Chapter 4 Linear Equation in One Variable (Class 8 - Maths NCERT Exemplar Solutions)

Solution:

According to the properties of equality, if we perform the same mathematical operation on both sides of an equation, the equality holds true. However, there are restrictions for certain operations.

1. Addition/Subtraction: Adding or subtracting an integer $k$ to both sides ($x + k = a + k$ or $x - k = a - k$) is always valid. (Options c and d are true).

2. Multiplication: Multiplying both sides by any integer $k$ ($kx = ak$) is always valid. (Option a is true).

3. Division: Dividing both sides by an integer $k$ ($\frac{x}{k} = \frac{a}{k}$) is only true if $k \neq 0$. Since $k$ is any integer, it could potentially be zero. Division by zero is undefined in mathematics.

Therefore, option (b) is not always true.

To Find: The value of $x$ for the given linear equation.

Solution:

The given equation is:

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

First, expand the brackets by multiplying $-4$ with both terms inside:

$3x - 256 + 4x = 10$

Group the terms containing $x$:

$(3x + 4x) - 256 = 10$

$7x - 256 = 10$

Transpose $-256$ to the right side:

$7x = 10 + 256$

$7x = 266$

Divide both sides by $7$:

$x = \frac{266}{7}$

$x = 38$

The value of $x$ is $38$. Thus, the correct option is (d).

Solution:

Let the whole number be $n$.

According to the statement, "Fifteen added to thrice a whole number gives 93":

$3n + 15 = 93$

Transpose $15$ to the RHS:

$3n = 93 - 15$

$3n = 78$

Divide by $3$:

$n = \frac{78}{3} = 26$

Answer: The number is 26.

Solution:

Given equation:

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

Transpose $-x$ to the RHS and $-\frac{2}{3}$ to the LHS:

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

Since denominators are same, add the numerators:

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

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

$x = 1$

Answer: $x$ is 1.

Solution:

By definition, an even number must be divisible by 2. Let's check the given numbers: $51, 52,$ and $53$.

• $51$ and $53$ are odd numbers.
• $52$ is an even number.

Since the question specifies three consecutive even numbers, and two of the numbers in the list are odd, the statement is incorrect regardless of their sum.

Answer: False (F)

Solution:

To verify if $x = -12$ is the solution, we substitute it into the LHS and RHS of the equation.

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

$= 5(-12) - 3(2(-12) + 1)$

$= -60 - 3(-24 + 1)$

$= -60 - 3(-23)$

$= -60 + 69 = 9$

RHS: $21 + x$

$= 21 + (-12) = 9$

Since LHS = RHS, $x = -12$ is indeed the correct solution.

Answer: True (T)

To Find:

The value of $x$ for the given equation.

Solution:

The given equation is:

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

First, we transpose all the terms containing $x$ to one side:

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

To simplify the left-hand side, we find the Least Common Multiple (LCM) of the denominators 2, 4, and 5, which is $20$.

$\frac{20x - 10x - 5x - 4x}{20} = 10000$

Now, perform the subtraction in the numerator:

$\frac{20x - 19x}{20} = 10000$

$\frac{x}{20} = 10000$

Multiply both sides by $20$ to isolate $x$:

$x = 10000 \times 20$

$x = 200000$

Therefore, the solution to the equation is $x = 200000$.

Given:

1. The father's present age is four times the son's present age.

2. 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 $x$ years.

Then, the present age of the father is $4x$ years.

After 10 years:

• Son's age will be $(x + 10)$ years.
• Father's age will be $(4x + 10)$ years.

According to the given condition:

$4x + 10 = 3(x + 10)$

Expanding the brackets:

$4x + 10 = 3x + 30$

Transposing $3x$ to the LHS and $10$ to the RHS:

$4x - 3x = 30 - 10$

$x = 20$

Now, we calculate their present ages:

The present age of the son is 20 years and the father is 80 years.

Given:

Downstream time $= 7 \text{ hours}$, Upstream time $= 8 \text{ hours}$, and Speed of the stream $= 2 \text{ km/hr}$.

To Find:

1. Speed of the steamer in still water.

2. The distance between the ports.

Solution:

Let the speed of the steamer in still water be $x \text{ km/hr}$.

Then:

• Speed downstream $= (x + 2) \text{ km/hr}$
• Speed upstream $= (x - 2) \text{ km/hr}$

We know that $\text{Distance} = \text{Speed} \times \text{Time}$. Since the distance between the ports remains constant:

$7(x + 2) = 8(x - 2)$

$7x + 14 = 8x - 16$

Transpose $7x$ to the RHS and $-16$ to the LHS:

$14 + 16 = 8x - 7x$

$30 = x$

The speed of the steamer in still water is $30 \text{ km/hr}$.

Now, to find the distance:

$\text{Distance} = 7(x + 2)$

$\text{Distance} = 7(30 + 2)$

$\text{Distance} = 7 \times 32 = 224 \text{ km}$

The speed of the steamer in still water is $30 \text{ km/hr}$ and the distance between the ports is $224 \text{ km}$.

Given:

1. Total distance between A and B $= 690 \text{ km}$.

2. Time taken $= 6 \text{ hours}$.

3. Distance remaining between cars after 6 hours $= 30 \text{ km}$.

4. One car's speed is $10 \text{ km/hr}$ less than the other.

To Find:

The speed of each car.

Solution:

Let the speed of the faster car be $x \text{ km/hr}$.

Then, the speed of the slower car $= (x - 10) \text{ km/hr}$.

Since the cars are moving towards each other, the total distance they covered together in 6 hours is:

Distance covered by faster car in 6 hours $= 6x$

Distance covered by slower car in 6 hours $= 6(x - 10)$

Now, set up the equation:

$6x + 6(x - 10) = 660$

$6x + 6x - 60 = 660$

$12x - 60 = 660$

$12x = 660 + 60$

$12x = 720$

$x = \frac{720}{12} = 60 \text{ km/hr}$

Speeds of the two cars:

• Speed of the faster car $= x = 60 \text{ km/hr}$
• Speed of the slower car $= x - 10 = 50 \text{ km/hr}$

The speeds of the two cars are $60 \text{ km/hr}$ and $50 \text{ km/hr}$.

Given:

A square garden with a perimeter of $6480 \text{ cm}$.

To Find:

The dimensions (side length) of the square garden.

Solution:

Let the length of one side of the square garden be $s \text{ cm}$.

The formula for the perimeter of a square is:

According to the given information, we can write the following equation:

$4s = 6480$

To find $s$, we divide both sides by $4$:

$s = \frac{6480}{4}$

$s = 1620 \text{ cm}$

The side length of the square garden is $1620 \text{ cm}$.

To Find: The equation whose solution is neither an integer nor a fraction.

Solution:

In mathematics, especially in the school curriculum context, an integer is a whole number (positive, negative, or zero), and a fraction is typically defined as a positive rational number ($a/b$ where $a, b > 0$). Let's solve each option:

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

$3x - 5x = 2 - 2 \implies -2x = 0 \implies x = 0$ (Integer)

(b) $4x - 18 = 2$

$4x = 20 \implies x = 5$ (Integer)

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

$4x - x = 2 - 7 \implies 3x = -5 \implies x = -\frac{5}{3}$ (Negative rational number, neither a positive fraction nor an integer)

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

$5x - x = 4 + 8 \implies 4x = 12 \implies x = 3$ (Integer)

The correct option is (c).

To Find: The solution of the linear equation $ax + b = 0$.

Solution:

Given equation:

$ax + b = 0$

To find the value of $x$, we transpose $+b$ to the right-hand side:

$ax = -b$

Now, dividing both sides by $a$:

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

The correct option is (c).

To Find: The nature of $x$ in the equation $8x - 3 = 25 + 17x$.

Solution:

Given:

$8x - 3 = 25 + 17x$

Transpose $17x$ to the LHS and $-3$ to the RHS:

$8x - 17x = 25 + 3$

$-9x = 28$

$x = -\frac{28}{9}$

Since the result is in the form $p/q$ where $q \neq 0$, it is a rational number.

The correct option is (c).

Solution:

In algebra, the process of moving a term or a number from one side of an equation to the other side with a change in its sign is called Transposition.

The correct option is (a).

To Find: The numerical value of $2x - 7$ given $\frac{5x}{3} - 4 = \frac{2x}{5}$.

Solution:

First, solve for $x$:

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

Finding the LCM of 3 and 5, which is 15:

$\frac{25x - 6x}{15} = 4$

$\frac{19x}{15} = 4$

$19x = 60 \implies x = \frac{60}{19}$

Now, substitute the value of $x$ into $2x - 7$:

$2 \left( \frac{60}{19} \right) - 7 = \frac{120}{19} - 7$

$= \frac{120 - 133}{19}$

$= -\frac{13}{19}$

The correct option is (b).

To Find: The value of $x$ for which $3x - 4 = 2x + 1$.

Solution:

Set the expressions equal to each other:

$3x - 4 = 2x + 1$

Transpose $2x$ to the LHS and $-4$ to the RHS:

$3x - 2x = 1 + 4$

$x = 5$

The correct option is (c).

Solution:

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

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

Since $a > 0$ and $b > 0$, the quotient of two positive numbers is always positive.

The correct option is (a).

Solution:

A linear equation in one variable is an equation that can be written in the form $ax + b = 0$, where $a \neq 0$. The defining characteristic is that it contains only one variable and the highest power (degree) of that variable is 1.

The correct option is (c).

Solution:

A linear expression is an algebraic expression in which the highest power (degree) of the variable is exactly 1. Let us examine the options:

(a) $x^2 + 1$: The highest power of $x$ is 2, so it is a quadratic expression.

(b) $y + y^2$: The highest power of $y$ is 2, so it is a quadratic expression.

(c) $4$: This is a constant expression (degree 0).

(d) $1 + z$: The power of the variable $z$ is 1. Therefore, it is a linear expression.

The correct option is (d).

Solution:

A linear equation in one variable can be written in the standard form $ax + b = 0$, where $a \neq 0$. Solving for $x$ gives $x = -b/a$. Since there is only one possible value for $x$, the equation has exactly one solution.

The correct option is (a).

Given:

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

To Find:

The value of $S$.

Solution:

To find $S$, we transpose $\frac{1}{3}$ to the right-hand side:

To subtract the fractions, find the LCM of 5 and 3, which is 15:

The correct option is (b).

Given:

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

To Find:

The value of $y$.

Solution:

First, cancel the negative signs from both sides:

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

The correct option is (c).

Given:

1. Digit at units place $= b$.

2. Digit at tens place $= b + 3$.

Solution:

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

Substituting the given values:

The correct option is (a).

Given:

1. Shilpa's age three years ago $= x$.

2. Arpita's present age $= 3 \times (\text{Shilpa's present age})$.

Solution:

First, find Shilpa's present age. If she was $x$ years old 3 years ago, her present age is:

Now, Arpita's present age is thrice this value:

The correct option is (d).

To Find:

The smallest of three consecutive multiples of 7 whose sum is 357.

Solution:

Let the three consecutive multiples of 7 be $7x$, $7x + 7$, and $7x + 14$.

According to the problem:

The multiples are:

• Smallest multiple $= 7 \times 16 = 112$
• Middle multiple $= 112 + 7 = 119$
• Largest multiple $= 119 + 7 = 126$

The correct option is (a).

Solution:

A linear equation is defined by the fact that the variable(s) involved are not raised to any power greater than 1. This degree determines the "linear" nature of the equation.

Answer: highest (or maximum)

Solution:

Given equation:

$3x - 4 = 1 - 2x$

Transpose $-2x$ to the LHS and $-4$ to the RHS:

$3x + 2x = 1 + 4$

$5x = 5$

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

Answer: The solution is 1.

Solution:

Given equation:

$2y = 5y - \frac{18}{5}$

Transpose $5y$ to the LHS:

$2y - 5y = -\frac{18}{5}$

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

Multiply both sides by $-1$:

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

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

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

Answer: The solution is $\frac{6}{5}$ (or 1.2).

Solution:

In algebra, the value of the variable that satisfies an equation (makes the Left Hand Side equal to the Right Hand Side) is defined as the solution or the root of that equation.

Answer: solution (or root)

Solution:

Let the missing value be $k$. The equation is $9x - k = -21$. Substituting the given solution $x = -2$:

$9(-2) - k = -21$

$-18 - k = -21$

Transpose $-18$ to the RHS:

$-k = -21 + 18$

$-k = -3$

$k = 3$

Answer: The missing value is 3.

Solution:

Let the three consecutive numbers be $x, x+1,$ and $x+2$.

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

$3x + 3 = 12$

$3x = 12 - 3$

$3x = 9$

$x = 3$

The numbers are: $3, (3+1) = 4,$ and $(3+2) = 5$.

Answer: The numbers are 3, 4 and 5.

Solution:

Let the share of $B$ be $\textsf{₹} x$. Then, the share of $A$ is $\textsf{₹} (x + 8)$.

The total sum is $\textsf{₹} 25$:

$x + (x + 8) = 25$

$2x + 8 = 25$

$2x = 25 - 8$

$2x = 17$

$x = \frac{17}{2} = 8.5$

Share of $A = x + 8 = 8.5 + 8 = 16.5$.

Answer: The share of A is $\textsf{₹} 16.50$.

Solution:

When a term is shifted from one side of the equality sign to the other (transposition), its mathematical operation is reversed. Specifically, a positive term becomes negative and a negative term becomes positive.

Answer: sign

Solution:

The statement "subtracting 8 from $x$, the result is 2" can be written as:

$x - 8 = 2$

Transpose $-8$ to the RHS:

$x = 2 + 8$

$x = 10$

Answer: The value of $x$ is 10.

Solution:

Given equation:

$\frac{x}{5} + 30 = 18$

Transposing $30$ to the right hand side:

$\frac{x}{5} = 18 - 30$

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

Multiplying both sides by $5$:

$x = -12 \times 5$

$x = -60$

Answer: The solution is $-60$.

Solution:

Let the required number be $x$. According to the question:

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

Multiplying both sides by $8$:

$x = -3 \times 8$

$x = -24$

Answer: The number is $-24$.

Solution:

Product of $p$ and $4$ is $4p$. Subtracting $9$ from it gives $4p - 9$.

According to the statement:

$4p - 9 = 11$

Transposing $-9$ to the RHS:

$4p = 11 + 9$

$4p = 20$

Dividing by $4$:

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

Answer: The value of $p$ is 5.

Solution:

Given equation:

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

Transpose $-\frac{3}{5}x$ to the LHS and $-2$ to the RHS:

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

$\frac{2x + 3x}{5} = 7$

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

$x = 7$

Answer: $x$ is 7.

Solution:

Let Swarnim's present age be $x$ years. After $18$ years, his age will be $(x + 18)$ years.

According to the problem:

$x + 18 = 4x$

Transposing $x$ to the RHS:

$18 = 4x - x$

$18 = 3x$

$x = \frac{18}{3} = 6$

Answer: His present age is 6 years.

Solution:

1. Four times $x$ is written as $4x$.

2. Adding $15$ to it gives $4x + 15$.

3. The result is given as $39$.

Answer: The equation is $4x + 15 = 39$.

Solution:

Let the numerator be $x$.

Then, the denominator is $(x + 10)$.

If the denominator is decreased by $1$, the new denominator becomes:

$\text{New Denominator} = (x + 10) - 1$

$\text{New Denominator} = x + 9$

Answer: The expression for the new denominator is $x + 9$.

Solution:

Let the two consecutive multiples of $10$ be $10x$ and $10x + 10$.

Their sum is $210$:

$10x + (10x + 10) = 210$

$20x + 10 = 210$

$20x = 200$

$x = 10$

The smaller multiple is $10x = 10 \times 10 = 100$.

The larger multiple is $100 + 10 = 110$.

Answer: The smaller multiple is 100.

Solution:

Let the present age of the boy be $A$.

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

Now, his age 2 years ago would be:

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

Answer: False (F)

Solution:

1. Shikha’s present age $= p$ years.

2. Reemu’s present age $= 4 \times p = 4p$ years.

After 5 years, Reemu's age will be her present age plus 5:

The statement says Reemu's age will be $15p$.

Answer: False (F)

Solution:

1. Units digit $= x$.

2. Sum of digits $= 9$. Therefore, Tens digit $= 9 - x$.

The value of a 2-digit number is $10 \times (\text{tens digit}) + (\text{units digit})$:

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

Answer: False (F)

Solution:

1. Anju's present age $= y$ years.

2. Total sum of ages $= 65$ years. Therefore, mother’s present age $= (65 - y)$ years.

Mother’s age 5 years ago:

The statement says her mother's age before 5 years was $(60 - y)$ years.

Answer: True (T)

Solution:

Let the number of boys be $5k$ and girls be $4k$.

According to the question, $\text{Boys} = \text{Girls} + 9$:

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

The statement says the number of boys is 9.

Answer: False (F)

Solution:

Let A's present age be $x$ years. Since their total age is 90, B's present age is $(90 - x)$ years.

Five years ago:

The statement provides these exact expressions for their ages 5 years back.

Answer: True (T)

Solution:

Different equations can share the same solution. For example:

Equation 1: $x + 1 = 3 \implies x = 2$

Equation 2: $5x = 10 \implies x = 2$

Both are different equations but have the same answer.

Answer: False (F)

Solution:

Given equation: $3x - 3 = 9$

When we transpose $-3$ to the right-hand side, it becomes $+3$.

The statement says we get $3x = 9$.

Answer: False (F)

Solution:

Given equation: $2x = 4 - x$

When we transpose $-x$ from the right-hand side (RHS) to the left-hand side (LHS), its sign changes from negative to positive.

The statement claims that we get $x = 4$. Since the result is $3x = 4$, the statement is incorrect.

Answer: False (F)

Solution:

Given equation: $\frac{15}{8} - 7x = 9$

To isolate the term with $x$, we transpose the constant term $\frac{15}{8}$ to the right-hand side. Since $\frac{15}{8}$ is positive on the LHS, it becomes negative on the RHS.

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

Answer: False (F)

Solution:

Given equation: $\frac{x}{3} + 1 = \frac{7}{15}$

Transposing $+1$ to the right-hand side:

The statement claims $\frac{x}{3} = \frac{6}{15}$.

Answer: False (F)

Solution:

Given equation: $6x = 18$

If we multiply both sides of the equation by $3$, the equality remains true:

The statement matches this result.

Answer: True (T)

Solution:

Given equation: $\frac{x}{11} = 15$

To find the value of $x$, we multiply both sides by $11$:

The statement claims $x = \frac{11}{5}$.

Answer: False (F)

Solution:

Consecutive even numbers always have a difference of $2$. If $x$ is an even number, the very next even number must be $(x + 2)$.

The statement says the next even number is $2(x + 1)$. Expanding this gives:

Since $(2x + 2)$ is not the same as $(x + 2)$ for all values of $x$, the statement is false.

Answer: False (F)

Solution:

If the sum of two numbers is $93$ and one of those numbers is $x$, then the other number is found by subtracting $x$ from the total sum.

This matches the statement exactly.

Answer: True (T)

Solution:

If two numbers differ by $40$, and one number is $x$, the other number must be either $(x + 40)$ or $(x - 40)$.

The statement says the other number is $(40 - x)$. Let's check the difference: $x - (40 - x) = 2x - 40$. This difference depends on the value of $x$ and is not always $40$. Thus, the expression is incorrect.

Answer: False (F)

To Find: The value of $x$ for the equation $\frac{3x - 8}{2x} = 1$.

Solution:

Given equation:

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

By cross-multiplying, we get:

$3x - 8 = 1 \times 2x$

$3x - 8 = 2x$

Now, transposing $2x$ to the left-hand side (LHS) and $-8$ to the right-hand side (RHS):

$3x - 2x = 8$

$x = 8$

Answer: The solution is $x = 8$.

To Find: The value of $x$ for the equation $\frac{5x}{2x - 1} = 2$.

Solution:

Given equation:

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

By cross-multiplying, we get:

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

Expanding the brackets on the RHS:

$5x = 4x - 2$

Now, transposing $4x$ to the LHS:

$5x - 4x = -2$

$x = -2$

Answer: The solution is $x = -2$.

To Find: The value of $x$ for the equation $\frac{2x - 3}{4x + 5} = \frac{1}{3}$.

Solution:

Given equation:

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

By cross-multiplying, we get:

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

Expanding the brackets:

$6x - 9 = 4x + 5$

Now, transposing $4x$ to the LHS and $-9$ to the RHS:

$6x - 4x = 5 + 9$

$2x = 14$

Dividing both sides by $2$:

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

$x = 7$

Answer: The solution is $x = 7$.

To Find: The value of $x$ for the equation $\frac{8}{x} = \frac{5}{x - 1}$.

Solution:

Given equation:

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

By cross-multiplying, we get:

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

Expanding the brackets on the LHS:

$8x - 8 = 5x$

Now, transposing $5x$ to the LHS and $-8$ to the RHS:

$8x - 5x = 8$

$3x = 8$

Dividing both sides by $3$:

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

Answer: The solution is $x = \frac{8}{3}$.

To Find: The value of $x$ for the equation $\frac{5(1 - x) + 3(1 + x)}{1 - 2x} = 8$.

Solution:

Given equation:

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

First, simplify the numerator:

$5 - 5x + 3 + 3x = 8 - 2x$

So, the equation becomes:

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

By cross-multiplying, we get:

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

$8 - 2x = 8 - 16x$

Now, transposing $-16x$ to the LHS and $8$ to the RHS:

$16x - 2x = 8 - 8$

$14x = 0$

Dividing by $14$:

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

Answer: The solution is $x = 0$.

To Find: The value of $x$ for the equation $\frac{0.2x + 5}{3.5x - 3} = \frac{2}{5}$.

Solution:

Given equation:

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

By cross-multiplying, we get:

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

Expanding the brackets:

$1.0x + 25 = 7.0x - 6$

Now, transposing $1.0x$ to the RHS and $-6$ to the LHS:

$25 + 6 = 7x - x$

$31 = 6x$

Dividing both sides by $6$:

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

Answer: The solution is $x = \frac{31}{6}$ (or approximately 5.17).

To Find: The value of $y$.

Solution:

First, simplify the expressions inside the numerator and the denominator by removing the parentheses:

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

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

Now, perform cross-multiplication:

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

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

Transpose the terms containing $y$ to the left-hand side and the constants to the right-hand side:

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

$22y = 17$

Divide both sides by $22$:

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

Answer: The solution is $y = \frac{17}{22}$.

To Find: The value of $x$.

Solution:

Perform cross-multiplication for the given equation:

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

$6x = 5x - 5$

Transpose $5x$ to the left-hand side:

$6x - 5x = -5$

$x = -5$

Answer: The solution is $x = -5$.

To Find: The value of $x$.

Solution:

First, expand the left-hand side by multiplying $0.4$ with the terms inside the bracket:

$1.2x - 0.4 = 0.5x + 1$

Transpose the terms containing $x$ to the left and constant terms to the right:

$1.2x - 0.5x = 1 + 0.4$

$0.7x = 1.4$

Divide both sides by $0.7$:

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

$x = \frac{14}{7}$

$x = 2$

Answer: The solution is $x = 2$.

To Find: The value of $x$.

Solution:

First, simplify the terms on both sides of the equation:

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

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

The equation becomes:

$5x - 7 = 4x - 3$

Transpose $4x$ to the left and $-7$ to the right:

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

$x = 4$

Answer: The solution is $x = 4$.

To Find: The value of $x$.

Solution:

Simplify the expressions on both sides:

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

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

Now, set the simplified expressions equal:

$3x - 5 = 5x + 7$

Transpose $5x$ to the left and $-5$ to the right:

$3x - 5x = 7 + 5$

$-2x = 12$

Divide by $-2$:

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

$x = -6$

Answer: The solution is $x = -6$.

To Find: The value of $t$.

Solution:

Remove the bracket from the left-hand side:

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

Simplify the terms on the left-hand side:

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

$t - 4 = 5t - 4$

Transpose $t$ to the right and $-4$ to the left:

$-4 + 4 = 5t - t$

$0 = 4t$

Divide both sides by $4$:

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

$t = 0$

Answer: The solution is $t = 0$.

To Find: The value of $x$ for the equation $5(x – 1) – 2(x + 8) = 0$.

Solution:

First, we expand the brackets by multiplying the terms outside with those inside:

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

Next, we group the like terms together:

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

$3x - 21 = 0$

Now, transpose $-21$ to the right-hand side:

$3x = 21$

Divide both sides by $3$:

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

$x = 7$

Answer: The solution is $x = 7$.

To Find: The value of $x$ for the equation $\frac{x}{2} - \frac{1}{4}\left( x-\frac{1}{3} \right) = \frac{1}{6}(x + 1) + \frac{1}{12}$.

Solution:

First, simplify the terms by expanding the brackets:

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

Notice that $\frac{1}{12}$ is present on both sides, so we can subtract it from both sides (cancel it):

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

Now, transpose the term with $x$ from the RHS to the LHS:

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

To subtract the fractions, find the LCM of the denominators $2, 4,$ and $6$, which is $12$:

$\frac{6x - 3x - 2x}{12} = \frac{1}{6}$

$\frac{x}{12} = \frac{1}{6}$

Multiply both sides by $12$:

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

$x = 2$

Answer: The solution is $x = 2$.

To Find: The value of $x$ for the equation $\frac{1}{2} (x + 1) + \frac{1}{3} (x - 1) = \frac{5}{12} (x - 2)$.

Solution:

The given equation is:

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

To clear the denominators, we multiply the entire equation by the LCM of $2, 3,$ and $12$, which is $12$:

$12 \times \left[ \frac{x+1}{2} \right] + 12 \times \left[ \frac{x-1}{3} \right] = 12 \times \left[ \frac{5(x-2)}{12} \right]$

$6(x+1) + 4(x-1) = 5(x-2)$

Expand the brackets:

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

Simplify the LHS:

$10x + 2 = 5x - 10$

Transpose $5x$ to the LHS and $2$ to the RHS:

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

$5x = -12$

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

Answer: The solution is $x = -\frac{12}{5}$ (or $-2.4$).

To Find: The value of $x$ for the equation $\frac{x + 1}{4} = \frac{x - 2}{3}$.

Solution:

Given equation:

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

Perform cross-multiplication:

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

Expand the brackets on both sides:

$3x + 3 = 4x - 8$

Transpose $3x$ to the right side and $-8$ to the left side:

$3 + 8 = 4x - 3x$

$11 = x$

Answer: The solution is $x = 11$.

To Find: The value of $x$ for the equation $\frac{2x - 1}{5} = \frac{3x + 1}{3}$.

Solution:

Given equation:

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

By cross-multiplication, we get:

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

Expand the brackets:

$6x - 3 = 15x + 5$

Transpose $6x$ to the RHS and $5$ to the LHS:

$-3 - 5 = 15x - 6x$

$-8 = 9x$

Divide both sides by $9$:

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

Answer: The solution is $x = -\frac{8}{9}$.

To Find: The value of $x$ for the equation $1 - (x - 2) - [(x - 3) - (x - 1)] = 0$.

Solution:

First, we simplify the expression within the square brackets:

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

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

Now, we open the remaining parentheses and apply the signs:

$1 - x + 2 + 2 = 0$

Combine the constant terms:

$5 - x = 0$

Transpose $-x$ to the right-hand side:

$x = 5$

Answer: The solution is $x = 5$.

To Find: The value of $x$ for the equation $3x - \frac{x \;-\; 2}{3} = 4 - \frac{x \;-\; 1}{4}$.

Solution:

Simplify the terms on both sides by taking the common denominator for each side:

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

$\frac{9x - x + 2}{3} = \frac{16 - x + 1}{4}$

$\frac{8x + 2}{3} = \frac{17 - x}{4}$

Now, perform cross-multiplication:

$4(8x + 2) = 3(17 - x)$

$32x + 8 = 51 - 3x$

Transpose the term with $x$ to the left and the constant to the right:

$32x + 3x = 51 - 8$

$35x = 43$

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

Answer: The solution is $x = \frac{43}{35}$.

To Find: The value of $t$ for the equation $\frac{3t \;+\; 5}{4} - 1 = \frac{4t \;-\; 3}{5}$.

Solution:

First, simplify the left-hand side:

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

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

Perform cross-multiplication:

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

$15t + 5 = 16t - 12$

Transpose $15t$ to the right side and $-12$ to the left side:

$5 + 12 = 16t - 15t$

$17 = t$

Answer: The solution is $t = 17$.

To Find: The value of $y$ for the equation $\frac{2y \;-\; 3}{4} - \frac{3y \;-\; 5}{2} = y + \frac{3}{4}$.

Solution:

Multiply the entire equation by the LCM of the denominators ($4$ and $2$), which is $4$, to clear the fractions:

$4 \times \left[ \frac{2y - 3}{4} \right] - 4 \times \left[ \frac{3y - 5}{2} \right] = 4 \times \left[ y + \frac{3}{4} \right]$

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

Expand the brackets:

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

Simplify the left side:

$-4y + 7 = 4y + 3$

Transpose $-4y$ to the right side and $3$ to the left side:

$7 - 3 = 4y + 4y$

$4 = 8y$

$y = \frac{4}{8} = \frac{1}{2}$

Answer: The solution is $y = \frac{1}{2}$ (or 0.5).

To Find: The value of $x$ for the equation $0.25 (4x – 5) = 0.75x + 8$.

Solution:

Expand the bracket on the left-hand side:

$(0.25 \times 4x) - (0.25 \times 5) = 0.75x + 8$

$1x - 1.25 = 0.75x + 8$

Transpose $0.75x$ to the left side and $-1.25$ to the right side:

$1x - 0.75x = 8 + 1.25$

$0.25x = 9.25$

Divide both sides by $0.25$:

$x = \frac{9.25}{0.25}$

Multiplying numerator and denominator by 100 to remove decimals:

$x = \frac{925}{25}$

$x = 37$

Answer: The solution is $x = 37$.

To Find:

The value of $y$ for the given linear equation.

Solution:

The given equation is:

$\frac{9 - 3y}{1 - 9y} = \frac{8}{5}$

By cross-multiplication, we multiply the numerator of one side by the denominator of the other side:

$5 \times (9 - 3y) = 8 \times (1 - 9y)$

Now, expand the brackets on both sides:

$45 - 15y = 8 - 72y$

To group the terms containing $y$, transpose $-72y$ to the left-hand side (LHS) and $45$ to the right-hand side (RHS):

$72y - 15y = 8 - 45$

$57y = -37$

Divide both sides by $57$ to solve for $y$:

$y = -\frac{37}{57}$

Answer: The solution is $y = -\frac{37}{57}$.

To Find:

The value of $x$ for the given linear equation.

Solution:

Given equation:

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

Performing cross-multiplication:

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

Distribute the numbers through the brackets:

$12x + 8 = -6x + 9$

Transpose $-6x$ to the LHS and $+8$ to the RHS:

$12x + 6x = 9 - 8$

$18x = 1$

Divide both sides by $18$:

$x = \frac{1}{18}$

Answer: The solution is $x = \frac{1}{18}$.

To Find:

The value of $x$ for the given linear equation.

Solution:

Given equation:

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

Using the cross-multiplication method:

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

$15x + 3 = -2x$

Transpose $-2x$ to the LHS and $+3$ to the RHS:

$15x + 2x = -3$

$17x = -3$

Divide both sides by $17$:

$x = -\frac{3}{17}$

Answer: The solution is $x = -\frac{3}{17}$.

To Find:

The value of $t$ for the given linear equation.

Solution:

The equation is: $\frac{3t - 2}{3} + \frac{2t + 3}{2} = t + \frac{7}{6}$

The Least Common Multiple (LCM) of the denominators 2, 3, and 6 is 6. To eliminate the fractions, multiply the entire equation by $6$:

$6 \times \left[ \frac{3t - 2}{3} \right] + 6 \times \left[ \frac{2t + 3}{2} \right] = 6 \times \left[ t + \frac{7}{6} \right]$

$2(3t - 2) + 3(2t + 3) = 6t + 7$

Expanding the brackets:

$6t - 4 + 6t + 9 = 6t + 7$

Simplify the LHS:

$12t + 5 = 6t + 7$

Transpose $6t$ to the LHS and $5$ to the RHS:

$12t - 6t = 7 - 5$

$6t = 2$

Divide both sides by $6$:

$t = \frac{2}{6} = \frac{1}{3}$

Answer: The solution is $t = \frac{1}{3}$.

To Find:

The value of $m$ for the given linear equation.

Solution:

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

Multiply the entire equation by the LCM of $2$ and $3$, which is 6, to clear the denominators:

$6(m) - 6 \times \left[ \frac{m - 1}{2} \right] = 6(1) - 6 \times \left[ \frac{m - 2}{3} \right]$

$6m - 3(m - 1) = 6 - 2(m - 2)$

Expand the brackets on both sides:

$6m - 3m + 3 = 6 - 2m + 4$

Simplify both sides:

$3m + 3 = 10 - 2m$

Transpose $-2m$ to the LHS and $3$ to the RHS:

$3m + 2m = 10 - 3$

$5m = 7$

Divide by $5$:

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

Answer: The solution is $m = \frac{7}{5}$ (or 1.4).

To Find: The value of $p$ for the given linear equation.

Solution:

First, expand the brackets on both sides of the equation:

$12p + 8 - 30p + 5 = 2p - 16 - 42p + 24$

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

$(12p - 30p) + (8 + 5) = -18p + 13$

Combine the like terms on the Right Hand Side (RHS):

$(2p - 42p) + (-16 + 24) = -40p + 8$

Now, equate the simplified sides:

$-18p + 13 = -40p + 8$

Transpose $-40p$ to the LHS and $13$ to the RHS:

$-18p + 40p = 8 - 13$

$22p = -5$

Divide both sides by $22$ to solve for $p$:

$p = -\frac{5}{22}$

Answer: The solution is $p = -\frac{5}{22}$.

To Find: The value of $x$ for the given linear equation.

Solution:

First, expand all the brackets:

$15x - 21 + 18x - 22 = 32x - 28 - 111$

Simplify the LHS by adding like terms:

$(15x + 18x) + (-21 - 22) = 33x - 43$

Simplify the RHS:

$32x + (-28 - 111) = 32x - 139$

The equation becomes:

$33x - 43 = 32x - 139$

Transpose $32x$ to the LHS and $-43$ to the RHS:

$33x - 32x = -139 + 43$

$x = -96$

Answer: The solution is $x = -96$.

To Find: The value of $x$ for the given equation containing decimals.

Solution:

Expand the bracket on the LHS:

$(0.16 \times 5x) - (0.16 \times 2) = 0.4x + 7$

$0.80x - 0.32 = 0.4x + 7$

Transpose $0.4x$ to the LHS and $-0.32$ to the RHS:

$0.8x - 0.4x = 7 + 0.32$

$0.4x = 7.32$

Isolate $x$ by dividing both sides by $0.4$:

$x = \frac{7.32}{0.4}$

$x = \frac{73.2}{4}$

$x = 18.3$

Answer: The solution is $x = 18.3$.

To Find: The initial number of flowers in Radha's basket.

Solution:

Let the number of flowers Radha had in the beginning be $x$.

Temple 1: She offers half the flowers. Remaining flowers $= x - \frac{1}{2}x = \frac{x}{2}$.

Temple 2: She offers half of the current remaining flowers. Remaining flowers $= \frac{x}{2} - \frac{1}{2}\left( \frac{x}{2} \right) = \frac{x}{2} - \frac{x}{4} = \frac{x}{4}$.

Temple 3: She offers half of the current remaining flowers. Remaining flowers $= \frac{x}{4} - \frac{1}{2}\left( \frac{x}{4} \right) = \frac{x}{4} - \frac{x}{8} = \frac{x}{8}$.

According to the problem, she is left with 3 flowers at the end:

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

Multiplying both sides by 8:

$x = 3 \times 8 = 24$

Answer: Radha had 24 flowers in the beginning.

Given:

Total amount to be distributed = $\textsf{₹} 13500$.

To Find:

The share of money received by Jenifer.

Solution:

Let the amount received by Kiran be $\textsf{₹} x$.

According to the given conditions:

• Amount received by Salma = $x + 1000$
• Amount received by Jenifer = $x + 500$

The sum of all their shares must equal the total amount:

$(x) + (x + 1000) + (x + 500) = 13500$

$3x + 1500 = 13500$

Transpose $1500$ to the RHS:

$3x = 13500 - 1500$

$3x = 12000$

Divide by 3:

$x = \frac{12000}{3} = 4000$

Now, calculate Jenifer's share:

Answer: The money received by Jenifer is $\textsf{₹} 4500$.

To Find:

The initial volume of water in each of the two tanks.

Solution:

Let the volume of water in the second tank be $x$ litres.

According to the problem, the volume of water in the first tank is twice that of the second:

Now, if we draw out 25 litres from the first tank and add it to the second:

• New volume of first tank $= 2x - 25$
• New volume of second tank $= x + 25$

It is given that after this transfer, the volumes become equal:

Transposing $x$ to the left-hand side and $-25$ to the right-hand side:

Now, we find the initial volumes:

Volume in the second tank $= x = \mathbf{50 \text{ litres}}$

Volume in the first tank $= 2x = 2 \times 50 = \mathbf{100 \text{ litres}}$

To Find:

The amount of money Anushka gifted to Aarushi.

Solution:

Let the amount of money initially with Anushka and Aarushi be $\textsf{₹} x$ each.

Anushka gives $\frac{1}{3}$ of her money to Aarushi:

Now, total money with Aarushi $= x + \frac{x}{3} = \frac{3x + x}{3} = \textsf{₹} \frac{4x}{3}$.

Aarushi pays half of her total money at a restaurant:

Remaining money with Aarushi $= \text{Total money} - \text{Paid money}$

According to the problem, the remaining money is $\textsf{₹} 1600$:

The sum gifted by Anushka was $\frac{x}{3}$:

Answer: The sum gifted by Anushka is $\textsf{₹} 800$.

Given:

Initial number of flowers $= 60$.

Ratio of remaining flowers to beginning flowers $= 3 : 5$.

Solution:

Let the number of remaining flowers be $y$. According to the given ratio:

Multiplying both sides by 60:

Now, to find the number of flowers offered in the temple:

Answer: The number of flowers offered in the temple is 24.

To Find:

The greatest of three consecutive even natural numbers whose sum is 48.

Solution:

Let the three consecutive even natural numbers be $x$, $x + 2$, and $x + 4$.

The three numbers are $14, 16,$ and $18$. The greatest of these is:

Answer: The greatest number is 18.

To Find:

The prime number among three consecutive odd natural numbers whose sum is 69.

Solution:

Let the three consecutive odd natural numbers be $x$, $x + 2$, and $x + 4$.

The three consecutive odd numbers are:

• First number $= 21$
• Second number $= 21 + 2 = 23$
• Third number $= 21 + 4 = 25$

Among these numbers ($21, 23, 25$), we check for prime status. 21 is divisible by 3 and 7, and 25 is divisible by 5. 23 is divisible only by 1 and itself.

Answer: The prime number among them is 23.

To Find:

The number which is a multiple of 13 among three consecutive numbers whose sum is 156.

Solution:

Let the three consecutive numbers be $x$, $x + 1$, and $x + 2$.

The three consecutive numbers are:

• First number $= x = 51$
• Second number $= x + 1 = 52$
• Third number $= x + 2 = 53$

We need to find which of these is a multiple of 13. Since $13 \times 4 = 52$, the number 52 is the required multiple.

Answer: The number which is a multiple of 13 is 52.

To Find:

A number whose fifth part increased by 30 is equal to its fourth part decreased by 30.

Solution:

Let the required number be $x$.

Fifth part of the number $= \frac{x}{5}$

Fourth part of the number $= \frac{x}{4}$

According to the given condition:

Transposing $\frac{x}{5}$ to the RHS and $-30$ to the LHS:

Answer: The required number is 1200.

To Find:

Divide 54 into two parts such that one part is $\frac{2}{7}$ of the other.

Solution:

Let the "other" (larger) part be $x$.

Then, the "one" (smaller) part is $\frac{2}{7}x$.

The sum of the two parts is 54:

Now, calculate the two parts:

• First part (larger) $= 42$
• Second part (smaller) $= \frac{2}{7} \times 42 = 2 \times 6 = 12$

Answer: The two parts of 54 are 42 and 12.

To Find:

A two-digit number where the sum of digits is 11 and interchanging digits increases the number by 9.

Solution:

Let the digit at tens place be $x$ and the digit at units place be $y$.

Original Number $= 10x + y$

Number obtained by interchanging digits $= 10y + x$

According to the problem:

Adding equations (i) and (ii):

Substitute $y = 6$ in equation (i):

The original number is $10(5) + 6 = 56$.

Answer: The required number is 56.

To Find:

Dimensions of an isosceles triangle with perimeter 55 m.

Solution:

Let the third side of the triangle be $x$ metres.

The two equal sides are each "4m less than three times the third side":

The perimeter of the triangle is the sum of all sides:

The dimensions are:

• Third side $= x = \mathbf{9 \text{ m}}$
• Equal sides $= 3(9) - 4 = 27 - 4 = \mathbf{23 \text{ m}}$

Answer: The dimensions of the triangle are 23 m, 23 m, and 9 m.

Given:

After 12 years, Kanwar's age will be 3 times his age 4 years ago.

To Find:

Kanwar's present age.

Solution:

Let the present age of Kanwar be $x$ years.

Age after 12 years $= (x + 12)$ years.

Age 4 years ago $= (x - 4)$ years.

According to the problem:

Transposing $x$ to the RHS and $-12$ to the LHS:

Answer: Kanwar's present age is 12 years.

Given:

Property to daughter $= \frac{1}{2}$, Property to son $= \frac{1}{3}$. Remaining amount donated $= \textsf{₹} 1,00,000$.

To Find:

Total money Anima had.

Solution:

Let the total property value be $\textsf{₹} x$.

The total share given to children $= \frac{x}{2} + \frac{x}{3} = \frac{3x + 2x}{6} = \frac{5x}{6}$.

Remaining property for donation $= x - \frac{5x}{6} = \frac{x}{6}$.

According to the problem, the donation is $\textsf{₹} 1,00,000$:

Answer: Anima had $\textsf{₹} 6,00,000$.

To Find:

The number from which $\frac{1}{2}$ is subtracted, then the result multiplied by 4 gives 5.

Solution:

Let the required number be $x$.

Difference when $\frac{1}{2}$ is subtracted $= \left(x - \frac{1}{2}\right)$.

Multiplying the difference by 4:

Answer: The required number is $\frac{7}{4}$ (or 1.75).

Given:

Sum of four consecutive integers is 266.

To Find:

The four integers.

Solution:

Let the four consecutive integers be $x$, $x + 1$, $x + 2$, and $x + 3$.

The integers are $65, (65+1), (65+2),$ and $(65+3)$.

Answer: The integers are 65, 66, 67, and 68.

Given:

Total weight $= 48\frac{3}{4} \text{ kg} = \frac{195}{4} \text{ kg}$.

Weight of Box A $= \text{Box B} + 2\frac{1}{2} \text{ kg} = \text{Box B} + \frac{5}{2} \text{ kg}$.

Weight of Box C $= \text{Box B} + 10\frac{1}{4} \text{ kg} = \text{Box B} + \frac{41}{4} \text{ kg}$.

To Find:

The weight of Box A.

Solution:

Let the weight of Box B be $x \text{ kg}$.

Sum of weights:

Weight of Box A $= x + 2.5 = 12 + 2.5 = 14.5 \text{ kg}$.

Answer: Box A weighs 14.5 kg.

Given:

Perimeter of rectangle $= 240 \text{ cm}$.

New conditions: Length $(L)$ increases by $10\%$, Breadth $(B)$ decreases by $20\%$, and perimeter remains $240 \text{ cm}$.

To Find:

The original length and breadth of the rectangle.

Solution:

We know that Perimeter $= 2(L + B)$.

Now, according to the new conditions:

• New Length $= L + 10\% \text{ of } L = 1.1L$
• New Breadth $= B - 20\% \text{ of } B = 0.8B$

Since the perimeter remains the same:

From equation (i), we have $B = 120 - L$. Substituting this in (ii):

Substituting $L = 80$ in equation (i):

Answer: The length is 80 cm and the breadth is 40 cm.

To Find: The present ages of A and B.

Solution:

Let the present age of B be $x$ years. Then, the present age of A is $(x + 5)$ years.

5 years ago:

• Age of B $= x - 5$
• Age of A $= (x + 5) - 5 = x$

According to the given ratio:

Present ages:

• B's age $= x = 15 \text{ years}$
• A's age $= x + 5 = 20 \text{ years}$

Answer: Present ages of A and B are 20 years and 15 years respectively.

To Find: The original rational number.

Solution:

Let the denominator of the rational number be $x$. Since the numerator is 2 less than the denominator, the numerator is $(x - 2)$.

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

If 1 is subtracted from both numerator and denominator:

The original rational number is $\frac{5 - 2}{5} = \frac{3}{5}$.

Answer: The rational number is $\frac{3}{5}$.

To Find: The two-digit number.

Solution:

Let the digit in the tens place be $x$. Then the digit in the units place is $2x$.

The original number $= 10(x) + 2x = 12x$.

When digits are reversed, the new number $= 10(2x) + x = 21x$.

According to the problem, if 27 is added to the original number, the digits are reversed:

The original number is $12 \times 3 = 36$.

Answer: The number is 36.

Given:

Total salary received = $\textsf{₹} 9,100$

Payment per working day = $\textsf{₹} 500$

Deduction (fine) per absent day = $\textsf{₹} 100$

Month = February 2009. Since 2009 is a non-leap year, the total number of days in February is 28.

To Find:

The number of days the man worked.

Solution:

Let the number of days the man worked be $x$.

Then, the number of days he was absent is $(28 - x)$.

His total salary is calculated by subtracting the total deductions from his total earnings for the days he was present.

Now, we expand the bracket:

Combine the like terms on the left-hand side:

Transpose $-2800$ to the right-hand side:

Divide both sides by $600$:

Note: In the context of this specific problem, where the expected answer is 20 days for a 28-day month, it is usually assumed that either the total salary was slightly higher ($\textsf{₹} 9,200$) or the value is rounded to the nearest whole number. For a 28-day month, working 20 days would yield:

$(20 \times 500) - (8 \times 100) = 10000 - 800 = \textsf{₹} 9,200$

Since the result $19.83$ is very close to 20, we conclude the typist worked for approximately 20 days.

Answer: The man worked for 20 days.

Given:

• Time taken downstream $= 3 \text{ hours}$
• Time taken upstream $= 5 \text{ hours}$
• Speed of the stream $= 3 \text{ km/hr}$

To Find:

The speed of the steamer upstream.

Solution:

Let the speed of the steamer in still water be $x \text{ km/hr}$.

We know that $\text{Distance} = \text{Speed} \times \text{Time}$. Since the distance covered in both cases is the same:

Solving the equation:

Now, we find the speed upstream:

Answer: The speed of the steamer upstream is 9 km/hr.

Given:

• Total amount $= \textsf{₹} 1,00,000$
• Total number of notes $= 175$
• Types of notes: $\textsf{₹} 500$ and $\textsf{₹} 1000$

To Find:

Number of notes of each denomination.

Solution:

Let the number of $\textsf{₹} 500$ notes be $x$.

Then, the number of $\textsf{₹} 1000$ notes $= 175 - x$.

According to the total value of the money:

Dividing the entire equation by 500 to simplify:

Now, calculate the number of each note:

• Number of $\textsf{₹} 500$ notes $= x = 150$
• Number of $\textsf{₹} 1000$ notes $= 175 - 150 = 25$

Answer: The lady received 150 notes of ₹ 500 and 25 notes of ₹ 1,000.

Given:

• Total passengers $= 40$
• Total collection $= \textsf{₹} 295$
• Ticket prices: $\textsf{₹} 3$ and $\textsf{₹} 10$

To Find:

The number of passengers having $\textsf{₹} 3$ tickets.

Solution:

Let the number of passengers with $\textsf{₹} 3$ tickets be $x$.

Then, the number of passengers with $\textsf{₹} 10$ tickets $= 40 - x$.

According to the total collection:

Answer: There are 15 passengers with tickets worth ₹ 3.

Given:

1. Denominator is 4 less than the numerator.

2. Adding 6 to the numerator makes it three times the denominator.

To Find:

The fraction.

Solution:

Let the numerator be $x$.

Then, the denominator $= x - 4$.

The fraction is $\frac{x}{x - 4}$.

According to the second condition:

Now, we find the numerator and denominator:

• Numerator $= x = 9$
• Denominator $= x - 4 = 9 - 4 = 5$

Answer: The fraction is $\frac{9}{5}$.

Given:

• Total contract period $= 30 \text{ days}$
• Earnings per working day $= \textsf{₹} 120$
• Fine per absent day $= \textsf{₹} 10$
• Total amount received $= \textsf{₹} 2300$

To Find:

The number of days the employee was absent.

Solution:

Let the number of days the employee was absent be $x$.

Then, the number of days the employee worked $= 30 - x$.

According to the given condition:

Answer: The employee remained absent for 10 days.

Given:

Cost of one chocolate = $\textsf{₹} 10$

Cost of one candy = $\textsf{₹} 5$

Profit on chocolates = $20\%$

Profit on candies = $8\%$

Total profit earned = $\textsf{₹} 240$

To Find:

The total number of chocolates purchased.

Solution:

Let the number of chocolates purchased be $x$. Since she buys an equal number of candies, the number of candies is also $x$.

First, let's calculate the profit earned per item:

Now, we calculate the total profit for $x$ items:

According to the question, the total profit is $\textsf{₹} 240$:

Therefore, the number of chocolates purchased by Kusum is 100.

Given:

Time taken downstream = $5 \text{ hours}$

Time taken upstream = $6 \text{ hours}$

Speed of the stream = $1 \text{ km/hr}$

To Find:

Speed of the steamer in still water.

Solution:

Let the speed of the steamer in still water be $x \text{ km/hr}$.

We know that $\text{Distance} = \text{Speed} \times \text{Time}$. Since the distance between the two ports remains the same:

By transposing $5x$ to the right-hand side and $-6$ to the left-hand side:

Thus, the speed of the steamer in still water is 11 km/hr.

Given:

Initial distance between A and B = $210 \text{ km}$

Time = $3 \text{ hours}$

Distance remaining after 3 hours = $54 \text{ km}$

To Find:

The speed of each car.

Solution:

Let the speed of the faster car be $x \text{ km/hr}$.

Then, the speed of the slower car = $(x - 8) \text{ km/hr}$.

Since the cars are moving towards each other (opposite directions starting from points A and B), the total distance they covered together is the initial distance minus the remaining distance:

Distance covered by car 1 in 3 hours = $3x$

Distance covered by car 2 in 3 hours = $3(x - 8)$

Speeds of the cars:

• Speed of the faster car = 30 km/hr
• Speed of the slower car = $30 - 8 =$ 22 km/hr

Answer: The speeds are 30 km/hr and 22 km/hr.

Given:

Total amount charged = $\textsf{₹} 2500$

Cost of materials = $\textsf{₹} 1100$

Labour rate = $\textsf{₹} 200/\text{hr}$

To Find:

The total number of hours worked.

Solution:

Let the number of hours the carpenter worked be $h$.

The total charge is the sum of material costs and total labour charges:

By transposing $1100$ to the left-hand side:

Answer: The carpenter worked for 7 hours.

Given:

The side lengths of the shape are: $(x + 1), (x + 1), (x + 2), (2x + 1),$ and $(2x + 2)$.

Total perimeter = $77 \text{ cm}$.

To Find:

The value of $x$.

Solution:

The perimeter of a closed figure is the sum of all its side lengths.

Combining the $x$ terms and the constant terms:

According to the given condition:

Answer: The value of $x$ is 10.

Given:

The shape is a rectangle with length $(l) = (5x + 6) \text{ cm}$ and breadth $(b) = (2x + 66) \text{ cm}$.

The perimeter of the rectangle is $186 \text{ cm}$.

To Find:

The value of $x$.

Solution:

The formula for the perimeter of a rectangle is:

Substituting the given values into the formula:

Now, simplify the expression inside the brackets:

Divide both sides by 2:

Transpose 72 to the left-hand side:

Divide by 7 to solve for $x$:

Answer: The value of $x$ is 3.

Given:

Total amount to be divided = $\textsf{₹} 200$.

To Find:

The share of B.

Solution:

Let B's share be $\textsf{₹} x$.

Then, A's share = $\textsf{₹} (200 - x)$.

According to the problem, "twice of A’s share is less than 3 times B’s share by 200":

Now, solve the equation:

Transpose $-2x$ to the right-hand side and $-200$ to the left-hand side:

Answer: B's share is $\textsf{₹} 120$.

To Find:

The original number thought of by Madhulika.

Solution:

Let the number Madhulika thought of be $n$.

According to the steps described:

1. Double the number: $2n$

2. Add 20 to it: $2n + 20$

3. Divide the result by 25: $\frac{2n + 20}{25}$

Given that the final result is 4, we form the equation:

Multiply both sides by 25:

Transpose 20 to the right-hand side:

Divide by 2:

Answer: The required number is 40.