CBSE Class 10 Mathematics (Standard) Sample Paper 2026-27: Solutions

The greatest number that divides 134 and 188 leaving remainders 4 and 6.

1. Take away the remainders: the number divides 134 − 4 = 130 and 188 − 6 = 182 exactly.
2. 130 = 2 × 5 × 13 and 182 = 2 × 7 × 13.
3. Common factors: 2 × 13 = 26, so HCF(130, 182) = 26.

Answer: 26 — option (B).

f(x) = px² + qx + r with p ≠ 0 and p + r = q. Find one zero of f(x).

1. Put x = −1: f(−1) = p − q + r = (p + r) − q = 0, so −1 is a zero.
2. Product of the zeroes = r/p, so the other zero is (r/p) ÷ (−1) = −r/p.
3. Among the options, −r/p is listed.

Answer: −r/p — option (C).

A pair of lines meets only at (5, −1); one line is x − y = 6. Which of I: 3x − 3y = 18, II: 2x − 3y = 13, III: 2x − 3y = 16 could be the other line?

1. The other line must pass through (5, −1) and must not be the same line as x − y = 6.
2. I: 3(5) − 3(−1) = 18 — it passes through the point, but 3x − 3y = 18 is just 3 × (x − y = 6), the same line, so the lines would meet everywhere, not at one point.
3. II: 2(5) − 3(−1) = 13 ✓, and its slope (2/3) differs from 1, so it meets x − y = 6 at exactly one point.
4. III: 2(5) − 3(−1) = 13 ≠ 16, so it does not pass through (5, −1).

Answer: II only — option (B).

(1 − p) is a root of x² + px + 1 − p = 0. Find its roots.

1. Substitute x = 1 − p: (1 − p)² + p(1 − p) + (1 − p) = 0.
2. Take (1 − p) common: (1 − p)[(1 − p) + p + 1] = (1 − p) × 2 = 0, so p = 1.
3. The equation becomes x² + x = 0, i.e. x(x + 1) = 0.

Answer: 0 and −1 — option (C).

The middle term of the A.P. 10, 7, 4, …, −62.

1. a = 10, d = −3. Number of terms: −62 = 10 + (n − 1)(−3) gives n − 1 = 24, so n = 25.
2. With 25 terms the middle one is the 13th.
3. a₁₃ = 10 + 12 × (−3) = −26.

Answer: −26 — option (A).

Two similar triangles have perimeters 56 cm and 70 cm. A side of the first is 14 cm; find the corresponding side of the second.

1. For similar triangles, perimeters are in the same ratio as corresponding sides.
2. 14 / x = 56 / 70, so x = 14 × 70 / 56 = 17.5 cm.

Answer: 17.5 cm — option (D).

Which point lies on the perpendicular bisector of the segment joining A(−3, −4) and B(3, 4)?

1. The perpendicular bisector passes through the midpoint of AB.
2. Midpoint = ((−3 + 3)/2, (−4 + 4)/2) = (0, 0).

Answer: (0, 0) — option (A).

tan(A + B) = √3 and tan(A − B) = 1/√3, with 0° < A + B < 90° and A > B. Find A and B.

1. tan 60° = √3, so A + B = 60°.
2. tan 30° = 1/√3, so A − B = 30°.
3. Adding: 2A = 90°, A = 45°; then B = 15°.

Answer: A = 45°, B = 15° — option (C).

T-shirts numbered 4 to 99 are in a box; one is drawn at random. Probability that its number is a perfect square or a perfect cube.

1. Numbers 4 to 99: 99 − 4 + 1 = 96 outcomes.
2. Perfect squares: 4, 9, 16, 25, 36, 49, 64, 81 (8 numbers). Perfect cubes: 8, 27, 64 (3 numbers).
3. 64 is in both lists, so the favourable count is 8 + 3 − 1 = 10.
4. P = 10/96 = 5/48.

Answer: 5/48 — option (D).

A car wheel has diameter 21 cm. How many complete revolutions does it make in 66 km?

1. Circumference = πd = (22/7) × 21 = 66 cm: one revolution covers 66 cm.
2. 66 km = 66 × 1,00,000 cm = 66,00,000 cm.
3. Revolutions = 66,00,000 ÷ 66 = 1,00,000 = 10⁵.

Answer: 10⁵ (1,00,000) revolutions — option (B).

Two cubes of volume 64 cm³ each are joined end to end. Surface area of the cuboid formed.

1. Side of each cube = ∛64 = 4 cm.
2. The cuboid is 8 cm × 4 cm × 4 cm.
3. Surface area = 2(lb + bh + hl) = 2(32 + 16 + 32) = 160 cm².

Answer: 160 cm² — option (B).

A group's mean age is 35 years; men average 38 and women 30. What percentage of the group are women?

1. Let there be m men and w women: (38m + 30w)/(m + w) = 35.
2. 38m + 30w = 35m + 35w gives 3m = 5w, so m : w = 5 : 3.
3. Women = 3 of every 8 people = 3/8 × 100 = 37.5%.

Answer: 37.5% — option (D).

Two dice are rolled. Probability that both show a number less than 4.

1. Total outcomes = 6 × 6 = 36.
2. Each die must show 1, 2 or 3: 3 × 3 = 9 favourable outcomes.
3. P = 9/36 = 1/4.

Answer: 1/4 — option (A).

PM is a median of triangle PQR with P(5, −6), Q(6, 4), R(0, 0). Find PM.

1. M is the midpoint of QR: ((6 + 0)/2, (4 + 0)/2) = (3, 2).
2. PM = √[(5 − 3)² + (−6 − 2)²] = √(4 + 64) = √68 = 2√17 units.

Answer: 2√17 units — option (A).

An arc of length 6 cm bounds a sector of area 24 cm². Find the radius.

1. Area of a sector = ½ × arc length × radius.
2. 24 = ½ × 6 × r, so r = 8 cm.

Answer: 8 cm — option (B).

Mean = 35.5 and median = 32. Find the mode.

1. Empirical relation: Mode = 3 × Median − 2 × Mean.
2. Mode = 3 × 32 − 2 × 35.5 = 96 − 71 = 25.

Answer: 25 — option (C).

A kite string 52 m long makes angle θ with the ground, tan θ = 12/5. Height of the kite.

1. tan θ = 12/5 means opposite : adjacent = 12 : 5, so the hypotenuse is √(144 + 25) = 13 parts.
2. sin θ = 12/13. The string is the hypotenuse, so height = 52 × 12/13 = 48 m.

Answer: 48 m — option (C).

2 diamonds and 4 spades are missing from a pack of 52. Probability that a card drawn is a heart.

1. Cards left = 52 − 6 = 46. All 13 hearts are still there.
2. P(heart) = 13/46.

Answer: 13/46 — option (B).

Assertion: three vertices of a parallelogram in order are (−1, −6), (2, −5), (7, 2), so the fourth is (4, 1). Reason: diagonals of a parallelogram bisect each other.

1. Call the vertices A(−1, −6), B(2, −5), C(7, 2) and D(x, y). The diagonals AC and BD have the same midpoint.
2. Midpoint of AC = (3, −2). So (2 + x)/2 = 3 and (−5 + y)/2 = −2, giving D = (4, 1). The assertion is true.
3. The reason is a true property, and it is exactly what we used.

Answer: Both true, and the reason explains the assertion — option (A).

Assertion: if the zeroes of (2k − 1)x² + 4x − 3 are reciprocals of each other, k = −1. Reason: if a = c, the zeroes of ax² + bx + c are reciprocals.

1. Zeroes α and 1/α have product 1, and the product of zeroes is c/a.
2. So −3 / (2k − 1) = 1, i.e. 2k − 1 = −3, k = −1. The assertion is true.
3. The reason is true (c/a = 1 when a = c) and it is the rule used above.

Answer: Both true, and the reason explains the assertion — option (A).

A hall is 9.75 m × 6.75 m × 5.25 m. Longest ruler that measures all three dimensions exactly.

1. In centimetres: 975, 675 and 525. The ruler's length is their HCF.
2. 975 = 3 × 5² × 13; 675 = 3³ × 5²; 525 = 3 × 5² × 7.
3. HCF = 3 × 5² = 75.

Answer: 75 cm (0.75 m).

Three bells toll every 5, 6 and 8 minutes and toll together at 10:00 a.m. How many more times do they toll together by 7:00 p.m.?

1. They toll together every LCM(5, 6, 8) minutes. 5 = 5, 6 = 2 × 3, 8 = 2³, so LCM = 2³ × 3 × 5 = 120 minutes = 2 hours.
2. After 10:00 a.m.: 12:00 noon, 2:00 p.m., 4:00 p.m., 6:00 p.m. The next, 8:00 p.m., is after closing.

Answer: 4 more times.

ABCD is a trapezium with AB ∥ DC; the diagonals meet at O. Show that AO/BO = CO/DO.

1. In triangles AOB and COD: ∠OAB = ∠OCD and ∠OBA = ∠ODC (alternate angles, since AB ∥ DC).
2. So triangle AOB ~ triangle COD (AA similarity).
3. Corresponding sides are proportional: AO/CO = BO/DO.
4. Rearranging, AO/BO = CO/DO.

Answer: Proved: AO/BO = CO/DO.

In CBSE's figure, ∠PQR = ∠QSP, PQ = 6 cm and PS = 3 cm. Find PR.

Uses a figure printed in CBSE's paper — open the paper above to see it.

1. Compare triangles PQR and PSQ: ∠PQR = ∠PSQ (given) and ∠QPR = ∠SPQ (the same angle at P).
2. So triangle PQR ~ triangle PSQ (AA), and PQ/PS = PR/PQ.
3. 6/3 = PR/6, so PR = 12 cm.

Answer: PR = 12 cm.

If tan θ = 1/√5, find (cosec²θ − sec²θ)/(cosec²θ + sec²θ).

1. sec²θ = 1 + tan²θ = 1 + 1/5 = 6/5.
2. cot θ = √5, so cosec²θ = 1 + cot²θ = 1 + 5 = 6.
3. (6 − 6/5)/(6 + 6/5) = (24/5)/(36/5) = 24/36 = 2/3.

Answer: 2/3.

Area of a sector of a circle of diameter 28 cm whose arc is 22 cm long.

1. Radius = 28/2 = 14 cm.
2. Area of sector = ½ × arc length × radius = ½ × 22 × 14 = 154 cm².

Answer: 154 cm².

Two dice are thrown. Probability that the product of the numbers is a prime number.

1. A product of two whole numbers is prime only when one number is 1 and the other is prime (2, 3 or 5).
2. Favourable: (1,2), (2,1), (1,3), (3,1), (1,5), (5,1) — 6 outcomes out of 36.
3. P = 6/36 = 1/6.

Answer: 1/6.

Given that √3 is irrational, prove that 2 + 3√3 is irrational.

1. Suppose, to the contrary, that 2 + 3√3 is rational: 2 + 3√3 = a/b with a, b integers and b ≠ 0.
2. Then 3√3 = a/b − 2, so √3 = (a − 2b)/(3b).
3. a − 2b and 3b are integers and 3b ≠ 0, so the right side is rational — that would make √3 rational.
4. This contradicts the given fact that √3 is irrational. So the assumption is false.

Answer: 2 + 3√3 is irrational.

α and β are the zeroes of 2x² − 8x + 5. Find (α + 1/β)(β + 1/α).

1. α + β = −(−8)/2 = 4 and αβ = 5/2.
2. Expand: (α + 1/β)(β + 1/α) = αβ + 1 + 1 + 1/(αβ) = αβ + 2 + 1/(αβ).
3. = 5/2 + 2 + 2/5 = 25/10 + 20/10 + 4/10 = 49/10.

Answer: 49/10.

A polygon's smallest exterior angle is 8° and each next exterior angle is 4° more. How many sides does it have?

1. The exterior angles form an A.P. with a = 8 and d = 4, and the exterior angles of any polygon add up to 360°.
2. Sₙ = n/2 [2(8) + (n − 1)4] = n/2 (4n + 12) = 2n² + 6n = 360.
3. n² + 3n − 180 = 0, i.e. (n + 15)(n − 12) = 0.
4. n cannot be negative, so n = 12. (Check: the largest angle is 8 + 11 × 4 = 52°, still less than 180°.)

Answer: 12 sides.

Sum of the integers between 1 and 400 that are multiples of both 4 and 5.

1. Multiples of both 4 and 5 are multiples of 20: 20, 40, …, 380 (400 is not 'between' 1 and 400).
2. a = 20, d = 20, last term 380: 380 = 20 + (n − 1)20 gives n = 19.
3. S₁₉ = 19/2 × (20 + 380) = 19 × 200 = 3800.

Answer: 3800.

Prove that a parallelogram circumscribing a circle is a rhombus.

1. Let parallelogram ABCD touch the circle at P on AB, Q on BC, R on CD and S on DA.
2. Tangents from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
3. Add them: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), i.e. AB + CD = AD + BC.
4. In a parallelogram AB = CD and AD = BC, so 2AB = 2AD, AB = AD.
5. A parallelogram with two adjacent sides equal has all four sides equal.

Answer: ABCD is a rhombus.

In what ratio does the x-axis divide the segment joining (−4, −6) and (−1, 7)? Find the point of division.

1. Let the ratio be k : 1. The point is ((−k − 4)/(k + 1), (7k − 6)/(k + 1)).
2. On the x-axis the y-coordinate is 0: 7k − 6 = 0, k = 6/7, so the ratio is 6 : 7.
3. x = (6 × (−1) + 7 × (−4))/(6 + 7) = (−6 − 28)/13 = −34/13.

Answer: Ratio 6 : 7; point (−34/13, 0).

Jyoti simplified cos θ/(1 − sin θ) + (1 − sin θ)/cos θ in four steps (shown in CBSE's paper). Find her error and the correct answer.

1. Step 1 (common denominator) is right: [cos²θ + (1 − sin θ)²] / [(1 − sin θ) cos θ].
2. Step 2 is wrong: she replaced (1 − sin θ)² by cos²θ. But (1 − sin θ)² = 1 − 2 sin θ + sin²θ, while cos²θ = 1 − sin²θ.
3. Correct working: numerator = cos²θ + 1 − 2 sin θ + sin²θ = 2 − 2 sin θ = 2(1 − sin θ).
4. So the expression = 2(1 − sin θ) / [(1 − sin θ) cos θ] = 2/cos θ = 2 sec θ.

Answer: The error is in step 2; the correct answer is 2 sec θ.

If 1/(sin θ − cos θ) = cosec θ/√2, prove that [1/(sin θ + cos θ)]² = sec²θ/2.

1. cosec θ/√2 = 1/(√2 sin θ), so the given line says sin θ − cos θ = √2 sin θ.
2. Square both sides: sin²θ + cos²θ − 2 sin θ cos θ = 2 sin²θ, so 2 sin θ cos θ = 1 − 2 sin²θ.
3. Left side to prove: 1/(sin θ + cos θ)² = 1/(1 + 2 sin θ cos θ) = 1/(1 + 1 − 2 sin²θ) = 1/(2(1 − sin²θ)).
4. = 1/(2 cos²θ) = sec²θ/2.

Answer: Proved.

A 50 m × 40 m park has a rectangular pool in the middle with a grass strip of uniform width around it. The strip's area is 1184 m². Find the pool's dimensions.

1. Let the strip be x m wide. The pool is (50 − 2x) m by (40 − 2x) m.
2. Park area − pool area = strip area: 2000 − (50 − 2x)(40 − 2x) = 1184, so (50 − 2x)(40 − 2x) = 816.
3. Expand: 2000 − 180x + 4x² = 816, so 4x² − 180x + 1184 = 0, i.e. x² − 45x + 296 = 0.
4. (x − 8)(x − 37) = 0, so x = 8 or 37. x = 37 would make the pool's sides negative, so x = 8.
5. Pool: 50 − 16 = 34 m by 40 − 16 = 24 m. (Check: 34 × 24 = 816 m² and 2000 − 816 = 1184 m².)

Answer: The pool is 34 m by 24 m.

Prove the basic proportionality theorem (a line parallel to one side of a triangle divides the other two sides in the same ratio). Then: PQRS is a trapezium with PQ ∥ SR; X on PS and Y on QR with XY ∥ PQ. Show PX/XS = QY/YR.

1. Theorem. In triangle ABC let DE ∥ BC with D on AB and E on AC. To prove: AD/DB = AE/EC.
2. Join BE and CD, and draw EM ⊥ AB and DN ⊥ AC.
3. ar(ADE) = ½ × AD × EM and ar(BDE) = ½ × DB × EM, so ar(ADE)/ar(BDE) = AD/DB.
4. Likewise ar(ADE) = ½ × AE × DN and ar(DEC) = ½ × EC × DN, so ar(ADE)/ar(DEC) = AE/EC.
5. Triangles BDE and DEC stand on the same base DE between the same parallels DE and BC, so ar(BDE) = ar(DEC). Hence AD/DB = AE/EC.
6. Application. Join PR; let it cut XY at M. Since XY ∥ PQ ∥ SR, XM ∥ SR and MY ∥ PQ.
7. In triangle PSR, XM ∥ SR, so PX/XS = PM/MR (theorem).
8. In triangle RPQ, MY ∥ PQ, so RM/MP = RY/YQ, i.e. PM/MR = QY/YR.
9. So PX/XS = QY/YR.

Answer: Proved.

A tent is a cylinder (height 3 m, radius 14 m) with a cone on top; total height 13.5 m. Canvas needed, with 26 m² extra for stitching and wastage, and its cost at ₹250 per m².

1. Height of the cone = 13.5 − 3 = 10.5 m. Slant height l = √(14² + 10.5²) = √(196 + 110.25) = √306.25 = 17.5 m.
2. Curved surface of the cone = πrl = (22/7) × 14 × 17.5 = 770 m².
3. Curved surface of the cylinder = 2πrh = 2 × (22/7) × 14 × 3 = 264 m².
4. Canvas = 770 + 264 + 26 = 1060 m² (no canvas for the floor).
5. Cost = 1060 × 250 = ₹2,65,000.

Answer: 1060 m² of canvas; cost ₹2,65,000.

A solid toy is a hemisphere of radius 3.5 cm with a cone of the same radius on top; total wood 166⅚ cm³. Height of the cone, and the cost of painting the hemisphere at ₹15 per cm².

1. Volume = cone + hemisphere: 1001/6 = ⅓πr²h + ⅔πr³ with r = 3.5 and π = 22/7.
2. ⅔πr³ = ⅔ × (22/7) × 42.875 = 539/6 cm³; ⅓πr² = ⅓ × (22/7) × 12.25 = 77/6 cm².
3. So 1001/6 = (77/6)h + 539/6, giving 77h = 462 and h = 6 cm.
4. Curved surface of the hemisphere = 2πr² = 2 × (22/7) × 12.25 = 77 cm².
5. Cost = 77 × 15 = ₹1,155.

Answer: Cone height 6 cm; painting cost ₹1,155.

Manish on the ground sees a kite 200 m away at an elevation of 30°. Mahesh, on a 50 m high roof on the other side, sees it at 45°. Distance between the kite and Mahesh.

1. The 200 m from Manish to the kite is the slant line, so the kite's height = 200 × sin 30° = 100 m.
2. Mahesh is 50 m up, so the kite is 100 − 50 = 50 m above his eye level.
3. Along his line of sight, sin 45° = 50 / d, so d = 50 / (1/√2) = 50√2 m (about 70.7 m).

Answer: 50√2 m.

From the top of a lighthouse the angles of depression of two ships on the same side are 45° and 30°; the ships are 200 m apart, one behind the other. Height of the lighthouse.

1. Let the height be h. The nearer ship (45°) is h / tan 45° = h metres from the foot.
2. The farther ship (30°) is h / tan 30° = h√3 metres from the foot.
3. h√3 − h = 200, so h = 200/(√3 − 1) = 200(√3 + 1)/2 = 100(√3 + 1) m (about 273.2 m).

Answer: 100(√3 + 1) m.

Small sanitizer bottles cost ₹10 and large ₹15; June sales were ₹12,750. Write the linear equation for June's sales.

1. Let x small and y large bottles be sold in June.
2. Sales: 10x + 15y = 12750 (or, dividing by 5, 2x + 3y = 2550).

Answer: 10x + 15y = 12750.

Both prices went up by ₹2 and July sales were ₹34,250. Write the equation for July's sales.

1. New prices: ₹12 and ₹17. With x small and y large bottles sold in July:
2. 12x + 17y = 34250.

Answer: 12x + 17y = 34250.

June: 1000 bottles were sold for ₹12,750. How many of each size?

1. x + y = 1000 and 10x + 15y = 12750.
2. From the first, x = 1000 − y. Then 10(1000 − y) + 15y = 12750, so 5y = 2750 and y = 550.
3. x = 1000 − 550 = 450.

Answer: 450 small and 550 large bottles.

July: 2500 bottles were sold for ₹34,250. How many of each size?

1. x + y = 2500 and 12x + 17y = 34250.
2. x = 2500 − y, so 12(2500 − y) + 17y = 34250, 30000 + 5y = 34250, y = 850.
3. x = 2500 − 850 = 1650.

Answer: 1650 small and 850 large bottles.

Tangents PA and PB are drawn from P to a circle with centre O (radius 9 m, OP = 18 m). Find ∠OAP.

Uses a figure printed in CBSE's paper — open the paper above to see it.

1. A tangent is perpendicular to the radius at the point of contact.

Answer: ∠OAP = 90°.

Find the length of the tangent PA.

Uses a figure printed in CBSE's paper — open the paper above to see it.

1. In right triangle OAP: PA² = OP² − OA² = 18² − 9² = 324 − 81 = 243.
2. PA = √243 = 9√3 m (about 15.6 m).

Answer: 9√3 m.

Find ∠AOB.

Uses a figure printed in CBSE's paper — open the paper above to see it.

1. In right triangle OAP, cos ∠AOP = OA/OP = 9/18 = 1/2, so ∠AOP = 60°.
2. By symmetry ∠BOP = 60° too, so ∠AOB = 120°.

Answer: ∠AOB = 120°.

If PA and PB were inclined to each other at 60°, what would their length be?

Uses a figure printed in CBSE's paper — open the paper above to see it.

1. OP bisects the angle between the tangents, so ∠OPA = 30°.
2. In right triangle OAP, tan 30° = OA/PA = 9/PA, so PA = 9/(1/√3) = 9√3 m.

Answer: 9√3 m.

Marks (out of 50) grouped 0-10, 10-20, 20-30, 30-40, 40-50 with 3, 6, 12, 15, 14 students. Which class is the most common performance range?

1. The most common range is the class with the highest frequency: 15 students in 30-40.

Answer: 30-40 (the modal class).

Find the class interval containing the median.

1. Cumulative frequencies: 3, 9, 21, 36, 50. N = 50, so N/2 = 25.
2. The first cumulative frequency above 25 is 36, for 30-40.

Answer: 30-40.

Find the mean (average) marks.

1. Class marks: 5, 15, 25, 35, 45. fx: 15, 90, 300, 525, 630; Σfx = 1560.
2. Mean = Σfx / Σf = 1560 / 50 = 31.2.

Answer: 31.2 marks.

Find the mode of the data.

1. Modal class 30-40: l = 30, h = 10, f₁ = 15, f₀ = 12 (class before), f₂ = 14 (class after).
2. Mode = l + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h = 30 + 3/(30 − 26) × 10 = 30 + 7.5 = 37.5.

Answer: 37.5 marks.