = 88$

Divide both sides by $22$:

$x = \frac{88}{22}$

$x = 4$

Checking the validity of digits:

Unit digit $= x = 4$. This is a valid digit (0-9).

Tens digit $= x + 3 = 4 + 3 = 7$. This is a valid digit (0-9).

Since both digits are valid, this case gives a possible original number.

The original number is $(7 \times 10) + 4 = 70 + 4 = 74$.

Case 2: Tens digit is $x - 3$

Unit digit $= x$

Tens digit $= x - 3$

For a two-digit number, the tens digit cannot be zero or negative if the unit digit is positive and greater than or equal to 3. Let's see what the value of $x$ is first.

The original two-digit number is:

Original number $= 10(x - 3) + x$

Original number $= 10x - 30 + x$

Original number $= 11x - 30$

When the digits are interchanged:

New unit digit $= x - 3$

New tens digit $= x$

The number obtained by interchanging the digits is:

Interchanged number $= 10x + (x - 3)$

Interchanged number $= 10x + x - 3$

Interchanged number $= 11x - 3$

The sum of the original number and the interchanged number is $121$.

Framing the Equation (Case 2):

(Original number) $+$ (Interchanged number) $= 121$

$(11x - 30) + (11x - 3) = 121$

Solving the Equation (Case 2):

Combine like terms on the LHS:

$11x + 11x - 30 - 3 = 121$

$22x - 33 = 121$

Transpose the constant term $-33$ to the RHS:

$22x = 121 + 33$

$22x = 154$

Divide both sides by $22$:

$x = \frac{154}{22}$

$x = 7$

Checking the validity of digits:

Unit digit $= x = 7$. This is a valid digit (0-9).

Tens digit $= x - 3 = 7 - 3 = 4$. This is a valid digit (0-9).

Since both digits are valid, this case also gives a possible original number.

The original number is $(4 \times 10) + 7 = 40 + 7 = 47$.

Conclusion:

Both cases yield valid original numbers that satisfy the given conditions.

If the unit digit is $4$ and the tens digit is $7$, the original number is $74$. The interchanged number is $47$, and $74 + 47 = 121$.

If the unit digit is $7$ and the tens digit is $4$, the original number is $47$. The interchanged number is $74$, and $47 + 74 = 121$.

The problem asks for "the original number". Both $74$ and $47$ fit the description of the original number depending on which digit is larger. Without further constraints, there are two possible numbers.

The possible original numbers are $74$ and $47$.