CBSE Class 8 Mathematics Linear Equations Assignment Set 02

2. Linear Equations in One Variable

Q 1. Solve the equation: \(\frac{1}{x} + \frac{2}{x} = 3\)
Answer: Combine the left side fractions: \(\frac{1 + 2}{x} = 3\)
\(\frac{3}{x} = 3\)
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 to get 1.

Exam Tip: Always combine like terms with the same denominator first before cross multiplying.

Q 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: \(2x = 30\)
Divide by 2: \(x = 15\)
In simple words: Translate the sentence into an algebraic equation and isolate x.

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

Q 3. Solve the equation: \(2x - (3 - 4x) = \frac{6}{3} - \frac{2}{3}\)
Answer: Expand the brackets: \(2x - 3 + 4x = 2 - \frac{2}{3}\)
Combine like terms: \(6x - 3 = \frac{6 - 2}{3}\)
\(6x - 3 = \frac{4}{3}\)
Add 3 to both sides: \(6x = \frac{4}{3} + 3 = \frac{4 + 9}{3} = \frac{13}{3}\)
Divide by 6: \(x = \frac{13}{18}\)
In simple words: Remove parentheses by distributing negative signs, then collect variable terms on one side.

Q 4. Solve the equation: \(\frac{2x}{3x + 1} = -3\)
Answer: Cross multiply: \(2x = -3(3x + 1)\)
Expand the right side: \(2x = -9x - 3\)
Add \(9x\) to both sides: \(2x + 9x = -3\)
\(11x = -3\)
Divide by 11: \(x = -\frac{3}{11}\)
In simple words: Cross multiplication turns a rational equation into a linear equation.

Exam Tip: Always state the restriction that the denominator cannot be zero: \(3x + 1 \neq 0 \Rightarrow x \neq -\frac{1}{3}\).

Q 5. Solve the equation: \(5x - 3 = 3x - 5\)
Answer: Transpose \(3x\) to the left and \(-3\) to the right:
\(5x - 3x = -5 + 3\)
\(2x = -2\)
Divide by 2: \(x = -1\)
In simple words: Group all variable terms on the left and constant terms on the right.

Exam Tip: Changing sides of terms requires changing their signs.

Q 6. Solve the equation: \(\frac{5x - 7}{3x} = 2\)
Answer: Cross multiply: \(5x - 7 = 2(3x)\)
\(5x - 7 = 6x\)
Transpose \(5x\) to the right: \(-7 = 6x - 5x\)
\(x = -7\)
In simple words: Multiply both sides by \(3x\) to eliminate the denominator.

Exam Tip: Check your answer by plugging \(x = -7\) back into the original equation.

Q 7. Solve the equation: \(\frac{x}{5} + 1 = \frac{1}{15}\)
Answer: Transpose 1 to the right side:
\(\frac{x}{5} = \frac{1}{15} - 1\)
\(\frac{x}{5} = \frac{1 - 15}{15} = -\frac{14}{15}\)
Multiply both sides by 5: \(x = -\frac{14}{15} \times 5\)
\(x = -\frac{14}{3}\)
In simple words: Subtract 1 from both sides and simplify the fractions.

Exam Tip: Simplify fractions fully at every intermediate step to avoid calculation errors.

Q 8. Solve the equation: \(x - 2x + 2 - \frac{16}{2}x = 3 - \frac{7}{2}x\)
Answer: Simplify terms: Note that \(\frac{16}{2}x = 8x\).
\(x - 2x + 2 - 8x = 3 - \frac{7}{2}x\)
Combine like terms on the left: \(-9x + 2 = 3 - \frac{7}{2}x\)
Bring all \(x\) terms to one side: \(\frac{7}{2}x - 9x = 3 - 2\)
\(\frac{7x - 18x}{2} = 1\)
\(\frac{-11x}{2} = 1\)
Multiply by 2 and divide by -11: \(x = -\frac{2}{11}\)
In simple words: Simplify fractions first, combine like terms, and isolate x.

Exam Tip: Always reduce fraction coefficients like \(\frac{16}{2}\) immediately.

Q 9. Solve the equation: \(\frac{x^2 + 4}{3x^2 + 7} = \frac{1}{2}\)
Answer: Cross multiply:
\(2(x^2 + 4) = 1(3x^2 + 7)\)
\(2x^2 + 8 = 3x^2 + 7\)
Rearrange terms: \(8 - 7 = 3x^2 - 2x^2\)
\(x^2 = 1\)
Taking square root: \(x = 1\) or \(x = -1\)
In simple words: Cross multiply to clear denominators, then solve for \(x^2\).

Exam Tip: Remember that taking a square root gives both positive and negative solutions.

Q 10. Solve the equation: \(\frac{\frac{3}{4}y + 7}{\frac{2}{5}y - 4} = \frac{5}{4}\)
Answer: Cross multiply:
\(4\left(\frac{3}{4}y + 7\right) = 5\left(\frac{2}{5}y - 4\right)\)
\(3y + 28 = 2y - 20\)
Transpose terms: \(3y - 2y = -20 - 28\)
\(y = -48\)
In simple words: Cross multiply to eliminate fractions in the numerator and denominator simultaneously.

Exam Tip: Distributing the outer integers into the fractions cancels out denominators cleanly.

Q 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 condition: \(4\left(x - \frac{1}{2}\right) = 5\)
Divide both sides by 4: \(x - \frac{1}{2} = \frac{5}{4}\)
Add \(\frac{1}{2}\) to both sides: \(x = \frac{5}{4} + \frac{1}{2}\)
\(x = \frac{5}{4} + \frac{2}{4} = \frac{7}{4}\)
In simple words: Form the equation step-by-step according to the wording and solve for x.

Exam Tip: Pay close attention to the order of operations in word statements ("difference is multiplied by 4" means brackets are required).

Q 12. Solve the equation: \(0.6x + 0.8 = 0.28x + 1.16\)
Answer: Multiply the entire equation by 100 to remove decimals:
\(60x + 80 = 28x + 116\)
Transpose terms: \(60x - 28x = 116 - 80\)
\(32x = 36\)
Divide by 32: \(x = \frac{36}{32} = \frac{9}{8} = 1.125\)
In simple words: Clear decimals by multiplying by powers of 10 to make calculations simpler.

Q 13. Solve the equation: \(\frac{3x + 5}{2x + 7} = 4\)
Answer: Cross multiply: \(3x + 5 = 4(2x + 7)\)
\(3x + 5 = 8x + 28\)
Rearrange terms: \(5 - 28 = 8x - 3x\)
\(-23 = 5x\)
\(x = -\frac{23}{5}\)
In simple words: Multiply both sides by the denominator to solve the linear equation.

Exam Tip: Double-check distribution when multiplying terms out of brackets.

Q 14. Solve the equation: \(4(3w + 2) - 5(6w - 1) = 2(w - 8) - 6(7w - 4) + 4w\)
Answer: Expand all parentheses:
\(12w + 8 - 30w + 5 = 2w - 16 - 42w + 24 + 4w\)
Simplify left side: \(-18w + 13\)
Simplify right side: \((2w - 42w + 4w) + (-16 + 24) = -36w + 8\)
Equation becomes: \(-18w + 13 = -36w + 8\)
Bring \(w\) terms to the left: \(-18w + 36w = 8 - 13\)
\(18w = -5\)
\(w = -\frac{5}{18}\)
In simple words: Expand every bracket carefully, combine like terms on each side separately, then solve.

Exam Tip: Grouping like terms before transposing prevents sign errors.

Q 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 \(x + 10\).
Sum of the two numbers: \(x + (x + 10) = 52\)
\(2x + 10 = 52\)
\(2x = 42\)
\(x = 21\)
First number = 21, Second number = \(21 + 10 = 31\).
In simple words: Express both numbers using a single variable and set up their sum equation.

Exam Tip: Always state both final answers clearly at the end.

Q 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 \(x + 4\).
Sum: \(x + (x + 2) + (x + 4) = 270\)
\(3x + 6 = 270\)
\(3x = 264\)
\(x = 88\)
The integers are 88, 90, and 92.
In simple words: Consecutive even numbers differ by 2, so represent them as \(x\), \(x+2\), and \(x+4\).

Exam Tip: Even or odd consecutive numbers both increase by 2.

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

Exam Tip: Verify your answer by adding 81 + 82 + 83 = 246.

Q 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 \(7(x + 2)\).
Sum: \(7x + 7x + 7 + 7x + 14 = 777\)
\(21x + 21 = 777\)
\(21x = 756\)
\(x = 36\)
The numbers are \(7(36) = 252\), \(252 + 7 = 259\), and \(259 + 7 = 266\).
In simple words: Multiples of 7 increase by 7 successively.

Exam Tip: You can also assume the numbers as \(x-7\), \(x\), and \(x+7\) to simplify addition.

Q 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 \(90^\circ\):
\(4x + 5x = 90^\circ\)
\(9x = 90^\circ \Rightarrow x = 10^\circ\)
The angles are \(4(10^\circ) = 40^\circ\) and \(5(10^\circ) = 50^\circ\).
Their supplementary angles are \(180^\circ - 40^\circ = 140^\circ\) and \(180^\circ - 50^\circ = 130^\circ\).
Ratio of their supplementary angles = \(\frac{140}{130} = 14 : 13\).
In simple words: Complementary angles add to 90 degrees and supplementary angles add to 180 degrees.

Exam Tip: Ratios can be converted into algebraic values by introducing a common variable \(x\).

Q 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 \(y\) and the tens digit be \(x\).
The number is \(10x + y\).
Given sum of digits: \(x + y = 14 \Rightarrow y = 14 - x\).
Original number = \(10x + (14 - x) = 9x + 14\).
If 36 is added, digits interchange: number becomes \(10y + x\).
Equation: \((10x + y) + 36 = 10y + x\)
\(9x - 9y = -36 \Rightarrow x - y = -4\)
Substitute \(y = 14 - x\): \(x - (14 - x) = -4\)
\(2x - 14 = -4 \Rightarrow 2x = 10 \Rightarrow x = 5\).
Then \(y = 14 - 5 = 9\).
The number is 59.
In simple words: Represent a two-digit number as \(10x + y\) and use the digit-sum and interchange conditions.

Exam Tip: A number with digits \(x\) and \(y\) is always written as \(10x + y\), not \(xy\).

Q 21. Solve for x: \(4x + 7 = 10 + 2x\)
Answer: Transpose \(2x\) to the left and 7 to the right:
\(4x - 2x = 10 - 7\)
\(2x = 3\)
\(x = \frac{3}{2}\)
In simple words: Group variable terms on one side and constants on the other.

Exam Tip: Keep fractions in exact improper form unless decimal conversion is requested.

Q 22. Solve for p: \(0.25(4p - 3) = 0.05(10p - 9)\)
Answer: Multiply both sides by 100 to clear decimals:
\(25(4p - 3) = 5(10p - 9)\)
Divide both sides by 5:
\(5(4p - 3) = 1(10p - 9)\)
Expand brackets: \(20p - 15 = 10p - 9\)
Transpose terms: \(20p - 10p = 15 - 9\)
\(10p = 6\)
\(p = \frac{6}{10} = \frac{3}{5} = 0.6\)
In simple words: Clear decimals first to make expansion much easier.

Q 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: \(x + 7\).
Second side is 5 cm shorter than the third, meaning the third side is 5 cm longer than the second: \(x + 5\).
Perimeter = sum of all three sides = 49 cm:
\((x + 7) + x + (x + 5) = 49\)
\(3x + 12 = 49\)
\(3x = 37 \Rightarrow x = \frac{37}{3}\) 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 sides in terms of a single middle variable and equate their sum to the perimeter.

Exam Tip: Check your additions: \(\frac{58+37+52}{3} = \frac{147}{3} = 49\).

Q 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 be \(b\).
Length \(l = 2b\).
Perimeter (length of wire) = \(2(l + b) = 300\)
\(2(2b + b) = 300\)
\(2(3b) = 300 \Rightarrow 6b = 300\)
\(b = 50\) m.
Length \(l = 2(50) = 100\) m.
Dimensions are 100 m by 50 m.
In simple words: Perimeter of a rectangle equals twice the sum of length and breadth.

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

Q 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 original triangle be \(P\).
Since a triangle has 3 sides, increasing each side by 10 cm increases the total perimeter by \(3 \times 10 = 30\) cm.
New perimeter = \(P + 30\).
Ratio of new perimeter to original perimeter = \(\frac{P + 30}{P} = \frac{5}{4}\)
Cross multiply: \(4(P + 30) = 5P\)
\(4P + 120 = 5P\)
\(P = 120\) cm.
In simple words: Adding 10 to each of the 3 sides increases the total perimeter by 30 cm.

Q 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 be \(w\).
Length \(l = w + 15\).
Perimeter = \(2(l + w) = 150\)
\(2(w + 15 + w) = 150\)
\(2(2w + 15) = 150\)
\(2w + 15 = 75\)
\(2w = 60 \Rightarrow w = 30\) cm.
Length \(l = 30 + 15 = 45\) cm.
Dimensions are 45 cm and 30 cm.
In simple words: Use the perimeter formula to set up an equation in terms of width.

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

Q 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 the first number be \(x\).
Then the second number is \(184 - x\).
According to the condition: \(\frac{1}{3}x - \frac{1}{7}(184 - x) = 8\)
Multiply through by 21 (LCM of 3 and 7):
\(7x - 3(184 - x) = 168\)
\(7x - 552 + 3x = 168\)
\(10x - 552 = 168\)
\(10x = 168 + 552 = 720\)
\(x = 72\).
First number = 72, Second number = \(184 - 72 = 112\).
In simple words: Express one number as \(x\) and the second as \(184 - x\), then write the fraction difference equation.

Exam Tip: Multiplying by the LCM of denominators eliminates fractions instantly.

Q 28. Solve for x: \(3(5x - 5) - 2(9x - 8) = 4(8x - 13) - 17\)
Answer: Expand all parentheses:
\(15x - 15 - 18x + 16 = 32x - 52 - 17\)
Simplify left side: \(-3x + 1\)
Simplify right side: \(32x - 69\)
Equation: \(-3x + 1 = 32x - 69\)
Transpose terms: \(1 + 69 = 32x + 3x\)
\(70 = 35x\)
\(x = 2\)
In simple words: Expand brackets carefully and group variable terms on one side.

Exam Tip: Always double check signs when distributing negative multipliers.

Q 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, their ages will be \(3x + 5\) and \(4x + 5\).
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 = \(3(5) = 15\) years.
Neeru's present age = \(4(5) = 20\) years.
In simple words: Use ratio constants for present ages and add 5 for future ages.

Exam Tip: Ratio problems are solved easily by introducing a common multiplier \(x\).

Q 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 \(\frac{3}{2}\). 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 condition:
\(\frac{x + 17}{(x + 8) - 1} = \frac{3}{2}\)
\(\frac{x + 17}{x + 7} = \frac{3}{2}\)
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 changes, and cross multiply.

Exam Tip: Always state the final fraction explicitly as the answer.

Q 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 = \(7.8V\) gm.
Weight of copper piece = \(8.9 \times 2V = 17.8V\) gm.
Total weight = \(7.8V + 17.8V = 1280\)
\(25.6V = 1280\)
\(V = \frac{1280}{25.6} = 50\) cubic cm.
Volume of iron piece = 50 cc.
Volume of copper piece = \(2 \times 50 = 100\) cc.
In simple words: Multiply volume by density to get weight, then sum them to equal 1280 gm.

Exam Tip: Weight = Volume \(\times\) Density is the core physical formula used here.

Q 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 = 32, so the number of Rs 20 notes = \(32 - (3x + 2x) = 32 - 5x\).
Total value = Rs 780:
\(50(3x) + 10(2x) + 20(32 - 5x) = 780\)
\(150x + 20x + 640 - 100x = 780\)
\(70x + 640 = 780\)
\(70x = 140 \Rightarrow x = 2\)
Number of Rs 50 notes = \(3(2) = 6\).
Number of Rs 10 notes = \(2(2) = 4\).
Number of Rs 20 notes = \(32 - 5(2) = 22\).
In simple words: Express notes in terms of ratio variable \(x\), find total value, and solve for \(x\).

Exam Tip: Verify total notes: \(6 + 4 + 22 = 32\) and total value: \(50(6) + 10(4) + 20(22) = 300 + 40 + 440 = 780\).

Q 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 \(x\).
Amount given to daughter = \(\frac{x}{2} + 10000\).
Amount given to son = \(\frac{x}{3} + 3000\).
Given that daughter's share is twice the son's share:
\(\frac{x}{2} + 10000 = 2\left(\frac{x}{3} + 3000\right)\)
\(\frac{x}{2} + 10000 = \frac{2x}{3} + 6000\)
Rearrange terms: \(10000 - 6000 = \frac{2x}{3} - \frac{x}{2}\)
\(4000 = \frac{4x - 3x}{6}\)
\(4000 = \frac{x}{6}\)
\(x = 24000\) Rs.
In simple words: Set up expressions for daughter and son amounts, then set daughter = 2 \(\times\) son.

Exam Tip: Read carefully to ensure the multiplier 2 is applied correctly to the son's total expression.

Q 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 \(\frac{3}{2}\). Find the rational number.
Answer: Let the numerator be \(x\).
Denominator = \(x + 8\).
According to the condition:
\(\frac{x + 17}{x + 8 - 1} = \frac{3}{2}\)
\(\frac{x + 17}{x + 7} = \frac{3}{2}\)
Cross multiply: \(2(x + 17) = 3(x + 7)\)
\(2x + 34 = 3x + 21\)
\(x = 13\)
Numerator = 13, Denominator = 21.
The rational number is \(\frac{13}{21}\).
In simple words: This is identical to Question 30; form the equation and cross multiply.

Exam Tip: Re-verify your steps if a question appears multiple times in assignment sets.

Q 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 present ages be \(5x\) and \(6x\).
Five years before, their ages were \(5x - 5\) and \(6x - 5\).
Given ratio 5 years ago 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 = \(5(5) = 25\) years.
Neera's present age = \(6(5) = 30\) years.
In simple words: Use present age ratios as \(5x\) and \(6x\), subtract 5 for past ages, and cross multiply.

Exam Tip: Assign variables to present ages rather than past ages to make equations cleaner.

Q 36. The sum of three consecutive multiples of 8 is 792. Find these multiples.
Answer: Let the three consecutive multiples of 8 be \(8x\), \(8(x + 1)\), and \(8(x + 2)\).
Sum: \(8x + 8x + 8 + 8x + 16 = 792\)
\(24x + 24 = 792\)
\(24x = 768\)
\(x = 32\)
The multiples are \(8(32) = 256\), \(256 + 8 = 264\), and \(264 + 8 = 272\).
In simple words: Multiples of 8 increase by 8 successively.

Exam Tip: Sum check: \(256 + 264 + 272 = 792\).

Q 37. The sum of three consecutive multiples of 9 is 795. Find these multiples.
Answer: Let the three consecutive multiples of 9 be \(9x\), \(9(x + 1)\), and \(9(x + 2)\).
Sum: \(9x + 9x + 9 + 9x + 18 = 795\)
\(27x + 27 = 795\)
\(27x = 768\)
\(x = \frac{768}{27} = \frac{256}{9}\) (Note: Typo in source data numbers, but solved algebraically).
In simple words: Express as multiples of 9 and equate sum to 795.

Exam Tip: Even if numbers do not divide cleanly, follow exact algebraic steps.

Q 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 Q 35). Let present ages be \(5x\) and \(6x\).
\(\frac{5x - 5}{6x - 5} = \frac{4}{5} \Rightarrow 25x - 25 = 24x - 20 \Rightarrow x = 5\).
Present ages are 25 years and 30 years.
In simple words: Use ratio variables for present ages and solve.

Exam Tip: Double check calculations for age problems.

Q 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 Q 29). Let ages be \(3x\) and \(4x\).
\(\frac{3x + 5}{4x + 5} = \frac{4}{5} \Rightarrow 15x + 25 = 16x + 20 \Rightarrow x = 5\).
Present ages are 15 years and 20 years.
In simple words: Future age ratios lead directly to linear equations in one variable.

Exam Tip: Always state final ages with units.

Very Short Answer type Questions

Question. Sum of two numbers is 30. If one number is twice the other, form an equation for finding the numbers.
Answer : 2x + x = 30.

Question. Find x if 2x + 5 = x + 25.
Answer : 20.

Question. Find the value of x/2 + x/3 + x/4
Answer : 13x/12

Question. Give a number which is always even in terms of variable x.
Answer : 2x

Question. Preeti has three more dolls than Renu. If there are 11 dolls in all, how many dolls does each have.
Answer : 7, 4

Question. If x = n + 1 then find the value of (2x + 5)
Answer : 2n + 7

Question. Form an equation : 5 added to twice a number is 135.
Answer : 2x + 5 = 135

Question. Ratio of three angles of a triangle is 1 : 2 : 3. Find the angles.
Answer : 30°, 60°, 90°

Question. Ankit covered (1/2) of the distance by metro train, (1/3)^rd of the distance by bus and rest of 6 km by car for moving from Dwarka to South Extension. Find the total distance covered?
Answer : 36 km

Question. If 2x + 9 = 47. Find x.
Answer : 19

Question. What type of number is (2x + 1). [odd or even].
Answer : Odd

Question. If the length of a rectangle is 5 more than its breadth (x). What is the perimeter of the rectangle?
Answer : 4x + 10.

Question. 3(x + 4) = x + 38 find x.
Answer : 13

Question. I guessed a number (x) then added 10 to it. Give the expression for double of it.
Answer : 2(x + 10)

Question. If x = n/3 + 2 than 3x - 1 = __________.
Answer : n + 5.

1. The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.

2. The ages of Rahul and Anshul are in ratio 5:7. Four years later the sum of these ages will be 56 years .What are their present ages?

3. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi‘s present age?

4. Rahul’s father is 26 years younger than Rahul’s grandfather and 29 years older than Rahul. The sum of the ages of all the three is 135 years. What is the age of each one of them ?

5. Shobo’s mother’s present age is 6 times Shobo’s present age. Shobo’s age five years from now will be one-third of his mother’s present age. What are their present ages?

6. A grand-father is 10 times older than his grand-daughter. He is also 54 years older than her. What is his present age ?

7. Aman’s age is three times his son’s age.10 years ago he was 5 times his son’s age. Find their present ages.

8. Two years ago Dilip was three times as old is his son and two years hence twice his age will be equal to five times that of his son. Find their present ages.

9. Present ages of Annu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.

10. Ages of Mohan and Sohan are in the ratio 3:4. Find their present ages if sum of their ages is 35 years.

11. Three consecutive integers add up to 51. What are these integers?

12. Sum of three consecutive multiples of 11 is 363. Find these multiples.

13. The sum of three consecutive multiples of 8 is 888. Find the multiples.

14. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

15. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

16. Sum of two numbers is 95. If one exceed the other by 15. Find the numbers.

17. The no. of boys and girls in a class is in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

18. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number, Find the numbers?

19. Lakshmi is a cashier in a bank. She has currency notes of denomination Rs 100, Rs 50 and Rs 10 respectively, the ratio of these notes is 2:3:5. The total cash with Lakshmi is Rs. 4, 00,000. How many notes of each denomination does she have?

20. Devasi has a total of Rs 590 as currency notes in the denominations of Rs 50, Rs 20 and Rs 10. The ratio of the number of Rs 50 notes and Rs 20 notes is 3:5. If she has a total of 25 notes in all, how many notes of each denomination does she have?

21. I have a total of Rs. 300 in coins of denomination Rs 1, Rs 2, Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many of each denomination are with me?

22. The organizers of an essay competition decide that the winner in the competition gets a prize of Rs 100 and a participant who does not win will get Rs 25 . The total prize money distributed is Rs. 3,000. Find the no. of winners, if the total no. of participants is 63.

23. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original by 27. What is that two –digit number?

24. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

25. Half a herd of deer are grazing in the field and three fourth of them are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

26. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length & breadth of the plot are in the ratio 11:4. At the rate of Rs. 100/metre it will cost the village panchayat Rs 75,000 to fence. What are the dimensions of the plot?

27. Hasaan buys two kinds of cloth material for school uniforms. Shirt material that costs him Rs 50 per meter and trouser material that costs him Rs 90 per meter. For every 3 meters of the shirt material, he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. If his total sale is Rs 36600, how much trouser material did he buy?

28. 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. 

29. The sum of the digits of a two digit numbers is 10. If the number formed by reversing the digits is less than the original number by 36, find the original number.

Questions carrying 1 Mark each :-

Q.1 By which sign the algebraic expressions are connected to form an equation?

Q.2 Find the solution of 2x+5= 7.

Questions carrying 2 Marks each :

Q.3 Solve for y :-
ÿ/xx = 2/5

Q.4 Find the solution of
0/2 + 55550 + x5 = 300

Q.5 Anumber is seven times the other number. If their sum is 96, find the numbers.

Questions carrying 3 marks each:-

Q.6 If the same number is added to both the numerator and denominator of a fraction x/5 ,then the result is 1/4. Find the number.

Q.7 The perimeter of a rectangle is 52cm. If its widthis 2cm more than one-third of its length, find the dimensions of the rectangle.

Q.8 The sum of three consecutive multiplies of 7 is 777. Find these multiples.

Q.9 Meena's mother is four times as old as Meena. After five years, her mother will be three times as old as she will be then. What are their present ages?

Questions carrying 6 marks each :-

Q.10 The sum of the digits of a two digit number is 10. If the number formed by reversing the digits is less than the original number by 36, find the number.

Q.11 A bag contains coins of denomination of Rs. 5 and Rs. 2. The total value of these coins is Rs. 1272. If the number of two rupee coins is 15% of the five rupee coins, find the number of coins of each denomination.

Multiple choice Questions carrying 1 mark each :-

Q.12 In a linear equation, the highest power of the variableis :-
a) One
b) two
c) Three
d) Zero

Q.13 An equation of the form ax + b = c, where a, b and c are numbers, a ÿ 0 and x is the variable; represents a
a) linear equation
b) linear equation in one variable
c) linear equation in two variables
d) None of these

Questions carrying 1 Mark each :-