Chapter 4 Linear Equation in One Variable

Given:

$x = a$, where $k$ is an integer.

To Determine:

Which statement is not always true for integer $k$.

Solution:

Given $x=a$. Let's check each option:

(a) $kx = ak$. Substituting $x=a$, we get $ka = ak$, which is always true.

(c) $x - k = a - k$. Substituting $x=a$, we get $a - k = a - k$, which is always true.

(d) $x + k = a + k$. Substituting $x=a$, we get $a + k = a + k$, which is always true.

(b) $\frac{x}{k} = \frac{a}{k}$. Substituting $x=a$, we get $\frac{a}{k} = \frac{a}{k}$.

This equality holds only when $k \neq 0$, because division by zero is undefined.

If $k = 0$, the expression $\frac{a}{k}$ is undefined.

Since the statement must be true for any integer $k$, and it is not true for $k=0$, this statement is not always true.

The statement which is not always true for an integer $k$ is $\frac{x}{k} = \frac{a}{k}$.

The correct answer is (b).

Given:

$3x - 4(64 - x) = 10$

To Find:

The value of $x$.

Solution:

Solve the equation for $x$:

$3x - 4(64 - x) = 10$

Distribute the $-4$:

$3x - 256 + 4x = 10$

Combine like terms:

$7x - 256 = 10$

Add $256$ to both sides:

$7x = 10 + 256$

$7x = 266$

Divide both sides by $7$:

The value of $x$ is $38$.

The correct answer is (d) 38.

Given:

Fifteen added to thrice a whole number equals 93.

To Find:

The whole number.

Solution:

Let the whole number be $x$.

Thrice the number is $3x$.

Fifteen added to thrice the number is $3x + 15$.

According to the problem, this equals 93.

So, we have the equation:

Subtract 15 from both sides:

Divide both sides by 3:

The number is 26, which is a whole number.

The number is 26.

Given:

The equation: $\frac{1}{3} - x = -\frac{2}{3}$

To Find:

The value of $x$.

Solution:

We need to solve for $x$ in the given equation:

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

Add $x$ to both sides:

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

Add $\frac{2}{3}$ to both sides:

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

Combine the fractions on the left side:

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

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

The value of $x$ is 1.

Given Statement:

Three consecutive even numbers whose sum is 156 are 51, 52 and 53.

To Determine:

Whether the statement is True (T) or False (F).

Solution:

The statement claims that 51, 52, and 53 are three consecutive even numbers.

Let's examine the numbers:

• 51 is an odd number.
• 52 is an even number.
• 53 is an odd number.

For numbers to be "three consecutive even numbers", all three numbers must be even.

Since 51 and 53 are not even numbers, the set {51, 52, 53} cannot be "three consecutive even numbers".

Although their sum is $51 + 52 + 53 = 156$, the condition that they are "even numbers" is not met.

Therefore, the statement is false.

The statement is False (F).

Given Statement:

$x = -12$ is the solution of the equation $5x - 3(2x + 1) = 21 + x$.

To Determine:

Whether the statement is True (T) or False (F).

Solution:

To check if $x = -12$ is the solution, we substitute $x = -12$ into both sides of the equation and see if the Left Hand Side (LHS) equals the Right Hand Side (RHS).

The given equation is:

Consider the LHS of the equation:

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

Substitute $x = -12$:

Calculate the values:

Consider the RHS of the equation:

RHS $= 21 + x$

Substitute $x = -12$:

Comparing the values of LHS and RHS from (ii) and (iii):

LHS $= 9$ and RHS $= 9$

Since the LHS equals the RHS when $x = -12$, the statement is true.

The statement is True (T).

Given:

The equation: $\frac{x}{2} + \frac{x}{4} + \frac{x}{5} + 10000 = x$

To Find:

The value of $x$ that satisfies the equation.

Solution:

We need to solve the given linear equation for $x$.

The equation is:

Rearrange the equation to get all terms involving $x$ on one side and the constant term on the other side.

Subtract $x$ from both sides and subtract $10000$ from both sides, or simply move $x$ to LHS and $10000$ to RHS:

$\frac{x}{2} + \frac{x}{4} + \frac{x}{5} - x = -10000$

Alternatively, move the fractional terms to the RHS:

Find the Least Common Multiple (LCM) of the denominators (2, 4, 5). The LCM is 20.

Rewrite the terms on the RHS with the common denominator 20:

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

$\frac{x}{2} = \frac{10x}{20}$

$\frac{x}{4} = \frac{5x}{20}$

$\frac{x}{5} = \frac{4x}{20}$

Substitute these back into the equation:

Combine the terms on the RHS:

Multiply both sides of the equation by 20 to solve for $x$:

The value of $x$ is 200000.

Given:

• The present age of the father is four times the age of his son.
• After 10 years, the father's age will be three times the son's age.

To Find:

The present ages of the father and the son.

Solution:

Let the present age of the son be $S$ years.

Let the present age of the father be $F$ years.

According to the first condition:

After 10 years:

Son's age will be $S + 10$ years.

Father's age will be $F + 10$ years.

According to the second condition:

The father's age after 10 years will be three times the son's age after 10 years.

Now we have a system of two linear equations. Substitute the expression for $F$ from equation (i) into equation (ii):

Simplify and solve for $S$:

Subtract $3S$ from both sides:

Subtract 10 from both sides:

Now, substitute the value of $S$ back into equation (i) to find $F$:

The present age of the son is 20 years.

The present age of the father is 80 years.

Given:

Downstream time $t_D = 7$ hr, Upstream time $t_U = 8$ hr, Stream speed $v_c = 2$ km/hr.

Distance downstream = Distance upstream.

To Find:

Speed of steamer in still water ($v_s$) and the distance ($d$).

Solution:

Let $v_s$ be the speed of the steamer in still water (km/hr).

Speed downstream = $v_s + v_c = v_s + 2$ km/hr.

Speed upstream = $v_s - v_c = v_s - 2$ km/hr.

Using Distance = Speed $\times$ Time:

Distance downstream $d = (v_s + 2) \times 7$

Distance upstream $d = (v_s - 2) \times 8$

Equating the distances (i) and (ii):

$7v_s + 14 = 8v_s - 16$

Solve for $v_s$:

$14 + 16 = 8v_s - 7v_s$

Speed of steamer in still water is $v_s = 30$ km/hr.

Substitute $v_s = 30$ into equation (i) to find the distance $d$:

$d = 7(30) + 14$

$d = 210 + 14$

The distance between the ports is 224 km.

Speed of the steamer in still water: 30 km/hr.

Distance between the ports: 224 km.

Given:

• Total distance between A and B = 690 km.
• Time traveled = 6 hours.
• Distance between cars after 6 hours = 30 km.
• Difference in speeds = 10 km/hr.

To Find:

The speed of each car.

Solution:

Let the speed of the faster car be $v_1$ km/hr and the speed of the slower car be $v_2$ km/hr.

According to the problem:

When two cars move towards each other, their relative speed is $v_1 + v_2$.

In 6 hours, the total distance covered by both cars together is the initial distance minus the remaining distance:

Distance covered = $690 - 30 = 660$ km.

Using Distance = Relative Speed $\times$ Time:

Divide equation (ii) by 6:

Now we have a system of two linear equations:

$v_1 - v_2 = 10$ (from i)

$v_1 + v_2 = 110$ (from iii)

Add equation (i) and equation (iii):

$(v_1 - v_2) + (v_1 + v_2) = 10 + 110$

$2v_1 = 120$

Divide by 2:

Substitute the value of $v_1$ into equation (i):

$60 - v_2 = 10$

Subtract 10 from both sides and add $v_2$ to both sides:

The speeds of the two cars are 60 km/hr and 50 km/hr.

Given:

• The garden is square shaped.
• Perimeter of the square garden = 6480 cm.

To Find:

The dimensions (side length) of the garden.

Solution:

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

The formula for the perimeter of a square is $4 \times \text{side length}$.

So, the perimeter of the garden is $4s$ cm.

We are given that the perimeter is 6480 cm.

Thus, we can write the equation:

To find the dimension $s$, we need to solve this equation. Divide both sides by 4:

Perform the division:

The side length of the square garden is 1620 cm.

The dimension of the square garden is 1620 cm by 1620 cm.

Given:

Four linear equations.

To Determine:

Which equation has a solution that is neither a fraction nor an integer.

Solution:

Let's solve each equation:

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

$3x - 5x = 2 - 2$

$-2x = 0$

$x = 0$ (Integer)

(b) $4x - 18 = 2$

$4x = 2 + 18$

$4x = 20$

$x = \frac{20}{4} = 5$ (Integer)

(c) $4x + 7 = x + 2$

$4x - x = 2 - 7$

$3x = -5$

$-\frac{5}{3}$ is a fraction (rational number) but not an integer.

(d) $5x - 8 = x + 4$

$5x - x = 4 + 8$

$4x = 12$

$x = \frac{12}{4} = 3$ (Integer)

Given the standard definitions, an integer is a type of fraction (rational number where the denominator is 1). The phrasing "neither a fraction nor an integer" for a linear equation with rational coefficients yielding a rational solution is unusual.

However, among the common interpretations or potentially slightly imprecise wording, the solution $x = -\frac{5}{3}$ is the only one that is a rational number but not an integer.

If interpreted as "whose solution is not an integer", then option (c) fits.

The solution to equation (c) is $x = -\frac{5}{3}$, which is a fraction but not an integer. Assuming the question intends to distinguish between integers and non-integer fractions, option (c) is the correct choice.

The correct answer is (c).

Given:

The linear equation: $ax + b = 0$

To Find:

The solution for $x$.

Solution:

We need to solve the equation $ax + b = 0$ for $x$.

First, isolate the term containing $x$ by subtracting $b$ from both sides of the equation:

Now, assuming $a \neq 0$, divide both sides of the equation by $a$ to find the value of $x$:

If $a=0$, the original equation becomes $0 \cdot x + b = 0$, or $b=0$. If $b=0$, any value of $x$ is a solution. If $b \neq 0$, there is no solution. However, the options provided assume a unique solution, which implies $a \neq 0$.

Comparing our solution $x = \frac{-b}{a}$ with the given options, we find that it matches option (c).

The solution of the equation $ax + b = 0$ is $x = \frac{-b}{a}$.

The correct answer is (c).

Given:

The equation: $8x - 3 = 25 + 17x$

To Determine:

The nature of the solution $x$ among the given options.

Solution:

We need to solve the given linear equation for $x$.

The equation is:

$8x - 3 = 25 + 17x$

Move the terms involving $x$ to one side and constant terms to the other side.

Subtract $8x$ from both sides:

Subtract 25 from both sides:

Divide both sides by 9:

Now, let's classify the solution $x = -\frac{28}{9}$ based on the given options:

• A number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, is called a rational number.
• An integer is a whole number (positive, negative, or zero: ..., -2, -1, 0, 1, 2, ...).
• A fraction is typically understood as a number expressed in the form $\frac{p}{q}$.

Our solution is $x = -\frac{28}{9}$.

• Is it a fraction? Yes, it is in the form $\frac{p}{q}$.
• Is it an integer? No, because $-28$ is not divisible by $9$ to give a whole number.
• Is it a rational number? Yes, because $-28$ and $9$ are integers and $9 \neq 0$.
• Can it be solved? Yes, we found a unique value for $x$.

The solution $x = -\frac{28}{9}$ is both a fraction and a rational number. However, the term "rational number" is the formal mathematical classification for numbers that can be written as a fraction $\frac{p}{q}$ (including integers, which can be written as $\frac{p}{1}$). Among the given options, "a rational number" is the most encompassing correct description.

The correct answer is (c) a rational number.

Given:

The process of shifting a number from one side of an equation to the other.

To Determine:

The correct term for this process among the given options.

Solution:

Let's examine the given options:

• Transposition: This is the process of moving a term from one side of an equation to the other side by changing its sign. For example, in $x + 5 = 10$, moving 5 to the RHS gives $x = 10 - 5$.
• Distributivity: This is a property of multiplication over addition or subtraction, e.g., $a(b+c) = ab + ac$.
• Commutativity: This property states that the order of operands does not matter for certain operations, e.g., $a+b = b+a$.
• Associativity: This property states that the grouping of operands does not matter for certain operations, e.g., $(a+b)+c = a+(b+c)$.

The description in the question, "The shifting of a number from one side of an equation to other", precisely matches the definition of Transposition.

The correct answer is (a).

Given:

The equation: $\frac{5x}{3} - 4 = \frac{2x}{5}$

To Find:

The numerical value of $2x - 7$.

Solution:

First, solve the given equation for $x$.

$\frac{5x}{3} - 4 = \frac{2x}{5}$

Multiply the entire equation by the LCM of the denominators (3 and 5), which is 15:

$15 \left(\frac{5x}{3}\right) - 15(4) = 15 \left(\frac{2x}{5}\right)$

$5(5x) - 60 = 3(2x)$

$25x - 60 = 6x$

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

$25x - 6x = 60$

$19x = 60$

Divide by 19:

Now, find the value of the expression $2x - 7$ using the value of $x$ from (i):

$2x - 7 = 2\left(\frac{60}{19}\right) - 7$

$2x - 7 = \frac{120}{19} - 7$

To subtract 7, express it as a fraction with denominator 19:

$2x - 7 = \frac{120}{19} - \frac{7 \times 19}{19}$

$2x - 7 = \frac{120}{19} - \frac{133}{19}$

Combine the fractions:

$2x - 7 = \frac{120 - 133}{19}$

The numerical value of $2x - 7$ is $-\frac{13}{19}$.

The correct answer is (b).

Given:

Two expressions: $3x - 4$ and $2x + 1$.

To Find:

The value of $x$ for which the expressions are equal.

Solution:

To find the value of $x$ for which the expressions are equal, we set them equal to each other:

$3x - 4 = 2x + 1$

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

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

$x - 4 = 1$

Add 4 to both sides of the equation:

$x - 4 + 4 = 1 + 4$

The value of $x$ for which the expressions are equal is 5.

The correct answer is (c).

Given:

The equation $ax = b$, where $a$ and $b$ are positive integers.

To Determine:

The nature of the solution for $x$.

Solution:

We need to solve the equation $ax = b$ for $x$.

The equation is:

Since $a$ is a positive integer, $a \neq 0$. We can divide both sides of the equation by $a$:

We are given that $a$ and $b$ are positive integers.

A positive integer is a number greater than 0.

• $b > 0$
• $a > 0$

When a positive number is divided by another positive number, the result is always positive.

Therefore, the solution $x = \frac{b}{a}$ must be positive.

The solution of the equation $ax = b$, where $a$ and $b$ are positive integers, is always positive.

The correct answer is (a).

Given:

The term "Linear equation in one variable".

To Determine:

The correct characteristic of a linear equation in one variable among the given options.

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, and $a$ and $b$ are constants with $a \neq 0$.

Let's analyze the characteristics of such an equation and compare them with the options:

• It involves only one type of variable (e.g., only $x$, or only $y$, etc.).
• The highest power of the variable is 1.
• It may contain one or more terms involving the variable (e.g., $2x + 3x = 10$).
• It contains at least one term with the variable (since $a \neq 0$).

Let's check the options:

• (a) "only one variable with any power." - Incorrect. The power must be 1.
• (b) "only one term with a variable." - Incorrect. Example: $2x + 3x = 10$ has two terms with the variable.
• (c) "only one variable with power 1." - Correct. It has only one *type* of variable, and its highest power is 1.
• (d) "only constant term." - Incorrect. A linear equation must have a variable term.

A linear equation in one variable has only one variable (type) and its highest power is 1.

The correct answer is (c).

Given:

Four algebraic expressions.

To Determine:

Which expression is a linear expression.

Solution:

A linear expression is an algebraic expression in which the highest power of the variable is 1.

Let's examine each option:

(a) $x^2 + 1$

The variable is $x$, and its highest power is 2 ($x^2$). This is a quadratic expression.

(b) $y + y^2$

The variable is $y$, and its highest power is 2 ($y^2$). This is a quadratic expression.

(c) 4

This is a constant expression. It can be considered a linear expression where the coefficient of the variable is zero (e.g., $0x + 4$), but among the options provided, one clearly has a variable with power 1.

(d) $1 + z$

The variable is $z$, and its power is 1 (since $z = z^1$). This fits the definition of a linear expression.

Comparing the options, $1 + z$ is the only expression where the highest power of the variable is 1.

The correct answer is (d).

Given:

A linear equation in one variable.

To Determine:

The number of solutions a linear equation in one variable has.

Solution:

A linear equation in one variable can typically be written in the form $ax + b = 0$, where $a$ and $b$ are constants and $a \neq 0$.

To find the solution, we solve for $x$:

$ax = -b$

Since $a \neq 0$, we can divide by $a$:

This process yields exactly one value for $x$, provided $a \neq 0$. This unique value is the solution to the equation.

Consider the edge case $a=0$. If $a=0$, the equation becomes $0 \cdot x + b = 0$, which simplifies to $b = 0$.

• If $a=0$ and $b=0$, the equation is $0 \cdot x + 0 = 0$, which is $0=0$. This is true for any value of $x$, meaning there are infinitely many solutions.
• If $a=0$ and $b \neq 0$, the equation is $0 \cdot x + b = 0$, which is $b=0$. This is a contradiction (since $b \neq 0$), meaning there is no solution.

However, when defining a "linear equation in one variable" in the standard context where a unique solution is expected, it is typically assumed that the coefficient of the variable is non-zero (i.e., $a \neq 0$ in $ax+b=0$). This ensures that the equation is indeed "linear" and "in one variable" in the sense of having a degree of 1 for the variable.

Given the multiple-choice options, the most appropriate answer for a typical linear equation in one variable ($a \neq 0$) is that it has only one solution.

Assuming $a \neq 0$, a linear equation in one variable has a unique solution.

The correct answer is (a) Only one solution.

Given:

The equation: $\frac{1}{3} + S = \frac{2}{5}$

To Find:

The value of $S$.

Solution:

We need to solve the given equation for $S$.

The equation is:

Subtract $\frac{1}{3}$ from both sides of the equation to isolate $S$:

Find a common denominator for the fractions on the RHS. The LCM of 5 and 3 is 15.

Rewrite the fractions with the common denominator 15:

$S = \frac{2 \times 3}{5 \times 3} - \frac{1 \times 5}{3 \times 5}$

$S = \frac{6}{15} - \frac{5}{15}$

Combine the fractions by subtracting the numerators:

The value of $S$ is $\frac{1}{15}$.

Comparing the solution $S = \frac{1}{15}$ with the given options, we see it matches option (b).

The correct answer is (b).

Given:

The equation: $\frac{-4}{3} y = -\frac{3}{4}$

To Find:

The value of $y$.

Solution:

We need to solve the given equation for $y$.

The equation is:

To isolate $y$, multiply both sides of the equation by the reciprocal of $-\frac{4}{3}$, which is $-\frac{3}{4}$.

On the LHS, the terms cancel out:

$\left(\cancel{-\frac{3}{4}}\right) \times \left(\cancel{-\frac{4}{3}} y\right) = y$

On the RHS, multiply the fractions:

$\left(-\frac{3}{4}\right) \times \left(-\frac{3}{4}\right) = \frac{(-3) \times (-3)}{4 \times 4} = \frac{9}{16}$

So, the equation becomes:

Now, let's express $\frac{9}{16}$ in terms of squares to match the options.

$\frac{9}{16} = \frac{3^2}{4^2} = \left(\frac{3}{4}\right)^2$

Comparing the solution $y = \left(\frac{3}{4}\right)^2$ with the given options, we see it matches option (c).

The correct answer is (c).

Given:

• A two-digit number.
• The digit in the tens place is 3 more than the digit in the units place.
• The digit in the units place is $b$.

To Find:

The expression for the number in terms of $b$.

Solution:

Let the digit in the units place be $u$ and the digit in the tens place be $t$.

We are given that the digit at the units place is $b$. So, $u = b$.

We are given that the digit in the tens place is 3 more than the digit in the units place. So, $t = u + 3$.

Substituting $u = b$, we get the digit in the tens place as $t = b + 3$.

A two-digit number can be expressed as $10 \times (\text{tens digit}) + 1 \times (\text{units digit})$.

So, the number is $10 \times t + 1 \times u$.

Substitute the expressions for $t$ and $u$ in terms of $b$:

Distribute the 10:

Number $= 10b + 30 + b$

Combine the terms involving $b$:

The expression for the number is $11b + 30$.

Comparing our expression with the given options, we see it matches option (a).

The correct answer is (a).

Given:

• Arpita's present age is thrice Shilpa's present age.
• Shilpa's age three years ago was $x$ years.

To Find:

Arpita's present age in terms of $x$.

Solution:

Let Shilpa's age three years ago be $S_{ago}$. We are given $S_{ago} = x$ years.

Let Shilpa's present age be $S_{present}$. Shilpa's present age is her age three years ago plus 3 years.

Substitute $S_{ago} = x$ into equation (i):

Let Arpita's present age be $A_{present}$. We are given that Arpita's present age is thrice Shilpa's present age.

Substitute the expression for $S_{present}$ into equation (ii):

We can also distribute the 3:

$A_{present} = 3x + 3 \times 3$

The expression for Arpita's present age is $3(x + 3)$ or $3x + 9$.

Comparing our result with the given options, we see that option (d) matches the form $3(x+3)$.

The correct answer is (d).

Given:

Sum of three consecutive multiples of 7 is 357.

To Find:

The smallest of the three multiples.

Solution:

Let the three consecutive multiples of 7 be $7n$, $7n+7$, and $7n+14$, where $n$ is an integer.

Their sum is given as 357:

$7n + (7n + 7) + (7n + 14) = 357$

Combine like terms:

$(7n + 7n + 7n) + (7 + 14) = 357$

$21n + 21 = 357$

Subtract 21 from both sides:

$21n = 357 - 21$

$21n = 336$

Divide by 21:

The smallest multiple is $7n$. Substitute $n=16$:

Smallest multiple $= 7 \times 16$

The three multiples are $7 \times 16 = 112$, $112+7 = 119$, and $119+7 = 126$.

Check sum: $112 + 119 + 126 = 357$.

The smallest multiple is 112.

The correct answer is (a).

Given Statement:

In a linear equation, the _________ power of the variable appearing in the equation is one.

To Fill:

The blank in the given statement.

Solution:

A linear equation is defined as an equation where the highest power of the variable is 1.

For example, in the equation $2x + 3 = 7$, the variable is $x$ and its power is 1. This is a linear equation.

In the expression $x^2 + x = 5$, the highest power of the variable $x$ is 2, so this is not a linear equation.

Therefore, the blank should be filled with the word describing the power that is equal to one in a linear equation.

In a linear equation, the highest power of the variable appearing in the equation is one.

The complete statement is:

In a linear equation, the highest power of the variable appearing in the equation is one.

Given Statement:

The solution of the equation $3x - 4 = 1 - 2x$ is _________.

To Find:

The solution of the given equation.

Solution:

We need to solve the equation $3x - 4 = 1 - 2x$ for $x$.

The equation is:

Move the terms involving $x$ to the left side and constant terms to the right side.

Add $2x$ to both sides:

Add 4 to both sides:

Divide both sides by 5:

The solution of the equation is $x = 1$.

The solution of the equation $3x - 4 = 1 - 2x$ is 1.

The complete statement is:

The solution of the equation $3x – 4 = 1 – 2 x$ is 1.

Given Statement:

The solution of the equation $2y = 5y – \frac{18}{5}$ is ________.

To Find:

The solution of the given equation.

Solution:

We need to solve the equation $2y = 5y - \frac{18}{5}$ for $y$.

The equation is:

Move the terms involving $y$ to one side and the constant term to the other side.

Subtract $5y$ from both sides:

Multiply both sides by $-1$ to make both sides positive:

$3y = \frac{18}{5}$

Divide both sides by 3 (or multiply by $\frac{1}{3}$):

Simplify the expression:

The solution of the equation is $y = \frac{6}{5}$.

The solution of the equation $2y = 5y – \frac{18}{5}$ is $\frac{6}{5}$.

The complete statement is:

The solution of the equation $2y = 5y – \frac{18}{5}$ is $\frac{6}{5}$.

Given Statement:

Any value of the variable which makes both sides of an equation equal is known as a _________ of the equation.

To Fill:

The blank in the given statement.

Solution:

By definition, a value of the variable that satisfies an equation (makes both sides equal) is called a solution or a root of the equation.

For example, in the equation $x + 2 = 5$, if we substitute $x = 3$, we get $3 + 2 = 5$, which simplifies to $5 = 5$. Since both sides are equal, $x = 3$ is the solution of the equation.

Any value of the variable which makes both sides of an equation equal is known as a solution (or root) of the equation.

The complete statement is:

Any value of the variable which makes both sides of an equation equal is known as a solution of the equation.

Given:

• Equation: $9x - \text{blank} = -21$
• Solution: $x = -2$

To Find:

The number that fills the blank.

Solution:

Let the number in the blank be represented by $y$. The equation is $9x - y = -21$.

We are given that $x = -2$ is the solution. This means when we substitute $x = -2$ into the equation, the equation should be true.

Substitute $x = -2$ into the equation:

Multiply 9 by -2:

Now, solve for $y$. Add 18 to both sides of the equation:

Multiply both sides by -1 to find the value of $y$:

The number that fills the blank is 3.

9x – 3 = –21 has the solution (–2).

The complete statement is:

9x – 3 = –21 has the solution (–2)

Given Statement:

Three consecutive numbers whose sum is 12 are _________, _________ and _________.

To Find:

The three consecutive numbers.

Solution:

Let the three consecutive numbers be $n$, $n+1$, and $n+2$, where $n$ is an integer.

The sum of these three numbers is given as 12.

So, we can write the equation:

$n + (n+1) + (n+2) = 12$

Combine like terms:

$(n + n + n) + (1 + 2) = 12$

$3n + 3 = 12$

Subtract 3 from both sides:

$3n = 12 - 3$

$3n = 9$

Divide by 3:

The smallest number is $n=3$.

The three consecutive numbers are:

• First number: $n = 3$
• Second number: $n + 1 = 3 + 1 = 4$
• Third number: $n + 2 = 3 + 2 = 5$

The three consecutive numbers are 3, 4, and 5.

Check their sum: $3 + 4 + 5 = 12$. This is correct.

Three consecutive numbers whose sum is 12 are 3, 4 and 5.

The complete statement is:

Three consecutive numbers whose sum is 12 are 3, 4 and 5.

Given:

• Total amount to be divided between A and B = $\textsf{₹}$ 25.
• A gets $\textsf{₹}$ 8 more than B.

To Find:

The share of A.

Solution:

Let B's share be $\textsf{₹} b$.

Since A gets $\textsf{₹}$ 8 more than B, A's share is $\textsf{₹} (b + 8)$.

The total amount is the sum of A's share and B's share.

Total amount = A's share + B's share

$25 = (b + 8) + b$

Combine the terms involving $b$:

$25 = 2b + 8$

Subtract 8 from both sides:

$25 - 8 = 2b$

$17 = 2b$

Divide by 2:

Now, find A's share, which is $b + 8$:

A's share $= 8.50 + 8$

So, A's share is $\textsf{₹}$ 16.50.

The share of A is 16.50.

The complete statement is:

The share of A when $\textsf{₹}$ 25 are divided between A and B so that A gets $\textsf{₹}$. 8 more than B is 16.50.

Given Statement:

A term of an equation can be transposed to the other side by changing its _________.

To Fill:

The blank in the given statement.

Solution:

Transposition is a method used to solve equations. When a term is moved from one side of the equals sign to the other, its operation is reversed, which effectively means changing its sign (positive becomes negative, and negative becomes positive).

For example, consider the equation:

$x + 5 = 10$

To transpose the term $+5$ to the other side, we subtract 5 from both sides:

$x + 5 - 5 = 10 - 5$

$x = 10 - 5$

The term $+5$ from the LHS appears as $-5$ on the RHS. The sign has changed.

Similarly, consider the equation:

$x - 3 = 7$

To transpose the term $-3$ to the other side, we add 3 to both sides:

$x - 3 + 3 = 7 + 3$

$x = 7 + 3$

The term $-3$ from the LHS appears as $+3$ on the RHS. The sign has changed.

Therefore, transposing a term involves changing its sign.

A term of an equation can be transposed to the other side by changing its sign.

The complete statement is:

A term of an equation can be transposed to the other side by changing its sign.

Given Statement:

On subtracting 8 from $x$, the result is 2. The value of $x$ is _________.

To Find:

The value of $x$.

Solution:

"Subtracting 8 from $x$" can be written as $x - 8$.

"the result is 2" means the expression equals 2.

So, we can write the equation:

To solve for $x$, add 8 to both sides of the equation:

The value of $x$ is 10.

On subtracting 8 from $x$, the result is 2. The value of $x$ is 10.

The complete statement is:

On subtracting 8 from x, the result is 2. The value of x is 10.

Given Statement:

$\frac{x}{5} + 30 = 18$ has the solution as _________.

To Find:

The solution of the given equation.

Solution:

We need to solve the equation $\frac{x}{5} + 30 = 18$ for $x$.

The equation is:

Subtract 30 from both sides of the equation:

Multiply both sides by 5 to solve for $x$:

The solution of the equation is $x = -60$.

$\frac{x}{5}$ + 30 = 18 has the solution as -60.

The complete statement is:

$\frac{x}{5}$ + 30 = 18 has the solution as -60.

Given Statement:

When a number is divided by 8, the result is –3. The number is _________.

To Find:

The number.

Solution:

Let the unknown number be $n$.

"When a number is divided by 8" can be written as $\frac{n}{8}$.

"the result is -3" means the expression equals -3.

So, we can write the equation:

To solve for $n$, multiply both sides of the equation by 8:

The number is -24.

When a number is divided by 8, the result is –3. The number is -24.

The complete statement is:

When a number is divided by 8, the result is –3. The number is -24.

Given Statement:

9 is subtracted from the product of $p$ and 4, the result is 11. The value of $p$ is _________.

To Find:

The value of $p$.

Solution:

"The product of $p$ and 4" can be written as $p \times 4$ or $4p$.

"9 is subtracted from the product of $p$ and 4" can be written as $4p - 9$.

"the result is 11" means the expression equals 11.

So, we can write the equation:

To solve for $p$, add 9 to both sides of the equation:

Divide both sides by 4:

The value of $p$ is 5.

9 is subtracted from the product of p and 4, the result is 11. The value of p is 5.

The complete statement is:

9 is subtracted from the product of p and 4, the result is 11. The value of p is 5.

Given Statement:

If $\frac{2}{5} x - 2 = 5 - \frac{3}{5} x$, then $x$ = _________.

To Find:

The value of $x$.

Solution:

We need to solve the equation $\frac{2}{5} x - 2 = 5 - \frac{3}{5} x$ for $x$.

The equation is:

Move the terms involving $x$ to the left side and constant terms to the right side.

Add $\frac{3}{5} x$ to both sides:

Combine the terms with $x$:

Add 2 to both sides:

The value of $x$ is 7.

If $\frac{2}{5}$ x - 2 = 5 - $\frac{3}{5}$ x , then x = 7.

The complete statement is:

If $\frac{2}{5}$ x - 2 = 5 - $\frac{3}{5}$ x , then x = 7.

Given Statement:

After 18 years, Swarnim will be 4 times as old as he is now. His present age is _________.

To Find:

Swarnim's present age.

Solution:

Let Swarnim's present age be $A$ years.

After 18 years, Swarnim's age will be $A + 18$ years.

According to the problem, his age after 18 years will be 4 times his present age.

So, we can write the equation:

Subtract $A$ from both sides of the equation to isolate the terms with $A$:

Divide both sides by 3:

So, Swarnim's present age is 6 years.

Check: Present age = 6. After 18 years, age = $6 + 18 = 24$. $4 \times \text{present age} = 4 \times 6 = 24$. The condition is satisfied.

After 18 years, Swarnim will be 4 times as old as he is now. His present age is 6 years.

The complete statement is:

After 18 years, Swarnim will be 4 times as old as he is now. His present age is 6 years.

Given Statement:

Adding 15 to 4 times $x$ is 39.

To Convert:

The given statement into an equation.

Solution:

Let's break down the statement into parts:

• "4 times $x$" can be written as $4 \times x$ or $4x$.
• "Adding 15 to 4 times $x$" can be written as $4x + 15$ or $15 + 4x$.
• "is 39" means the expression equals 39.

Combining these parts, we get the equation:

We can also write it as $15 + 4x = 39$, which is equivalent.

Convert the statement Adding 15 to 4 times x is 39 into an equation $4x + 15 = 39$.

The complete statement is:

Convert the statement Adding 15 to 4 times x is 39 into an equation $4x + 15 = 39$.

Given:

• A rational number $\frac{N}{D}$.
• The denominator is 10 more than the numerator ($D = N + 10$).
• The numerator is increased by 1 (new numerator $N' = N + 1$).
• The denominator is decreased by 1 (new denominator $D' = D - 1$).

To Find:

The expression for the new denominator ($D'$) in terms of $N$ (the original numerator).

Solution:

Let the original numerator be $N$.

According to the problem, the original denominator $D$ is 10 greater than the numerator:

The new numerator is the original numerator increased by 1:

The new denominator is the original denominator decreased by 1:

We need to find the expression for the new denominator ($D'$) in terms of the original numerator ($N$).

Substitute the expression for the original denominator $D$ (in terms of $N$) into the expression for the new denominator $D'$:

Simplify the expression for $D'$:

The expression for the new denominator is $N + 9$.

The expression for the new denominator is N + 9.

The complete statement is:

The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1 the and denominator is decreased by 1, then expression for new denominator is N + 9.

Given Statement:

The sum of two consecutive multiples of 10 is 210. The smaller multiple is _________.

To Find:

The smaller of the two consecutive multiples of 10.

Solution:

Let the smaller of the two consecutive multiples of 10 be $10n$, where $n$ is an integer.

The next consecutive multiple of 10 will be $10n + 10$.

The sum of these two multiples is given as 210.

So, we can write the equation:

Combine like terms on the left side:

Subtract 10 from both sides:

Divide both sides by 20:

The smaller multiple is $10n$. Substitute $n=10$:

The two consecutive multiples are $100$ and $100+10 = 110$. Their sum is $100 + 110 = 210$, which matches the given information.

The smaller multiple is 100.

The sum of two consecutive multiples of 10 is 210. The smaller multiple is 100.

The complete statement is:

The sum of two consecutive multiples of 10 is 210. The smaller multiple is 100.

Given Statement:

3 years ago, the age of a boy was $y$ years. His age 2 years ago was $(y – 2)$ years.

To Determine:

Whether the statement is True (T) or False (F).

Solution:

Let the present age of the boy be $P$ years.

According to the first part of the statement, his age 3 years ago was $y$ years.

From this, we can express his present age in terms of $y$:

Now, let's find his age 2 years ago. His age 2 years ago would be his present age minus 2 years:

Substitute the expression for $P$ from the previous step into equation (i):

The statement claims that his age 2 years ago was $(y - 2)$ years.

Our calculation shows his age 2 years ago was $(y + 1)$ years.

Since $(y + 1)$ is not equal to $(y - 2)$, the statement is false.

The statement is False (F).

Given Statement:

Shikha’s present age is $p$ years. Reemu’s present age is 4 times the present age of Shikha. After 5 years Reemu’s age will be $15p$ years.

To Determine:

Whether the statement is True (T) or False (F).

Solution:

Let Shikha's present age be $S_{present}$ years. We are given $S_{present} = p$ years.

Let Reemu's present age be $R_{present}$ years. We are given that Reemu's present age is 4 times Shikha's present age.

Now consider their ages after 5 years.

Shikha's age after 5 years will be $S_{present} + 5 = p + 5$ years.

Reemu's age after 5 years will be $R_{present} + 5 = 4p + 5$ years.

The statement claims that Reemu's age after 5 years will be $15p$ years.

We found Reemu's age after 5 years to be $4p + 5$.

For the statement to be true, $4p + 5$ must be equal to $15p$ for all possible values of $p$ (where $p$ represents an age, so $p \geq 0$).

Let's check if $4p + 5 = 15p$ is always true.

$5 = 15p - 4p$

$5 = 11p$

$p = \frac{5}{11}$

The equality $4p + 5 = 15p$ is only true when $p = \frac{5}{11}$. It is not true for all values of $p$. For example, if Shikha's present age is $p=1$ year, Reemu's present age is $4 \times 1 = 4$ years. After 5 years, Reemu's age will be $4+5=9$ years. The statement claims it would be $15p = 15 \times 1 = 15$ years. Since $9 \neq 15$, the statement is false.

The statement is False (F).

Given Statement:

In a 2 digit number, the units place digit is $x$. If the sum of digits be 9, then the number is $(10x – 9)$.

To Determine:

Whether the statement is True (T) or False (F).

Solution:

Let the units place digit be $u$ and the tens place digit be $t$.

We are given that the units place digit is $x$.

We are given that the sum of the digits is 9.

Substitute $u = x$ into the sum of digits equation:

Solve for the tens place digit $t$:

A two-digit number is formed by $10 \times (\text{tens digit}) + 1 \times (\text{units digit})$.

So, the number is $10 \times t + u$.

Substitute the expressions for $t$ and $u$ in terms of $x$:

Simplify the expression:

The statement claims that the number is $(10x - 9)$.

Our calculation shows the number is $(90 - 9x)$.

Let's test with an example. If the units digit $x = 1$, and the sum of digits is 9, then the tens digit is $9 - 1 = 8$. The number is 81.

According to the statement, the number would be $10x - 9 = 10(1) - 9 = 10 - 9 = 1$.

Since $81 \neq 1$, the statement is false.

The correct expression for the number is $90 - 9x$.

The statement is False (F).

Given Statement:

Sum of the ages of Anju and her mother is 65 years. If Anju’s present age is $y$ years then her mother’s age before 5 years is $(60 – y)$ years.

To Determine:

Whether the statement is True (T) or False (F).

Solution:

Let Anju's present age be $A_{present}$ and her mother's present age be $M_{present}$.

We are given that Anju's present age is $y$ years.

We are given that the sum of their present ages is 65 years.

Substitute $A_{present} = y$ into the sum of ages equation:

Solve for the mother's present age $M_{present}$:

Now, we need to find the mother's age before 5 years. This would be her present age minus 5 years.

Substitute the expression for $M_{present}$ from equation (i) into equation (ii):

The statement claims that her mother’s age before 5 years is $(60 – y)$ years.

Our calculation also shows that her mother's age before 5 years is $(60 - y)$ years.

Since our result matches the claim, the statement is true.

The statement is True (T).

Given Statement:

The ratio of boys to girls is 5:4. The number of boys is 9 more than the number of girls. The number of boys is 9.

To Determine:

Whether the statement is True (T) or False (F).

Solution:

Let the number of boys be $B$ and the number of girls be $G$.

The ratio of boys to girls is given as 5:4. This can be written as:

This implies that we can write $B = 5k$ and $G = 4k$ for some common positive factor $k$.

We are also given that the number of boys is 9 more than the number of girls:

Substitute the expressions for $B$ and $G$ in terms of $k$ into this equation:

Subtract $4k$ from both sides to solve for $k$:

Now that we have the value of $k$, we can find the actual number of boys ($B$) and girls ($G$).

Number of boys $B = 5k = 5 \times 9 = 45$.

Number of girls $G = 4k = 4 \times 9 = 36$.

Let's check if these numbers satisfy the given conditions:

• Ratio: $\frac{B}{G} = \frac{45}{36} = \frac{5 \times 9}{4 \times 9} = \frac{5}{4}$. (Correct)
• Difference: $B - G = 45 - 36 = 9$. (Correct)

Our calculation shows the number of boys is 45.

The statement claims that the number of boys is 9.

Since $45 \neq 9$, the statement is false.

The statement is False (F).

Let the current age of A be $x$ years.

Given that the sum of the current ages of A and B is 90 years.

So, the current age of B is $(90 - x)$ years.

To find the ages five years ago, we subtract 5 from their current ages.

A's age five years ago $= (\text{Current age of A}) - 5$

A's age five years ago $= x - 5$ years.

B's age five years ago $= (\text{Current age of B}) - 5$

B's age five years ago $= (90 - x) - 5$ years.

B's age five years ago $= 90 - x - 5$ years.

B's age five years ago $= 85 - x$ years.

The statement claims: "Hence, the ages of A and B five years back would be (x – 5) years and (85 – x) years respectively."

Our calculation shows that if A's current age is $x$ and their combined current age is 90, then their ages five years ago are indeed $(x - 5)$ years and $(85 - x)$ years.

The condition "Five years ago A was thrice as old as B was" means $x-5 = 3(85-x)$. This condition allows us to find the specific value of $x$ (which is $x=65$), but the statement is simply about expressing the ages five years back in terms of $x$ given the sum of current ages, which is correct.

Therefore, the given statement is correct.

The correct answer is True (T).

The statement claims that two different equations can never have the same answer (solution).

Let's consider some examples of simple equations.

Consider the equation: $x + 3 = 7$

Solving for $x$: $x = 7 - 3 = 4$.

The solution to this equation is $x = 4$.

Now consider a different equation: $2x = 8$

Solving for $x$: $x = \frac{8}{2} = 4$.

The solution to this equation is also $x = 4$.

These two equations, $x + 3 = 7$ and $2x = 8$, are clearly different equations.

However, they both have the same solution, which is $x = 4$.

Another example:

Equation 1: $x^2 = 9$. Solutions are $x = 3$ and $x = -3$.

Equation 2: $|x| = 3$. Solutions are $x = 3$ and $x = -3$.

These are different equations, but they share the same set of solutions.

Since we can find examples of different equations that have the same solution(s), the statement is incorrect.

Therefore, the given statement is false.

The correct answer is False (F).

The given equation is: $3x - 3 = 9$.

The process of transposing a term involves moving it from one side of the equation to the other while changing its sign.

In this equation, we are transposing the term $-3$ from the Left Hand Side (LHS) to the Right Hand Side (RHS).

When $-3$ is moved from LHS to RHS, its sign changes from $-$ to $+$.

So, transposing $-3$ to the RHS of the equation $3x - 3 = 9$, we get:

$3x = 9 + 3$

Simplifying the RHS:

$3x = 12$

The statement claims that transposing $-3$ to the RHS gives $3x = 9$.

Our calculation shows that transposing $-3$ to the RHS gives $3x = 12$.

Since $3x = 12$ is not the same as $3x = 9$, the statement is incorrect.

Therefore, the given statement is false.

The correct answer is False (F).

The given equation is: $2x = 4 - x$.

We need to transpose the term $-x$ from the Right Hand Side (RHS) to the Left Hand Side (LHS).

When a term is transposed from one side of the equation to the other, its sign changes.

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

Starting with the equation $2x = 4 - x$, we add $x$ to both sides (or transpose $-x$):

$2x + x = 4$

Now, simplify the LHS by combining like terms:

$3x = 4$

The statement claims that transposing $-x$ to LHS gives $x = 4$.

Our calculation shows that transposing $-x$ to LHS gives $3x = 4$.

To get $x$ from $3x = 4$, we would divide both sides by 3, which gives $x = \frac{4}{3}$. This is not equal to 4.

Since $3x = 4$ is not the same as $x = 4$, the statement is incorrect regarding the result after transposing $-x$ to the LHS.

Therefore, the given statement is false.

The correct answer is False (F).

The given equation is: $\frac{15}{8} - 7x = 9$.

We need to analyze the step of transposing $\frac{15}{8}$ from the Left Hand Side (LHS) to the Right Hand Side (RHS).

The term $\frac{15}{8}$ on the LHS has a positive sign (it is $+\frac{15}{8}$).

When a term is transposed from one side of the equation to the other, its sign must be changed.

Transposing $+\frac{15}{8}$ from LHS to RHS means we subtract $\frac{15}{8}$ from both sides of the equation.

Starting with $\frac{15}{8} - 7x = 9$, transpose $\frac{15}{8}$:

$-7x = 9 - \frac{15}{8}$.

The statement claims that transposing $\frac{15}{8}$ results in $-7x = 9 + \frac{15}{8}$.

Comparing our correct result ($-7x = 9 - \frac{15}{8}$) with the statement's claim ($-7x = 9 + \frac{15}{8}$), we see that the sign of $\frac{15}{8}$ is different.

The statement incorrectly keeps the positive sign when transposing the term $\frac{15}{8}$.

Therefore, the given statement is false.

The correct answer is False (F).

The given equation is: $\frac{x}{3} + 1 = \frac{7}{15}$.

We want to isolate the term $\frac{x}{3}$ on the Left Hand Side (LHS).

To do this, we need to transpose the constant term $+1$ from the LHS to the Right Hand Side (RHS).

When transposing a term from one side of the equation to the other, we change its sign.

So, $+1$ on the LHS becomes $-1$ on the RHS.

Transposing 1 to the RHS, we get:

$\frac{x}{3} = \frac{7}{15} - 1$

Now, we need to simplify the expression on the RHS. To subtract 1 from $\frac{7}{15}$, we write 1 with a denominator of 15:

$1 = \frac{15}{15}$

So, the equation becomes:

$\frac{x}{3} = \frac{7}{15} - \frac{15}{15}$

$\frac{x}{3} = \frac{7 - 15}{15}$

$\frac{x}{3} = \frac{-8}{15}$

The statement claims that $\frac{x}{3} = \frac{6}{15}$ after transposing 1.

Our calculation shows that $\frac{x}{3} = \frac{-8}{15}$.

Since $\frac{-8}{15}$ is not equal to $\frac{6}{15}$, the statement is incorrect.

Therefore, the given statement is false.

The correct answer is False (F).

We are given the equation: $6x = 18$.

We need to determine if the statement "$18x = 54$" is true, given that $6x = 18$ is true.

From the equation $6x = 18$, we can find the value of $x$ by dividing both sides by 6:

$\frac{6x}{6} = \frac{18}{6}$

$x = 3$

Now, we substitute the value of $x = 3$ into the expression $18x$ to see what its value is:

$18x = 18 \times 3$

$18x = 54$

Our calculation shows that if $6x = 18$, then $18x = 54$.

This matches the claim made in the statement.

Alternatively, we can notice that $18x$ is 3 times $6x$.

Given $6x = 18$, we can multiply both sides of this equation by 3:

$3 \times (6x) = 3 \times 18$

$(3 \times 6)x = 54$

$18x = 54$

This also confirms the statement.

Therefore, the given statement is true.

The correct answer is True (T).

The given equation is: $\frac{x}{11} = 15$.

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

The variable $x$ is being divided by 11 on the Left Hand Side (LHS).

To undo division by 11, we multiply both sides of the equation by 11.

Multiply both sides by 11:

$\frac{x}{11} \times 11 = 15 \times 11$

On the LHS, the 11 in the numerator and denominator cancel out:

$x = 15 \times 11$

$x = 165$

The statement claims that if $\frac{x}{11} = 15$, then $x = \frac{11}{5}$.

Our calculation shows that if $\frac{x}{11} = 15$, then $x = 165$.

Since $165$ is not equal to $\frac{11}{5}$, the statement is incorrect.

Therefore, the given statement is false.

The correct answer is False (F).

Let $x$ be an even number.

Even numbers are integers that are divisible by 2, like 2, 4, 6, 8, -2, 0, etc.

The difference between consecutive even numbers is always 2.

To find the next even number after a given even number $x$, we need to add 2 to $x$.

So, the next even number after $x$ is $x + 2$.

The statement claims that the next even number is $2(x + 1)$.

Let's simplify the expression given in the statement: $2(x + 1) = 2x + 2$.

Now we compare the correct expression for the next even number ($x+2$) with the expression claimed in the statement ($2x+2$).

We need to determine if $x + 2$ is always equal to $2x + 2$ for any even number $x$.

If $x + 2 = 2x + 2$, subtracting 2 from both sides gives $x = 2x$.

Subtracting $x$ from both sides gives $0 = x$.

This means the equality $x+2 = 2x+2$ is true only when $x=0$. For any other even number (positive or negative), the equality does not hold.

Let's test with an example where $x \neq 0$. Let $x = 6$ (which is an even number).

The next even number after 6 is $6 + 2 = 8$.

According to the statement, the next even number is $2(x + 1) = 2(6 + 1) = 2(7) = 14$.

Since $8 \neq 14$, the statement's formula $2(x+1)$ does not give the next even number for $x=6$.

The correct way to represent the next even number after an even number $x$ is $x+2$. The expression $2(x+1)$ simplifies to $2x+2$, which is generally not the same as $x+2$ unless $x=0$.

Therefore, the given statement is false.

The correct answer is False (F).

Let the two numbers be $a$ and $b$.

We are given that the sum of these two numbers is 93.

So, $a + b = 93$.

We are also given that one of the numbers is $x$. Let's assume $a = x$.

Substitute $a = x$ into the sum equation:

$x + b = 93$.

To find the other number, $b$, we subtract $x$ from both sides of the equation:

$b = 93 - x$.

So, if one number is $x$ and their sum is 93, the other number must be $93 - x$.

The fact that the numbers are consecutive means that $|x - (93 - x)| = 1$. This condition allows us to find the specific value of $x$ (which is 46 or 47), but the statement simply describes the relationship between the two numbers given their sum, which is algebraically correct.

Therefore, the given statement is true.

The correct answer is True (T).

Let the two numbers be $a$ and $b$.

The first part of the statement says that the two numbers differ by 40.

This means that the absolute difference between the two numbers is 40.

So, $|a - b| = 40$.

This implies that one number is 40 greater than the other.

If $a$ is the greater number, then $a - b = 40$, which means $a = b + 40$.

If $b$ is the greater number, then $b - a = 40$, which means $b = a + 40$.

The statement claims: "If one number is $x$, then the other number is $(40 - x)$."

Let one number be $x$. According to the condition that the two numbers differ by 40, the other number must be either $x + 40$ or $x - 40$.

Let's check if the difference between $x$ and $(40 - x)$ is 40.

The difference is $|x - (40 - x)| = |x - 40 + x| = |2x - 40|$.

For the numbers to differ by 40, we require $|2x - 40| = 40$.

This equation is true only if $2x - 40 = 40$ or $2x - 40 = -40$.

$2x = 80 \implies x = 40$

$2x = 0 \implies x = 0$

So, the other number is $(40 - x)$ only if $x=0$ (the numbers are 0 and 40) or if $x=40$ (the numbers are 40 and 0). The expression $(40-x)$ does not generally represent the other number when two numbers differ by 40.

For example, if one number is $x=10$, and they differ by 40, the other number must be $10+40=50$ or $10-40=-30$. The statement claims the other number is $40-x = 40-10 = 30$. The pair $(10, 30)$ differs by $|10-30|=20$, not 40.

The condition about increasing each number by 8 and the resulting relationship is part of setting up an equation to solve for $x$, but it does not affect the initial claim about the relationship between the two numbers based on their difference.

Based on the information that the two numbers differ by 40, if one number is $x$, the other number is $x+40$ or $x-40$, not $40-x$ in general.

Therefore, the given statement is false.

The correct answer is False (F).

The given equation is:

$\frac{3x - 8}{2x} = 1$

To solve for $x$, we need to eliminate the denominator. We can do this by multiplying both sides of the equation by $2x$. It's important to note that the denominator cannot be zero, so $2x \neq 0$, which means $x \neq 0$.

Multiply both sides by $2x$:

$(2x) \times \frac{3x - 8}{2x} = 1 \times (2x)$

This simplifies to:

$3x - 8 = 2x$

Now, we need to gather the terms containing $x$ on one side of the equation and the constant terms on the other side. We can transpose the term $2x$ from the Right Hand Side (RHS) to the Left Hand Side (LHS).

When transposing $2x$, its sign changes from positive to negative:

$3x - 2x - 8 = 0$

Next, transpose the constant term $-8$ from the LHS to the RHS. Its sign changes from negative to positive:

$3x - 2x = 8$

Combine the like terms on the LHS:

$(3 - 2)x = 8$

$1x = 8$

$x = 8$

We found the value of $x$ to be 8. This value is not 0, which satisfies the condition $x \neq 0$ derived from the original equation's denominator.

The solution to the equation is $x = 8$.

The value of $x$ is 8.

The given equation is:

$\frac{5x}{2x - 1} = 2$

To solve for $x$, we first need to eliminate the denominator $(2x - 1)$. We can do this by multiplying both sides of the equation by $(2x - 1)$. It's important to note that the denominator cannot be zero, so $2x - 1 \neq 0$, which means $x \neq \frac{1}{2}$.

Multiply both sides by $(2x - 1)$:

$(2x - 1) \times \frac{5x}{2x - 1} = 2 \times (2x - 1)$

This simplifies to:

$5x = 2(2x - 1)$

Now, distribute the 2 on the Right Hand Side (RHS):

$5x = (2 \times 2x) - (2 \times 1)$

$5x = 4x - 2$

Next, we need to collect the terms involving $x$ on one side of the equation. Let's transpose the term $4x$ from the RHS to the Left Hand Side (LHS).

When transposing $4x$, its sign changes from positive to negative:

$5x - 4x = -2$

Combine the like terms on the LHS:

$(5 - 4)x = -2$

$1x = -2$

$x = -2$

We check if our solution $x=-2$ violates the initial constraint $x \neq \frac{1}{2}$. Since $-2 \neq \frac{1}{2}$, the solution is valid.

The solution to the equation is $x = -2$.

The value of $x$ is -2.

The given equation is:

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

To solve for $x$, we can use cross-multiplication. This involves multiplying the numerator of one fraction by the denominator of the other fraction across the equals sign.

Multiply $(2x - 3)$ by 3 and $(4x + 5)$ by 1:

$3 \times (2x - 3) = 1 \times (4x + 5)$

Now, distribute the numbers on both sides of the equation:

$3 \times 2x - 3 \times 3 = 1 \times 4x + 1 \times 5$

$6x - 9 = 4x + 5$

Next, we need to collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the Left Hand Side (LHS) and constant terms to the Right Hand Side (RHS).

Transpose $4x$ from the RHS to the LHS (change its sign):

$6x - 4x - 9 = 5$

Transpose $-9$ from the LHS to the RHS (change its sign):

$6x - 4x = 5 + 9$

Combine the like terms on both sides:

$(6 - 4)x = 14$

$2x = 14$

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

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

$x = 7$

We should also check the denominator in the original equation. $4x+5 \neq 0$. If $x=7$, $4(7)+5 = 28+5 = 33 \neq 0$. So the solution is valid.

The solution to the equation is $x = 7$.

The value of $x$ is 7.

The given equation is:

$\frac{8}{x} = \frac{5}{x - 1}$

To solve for $x$, we can use cross-multiplication. This involves multiplying the numerator of each fraction by the denominator of the other fraction across the equals sign. Before we start, we should note that the denominators cannot be zero, so $x \neq 0$ and $x - 1 \neq 0 \implies x \neq 1$.

Multiply 8 by $(x - 1)$ and 5 by $x$:

$8 \times (x - 1) = 5 \times x$

$8(x - 1) = 5x$

Now, distribute the 8 on the Left Hand Side (LHS):

$8 \times x - 8 \times 1 = 5x$

$8x - 8 = 5x$

Next, we need to collect the terms containing $x$ on one side. Let's transpose the term $5x$ from the Right Hand Side (RHS) to the LHS.

When transposing $5x$, its sign changes from positive to negative:

$8x - 5x - 8 = 0$

Now, transpose the constant term $-8$ from the LHS to the RHS. Its sign changes from negative to positive:

$8x - 5x = 8$

Combine the like terms on the LHS:

$(8 - 5)x = 8$

$3x = 8$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 3:

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

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

We check if our solution $x=\frac{8}{3}$ violates the initial constraints $x \neq 0$ and $x \neq 1$. Since $\frac{8}{3} \neq 0$ and $\frac{8}{3} \neq 1$, the solution is valid.

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

The value of $x$ is $\frac{8}{3}$.

The given equation is:

$\frac{5(1 - x) + 3 (1 + x)}{1 - 2x} = 8$

First, simplify the numerator on the Left Hand Side (LHS) by distributing the numbers:

Numerator $= 5(1 - x) + 3(1 + x)$

Numerator $= (5 \times 1 - 5 \times x) + (3 \times 1 + 3 \times x)$

Numerator $= 5 - 5x + 3 + 3x$

Combine the constant terms and the terms involving $x$ in the numerator:

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

Numerator $= 8 - 2x$

Substitute the simplified numerator back into the original equation:

$\frac{8 - 2x}{1 - 2x} = 8$

To eliminate the denominator $(1 - 2x)$, multiply both sides of the equation by $(1 - 2x)$. Note that the denominator cannot be zero, so $1 - 2x \neq 0$, which means $x \neq \frac{1}{2}$.

$(1 - 2x) \times \frac{8 - 2x}{1 - 2x} = 8 \times (1 - 2x)$

$8 - 2x = 8(1 - 2x)$

Distribute the 8 on the Right Hand Side (RHS):

$8 - 2x = (8 \times 1) - (8 \times 2x)$

$8 - 2x = 8 - 16x$

Now, we need to collect the terms involving $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and constant terms to the RHS.

Transpose $-16x$ from the RHS to the LHS (change its sign):

$8 - 2x + 16x = 8$

Transpose the constant term 8 from the LHS to the RHS (change its sign):

$-2x + 16x = 8 - 8$

Combine the like terms on both sides:

$(-2 + 16)x = 0$

$14x = 0$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 14:

$\frac{14x}{14} = \frac{0}{14}$

$x = 0$

We check if our solution $x=0$ violates the initial constraint $x \neq \frac{1}{2}$. Since $0 \neq \frac{1}{2}$, the solution is valid.

The solution to the equation is $x = 0$.

The value of $x$ is 0.

The given equation is:

$\frac{0.2x + 5}{3.5x - 3} = \frac{2}{5}$

To solve for $x$, we use cross-multiplication.

Multiply the numerator of the LHS by the denominator of the RHS and the numerator of the RHS by the denominator of the LHS.

$5 \times (0.2x + 5) = 2 \times (3.5x - 3)$

Distribute the numbers on both sides of the equation:

$5 \times 0.2x + 5 \times 5 = 2 \times 3.5x - 2 \times 3$

$1.0x + 25 = 7.0x - 6$

$x + 25 = 7x - 6$

Now, we need to collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the Right Hand Side (RHS) and constant terms to the Left Hand Side (LHS).

Transpose $7x$ from the RHS to the LHS (change its sign):

$x - 7x + 25 = -6$

Transpose the constant term 25 from the LHS to the RHS (change its sign):

$x - 7x = -6 - 25$

Combine the like terms on both sides:

$(1 - 7)x = -31$

$-6x = -31$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is -6:

$\frac{-6x}{-6} = \frac{-31}{-6}$

$x = \frac{31}{6}$

We should also check the denominator in the original equation. $3.5x - 3 \neq 0$. If $x=\frac{31}{6}$, $3.5(\frac{31}{6}) - 3 = \frac{7}{2} \times \frac{31}{6} - 3 = \frac{217}{12} - \frac{36}{12} = \frac{181}{12} \neq 0$. So the solution is valid.

The solution to the equation is $x = \frac{31}{6}$.

The value of $x$ is $\frac{31}{6}$.

The given equation is:

$\frac{y - (4 - 3y)}{2y - (3 + 4y)} = \frac{1}{5}$

First, simplify the numerator and the denominator on the Left Hand Side (LHS) by removing the parentheses. Remember that a minus sign before parentheses changes the sign of each term inside.

Numerator: $y - (4 - 3y) = y - 4 + 3y = (y + 3y) - 4 = 4y - 4$

Denominator: $2y - (3 + 4y) = 2y - 3 - 4y = (2y - 4y) - 3 = -2y - 3$

Substitute the simplified expressions back into the equation:

$\frac{4y - 4}{-2y - 3} = \frac{1}{5}$

Now, use cross-multiplication to eliminate the denominators. Note that the denominator cannot be zero, so $-2y - 3 \neq 0 \implies 2y \neq -3 \implies y \neq -\frac{3}{2}$.

Multiply $5$ by the numerator $(4y - 4)$ and $1$ by the denominator $(-2y - 3)$:

$5 \times (4y - 4) = 1 \times (-2y - 3)$

$5(4y - 4) = -2y - 3$

Distribute the 5 on the LHS:

$5 \times 4y - 5 \times 4 = -2y - 3$

$20y - 20 = -2y - 3$

Next, collect the terms involving $y$ on one side and the constant terms on the other side. Let's move the $y$ terms to the LHS and constant terms to the Right Hand Side (RHS).

Transpose $-2y$ from the RHS to the LHS (change its sign):

$20y + 2y - 20 = -3$

Transpose $-20$ from the LHS to the RHS (change its sign):

$20y + 2y = -3 + 20$

Combine the like terms on both sides:

$(20 + 2)y = 17$

$22y = 17$

Finally, to find $y$, divide both sides of the equation by the coefficient of $y$, which is 22:

$\frac{22y}{22} = \frac{17}{22}$

$y = \frac{17}{22}$

We check if our solution $y=\frac{17}{22}$ violates the initial constraint $y \neq -\frac{3}{2}$. Since $\frac{17}{22} \neq -\frac{3}{2}$, the solution is valid.

The solution to the equation is $y = \frac{17}{22}$.

The value of $y$ is $\frac{17}{22}$.

The given equation is:

$\frac{x}{5} = \frac{x - 1}{6}$

To solve for $x$, we can use cross-multiplication. This involves multiplying the numerator of one fraction by the denominator of the other fraction across the equals sign.

Multiply $x$ by 6 and $(x - 1)$ by 5:

$6 \times x = 5 \times (x - 1)$

$6x = 5(x - 1)$

Now, distribute the 5 on the Right Hand Side (RHS):

$6x = 5 \times x - 5 \times 1$

$6x = 5x - 5$

Next, we need to collect the terms containing $x$ on one side of the equation. Let's move the $x$ terms to the Left Hand Side (LHS).

Transpose $5x$ from the RHS to the LHS (change its sign):

$6x - 5x = -5$

Combine the like terms on the LHS:

$(6 - 5)x = -5$

$1x = -5$

$x = -5$

The solution to the equation is $x = -5$.

The value of $x$ is -5.

The given equation is:

$0.4(3x - 1) = 0.5x + 1$

First, we distribute 0.4 on the Left Hand Side (LHS):

$0.4 \times 3x - 0.4 \times 1 = 0.5x + 1$

$1.2x - 0.4 = 0.5x + 1$

Now, we collect the terms containing $x$ on one side and the constant terms on the other side. Let's transpose the term $0.5x$ from the Right Hand Side (RHS) to the LHS.

When transposing $0.5x$, its sign changes from positive to negative:

$1.2x - 0.5x - 0.4 = 1$

Next, transpose the constant term $-0.4$ from the LHS to the RHS. Its sign changes from negative to positive:

$1.2x - 0.5x = 1 + 0.4$

Combine the like terms on both sides:

$(1.2 - 0.5)x = 1.4$

$0.7x = 1.4$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 0.7:

$\frac{0.7x}{0.7} = \frac{1.4}{0.7}$

$x = \frac{1.4}{0.7}$

To perform the division $\frac{1.4}{0.7}$, we can multiply the numerator and the denominator by 10 to remove the decimals:

$x = \frac{1.4 \times 10}{0.7 \times 10} = \frac{14}{7}$

$x = 2$

The solution to the equation is $x = 2$.

The value of $x$ is 2.

The given equation is:

$8x - 7 - 3x = 6x - 2x - 3$

First, simplify both sides of the equation by combining like terms.

On the Left Hand Side (LHS), combine the terms involving $x$:

$8x - 3x - 7 = (8 - 3)x - 7 = 5x - 7$

On the Right Hand Side (RHS), combine the terms involving $x$:

$6x - 2x - 3 = (6 - 2)x - 3 = 4x - 3$

The simplified equation is:

$5x - 7 = 4x - 3$

Now, we collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and constant terms to the RHS.

Transpose $4x$ from the RHS to the LHS (change its sign):

$5x - 4x - 7 = -3$

Transpose $-7$ from the LHS to the RHS (change its sign):

$5x - 4x = -3 + 7$

Combine the like terms on both sides:

$(5 - 4)x = 4$

$1x = 4$

$x = 4$

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

The value of $x$ is 4.

The given equation is:

$10x - 5 - 7x = 5x + 15 - 8$

First, simplify both sides of the equation by combining like terms.

On the Left Hand Side (LHS), combine the terms involving $x$ and the constant terms:

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

On the Right Hand Side (RHS), combine the constant terms:

$5x + 15 - 8 = 5x + (15 - 8) = 5x + 7$

The simplified equation is:

$3x - 5 = 5x + 7$

Now, we collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the RHS and constant terms to the LHS.

Transpose $5x$ from the RHS to the LHS (change its sign):

$3x - 5x - 5 = 7$

Transpose $-5$ from the LHS to the RHS (change its sign):

$3x - 5x = 7 + 5$

Combine the like terms on both sides:

$(3 - 5)x = 12$

$-2x = 12$

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

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

$x = -6$

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

The value of $x$ is -6.

The given equation is:

$4t - 3 - (3t + 1) = 5t - 4$

First, simplify the Left Hand Side (LHS) by removing the parentheses. Remember that the minus sign before the parentheses changes the sign of each term inside.

$4t - 3 - 3t - 1 = 5t - 4$

Combine the like terms on the LHS:

$(4t - 3t) + (-3 - 1) = 5t - 4$

$t - 4 = 5t - 4$

Now, we collect the terms containing $t$ on one side and the constant terms on the other side. Let's move the $t$ terms to the Right Hand Side (RHS) and the constant terms to the LHS.

Transpose $5t$ from the RHS to the LHS (change its sign):

$t - 5t - 4 = -4$

Transpose $-4$ from the LHS to the RHS (change its sign):

$t - 5t = -4 + 4$

Combine the like terms on both sides:

$(1 - 5)t = 0$

$-4t = 0$

Finally, to find $t$, divide both sides of the equation by the coefficient of $t$, which is -4:

$\frac{-4t}{-4} = \frac{0}{-4}$

$t = 0$

The solution to the equation is $t = 0$.

The value of $t$ is 0.

The given equation is:

$5(x - 1) - 2(x + 8) = 0$

First, distribute the constants outside the parentheses on the Left Hand Side (LHS):

$5 \times x - 5 \times 1 - (2 \times x + 2 \times 8) = 0$

$5x - 5 - (2x + 16) = 0$

Now, remove the parentheses, remembering that the minus sign before $(2x + 16)$ changes the sign of each term inside:

$5x - 5 - 2x - 16 = 0$

Combine the like terms on the LHS (terms with $x$ and constant terms):

$(5x - 2x) + (-5 - 16) = 0$

$3x - 21 = 0$

Next, transpose the constant term $-21$ from the LHS to the Right Hand Side (RHS). Its sign changes from negative to positive:

$3x = 21$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 3:

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

$x = 7$

The solution to the equation is $x = 7$.

The value of $x$ is 7.

The given equation is:

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

First, distribute the fractions on both sides of the equation:

LHS: $\frac{x}{2} - \left( \frac{1}{4} \times x - \frac{1}{4} \times \frac{1}{3} \right) = \frac{x}{2} - \left( \frac{x}{4} - \frac{1}{12} \right) = \frac{x}{2} - \frac{x}{4} + \frac{1}{12}$

RHS: $\frac{1}{6}(x + 1) + \frac{1}{12} = \frac{1}{6} \times x + \frac{1}{6} \times 1 + \frac{1}{12} = \frac{x}{6} + \frac{1}{6} + \frac{1}{12}$

The equation now becomes:

$\frac{x}{2} - \frac{x}{4} + \frac{1}{12} = \frac{x}{6} + \frac{1}{6} + \frac{1}{12}$

To eliminate the fractions, we can multiply every term in the equation by the Least Common Multiple (LCM) of the denominators (2, 4, 12, 6, 12). The denominators are 2, 4, 6, and 12.

The LCM of 2, 4, 6, and 12 is 12.

$\begin{array}{c|cc} 2 & 2 \;, & 4 \;, & 6 \;, & 12 \\ \hline 2 & 1 \; , & 2 \; , & 3 \; , & 6 \\ \hline 3 & 1 \; , & 1 \; , & 3 \; , & 3 \\ \hline & 1 \; , & 1 \; , & 1 \; , & 1 \end{array}$

LCM = $2 \times 2 \times 3 = 12$.

Multiply each term by 12:

$12 \times \frac{x}{2} - 12 \times \frac{x}{4} + 12 \times \frac{1}{12} = 12 \times \frac{x}{6} + 12 \times \frac{1}{6} + 12 \times \frac{1}{12}$

Simplify each term:

$\cancel{12}^6 \times \frac{x}{\cancel{2}} - \cancel{12}^3 \times \frac{x}{\cancel{4}} + \cancel{12} \times \frac{1}{\cancel{12}} = \cancel{12}^2 \times \frac{x}{\cancel{6}} + \cancel{12}^2 \times \frac{1}{\cancel{6}} + \cancel{12} \times \frac{1}{\cancel{12}}$

$6x - 3x + 1 = 2x + 2 + 1$

Combine like terms on both sides:

$(6 - 3)x + 1 = 2x + (2 + 1)$

$3x + 1 = 2x + 3$

Now, collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and constant terms to the RHS.

Transpose $2x$ from the RHS to the LHS (change its sign):

$3x - 2x + 1 = 3$

Transpose 1 from the LHS to the RHS (change its sign):

$3x - 2x = 3 - 1$

Combine the like terms on both sides:

$(3 - 2)x = 2$

$1x = 2$

$x = 2$

The solution to the equation is $x = 2$.

The value of $x$ is 2.

The given equation is:

$\frac{1}{2} (x + 1) + \frac{1}{3} (x - 1) = \frac{5}{12} (x - 2)$

First, distribute the fractions on both sides of the equation:

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

RHS: $\frac{5}{12} \times x - \frac{5}{12} \times 2 = \frac{5x}{12} - \frac{10}{12}$

The equation now becomes:

$\frac{x}{2} + \frac{1}{2} + \frac{x}{3} - \frac{1}{3} = \frac{5x}{12} - \frac{10}{12}$

To eliminate the fractions, we can multiply every term in the equation by the Least Common Multiple (LCM) of the denominators (2, 3, 12). The denominators are 2, 3, and 12.

The LCM of 2, 3, and 12 is 12.

$\begin{array}{c|cc} 2 & 2 \;, & 3 \;, & 12 \\ \hline 3 & 1 \; , & 3 \; , & 6 \\ \hline & 1 \; , & 1 \; , & 2 \end{array}$

LCM = $2 \times 3 \times 2 = 12$.

Multiply each term by 12:

$12 \times \frac{x}{2} + 12 \times \frac{1}{2} + 12 \times \frac{x}{3} - 12 \times \frac{1}{3} = 12 \times \frac{5x}{12} - 12 \times \frac{10}{12}$

Simplify each term:

$\cancel{12}^6 \times \frac{x}{\cancel{2}} + \cancel{12}^6 \times \frac{1}{\cancel{2}} + \cancel{12}^4 \times \frac{x}{\cancel{3}} - \cancel{12}^4 \times \frac{1}{\cancel{3}} = \cancel{12} \times \frac{5x}{\cancel{12}} - \cancel{12} \times \frac{10}{\cancel{12}}$

$6x + 6 + 4x - 4 = 5x - 10$

Combine like terms on both sides:

$(6x + 4x) + (6 - 4) = 5x - 10$

$10x + 2 = 5x - 10$

Now, collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and constant terms to the RHS.

Transpose $5x$ from the RHS to the LHS (change its sign):

$10x - 5x + 2 = -10$

Transpose 2 from the LHS to the RHS (change its sign):

$10x - 5x = -10 - 2$

Combine the like terms on both sides:

$(10 - 5)x = -12$

$5x = -12$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 5:

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

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

The solution to the equation is $x = -\frac{12}{5}$.

The value of $x$ is $-\frac{12}{5}$.

The given equation is:

$\frac{x + 1}{4} = \frac{x - 2}{3}$

To solve for $x$, we can use cross-multiplication. This involves multiplying the numerator of one fraction by the denominator of the other fraction across the equals sign.

Multiply $(x + 1)$ by 3 and $(x - 2)$ by 4:

$3 \times (x + 1) = 4 \times (x - 2)$

$3(x + 1) = 4(x - 2)$

Now, distribute the constants on both sides of the equation:

$3 \times x + 3 \times 1 = 4 \times x - 4 \times 2$

$3x + 3 = 4x - 8$

Next, we need to collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the Right Hand Side (RHS) and constant terms to the Left Hand Side (LHS).

Transpose $4x$ from the RHS to the LHS (change its sign):

$3x - 4x + 3 = -8$

Transpose 3 from the LHS to the RHS (change its sign):

$3x - 4x = -8 - 3$

Combine the like terms on both sides:

$(3 - 4)x = -11$

$-1x = -11$

$-x = -11$

To find $x$, divide both sides of the equation by -1:

$\frac{-x}{-1} = \frac{-11}{-1}$

$x = 11$

The solution to the equation is $x = 11$.

The value of $x$ is 11.

The given equation is:

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

To solve for $x$, we use cross-multiplication. This involves multiplying the numerator of one fraction by the denominator of the other fraction across the equals sign.

Multiply $(2x - 1)$ by 3 and $(3x + 1)$ by 5:

$3 \times (2x - 1) = 5 \times (3x + 1)$

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

Now, distribute the constants on both sides of the equation:

$3 \times 2x - 3 \times 1 = 5 \times 3x + 5 \times 1$

$6x - 3 = 15x + 5$

Next, we need to collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the Right Hand Side (RHS) and constant terms to the Left Hand Side (LHS).

Transpose $15x$ from the RHS to the LHS (change its sign):

$6x - 15x - 3 = 5$

Transpose $-3$ from the LHS to the RHS (change its sign):

$6x - 15x = 5 + 3$

Combine the like terms on both sides:

$(6 - 15)x = 8$

$-9x = 8$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is -9:

$\frac{-9x}{-9} = \frac{8}{-9}$

$x = -\frac{8}{9}$

The solution to the equation is $x = -\frac{8}{9}$.

The value of $x$ is $-\frac{8}{9}$.

The given equation is:

$1 - (x - 2) - [(x - 3) - (x - 1)] = 0$

We need to simplify the expression by removing the parentheses and square brackets.

First, remove the innermost parentheses:

$1 - (x - 2) - [x - 3 - x + 1] = 0$

Simplify the expression inside the square brackets:

$[x - 3 - x + 1] = (x - x) + (-3 + 1) = 0 + (-2) = -2$

Substitute this back into the equation:

$1 - (x - 2) - [-2] = 0$

$1 - (x - 2) + 2 = 0$

Now, remove the remaining parentheses. Remember the minus sign before $(x - 2)$ changes the sign of each term inside:

$1 - x + 2 + 2 = 0$

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

$1 + 2 + 2 - x = 0$

$5 - x = 0$

To solve for $x$, transpose the term $-x$ from the LHS to the Right Hand Side (RHS). Its sign changes from negative to positive:

$5 = x$

Alternatively, transpose 5 from the LHS to the RHS:

$-x = 0 - 5$

$-x = -5$

Multiply or divide both sides by -1:

$x = 5$

The solution to the equation is $x = 5$.

The value of $x$ is 5.

The given equation is:

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

To eliminate the fractions, we multiply every term in the equation by the Least Common Multiple (LCM) of the denominators (3 and 4). The denominators are 3 and 4.

The LCM of 3 and 4 is 12.

$\begin{array}{c|cc} & 3 \;, & 4 \\ \hline 2 & 3 \; , & 2 \\ \hline 2 & 3 \; , & 1 \\ \hline 3 & 1 \; , & 1 \end{array}$

LCM = $2 \times 2 \times 3 = 12$.

Multiply each term by 12:

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

Simplify each term:

$36x - \cancel{12}^4 \times \frac{x - 2}{\cancel{3}} = 48 - \cancel{12}^3 \times \frac{x - 1}{\cancel{4}}$

$36x - 4(x - 2) = 48 - 3(x - 1)$

Now, distribute the constants into the parentheses on both sides:

$36x - (4 \times x - 4 \times 2) = 48 - (3 \times x - 3 \times 1)$

$36x - (4x - 8) = 48 - (3x - 3)$

Remove the parentheses, remembering that a minus sign before parentheses changes the sign of each term inside:

$36x - 4x + 8 = 48 - 3x + 3$

Combine like terms on both sides:

$(36x - 4x) + 8 = (48 + 3) - 3x$

$32x + 8 = 51 - 3x$

Now, collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and constant terms to the RHS.

Transpose $-3x$ from the RHS to the LHS (change its sign):

$32x + 3x + 8 = 51$

Transpose 8 from the LHS to the RHS (change its sign):

$32x + 3x = 51 - 8$

Combine the like terms on both sides:

$(32 + 3)x = 43$

$35x = 43$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 35:

$\frac{35x}{35} = \frac{43}{35}$

$x = \frac{43}{35}$

The solution to the equation is $x = \frac{43}{35}$.

The value of $x$ is $\frac{43}{35}$.

The given equation is:

$\frac{3t + 5}{4} - 1 = \frac{4t - 3}{5}$

To eliminate the fractions, we multiply every term in the equation by the Least Common Multiple (LCM) of the denominators (4 and 5). The denominators are 4 and 5.

The LCM of 4 and 5 is 20.

Multiply each term by 20:

$20 \times \frac{3t + 5}{4} - 20 \times 1 = 20 \times \frac{4t - 3}{5}$

Simplify each term by cancelling the denominators:

$\cancel{20}^5 \times \frac{3t + 5}{\cancel{4}} - 20 = \cancel{20}^4 \times \frac{4t - 3}{\cancel{5}}$

$5(3t + 5) - 20 = 4(4t - 3)$

Now, distribute the constants into the parentheses on both sides:

$5 \times 3t + 5 \times 5 - 20 = 4 \times 4t - 4 \times 3$

$15t + 25 - 20 = 16t - 12$

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

$15t + (25 - 20) = 16t - 12$

$15t + 5 = 16t - 12$

Now, we collect the terms containing $t$ on one side and the constant terms on the other side. Let's move the $t$ terms to the Right Hand Side (RHS) and constant terms to the LHS.

Transpose $16t$ from the RHS to the LHS (change its sign):

$15t - 16t + 5 = -12$

Transpose 5 from the LHS to the RHS (change its sign):

$15t - 16t = -12 - 5$

Combine the like terms on both sides:

$(15 - 16)t = -17$

$-1t = -17$

$-t = -17$

To find $t$, multiply or divide both sides of the equation by -1:

$\frac{-t}{-1} = \frac{-17}{-1}$

$t = 17$

The solution to the equation is $t = 17$.

The value of $t$ is 17.

The given equation is:

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

To eliminate the fractions, we multiply every term in the equation by the Least Common Multiple (LCM) of the denominators (4 and 2). The denominators are 4 and 2.

The LCM of 4 and 2 is 4.

$\begin{array}{c|cc} 2 & 4 \;, & 2 \\ \hline 2 & 2 \; , & 1 \\ \hline & 1 \; , & 1 \end{array}$

LCM = $2 \times 2 = 4$.

Multiply each term by 4:

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

Simplify each term by cancelling the denominators:

$\cancel{4} \times \frac{2y - 3}{\cancel{4}} - \cancel{4}^2 \times \frac{3y - 5}{\cancel{2}} = 4y + \cancel{4} \times \frac{3}{\cancel{4}}$

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

Now, remove the parentheses and distribute the constants. Remember the minus sign before the second parenthesis on the LHS.

$2y - 3 - (2 \times 3y - 2 \times 5) = 4y + 3$

$2y - 3 - (6y - 10) = 4y + 3$

$2y - 3 - 6y + 10 = 4y + 3$

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

$(2y - 6y) + (-3 + 10) = 4y + 3$

$-4y + 7 = 4y + 3$

Now, we collect the terms containing $y$ on one side and the constant terms on the other side. Let's move the $y$ terms to the Right Hand Side (RHS) and constant terms to the LHS.

Transpose $4y$ from the RHS to the LHS (change its sign):

$-4y - 4y + 7 = 3$

Transpose 7 from the LHS to the RHS (change its sign):

$-4y - 4y = 3 - 7$

Combine the like terms on both sides:

$(-4 - 4)y = -4$

$-8y = -4$

Finally, to find $y$, divide both sides of the equation by the coefficient of $y$, which is -8:

$\frac{-8y}{-8} = \frac{-4}{-8}$

$y = \frac{4}{8}$

Simplify the fraction:

$y = \frac{\cancel{4}^1}{\cancel{8}_2} = \frac{1}{2}$

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

The solution to the equation is $y = \frac{1}{2}$.

The value of $y$ is $\frac{1}{2}$.

The given equation is:

$0.25 (4x - 5) = 0.75x + 8$

We can solve this equation using decimals or by converting the decimals to fractions.

Method 1: Using Decimals

First, distribute 0.25 on the Left Hand Side (LHS):

$0.25 \times 4x - 0.25 \times 5 = 0.75x + 8$

$1.00x - 1.25 = 0.75x + 8$

$x - 1.25 = 0.75x + 8$

Collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and constant terms to the Right Hand Side (RHS).

Transpose $0.75x$ from the RHS to the LHS (change its sign):

$x - 0.75x - 1.25 = 8$

Transpose $-1.25$ from the LHS to the RHS (change its sign):

$x - 0.75x = 8 + 1.25$

Combine the like terms on both sides:

$(1 - 0.75)x = 9.25$

$0.25x = 9.25$

Finally, to find $x$, divide both sides by 0.25:

$x = \frac{9.25}{0.25}$

To perform the division, multiply the numerator and denominator by 100 to remove decimals:

$x = \frac{9.25 \times 100}{0.25 \times 100} = \frac{925}{25}$

Divide 925 by 25:

$x = 37$

Method 2: Converting to Fractions

Convert the decimals to fractions:

$0.25 = \frac{25}{100} = \frac{1}{4}$

$0.75 = \frac{75}{100} = \frac{3}{4}$

Substitute the fractions back into the equation:

$\frac{1}{4}(4x - 5) = \frac{3}{4}x + 8$

Distribute $\frac{1}{4}$ on the LHS:

$\frac{1}{4} \times 4x - \frac{1}{4} \times 5 = \frac{3}{4}x + 8$

$x - \frac{5}{4} = \frac{3}{4}x + 8$

To eliminate the fractions, multiply every term by the LCM of the denominators (4), which is 4:

$4 \times x - 4 \times \frac{5}{4} = 4 \times \frac{3}{4}x + 4 \times 8$

$4x - 5 = 3x + 32$

Collect $x$ terms on the LHS and constant terms on the RHS:

$4x - 3x = 32 + 5$

$x = 37$

Both methods yield the same solution.

The solution to the equation is $x = 37$.

The value of $x$ is 37.

The given equation is:

$\frac{9 - 3y}{1 - 9y} = \frac{8}{5}$

To solve for $y$, we use cross-multiplication. This involves multiplying the numerator of one fraction by the denominator of the other fraction across the equals sign. Note that the denominator cannot be zero, so $1 - 9y \neq 0 \implies 9y \neq 1 \implies y \neq \frac{1}{9}$.

Multiply $(9 - 3y)$ by 5 and $(1 - 9y)$ by 8:

$5 \times (9 - 3y) = 8 \times (1 - 9y)$

$5(9 - 3y) = 8(1 - 9y)$

Now, distribute the constants on both sides of the equation:

$5 \times 9 - 5 \times 3y = 8 \times 1 - 8 \times 9y$

$45 - 15y = 8 - 72y$

Next, we need to collect the terms containing $y$ on one side and the constant terms on the other side. Let's move the $y$ terms to the Left Hand Side (LHS) and constant terms to the Right Hand Side (RHS).

Transpose $-72y$ from the RHS to the LHS (change its sign):

$45 - 15y + 72y = 8$

Transpose 45 from the LHS to the RHS (change its sign):

$-15y + 72y = 8 - 45$

Combine the like terms on both sides:

$(-15 + 72)y = -37$

$57y = -37$

Finally, to find $y$, divide both sides of the equation by the coefficient of $y$, which is 57:

$\frac{57y}{57} = \frac{-37}{57}$

$y = -\frac{37}{57}$

We check if our solution $y=-\frac{37}{57}$ violates the initial constraint $y \neq \frac{1}{9}$. Since $-\frac{37}{57} \neq \frac{1}{9}$, the solution is valid.

The solution to the equation is $y = -\frac{37}{57}$.

The value of $y$ is $-\frac{37}{57}$.

The given equation is:

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

To solve for $x$, we use cross-multiplication. Note that the denominator cannot be zero, so $2x - 3 \neq 0 \implies 2x \neq 3 \implies x \neq \frac{3}{2}$.

Multiply $(3x + 2)$ by 4 and $(2x - 3)$ by -3:

$4 \times (3x + 2) = -3 \times (2x - 3)$

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

Now, distribute the constants on both sides of the equation:

$4 \times 3x + 4 \times 2 = -3 \times 2x - 3 \times (-3)$

$12x + 8 = -6x + 9$

Next, we need to collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the Left Hand Side (LHS) and constant terms to the Right Hand Side (RHS).

Transpose $-6x$ from the RHS to the LHS (change its sign):

$12x + 6x + 8 = 9$

Transpose 8 from the LHS to the RHS (change its sign):

$12x + 6x = 9 - 8$

Combine the like terms on both sides:

$(12 + 6)x = 1$

$18x = 1$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 18:

$\frac{18x}{18} = \frac{1}{18}$

$x = \frac{1}{18}$

We check if our solution $x=\frac{1}{18}$ violates the initial constraint $x \neq \frac{3}{2}$. Since $\frac{1}{18} \neq \frac{3}{2}$, the solution is valid.

The solution to the equation is $x = \frac{1}{18}$.

The value of $x$ is $\frac{1}{18}$.

The given equation is:

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

To solve for $x$, we use cross-multiplication. Note that the denominator cannot be zero, so $2x \neq 0 \implies x \neq 0$.

Multiply $(5x + 1)$ by 3 and $2x$ by -1:

$3 \times (5x + 1) = -1 \times (2x)$

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

Now, distribute the 3 on the Left Hand Side (LHS):

$3 \times 5x + 3 \times 1 = -2x$

$15x + 3 = -2x$

Next, we need to collect the terms containing $x$ on one side. Let's move the $x$ terms to the LHS.

Transpose $-2x$ from the Right Hand Side (RHS) to the LHS (change its sign):

$15x + 2x + 3 = 0$

Transpose the constant term 3 from the LHS to the RHS (change its sign):

$15x + 2x = -3$

Combine the like terms on the LHS:

$(15 + 2)x = -3$

$17x = -3$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 17:

$\frac{17x}{17} = \frac{-3}{17}$

$x = -\frac{3}{17}$

We check if our solution $x=-\frac{3}{17}$ violates the initial constraint $x \neq 0$. Since $-\frac{3}{17} \neq 0$, the solution is valid.

The solution to the equation is $x = -\frac{3}{17}$.

The value of $x$ is $-\frac{3}{17}$.

The given equation is:

$\frac{3t - 2}{3} + \frac{2t + 3}{2} = t + \frac{7}{6}$

To eliminate the fractions, we multiply every term in the equation by the Least Common Multiple (LCM) of the denominators (3, 2, and 6). The denominators are 3, 2, and 6.

The LCM of 3, 2, and 6 is 6.

$\begin{array}{c|cc} 2 & 3 \;, & 2 \;, & 6 \\ \hline 3 & 3 \; , & 1 \; , & 3 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

LCM = $2 \times 3 = 6$.

Multiply each term by 6:

$6 \times \frac{3t - 2}{3} + 6 \times \frac{2t + 3}{2} = 6 \times t + 6 \times \frac{7}{6}$

Simplify each term by cancelling the denominators:

$\cancel{6}^2 \times \frac{3t - 2}{\cancel{3}} + \cancel{6}^3 \times \frac{2t + 3}{\cancel{2}} = 6t + \cancel{6} \times \frac{7}{\cancel{6}}$

$2(3t - 2) + 3(2t + 3) = 6t + 7$

Now, distribute the constants into the parentheses on the Left Hand Side (LHS):

$(2 \times 3t - 2 \times 2) + (3 \times 2t + 3 \times 3) = 6t + 7$

$(6t - 4) + (6t + 9) = 6t + 7$

$6t - 4 + 6t + 9 = 6t + 7$

Combine the like terms on the LHS:

$(6t + 6t) + (-4 + 9) = 6t + 7$

$12t + 5 = 6t + 7$

Now, we collect the terms containing $t$ on one side and the constant terms on the other side. Let's move the $t$ terms to the LHS and constant terms to the Right Hand Side (RHS).

Transpose $6t$ from the RHS to the LHS (change its sign):

$12t - 6t + 5 = 7$

Transpose 5 from the LHS to the RHS (change its sign):

$12t - 6t = 7 - 5$

Combine the like terms on both sides:

$(12 - 6)t = 2$

$6t = 2$

Finally, to find $t$, divide both sides of the equation by the coefficient of $t$, which is 6:

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

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

Simplify the fraction:

$t = \frac{\cancel{2}^1}{\cancel{6}_3} = \frac{1}{3}$

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

The solution to the equation is $t = \frac{1}{3}$.

The value of $t$ is $\frac{1}{3}$.

The given equation is:

$m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3}$

To eliminate the fractions, we multiply every term in the equation by the Least Common Multiple (LCM) of the denominators (2 and 3). The denominators are 2 and 3.

The LCM of 2 and 3 is 6.

Multiply each term by 6:

$6 \times m - 6 \times \frac{m - 1}{2} = 6 \times 1 - 6 \times \frac{m - 2}{3}$

Simplify each term by cancelling the denominators:

$6m - \cancel{6}^3 \times

Combine the like terms on both sides:

$(3 + 2)m = 7$

$5m = 7$

Finally, to find $m$, divide both sides of the equation by the coefficient of $m$, which is 5:

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

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

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

The value of $m$ is $\frac{7}{5}$.

The given equation is:

$4(3p + 2) - 5(6p - 1) = 2(p - 8) - 6(7p - 4)$

First, distribute the constants into the parentheses on both sides of the equation.

Left Hand Side (LHS):

$4(3p + 2) - 5(6p - 1) = (4 \times 3p + 4 \times 2) - (5 \times 6p - 5 \times 1)$

$= (12p + 8) - (30p - 5)$

$= 12p + 8 - 30p + 5$

Right Hand Side (RHS):

$2(p - 8) - 6(7p - 4) = (2 \times p - 2 \times 8) - (6 \times 7p - 6 \times 4)$

$= (2p - 16) - (42p - 24)$

$= 2p - 16 - 42p + 24$

Now, combine the like terms on each side of the equation.

LHS: $(12p - 30p) + (8 + 5) = -18p + 13$

RHS: $(2p - 42p) + (-16 + 24) = -40p + 8$

The equation becomes:

$-18p + 13 = -40p + 8$

Next, collect the terms containing $p$ on one side and the constant terms on the other side. Let's move the $p$ terms to the LHS and the constant terms to the RHS.

Transpose $-40p$ from the RHS to the LHS (change its sign):

$-18p + 40p + 13 = 8$

Transpose 13 from the LHS to the RHS (change its sign):

$-18p + 40p = 8 - 13$

Combine the like terms on both sides:

$(-18 + 40)p = -5$

$22p = -5$

Finally, to find $p$, divide both sides of the equation by the coefficient of $p$, which is 22:

$\frac{22p}{22} = \frac{-5}{22}$

$p = -\frac{5}{22}$

The solution to the equation is $p = -\frac{5}{22}$.

The value of $p$ is $-\frac{5}{22}$.

The given equation is:

$3(5x - 7) + 2(9x - 11) = 4(8x - 7) - 111$

First, distribute the constants into the parentheses on both sides of the equation.

Left Hand Side (LHS):

$3(5x - 7) + 2(9x - 11) = (3 \times 5x - 3 \times 7) + (2 \times 9x - 2 \times 11)$

$= (15x - 21) + (18x - 22)$

$= 15x - 21 + 18x - 22$

Right Hand Side (RHS):

$4(8x - 7) - 111 = (4 \times 8x - 4 \times 7) - 111$

$= (32x - 28) - 111$

$= 32x - 28 - 111$

Now, combine the like terms on each side of the equation.

LHS: $(15x + 18x) + (-21 - 22) = 33x - 43$

RHS: $32x + (-28 - 111) = 32x - 139$

The equation becomes:

$33x - 43 = 32x - 139$

Next, collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and the constant terms to the RHS.

Transpose $32x$ from the RHS to the LHS (change its sign):

$33x - 32x - 43 = -139$

Transpose $-43$ from the LHS to the RHS (change its sign):

$33x - 32x = -139 + 43$

Combine the like terms on both sides:

$(33 - 32)x = -96$

$1x = -96$

$x = -96$

The solution to the equation is $x = -96$.

The value of $x$ is -96.

The given equation is:

$0.16 (5x - 2) = 0.4x + 7$

First, distribute 0.16 on the Left Hand Side (LHS):

$0.16 \times 5x - 0.16 \times 2 = 0.4x + 7$

$0.80x - 0.32 = 0.4x + 7$

$0.8x - 0.32 = 0.4x + 7$

Now, we collect the terms containing $x$ on one side and the constant terms on the other side. Let's move the $x$ terms to the LHS and constant terms to the Right Hand Side (RHS).

Transpose $0.4x$ from the RHS to the LHS (change its sign):

$0.8x - 0.4x - 0.32 = 7$

Transpose $-0.32$ from the LHS to the RHS (change its sign):

$0.8x - 0.4x = 7 + 0.32$

Combine the like terms on both sides:

$(0.8 - 0.4)x = 7.32$

$0.4x = 7.32$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is 0.4:

$x = \frac{7.32}{0.4}$

To perform the division, multiply the numerator and denominator by 10 to remove the decimal from the denominator (or 100 to remove all decimals):

$x = \frac{7.32 \times 10}{0.4 \times 10} = \frac{73.2}{4}$

Now, perform the division:

$x = 18.3$

The solution to the equation is $x = 18.3$.

The value of $x$ is 18.3.

Given:

Radha offers one half of the flowers at each of the three temples she visits.

She is left with 3 flowers at the end, after visiting the third temple.

To Find:

The number of flowers Radha had in the beginning.

Solution:

Let the number of flowers Radha had in the beginning be $x$.

At the first temple, she offers half of the flowers ($x$).

Number of flowers offered $= \frac{1}{2}x$

Number of flowers remaining after the first temple $= x - \frac{1}{2}x = \frac{1}{2}x$

At the second temple, she offers half of the remaining flowers ($\frac{1}{2}x$).

Number of flowers offered $= \frac{1}{2} \left(\frac{1}{2}x\right) = \frac{1}{4}x$

Number of flowers remaining after the second temple $= \frac{1}{2}x - \frac{1}{4}x = \frac{2x - x}{4} = \frac{1}{4}x$

At the third temple, she offers half of the remaining flowers ($\frac{1}{4}x$).

Number of flowers offered $= \frac{1}{2} \left(\frac{1}{4}x\right) = \frac{1}{8}x$

Number of flowers remaining after the third temple $= \frac{1}{4}x - \frac{1}{8}x = \frac{2x - x}{8} = \frac{1}{8}x$

According to the problem, she is left with 3 flowers at the end. So, the number of flowers remaining after the third temple is 3.

We can set up the equation:

$\frac{1}{8}x = 3$

To solve for $x$, multiply both sides of the equation by 8:

$8 \times \frac{1}{8}x = 8 \times 3$

$x = 24$

Alternate Solution (Working Backwards):

Let the number of flowers left at the end (after the 3rd temple) be $F_3 = 3$.

Before the 3rd temple, she had $F_2$ flowers. She offered $\frac{1}{2}F_2$ and was left with $\frac{1}{2}F_2 = F_3$.

$\frac{1}{2}F_2 = 3$

$F_2 = 3 \times 2 = 6$ flowers (before the 3rd temple).

Before the 2nd temple, she had $F_1$ flowers. She offered $\frac{1}{2}F_1$ and was left with $\frac{1}{2}F_1 = F_2$.

$\frac{1}{2}F_1 = 6$

$F_1 = 6 \times 2 = 12$ flowers (before the 2nd temple).

Before the 1st temple, she had $F_0$ flowers. She offered $\frac{1}{2}F_0$ and was left with $\frac{1}{2}F_0 = F_1$.

$\frac{1}{2}F_0 = 12$

$F_0 = 12 \times 2 = 24$ flowers (in the beginning).

Both methods show that Radha had 24 flowers in the beginning.

The number of flowers she had in the beginning was 24.

Given:

Total amount to be distributed = $\textsf{₹}$ 13500.

Salma gets $\textsf{₹}$ 1000 more than Kiran.

Jenifer gets $\textsf{₹}$ 500 more than Kiran.

To Find:

The amount of money received by Jenifer.

Solution:

Let the amount of money received by Kiran be $\textsf{₹} x$.

According to the problem, Salma gets $\textsf{₹}$ 1000 more than Kiran.

Amount received by Salma = (Amount received by Kiran) + $\textsf{₹}$ 1000

Amount received by Salma = $\textsf{₹} (x + 1000)$.

According to the problem, Jenifer gets $\textsf{₹}$ 500 more than Kiran.

Amount received by Jenifer = (Amount received by Kiran) + $\textsf{₹}$ 500

Amount received by Jenifer = $\textsf{₹} (x + 500)$.

The total amount distributed among Salma, Kiran, and Jenifer is $\textsf{₹}$ 13500.

(Amount by Salma) + (Amount by Kiran) + (Amount by Jenifer) = $\textsf{₹}$ 13500

$(x + 1000) + x + (x + 500) = 13500$

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

$x + 1000 + x + x + 500 = 13500$

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

$(x + x + x) + (1000 + 500) = 13500$

$3x + 1500 = 13500$

Transpose the constant term 1500 from the LHS to the Right Hand Side (RHS). Its sign changes from positive to negative:

$3x = 13500 - 1500$

$3x = 12000$

To find $x$, divide both sides of the equation by the coefficient of $x$, which is 3:

$\frac{3x}{3} = \frac{12000}{3}$

$x = 4000$

So, the amount received by Kiran is $\textsf{₹}$ 4000.

The question asks for the money received by Jenifer.

Amount received by Jenifer $= x + 500$

Amount received by Jenifer $= 4000 + 500 = 4500$

The amount of money received by Jenifer is $\textsf{₹}$ 4500.

Given:

The volume of water in one tank is twice that in the other tank.

25 litres are drawn from the first tank and added to the other.

After the transfer, the volumes in both tanks are equal.

To Find:

The initial volume of water in each tank.

Solution:

Let the initial volume of water in the smaller tank be $x$ litres.

According to the problem, the volume of water in the other tank (the larger one) is twice that in the smaller tank.

Initial volume of water in the larger tank $= 2x$ litres.

Now, consider the change in volumes after drawing out 25 litres from the first (larger) tank and adding it to the other (smaller) tank.

Volume of water in the larger tank after drawing out 25 litres $= (2x - 25)$ litres.

Volume of water in the smaller tank after adding 25 litres $= (x + 25)$ litres.

According to the problem, after this transfer, the volumes of water in each tank become the same.

So, we can set up the equation:

$2x - 25 = x + 25$

Now, we solve this linear equation for $x$. Collect the terms containing $x$ on one side and the constant terms on the other side.

Transpose $x$ from the Right Hand Side (RHS) to the Left Hand Side (LHS) (change its sign):

$2x - x - 25 = 25$

Transpose $-25$ from the LHS to the RHS (change its sign):

$2x - x = 25 + 25$

Combine the like terms on both sides:

$(2 - 1)x = 50$

$x = 50$

So, the initial volume of water in the smaller tank is 50 litres.

The initial volume of water in the larger tank is $2x = 2 \times 50 = 100$ litres.

Let's verify the solution:

Initial volumes are 100 litres and 50 litres.

Draw 25 litres from the first (100L) $\implies 100 - 25 = 75$ litres.

Add 25 litres to the other (50L) $\implies 50 + 25 = 75$ litres.

The volumes are indeed the same (75 litres).

The initial volumes of water in the tanks are 100 litres and 50 litres.

Given:

Anushka and Aarushi initially have equal amounts of money.

Anushka gave $\frac{1}{3}$ of her money to Aarushi.

Aarushi paid half of her total money for a party.

Aarushi is left with $\textsf{₹}$1600.

To Find:

The amount of money gifted by Anushka to Aarushi.

Solution:

Let the initial amount of money with Anushka and Aarushi be $\textsf{₹} x$ each.

Anushka gave $\frac{1}{3}$ of her money ($x$) to Aarushi.

Amount gifted by Anushka = $\frac{1}{3}x$.

After receiving the gift, the total money Aarushi has is her initial amount plus the gifted amount.

Total money with Aarushi after receiving gift $= x + \frac{1}{3}x = \frac{3x + x}{3} = \frac{4x}{3}$.

Aarushi paid half of the total money she had for the party.

Amount paid for the party $= \frac{1}{2} \times (\text{Total money with Aarushi})$

Amount paid for the party $= \frac{1}{2} \times \frac{4x}{3} = \frac{\cancel{4}^2x}{\cancel{2} \times 3} = \frac{2x}{3}$.

The money remaining with Aarushi is the total money she had minus the amount paid for the party.

Remaining money with Aarushi $= \frac{4x}{3} - \frac{2x}{3} = \frac{4x - 2x}{3} = \frac{2x}{3}$.

According to the problem, the remaining money with Aarushi is $\textsf{₹}$1600.

So, we set up the equation:

$\frac{2x}{3} = 1600$

To solve for $x$, multiply both sides by 3, then divide by 2.

$2x = 1600 \times 3$

$2x = 4800$

$x = \frac{4800}{2}$

$x = 2400$

So, the initial amount of money with Anushka and Aarushi was $\textsf{₹}$2400 each.

The question asks for the sum gifted by Anushka.

Amount gifted by Anushka $= \frac{1}{3}x$

Amount gifted by Anushka $= \frac{1}{3} \times 2400 = \frac{2400}{3}$

Amount gifted by Anushka $= 800$

The sum gifted by Anushka was $\textsf{₹}$800.

Verification:

Initial money: Anushka $\textsf{₹}$2400, Aarushi $\textsf{₹}$2400.

Anushka gifts $\frac{1}{3} \times 2400 = \textsf{₹}800$.

Money with Anushka after gifting = $2400 - 800 = \textsf{₹}1600$.

Money with Aarushi after receiving gift = $2400 + 800 = \textsf{₹}3200$.

Aarushi pays half for the party = $\frac{1}{2} \times 3200 = \textsf{₹}1600$.

Remaining money with Aarushi = $3200 - 1600 = \textsf{₹}1600$.

This matches the information given in the problem.

The sum gifted by Anushka was $\textsf{₹}$800.

Given:

Total number of flowers Kaustubh had in the beginning = 60.

Ratio of remaining flowers to flowers in the beginning = 3 : 5.

To Find:

The number of flowers offered by Kaustubh in the temple.

Solution:

Let the number of flowers offered by Kaustubh in the temple be $x$.

The number of flowers remaining after offering some in the temple is the initial number minus the number offered.

Number of remaining flowers $= 60 - x$.

The ratio of the number of remaining flowers to that of flowers in the beginning is given as 3 : 5.

So, we can write the ratio as a fraction:

$\frac{\text{Number of remaining flowers}}{\text{Number of flowers in the beginning}} = \frac{3}{5}$

Substitute the expressions for the number of flowers:

$\frac{60 - x}{60} = \frac{3}{5}$

Now, we solve this linear equation for $x$. We can use cross-multiplication.

Multiply $(60 - x)$ by 5 and 60 by 3:

$5 \times (60 - x) = 3 \times 60$

$5(60 - x) = 180$

Distribute the 5 on the Left Hand Side (LHS):

$5 \times 60 - 5 \times x = 180$

$300 - 5x = 180$

Now, collect the constant terms on the Right Hand Side (RHS). Transpose 300 from the LHS to the RHS (change its sign):

$-5x = 180 - 300$

Combine the constant terms on the RHS:

$-5x = -120$

Finally, to find $x$, divide both sides of the equation by the coefficient of $x$, which is -5:

$\frac{-5x}{-5} = \frac{-120}{-5}$

$x = 24$

So, the number of flowers offered by him in the temple is 24.

Verification:

Initial flowers = 60.

Flowers offered = 24.

Remaining flowers = $60 - 24 = 36$.

Ratio of remaining flowers to initial flowers = $\frac{36}{60} = \frac{36 \div 12}{60 \div 12} = \frac{3}{5}$.

This matches the given ratio 3:5.

The number of flowers offered by him in the temple is 24.

Given:

The sum of three consecutive even natural numbers is 48.

To Find:

The greatest of these three numbers.

Solution:

Let the first even natural number be $x$.

Since the numbers are consecutive even natural numbers, the next even number is obtained by adding 2 to the previous one.

The second consecutive even natural number is $x + 2$.

The third consecutive even natural number is $(x + 2) + 2 = x + 4$.

The sum of these three numbers is given as 48.

So, we set up the equation:

$x + (x + 2) + (x + 4) = 48$

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

$(x + x + x) + (2 + 4) = 48$

$3x + 6 = 48$

Transpose the constant term 6 from the LHS to the Right Hand Side (RHS). Its sign changes from positive to negative:

$3x = 48 - 6$

$3x = 42$

To find $x$, divide both sides of the equation by the coefficient of $x$, which is 3:

$\frac{3x}{3} = \frac{42}{3}$

$x = 14$

So, the first even natural number is 14.

The three consecutive even natural numbers are:

First number $= x = 14$

Second number $= x + 2 = 14 + 2 = 16$

Third number $= x + 4 = 14 + 4 = 18$

The greatest of these three numbers is the third number, which is 18.

Verification:

The sum of the three numbers is $14 + 16 + 18 = 48$. This matches the given sum.

The numbers 14, 16, and 18 are consecutive even natural numbers.

The greatest of these numbers is 18.

Solution:

Given:

The sum of three consecutive odd natural numbers is 69.

To Find:

The prime number among these three numbers.

Let the first odd natural number be $x$.

Since the numbers are consecutive odd natural numbers, the next odd natural number will be $x + 2$, and the third consecutive odd natural number will be $(x+2) + 2 = x + 4$.

The three consecutive odd natural numbers are $x$, $x+2$, and $x+4$.

According to the problem, the sum of these three numbers is 69.

So, we can write the equation:

$x + (x+2) + (x+4) = 69$

Combine the like terms:

$(x+x+x) + (2+4) = 69$

$3x + 6 = 69$

Subtract 6 from both sides of the equation:

$3x + 6 - 6 = 69 - 6$

$3x = 63$

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

$\frac{3x}{3} = \frac{63}{3}$

$x = 21$

So, the first odd natural number is $x = 21$.

The second consecutive odd natural number is $x+2 = 21+2 = 23$.

The third consecutive odd natural number is $x+4 = 21+4 = 25$.

The three consecutive odd natural numbers are 21, 23, and 25.

Now we need to find the prime number among these three numbers.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's examine each number:

1. 21: The divisors of 21 are 1, 3, 7, and 21. Since 21 has divisors other than 1 and 21 (namely 3 and 7), 21 is not a prime number. It is a composite number.

2. 23: To check if 23 is prime, we test for divisibility by prime numbers less than or equal to $\sqrt{23}$. $\sqrt{23}$ is approximately 4.8. The prime numbers less than 4.8 are 2 and 3. 23 is not divisible by 2 (it's odd). The sum of the digits of 23 is $2+3=5$, which is not divisible by 3, so 23 is not divisible by 3. Since 23 is not divisible by any prime less than or equal to its square root, 23 is a prime number.

3. 25: The divisors of 25 are 1, 5, and 25. Since 25 has a divisor other than 1 and 25 (namely 5), 25 is not a prime number. It is a composite number.

Out of the three numbers (21, 23, and 25), only 23 is a prime number.

Final Answer: The prime number out of the given three consecutive odd natural numbers is 23.

Solution:

Given:

The sum of three consecutive numbers is 156.

To Find:

The number which is a multiple of 13 out of these three numbers.

Let the first of the three consecutive numbers be $n$.

Since the numbers are consecutive, the next two numbers will be $n+1$ and $n+2$.

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

According to the problem, the sum of these three numbers is 156.

So, we can write the equation:

$n + (n+1) + (n+2) = 156$

Combine the like terms:

$(n+n+n) + (1+2) = 156$

$3n + 3 = 156$

Subtract 3 from both sides of the equation:

$3n + 3 - 3 = 156 - 3$

$3n = 153$

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

$\frac{3n}{3} = \frac{153}{3}$

$n = 51$

So, the first number is $n = 51$.

The second consecutive number is $n+1 = 51+1 = 52$.

The third consecutive number is $n+2 = 51+2 = 53$.

The three consecutive numbers are 51, 52, and 53.

Now we need to find which of these numbers is a multiple of 13.

A multiple of 13 is a number that can be divided by 13 with no remainder.

Let's check each number:

1. 51: Divide 51 by 13. $51 \div 13$. $13 \times 3 = 39$, $13 \times 4 = 52$. 51 is not a multiple of 13.

2. 52: Divide 52 by 13. $52 \div 13$. $13 \times 4 = 52$. 52 is a multiple of 13.

3. 53: Divide 53 by 13. $53 \div 13$. $13 \times 4 = 52$, leaving a remainder of 1. 53 is not a multiple of 13.

Out of the three numbers (51, 52, and 53), the number which is a multiple of 13 is 52.

Final Answer: The number which is a multiple of 13 is 52.

Solution:

Given:

A statement describing a relationship between a number, its fifth part, its fourth part, and the constants 30.

To Find:

The number that satisfies the given condition.

Let the unknown number be $x$.

According to the problem statement:

"whose fifth part increased by 30" can be written as $\frac{x}{5} + 30$.

"its fourth part decreased by 30" can be written as $\frac{x}{4} - 30$.

These two expressions are equal.

So, the equation is:

$\frac{x}{5} + 30 = \frac{x}{4} - 30$

To solve for $x$, we can first move the constant terms to one side and the terms involving $x$ to the other side.

Add 30 to both sides of the equation:

$\frac{x}{5} + 30 + 30 = \frac{x}{4} - 30 + 30$

$\frac{x}{5} + 60 = \frac{x}{4}$

Subtract $\frac{x}{5}$ from both sides of the equation:

$60 = \frac{x}{4} - \frac{x}{5}$

Find a common denominator for the fractions on the right side. The least common multiple (LCM) of 4 and 5 is 20.

$60 = \frac{5x}{20} - \frac{4x}{20}$

Combine the fractions on the right side:

$60 = \frac{5x - 4x}{20}$

$60 = \frac{x}{20}$

Multiply both sides of the equation by 20 to solve for $x$:

$60 \times 20 = \frac{x}{20} \times 20$

$1200 = x$

So, the number is 1200.

Check the answer:

Fifth part of 1200 increased by 30: $\frac{1200}{5} + 30 = 240 + 30 = 270$.

Fourth part of 1200 decreased by 30: $\frac{1200}{4} - 30 = 300 - 30 = 270$.

Since $270 = 270$, the number 1200 satisfies the condition.

Final Answer: The number is 1200.

Solution:

Given:

A number 54 is to be divided into two parts.

One part is $\frac{2}{7}$ of the other part.

To Find:

The two parts into which 54 is divided.

Let the two parts be $x$ and $y$.

According to the problem, the sum of the two parts is 54.

$x + y = 54$

Also, one part is $\frac{2}{7}$ of the other. Let's assume the first part $x$ is $\frac{2}{7}$ of the second part $y$.

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

Now we have a system of two linear equations:

1) $x + y = 54$

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

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

$\left(\frac{2}{7}y\right) + y = 54$

Combine the terms involving $y$ on the left side. The coefficient of $y$ is 1, which can be written as $\frac{7}{7}$.

$\frac{2}{7}y + \frac{7}{7}y = 54$

$\frac{2y + 7y}{7} = 54$

$\frac{9y}{7} = 54$

To solve for $y$, multiply both sides of the equation by 7:

$9y = 54 \times 7$

$9y = 378$

Now, divide both sides by 9 to find the value of $y$:

$y = \frac{378}{9}$

$y = 42$

Now that we have the value of $y$, we can find the value of $x$ using equation (2):

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

$x = \frac{2}{7} \times 42$

$x = 2 \times \left(\frac{42}{7}\right)$

$x = 2 \times 6$

$x = 12$

The two parts are 12 and 42.

Check:

Sum of the parts: $12 + 42 = 54$. (Correct)

Is one part $\frac{2}{7}$ of the other? Let's check if $12 = \frac{2}{7} \times 42$.

$\frac{2}{7} \times 42 = 2 \times 6 = 12$. (Correct)

Final Answer: The two parts into which 54 is divided are 12 and 42.

Solution:

Given:

1. The sum of the digits of a two-digit number is 11.

2. The given number is less than the number obtained by interchanging the digits by 9.

To Find:

The two-digit number.

Let the digit in the tens place of the two-digit number be $t$ and the digit in the units place be $u$.

The two-digit number can be expressed as $10t + u$.

The number obtained by interchanging the digits will have $u$ in the tens place and $t$ in the units place. This number can be expressed as $10u + t$.

According to the first condition, the sum of the digits is 11:

According to the second condition, the given number ($10t + u$) is less than the number obtained by interchanging the digits ($10u + t$) by 9. This means the difference between the interchanged number and the original number is 9.

$(10u + t) - (10t + u) = 9$

Simplify the equation:

$10u + t - 10t - u = 9$

$(10u - u) + (t - 10t) = 9$

$9u - 9t = 9$

Divide the entire equation by 9:

Now we have a system of two linear equations:

(i) $t + u = 11$

(ii) $-t + u = 1$

We can solve this system by adding equation (i) and equation (ii):

$(t + u) + (-t + u) = 11 + 1$

$t + u - t + u = 12$

$2u = 12$

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

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

$u = 6$

Now substitute the value of $u=6$ into equation (i):

$t + u = 11$

$t + 6 = 11$

$t = 11 - 6$

$t = 5$

The digit in the tens place is $t=5$, and the digit in the units place is $u=6$.

The original two-digit number is $10t + u = 10(5) + 6 = 50 + 6 = 56$.

Check:

1. Sum of digits: $5 + 6 = 11$. (Correct)

2. Number with interchanged digits: $10(6) + 5 = 60 + 5 = 65$.

Difference between interchanged number and original number: $65 - 56 = 9$. (Correct)

Final Answer: The number is 56.

Solution:

Given:

1. A triangle has two equal sides (it is an isosceles triangle).

2. Each of the equal sides is 4 m less than three times the third side.

3. The perimeter of the triangle is 55 m.

To Find:

The lengths of the three sides (dimensions) of the triangle.

Let the length of the third side of the triangle be $x$ metres.

According to the problem, each of the two equal sides is 4 m less than three times the third side.

Three times the third side is $3x$.

4 m less than three times the third side is $3x - 4$.

So, the lengths of the two equal sides are each $(3x - 4)$ metres.

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

Perimeter = Length of third side + Length of first equal side + Length of second equal side

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

We are given that the perimeter is 55 m.

So, we can write the equation:

$x + (3x - 4) + (3x - 4) = 55$

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

$x + 3x - 4 + 3x - 4 = 55$

$(x + 3x + 3x) + (-4 - 4) = 55$

$7x - 8 = 55$

Add 8 to both sides of the equation:

$7x - 8 + 8 = 55 + 8$

$7x = 63$

Divide both sides by 7 to solve for $x$:

$\frac{7x}{7} = \frac{63}{7}$

$x = 9$

Now we can find the lengths of the three sides:

Length of the third side = $x = 9$ metres.

Length of each of the equal sides = $3x - 4 = 3(9) - 4 = 27 - 4 = 23$ metres.

The dimensions of the triangle are 9 m, 23 m, and 23 m.

Check:

Sum of the sides = $9 + 23 + 23 = 9 + 46 = 55$ metres.

This matches the given perimeter of 55 m.

Also, the equal sides (23 m) are 4 m less than three times the third side (3 * 9 = 27 m). $27 - 4 = 23$ m. This condition is also satisfied.

Final Answer: The dimensions of the triangle are 9 m, 23 m, and 23 m.

Solution:

Given:

A condition relating Kanwar's age in the future and his age in the past.

To Find:

Kanwar's present age.

Let Kanwar's present age be $x$ years.

Kanwar's age after 12 years will be $x + 12$ years.

Kanwar's age 4 years ago was $x - 4$ years.

According to the problem statement, after 12 years, Kanwar shall be 3 times as old as he was 4 years ago.

So, we can set up the equation:

Now, we solve this linear equation for $x$. First, distribute the 3 on the right side:

$x + 12 = 3x - 12$

To gather the terms involving $x$ on one side and constant terms on the other, subtract $x$ from both sides:

$x - x + 12 = 3x - x - 12$

$12 = 2x - 12$

Now, add 12 to both sides of the equation:

$12 + 12 = 2x - 12 + 12$

$24 = 2x$

Finally, divide both sides by 2 to find the value of $x$:

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

$12 = x$

So, Kanwar's present age is 12 years.

Check:

Present age = 12 years.

Age after 12 years = $12 + 12 = 24$ years.

Age 4 years ago = $12 - 4 = 8$ years.

Is 24 years equal to 3 times 8 years? $3 \times 8 = 24$. Yes, it is.

The condition is satisfied.

Final Answer: Kanwar's present age is 12 years.

Solution:

Given:

1. Share of property left to daughter = $\frac{1}{2}$ of the total property.

2. Share of property left to son = $\frac{1}{3}$ of the total property.

3. The rest was donated to an educational institute.

4. The value of the donation was $\textsf{₹}$ 1,00,000.

To Find:

The total amount of money Anima had (the total worth of her property).

Let the total amount of money Anima had be $\textsf{₹} x$.

Amount given to her daughter = $\frac{1}{2}$ of $x = \frac{x}{2}$.

Amount given to her son = $\frac{1}{3}$ of $x = \frac{x}{3}$.

The total amount given to her daughter and son is the sum of their shares:

Amount given to daughter and son = $\frac{x}{2} + \frac{x}{3}$

To add these fractions, find a common denominator. The LCM of 2 and 3 is 6.

$\frac{x}{2} + \frac{x}{3} = \frac{3x}{6} + \frac{2x}{6} = \frac{3x + 2x}{6} = \frac{5x}{6}$

So, the total amount given to the daughter and son is $\frac{5}{6}$ of the total property.

The rest of the property was donated to the educational institute.

The rest = Total property - Amount given to daughter and son

The rest = $x - \frac{5x}{6}$

To subtract the fraction, write $x$ as $\frac{6x}{6}$.

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

So, the amount donated to the educational institute is $\frac{1}{6}$ of the total property.

We are given that the value of the donation was $\textsf{₹}$ 1,00,000.

Therefore, we can set up the equation:

To find the total amount $x$, multiply both sides of the equation by 6:

$x = 1,00,000 \times 6$

$x = 6,00,000$

So, Anima had $\textsf{₹}$ 6,00,000.

Check:

Total property = $\textsf{₹}$ 6,00,000.

Amount to daughter = $\frac{1}{2} \times 6,00,000 = \textsf{₹}$ 3,00,000.

Amount to son = $\frac{1}{3} \times 6,00,000 = \textsf{₹}$ 2,00,000.

Total amount given to daughter and son = $\textsf{₹}$ 3,00,000 + $\textsf{₹}$ 2,00,000 = $\textsf{₹}$ 5,00,000.

Amount donated = Total property - Amount given to daughter and son

Amount donated = $\textsf{₹}$ 6,00,000 - $\textsf{₹}$ 5,00,000 = $\textsf{₹}$ 1,00,000.

This matches the given donation amount.

Final Answer: Anima had $\textsf{₹}$ 6,00,000.

Solution:

Given:

A condition where subtracting $\frac{1}{2}$ from a number and then multiplying the result by 4 gives 5.

To Find:

The number that satisfies the given condition.

Let the unknown number be $x$.

According to the problem statement:

"$\frac{1}{2}$ is subtracted from a number": $x - \frac{1}{2}$.

"the difference is multiplied by 4": $4 \times \left(x - \frac{1}{2}\right)$.

"the result is 5": $4\left(x - \frac{1}{2}\right) = 5$.

Now we solve the equation for $x$:

$4\left(x - \frac{1}{2}\right) = 5$

Distribute the 4 on the left side:

$4x - 4 \times \frac{1}{2} = 5$

$4x - 2 = 5$

Add 2 to both sides of the equation:

$4x - 2 + 2 = 5 + 2$

$4x = 7$

Divide both sides by 4 to find the value of $x$:

$\frac{4x}{4} = \frac{7}{4}$

$x = \frac{7}{4}$

The number is $\frac{7}{4}$.

Check:

Subtract $\frac{1}{2}$ from the number: $\frac{7}{4} - \frac{1}{2} = \frac{7}{4} - \frac{2}{4} = \frac{7-2}{4} = \frac{5}{4}$.

Multiply the difference by 4: $4 \times \frac{5}{4} = \frac{\cancel{4} \times 5}{\cancel{4}} = 5$.

The result is 5, which matches the given condition.

Final Answer: The number is $\frac{7}{4}$.

Solution:

Given:

The sum of four consecutive integers is 266.

To Find:

The four consecutive integers.

Let the first integer be $n$.

Since the integers are consecutive, the next three integers will be $n+1$, $n+2$, and $n+3$.

The four consecutive integers are $n$, $n+1$, $n+2$, and $n+3$.

According to the problem, the sum of these four integers is 266.

So, we can write the equation:

$n + (n+1) + (n+2) + (n+3) = 266$

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

$(n+n+n+n) + (1+2+3) = 266$

$4n + 6 = 266$

Subtract 6 from both sides of the equation:

$4n + 6 - 6 = 266 - 6$

$4n = 260$

Divide both sides by 4 to find the value of $n$:

$\frac{4n}{4} = \frac{260}{4}$

$n = 65$

So, the first integer is $n = 65$.

The second integer is $n+1 = 65+1 = 66$.

The third integer is $n+2 = 65+2 = 67$.

The fourth integer is $n+3 = 65+3 = 68$.

The four consecutive integers are 65, 66, 67, and 68.

Check:

Sum of the integers = $65 + 66 + 67 + 68$.

$65 + 66 = 131$

$67 + 68 = 135$

$131 + 135 = 266$.

The sum is 266, which matches the given condition.

Final Answer: The four consecutive integers are 65, 66, 67, and 68.

Solution:

Given:

1. Box A weighs $2\frac{1}{2}$ kg more than Box B.

2. Box C weighs $10\frac{1}{4}$ kg more than Box B.

3. The total weight of the three boxes is $48\frac{3}{4}$ kg.

To Find:

The weight of Box A in kilograms.

Let the weight of Box B be $x$ kg.

According to the problem, the weight of Box A is $2\frac{1}{2}$ kg more than Box B.

Weight of Box A = $x + 2\frac{1}{2}$ kg.

Convert the mixed number to an improper fraction: $2\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2}$.

So, Weight of Box A = $x + \frac{5}{2}$ kg.

The weight of Box C is $10\frac{1}{4}$ kg more than Box B.

Weight of Box C = $x + 10\frac{1}{4}$ kg.

Convert the mixed number to an improper fraction: $10\frac{1}{4} = \frac{10 \times 4 + 1}{4} = \frac{41}{4}$.

So, Weight of Box C = $x + \frac{41}{4}$ kg.

The total weight of the three boxes is $48\frac{3}{4}$ kg.

Convert the total weight to an improper fraction: $48\frac{3}{4} = \frac{48 \times 4 + 3}{4} = \frac{192 + 3}{4} = \frac{195}{4}$.

Total weight = $\frac{195}{4}$ kg.

The sum of the weights of the three boxes is the total weight:

(Weight of Box A) + (Weight of Box B) + (Weight of Box C) = Total weight

$\left(x + \frac{5}{2}\right) + x + \left(x + \frac{41}{4}\right) = \frac{195}{4}$

Combine the terms involving $x$ and the constant terms on the left side:

$(x + x + x) + \left(\frac{5}{2} + \frac{41}{4}\right) = \frac{195}{4}$

$3x + \left(\frac{5}{2} + \frac{41}{4}\right) = \frac{195}{4}$

Find a common denominator for the fractions inside the parentheses. The LCM of 2 and 4 is 4.

$\frac{5}{2} = \frac{5 \times 2}{2 \times 2} = \frac{10}{4}$

$3x + \left(\frac{10}{4} + \frac{41}{4}\right) = \frac{195}{4}$

$3x + \frac{10 + 41}{4} = \frac{195}{4}$

$3x + \frac{51}{4} = \frac{195}{4}$

Subtract $\frac{51}{4}$ from both sides of the equation:

$3x = \frac{195}{4} - \frac{51}{4}$

$3x = \frac{195 - 51}{4}$

$3x = \frac{144}{4}$

$3x = 36$

Divide both sides by 3 to find the value of $x$ (the weight of Box B):

$\frac{3x}{3} = \frac{36}{3}$

$x = 12$

The weight of Box B is 12 kg.

Now we need to find the weight of Box A, which is $x + \frac{5}{2}$.

Weight of Box A = $12 + \frac{5}{2}$

Convert 12 to a fraction with denominator 2: $12 = \frac{12 \times 2}{2} = \frac{24}{2}$.

Weight of Box A = $\frac{24}{2} + \frac{5}{2}$

Weight of Box A = $\frac{24 + 5}{2}$

Weight of Box A = $\frac{29}{2}$ kg.

The weight of Box A can also be expressed as a mixed number: $\frac{29}{2} = 14\frac{1}{2}$ kg.

Check:

Weight of Box B ($x$) = 12 kg.

Weight of Box A ($x + \frac{5}{2}$) = $12 + \frac{5}{2} = \frac{24+5}{2} = \frac{29}{2}$ kg.

Weight of Box C ($x + \frac{41}{4}$) = $12 + \frac{41}{4} = \frac{48+41}{4} = \frac{89}{4}$ kg.

Total weight = $\frac{29}{2} + 12 + \frac{89}{4} = \frac{58}{4} + \frac{48}{4} + \frac{89}{4} = \frac{58+48+89}{4} = \frac{195}{4}$ kg.

$\frac{195}{4} = 48\frac{3}{4}$ kg, which matches the given total weight.

Final Answer: Box A weighs $\frac{29}{2}$ kg (or $14\frac{1}{2}$ kg).

Solution:

Given:

1. Perimeter of the original rectangle = 240 cm.

2. New length is 10% more than the original length.

3. New breadth is 20% less than the original breadth.

4. The perimeter of the new rectangle is the same as the original perimeter (240 cm).

To Find:

The original length and breadth of the rectangle.

Let the original length of the rectangle be $l$ cm and the original breadth be $b$ cm.

The formula for the perimeter of a rectangle is $2 \times (\text{length} + \text{breadth})$.

Using the given perimeter of the original rectangle:

Divide both sides by 2:

Now, consider the changes in length and breadth.

The length is increased by 10%.

New length ($l'$) = $l + 10\% \text{ of } l = l + \frac{10}{100}l = l + 0.10l = 1.10l$ cm.

The breadth is decreased by 20%.

New breadth ($b'$) = $b - 20\% \text{ of } b = b - \frac{20}{100}b = b - 0.20b = 0.80b$ cm.

The perimeter of the new rectangle is the same as the original perimeter, which is 240 cm.

$2(l' + b') = 240$

Substitute the expressions for $l'$ and $b'$:

$2(1.10l + 0.80b) = 240$

Divide both sides by 2:

Now we have a system of two linear equations:

(i) $l + b = 120$

(ii) $1.1l + 0.8b = 120$

From equation (i), we can express $l$ in terms of $b$:

$l = 120 - b$

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

$1.1(120 - b) + 0.8b = 120$

Distribute 1.1:

$1.1 \times 120 - 1.1b + 0.8b = 120$

$132 - 1.1b + 0.8b = 120$

Combine the terms involving $b$:

$132 - 0.3b = 120$

Subtract 132 from both sides:

$-0.3b = 120 - 132$

$-0.3b = -12$

Divide both sides by -0.3:

$b = \frac{-12}{-0.3}$

$b = \frac{12}{0.3} = \frac{120}{3}$

$b = 40$

The original breadth is 40 cm.

Now substitute the value of $b = 40$ into equation (i) to find $l$:

$l + b = 120$

$l + 40 = 120$

$l = 120 - 40$

$l = 80$

The original length is 80 cm.

Check:

Original perimeter = $2(80 + 40) = 2(120) = 240$ cm. (Correct)

New length ($l'$) = $1.1 \times 80 = 88$ cm.

New breadth ($b'$) = $0.8 \times 40 = 32$ cm.

New perimeter = $2(l' + b') = 2(88 + 32) = 2(120) = 240$ cm. (Correct)

The new perimeter is the same as the original perimeter.

Final Answer: The original length of the rectangle is 80 cm and the original breadth is 40 cm.

Solution:

Given:

1. The age of A is 5 years more than the age of B.

2. 5 years ago, the ratio of A's age to B's age was 3 : 2.

To Find:

The present ages of A and B.

Let the present age of B be $x$ years.

According to the first condition, the present age of A is 5 years more than that of B.

Present age of A = $(x + 5)$ years.

Now consider their ages 5 years ago:

Age of B 5 years ago = Present age of B - 5 = $(x - 5)$ years.

Age of A 5 years ago = Present age of A - 5 = $(x + 5) - 5 = x$ years.

According to the second condition, 5 years ago, the ratio of A's age to B's age was 3 : 2.

Ratio of (Age of A 5 years ago) to (Age of B 5 years ago) = 3 : 2

$\frac{\text{Age of A 5 years ago}}{\text{Age of B 5 years ago}} = \frac{3}{2}$

Substitute the expressions for their ages 5 years ago:

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

To solve for $x$, cross-multiply the equation:

$2 \times x = 3 \times (x - 5)$

$2x = 3x - 15$

To isolate $x$, subtract $2x$ from both sides and add 15 to both sides:

$15 = 3x - 2x$

$15 = x$

So, the present age of B is $x = 15$ years.

The present age of A is $x + 5 = 15 + 5 = 20$ years.

Check:

Present age of A = 20 years.

Present age of B = 15 years.

Is A's age 5 years more than B's age? $15 + 5 = 20$. Yes.

A's age 5 years ago = $20 - 5 = 15$ years.

B's age 5 years ago = $15 - 5 = 10$ years.

Ratio of their ages 5 years ago = $\frac{15}{10} = \frac{3 \times 5}{2 \times 5} = \frac{3}{2}$. Yes.

Both conditions are satisfied.

Final Answer: The present age of A is 20 years and the present age of B is 15 years.

Solution:

Given:

1. In a rational number, the numerator is 2 less than the denominator.

2. When 1 is subtracted from both the numerator and the denominator, the resulting fraction simplifies to $\frac{1}{2}$.

To Find:

The original rational number.

Let the denominator of the rational number be $x$.

According to the first condition, the numerator is 2 less than the denominator.

Numerator = $x - 2$.

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

According to the second condition, when 1 is subtracted from both the numerator and the denominator, the resulting fraction is $\frac{1}{2}$.

New numerator = (Original numerator) - 1 = $(x - 2) - 1 = x - 3$.

New denominator = (Original denominator) - 1 = $x - 1$.

The new fraction is $\frac{x-3}{x-1}$.

We are given that this new fraction in its simplest form is equal to $\frac{1}{2}$.

So, we can set up the equation:

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

To solve for $x$, we cross-multiply:

$2 \times (x - 3) = 1 \times (x - 1)$

$2x - 6 = x - 1$

Now, we rearrange the terms to solve for $x$. Subtract $x$ from both sides:

$2x - x - 6 = x - x - 1$

$x - 6 = -1$

Add 6 to both sides:

$x - 6 + 6 = -1 + 6$

$x = 5$

The denominator of the original rational number is $x = 5$.

The numerator of the original rational number is $x - 2 = 5 - 2 = 3$.

The original rational number is $\frac{\text{Numerator}}{\text{Denominator}} = \frac{3}{5}$.

Check:

1. Is the numerator (3) 2 less than the denominator (5)? $5 - 2 = 3$. Yes, the first condition is met.

2. Subtract 1 from the numerator and denominator: $\frac{3 - 1}{5 - 1} = \frac{2}{4}$.

Simplify the resulting fraction: $\frac{2}{4} = \frac{\cancel{2}^{1}}{\cancel{4}_{2}} = \frac{1}{2}$. Yes, the second condition is met.

Final Answer: The rational number is $\frac{3}{5}$.

Solution:

Given:

1. In a two-digit number, the digit in the units place is twice the digit in the tens place.

2. If 27 is added to the number, the digits are reversed.

To Find:

The two-digit number.

Let the digit in the tens place be $t$ and the digit in the units place be $u$.

The value of the original two-digit number is $10 \times t + u = 10t + u$.

The value of the number obtained by reversing the digits is $10 \times u + t = 10u + t$.

According to the first condition, the digit in the units place ($u$) is twice the digit in the tens place ($t$).

According to the second condition, if 27 is added to the original number, the digits are reversed.

(Original number) + 27 = (Number with reversed digits)

Now, substitute the value of $u$ from equation (i) into this equation:

$10t + (2t) + 27 = 10(2t) + t$

$12t + 27 = 20t + t$

$12t + 27 = 21t$

Subtract $12t$ from both sides of the equation:

$27 = 21t - 12t$

$27 = 9t$

Divide both sides by 9 to find the value of $t$:

$\frac{27}{9} = t$

$t = 3$

Now substitute the value of $t=3$ back into equation (i) to find the value of $u$:

$u = 2t$

$u = 2 \times 3$

$u = 6$

The digit in the tens place is 3, and the digit in the units place is 6.

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

Check:

1. Is the units digit (6) twice the tens digit (3)? $2 \times 3 = 6$. Yes, the first condition is met.

2. Original number = 36. Number with reversed digits = 63.

Is $36 + 27 = 63$? $36 + 27 = 63$. Yes, the second condition is met.

Final Answer: The number is 36.

Solution:

Given: Daily wage = $\textsf{₹}$ 500, Deduction per day absent = $\textsf{₹}$ 100, Total salary = $\textsf{₹}$ 9,100, Month = February 2009 (28 days).

To Find: Number of days worked.

Let the number of days the typist worked be $w$.

Number of days absent = $28 - w$.

Total salary = (Earnings from working days) - (Deductions from absent days)

$9100 = 500w - 100(28 - w)$

Solve for $w$:

$9100 = 500w - 2800 + 100w$

$9100 = 600w - 2800$

$9100 + 2800 = 600w$

$11900 = 600w$

$w = \frac{11900}{600}$

$w = \frac{119}{6}$

Based on the given numbers, the number of days worked is $\frac{119}{6}$. Note that this is not an integer, suggesting the problem statement might contain a typo in the numbers.

Final Answer: The typist worked for $\frac{119}{6}$ days.

Given:

Time taken by the steamer to go downstream ($t_{down}$) = 3 hours

Time taken by the steamer to go upstream ($t_{up}$) = 5 hours

Speed of the stream ($v_c$) = 3 km/hr

The distance between the two ports is the same for both journeys.

To Find:

The speed of the steamer upstream.

Solution:

Let the distance between the two ports be $D$ km.

Let the speed of the steamer in still water be $v_s$ km/hr.

The speed of the stream is given as $v_c = 3$ km/hr.

When the steamer goes downstream, its speed is the sum of its speed in still water and the speed of the stream.

Speed downstream $= v_s + v_c = v_s + 3$ km/hr.

The distance covered downstream is given by Speed $\times$ Time.

When the steamer goes upstream, its speed is the difference between its speed in still water and the speed of the stream (assuming $v_s > v_c$).

Speed upstream $= v_s - v_c = v_s - 3$ km/hr.

The distance covered upstream is given by Speed $\times$ Time.

Since the distance $D$ is the same in both cases, we can equate the expressions for $D$ from equations (i) and (ii).

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

$3v_s + 9 = 5v_s - 15$

Subtract $3v_s$ from both sides:

$9 = 2v_s - 15$

Add 15 to both sides:

$24 = 2v_s$

Divide by 2:

$v_s = \frac{24}{2}$

$v_s = 12$ km/hr

So, the speed of the steamer in still water is 12 km/hr.

The question asks for the speed of the steamer upstream.

Speed upstream $= v_s - v_c$

Speed upstream $= 12 - 3$

Speed upstream $= 9$ km/hr

The speed of the steamer upstream is 9 km/hr.

Given:

Total amount of money = $\textsf{₹} 1,00,000$

Denominations of currency notes = $\textsf{₹} 500$ and $\textsf{₹} 1,000$

Total number of currency notes received = 175

To Find:

The number of $\textsf{₹} 500$ currency notes and the number of $\textsf{₹} 1,000$ currency notes.

Solution:

Let $x$ be the number of $\textsf{₹} 500$ currency notes.

Let $y$ be the number of $\textsf{₹} 1,000$ currency notes.

According to the problem, the total number of notes is 175.

The total value of the $\textsf{₹} 500$ notes is $500x$.

The total value of the $\textsf{₹} 1,000$ notes is $1000y$.

The total value of the money is $\textsf{₹} 1,00,000$.

We have a system of two linear equations:

$x + y = 175$

$500x + 1000y = 100000$

Let's simplify equation (2) by dividing by 500:

Now we solve the system of equations (1) and (3):

$x + y = 175$

$x + 2y = 200$

Subtract equation (1) from equation (3):

$x + 2y - x - y = 25$

$y = 25$

So, the number of $\textsf{₹} 1,000$ notes is 25.

Substitute the value of $y$ into equation (1) to find $x$:

Subtract 25 from both sides:

$x = 150$

So, the number of $\textsf{₹} 500$ notes is 150.

Check the answer:

Total number of notes = $150 + 25 = 175$ (Correct)

Total value = $(150 \times 500) + (25 \times 1000) = 75000 + 25000 = 100000$ (Correct)

The number of $\textsf{₹} 500$ currency notes is 150 and the number of $\textsf{₹} 1,000$ currency notes is 25.

Given:

Total number of passengers = 40

Cost of one type of ticket = $\textsf{₹} 3$

Cost of the other type of ticket = $\textsf{₹} 10$

Total collection from passengers = $\textsf{₹} 295$

To Find:

The number of passengers with $\textsf{₹} 3$ tickets.

Solution:

Let $x$ be the number of passengers with $\textsf{₹} 3$ tickets.

Let $y$ be the number of passengers with $\textsf{₹} 10$ tickets.

According to the problem, the total number of passengers is 40.

The total collection from passengers is $\textsf{₹} 295$.

The value collected from $\textsf{₹} 3$ tickets is $3x$.

The value collected from $\textsf{₹} 10$ tickets is $10y$.

We have a system of two linear equations:

$x + y = 40$

$3x + 10y = 295$

From equation (1), we can express $y$ in terms of $x$:

Substitute this expression for $y$ into equation (2):

Now, solve for $x$:

$3x + 400 - 10x = 295$

$3x - 10x + 400 = 295$

$-7x + 400 = 295$

Subtract 400 from both sides:

$-7x = -105$

Divide by -7:

$x = 15$

So, the number of passengers with $\textsf{₹} 3$ tickets is 15.

We can find the number of passengers with $\textsf{₹} 10$ tickets using equation (3):

$y = 40 - x = 40 - 15 = 25$

Number of $\textsf{₹} 10$ tickets is 25.

Check the answer:

Total passengers = $15 + 25 = 40$ (Correct)

Total collection = $(15 \times 3) + (25 \times 10) = 45 + 250 = 295$ (Correct)

The number of passengers who have tickets worth $\textsf{₹} 3$ is 15.

Given:

A fraction whose denominator is 4 less than its numerator.

When 6 is added to the numerator, the new numerator is thrice the original denominator.

To Find:

The fraction.

Solution:

Let the numerator of the fraction be $n$.

Let the denominator of the fraction be $d$.

The fraction is $\frac{n}{d}$.

According to the first condition, the denominator is 4 less than the numerator:

According to the second condition, if 6 is added to the numerator, the new numerator ($n+6$) is thrice the denominator ($3d$):

Now we have a system of two linear equations:

$d = n - 4$

$n + 6 = 3d$

We can substitute the expression for $d$ from equation (1) into equation (2):

Now, solve the equation for $n$:

$n + 6 = 3n - 12$

Subtract $n$ from both sides:

$6 = 2n - 12$

Add 12 to both sides:

$18 = 2n$

Divide by 2:

$n = 9$

So, the numerator is 9.

Now, substitute the value of $n$ into equation (1) to find $d$:

$d = 5$

So, the denominator is 5.

The fraction is $\frac{n}{d} = \frac{9}{5}$.

Check the conditions:

Is the denominator 4 less than the numerator? $5 = 9 - 4$ (Yes)

If 6 is added to the numerator (9+6=15), is it thrice the denominator (5)? $15 = 3 \times 5$ (Yes)

Both conditions are satisfied.

The fraction is $\frac{9}{5}$.

Given:

Total contract duration = 30 days

Earnings per day worked = $\textsf{₹} 120$

Fine per day absent = $\textsf{₹} 10$

Total amount received by the employee = $\textsf{₹} 2300$

To Find:

The number of days the employee remained absent.

Solution:

Let $x$ be the number of days the employee worked.

Let $y$ be the number of days the employee was absent.

The total number of days in the contract is 30.

For each day worked, the employee earns $\textsf{₹} 120$. So, total earnings from working days $= 120x$.

For each day absent, the employee is fined $\textsf{₹} 10$. So, total fine from absent days $= 10y$.

The total amount received by the employee is the total earnings minus the total fine.

We have a system of two linear equations:

$x + y = 30$

$120x - 10y = 2300$

From equation (1), we can express $x$ in terms of $y$ (since we need to find $y$, the number of absent days):

Substitute this expression for $x$ into equation (2):

Now, solve the equation for $y$:

$3600 - 120y - 10y = 2300$

$3600 - 130y = 2300$

Subtract 3600 from both sides:

$-130y = -1300$

Divide by -130:

$y = 10$

So, the number of days the employee was absent is 10.

We can find the number of days worked using equation (3):

$x = 30 - y = 30 - 10 = 20$

Number of days worked is 20.

Check the answer:

Total days = $20 \text{ (worked)} + 10 \text{ (absent)} = 30$ (Correct)

Total received = $(20 \text{ days} \times \textsf{₹} 120/\text{day}) - (10 \text{ days} \times \textsf{₹} 10/\text{day}) = \textsf{₹} 2400 - \textsf{₹} 100 = \textsf{₹} 2300$ (Correct)

The employee remained absent for 10 days.

Given:

Cost price of 1 chocolate = $\textsf{₹} 10$

Cost price of 1 candy = $\textsf{₹} 5$

Number of chocolates purchased = Number of candies purchased

Profit percentage on chocolates = 20%

Profit percentage on candies = 8%

Total profit = $\textsf{₹} 240$

To Find:

The number of chocolates purchased.

Solution:

Let the number of chocolates purchased be $n$.

Since an equal number of candies were purchased, the number of candies purchased is also $n$.

Calculate the cost price for the total number of chocolates and candies:

Total cost price of chocolates $= \text{Number of chocolates} \times \text{Cost per chocolate}$

Total cost price of chocolates $= n \times \textsf{₹} 10 = \textsf{₹} 10n$

Total cost price of candies $= \text{Number of candies} \times \text{Cost per candy}$

Total cost price of candies $= n \times \textsf{₹} 5 = \textsf{₹} 5n$

Calculate the profit earned on chocolates and candies:

Profit on chocolates $= 20\%$ of Total cost price of chocolates

Profit on chocolates $= 20\% \times 10n = \frac{20}{100} \times 10n = 0.20 \times 10n = 2n$

Profit on candies $= 8\%$ of Total cost price of candies

Profit on candies $= 8\% \times 5n = \frac{8}{100} \times 5n = 0.08 \times 5n = 0.4n$

The total profit is the sum of the profit on chocolates and the profit on candies.

We are given that the total profit is $\textsf{₹} 240$.

Now, we solve the equation for $n$:

$2.4n = 240$

Divide both sides by 2.4:

To simplify the division, multiply the numerator and the denominator by 10:

$n = \frac{240 \times 10}{2.4 \times 10} = \frac{2400}{24}$

$n = 100$

The number of chocolates purchased is 100.

The number of candies purchased is also 100.

We can verify the result:

Cost of 100 chocolates = $100 \times \textsf{₹} 10 = \textsf{₹} 1000$

Profit on chocolates = 20% of $\textsf{₹} 1000 = \frac{20}{100} \times 1000 = \textsf{₹} 200$

Cost of 100 candies = $100 \times \textsf{₹} 5 = \textsf{₹} 500$

Profit on candies = 8% of $\textsf{₹} 500 = \frac{8}{100} \times 500 = \textsf{₹} 40$

Total profit = $\textsf{₹} 200 + \textsf{₹} 40 = \textsf{₹} 240$ (This matches the given total profit).

The number of chocolates purchased is 100.

Given:

Time taken by the steamer to go downstream ($t_{down}$) = 5 hours

Time taken by the steamer to go upstream ($t_{up}$) = 6 hours

Speed of the stream ($v_c$) = 1 km/hr

The distance between the two ports is the same for both journeys.

To Find:

The speed of the steamer in still water ($v_s$).

Solution:

Let the distance between the two ports be $D$ km.

Let the speed of the steamer in still water be $v_s$ km/hr.

The speed of the stream is given as $v_c = 1$ km/hr.

The speed of the steamer when going downstream is the sum of its speed in still water and the speed of the stream.

Speed downstream $= v_s + v_c = v_s + 1$ km/hr.

The distance covered downstream is given by Speed $\times$ Time.

The speed of the steamer when going upstream is the difference between its speed in still water and the speed of the stream (assuming the steamer's speed in still water is greater than the stream's speed, $v_s > v_c$).

Speed upstream $= v_s - v_c = v_s - 1$ km/hr.

The distance covered upstream is given by Speed $\times$ Time.

Since the distance $D$ is the same in both cases, we can equate the expressions for $D$ from equations (i) and (ii).

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

$5v_s + 5 = 6v_s - 6$

Rearrange the terms to group $v_s$ on one side and constants on the other side:

$11 = v_s$

$v_s = 11$ km/hr

So, the speed of the steamer in still water is 11 km/hr.

We assumed $v_s > v_c$, which means $11 > 1$. This is true.

The speed of the steamer in still water is 11 km/hr.

Given:

Initial distance between places A and B = 210 km

Time of travel for both cars = 3 hours

Distance between the cars after 3 hours = 54 km

Speed of one car is less than the speed of the other car by 8 km/hr.

To Find:

The speed of each car.

Solution:

Let the speed of the faster car be $v_1$ km/hr.

Let the speed of the slower car be $v_2$ km/hr.

According to the problem, the speed of one car is 8 km/hr less than the other. Assuming $v_1$ is the speed of the faster car:

The cars are moving in opposite directions (towards each other). Their relative speed is the sum of their individual speeds.

Relative speed = $v_1 + v_2$ km/hr.

The cars travel for 3 hours. In this time, the total distance they have covered together is the initial distance minus the remaining distance between them.

Initial distance = 210 km

Distance after 3 hours = 54 km

Combined distance covered in 3 hours = Initial distance - Distance after 3 hours

Combined distance covered = $210 \text{ km} - 54 \text{ km} = 156$ km.

The combined distance covered is also equal to the relative speed multiplied by the time.

Divide both sides of equation (2) by 3:

Now we have a system of two linear equations:

Equation (1): $v_1 - v_2 = 8$

Equation (3): $v_1 + v_2 = 52$

We can solve this system by adding the two equations:

$v_1 - v_2 + v_1 + v_2 = 60$

$2v_1 = 60$

Divide by 2 to find $v_1$:

$v_1 = 30$ km/hr

Now substitute the value of $v_1 = 30$ into equation (1) to find $v_2$:

Subtract 30 from both sides:

$-v_2 = 8 - 30$

$-v_2 = -22$

Multiply by -1:

$v_2 = 22$ km/hr

So, the speed of the faster car is 30 km/hr and the speed of the slower car is 22 km/hr.

Check:

Speed difference: $30 - 22 = 8$ km/hr (Correct)

Distance covered by faster car in 3 hours = $30 \times 3 = 90$ km

Distance covered by slower car in 3 hours = $22 \times 3 = 66$ km

Total distance covered = $90 + 66 = 156$ km

Remaining distance = $210 - 156 = 54$ km (Correct)

The speed of one car is 30 km/hr and the speed of the other car is 22 km/hr.

Given:

Total charge for making the bed = $\textsf{₹} 2500$

Cost of materials used = $\textsf{₹} 1100$

Labour charges per hour = $\textsf{₹} 200$

To Find:

The number of hours the carpenter worked.

Solution:

The total charge for making the bed consists of two parts: the cost of materials and the labour charges.

Let the number of hours the carpenter worked be $h$ hours.

The labour charges are calculated by multiplying the labour rate per hour by the number of hours worked.

Labour charges = Labour rate per hour $\times$ Number of hours worked

Labour charges = $\textsf{₹} 200/hr \times h \text{ hours} = \textsf{₹} 200h$

Substitute the given values into the total charge equation:

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

Subtract the cost of materials from the total charge to find the total labour charges:

$1400 = 200h$

Divide both sides by 200 to find the number of hours worked:

Simplify the fraction:

$h = \frac{14}{2}$

$h = 7$

The number of hours the carpenter worked is 7 hours.

Check the answer:

Cost of materials = $\textsf{₹} 1100$

Labour charges = 7 hours $\times$ $\textsf{₹} 200$/hr = $\textsf{₹} 1400$

Total charge = $\textsf{₹} 1100 + \textsf{₹} 1400 = \textsf{₹} 2500$ (This matches the given total charge).

The carpenter worked for 7 hours.

Given:

The perimeter of the given shape is 77 cm.

The side lengths of the shape, as observed from the image, are:

$x+5$ cm

$x+3$ cm

$3x-2$ cm

$x-1$ cm

6 cm

$2x$ cm

To Find:

The value of $x$.

Solution:

The perimeter of a polygon is the sum of the lengths of all its sides.

Perimeter = Sum of all side lengths

Substitute the given side lengths into the perimeter equation:

We are given that the perimeter is 77 cm. So, we set the sum of the side lengths equal to 77:

Now, simplify the equation by combining like terms (terms with $x$ and constant terms):

Combine the $x$ terms:

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

Combine the constant terms:

$5 + 3 - 2 - 1 + 6 = 8 - 2 - 1 + 6 = 6 - 1 + 6 = 5 + 6 = 11$

So, the equation simplifies to:

Now, solve the linear equation for $x$. Subtract 11 from both sides of the equation:

$8x = 66$

Divide both sides by 8:

Simplify the fraction:

$x = \frac{\cancel{66}^{33}}{\cancel{8}_{4}}$

$x = \frac{33}{4}$

We can also express this as a decimal:

$x = 8.25$

So, the value of $x$ is $\frac{33}{4}$ cm or 8.25 cm.

The value of $x$ for which the perimeter of the shape is 77 cm is $\frac{33}{4}$ cm or 8.25 cm.

Given:

The perimeter of the given shape is 186 cm.

The side lengths of the shape, as observed from the image, are:

$2x-5$ cm

$x+5$ cm

$3x$ cm

$2x+1$ cm

$3x-2$ cm

$x$ cm

To Find:

The value of $x$.

Solution:

The perimeter of a polygon is the sum of the lengths of all its sides.

Perimeter = Sum of all side lengths

Substitute the given side lengths into the perimeter equation:

We are given that the perimeter is 186 cm. So, we set the sum of the side lengths equal to 186:

Now, simplify the equation by combining like terms (terms with $x$ and constant terms):

Combine the $x$ terms:

$2x + x + 3x + 2x + 3x + x = (2+1+3+2+3+1)x = 12x$

Combine the constant terms:

$-5 + 5 + 1 - 2 = 0 + 1 - 2 = -1$

So, the equation simplifies to:

Now, solve the linear equation for $x$. Add 1 to both sides of the equation:

$12x = 187$

Divide both sides by 12:

The value of $x$ is $\frac{187}{12}$ cm.

The value of $x$ for which the perimeter of the shape is 186 cm is $\frac{187}{12}$ cm.

Given:

Total amount to be divided = $\textsf{₹} 200$

The amount is divided between A and B.

Condition: Twice of A's share is $\textsf{₹} 200$ less than 3 times B's share.

To Find:

B's share.

Solution:

Let A's share be $\textsf{₹} a$.

Let B's share be $\textsf{₹} b$.

According to the problem, the total amount is $\textsf{₹} 200$. So, the sum of A's share and B's share is $\textsf{₹} 200$.

According to the second condition, twice of A's share ($2a$) is $\textsf{₹} 200$ less than 3 times B's share ($3b$).

This means that $2a$ is obtained by subtracting 200 from $3b$.

We have a system of two linear equations:

$a + b = 200$

$2a = 3b - 200$

From equation (1), we can express $a$ in terms of $b$:

Substitute this expression for $a$ from equation (3) into equation (2):

Now, solve this equation for $b$:

Distribute the 2 on the left side:

$400 - 2b = 3b - 200$

Add $2b$ to both sides of the equation:

$400 = 5b - 200$

Add 200 to both sides of the equation:

$600 = 5b$

Divide both sides by 5:

$b = 120$

So, B's share is $\textsf{₹} 120$.

We can also find A's share using equation (3):

A's share is $\textsf{₹} 80$.

Check the condition: Is twice A's share less than 3 times B's share by $\textsf{₹} 200$?

Twice A's share = $2 \times \textsf{₹} 80 = \textsf{₹} 160$

Three times B's share = $3 \times \textsf{₹} 120 = \textsf{₹} 360$

The difference is $\textsf{₹} 360 - \textsf{₹} 160 = \textsf{₹} 200$. The condition is satisfied.

B's share is $\textsf{₹} 120$.

Given:

A process involving a number: doubled, added 20, and then divided by 25.

The final result after the process is 4.

To Find:

The original number Madhulika thought of.

Solution:

Let the number Madhulika thought of be $x$.

According to the problem, the steps taken are:

1. The number is doubled: $2x$

2. 20 is added to the result: $2x + 20$

3. The resulting number is divided by 25: $\frac{2x + 20}{25}$

The final result after these steps is 4. So, we can form the equation:

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

Multiply both sides of the equation by 25 to eliminate the denominator:

$2x + 20 = 100$

Subtract 20 from both sides of the equation to isolate the term with $x$:

$2x = 80$

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

$x = 40$

So, the number Madhulika thought of is 40.

Check the steps with $x=40$:

1. Doubled: $2 \times 40 = 80$

2. Added 20: $80 + 20 = 100$

3. Divided by 25: $\frac{100}{25} = 4$

The final result is 4, which matches the given information.

The number Madhulika thought of is 40.