Class 9 maths · Original EZT practice
ML Aggarwal Class 9: original worked practice
If you know the formula but still cannot finish the question, work through one method at a time. Start with a topic below, attempt the example and compare the reasoning, not just the final line.
This page gives Class 9 students 14 original maths examples with stepwise solutions, a check and a second question to try. Use them alongside your ML Aggarwal book to practise a method, not as an answer key to its exercises.
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.
APC Books lists Understanding ICSE Mathematics Class IX by M L Aggarwal, published by Avichal Publishing Company, 18th edition, ISBN 978-93-6333-744-2. That source identifies the book; it does not verify the order of this practice index. EZ Tuitions is not affiliated with the author or publisher.
Try first. Read one step. Check the result.
Write what is given, what you need to find and the rule you plan to use. Then attempt the example before opening its solution. If a step differs from yours, pause there. Use the check to test the result and finish with the second question without copying the method line for line.
Real numbers and surds
Simplify before combining like surds.
EZT practice question
Simplify 3√12 - √27 + 2√3.
Show the worked solution
- Write √12 = 2√3 and √27 = 3√3.
- Substitute: 3(2√3) - 3√3 + 2√3.
- Combine the coefficients: (6 - 3 + 2)√3 = 5√3.
Answer: 5√3
Check: Every term is now a multiple of √3; only their coefficients are added or subtracted.
Watch for this: √12 - √27 is not √(12 - 27). Square roots do not distribute over subtraction.
Now try it yourself
Rationalise 5/(√7 + √2).
Check your answer
√7 - √2
Multiply numerator and denominator by √7 - √2. The denominator becomes 7 - 2 = 5, which cancels the numerator factor 5.
Compound interest
Track the growing amount and the compounding period.
EZT practice question
A fictional account starts with ₹5,000 at 10% per year, compounded annually. Find the amount and compound interest after two years. Assume no deposits, withdrawals or charges.
Show the worked solution
- After year 1: ₹5,000 × 1.10 = ₹5,500.
- After year 2: ₹5,500 × 1.10 = ₹6,050.
- Compound interest = amount - principal = ₹6,050 - ₹5,000 = ₹1,050.
Answer: Amount ₹6,050; compound interest ₹1,050
Check: The second year earns ₹550, not ₹500, because interest is calculated on ₹5,500.
Watch for this: Using ₹500 for both years gives simple interest, not the stated annual compounding.
Now try it yourself
A fictional ₹8,000 balance earns 10% per year, compounded half-yearly, for one year. Find the amount and compound interest, with no other transactions or charges.
Check your answer
Amount ₹8,820; compound interest ₹820
Each half-year uses 5%. Amount = 8,000 × 1.05 × 1.05 = 8,820. Subtract 8,000 to get 820. These are invented arithmetic assumptions, not a real product rate.
Algebraic expansions
Keep the middle term in a squared expression.
EZT practice question
Expand (2x - 3y)².
Show the worked solution
- Use (a - b)² = a² - 2ab + b².
- Here a = 2x and b = 3y.
- The three terms are 4x², -12xy and 9y².
Answer: 4x² - 12xy + 9y²
Check: At x = 3 and y = 1, the original is (6 - 3)² = 9. The expansion gives 36 - 36 + 9 = 9.
Watch for this: Squaring each term alone misses -12xy.
Now try it yourself
Expand (x + 2)(x - 5).
Check your answer
x² - 3x - 10
Multiply each term: x² - 5x + 2x - 10. Combine -5x and 2x.
Factorisation
Find a pair that matches both the product and the sum.
EZT practice question
Factorise 6x² + x - 2.
Show the worked solution
- Multiply the first and last coefficients: 6 × (-2) = -12.
- Use 4 and -3: their product is -12 and their sum is 1.
- Split and group: 6x² + 4x - 3x - 2 = 2x(3x + 2) - (3x + 2).
- Take the common bracket: (3x + 2)(2x - 1).
Answer: (3x + 2)(2x - 1)
Check: Expanding gives 6x² - 3x + 4x - 2 = 6x² + x - 2.
Watch for this: A pair whose product is -12 but whose sum is not 1 does not split the middle term correctly.
Now try it yourself
Factorise 8a³ - 27b³.
Check your answer
(2a - 3b)(4a² + 6ab + 9b²)
Apply u³ - v³ = (u - v)(u² + uv + v²), with u = 2a and v = 3b.
Simultaneous linear equations
Use both equations and check the pair in both.
EZT practice question
Solve 2x + 3y = 17 and 3x - 2y = 6.
Show the worked solution
- Multiply the first equation by 2: 4x + 6y = 34.
- Multiply the second by 3: 9x - 6y = 18.
- Add: 13x = 52, so x = 4.
- Substitute into 2x + 3y = 17: 8 + 3y = 17, so y = 3.
Answer: x = 4, y = 3
Check: 2(4) + 3(3) = 17 and 3(4) - 2(3) = 6.
Watch for this: Solving one equation alone leaves many possible pairs; the answer must satisfy both.
Now try it yourself
In a fictional stationery purchase, two pens and three notebooks cost ₹81. Four pens and one notebook cost ₹67. Find each price, assuming fixed prices and no other charges.
Check your answer
Pen ₹12; notebook ₹19
Let the prices be p and n. From 2p + 3n = 81 and 4p + n = 67, double the first and subtract the second: 5n = 95. Thus n = 19 and p = 12.
Indices and logarithms
Connect a logarithm to its equivalent power statement.
EZT practice question
Solve 3^(x + 1) = 81.
Show the worked solution
- Write 81 as 3⁴.
- The bases are equal and 3 is positive and not 1, so x + 1 = 4.
- Subtract 1: x = 3.
Answer: x = 3
Check: 3^(3 + 1) = 3⁴ = 81.
Watch for this: Do not divide the exponents as though they were coefficients.
Now try it yourself
Evaluate log₂ 32 - log₃ 9.
Check your answer
3
Since 2⁵ = 32, log₂ 32 = 5. Since 3² = 9, log₃ 9 = 2. The difference is 5 - 2 = 3.
Triangle reasoning
Use the given equal sides, not the appearance of a sketch.
EZT practice question
In triangle ABC, AB = AC and angle A = 40°. Find angles B and C.
Show the worked solution
- AB = AC makes the angles opposite them equal: angle C = angle B.
- The three interior angles total 180°, so B + C = 180° - 40° = 140°.
- Split 140° equally: B = C = 70°.
Answer: Angle B = angle C = 70°
Check: 40° + 70° + 70° = 180°.
Watch for this: The equal angles are opposite the equal sides. Do not choose the apex angle just because it looks central.
Now try it yourself
Can lengths 4 cm, 7 cm and 12 cm form a triangle?
Check your answer
No
The sum of the two smaller lengths is 4 + 7 = 11 cm, less than 12 cm. A triangle needs the sum of any two sides to be greater than the third.
Midpoint theorem and Pythagoras
Identify the theorem conditions before substituting lengths.
EZT practice question
D and E are the midpoints of AB and AC in triangle ABC. If BC = 14 cm, find DE and its relation to BC.
Show the worked solution
- D and E divide two sides of the same triangle into equal halves.
- By the midpoint theorem, DE is parallel to BC and half its length.
- DE = 14 ÷ 2 = 7 cm.
Answer: DE = 7 cm and DE is parallel to BC
Check: The midpoint segment has half the third-side length; it is not a new side equal to BC.
Watch for this: The half-length result needs both points to be midpoints. It does not follow from an arbitrary line across a triangle.
Now try it yourself
A right triangle has perpendicular sides 9 cm and 12 cm. Find the hypotenuse.
Check your answer
15 cm
c² = 9² + 12² = 81 + 144 = 225, so the positive length is c = 15 cm.
Parallelograms and area
Use the perpendicular height, not a sloping side.
EZT practice question
A parallelogram has base 12 cm and perpendicular height 7 cm. Find its area and the area of a triangle on the same base between the same parallels.
Show the worked solution
- Parallelogram area = base × perpendicular height = 12 × 7 = 84 cm².
- The triangle has the same base and height.
- Triangle area = ½ × 12 × 7 = 42 cm².
Answer: Parallelogram 84 cm²; triangle 42 cm²
Check: The triangle is half the area because both use the same base and perpendicular height.
Watch for this: A sloping side cannot replace the height unless it is perpendicular to the base.
Now try it yourself
The diagonals of a parallelogram intersect at O. If AO = 5 cm and BO = 4 cm, find AC and BD.
Check your answer
AC = 10 cm; BD = 8 cm
Parallelogram diagonals bisect each other. Thus AC = 2AO and BD = 2BO; they need not be equal to each other.
Circle chords
Draw the radius and use the perpendicular-to-chord fact.
EZT practice question
A circle has radius 13 cm. A chord is 5 cm from the centre. Find the chord length.
Show the worked solution
- Let O be the centre, AB the chord and M the perpendicular foot from O to AB.
- The perpendicular from the centre bisects the chord, so AM is half the chord.
- In right triangle OMA, AM² = 13² - 5² = 169 - 25 = 144.
- AM = 12 cm, so AB = 24 cm.
Answer: 24 cm
Check: Half the chord, 12 cm, is shorter than the 13 cm radius; the full chord is shorter than the 26 cm diameter.
Watch for this: The Pythagoras result gives half the chord, so remember to double it.
Now try it yourself
A chord in a circle of radius 10 cm is 12 cm long. Find its perpendicular distance from the centre.
Check your answer
8 cm
Half the chord is 6 cm. The perpendicular distance d satisfies d² = 10² - 6² = 64, so d = 8 cm.
Mensuration
Keep area, surface area and volume in different units.
EZT practice question
A closed cuboid measures 8 cm by 5 cm by 3 cm. Find its volume and total surface area.
Show the worked solution
- Volume = length × breadth × height = 8 × 5 × 3 = 120 cm³.
- Total surface area = 2(lb + bh + hl).
- Substitute: 2(40 + 15 + 24) = 2 × 79 = 158 cm².
Answer: Volume 120 cm³; total surface area 158 cm²
Check: The surface area counts all six faces. A closed cuboid includes its top as well as its base.
Watch for this: Do not report volume in cm² or use a closed-box formula for an open box without adjusting it.
Now try it yourself
Find the area of a triangle with sides 5 cm, 12 cm and 13 cm.
Check your answer
30 cm²
The semiperimeter is 15 cm. Heron: area = √[15(15 - 5)(15 - 12)(15 - 13)] = √900 = 30 cm². The lengths also form a right triangle.
Trigonometry
Name the reference angle before labelling opposite and adjacent.
EZT practice question
In a right triangle, for acute angle θ, the opposite side is 5 cm and the hypotenuse is 13 cm. Find sin θ, cos θ and tan θ.
Show the worked solution
- Find the adjacent side using Pythagoras: √(13² - 5²) = √144 = 12 cm.
- sin θ = opposite/hypotenuse = 5/13.
- cos θ = adjacent/hypotenuse = 12/13.
- tan θ = opposite/adjacent = 5/12.
Answer: sin θ = 5/13; cos θ = 12/13; tan θ = 5/12
Check: sin² θ + cos² θ = (25 + 144)/169 = 1.
Watch for this: Opposite and adjacent depend on the chosen angle, while the hypotenuse stays opposite the right angle.
Now try it yourself
For acute angle A in a right triangle, the adjacent side is 8 cm and the hypotenuse is 17 cm. Find tan A.
Check your answer
15/8
The opposite side is √(17² - 8²) = √225 = 15 cm. Thus tan A = opposite/adjacent = 15/8.
Coordinate geometry
Read horizontal movement first, then vertical movement.
EZT practice question
Point P is (-3, 4). Reflect it in the y-axis to get Q. State the coordinates and quadrant of each point.
Show the worked solution
- P has a negative x-coordinate and positive y-coordinate, so it is in Quadrant II.
- Reflection in the y-axis changes the sign of x and keeps y unchanged.
- Q = (3, 4), which has two positive coordinates and is in Quadrant I.
Answer: P = (-3, 4), Quadrant II; Q = (3, 4), Quadrant I
Check: Both points have y = 4 and lie 3 units from the y-axis, on opposite sides.
Watch for this: Changing both signs reflects through the origin, not in the y-axis.
Now try it yourself
Reflect R = (5, -2) in the x-axis. State the new point and its quadrant.
Check your answer
(5, 2), Quadrant I
For reflection in the x-axis, keep x and change the sign of y. Both new coordinates are positive.
Mean and median
Order the data before finding the middle value.
EZT practice question
The fictional daily puzzle scores of five students are 12, 4, 7, 10 and 7. Find the mean and median.
Show the worked solution
- Add the five values: 12 + 4 + 7 + 10 + 7 = 40.
- Mean = 40 ÷ 5 = 8.
- Order the values: 4, 7, 7, 10, 12. The middle (third) value is 7, so the median is 7.
Answer: Mean 8; median 7
Check: The mean lies between the smallest and largest scores. The median is the middle value after ordering, not before.
Watch for this: Do not delete the repeated 7. It represents a second observation.
Now try it yourself
Find the mean and median of 3, 5, 8 and 12.
Check your answer
Mean 7; median 6.5
Mean = (3 + 5 + 8 + 12)/4 = 7. There are two middle values, so median = (5 + 8)/2 = 6.5.
Need help choosing the right method?
Bring your ICSE maths book, your school's topic list and one attempt you could not finish. Ask a tutor to work on the missing idea, then check that you can solve a different question without prompts.
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Questions about using these solutions
- Are these the official ML Aggarwal Class 9 solutions?
- No. These are original EZT worked examples and self-check questions for topic practice alongside your own book. They do not reproduce textbook exercises, provide a full answer key or claim an official chapter order. Match the topic to your own book; editions vary.
- Which book does this page refer to?
- The publisher APC Books identifies Understanding ICSE Mathematics Class IX by M L Aggarwal, published by Avichal Publishing Company. Its listing shows the 18th edition and ISBN 978-93-6333-744-2. We use that listing for book identity, not as proof of a chapter or exercise map.
- Can I use the topic index as my book contents page?
- No. This index groups our practice by skill. It is not an official contents page or a claim about the order in your edition. Use your own contents page and your teacher syllabus to locate the relevant topic.
- Is ML Aggarwal the same as an NCERT mathematics book?
- No. The APC listing identifies this title as Understanding ICSE Mathematics. NCERT is a different textbook publisher. Use the book and course requirements set by your school; an example on a shared topic does not establish identical syllabus, chapter order or assessment coverage.
- What should I do after checking an answer?
- Explain the method without looking, substitute back when possible and note the step that caused an error. Then try a different question from your own book. If the difficulty is in choosing the method rather than doing arithmetic, bring both attempts to a tutor.