RBSE Solutions Class 8 Maths Chapter 11 Linear Equations with One Variable More Ques

Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Additional Questions

I. Objective Type Questions

Question 1. The necessary condition for a linear equation is
(a) highest power of variable is 1.
(b) highest power of variable is 2.
(c) highest power of variable is 3.
(d) None of the options
Answer: (a) highest power of variable is 1.
In simple words: For an equation to be called "linear", the variable in it (like x or y) must have a power of 1. If the power is 2 (like \( x^2 \)) or more, it's not a linear equation.

🎯 Exam Tip: Remember that "linear" means it forms a straight line when graphed, which only happens when the variable's highest power is 1.

Question 2. On transposing the term with sign changed
(a) in 'x'
(b) in \( + \)
(c) in \( - \)
(d) remains same
Answer: (c) in \( - \)
In simple words: When you move a term from one side of an equation to the other, its mathematical sign changes. If it was positive, it becomes negative; if it was negative, it becomes positive. This is a fundamental rule for balancing equations.

🎯 Exam Tip: Always remember to change the sign of a term when moving it across the equals sign to avoid errors in solving equations.

Question 3. The value of x in 3x = 21 is
(a) 7
(b) 24
(c) 18
(d) 63
Answer: (a) 7
In simple words: To find x, divide both sides of the equation by 3. This gives 21 divided by 3, which is 7.

🎯 Exam Tip: When a number is multiplied by a variable, divide by that number on both sides to isolate the variable.

Question 5. The solution of x - 4 = 5 is
(a) 4x5
(b) 5x4
(c) 5-4
(d) 5+4
Answer: (d) 5+4
In simple words: To find x, you need to move the -4 from the left side to the right side of the equation. When -4 moves, its sign changes to +4, so x becomes 5 plus 4.

🎯 Exam Tip: Always perform the inverse operation to isolate the variable; if a number is subtracted, add it to both sides.

Question 6. In the following the linear equation is
(a) \( \frac {x}{4} =\frac{4}{x} \)
(b) \( \frac {1}{x}+\frac {1}{x-1} =1 \)
(c) \( \frac {x}{2}+\frac {x}{3} =\frac {1}{4} \)
(d) \( x^2 + 2x + 3 = 0 \)
Answer: (c) \( \frac {x}{2}+\frac {x}{3} =\frac {1}{4} \)
In simple words: A linear equation is one where the highest power of the variable is 1. Option (c) simplifies to \( \frac{5x}{6} = \frac{1}{4} \), which means \( x \) has a power of 1. The other options involve \( x^2 \) or \( x \) in the denominator, which makes them non-linear.

🎯 Exam Tip: To identify a linear equation, clear any denominators by multiplying by the LCM and ensure the highest power of the variable is exactly 1.

Question 7. The degree of (x - 1)² = x² - 3 is
(a) 1
(b) 2
(c) 0
(d) 3
Answer: (a) 1
In simple words: If you expand the equation \( (x - 1)^2 = x^2 - 3 \), it becomes \( x^2 - 2x + 1 = x^2 - 3 \). When you cancel \( x^2 \) from both sides, you are left with \( -2x + 1 = -3 \), where the highest power of x is 1. Therefore, its degree is 1.

🎯 Exam Tip: Always simplify both sides of an equation before determining its degree, as higher power terms might cancel out.

Question 8. The solution of \( \frac {5}{x}=2 \) is
(a) 10
(b) \( \frac {2}{5} \)
(c) \( \frac {5}{2} \)
(d) \( \frac {1}{10} \)
Answer: (c) \( \frac {5}{2} \)
In simple words: To solve for x, multiply both sides by x, which gives \( 5 = 2x \). Then, divide both sides by 2 to find x, which is \( \frac{5}{2} \).

🎯 Exam Tip: To solve equations with variables in the denominator, multiply both sides by the variable to move it to the numerator.

II. Fill in the blanks

Question 1. Sign = is always used in equations.
Answer: =
In simple words: An equation always has an equals sign \( (=) \) to show that two expressions are the same. This sign balances the two sides of the equation.

🎯 Exam Tip: The equals sign is crucial in an equation, as it signifies a balance or equality between the expressions on either side.

Question 2. Variables can be transposed from one side to another side as numbers.
Answer: transposed
In simple words: Just like numbers, variables can be moved from one side of an equation to the other by changing their sign. This process is called transposing.

🎯 Exam Tip: Transposing terms (variables or constants) is a key step in rearranging equations to solve for unknowns.

Question 3. An equation involving only linear polynomials is called a linear equation.
Answer: linear
In simple words: A linear equation only has variables raised to the power of one. If there are terms like \( x^2 \) or \( y^3 \), it is not a linear equation.

🎯 Exam Tip: Ensure all variable terms in an equation have a degree of 1 for it to be classified as a linear equation.

Question 4. A value of the variable which makes the equation a true statement, is called a solution or a root of the equation.
Answer: solution, root
In simple words: When you find a number that makes an equation correct after you put it in place of the variable, that number is called the solution or root of the equation. It is the specific value that makes both sides of the equation equal.

🎯 Exam Tip: Always check your solution by substituting the value back into the original equation to ensure it yields a true statement.

III. True/False Type Questions

Question 1. The root of \( z \div 4 = 8 \) is 32.
Answer: True
In simple words: If you multiply both sides of the equation \( z \div 4 = 8 \) by 4, you get \( z = 8 \times 4 \), which means \( z = 32 \). So, the statement that the root is 32 is correct.

🎯 Exam Tip: To solve for a variable that is being divided, multiply both sides of the equation by the divisor.

Question 2. The root of \( 3x = \frac{20}{7}-x \) is \( \frac{5}{7} \)
Answer: True
In simple words: To check this, add x to both sides to get \( 4x = \frac{20}{7} \). Then, divide by 4: \( x = \frac{20}{7 \times 4} = \frac{5}{7} \). Since the calculation matches the given root, the statement is true.

🎯 Exam Tip: When checking a root, substitute the given value into the equation and verify if both sides are equal.

IV. Matching Type Questions

Answer:
1. \( \Leftrightarrow \) (c) 21
2. \( \Leftrightarrow \) (d) 1
3. \( \Leftrightarrow \) (a) 0
4. \( \Leftrightarrow \) (b) \( \frac{3}{5} \)
In simple words: We match each statement in Part 1 with its correct answer from Part 2. For example, a number with a one's digit of 1 and a ten's digit of 2 is 21. A linear equation has a degree of 1. A single linear equation typically has one unique solution, but if you consider the number of *distinct* solutions, it can be zero in certain contexts, or one. If we consider maximum solutions for a *single variable* linear equation, it is 1. If we consider a system of equations, it can be infinite or zero. Given the options, 0 for maximum solutions is an unusual interpretation, but is the provided match. A fraction with numerator 3 and denominator 5 is \( \frac{3}{5} \).

🎯 Exam Tip: For matching questions, carefully evaluate each statement and its potential matches. Double-check number formation, definitions (like degree), and basic fraction rules.

V. Very Short Answer Type Questions

Question 2. Which type of equations can be solved by linear equations?
Answer: Linear equations can be used to solve many kinds of real-world problems. This includes problems about age, numbers, areas, and other situations where quantities relate in a straightforward way. It helps to model and find unknown values in these scenarios.
In simple words: Linear equations are used for problems about age, numbers, and areas where things change in a simple, direct line.

🎯 Exam Tip: Practice converting word problems into linear equations by identifying the unknown variables and establishing clear relationships between them.

Question 3. Solve 2x - 3 = 7.
Answer: We need to find the value of x that makes the equation true.
\( 2x - 3 = 7 \)
To isolate the term with x, add 3 to both sides of the equation:
\( \implies 2x = 7 + 3 \)
\( \implies 2x = 10 \)
Now, divide both sides by 2 to find x:
\( \implies x = \frac{10}{2} \)
\( \implies x = 5 \)
In simple words: First, move the number 3 to the other side by adding it. This makes \( 2x = 10 \). Then, divide by 2 to get \( x = 5 \).

🎯 Exam Tip: Always perform operations in the reverse order of PEMDAS/BODMAS to solve for a variable: first addition/subtraction, then multiplication/division.

Question 4. Solve 2x - 3 = x + 2.
Answer: We want to find the value of x.
\( 2x - 3 = x + 2 \)
First, bring all terms with x to one side and constants to the other. Subtract x from both sides:
\( \implies 2x - x = 2 + 3 \)
Now, combine the like terms:
\( \implies x = 5 \)
In simple words: Move all the 'x' terms to one side and the numbers to the other. So, \( 2x - x \) on one side and \( 2 + 3 \) on the other. This gives \( x = 5 \).

🎯 Exam Tip: When a variable appears on both sides of an equation, collect all variable terms on one side and constant terms on the other to simplify.

Question 5. Solve the equation \( \frac {x}{3}+1=\frac{7}{15} \)
Answer: We need to find the value of x that satisfies the equation.
\( \frac{x}{3}+1 = \frac{7}{15} \)
First, subtract 1 from both sides to isolate the term with x:
\( \implies \frac{x}{3} = \frac{7}{15} - 1 \)
To subtract 1, write it as a fraction with denominator 15:
\( \implies \frac{x}{3} = \frac{7}{15} - \frac{15}{15} \)
\( \implies \frac{x}{3} = \frac{7 - 15}{15} \)
\( \implies \frac{x}{3} = \frac{-8}{15} \)
Finally, multiply both sides by 3 to solve for x:
\( \implies x = \frac{-8 \times 3}{15} \)
\( \implies x = \frac{-24}{15} \)
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 3:
\( \implies x = \frac{-8}{5} \)
In simple words: First, take 1 to the other side and subtract it from \( \frac{7}{15} \). This leaves \( \frac{x}{3} = \frac{-8}{15} \). Then multiply both sides by 3 to get \( x = \frac{-24}{15} \), which simplifies to \( \frac{-8}{5} \).

🎯 Exam Tip: When dealing with fractions in an equation, always find a common denominator or multiply by the LCM to clear fractions for easier solving.

VI. Short Answer Type Questions

Question 1. Solve the equation \( 4x - [2 + \{x - (3 - x)\}] = 3x + 6 \)
Answer: We need to simplify and solve the equation for x.
\( 4x - [2 + \{x - (3 - x)\}] = 3x + 6 \)
First, simplify the innermost parenthesis:
\( \implies 4x - [2 + \{x - 3 + x\}] = 3x + 6 \)
Next, simplify the curly braces:
\( \implies 4x - [2 + 2x - 3] = 3x + 6 \)
Now, simplify the square brackets:
\( \implies 4x - (2x - 1) = 3x + 6 \)
Distribute the negative sign:
\( \implies 4x - 2x + 1 = 3x + 6 \)
Combine like terms on the left side:
\( \implies 2x + 1 = 3x + 6 \)
Move all terms with x to one side and constants to the other. Subtract 3x from both sides and subtract 1 from both sides:
\( \implies 2x - 3x = 6 - 1 \)
\( \implies -x = 5 \)
Multiply both sides by -1 to find x:
\( \implies x = -5 \)
In simple words: Start by clearing the inner brackets, then the outer ones. This simplifies the equation to \( 2x + 1 = 3x + 6 \). Finally, gather 'x' terms on one side and numbers on the other to get \( -x = 5 \), so \( x = -5 \).

🎯 Exam Tip: When solving equations with multiple brackets, always simplify from the innermost bracket outwards (parentheses, then curly braces, then square brackets).

Question 2. Solve \( \sqrt{3}x - 2 = 2\sqrt{3} + 4 \).
Answer: We need to find the value of x.
\( \sqrt{3}x - 2 = 2\sqrt{3} + 4 \)
First, add 2 to both sides of the equation to move the constant term:
\( \implies \sqrt{3}x = 2\sqrt{3} + 4 + 2 \)
\( \implies \sqrt{3}x = 2\sqrt{3} + 6 \)
Factor out \( \sqrt{3} \) from the right side. Note that \( 6 = 2 \times 3 = 2 \times \sqrt{3} \times \sqrt{3} \):
\( \implies \sqrt{3}x = \sqrt{3} (2 + 2\sqrt{3}) \)
Wait, the source's factorization is \( 2\sqrt{3}(1+\sqrt{3}) \). Let's use that:
\( \implies \sqrt{3}x = 2\sqrt{3}(1+\sqrt{3}) \)
Now, divide both sides by \( \sqrt{3} \):
\( \implies x = \frac{2\sqrt{3}(1+\sqrt{3})}{\sqrt{3}} \)
Cancel out \( \sqrt{3} \):
\( \implies x = 2(1 + \sqrt{3}) \)
In simple words: Move the number -2 to the right side, so it becomes \( \sqrt{3}x = 2\sqrt{3} + 6 \). Then, divide both sides by \( \sqrt{3} \) to solve for x. Simplify the right side to get \( x = 2(1 + \sqrt{3}) \).

🎯 Exam Tip: When dealing with square roots, remember to simplify terms or factor out common square roots to isolate the variable efficiently.

Question 3. Solve the equation \( \frac{8x+3}{2x-4 } =\frac{4x}{x-5 } \)
Answer: We will solve this rational equation using cross-multiplication.
\( \frac{8x+3}{2x-4} = \frac{4x}{x-5} \)
Cross-multiply the terms:
\( \implies (8x + 3)(x - 5) = 4x(2x - 4) \)
Expand both sides of the equation:
\( \implies 8x^2 - 40x + 3x - 15 = 8x^2 - 16x \)
Combine like terms on the left side:
\( \implies 8x^2 - 37x - 15 = 8x^2 - 16x \)
Subtract \( 8x^2 \) from both sides:
\( \implies -37x - 15 = -16x \)
Move all terms with x to one side and constants to the other. Add \( 37x \) to both sides:
\( \implies -15 = -16x + 37x \)
\( \implies -15 = 21x \)
Divide both sides by 21:
\( \implies x = \frac{-15}{21} \)
Simplify the fraction by dividing the numerator and denominator by 3:
\( \implies x = \frac{-5}{7} \)
In simple words: Multiply diagonally across the equals sign. Expand both sides. Cancel out \( 8x^2 \). Then, bring all 'x' terms to one side and numbers to the other to solve for x. The answer is \( \frac{-5}{7} \).

🎯 Exam Tip: Always remember to distribute correctly when expanding products of binomials or monomials, and double-check for errors when combining like terms.

Question 4. A bag contains 15 coins of two-rupee and five-rupee. The total amount of the coins is Rs 45. How many coins of each kind are there in the bag.
Answer: Let's find the number of each type of coin.
Let the number of 2-rupee coins be \( x \).
Since there are 15 coins in total, the number of 5-rupee coins will be \( 15 - x \).
The total value of the 2-rupee coins is \( 2 \times x = 2x \) Rs.
The total value of the 5-rupee coins is \( 5 \times (15 - x) \) Rs.
The total amount in the bag is given as Rs 45. So, we can form the equation:
\( 2x + 5(15 - x) = 45 \)
Distribute the 5:
\( \implies 2x + 75 - 5x = 45 \)
Combine the x-terms:
\( \implies -3x + 75 = 45 \)
Subtract 75 from both sides:
\( \implies -3x = 45 - 75 \)
\( \implies -3x = -30 \)
Divide by -3 to find x:
\( \implies x = \frac{-30}{-3} \)
\( \implies x = 10 \)
So, there are 10 two-rupee coins.
The number of 5-rupee coins is \( 15 - x = 15 - 10 = 5 \).
Thus, there are 10 two-rupee coins and 5 five-rupee coins in the bag. These values ensure the total number of coins is 15 and the total value is Rs 45.
In simple words: Let x be the number of 2-rupee coins. Then \( 15 - x \) is the number of 5-rupee coins. Set up an equation for the total value: \( 2x + 5(15 - x) = 45 \). Solve for x to find there are 10 two-rupee coins and 5 five-rupee coins.

🎯 Exam Tip: In word problems, clearly define your variables, then translate each piece of information into an algebraic expression to form the equation.

Question 5. Find the three consecutive numbers where the sum of twice of first number, thrice of second number and four times of third number is 182.
Answer: Let's find these consecutive numbers.
Let the three consecutive numbers be \( x, x+1, \) and \( x+2 \).
According to the problem statement, the sum of twice the first number, thrice the second number, and four times the third number is 182:
\( 2(x) + 3(x+1) + 4(x+2) = 182 \)
Distribute the multipliers into the parentheses:
\( \implies 2x + 3x + 3 + 4x + 8 = 182 \)
Combine all the x-terms and all the constant terms on the left side:
\( \implies (2x + 3x + 4x) + (3 + 8) = 182 \)
\( \implies 9x + 11 = 182 \)
Subtract 11 from both sides to isolate the x-term:
\( \implies 9x = 182 - 11 \)
\( \implies 9x = 171 \)
Divide by 9 to solve for x:
\( \implies x = \frac{171}{9} \)
\( \implies x = 19 \)
Now, find the three consecutive numbers:
First number \( = x = 19 \)
Second number \( = x + 1 = 19 + 1 = 20 \)
Third number \( = x + 2 = 19 + 2 = 21 \)
So, the three consecutive numbers are 19, 20, and 21. These numbers satisfy the conditions of the problem.
In simple words: Call the numbers \( x, x+1, \) and \( x+2 \). Set up the equation: \( 2x + 3(x+1) + 4(x+2) = 182 \). Solve for x, which is 19. The numbers are then 19, 20, and 21.

🎯 Exam Tip: For problems involving consecutive numbers, represent them as \( x, x+1, x+2 \) (for integers) or \( x, x+2, x+4 \) (for consecutive even/odd numbers) to simplify equation setup.

Question 6. Ramesh's father is 27 years older than Ramesh. After 5 years, the ratio of ages of Ramesh and his father would be 2: 3. Find their present age.
Answer: Let's find their current ages.
Let Ramesh's present age be \( x \) years.
Since Ramesh's father is 27 years older, his father's present age is \( (x + 27) \) years.
After 5 years from now:
Ramesh's age will be \( (x + 5) \) years.
His father's age will be \( (x + 27 + 5) = (x + 32) \) years.
The problem states that after 5 years, the ratio of their ages will be 2: 3. We can write this as an equation:
\( \frac{x+5}{x+32} = \frac{2}{3} \)
Cross-multiply to solve the equation:
\( \implies 3(x + 5) = 2(x + 32) \)
Distribute the numbers on both sides:
\( \implies 3x + 15 = 2x + 64 \)
Move all x-terms to one side and constants to the other. Subtract \( 2x \) from both sides and subtract 15 from both sides:
\( \implies 3x - 2x = 64 - 15 \)
\( \implies x = 49 \)
So, Ramesh's present age is 49 years.
Ramesh's father's present age is \( x + 27 = 49 + 27 = 76 \) years.
In simple words: Let Ramesh's age be x. His father's age is \( x+27 \). After 5 years, their ages are \( x+5 \) and \( x+32 \). Set up the ratio equation \( \frac{x+5}{x+32} = \frac{2}{3} \). Solve for x to find Ramesh is 49 years old.

🎯 Exam Tip: When dealing with age problems involving ratios, always express current ages, then ages after/before a certain time, before setting up the ratio equation.

Question 7. Solve \( \frac{{6x}^{2}+13x-4}{2x+5} = \frac{{12x}^{2}+5x-2}{4x+3} \)
Answer: We will solve this equation by cross-multiplication and simplification.
Given the equation:
\( \frac{6x^2+13x-4}{2x+5} = \frac{12x^2+5x-2}{4x+3} \)
Perform cross-multiplication:
\( \implies (6x^2+13x-4)(4x+3) = (12x^2+5x-2)(2x+5) \)
Expand both sides of the equation:
LHS: \( (6x^2+13x-4)(4x+3) = 6x^2(4x) + 6x^2(3) + 13x(4x) + 13x(3) - 4(4x) - 4(3) \)
\( = 24x^3 + 18x^2 + 52x^2 + 39x - 16x - 12 \)
\( = 24x^3 + 70x^2 + 23x - 12 \)
RHS: \( (12x^2+5x-2)(2x+5) = 12x^2(2x) + 12x^2(5) + 5x(2x) + 5x(5) - 2(2x) - 2(5) \)
\( = 24x^3 + 60x^2 + 10x^2 + 25x - 4x - 10 \)
\( = 24x^3 + 70x^2 + 21x - 10 \)
Now, equate the expanded forms of LHS and RHS:
\( \implies 24x^3 + 70x^2 + 23x - 12 = 24x^3 + 70x^2 + 21x - 10 \)
Subtract \( 24x^3 \) and \( 70x^2 \) from both sides. These terms cancel out:
\( \implies 23x - 12 = 21x - 10 \)
Move all x-terms to one side and constants to the other. Subtract \( 21x \) from both sides and add 12 to both sides:
\( \implies 23x - 21x = -10 + 12 \)
\( \implies 2x = 2 \)
Divide by 2:
\( \implies x = 1 \)
So, the required solution for the given equation is \( x = 1 \). This ensures the equation holds true.
In simple words: Cross-multiply and expand both sides. Notice that \( 24x^3 \) and \( 70x^2 \) terms cancel out from both sides. This simplifies the equation to \( 23x - 12 = 21x - 10 \). Solving this simpler linear equation gives \( x = 1 \).

🎯 Exam Tip: When dealing with complex algebraic fractions, cross-multiplication is usually the first step. Carefully expand and combine like terms to reduce the equation to a simpler form.

Question 8. Denominator of a rational number exceeds its numerator by 5. If 2 is added to its numerator and denominator then we get \( \frac{1}{2} \). Find the rational number.
Answer: Let's find the original rational number.
Let the numerator of the rational number be \( x \).
According to the problem, the denominator exceeds the numerator by 5. So, the denominator is \( x + 5 \).
The original rational number is \( \frac{x}{x+5} \).
Now, if 2 is added to both its numerator and denominator, the new fraction is \( \frac{1}{2} \).
So, the new fraction is \( \frac{x+2}{(x+5)+2} = \frac{x+2}{x+7} \).
Set this equal to \( \frac{1}{2} \):
\( \frac{x+2}{x+7} = \frac{1}{2} \)
Cross-multiply to solve for x:
\( \implies 2(x + 2) = 1(x + 7) \)
\( \implies 2x + 4 = x + 7 \)
Move the x-terms to one side and the constants to the other. Subtract x from both sides and subtract 4 from both sides:
\( \implies 2x - x = 7 - 4 \)
\( \implies x = 3 \)
So, the numerator \( x \) is 3.
The denominator is \( x + 5 = 3 + 5 = 8 \).
Therefore, the required rational number is \( \frac{3}{8} \). This fraction satisfies all the given conditions.
In simple words: Let the numerator be x, so the denominator is \( x+5 \). When 2 is added to both, the fraction becomes \( \frac{x+2}{x+7} = \frac{1}{2} \). Solve this equation to get \( x=3 \). This means the number is \( \frac{3}{8} \).

🎯 Exam Tip: For problems involving fractions and their parts, clearly define the numerator and denominator using a single variable, then form the equation based on the given conditions.

Question 9. A rectangle has a breadth of \( x \) metres and a length of \( (3+x) \) metres. If its perimeter is 54 metres, find its length and breadth.
Answer: Let's find the dimensions of the rectangle.
Given:
Length of the rectangle \( (L) = (3 + x) \) metres.
Breadth of the rectangle \( (B) = x \) metres.
The formula for the perimeter of a rectangle is \( P = 2(L + B) \).
Substitute the given length and breadth into the perimeter formula:
\( P = 2((3 + x) + x) \)
\( P = 2(3 + 2x) \) metres.
We are given that the perimeter \( (P) \) is 54 metres.
So, we can set up the equation:
\( 2(3 + 2x) = 54 \)
Divide both sides by 2:
\( \implies 3 + 2x = \frac{54}{2} \)
\( \implies 3 + 2x = 27 \)
Subtract 3 from both sides:
\( \implies 2x = 27 - 3 \)
\( \implies 2x = 24 \)
Divide by 2 to solve for x:
\( \implies x = \frac{24}{2} \)
\( \implies x = 12 \)
Now, find the length and breadth using the value of x:
Breadth \( = x = 12 \) metres.
Length \( = 3 + x = 3 + 12 = 15 \) metres.
So, the length of the rectangle is 15 metres and the breadth is 12 metres. These dimensions satisfy the given perimeter.
In simple words: Use the formula for the perimeter of a rectangle: \( P = 2(L+B) \). Substitute \( L = 3+x \) and \( B = x \), and \( P = 54 \). Solve the equation \( 2(3+2x) = 54 \) to get \( x=12 \). So, the breadth is 12m and the length is 15m.

🎯 Exam Tip: Always write down the relevant formulas (like perimeter, area) when solving geometry-related word problems to ensure you set up the equation correctly.

Question 10. Find three consecutive multiples of 11, whose sum is 363.
Answer: Let's find these three multiples of 11.
Let the first multiple of 11 be \( 11x \).
Since the numbers are consecutive multiples of 11, the next two will be \( 11(x+1) \) and \( 11(x+2) \).
The sum of these three multiples is given as 363:
\( 11x + 11(x+1) + 11(x+2) = 363 \)
Expand the terms:
\( \implies 11x + 11x + 11 + 11x + 22 = 363 \)
Combine all the x-terms and all the constant terms:
\( \implies (11x + 11x + 11x) + (11 + 22) = 363 \)
\( \implies 33x + 33 = 363 \)
Subtract 33 from both sides to isolate the x-term:
\( \implies 33x = 363 - 33 \)
\( \implies 33x = 330 \)
Divide by 33 to solve for x:
\( \implies x = \frac{330}{33} \)
\( \implies x = 10 \)
Now, find the three consecutive multiples of 11:
First multiple \( = 11x = 11 \times 10 = 110 \)
Second multiple \( = 11(x+1) = 11(10+1) = 11 \times 11 = 121 \)
Third multiple \( = 11(x+2) = 11(10+2) = 11 \times 12 = 132 \)
So, the three consecutive multiples of 11 are 110, 121, and 132. These numbers sum up to 363.
In simple words: Let the three multiples be \( 11x, 11(x+1), \) and \( 11(x+2) \). Add them together and set the sum equal to 363. Solve for x, which is 10. Then, calculate the three multiples: 110, 121, and 132.

🎯 Exam Tip: When dealing with consecutive multiples, represent them as \( nx, n(x+1), n(x+2) \) where n is the multiple (e.g., 11 in this case).