Class 9 ICSE maths · Original EZT practice
Selina Concise Class 9: original error-spotting practice
Getting an answer wrong is useful when you can find the first step that failed. These short practice pairs focus on errors that are easy to miss: an unchanged interest base, the wrong angle or one extra face of a box.
This page gives Class 9 students 12 original error-spotting examples with repaired solutions and 12 fresh questions with explained answers. Use them alongside Selina Concise Mathematics to practise reasoning, not to look up textbook 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.
Selina Publishers lists Concise Mathematics for Standard 9. Its publisher cover reads I.C.S.E. Concise Mathematics IX Revised. This identifies the book, not a verified chapter sequence. EZ Tuitions is not affiliated with Selina Publishers.
Attempt it, spot the error, repair the method.
Try the question before reading the labelled wrong attempt. Decide why that line fails, then open the repair. The aim is to explain the missing rule in your own words. Finish with the second question and check its explanation only after writing your answer.
Surds: combine only like terms
EZT practice question
Simplify √50 + √8 - √18.
Deliberately wrong attempt - find the error
√50 + √8 - √18 = √40.
Show the error and repaired solution
Why it fails: The square-root sign cannot be taken outside an addition or subtraction in that way. First extract square factors from each term.
- √50 = 5√2, √8 = 2√2 and √18 = 3√2.
- Substitute to get 5√2 + 2√2 - 3√2.
- Combine the coefficients: (5 + 2 - 3)√2 = 4√2.
Answer: 4√2
Check: The original is about 7.071 + 2.828 - 4.243 = 5.656, which agrees with 4√2, not √40.
Now try it yourself
Rationalise 6/(√11 - √5).
Check your answer
√11 + √5
Multiply numerator and denominator by √11 + √5. The denominator is 11 - 5 = 6, so the factor 6 cancels.
Compound interest: keep the base changing
EZT practice question
A fictional balance of ₹10,000 grows at 5% per year, compounded annually, for two years. Find the compound interest. Assume no other transactions or charges.
Deliberately wrong attempt - find the error
Interest = 10,000 × 5 × 2/100 = ₹1,000.
Show the error and repaired solution
Why it fails: That calculation keeps the principal fixed for both years. Annual compounding earns the second year of interest on the first year-end amount.
- Year 1 amount = ₹10,000 × 1.05 = ₹10,500.
- Year 2 amount = ₹10,500 × 1.05 = ₹11,025.
- Compound interest = ₹11,025 - ₹10,000 = ₹1,025.
Answer: ₹1,025
Check: The second year adds ₹525. The extra ₹25 is interest on the ₹500 interest from year 1.
Now try it yourself
A fictional machine is valued at ₹20,000. Its value falls by 10% of the remaining value each year. Find the value after two years.
Check your answer
₹16,200
Each year retains 90%: 20,000 × 0.9 × 0.9 = 16,200. Subtracting 20% of the initial value would ignore the changing base. These are invented rates, not market data.
Expansions: track every sign
EZT practice question
Expand (3a + 2b)(3a - 2b).
Deliberately wrong attempt - find the error
(3a + 2b)(3a - 2b) = 9a² + 4b².
Show the error and repaired solution
Why it fails: The last product is (2b)(-2b), which is negative. The two middle products cancel rather than changing that sign.
- Multiply each term: 9a² - 6ab + 6ab - 4b².
- The middle terms total zero.
- The result is 9a² - 4b², matching (u + v)(u - v) = u² - v².
Answer: 9a² - 4b²
Check: At a = 2, b = 1, the product is 8 × 4 = 32. The result gives 36 - 4 = 32.
Now try it yourself
Expand (x - 4)³.
Check your answer
x³ - 12x² + 48x - 64
Use (u - v)³ = u³ - 3u²v + 3uv² - v³, with u = x and v = 4. The signs alternate.
Factorisation: check by multiplying back
EZT practice question
Factorise 2x² - 7x + 3.
Deliberately wrong attempt - find the error
2x² - 7x + 3 = (2x - 3)(x - 1).
Show the error and repaired solution
Why it fails: Those brackets give a middle coefficient of -5, not -7. Matching the first and last terms alone is not enough.
- The product of the first and last coefficients is 2 × 3 = 6.
- Choose -6 and -1: product 6, sum -7.
- Split and group: 2x² - 6x - x + 3 = 2x(x - 3) - (x - 3).
- Take the common bracket: (2x - 1)(x - 3).
Answer: (2x - 1)(x - 3)
Check: Multiplying gives 2x² - 6x - x + 3 = 2x² - 7x + 3.
Now try it yourself
Factorise 9p² - 24pq + 16q².
Check your answer
(3p - 4q)²
The first and last terms are squares, and the middle term is -2(3p)(4q). This matches u² - 2uv + v².
Linear equations: use both conditions
EZT practice question
Solve x + 2y = 11 and 2x - y = 7.
Deliberately wrong attempt - find the error
x = 7 and y = 2, because 7 + 2(2) = 11.
Show the error and repaired solution
Why it fails: This pair fits the first equation but gives 2(7) - 2 = 12 in the second. A simultaneous solution must satisfy both equations.
- Double the second equation: 4x - 2y = 14.
- Add the first equation: 5x = 25, so x = 5.
- Substitute into x + 2y = 11: 5 + 2y = 11, so y = 3.
Answer: x = 5, y = 3
Check: 5 + 2(3) = 11 and 2(5) - 3 = 7.
Now try it yourself
The sum of two numbers is 23 and their difference is 7. Find the numbers.
Check your answer
15 and 8
Let u be larger: u + v = 23 and u - v = 7. Add to get 2u = 30, so u = 15 and v = 8.
Powers and logs: name the operation
EZT practice question
Simplify (2³ × 2⁴)/2⁵.
Deliberately wrong attempt - find the error
(2³ × 2⁴)/2⁵ = 2^(3 × 4 - 5) = 2⁷.
Show the error and repaired solution
Why it fails: Multiplying powers with the same base adds their exponents. It does not multiply the exponents.
- In the numerator, 2³ × 2⁴ = 2^(3 + 4) = 2⁷.
- Divide by 2⁵: subtract exponents to get 2^(7 - 5) = 2².
- Evaluate 2² = 4.
Answer: 4
Check: The numerical fraction is (8 × 16)/32 = 128/32 = 4.
Now try it yourself
Evaluate log₅ 125 + log₁₀ 0.01.
Check your answer
1
5³ = 125, so log₅ 125 = 3. Also 10^(-2) = 0.01, so log₁₀ 0.01 = -2. Their sum is 1.
Triangles: read the equal-side information
EZT practice question
In triangle PQR, PQ = PR. The exterior angle at P, formed by extending QP beyond P, is 130°. Find the interior angles.
Deliberately wrong attempt - find the error
Angle P = 130°, so angles Q and R are each 25°.
Show the error and repaired solution
Why it fails: The 130° angle is exterior, not the interior apex angle. It forms a straight angle with the interior angle P.
- Interior angle P = 180° - 130° = 50°.
- Since PQ = PR, the opposite angles R and Q are equal.
- Q + R = 180° - 50° = 130°, so Q = R = 65°.
Answer: P = 50°, Q = R = 65°
Check: The interior angles sum to 180°. The exterior angle also equals Q + R = 130°.
Now try it yourself
A triangle has sides 6 cm, 10 cm and 13 cm. Which side is opposite its largest angle?
Check your answer
The 13 cm side
In a triangle the greater side is opposite the greater angle. All three lengths satisfy the triangle inequality, including 6 + 10 > 13.
Parallelograms: distinguish opposite and adjacent
EZT practice question
Adjacent angles A and B of a parallelogram are (3x + 10)° and (2x + 20)°. Find x and all four angles.
Deliberately wrong attempt - find the error
3x + 10 = 2x + 20, so x = 10.
Show the error and repaired solution
Why it fails: Opposite angles are equal; adjacent angles are supplementary. These two angles must add to 180°.
- (3x + 10) + (2x + 20) = 180.
- 5x + 30 = 180, so x = 30.
- A = 100° and B = 80°. Opposite angles give C = 100° and D = 80°.
Answer: x = 30; A = C = 100°, B = D = 80°
Check: Each adjacent pair totals 180°, and all four angles total 360°.
Now try it yourself
In a rhombus, the diagonals are 16 cm and 30 cm. Find the side length.
Check your answer
17 cm
Rhombus diagonals bisect at right angles. A side is the hypotenuse of a right triangle with legs 8 and 15, so its length is √(64 + 225) = 17 cm.
Circles: split a chord before using Pythagoras
EZT practice question
A chord is 16 cm long in a circle of radius 10 cm. Find the perpendicular distance from the centre to the chord.
Deliberately wrong attempt - find the error
Distance² = 16² - 10².
Show the error and repaired solution
Why it fails: The radius is the hypotenuse, and the perpendicular from the centre bisects the chord. The other leg is half the chord, not the full 16 cm.
- Half the chord is 16 ÷ 2 = 8 cm.
- Let the perpendicular distance be d. Then d² + 8² = 10².
- d² = 100 - 64 = 36; the positive distance is d = 6 cm.
Answer: 6 cm
Check: The right triangle has sides 6, 8 and 10 cm. The distance is less than the radius.
Now try it yourself
A chord is 12 cm from the centre of a circle of radius 15 cm. Find its length.
Check your answer
18 cm
Half the chord is √(15² - 12²) = √81 = 9 cm. Double this to get 18 cm.
Mensuration: count the faces you actually have
EZT practice question
A rectangular box has no lid. Its outside dimensions are 10 cm long, 6 cm wide and 4 cm high. Ignoring thickness, find the outside area to be painted, including its base.
Deliberately wrong attempt - find the error
Area = 2(10 × 6 + 6 × 4 + 4 × 10) = 248 cm².
Show the error and repaired solution
Why it fails: That formula counts six faces, including a lid. This box has only a base and four side faces.
- The base area is 10 × 6 = 60 cm².
- Two long sides total 2 × 10 × 4 = 80 cm²; two short sides total 2 × 6 × 4 = 48 cm².
- Add the five faces: 60 + 80 + 48 = 188 cm².
Answer: 188 cm²
Check: A closed box has area 248 cm². Removing the 60 cm² lid leaves 188 cm².
Now try it yourself
An isosceles triangle has equal sides 10 cm and base 12 cm. Find its area.
Check your answer
48 cm²
The altitude bisects the base into 6 cm halves. Height = √(10² - 6²) = 8 cm. Area = ½ × 12 × 8 = 48 cm².
Trigonometry: keep the reference angle fixed
EZT practice question
For acute angle A in a right triangle, tan A = 3/4. Find sin A and cos A.
Deliberately wrong attempt - find the error
sin A = 3/4 and cos A = 4/3.
Show the error and repaired solution
Why it fails: Tangent compares opposite with adjacent. Sine and cosine use the hypotenuse, which must first be found in the same ratio triangle.
- Take opposite = 3k and adjacent = 4k, with k positive.
- The hypotenuse is √(9k² + 16k²) = 5k.
- sin A = 3k/5k = 3/5 and cos A = 4k/5k = 4/5.
Answer: sin A = 3/5; cos A = 4/5
Check: Both ratios are below 1, and (3/5)² + (4/5)² = 1.
Now try it yourself
For acute angle B, cos B = 5/13. Find tan B.
Check your answer
12/5
Let adjacent = 5k and hypotenuse = 13k. Opposite = √(169k² - 25k²) = 12k. Thus tan B = opposite/adjacent = 12/5.
Statistics: sort before finding the middle
EZT practice question
For the fictional values 9, 2, 6, 14, 6 and 5, find the median and mean.
Deliberately wrong attempt - find the error
Median = (6 + 14)/2 = 10, using the third and fourth values as listed.
Show the error and repaired solution
Why it fails: The middle positions must be taken after sorting. The order in which observations were recorded is not necessarily numerical order.
- Order the values: 2, 5, 6, 6, 9, 14.
- There are six values. Median = (third + fourth)/2 = (6 + 6)/2 = 6.
- The total is 42. Mean = 42 ÷ 6 = 7.
Answer: Median 6; mean 7
Check: The two middle ordered values are both 6. Each recorded value, including the repeated 6, contributes to the mean.
Now try it yourself
The mean of five values is 9. Four of them are 4, 8, 10 and 11. Find the fifth.
Check your answer
12
The total must be 5 × 9 = 45. The known values sum to 33, so the missing value is 45 - 33 = 12.
Keep losing a mark at the same step?
Bring one question you got wrong and leave the working visible. Ask an ICSE maths tutor to help you name the error, correct it and test the same idea with a different question. The second attempt matters more than copying a perfect first solution.
For subject-focused support, see maths tuition. Agree the topics, schedule and fee with your chosen tutor.
Questions about this practice page
- Does this page give Selina Concise Class 9 textbook answers?
- It gives original EZT practice alongside the book, not a reproduced exercise answer key. The worked questions, wrong attempts and self-check questions were written for this page. Match each skill to the topic in your own book; editions vary.
- Is this the official chapter order for my Selina edition?
- No. The index groups our error-spotting practice by skill, not by textbook chapter or exercise number. We have not verified a current publisher contents page, so we do not claim edition-specific chapter coverage or a current syllabus year.
- Which Selina book is being referred to?
- Selina Publishers lists Concise Mathematics for Standard 9. Its publisher cover identifies I.C.S.E. Concise Mathematics IX Revised. Check your own copy before matching topics; a book identity is not a verified chapter map.
- How should I use the deliberately wrong answers?
- Attempt the question yourself first. Then identify the first incorrect step in the labelled wrong attempt. Open the repair, explain why the rule applies and solve the second question without looking. Do not memorise the incorrect line as a formula.
- Can a tutor help if I keep making the same mistake?
- Bring your attempt and point to the step where you are unsure. Ask the tutor to test whether the gap is in the rule, the diagram or the arithmetic, then give you a fresh question. Agree your topics, timetable and fee directly before starting lessons.