JKBOSE Class 10 Maths Important Questions 2026–27 – All Units with Detailed Solutions
These are chapter-wise important practice questions for all seven units of Class 10 Mathematics. The solutions are written in simple board-exam style. In the examination, write the formula or theorem first, show the working step by step, and clearly mark the final answer.
Quick Unit Index
- Unit I – Number Systems / Real Numbers
- Unit II – Algebra
- Unit III – Coordinate Geometry
- Unit IV – Geometry
- Unit V – Trigonometry
- Unit VI – Mensuration
- Unit VII – Statistics and Probability
Unit I – Number Systems / Real Numbers
Q. 1 Find the HCF of 867 and 255 using Euclid’s division algorithm.
Solution
Apply Euclid’s division algorithm:
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
The last non-zero remainder is 51.
∴ HCF = 51
Q. 2 Prove that √3 is irrational.
Solution
Assume that √3 is rational.
√3 = p/q
where p and q are coprime integers and q ≠ 0.
Squaring both sides:
p² = 3q²
Therefore p² is divisible by 3, so p is divisible by 3. Put p = 3k.
9k² = 3q²
q² = 3k²
Hence q is also divisible by 3. This contradicts the fact that p and q are coprime.
Therefore √3 is irrational.
Q. 3 Without actual division, determine whether 13/3125 has a terminating decimal expansion.
Solution
Factorise the denominator:
3125 = 5⁵
A rational number has a terminating decimal expansion when its denominator in lowest form contains only the prime factors 2 and/or 5.
Here the denominator contains only 5.
Therefore 13/3125 has a terminating decimal expansion.
Q. 4 Find the HCF and LCM of 24 and 36 by prime factorisation.
Solution
24 = 2³ × 3
36 = 2² × 3²
Take the smallest powers for HCF:
HCF = 2² × 3 = 12
Take the greatest powers for LCM:
LCM = 2³ × 3² = 72
∴ HCF = 12 and LCM = 72.
Unit II – Algebra
Q. 5 Find the zeroes of p(x) = x² − 5x + 6 and verify the relationship between zeroes and coefficients.
Solution
Factorise the polynomial:
x² − 5x + 6 = x² − 2x − 3x + 6
= x(x − 2) − 3(x − 2)
= (x − 2)(x − 3)
Therefore the zeroes are 2 and 3.
For ax² + bx + c, sum of zeroes = −b/a and product = c/a.
Sum = 2 + 3 = 5 = −(−5)/1
Product = 2 × 3 = 6 = 6/1
Both relationships are verified.
Q. 6 Solve the pair of linear equations: 2x + 3y = 13 and x − y = 1.
Solution
From x − y = 1,
x = y + 1
Substitute in the first equation:
2(y + 1) + 3y = 13
2y + 2 + 3y = 13
5y = 11
y = 11/5
Therefore,
x = 11/5 + 1 = 16/5
∴ x = 16/5, y = 11/5.
Q. 7 Solve x² − 7x + 12 = 0 by factorisation.
Solution
We need two numbers whose product is 12 and sum is −7.
−3 × −4 = 12
−3 + −4 = −7
x² − 7x + 12 = (x − 3)(x − 4)
Therefore,
x − 3 = 0 or x − 4 = 0
∴ x = 3 or x = 4.
Q. 8 Find the 20th term of the AP 3, 7, 11, 15, ...
Solution
Here a = 3, d = 7 − 3 = 4 and n = 20.
Use:
aₙ = a + (n − 1)d
a₂₀ = 3 + 19 × 4
= 3 + 76 = 79
∴ 20th term = 79.
Q. 9 Find the sum of the first 20 terms of the AP 3, 7, 11, 15, ...
Solution
Here a = 3, d = 4 and n = 20.
Use:
Sₙ = n/2 [2a + (n − 1)d]
S₂₀ = 20/2 [2(3) + 19(4)]
= 10[6 + 76]
= 10 × 82 = 820
∴ S₂₀ = 820.
Unit III – Coordinate Geometry
Q. 10 Find the distance between A(2, 3) and B(8, 11).
Solution
Use the distance formula:
AB = √[(x₂ − x₁)² + (y₂ − y₁)²]
= √[(8 − 2)² + (11 − 3)²]
= √[6² + 8²]
= √100 = 10
∴ AB = 10 units.
Q. 11 Find the midpoint of the line segment joining (−2, 5) and (6, −3).
Solution
Use the midpoint formula:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
M = ((−2 + 6)/2, (5 − 3)/2)
M = (4/2, 2/2)
∴ M = (2, 1).
Q. 12 Find the area of the triangle whose vertices are (1, −1), (−4, 6) and (−3, −5).
Solution
Use the coordinate-area formula:
Area = 1/2 |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
= 1/2 |1(6+5) + (−4)(−5+1) + (−3)(−1−6)|
= 1/2 |11 + 16 + 21|
= 48/2 = 24
∴ Area = 24 square units.
Q. 13 Find the point which divides the line segment joining A(2, 3) and B(8, 9) internally in the ratio 1:2.
Solution
Use the section formula for ratio m:n:
P = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))
Here m = 1 and n = 2.
P = ((1×8 + 2×2)/3, (1×9 + 2×3)/3)
= (12/3, 15/3)
∴ P = (4, 5).
Unit IV – Geometry
Q. 14 State and prove the Basic Proportionality Theorem (BPT).
Solution
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides those two sides in the same ratio.
In △ABC, let D lie on AB and E lie on AC such that DE ∥ BC.
Because DE ∥ BC, △ADE and △ABC are similar.
AD/AB = AE/AC
Using AB = AD + DB and AC = AE + EC, the proportionality gives the corresponding side-division relation:
AD/DB = AE/EC
Hence the two sides are divided in the same ratio. Proved.
Q. 15 In two similar triangles, the corresponding sides are in the ratio 3:5. Find the ratio of their areas.
Solution
For similar triangles, the ratio of areas is equal to the square of the ratio of corresponding sides.
Area₁/Area₂ = (3/5)²
= 9/25
∴ Ratio of areas = 9:25.
Q. 16 Prove Pythagoras theorem.
Solution
Consider a right triangle ABC, right-angled at A. Let AB = c, AC = b and BC = a.
Draw the altitude from A to hypotenuse BC. By similarity of the smaller triangles with the original triangle, the standard relations give:
AB² = BC × BD
AC² = BC × CD
Adding:
AB² + AC² = BC(BD + CD)
Since BD + CD = BC,
AB² + AC² = BC²
Hence, in a right triangle, square of hypotenuse equals the sum of squares of the other two sides. Proved.
Q. 17 From an external point P, tangents PA and PB are drawn to a circle. Prove PA = PB.
Solution
Let O be the centre of the circle. Join OA, OB and OP.
Radius is perpendicular to tangent at the point of contact:
OA ⟂ PA and OB ⟂ PB
In right triangles OAP and OBP:
OA = OB
OP = OP
Therefore the two right triangles are congruent by RHS.
Corresponding tangent lengths are equal.
∴ PA = PB. Proved.
Unit V – Trigonometry
Q. 18 If sin θ = 3/5, find cos θ and tan θ for an acute angle θ.
Solution
Use the identity:
sin²θ + cos²θ = 1
(3/5)² + cos²θ = 1
9/25 + cos²θ = 25/25
cos²θ = 16/25
Since θ is acute, cos θ is positive.
cos θ = 4/5
tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4
∴ cos θ = 4/5 and tan θ = 3/4.
Q. 19 Prove that (1 − cos θ)(1 + cos θ) = sin²θ.
Solution
Multiply the left-hand side:
(1 − cos θ)(1 + cos θ) = 1 − cos²θ
Using sin²θ + cos²θ = 1:
1 − cos²θ = sin²θ
Therefore (1 − cos θ)(1 + cos θ) = sin²θ. Proved.
Q. 20 Evaluate: sin 30° cos 60° + cos 30° sin 60°.
Solution
Use the values:
sin 30° = 1/2, cos 60° = 1/2
cos 30° = √3/2, sin 60° = √3/2
Expression = 1/4 + 3/4
= 1
∴ Answer = 1.
Q. 21 From a point on the ground, the angle of elevation of the top of a tower is 45°. If the distance from the foot of the tower is 20 m, find the height of the tower.
Solution
Let the height of the tower be h.
Using tan θ = perpendicular/base:
tan 45° = h/20
1 = h/20
h = 20 m
∴ Height of tower = 20 m.
Unit VI – Mensuration
Q. 22 Find the area of a sector of a circle with radius 6 cm and central angle 60°.
Solution
Area of sector:
Area = θ/360° × πr²
= 60/360 × π × 6²
= 1/6 × 36π
∴ Area = 6π cm².
Q. 23 Find the area of a quadrant of a circle of radius 14 cm.
Solution
A quadrant is one-fourth of a circle.
Area = 1/4 × πr²
= 1/4 × 22/7 × 14 × 14
= 154 cm²
∴ Area of quadrant = 154 cm².
Q. 24 A cone has radius 7 cm and height 24 cm. Find its slant height and curved surface area.
Solution
Slant height:
l = √(r² + h²)
= √(7² + 24²)
= √625 = 25 cm
Curved surface area:
CSA = πrl
= 22/7 × 7 × 25
∴ CSA = 550 cm².
Q. 25 A cylinder has radius 7 cm and height 10 cm. Find its volume.
Solution
Use:
V = πr²h
= 22/7 × 7² × 10
= 1540 cm³
∴ Volume = 1540 cm³.
Unit VII – Statistics and Probability
Q. 26 Find the mean of 10, 15, 20, 25 and 30.
Solution
Mean = Sum of observations / Number of observations.
Mean = (10 + 15 + 20 + 25 + 30)/5
= 100/5
∴ Mean = 20.
Q. 27 Find the median of 7, 3, 10, 5, 8.
Solution
Arrange the observations in ascending order:
3, 5, 7, 8, 10
There are 5 observations. The middle observation is the third observation.
∴ Median = 7.
Q. 28 Find the probability of getting a king when one card is drawn from a well-shuffled deck of 52 cards.
Solution
Total number of cards = 52.
Number of kings = 4.
P(King) = 4/52
= 1/13
∴ Probability = 1/13.
Q. 29 A die is thrown once. Find the probability of getting a number greater than 4.
Solution
Possible outcomes:
1, 2, 3, 4, 5, 6
Numbers greater than 4 are 5 and 6. Favourable outcomes = 2.
P = 2/6 = 1/3
∴ Probability = 1/3.
Q. 30 Find the mean of the following grouped data using the direct method.
Class intervals: 0–10, 10–20, 20–30, 30–40.
Frequencies: 2, 3, 4, 1.
Find class marks:
xᵢ = 5, 15, 25, 35
Now calculate fᵢxᵢ:
2×5 + 3×15 + 4×25 + 1×35 = 10 + 45 + 100 + 35 = 190
Σfᵢ = 2 + 3 + 4 + 1 = 10
Mean = Σfᵢxᵢ / Σfᵢ = 190/10 = 19
∴ Mean = 19.
Board-exam writing point: For every numerical question, write the formula first, substitute the values on the next line, simplify step by step, and write the unit wherever the answer represents a length, area or volume.
Final Revision Checklist
- Revise formulas before solving the full paper.
- Practise Euclid’s algorithm and irrationality proofs carefully.
- For Algebra, show factorisation and substitution steps.
- For Coordinate Geometry, write the correct formula before substitution.
- For Geometry, write the theorem statement and justify each step.
- For Trigonometry, learn standard values and identities.
- For Mensuration, check units and whether the question asks for CSA, TSA, area or volume.
- For Statistics and Probability, show the complete calculation instead of writing only the final answer.
Conclusion
Revise these unit-wise questions as a practice set. In the board examination, focus on correct formulas, complete working, theorem statements, units and a clearly marked final answer.
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