Class 10 · 10th standard · 2026-27

Class 10 maths worksheets, free, with every step on the page

Three free Class 10 maths worksheets, with every question and every step printed on this page rather than behind a download. They cover the three chapters that carry the most board marks and are the most reliably improvable - quadratics, trigonometry and arithmetic progressions. All three are method rather than insight, which is exactly what a board year has time to fix. Every exercise opens with one question worked through in full.

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3 worksheets · 85 questions · every answer on the page · no sign-up · aligned to NCERT's Class 10 maths book is Mathematics, offered at Standard and Basic levels. These sheets are written to the ideas rather than to one book's chapter order and suit both.

What these practise

Solving quadratics by splitting the middle term, including with surds
The quadratic formula, and the sign of b
The discriminant and the nature of the roots
Forming a quadratic from a described situation
The standard trigonometric values, known cold
Proving identities by converting to sine and cosine
Complementary angles used to simplify
Heights and distances from a labelled diagram
The nth term and the sum of n terms of an AP
Finding a and d from two given terms, and n from the last term

Last updated 2026-07-29

Free, with answers

The worksheets themselves

Every question is on this page, and so is every answer. Nothing is behind a sign-up, and there is no file to download unless you want one - use the print button on any sheet to save it as a PDF.

practice · medium · 35 min

Quadratic equations

Class 10 Maths · Quadratic equations · 30 questions with answers

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Quadratics is the chapter Class 8 factorisation was built for, and it repays that work directly. Three things are asked: solve it, say what kind of roots it has, and turn a described situation into an equation. The third is where marks are actually lost, and it is lost in the first line rather than in the algebra.

Solves a quadratic by splitting the middle term
Applies the quadratic formula correctly, including the sign of b
Uses the discriminant to state the nature of the roots
Finds an unknown coefficient given a condition on the roots
Forms a quadratic equation from a described situation

Exercise 1 of 3 · 10 questions

Solve by factorisation.

Write the answer · Warm-up

Here’s one done for you

The last one looks alarming and works exactly like the others. The product to aim for is (first coefficient) × (last term) = √2 × 5√2 = 10, and the sum is 7, giving 5 and 2. Split the middle term as 5x + 2x and group. Surds in the coefficients change nothing about the method - only about how the arithmetic looks, and children who realise that stop being frightened by these.

1)x² - 5x + 6 = 0
2)2x² + x - 6 = 0
3)x² - 3x - 10 = 0
4)6x² - x - 2 = 0
5)√2x² + 7x + 5√2 = 0
6)x² - 7x + 12 = 0
7)3x² - 10x + 8 = 0
8)x² + 2x - 8 = 0
9)2x² - 5x - 3 = 0
10)x² - 16 = 0

Answers

1) (x - 2)(x - 3) = 0, so x = 2 or x = 3.
2) (2x - 3)(x + 2) = 0, so x = 3/2 or x = -2. Split as 2x² + 4x - 3x - 6.
3) (x - 5)(x + 2) = 0, so x = 5 or x = -2.
4) (3x - 2)(2x + 1) = 0, so x = 2/3 or x = -1/2. Split as 6x² - 4x + 3x - 2.
5) (√2x + 5)(x + √2) = 0, so x = -5/√2 or x = -√2. The product is √2 × 5√2 = 10 and the sum is 7, so split as 5x + 2x.
6) (x - 3)(x - 4) = 0, so x = 3 or x = 4.
7) (3x - 4)(x - 2) = 0, so x = 4/3 or x = 2.
8) (x + 4)(x - 2) = 0, so x = -4 or x = 2.
9) (2x + 1)(x - 3) = 0, so x = -1/2 or x = 3.
10) (x + 4)(x - 4) = 0, so x = 4 or x = -4.

Exercise 2 of 3 · 10 questions

Solve using the quadratic formula, then state the nature of the roots.

Write the answer · Guided

Here’s one done for you

For 2x² - 7x + 3, note that b = -7 and not 7, so -b is +7. That sign is the commonest error in the whole chapter, and it survives because the answer still looks like a number. Write a, b and c on their own line before substituting - three seconds, and it removes the error entirely.

1)2x² - 7x + 3 = 0
2)3x² - 5x + 2 = 0
3)x² + 4x - 5 = 0
4)2x² - 4x + 3 = 0
5)x² - 6x + 9 = 0
6)5x² - 6x + 1 = 0
7)4x² - 4x + 1 = 0
8)x² - 4x + 4 = 0
9)3x² + 2x + 5 = 0
10)2x² + 5x - 3 = 0

Answers

1) D = 49 - 24 = 25, so x = (7 ± 5)/4, giving x = 3 or x = 1/2. Two distinct real roots.
2) D = 25 - 24 = 1, so x = (5 ± 1)/6, giving x = 1 or x = 2/3. Two distinct real roots.
3) D = 16 + 20 = 36, so x = (-4 ± 6)/2, giving x = 1 or x = -5. Two distinct real roots.
4) D = 16 - 24 = -8, which is negative. No real roots.
5) D = 36 - 36 = 0, so x = 3 is a repeated root. Two equal real roots.
6) D = 36 - 20 = 16, so x = (6 ± 4)/10, giving x = 1 or x = 1/5. Two distinct real roots.
7) D = 16 - 16 = 0, so x = 4/8 = 1/2 twice. Two equal real roots.
8) D = 16 - 16 = 0, so x = 2 twice. Two equal real roots.
9) D = 4 - 60 = -56, which is negative, so there are no real roots.
10) D = 25 + 24 = 49, so x = (-5 ± 7)/4, giving x = 1/2 or x = -3. Two distinct real roots.

Exercise 3 of 3 · 10 questions

Form the equation and solve it. Check the answer against the question.

Write the answer · Practice

Here’s one done for you

The train question is the one worth studying, because the algebra is easy and the first line is not. Time equals distance over speed, so the two journey times are 360/x and 360/(x + 5), and the faster one takes an hour less - which gives 360/x - 360/(x + 5) = 1. Once that line is written the rest is routine. In every word problem here, the marks are decided by whether the child can write that first line, so it is worth practising the setting-up separately from the solving.

1)Find k so that kx² - 2kx + 6 = 0 has two equal roots.
2)The sum of a number and its reciprocal is 10/3. Find the number.
3)The product of two consecutive positive integers is 306. Find them.
4)A rectangular plot has an area of 528 m², and its length is one metre more than twice its breadth. Find both.
5)A train covers 360 km at a uniform speed. Had the speed been 5 km/h more, it would have taken one hour less. Find the speed.
6)Find the value of k for which x² + kx + 9 = 0 has equal roots.
7)The sum of the squares of two consecutive positive integers is 313. Find them.
8)The perimeter of a rectangle is 28 m and its area is 48 m². Find its sides.
9)A two-digit number is such that the product of its digits is 18, and when 27 is subtracted the digits interchange. Find the number.
10)The difference of two numbers is 3 and their product is 88. Find them.

Answers

1) Equal roots require D = 0, so 4k² - 24k = 0, giving 4k(k - 6) = 0 and k = 0 or 6. k = 0 makes it non-quadratic, so k = 6.
2) x + 1/x = 10/3 gives 3x² - 10x + 3 = 0. D = 64, so x = (10 ± 8)/6, giving x = 3 or x = 1/3.
3) n(n + 1) = 306 gives n² + n - 306 = 0. D = 1225 = 35², so n = 17 (rejecting the negative root). The integers are 17 and 18.
4) b(2b + 1) = 528 gives 2b² + b - 528 = 0. D = 4225 = 65², so b = 16 and the length is 33 m.
5) 360/x - 360/(x + 5) = 1 gives x² + 5x - 1800 = 0. D = 7225 = 85², so x = 40. The speed is 40 km/h.
6) Equal roots need D = 0, so k² - 36 = 0 and k = ±6.
7) n² + (n + 1)² = 313 gives 2n² + 2n - 312 = 0, so n² + n - 156 = 0 and (n + 13)(n - 12) = 0. Taking the positive value, the integers are 12 and 13.
8) Let the sides be l and 14 - l. Then l(14 - l) = 48, so l² - 14l + 48 = 0 and (l - 6)(l - 8) = 0. The sides are 6 m and 8 m.
9) Let the digits be x and y with xy = 18 and 10x + y - 27 = 10y + x, so x - y = 3. Then x = 6 and y = 3, giving 63. Check: 63 - 27 = 36.
10) Let them be n and n + 3. Then n(n + 3) = 88, so n² + 3n - 88 = 0 and (n + 11)(n - 8) = 0. The numbers are 8 and 11 (or -11 and -8).

While your child works

Write a, b and c on their own line before using the formula. The sign of b is the commonest error in the chapter.
Reject impossible roots explicitly - a negative length or a negative speed. Saying why earns the mark.
In word problems, practise writing the first line without solving. That is where the marks actually are.

Open the Quadratic equations worksheet on its own page

practice · challenging · 35 min

Trigonometry

Class 10 Maths · Trigonometry · 29 questions with answers

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Trigonometry is the chapter that most rewards being systematic. The standard values are a small table, the identities all come from three relationships, and heights and distances always start with a labelled diagram. Almost nothing here needs insight - it needs the table known cold and the diagram drawn first.

Recalls the standard values without hesitation
Evaluates an expression in standard angles
Proves an identity by converting everything to sine and cosine
Uses complementary angles to simplify
Solves a height and distance problem from a labelled diagram

Exercise 1 of 3 · 10 questions

Evaluate.

Write the answer · Warm-up

Here’s one done for you

In the last one, note that sec 30° = 2/√3, so sec² 30° = 4/3 rather than 2/√3. Squaring the whole ratio - not just part of it - is where this question is usually lost. Writing sec 30° as its value on a separate line before squaring takes two seconds and prevents it.

1)sin 30° cos 60° + cos 30° sin 60°
2)2 tan² 45° + cos² 30° - sin² 60°
3)(sin 30° + cos 60°) ÷ tan 45°
4)cos² 45° + sin² 45°
5)(5 cos² 60° + 4 sec² 30° - tan² 45°) ÷ (sin² 30° + cos² 30°)
6)2 sin 30° + 3 cos 60°
7)sin 60° cos 30° - cos 60° sin 30°
8)tan 45° + cot 45°
9)sin² 30° + cos² 30°
10)(tan 60° − tan 30°) ÷ (1 + tan 60° tan 30°)

Answers

1) (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
2) 2(1) + 3/4 - 3/4 = 2.
3) (1/2 + 1/2) ÷ 1 = 1.
4) 1/2 + 1/2 = 1, which is the identity sin²θ + cos²θ = 1 at 45°.
5) The denominator is 1. The numerator is 5(1/4) + 4(4/3) - 1 = 5/4 + 16/3 - 1 = 67/12.
6) 2(1/2) + 3(1/2) = 1 + 1.5 = 2.5.
7) (√3/2)(√3/2) - (1/2)(1/2) = 3/4 - 1/4 = 1/2. It is sin(60° - 30°) = sin 30°.
8) 1 + 1 = 2.
9) 1, since sin²θ + cos²θ = 1 for any angle.
10) (√3 - 1/√3) ÷ (1 + 1) = (2/√3) ÷ 2 = 1/√3. It is tan(60° - 30°) = tan 30°.

Exercise 2 of 3 · 8 questions

Prove each identity. Work on one side only.

Write the answer · Guided

Here’s one done for you

The fourth proof is the model for the whole chapter. Convert everything to sine and cosine, then simplify - (1 - sinθ)(1 + sinθ) becomes 1 - sin²θ, which is cos²θ, which cancels. When an identity looks impossible, converting to sine and cosine is almost always the move, and 'almost always' is close enough to always in a board paper.

1)(1 - cos²θ) cosec²θ = 1
2)(1 + tan²A) ÷ (1 + cot²A) = tan²A
3)(sin A + cos A)² + (sin A - cos A)² = 2
4)sec θ (1 - sin θ)(sec θ + tan θ) = 1
5)Prove (1 + cot²A) sin²A = 1.
6)Prove (sec A - tan A)(sec A + tan A) = 1.
7)Prove (1 - sin²θ) sec²θ = 1.
8)Prove tan θ + cot θ = sec θ cosec θ.

Answers

1) LHS = sin²θ × (1/sin²θ) = 1 = RHS, using 1 - cos²θ = sin²θ.
2) LHS = sec²A ÷ cosec²A = (1/cos²A) ÷ (1/sin²A) = sin²A/cos²A = tan²A = RHS.
3) Expanding gives sin²A + 2 sinA cosA + cos²A + sin²A - 2 sinA cosA + cos²A = 2(sin²A + cos²A) = 2.
4) LHS = (1/cosθ)(1 - sinθ)(1/cosθ + sinθ/cosθ) = (1 - sinθ)(1 + sinθ)/cos²θ = (1 - sin²θ)/cos²θ = cos²θ/cos²θ = 1.
5) 1 + cot²A = cosec²A, and cosec²A × sin²A = (1/sin²A) × sin²A = 1.
6) The left side is sec²A - tan²A, and sec²A - tan²A = 1 is a standard identity.
7) 1 - sin²θ = cos²θ, and cos²θ × sec²θ = cos²θ × (1/cos²θ) = 1.
8) tan θ + cot θ = sin/cos + cos/sin = (sin² + cos²)/(sin cos) = 1/(sin cos) = sec θ cosec θ.

Exercise 3 of 3 · 11 questions

Draw the diagram first, then solve. Use complementary angles where they help.

Write the answer · Practice

Here’s one done for you

For the kite, the height is opposite the angle and the string is the hypotenuse, so sine is the ratio to use - 60 = L sin 60°. Choosing the wrong ratio is the main way these questions go wrong, and a labelled diagram prevents it: mark which side is opposite, which is adjacent and which is the hypotenuse before writing anything. Thirty seconds of diagram saves the whole question.

1)Evaluate sin 18° ÷ cos 72°.
2)Evaluate tan 26° ÷ cot 64°.
3)Evaluate sin 25° cos 65° + cos 25° sin 65°.
4)The shadow of a tower is √3 times its height. Find the sun's angle of elevation.
5)A ladder 10 m long leans against a wall at 60° to the ground. How high up the wall does it reach?
6)A kite is flying at a height of 60 m with the string making 60° with the horizontal. Find the length of the string.
7)Evaluate cos 40° ÷ sin 50°.
8)Evaluate tan 15° tan 75°.
9)Evaluate sec 70° ÷ cosec 20°.
10)A pole 12 m high casts a shadow 12 m long. Find the sun's angle of elevation.
11)From the top of a 30 m building the angle of depression of a car is 30°. Find the distance of the car from the foot of the building.

Answers

1) 1. Since cos 72° = cos(90° - 18°) = sin 18°.
2) 1. Since cot 64° = cot(90° - 26°) = tan 26°.
3) 1. Since cos 65° = sin 25° and sin 65° = cos 25°, the expression becomes sin² 25° + cos² 25° = 1.
4) 30°. If the height is h, the shadow is √3h, so tan θ = h/(√3h) = 1/√3, giving θ = 30°.
5) 5√3 m, about 8.66 m. The height is 10 sin 60° = 10 × √3/2.
6) 40√3 m, about 69.3 m. The height is opposite the 60° angle, so 60 = L sin 60°, giving L = 60 ÷ (√3/2) = 120/√3.
7) 1. cos 40° = sin 50°, because they are complementary.
8) 1. tan 75° = cot 15°, and tan 15° cot 15° = 1.
9) 1. sec 70° = cosec 20°, because 70° and 20° are complementary.
10) tan θ = 12/12 = 1, so θ = 45°.
11) tan 30° = 30/d, so d = 30 ÷ (1/√3) = 30√3 m, about 51.96 m.

While your child works

The standard values table must be automatic. Everything else in the chapter depends on it, and hesitation there slows every question.
When an identity looks impossible, convert everything to sine and cosine. It works nearly every time.
For heights and distances, draw and label the diagram before writing any equation. Choosing the wrong ratio is the main error.

Open the Trigonometry worksheet on its own page

practice · medium · 30 min

Arithmetic progressions

Class 10 Maths · Arithmetic progressions · 26 questions with answers

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Arithmetic progressions is the most reliably scored chapter in the Class 10 paper: two formulas, a fixed method, and questions that repeat year after year. It is also where careless marks go - a common difference read as the second term, or n found and then not used. Both are habits rather than misunderstandings.

Identifies the first term and the common difference correctly
Finds the nth term, and finds which term a given value is
Finds the sum of the first n terms
Solves for a and d from two given terms
Applies an AP to a described situation

Exercise 1 of 3 · 10 questions

Find what is asked.

Write the answer · Warm-up

Here’s one done for you

The last question is the useful one. If solving for n does not give a positive whole number, the value is not a term of the AP - and saying that explicitly is the answer rather than a dead end. Children often assume they have made an arithmetic mistake and search for one. The non-integer is the result.

1)Find the 10th term of 2, 7, 12, ...
2)Which term of 3, 8, 13, ... is 78?
3)Find the common difference of 10, 7, 4, ...
4)Find the 15th term of the AP whose first term is -5 and common difference is 3.
5)Is 100 a term of the AP 3, 7, 11, ...?
6)Find the 12th term of 5, 9, 13, ...
7)Which term of 7, 13, 19, ... is 103?
8)Find the first term of an AP whose 5th term is 20 and common difference is 3.
9)Is 68 a term of the AP 4, 9, 14, ...?
10)Find the common difference of an AP whose 3rd term is 11 and 8th term is 31.

Answers

1) a = 2, d = 5, so a₁₀ = 2 + 9(5) = 47.
2) 3 + (n - 1)5 = 78 gives (n - 1)5 = 75 and n = 16. It is the 16th term.
3) d = 7 - 10 = -3.
4) a₁₅ = -5 + 14(3) = 37.
5) 3 + (n - 1)4 = 100 gives (n - 1)4 = 97, which is not a whole number when divided by 4. So 100 is not a term of this AP.
6) a = 5, d = 4, so a₁₂ = 5 + 11(4) = 49.
7) a = 7, d = 6, so 7 + (n - 1)6 = 103 gives n - 1 = 16 and n = 17.
8) a + 4(3) = 20, so a = 8.
9) 4 + (n - 1)5 = 68 gives (n - 1)5 = 64, which is not a whole number, so 68 is not a term.
10) a + 7d - (a + 2d) = 31 - 11, so 5d = 20 and d = 4.

Exercise 2 of 3 · 8 questions

Find the sum.

Write the answer · Guided

Here’s one done for you

When the last term is known, use S = (n/2)(a + l) rather than the longer formula - it is faster and there is less to get wrong. The step people skip is finding n first, using n = (l - a)/d + 1. That '+1' is the commonest slip in the chapter: from 12 to 99 in steps of 3 there are 29 gaps and 30 numbers.

1)The sum of the first 20 terms of 1, 3, 5, ...
2)The sum of the first 15 terms of the AP with a = 3 and d = 4.
3)How many two-digit numbers are divisible by 3?
4)Find 7 + 10.5 + 14 + ... + 84.
5)Find the sum of the first 25 terms of 3, 7, 11, ...
6)Find the sum of the first 12 terms of 2, 5, 8, ...
7)How many two-digit numbers are divisible by 4, and what is their sum?
8)Find the sum of all multiples of 7 lying between 100 and 200.

Answers

1) S₂₀ = (20/2)[2(1) + 19(2)] = 10(40) = 400.
2) S₁₅ = (15/2)[2(3) + 14(4)] = (15/2)(62) = 465.
3) They form the AP 12, 15, ..., 99. n = (99 - 12)/3 + 1 = 30. There are 30 such numbers.
4) a = 7, d = 3.5, last term 84. n = (84 - 7)/3.5 + 1 = 23. S = (23/2)(7 + 84) = (23/2)(91) = 1046.5.
5) S₂₅ = 25/2 [2(3) + 24(4)] = 25/2 × 102 = 1,275.
6) S₁₂ = 12/2 [2(2) + 11(3)] = 6 × 37 = 222.
7) They run 12, 16, ..., 96, so n = 22. Sum = 22/2 (12 + 96) = 11 × 108 = 1,188.
8) They run 105, 112, ..., 196, so n = 14. Sum = 14/2 (105 + 196) = 7 × 301 = 2,107.

Exercise 3 of 3 · 8 questions

Work these out.

Write the answer · Practice

Here’s one done for you

The second question is the one worth learning properly, because it looks unfamiliar and has a one-line method: the nth term is the sum of n terms minus the sum of (n-1) terms. Here that is S₁₀ - S₉ = 350 - 288 = 62. Any question giving you Sₙ as a formula is asking you to use that relationship, and recognising it turns a hard-looking question into an easy one.

1)The 5th term of an AP is 19 and the 9th is 35. Find the first term and the common difference.
2)The sum of the first n terms of an AP is 3n² + 5n. Find its 10th term.
3)A person saves ₹100 in the first month and ₹50 more each month than the month before. How much will they have saved in a year?
4)Find the number of terms in the AP 18, 15.5, 13, ..., -47.
5)The 4th term of an AP is 14 and the 10th is 32. Find the first term and the common difference.
6)The sum of the first 10 terms of an AP is 155 and the first term is 2. Find the common difference.
7)Find the middle term of the AP 6, 13, 20, ..., 216.
8)Which term of the AP 5, 2, -1, ... is -49?

Answers

1) a + 4d = 19 and a + 8d = 35. Subtracting gives 4d = 16, so d = 4 and a = 3.
2) a₁₀ = S₁₀ - S₉ = (300 + 50) - (243 + 45) = 350 - 288 = 62.
3) a = 100, d = 50, n = 12. S = (12/2)[200 + 11(50)] = 6(750) = ₹4,500.
4) d = -2.5. Then 18 + (n - 1)(-2.5) = -47 gives (n - 1)(-2.5) = -65, so n - 1 = 26 and n = 27.
5) a + 3d = 14 and a + 9d = 32, so 6d = 18, d = 3 and a = 5.
6) 155 = 10/2 [2(2) + 9d] = 5(4 + 9d), so 31 = 4 + 9d and d = 3.
7) a = 6, d = 7, and 6 + (n - 1)7 = 216 gives n = 31. The middle term is the 16th: 6 + 15(7) = 111.
8) a = 5, d = -3, and 5 + (n - 1)(-3) = -49 gives (n - 1)(-3) = -54, so n = 19.

While your child works

Check the common difference is second minus first, not the second term itself. That single slip ruins otherwise perfect work.
n = (l - a)/d + 1. The '+1' is the commonest error in the chapter - 29 gaps means 30 terms.
If solving for n gives a fraction, the value is not a term of the AP. That is the answer, not a mistake.

Open the Arithmetic progressions worksheet on its own page

Why these three chapters?

Because they carry a large share of the paper and they are the three most improvable in the time a board year actually has.

Quadratics rewards Class 8 factorisation directly. Trigonometry is a small table of values plus one technique - convert to sine and cosine - repeated. Arithmetic progressions is two formulas and a fixed method, with questions that repeat year after year.

None of the three needs insight. They need the method drilled until it is automatic, which is the one kind of improvement a board year can reliably deliver. The wider argument about what the paper rewards is on our Class 10 maths tuition page.

How should we use these at home?

Thirty to thirty-five minutes each, and then again under time pressure closer to the exam. Every exercise opens with one question worked through in full.

In quadratics, write a, b and c on their own line before using the formula. The sign of b is the commonest error.
In trigonometry, draw the diagram first for every height and distance question.
In AP, check the '+1' when counting terms. Twenty-nine gaps means thirty terms.
Let your child mark their own work. Every answer is on the page for exactly that reason.

Where do Class 10 children usually stall?

Three places, and all three are careless rather than conceptual - which is the good news.

The sign of b in the quadratic formula. For 2x² - 7x + 3, b is -7, so -b is +7. The wrong answer still looks like a number, which is why it survives.
Squaring only part of a trigonometric ratio. sec² 30° is 4/3, not 2/√3.
The first line of a word problem. The algebra is usually fine; translating the situation is not. Practise writing the equation without solving it.

All three respond to a fortnight of deliberate checking, which is why presentation and method are the first things worth working on in a board year - see the Class 10 board year.

Questions parents ask

Class 10 Maths worksheets - your questions

Are these Class 10 maths worksheets free?

Yes. Every question and every answer is on this page, with no sign-up, no email wall and no account. The printable version puts the answers on a separate final sheet.

Do these suit Maths Standard and Maths Basic?

Yes. Both levels cover the same syllabus at different depths, and these three chapters appear in both. The word problems are the part where Standard demands more.

What does the discriminant tell us?

Positive means two distinct real roots, zero means two equal real roots, negative means no real roots. It answers 'what kind of roots' without solving the equation at all.

How do I prove a trigonometric identity?

Work on one side only and convert everything to sine and cosine before simplifying. That one approach handles almost every identity in the Class 10 syllabus.

Why does my child lose marks in word problems?

Almost always in the first line rather than in the algebra. Practise writing the equation from the situation without solving it - the setting-up is the skill the marks are for.

Are these matched to NCERT?

They are written to the ideas the Class 10 paper tests rather than to one book's chapter order, so they work whether your school uses NCERT or its own text.

How long should a Class 10 child spend on one sheet?

Thirty to thirty-five minutes first time. Closer to the exam, do them again under time pressure - working to time is a separate skill from working correctly.

Can I print these worksheets?

Yes. They print cleanly on A4 with the answers on a separate final page, and nothing needs to be downloaded or signed up for.

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Class 10 maths worksheets, free, with every step on the page

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Quadratic equations

Trigonometry

Arithmetic progressions

Class 10 Maths - the guide

Why these three chapters?

How should we use these at home?

Where do Class 10 children usually stall?

Class 10 Maths worksheets - your questions

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