Class 9 maths · Original EZT practice
RS Aggarwal Class 9: original worked practice
A worked answer should help you choose a method, not just copy a result. These practice pairs start with the decision: what changes, which condition applies and what a sensible answer should look like.
This page gives Class 9 students 15 original worked examples and 15 fresh self-check questions with explained answers. Use the EZT practice-skill index alongside your RS Aggarwal book to learn a method, not to look up its exercise answers.
Original EZT practice, not copied textbook exercises or a full-solutions service. Match the topic to your own book; editions vary. This is not an official chapter index, exercise map or claim about your current syllabus. Follow your school's topic list.
Bharati Bhawan lists Secondary School Mathematics for Class 9 by R S Aggarwal and Veena Aggarwal. That source identifies the book, not the order of our practice topics. EZ Tuitions is not affiliated with the authors or publisher.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Rational numbers between two fractions
Choose the method: Use a common denominator large enough to leave integer numerators between the endpoints.
EZT practice question
Find three different rational numbers strictly between 2/5 and 3/5.
Show the worked solution
- Write the endpoints with denominator 20: 2/5 = 8/20 and 3/5 = 12/20.
- The integers 9, 10 and 11 lie strictly between 8 and 12.
- Therefore 9/20, 10/20 and 11/20 are suitable numbers. Simplifying 10/20 gives 1/2 if needed.
Answer: 9/20, 1/2 and 11/20
Check: Their decimal values are 0.45, 0.5 and 0.55. All are greater than 0.4 and less than 0.6.
Watch for this: The endpoints themselves do not count when the question says strictly between.
Now try it yourself
Find two different rational numbers strictly between 1/3 and 1/2.
Check your answer
3/8 and 5/12, for example
Use denominator 24: the endpoints are 8/24 and 12/24. Choose 9/24 = 3/8 and 10/24 = 5/12. Other valid choices are possible.
A linear model from a practical rule
Choose the method: Separate the fixed starting value from the amount added for each unit.
EZT practice question
In a fictional print shop, an order has a ₹25 setup charge plus ₹3 per page. Write the cost C for n pages and find the cost of 18 pages. Assume no tax or other charges.
Show the worked solution
- The setup charge is paid once, so it contributes 25.
- The page charge for n pages is 3n.
- The model is C = 3n + 25. For n = 18, C = 54 + 25 = ₹79.
Answer: C = 3n + 25; 18 pages cost ₹79
Check: One extra page increases the bill by ₹3. In the model, the fixed setup charge does not grow with n.
Watch for this: Multiplying the setup charge by the page count would count it more than once.
Now try it yourself
Using the same fictional rule, a bill is ₹97. How many pages were printed?
Check your answer
24 pages
Solve 3n + 25 = 97. Subtract 25 to get 3n = 72, then divide by 3. The whole-number answer fits a page count.
Arithmetic progressions: counting the gaps
Choose the method: An nth term has n - 1 equal jumps after the first term.
EZT practice question
A hall has 12 seats in its first row, 15 in its second and 18 in its third. If this pattern continues, how many seats are in row 10?
Show the worked solution
- The common difference is 3 seats per row.
- Use aₙ = a₁ + (n - 1)d with a₁ = 12, d = 3 and n = 10.
- a₁₀ = 12 + 9 × 3 = 39 seats.
Answer: 39 seats
Check: Row 1 uses zero jumps. Row 10 is nine jumps after row 1, not ten.
Watch for this: 12 + 10 × 3 gives row 11, an off-by-one error.
Now try it yourself
For the sequence 7, 11, 15, 19, ... find the first term that is greater than 50 and state its position.
Check your answer
51, the 12th term
aₙ = 7 + 4(n - 1). The inequality aₙ > 50 gives n > 11.75. The first integer is 12, and a₁₂ = 51; a₁₁ = 47.
Geometric progressions: repeated multiplication
Choose the method: Use a common ratio when each term is multiplied, not increased by a fixed amount.
EZT practice question
A pattern contains 3 tiles at stage 1. Each later stage has twice as many tiles as the previous stage. How many tiles are at stage 6?
Show the worked solution
- The terms begin 3, 6, 12, ... with common ratio 2.
- From stage 1 to stage 6 there are five doublings.
- a₆ = 3 × 2⁵ = 96 tiles.
Answer: 96 tiles
Check: The six stages are 3, 6, 12, 24, 48 and 96.
Watch for this: Adding 2 at each stage would create a different pattern. The rule here is doubling.
Now try it yourself
A sequence starts at 160 and each following term is half the previous term. Find its fifth term.
Check your answer
10
There are four halvings: 160, 80, 40, 20, 10. Equivalently a₅ = 160 × (1/2)⁴ = 10.
Identities and the terms that cancel
Choose the method: Expand both expressions separately before combining them.
EZT practice question
Simplify (x + 5)² - (x - 5)².
Show the worked solution
- Expand the first square: x² + 10x + 25.
- Expand the second square: x² - 10x + 25.
- Subtract the whole second expression: x² + 10x + 25 - x² + 10x - 25 = 20x.
Answer: 20x
Check: At x = 2, the original is 7² - (-3)² = 49 - 9 = 40. The result is 20 × 2 = 40.
Watch for this: The minus before the second bracket changes every sign inside it, including -10x.
Now try it yourself
Factorise y² + 10y + 21.
Check your answer
(y + 3)(y + 7)
The numbers 3 and 7 have sum 10 and product 21. Multiplying the brackets gives y² + 7y + 3y + 21.
Two equations for two unknown quantities
Choose the method: Use units to decide what each variable represents, then translate both conditions.
EZT practice question
In a fictional event, 2 adult tickets and 3 student tickets cost ₹440. Three adult tickets and one student ticket cost ₹380. Find each ticket price, with fixed prices and no other charges.
Show the worked solution
- Let adult price be a and student price be s, in rupees: 2a + 3s = 440 and 3a + s = 380.
- Multiply the second equation by 3: 9a + 3s = 1,140.
- Subtract the first: 7a = 700, so a = 100.
- Substitute into 3a + s = 380: 300 + s = 380, so s = 80.
Answer: Adult ₹100; student ₹80
Check: 2(100) + 3(80) = 440 and 3(100) + 80 = 380.
Watch for this: Using only the total number of tickets loses the distinction between the two prices.
Now try it yourself
Do the equations 2x + y = 9 and 4x + 2y = 20 have a simultaneous solution?
Check your answer
No solution
Doubling the first equation requires 4x + 2y = 18, not 20. The same left-hand expression cannot equal both values, so the lines are distinct and parallel.
Coordinates: distance and midpoint
Choose the method: Keep horizontal and vertical changes separate before using the distance formula.
EZT practice question
For A = (-2, 1) and B = (4, 9), find AB and the midpoint M.
Show the worked solution
- The changes from A to B are 4 - (-2) = 6 and 9 - 1 = 8.
- AB = √(6² + 8²) = √100 = 10 units.
- Average each coordinate for the midpoint: M = ((-2 + 4)/2, (1 + 9)/2) = (1, 5).
Answer: AB = 10 units; M = (1, 5)
Check: From A to M the changes are 3 and 4, so AM = 5 units. The same is true from M to B.
Watch for this: A midpoint uses coordinate averages, not the length of the segment as a coordinate.
Now try it yourself
The midpoint of P = (3, -1) and Q is (7, 2). Find Q.
Check your answer
Q = (11, 5)
From (3 + x)/2 = 7, x = 11. From (-1 + y)/2 = 2, y = 5.
Angles on a straight line
Choose the method: Identify the angle relationship before writing the equation.
EZT practice question
Two adjacent angles form a straight angle. Their measures are (4x + 8)° and (2x + 16)°. Find x and both angles.
Show the worked solution
- A straight angle is 180°, so (4x + 8) + (2x + 16) = 180.
- 6x + 24 = 180, giving 6x = 156 and x = 26.
- The angles are 112° and 68°.
Answer: x = 26; angles 112° and 68°
Check: 112° + 68° = 180°.
Watch for this: Adjacent angles do not always add to 180°. They do here because their outer arms form a straight line.
Now try it yourself
Two parallel lines are cut by a transversal. An interior angle is 72°. Find the interior angle on the same side of the transversal.
Check your answer
108°
Co-interior angles between parallel lines are supplementary, so the required angle is 180° - 72° = 108°.
Triangle congruence: the included angle
Choose the method: Check which sides contain the given angle; three measurements alone are not the name of a test.
EZT practice question
In triangles ABC and DEF, AB = DE, AC = DF and angle BAC = angle EDF. Are the triangles congruent? State the test and the corresponding side for BC.
Show the worked solution
- Angle BAC is between AB and AC. Angle EDF is between DE and DF.
- Two corresponding sides and their included angle are equal, so triangle ABC is congruent to triangle DEF by SAS.
- The correspondence is A↔D, B↔E, C↔F. Hence BC corresponds to EF and BC = EF.
Answer: Yes, SAS; BC corresponds to EF
Check: Both angles named in the data have the common vertex between the two named equal sides.
Watch for this: Replacing the included angle with an unrelated angle would not automatically satisfy SAS.
Now try it yourself
Does equality of all three corresponding angles alone guarantee two triangles are congruent?
Check your answer
No
An equilateral triangle of side 2 cm and one of side 5 cm both have angles 60°, 60°, 60°, but their side lengths differ. Equal angles alone do not fix size.
A parallelogram from its vertices
Choose the method: For consecutive vertices, opposite diagonals must have the same midpoint.
EZT practice question
A, B, C, D are consecutive vertices of a parallelogram. A = (1, 2), B = (6, 2) and C = (8, 5). Find D.
Show the worked solution
- Write D = (u, v). Since the diagonals bisect each other, the midpoint of AC equals the midpoint of BD.
- Equating x-coordinates gives (1 + 8)/2 = (6 + u)/2, so u = 3.
- Equating y-coordinates gives (2 + 5)/2 = (2 + v)/2, so v = 5. Thus D = (3, 5).
Answer: D = (3, 5)
Check: AB and DC are both horizontal with length 5. The displacement from B to C and from A to D is (2, 3).
Watch for this: The given order matters. Treating A, B and C as a different set of consecutive vertices can produce a different fourth point.
Now try it yourself
D and E are the midpoints of two sides of a triangle. The third side is 18 cm. Find DE.
Check your answer
9 cm
By the midpoint theorem, the segment joining the midpoints is parallel to the third side and half its length: 18/2 = 9 cm.
Cyclic quadrilaterals: opposite angles
Choose the method: Use supplementary opposite angles only after the quadrilateral is known to be cyclic.
EZT practice question
ABCD is a cyclic quadrilateral. Angle A = 82° and angle B = 104°. Find angles C and D.
Show the worked solution
- Opposite angles of a cyclic quadrilateral sum to 180°.
- C = 180° - A = 180° - 82° = 98°.
- D = 180° - B = 180° - 104° = 76°.
Answer: C = 98°; D = 76°
Check: 82° + 104° + 98° + 76° = 360°, and both opposite pairs total 180°.
Watch for this: Any quadrilateral has angle sum 360°, but not every quadrilateral has supplementary opposite angles.
Now try it yourself
A circle has diameter AB and C is any other point on the circle. Find angle ACB.
Check your answer
90°
The angle in a semicircle is a right angle. The condition that AB is a diameter is needed.
A sector as a fraction of a circle
Choose the method: The central angle tells you the fraction of the full circle.
EZT practice question
A sector has radius 7 cm and central angle 120°. Find its area and arc length. Use π = 22/7.
Show the worked solution
- The sector is 120/360 = 1/3 of a full circle.
- Area = (1/3) × πr² = (1/3) × (22/7) × 49 = 154/3 cm².
- Arc length = (1/3) × 2πr = (1/3) × 44 = 44/3 cm.
Answer: Area 154/3 cm²; arc length 44/3 cm
Check: Multiplying either result by 3 gives the full circle result: area 154 cm² and circumference 44 cm.
Watch for this: Arc length is only the curved part of the boundary. The full sector perimeter also includes two radii.
Now try it yourself
Using the same radius, angle and π value, find the perimeter of the sector.
Check your answer
86/3 cm
Add two radii to the arc length: 44/3 + 7 + 7 = 44/3 + 42/3 = 86/3 cm.
A cone: height is not slant height
Choose the method: For a right circular cone, radius, height and slant height form a right triangle.
EZT practice question
A right circular cone has radius 5 cm and perpendicular height 12 cm. Find its slant height and curved surface area, leaving the area in terms of π.
Show the worked solution
- Use l² = r² + h² = 5² + 12² = 169.
- The positive slant height is l = 13 cm.
- Curved surface area = πrl = π × 5 × 13 = 65π cm².
Answer: Slant height 13 cm; curved surface area 65π cm²
Check: The slant height is longer than the perpendicular height. Curved surface area excludes the circular base.
Watch for this: Using πrh would use the wrong length for the curved surface.
Now try it yourself
Find the volume of the same cone.
Check your answer
100π cm³
Volume = (1/3)πr²h = (1/3)π × 25 × 12 = 100π cm³. This formula uses perpendicular height, not slant height.
Weighted mean: count the observations
Choose the method: Multiply each group mean by its group size before combining the totals.
EZT practice question
In a fictional practice group, 8 students have a mean score of 6 and 12 students have a mean score of 9. Find the combined mean score.
Show the worked solution
- The first group has a total score of 8 × 6 = 48.
- The second group has a total score of 12 × 9 = 108.
- Combined mean = (48 + 108)/(8 + 12) = 156/20 = 7.8.
Answer: 7.8
Check: The result lies between 6 and 9 and is closer to 9 because that group has more students.
Watch for this: (6 + 9)/2 = 7.5 treats the groups as equal-sized, which they are not.
Now try it yourself
Three items cost ₹20 each and two items cost ₹35 each. Find the mean cost per item.
Check your answer
₹26
Total cost = 3 × 20 + 2 × 35 = ₹130. There are 5 items, so mean = 130/5 = ₹26.
Probability: state the equally likely outcomes
Choose the method: Count favourable outcomes within the full sample space, not just the outcomes mentioned in the question.
EZT practice question
One fair six-sided die numbered 1 to 6 is rolled once. Find the probability of a result greater than 4.
Show the worked solution
- The equally likely outcomes are 1, 2, 3, 4, 5 and 6.
- The favourable outcomes are 5 and 6: two outcomes.
- Probability = 2/6 = 1/3.
Answer: 1/3
Check: The probability lies between 0 and 1. The complementary event has four outcomes and probability 2/3.
Watch for this: A result being either greater than 4 or not does not make those two events equally likely.
Now try it yourself
In a fictional experiment, a spinner lands on blue 18 times in 60 spins. Find the empirical probability of blue. Does that prove the next spin will be blue?
Check your answer
3/10; no
The observed proportion is 18/60 = 3/10. This is an estimate from past spins, not a guarantee about a single next result or proof that the spinner is fair.
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Questions about this practice page
- Are these the official RS Aggarwal Class 9 solutions?
- No. These are original EZT practice questions, worked methods and explained self-checks to use alongside your book. They are not copied textbook exercises or a full exercise answer key. Match the topic to your own book; editions vary.
- Which RS Aggarwal Class 9 book is referred to here?
- Bharati Bhawan lists Secondary School Mathematics for Class 9 by R S Aggarwal and Veena Aggarwal. That publisher source identifies the book, but it does not verify the order or exercise numbers in our practice index.
- Is this an edition-specific chapter index?
- No. It is an EZT practice-skill index. We have not verified a primary contents preview for your edition, so use your own book contents and your teacher topic list to decide which sections to practise. No current syllabus-year or complete-coverage claim is made.
- Why practise sequences, coordinates and probability here?
- They offer different ways to explain a pattern, a position or a chance event. The examples here are selected practice, not a list of every topic in a particular book or course. Check your school plan rather than assuming that a topic on this page is assigned to you.
- How can I tell whether I understand a worked solution?
- Close the solution and explain why the first method fits the question. Then solve the second question with your own working. If you need the same prompt again, mark that step and bring both attempts to a tutor.