CBSE Class 8 Mathematics Linear Equations In One Variable Worksheet Set 05

Linear Equations in One Variable

Question 1. Solve the equation: \(\frac{1}{x} + \frac{2}{x} = 3\)
Answer: Combine the fractions on the left side:
\(\frac{1 + 2}{x} = 3\)
\(\frac{3}{x} = 3\)
Multiply both sides by \(x\):
\(3 = 3x\)
Divide by 3:
\(x = 1\)
In simple words: Add the terms with common denominators and solve for \(x\).

Exam Tip: Always check if the solution makes any denominator zero. Here \(x = 1\) is valid.

Question 2. The two third of a number increased by 9 equals 19. Find the number.
Answer: Let the number be \(x\).
According to the question:
\(\frac{2}{3}x + 9 = 19\)
Subtract 9 from both sides:
\(\frac{2}{3}x = 19 - 9\)
\(\frac{2}{3}x = 10\)
Multiply by 3 and divide by 2:
\(x = \frac{10 \times 3}{2} = 15\)
In simple words: Translate the sentence into an algebraic equation step by step.

Exam Tip: Define the variable clearly at the beginning of word problems.

Question 3. Solve the equation: \(2x - (3 - 4x) = \frac{6}{3}\)
Answer: Simplify both sides of the equation:
\(2x - 3 + 4x = 2\)
Combine like terms:
\(6x - 3 = 2\)
Add 3 to both sides:
\(6x = 5\)
\(x = \frac{5}{6}\)
In simple words: Open the bracket carefully by changing signs, then isolate \(x\).

Exam Tip: Pay close attention to negative signs outside brackets.

Question 4. Solve the equation: \(\frac{2x}{3x + 1} = -3\)
Answer: Multiply both sides by \(3x + 1\):
\(2x = -3(3x + 1)\)
Expand the right side:
\(2x = -9x - 3\)
Add \(9x\) to both sides:
\(11x = -3\)
\(x = -\frac{3}{11}\)
In simple words: Cross-multiply to remove the denominator, then solve normally.

Exam Tip: State the restriction that \(3x + 1 \neq 0\).

Question 5. Solve the equation: \(5x - 3 = 3x - 5\)
Answer: Transpose \(3x\) to the left side and \(-3\) to the right side:
\(5x - 3x = -5 + 3\)
\(2x = -2\)
Divide by 2:
\(x = -1\)
In simple words: Bring all variable terms to one side and constants to the other side.

Exam Tip: When moving terms across the equals sign, always change their signs.

Question 6. Solve the equation: \(\frac{5x - 7}{3x} = 2\)
Answer: Multiply both sides by \(3x\):
\(5x - 7 = 2(3x)\)
\(5x - 7 = 6x\)
Transpose \(5x\) to the right side:
\(-7 = 6x - 5x\)
\(x = -7\)
In simple words: Multiply both sides by the denominator to clear fractions.

Question 7. Solve the equation: \(\frac{x}{5} + 1 = \frac{1}{15}\)
Answer: Transpose 1 to the right side:
\(\frac{x}{5} = \frac{1}{15} - 1\)
Take LCM on the right side:
\(\frac{x}{5} = \frac{1 - 15}{15}\)
\(\frac{x}{5} = \frac{-14}{15}\)
Multiply both sides by 5:
\(x = \frac{-14 \times 5}{15}\)
\(x = -\frac{14}{3}\)
In simple words: Isolate the variable term first, simplify, and then solve for \(x\).

Exam Tip: Simplify fractions at the final step to keep numbers manageable.

Question 8. Solve the equation: \(x - 2x + 2 - \frac{16}{2}x = 3 - \frac{7}{2}x\)
Answer: Simplify the terms on the left side (\(\frac{16}{2}x = 8x\)):
\(x - 2x + 2 - 8x = 3 - \frac{7}{2}x\)
\(-9x + 2 = 3 - \frac{7}{2}x\)
Bring all \(x\) terms to the left and constants to the right:
\(\frac{7}{2}x - 9x = 3 - 2\)
Take LCM on the left side (\(\frac{7x - 18x}{2}\)):
\(\frac{-11x}{2} = 1\)
Multiply both sides by 2 and divide by \(-11\):
\(x = -\frac{2}{11}\)
In simple words: Simplify like terms first before gathering variables on one side.

Exam Tip: Be careful with fraction addition and subtraction.

Question 9. Solve the equation: \(\frac{x^2 + 4}{3x^2 + 7} = \frac{1}{2}\)
Answer: Cross-multiply both sides:
\(2(x^2 + 4) = 1(3x^2 + 7)\)
Expand both sides:
\(2x^2 + 8 = 3x^2 + 7\)
Rearrange terms:
\(8 - 7 = 3x^2 - 2x^2\)
\(x^2 = 1\)
\(x = \pm 1\)
In simple words: Cross-multiply to remove fractions, then solve for \(x^2\).

Exam Tip: Remember that taking square roots gives both positive and negative values.

Question 10. Solve the equation: \(\frac{\frac{3}{4}y + 7}{\frac{2}{5}y - 4} = \frac{5}{4}\)
Answer: Cross-multiply:
\(4\left(\frac{3}{4}y + 7\right) = 5\left(\frac{2}{5}y - 4\right)\)
Expand the brackets:
\(3y + 28 = 2y - 20\)
Transpose \(2y\) to the left and 28 to the right:
\(3y - 2y = -20 - 28\)
\(y = -48\)
In simple words: Cross-multiply and distribute the numbers to eliminate fractions.

Exam Tip: Distributing carefully prevents sign errors.

Question 11. If \(\frac{1}{2}\) is subtracted from a number and the difference is multiplied by 4, the result is 5. Find the number.
Answer: Let the number be \(x\).
According to the question:
\(4\left(x - \frac{1}{2}\right) = 5\)
Divide both sides by 4:
\(x - \frac{1}{2} = \frac{5}{4}\)
Add \(\frac{1}{2}\) to both sides:
\(x = \frac{5}{4} + \frac{1}{2}\)
\(x = \frac{5 + 2}{4} = \frac{7}{4}\)
In simple words: Form the equation exactly as stated and solve for the unknown.

Exam Tip: You can either expand 4 first or divide by 4; dividing first is quicker.

Question 12. Solve the equation: \(0.6x + 0.8 = 0.28x + 1.16\)
Answer: Subtract \(0.28x\) from both sides:
\(0.6x - 0.28x + 0.8 = 1.16\)
\(0.32x + 0.8 = 1.16\)
Subtract 0.8 from both sides:
\(0.32x = 1.16 - 0.8\)
\(0.32x = 0.36\)
Divide by 0.32:
\(x = \frac{0.36}{0.32} = \frac{36}{32} = \frac{9}{8} = 1.125\)
In simple words: Group decimal terms together and simplify.

Question 13. Solve the equation: \(\frac{3x + 5}{2x + 7} = 4\)
Answer: Multiply both sides by \(2x + 7\):
\(3x + 5 = 4(2x + 7)\)
\(3x + 5 = 8x + 28\)
Transpose terms:
\(5 - 28 = 8x - 3x\)
\(-23 = 5x\)
\(x = -\frac{23}{5}\)
In simple words: Clear the denominator by cross-multiplication and solve for \(x\).

Exam Tip: Always double-check your distribution steps.

Question 14. Solve the equation: \(4(3w + 2) - 5(6w - 1) = 2(w - 8) - 6(7w - 4) + 4w\)
Answer: Expand all brackets:
\(12w + 8 - 30w + 5 = 2w - 16 - 42w + 24 + 4w\)
Combine like terms on both sides:
Left side: \(-18w + 13\)
Right side: \(-36w + 8\)
Equation becomes:
\(-18w + 13 = -36w + 8\)
Add \(36w\) to both sides and subtract 13:
\(36w - 18w = 8 - 13\)
\(18w = -5\)
\(w = -\frac{5}{18}\)
In simple words: Expand every parenthesis carefully, combine like terms, and solve.

Exam Tip: Watch out for negative signs when distributing numbers into parentheses.

Question 15. The sum of two numbers is 52. The second number is 10 more than the first. Find the numbers.
Answer: Let the first number be \(x\).
Then the second number is \(x + 10\).
According to the question:
\(x + (x + 10) = 52\)
\(2x + 10 = 52\)
\(2x = 42\)
\(x = 21\)
First number = 21
Second number = \(21 + 10 = 31$
In simple words: Assume the first number is \(x\), express the second in terms of \(x\), and set up their sum.

Exam Tip: Always state both final answers clearly.

Question 16. The sum of three consecutive even integers is 270. Find the integers.
Answer: Let the three consecutive even integers be \(x\), \(x + 2\), and \(x + 4\).
According to the question:
\(x + (x + 2) + (x + 4) = 270\)
\(3x + 6 = 270\)
\(3x = 264\)
\(x = 88\)
The integers are 88, 90, and 92.
In simple words: Consecutive even integers differ by 2, so represent them as \(x\), \(x+2\), and \(x+4\).

Exam Tip: Consecutive integers (whether even, odd, or regular) always increase by adding 2 or 1 respectively.

Question 17. The sum of three consecutive integers is 246. Find the integers.
Answer: Let the three consecutive integers be \(x\), \(x + 1\), and \(x + 2\).
According to the question:
\(x + (x + 1) + (x + 2) = 246\)
\(3x + 3 = 246\)
\(3x = 243\)
\(x = 81\)
The integers are 81, 82, and 83.
In simple words: Consecutive integers differ by 1.

Exam Tip: Verify the sum: \(81 + 82 + 83 = 246\).

Question 18. The sum of three consecutive multiples of 7 is 777, find the numbers.
Answer: Let the three consecutive multiples of 7 be \(7x\), \(7(x + 1)\), and \(7(x + 2)\), or simply \(x\), \(x + 7$, and \(x + 14\).
Using \(x\), \(x + 7\), and \(x + 14\):
\(x + (x + 7) + (x + 14) = 777\)
\(3x + 21 = 777\)
\(3x = 756\)
\(x = 252\)
The numbers are 252, \(252 + 7 = 259\), and \(252 + 14 = 266$.
In simple words: Multiples of 7 increase by 7 each time.

Exam Tip: Ensure all three numbers are divisible by 7.

Question 19. The ratio of two complementary angles is 4 : 5. Find these angles and ratio of their supplementary angles.
Answer: Let the angles be \(4x$ and \(5x\).
Since they are complementary, their sum is 90 degrees:
\(4x + 5x = 90^\circ\)
\(9x = 90^\circ\)
\(x = 10^\circ\)
Angles are \(4(10^\circ) = 40^\circ\) and \(5(10^\circ) = 50^\circ\).
Their supplementary angles are:\br /> Supplement of 40 degrees = \(180^\circ - 40^\circ = 140^\circ\)
Supplement of 50 degrees = \(180^\circ - 50^\circ = 130^\circ\)
Ratio of their supplementary angles = \(\frac{140}{130} = 14 : 13\)
In simple words: Use the given ratio to express the angles, find their values using the 90 degree rule, then calculate their supplements.

Exam Tip: Complementary angles sum to 90 degrees and supplementary angles sum to 180 degrees.

Question 20. The sum of the digits of a two-digit number is 14. If 36 is added to the number, its digits are interchanged. Find the number.
Answer: Let the units digit be \(x\). Since the sum of digits is 14, the tens digit is \(14 - x\).
Original number = \(10(14 - x) + x = 140 - 10x + x = 140 - 9x\).
Number with interchanged digits = \(10x + (14 - x) = 9x + 14\).
According to the question, adding 36 to the original number gives the interchanged number:
\((140 - 9x) + 36 = 9x + 14\)
\(176 - 9x = 9x + 14\)
\(176 - 14 = 9x + 9x\)
\(162 = 18x\)
\(x = 9\)
Units digit = 9, Tens digit = \(14 - 9 = 5\).
The number is 59.
In simple words: Represent a two-digit number using its tens and units digits, form the equation based on digit reversal, and solve.

Exam Tip: A two-digit number is always written as \(10 \times (\text{tens digit}) + (\text{units digit})\).

Question 21. Solve for x: \(4x + 7 = 10 + 2x\)
Answer: Subtract \(2x$ from both sides:
\(4x - 2x + 7 = 10\)
\(2x + 7 = 10\)
Subtract 7 from both sides:
\(2x = 3\)
\(x = \frac{3}{2}\)
In simple words: Group variable terms on the left and constants on the right.

Exam Tip: Keep fractions in improper form unless decimals are requested.

Question 22. Solve for p: \(0.25(4p - 3) = 0.05(10p - 9)\)
Answer: Multiply both sides by 100 to remove decimals:
\(25(4p - 3) = 5(10p - 9)\)
Divide both sides by 5:
\(5(4p - 3) = 10p - 9\)
Expand brackets:
\(20p - 15 = 10p - 9\)
Transpose \(10p$ to the left and \(-15\) to the right:
\(20p - 10p = 15 - 9\)
\(10p = 6\)
\(p = \frac{6}{10} = \frac{3}{5} = 0.6\)
In simple words: Clear decimals first by multiplying by a power of 10 to make calculations easier.

Question 23. The perimeter of a triangle is 49 cm. The one side is 7 cm longer than the other and 5 cm shorter than the third. Find the length of each side of the triangle.
Answer: Let the second side be \(x\).
First side is 7 cm longer than the second side: \(x + 7\).
Second side is 5 cm shorter than the third side, meaning the third side is 5 cm longer than the second side: \(x + 5\).
Perimeter = Sum of all three sides = 49 cm.
\((x + 7) + x + (x + 5) = 49\)
\(3x + 12 = 49\)
\(3x = 37\)
\(x = \frac{37}{3} \text{ cm}\)
First side = \(\frac{37}{3} + 7 = \frac{58}{3}\) cm
Second side = \(\frac{37}{3}\) cm
Third side = \(\frac{37}{3} + 5 = \frac{52}{3}\) cm
In simple words: Express all three sides in terms of a single variable, then use the perimeter formula.

Exam Tip: Read carefully whether a side is longer or shorter relative to the chosen reference variable.

Question 24. Nisha has a rectangular plot of land that has been fenced with 300 m long wires. Find the dimensions of the plot, if its length is twice the breadth.
Answer: Let the breadth of the plot be \(x\) m.
Then the length is \(2x\) m.
Perimeter = Fence length = 300 m.
Formula for perimeter of rectangle = \(2(\text{length} + \text{breadth}) = 300$
\(2(2x + x) = 300\)
\(2(3x) = 300\)
\(6x = 300\)
\(x = 50\)
Breadth = 50 m
Length = \(2 \times 50 = 100\) m
In simple words: The length of the wire used for fencing represents the perimeter of the rectangle.

Exam Tip: Always include proper units (meters) in your final dimensions.

Question 25. Each side of a triangle is increased by 10 cm; if the ratio of the perimeter of the new triangle and the given triangle is 5 : 4, find the perimeter of the given triangle.
Answer: Let the perimeter of the given triangle be \(P\).
A triangle has 3 sides. When each side is increased by 10 cm, the total perimeter increases by \(3 \times 10 = 30\) cm.
Perimeter of the new triangle = \(P + 30\).
Given ratio of new perimeter to old perimeter = \(5 : 4$:
\(\frac{P + 30}{P} = \frac{5}{4}\)
Cross-multiply:
\(4(P + 30) = 5P\)
\(4P + 120 = 5P\)
\(120 = 5P - 4P\)
\(P = 120 \text{ cm}\)
In simple words: Increasing each of the 3 sides by 10 increases the total perimeter by 30. Use the given ratio to solve for the original perimeter.

Question 26. The length of a rectangle is 15 cm more than its width. The perimeter is 150 cm. Find the dimensions of the rectangle.
Answer: Let the width of the rectangle be \(x$ cm.
Length = \(x + 15\) cm.
Perimeter = \(2(\text{length} + \text{width}) = 150$
\(2(x + 15 + x) = 150\)
\(2(2x + 15) = 150\)
Divide by 2:
\(2x + 15 = 75\)
\(2x = 60\)
\(x = 30\)
Width = 30 cm
Length = \(30 + 15 = 45\) cm
In simple words: Set width as \(x\), write length in terms of \(x\), and apply the perimeter formula.

Exam Tip: Verify the perimeter: \(2(45 + 30) = 2(75) = 150\) cm.

Question 27. The sum of two numbers is 184. One-third of one number exceeds one-seventh of the other number by 8. Find the two numbers.
Answer: Let one number be \(x\).
Then the other number is \(184 - x\).
According to the question, one-third of the first number exceeds one-seventh of the second number by 8:
\(\frac{1}{3}x - \frac{1}{7}(184 - x) = 8\)
Take LCM of 3 and 7, which is 21:
\(\frac{7x - 3(184 - x)}{21} = 8\)
\(7x - 552 + 3x = 8 \times 21\)
\(10x - 552 = 168\)
\(10x = 168 + 552\)
\(10x = 720\)
\(x = 72\)
First number = 72
Second number = \(184 - 72 = 112$
In simple words: Define the numbers as \(x\) and \(184-x\), then set up the difference equation.

Exam Tip: Multiply the entire equation by the LCM to clear fractions quickly.

Question 28. Solve for x: \(3(5x - 5) - 2(9x - 8) = 4(8x - 13) - 17\)
Answer: Expand all parentheses:
\(15x - 15 - 18x + 16 = 32x - 52 - 17\)
Combine like terms on both sides:
\(-3x + 1 = 32x - 69\)
Transpose \(-3x\) to the right and \(-69\) to the left:
\(1 + 69 = 32x + 3x\)
\(70 = 35x\)
\(x = 2\)
In simple words: Expand carefully, combine like terms, and bring all variables to one side.

Exam Tip: Double-check negative signs when multiplying across brackets.

Question 29. The present ages of Veeru and Neeru are in the ratio 3:4. Five years from now, the ratio of their ages will be 4:5. Find their present ages.
Answer: Let Veeru's present age be \(3x\) and Neeru's present age be \(4x\).
Five years from now:
Veeru's age = \(3x + 5$
Neeru's age = \(4x + 5$
Given ratio after 5 years is 4 : 5:
\(\frac{3x + 5}{4x + 5} = \frac{4}{5}\)
Cross-multiply:
\(5(3x + 5) = 4(4x + 5)\)
\(15x + 25 = 16x + 20\)
Transpose terms:
\(25 - 20 = 16x - 15x\)
\(x = 5\)
Veeru's present age = \(3 \times 5 = 15\) years
Neeru's present age = \(4 \times 5 = 20\) years
In simple words: Use ratio variables \(3x\) and \(4x\), add 5 to both, and equate to the new ratio.

Exam Tip: Always check if the future ratio matches the calculated ages (\(15+5 = 20\), \(20+5 = 25\), ratio is \(20:25 = 4:5\)).

Question 30. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.
Answer: Let the numerator be \(x\).
Then the denominator is \(x + 8\).
Original rational number = \(\frac{x}{x + 8}\).
According to the question, when numerator is increased by 17 and denominator is decreased by 1, the fraction becomes \(\frac{3}{2}\):
\(\frac{x + 17}{(x + 8) - 1} = \frac{3}{2}\)
\(\frac{x + 17}{x + 7} = \frac{3}{2}\)
Cross-multiply:
\(2(x + 17) = 3(x + 7)\)
\(2x + 34 = 3x + 21\)
Transpose terms:
\(34 - 21 = 3x - 2x\)
\(x = 13\)
Numerator = 13
Denominator = \(13 + 8 = 21$
The rational number is \(\frac{13}{21}\).
In simple words: Set numerator as \(x\), denominator as \(x+8\), apply the given changes, and cross-multiply.

Exam Tip: Write the final answer in fractional form.

Question 31. The sum of the weights of an iron piece and of a copper piece is 1280 gm. The volume of the copper piece is twice that of the iron piece. If the weight of 1 cubic centimeter of iron is 7.8 gm and that of copper is 8.9 gm. Find the volume of each piece.
Answer: Let the volume of the iron piece be \(V\) cubic cm.
Then the volume of the copper piece is \(2V\) cubic cm.
Weight of iron piece = \(\text{Volume} \times \text{Density} = 7.8V\) gm.
Weight of copper piece = \(2V \times 8.9 = 17.8V\) gm.
Total weight = \(7.8V + 17.8V = 1280$
\(25.6V = 1280\)
\(V = \frac{1280}{25.6} = 50 \text{ cubic cm}\)
Volume of iron piece = 50 cubic cm.
Volume of copper piece = \(2 \times 50 = 100\) cubic cm.
In simple words: Calculate the weight of each piece by multiplying volume by density, then set their sum to 1280.

Exam Tip: \(\text{Weight} = \text{Volume} \times \text{Density}\).

Question 32. Srishti has a total of Rs 780 as currency notes in the denominations of Rs 10, Rs 20 and Rs 50. The ratio of the number of Rs 50 notes and Rs 10 notes is 3 : 2. If she has a total of 32 notes, how many notes of each denomination she has?
Answer: Let the number of Rs 50 notes be \(3x\) and the number of Rs 10 notes be \(2x\).
Total number of notes is 32. Therefore, the number of Rs 20 notes = \(32 - (3x + 2x) = 32 - 5x$.
Total amount = \(50(3x) + 10(2x) + 20(32 - 5x) = 780$
\(150x + 20x + 640 - 100x = 780\)
\(70x + 640 = 780\)
\(70x = 140\)
\(x = 2\)
Number of Rs 50 notes = \(3 \times 2 = 6$
Number of Rs 10 notes = \(2 \times 2 = 4$
Number of Rs 20 notes = \(32 - 5(2) = 22$
In simple words: Express note counts using ratio variables, find the remaining notes for Rs 20, and equate total value to Rs 780.

Exam Tip: Verify total value: \((6 \times 50) + (4 \times 10) + (22 \times 20) = 300 + 40 + 440 = 780\).

Question 33. Reeta received a certain amount of money on her retirement from her company. She gives half of this money and additional sum of Rs 10,000 to her daughter. She also gives one third of the money received and an additional sum of Rs 3000 to her son. If the daughter gets twice as much as her son, find the total amount of money Reeta received from her company.
Answer: Let the total amount received be Rs \(x\).
Amount given to daughter = \(\frac{x}{2} + 10000\).
Amount given to son = \(\frac{x}{3} + 3000\).
Given that the daughter gets twice as much as her son:
\(\frac{x}{2} + 10000 = 2\left(\frac{x}{3} + 3000\right)\)
\(\frac{x}{2} + 10000 = \frac{2x}{3} + 6000\)
Rearrange terms:
\(10000 - 6000 = \frac{2x}{3} - \frac{x}{2}\)
\(4000 = \frac{4x - 3x}{6}\)
\(4000 = \frac{x}{6}\)
\(x = 24000\)
Total amount received = Rs 24,000.
In simple words: Set up expressions for the daughter's and son's shares, then set daughter's share equal to twice the son's share.

Exam Tip: Carefully translate phrases like "half of this money and additional sum" into mathematical expressions.

Question 34. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.
Answer: Let the numerator be \(x\).
Denominator = \(x + 8\).
Original fraction = \(\frac{x}{x + 8}\).
According to the question:
\(\frac{x + 17}{(x + 8) - 1} = \frac{3}{2}\)
\(\frac{x + 17}{x + 7} = \frac{3}{2}\)
Cross-multiply:
\(2(x + 17) = 3(x + 7)\)
\(2x + 34 = 3x + 21\)
\(34 - 21 = 3x - 2x\)
\(x = 13\)
Numerator = 13, Denominator = \(13 + 8 = 21$
The rational number is \(\frac{13}{21}\).
In simple words: This is the same problem structure as Question 30; form the equation and cross-multiply.

Exam Tip: Double check substitution after solving.

Question 35. Five years before, the age of Neeraj and Neera was in the ratio 4:5. The ratio of their present ages is 5:6. Find their present ages.
Answer: Let the present ages of Neeraj and Neera be \(5x\) and \(6x\) respectively.
Five years before:
Neeraj's age = \(5x - 5$
Neera's age = \(6x - 5$
Given that 5 years ago the ratio of their ages was 4 : 5:
\(\frac{5x - 5}{6x - 5} = \frac{4}{5}\)
Cross-multiply:
\(5(5x - 5) = 4(6x - 5)\)
\(25x - 25 = 24x - 20\)
\(25x - 24x = 25 - 20\)
\(x = 5\)
Neeraj's present age = \(5 \times 5 = 25$ years
Neera's present age = \(6 \times 5 = 30$ years
In simple words: Use present age ratio as \(5x\) and \(6x\), subtract 5 for past ages, and set up the ratio equation.

Exam Tip: Past ages require subtracting the given number of years from present ages.

Question 36. The sum of three consecutive multiples of 8 is 792. Find these multiples.
Answer: Let the three consecutive multiples of 8 be \(x\), \(x + 8\), and \(x + 16\).
According to the question:
\(x + (x + 8) + (x + 16) = 792\)
\(3x + 24 = 792\)
\(3x = 768\)
\(x = 256\)
The multiples are 256, \(256 + 8 = 264\), and \(256 + 16 = 272\).
In simple words: Consecutive multiples of 8 differ by 8.

Exam Tip: Confirm your answers are all multiples of 8.

Question 37. The sum of three consecutive multiples of 9 is 795. Find these multiples.
Answer: Let the three consecutive multiples of 9 be \(x\), \(x + 9\), and \(x + 18\).
According to the question:
\(x + (x + 9) + (x + 18) = 795\)
\(3x + 27 = 795\)
\(3x = 768\)
\(x = 256\)
The numbers are 256, 265, and 274. (Note: 256 is not a multiple of 9. Let us re-evaluate: \(768 / 3 = 256\). Wait, let's set it as \(9x, 9(x+1), 9(x+2)\):
\(9(3x + 3) = 795\)
\(3x + 3 = \frac{795}{9} = 88.33\) (Source OCR has a slight numerical discrepancy in the original worksheet prompt, but solving strictly: \(3x + 27 = 795 \implies 3x = 768 \implies x = 256\). If strictly multiples of 9: \(9a, 9a+9, 9a+18 \implies 27a + 27 = 795 \implies 27a = 768 \implies\) non-integer. Standard curriculum question usually has clean numbers; we present the algebraic solution directly based on \(x + (x+9) + (x+18) = 795\)).
In simple words: Set up the consecutive multiples and solve the linear equation.

Exam Tip: Always check if the resulting values are correct multiples of the given base number.

Question 38. Five years before, the age of Neeraj and Neera was in the ratio 4:5. The ratio of their present ages is 5:6. Find their present ages.
Answer: (Identical to Question 35)
Present ages are 25 years and 30 years.
In simple words: Solved previously as Question 35.

Exam Tip: Repeated questions share identical methods.

Question 39. The present ages of Veeru and Neeru are in the ratio 3:4. Five years from now, the ratio of their ages will be 4:5. Find their present ages.
Answer: (Identical to Question 29)
Present ages are 15 years and 20 years.
In simple words: Solved previously as Question 29.

Exam Tip: Repeated questions share identical methods.

Most Important Questions

Question 1. Solve: \(-3(x + 2) = -12\)
Answer: Divide both sides by \(-3\):
\(x + 2 = \frac{-12}{-3}\)
\(x + 2 = 4\)
Subtract 2:
\(x = 2\)
In simple words: Divide first by \(-3\) to simplify the equation quickly.

Exam Tip: Dividing by a negative number turns it positive.

Question 2. \(5a + 5 = 20\)
Answer: Subtract 5 from both sides:
\(5a = 15\)
Divide by 5:
\(a = 3\)
In simple words: Isolate the term with \(a\) and solve.

Exam Tip: Basic linear equation isolation step.

Question 3. \(-2 + g = 7\)
Answer: Add 2 to both sides:
\(g = 7 + 2\)
\(g = 9\)
In simple words: Transpose \(-2\) to the right side.

Exam Tip: Negative term becomes positive when transposed.

Question 4. \(7a + 2 = -12\)
Answer: Subtract 2 from both sides:
\(7a = -14\)
Divide by 7:
\(a = -2\)
In simple words: Isolate \(7a\) and divide.

Exam Tip: Negative divided by positive is negative.

Question 5. If \(3x + 2 = 12\), then find the value of \(6x + 4\).
Answer: Method 1: Solve for \(3x\):
\(3x = 12 - 2 = 10\)
Notice that \(6x + 4 = 2(3x) + 4\):
\(= 2(10) + 4 = 20 + 4 = 24\)
In simple words: Instead of finding \(x\), notice that the target expression is twice the first equation plus 4.

Exam Tip: Look for algebraic shortcuts before solving fully.

Question 6. Solve the following equation: \(2(7b + 12) = 24\)
Answer: Divide both sides by 2:
\(7b + 12 = 12\)
Subtract 12:
\(7b = 0\)
\(b = 0\)
In simple words: Divide by 2 first to avoid unnecessary multiplication.

Exam Tip: Zero is a valid solution in linear equations.

Question 7. \(4 - 5x = -21\)
Answer: Subtract 4 from both sides:
\(-5x = -25\)
Divide by \(-5\):
\(x = 5\)
In simple words: Isolate the \(x\) term and divide by the coefficient.

Exam Tip: Two negatives make a positive during division.

Question 8. If \(11x - 7 = 26\), then find the value of the expression \((x - 3)\).
Answer: Solve for \(11x\):
\(11x = 26 + 7 = 33\)
\(x = \frac{33}{11} = 3\)
Now find \(x - 3\):
\(3 - 3 = 0\)
In simple words: Find the value of \(x\) first, then substitute into \((x - 3)\).

Exam Tip: Always evaluate the exact expression requested.

Question 9. Sum of three consecutive even integers is 270. Find the integers.
Answer: (Identical to Question 16)
Integers are 88, 90, and 92.
In simple words: Solved previously as Question 16.

Exam Tip: Repeated problems reinforce core concepts.

Question 10. Sum of two numbers is 52; if second number is 10 more than first, find the number.
Answer: (Identical to Question 15)
Numbers are 21 and 31.
In simple words: Solved previously as Question 15.

Exam Tip: State both required numbers clearly.

Question 11. Sum of three consecutive multiples of 7 is 777 find the numbers.
Answer: (Identical to Question 18)
Numbers are 252, 259, and 266.
In simple words: Solved previously as Question 18.

Question 12. The ratio of two complementary angles is 4 : 5. Find these angles and ratio of their supplementary angles.
Answer: (Identical to Question 19)
Angles are \(40^\circ\) and \(50^\circ\); ratio of supplements is \(14 : 13\).
In simple words: Solved previously as Question 19.

Exam Tip: Remember angle sum properties.

Question 13. Nisha has rectangular plot of land that has been fenced with 300m long wires. Find the dimensions of the plot, if its length is twice the breadth.
Answer: (Identical to Question 24)
Breadth = 50 m, Length = 100 m.
In simple words: Solved previously as Question 24.

Exam Tip: Perimeter equals total fence length.

Question 14. The length of a rectangle is 15 cm more than its width. The perimeter is 150 cm. Find the dimensions of the rectangle.
Answer: (Identical to Question 26)
Width = 30 cm, Length = 45 cm.
In simple words: Solved previously as Question 26.

Exam Tip: Check perimeter calculations.

Question 15. Naveen's present age is three times of age of his son. After 5 years the difference between the ages of both is 30. Find the present age of Naveen.
Answer: Let son's present age be \(x\).
Naveen's present age = \(3x\).
After 5 years:
Son's age = \(x + 5$
Naveen's age = \(3x + 5$
Difference between their ages after 5 years = 30:
\((3x + 5) - (x + 5) = 30\)
\(2x = 30\)
\(x = 15\)
Naveen's present age = \(3 \times 15 = 45$ years.
In simple words: The difference between two people's ages always remains constant over time.

Exam Tip: Age difference never changes as years pass.

Question 16. Solve the equation: \(5x - 3 = 3x - 5\).
Answer: (Identical to Question 5)
\(x = -1\)
In simple words: Solved previously as Question 5.

Question 17. Solve the equation: \(5x - \frac{3}{2} = 2x + \frac{7}{2}\)
Answer: Transpose \(2x\) to the left and \(-\frac{3}{2}\) to the right:
\(5x - 2x = \frac{7}{2} + \frac{3}{2}\)
\(3x = \frac{10}{2}\)
\(3x = 5\)
\(x = \frac{5}{3}\)
In simple words: Group variable terms and fraction terms together.

Exam Tip: Combine fractions with common denominators directly.

Question 18. Simplify : \(3t + (t - 2) = 2 - 5t\)
Answer: Remove brackets on the left:
\(3t + t - 2 = 2 - 5t\)
\(4t - 2 = 2 - 5t\)
Transpose \(-5t\) to the left and \(-2\) to the right:
\(4t + 5t = 2 + 2\)
\(9t = 4\)
\(t = \frac{4}{9}\)
In simple words: Combine like terms and solve for \(t\).

Exam Tip: Ensure all variable terms are gathered correctly.

Question 19. Solve the equation: \(\frac{3x}{2} - \frac{3}{2} = \frac{5x}{2} + \frac{7}{2}\)
Answer: Multiply the entire equation by 2 to clear denominators:
\(3x - 3 = 5x + 7\)
Transpose \(3x\) to the right and 7 to the left:
\(-3 - 7 = 5x - 3x\)
\(-10 = 2x\)
\(x = -5\)
In simple words: Since all terms share a denominator of 2, multiply everything by 2 to eliminate fractions instantly.

Exam Tip: Clearing denominators early simplifies working with equations.

Question 20. Simplify: \(m = \frac{4}{5}(m - 10)\)
Answer: Multiply both sides by 5:
\(5m = 4(m - 10)\)
Expand the bracket:
\(5m = 4m - 40\)
Subtract \(4m$ from both sides:
\(m = -40\)
In simple words: Multiply by 5 to remove the fraction, then expand and solve.

Exam Tip: Removing denominators first is always recommended.

Question 21. Simplify : \(0.03a + 2 = 8 - 0.02x + 2(0.6x + 1)\)
Answer: Expand the bracket on the right:
\(0.03a + 2 = 8 - 0.02x + 1.2x + 2\)
Combine like terms on the right side:
\(0.03a + 2 = 10 + 1.18x\)
Since this equation contains two variables (\(a$ and \(x\)), it represents a linear relationship between \(a\) and \(x\):
\(0.03a = 1.18x + 8\)
\(a = \frac{1.18x + 8}{0.03} = \frac{118x + 800}{3}\)
In simple words: Expand brackets, combine constants, and express one variable in terms of the other.

Exam Tip: Equations with two variables express one in terms of the other unless additional equations are given.

Question 22. Solve the equation \(0.6x + 0.8 = 0.28x + 1.16\).
Answer: (Identical to Question 12)
\(x = 1.125\)
In simple words: Solved previously as Question 12.

Exam Tip: Decimal equations can be solved by grouping decimal coefficients.

Question 23. Solve the equation: \(4(3w + 2) - 5(6w - 1) = 2(w - 8) - 6(7w - 4) + 4w\)
Answer: (Identical to Question 14)
\(w = -\frac{5}{18}\)
In simple words: Solved previously as Question 14.

Exam Tip: Careful expansion prevents errors.

Question 24. The sum of the digits of a two-digit number is 8. If 18 is subtracted from the number, its digits are interchanged. Find the number.
Answer: Let the units digit be \(x\). Tens digit = \(8 - x\).
Original number = \(10(8 - x) + x = 80 - 9x\).
Interchanged number = \(10x + (8 - x) = 9x + 8\).
According to the question, subtracting 18 from the original number gives the interchanged number:
\((80 - 9x) - 18 = 9x + 8\)
\(62 - 9x = 9x + 8\)
\(62 - 8 = 9x + 9x\)
\(54 = 18x\)
\(x = 3\)
Units digit = 3, Tens digit = \(8 - 3 = 5\).
The number is 53.
In simple words: Set up the number using its digits, subtract 18, and equate to the reversed-digit number.

Exam Tip: Two-digit numbers are represented as \(10t + u\).

Question 25. A steamer is going downstream in a river and cover a distance between two villages in 20 hours and same distance it covers in 25 hours when he return back in upstream if the speed of the river is 4 km/h. Find the distance between two villages.
Answer: Let the speed of the steamer in still water be \(v$ km/h.
Speed of the stream = 4 km/h.
Downstream speed = \(v + 4\) km/h.
Upstream speed = \(v - 4\) km/h.
Distance = \(\text{Speed} \times \text{Time}\). Since the distance is the same in both cases:
\((\text{Downstream speed}) \times 20 = (\text{Upstream speed}) \times 25\)
\((v + 4) \times 20 = (v - 4) \times 25\)
Divide both sides by 5:
\(4(v + 4) = 5(v - 4)\)
\(4v + 16 = 5v - 20\)
\(16 + 20 = 5v - 4v\)
\(v = 36 \text{ km/h}\)
Now, calculate the distance:
Distance = \((36 + 4) \times 20 = 40 \times 20 = 800$ km.
In simple words: Equate the distance calculated from downstream speed and time to the distance from upstream speed and time.

Exam Tip: Downstream speed is steamer speed plus stream speed; upstream speed is steamer speed minus stream speed.

Question 26. Solve the equation: \(\frac{1}{x} + \frac{2}{x} = 3\)
Answer: (Identical to Question 1)
\(x = 1\)
In simple words: Solved previously as Question 1.

Exam Tip: Combine like fraction terms.

Question 27. Solve the equation: \(\frac{\frac{3}{4}y + 7}{\frac{2}{5}y - 4} = \frac{5}{4}\)
Answer: (Identical to Question 10)
\(y = -48\)
In simple words: Solved previously as Question 10.

Exam Tip: Cross-multiply to eliminate fractions.

Question 28. Solve the equation: \(\frac{6x + 3}{2} + 1 = \frac{x - 2}{6}\)
Answer: Multiply the entire equation by the LCM of denominators (which is 6):
\(3(6x + 3) + 6(1) = 1(x - 2)\)
Expand brackets:
\(18x + 9 + 6 = x - 2\)
\(18x + 15 = x - 2\)
Transpose \(x$ to the left and 15 to the right:
\(18x - x = -2 - 15\)
\(17x = -17\)
\(x = -1\)
In simple words: Multiply every term by the LCM to clear fractions in one step.

Exam Tip: Always multiply every single term by the LCM, including whole numbers.

Question 29. If two third of a number is increased by 9, sum will be 19 find the number.
Answer: (Identical to Question 2)
Number = 15.
In simple words: Solved previously as Question 2.

Exam Tip: Translate statements directly into linear equations.

Question 30. Solve the equation : \(\frac{a}{2} - \frac{3a}{4} = 21 - \frac{5a}{6}\)
Answer: Find the LCM of denominators 2, 4, and 6, which is 12.
Multiply every term by 12:
\(12\left(\frac{a}{2}\right) - 12\left(\frac{3a}{4}\right) = 12(21) - 12\left(\frac{5a}{6}\right)\)
\(6a - 9a = 252 - 10a\)
\(-3a = 252 - 10a\)
Transpose \(-10a\) to the left:
\(-3a + 10a = 252\)
\(7a = 252\)
\(a = 36\)
In simple words: Multiply all terms by the LCM of denominators to clear fractions.

Exam Tip: Clearing denominators makes fraction equations much easier to solve.

Question 31. If \(\frac{1}{2}\) is subtracted from a number and difference is multiple by 4, the result is 5 find the number.
Answer: (Identical to Question 11)
Number = \(\frac{7}{4}\).
In simple words: Solved previously as Question 11.

Exam Tip: Set up parentheses correctly according to word placement.

Question 32. Solve the equation : \(x - 2x + 2 - \frac{16}{2}x = 3 - \frac{7}{2}x\)
Answer: (Identical to Question 8)
\(x = -\frac{2}{11}\)
In simple words: Solved previously as Question 8.

Exam Tip: Simplify constant coefficients before gathering variables.

Question 33. Solve the equation: \(y - \frac{y - 1}{2} = 1 - \frac{y - 2}{3}\)
Answer: Multiply the entire equation by the LCM of 2 and 3, which is 6:
\(6y - 3(y - 1) = 6 - 2(y - 2)\)
Expand brackets:
\(6y - 3y + 3 = 6 - 2y + 4\)
\(3y + 3 = 10 - 2y\)
Transpose \(-2y\) to the left and 3 to the right:
\(3y + 2y = 10 - 3\)
\(5y = 7\)
\(y = \frac{7}{5}\)
In simple words: Clear denominators by multiplying by the LCM, being careful with negative signs in numerators.

Exam Tip: Put brackets around numerators when multiplying by negative coefficients.

Question 34. The sum of the weights of an iron piece and of a copper piece is 1280 gm. The volume of the copper piece is twice that of the iron piece. If the weight of 1 cubic centimeter of iron is 7.8 gm and that of copper is 8.9 gm, find the volume of each piece.
Answer: (Identical to Question 31)
Iron volume = 50 cubic cm, Copper volume = 100 cubic cm.
In simple words: Solved previously as Question 31.

Exam Tip: Density times volume gives weight.

Question 35. Reeta receives a certain amount of money on her retirement from her company. She gives half of this money and additional sum of Rs. 10000 to her daughter. She also gives one third of the money received and an additional sum of Rs. 3000 to her son. If the daughter gets twice as much as her son, find the total amount of money Reeta received from her company.
Answer: (Identical to Question 33)
Total amount = Rs 24,000.
In simple words: Solved previously as Question 33.

Exam Tip: Read word problems carefully to set up correct equations.

Question 36. Present ages of Seema and Sharad are in the ratio 5 : 6. Twelve years from now the ratio of their ages will be 9 : 10. Find their present ages.
Answer: Let Seema's present age be \(5x\) and Sharad's present age be \(6x\).
Twelve years from now:
Seema's age = \(5x + 12$
Sharad's age = \(6x + 12$
Given ratio after 12 years is 9 : 10:
\(\frac{5x + 12}{6x + 12} = \frac{9}{10}\)
Cross-multiply:
\(10(5x + 12) = 9(6x + 12)\)
\(50x + 120 = 54x + 108\)
Transpose terms:
\(120 - 108 = 54x - 50x\)
\(12 = 4x\)
\(x = 3\)
Seema's present age = \(5 \times 3 = 15\) years
Sharad's present age = \(6 \times 3 = 18\) years
In simple words: Express present ages using ratio variables, add 12 for future ages, and equate to the given future ratio.

Exam Tip: Always verify future age ratios to check your work.