CBSE Class 10 Maths (Standard) Board Paper 2026 (30/1/1): Solutions

Find the HCF of 960 and 432.

1. Divide 960 by 432 to get a quotient of 2 and a remainder of 96.
2. Divide 432 by 96 to get a quotient of 4 and a remainder of 48.
3. Divide 96 by 48 to get a quotient of 2 and a remainder of 0.
4. Since the remainder is 0, the last non-zero remainder is the HCF, which is 48.

Answer: HCF = 48 — option (a).

Is the natural number 2 prime, composite, both, or neither?

1. A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.
2. The number 2 has only two factors, 1 and 2, so it is a prime number.

Answer: 2 is a prime number (its only factors are 1 and 2) — option (a).

For every natural number n, what is the last digit of 6^n?

1. Calculate the first few powers of 6 to observe the pattern: 6¹ = 6, 6² = 36, and 6³ = 216.
2. Since the last digit of 6 multiplied by 6 is always 6 (6 × 6 = 36), the last digit of 6ⁿ is 6 for every natural number n.

Answer: Every power of 6 ends in the digit 6 — option (b).

From the given graph of y = f(x), how many zeroes does f(x) have?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the wavy curve y = f(x) lies wholly above the x-axis (it dips near the origin but does not touch or cross the x-axis).

1. A zero of f(x) is a point where the graph of y = f(x) intersects the x-axis.
2. From the figure, the curve lies wholly above the x-axis and never touches or crosses it.
3. Since the curve does not meet the x-axis at any point, f(x) has 0 zeroes.

Answer: The curve never meets the x-axis, so f(x) has 0 zeroes — option (a).

Two linear equations in two variables are shown by coincident lines. How many solutions does the pair have?

1. Coincident lines represent the same linear equation, meaning every point on the line satisfies both equations.
2. Since there are infinitely many points on a line, the pair of equations has infinitely many solutions, making option (d) correct.

Answer: Coincident lines give infinitely many solutions — option (d).

Common difference of the AP √2, 2√2, 3√2, 4√2, ...

1. The common difference d is the difference between the second term and the first term.
2. d = 2√2 − √2 = √2 (≈ 1.4142)

Answer: d = 2√2 − √2 = √2 (≈ 1.4142) — option (a).

ΔABC ~ ΔDEF with DE = 2AB and BC = 8 cm. Find EF.

1. Since ΔABC ~ ΔDEF, the ratio of corresponding sides is equal, so DE/AB = EF/BC.
2. Given DE = 2AB, the ratio DE/AB = 2, which means EF = 2 × BC.
3. Substituting BC = 8 cm, we get EF = 2 × 8 = 16 cm.

Answer: Sides scale by 2, so EF = 2 × 8 = 16 cm — option (d).

Where does the mid-point of the segment from (5, −4) to (6, 4) lie?

1. Apply the midpoint formula: x = (5 + 6) / 2 = 5.5 and y = (−4 + 4) / 2 = 0.
2. Since the y-coordinate is 0, the point (5.5, 0) lies on the x-axis.

Answer: Mid-point is (5.5, 0), which lies on the x-axis — option (a).

If sin θ = a/b, express cos θ in terms of a and b.

1. Use the identity sin² θ + cos² θ = 1 to write cos² θ = 1 − sin² θ.
2. Substitute sin θ = a/b to get cos² θ = 1 − (a/b)² = (b² − a²)/b².
3. Take the square root to find cos θ = √(b² − a²)/b, which matches option (c).

Answer: cos θ = √(b² − a²)/b — option (c).

Given cos A = 1/2, evaluate sin²A + 2cos²A.

1. Since cos A = 1/2, we have cos²A = (1/2)² = 1/4.
2. Using the identity sin²A + cos²A = 1, we find sin²A = 1 - cos²A = 1 - 1/4 = 3/4.
3. Substitute the values into the expression: sin²A + 2cos²A = 3/4 + 2 × 1/4 = 3/4 + 1/2 = 5/4.

Answer: 3/4 + 2 × 1/4 = 5/4 — option (b).

A tower is 30 m high. Find the angle of elevation of its top from a car 10√3 m from its base.

1. Using the tangent ratio for the angle of elevation, tan θ = Height of tower / Distance from base.
2. Substituting the given values, tan θ = 30 / (10√3) = 3 / √3 = √3.
3. Since tan 60° = √3, the angle of elevation θ is 60°.

Answer: tan θ = 30/(10√3) = √3, so θ = 60° — option (d).

TP and TQ are tangents from T to a circle with centre O, and ∠POQ = 120°. Find ∠PTQ.

1. Since TP and TQ are tangents to the circle, the radii OP and OQ are perpendicular to them, so ∠OPT = ∠OQT = 90°.
2. In the quadrilateral OPTQ, the sum of interior angles is 360°, so ∠PTQ = 360° − (90° + 90° + 120°) = 60°.

Answer: ∠PTQ = 180° − 120° = 60° — option (a).

PA is a tangent at A from external point P to a circle with centre O; AOB is a diameter and ∠POB = 125°. Find ∠APO.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, PA touches the circle at A; A, O, B are collinear (AB is a diameter); P, O lie on a line and the angle between OP and OB at O is 125°.

1. Since A, O, and B are collinear, ∠AOP and ∠POB form a linear pair, so ∠AOP = 180° − 125° = 55°.
2. PA is a tangent at A, so the radius OA is perpendicular to PA, giving ∠OAP = 90°.
3. In triangle OAP, the sum of angles is 180°, so ∠APO = 180° − 90° − 55° = 35°.

Answer: ∠AOP = 180° − 125° = 55°, ∠OAP = 90°, so ∠APO = 35° — option (d).

Arc length of a sector with radius 21 cm and central angle 60°.

1. Use the formula for arc length: Arc = (θ/360) × 2πr.
2. Substitute θ = 60°, r = 21 cm, and π = 22/7: Arc = (60/360) × 2 × (22/7) × 21 = 22 cm.

Answer: Arc = (60/360) × 2 × (22/7) × 21 = 22 cm — option (a).

A clock's 7 cm hour hand: what angle does it turn through from 7:00 a.m. to 8:10 a.m.?

1. Calculate the total time elapsed from 7:00 a.m. to 8:10 a.m., which is 70 minutes.
2. Since the hour hand moves 0.5° per minute, the angle turned is 70 × 0.5° = 35°.

Answer: 70 minutes × 0.5° per minute = 35° — option (c).

Total surface area of a solid hemisphere whose diameter is 2d.

1. Given the diameter of the solid hemisphere is 2d, the radius r is half of the diameter, so r = d.
2. The total surface area (TSA) of a solid hemisphere is given by the formula 3πr².
3. Substituting r = d into the formula gives TSA = 3πd², which corresponds to option (a).

Answer: Radius = d, so TSA = 3πd² — option (a).

Data has mean 12 and mode 21. Find its median.

1. Use the empirical relationship between mean, median, and mode: 3 × Median = Mode + 2 × Mean.
2. Substitute the given values: 3 × Median = 21 + 2 × 12 = 21 + 24 = 45.
3. Solve for the median: Median = 45 ÷ 3 = 15.

Answer: Median = (mode + 2 × mean)/3 = 15 — option (c).

One die is thrown. Probability that the number shown is not 3.

1. Total outcomes when a die is thrown are 6, and the favorable outcomes for 'not 3' are 5.
2. Probability = Favorable outcomes / Total outcomes = 5/6, so option (c) is correct.

Answer: P = 5/6 — option (c).

Assertion: P(a leap year has 53 Mondays) = 2/7. Reason: P(a non-leap year has 53 Mondays) = 5/7.

1. A leap year has 366 days, which is 52 weeks and 2 extra days, so the probability of having 53 Mondays is 2/7, making the Assertion true.
2. A non-leap year has 365 days, which is 52 weeks and 1 extra day, so the probability of having 53 Mondays is 1/7, making the Reason false.
3. Since the Assertion is true and the Reason is false, the correct option is (c).

Answer: A leap year has 52 weeks + 2 extra days, so A (2/7) is true; a non-leap year has 1 extra day, so the probability is 1/7 and R is false — option (c).

Assertion: p(y) = y² + 4y + 3 has two zeroes. Reason: a quadratic polynomial has at most two zeroes.

1. Factorise: y² + 4y + 3 = (y + 1)(y + 3), so p(y) = 0 when y = −1 or y = −3.
2. So p(y) has two zeroes and Assertion (A) is true.
3. Reason (R) is true for every quadratic, but saying 'at most two' does not explain why this one has exactly two.
4. Both are true and R does not explain A, so the answer is option (b).

Answer: y² + 4y + 3 = (y + 1)(y + 3) has zeroes −1 and −3, so A is true; R is true but does not explain why this one has two — option (b).

α and β are zeroes of p(x) = x² − 3x − 1. Find 1/α + 1/β.

1. Using the sum of zeroes formula, α + β = -(-3)/1 = 3.
2. Using the product of zeroes formula, αβ = -1/1 = -1.
3. Expressing the required value as 1/α + 1/β = (α + β)/αβ.
4. Substituting the values gives 3/(-1) = -3.

Answer: α + β = 3, αβ = -1, so 1/α + 1/β = (α + β)/αβ = -3.

In ΔABC, DE ∥ BC with AD = x, DB = x − 2, AE = x + 2, EC = x − 1. Find x.

1. Since DE is parallel to BC, by the Basic Proportionality Theorem, AD/DB = AE/EC.
2. Substitute the given values into the proportion: x/(x - 2) = (x + 2)/(x - 1).
3. Cross-multiply to get x(x - 1) = (x + 2)(x - 2).
4. Expand both sides to obtain x² - x = x² - 4.
5. Subtract x² from both sides and solve for x to get x = 4.

Answer: x/(x − 2) = (x + 2)/(x − 1) gives x² − x = x² − 4, so x = 4.

ΔABC ~ ΔXYZ with AB = 4 cm, BC = 6 cm, AC = y; XY = x, YZ = 7.2 cm, XZ = 6 cm. Find x and y.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, ΔABC has AB = 4 cm, BC = 6 cm, AC = y; ΔXYZ has XY = x, YZ = 7.2 cm, XZ = 6 cm.

1. Since ΔABC ~ ΔXYZ, the ratio of corresponding sides is equal: AB/XY = BC/YZ = AC/XZ.
2. Substituting the given values, we get 4/x = 6/7.2 = y/6.
3. From 4/x = 6/7.2, we have 6x = 4 × 7.2, so 6x = 28.8, which gives x = 4.8 cm.
4. From y/6 = 6/7.2, we have 7.2y = 36, so y = 36/7.2 = 5 cm.

Answer: 4/x = 6/7.2 = y/6, so x = 4.8 cm, y = 5 cm.

A circle has centre (x − 7, 2x) and radius 5√2 and passes through (−9, 11). Find x.

1. The distance between the centre (x − 7, 2x) and the point (−9, 11) equals the radius 5√2.
2. Using the distance formula, (x − 7 + 9)² + (2x − 11)² = (5√2)².
3. Simplifying gives (x + 2)² + (2x − 11)² = 50.
4. Expanding and combining like terms results in 5x² − 40x + 125 = 50.
5. Subtracting 50 and dividing by 5 gives x² − 8x + 15 = 0.
6. Factoring the equation yields (x − 3)(x − 5) = 0.
7. Thus, x = 3 or x = 5.

Answer: (x + 2)² + (2x − 11)² = 50 gives x² − 8x + 15 = 0, so x = 3 or 5.

Given tan θ = 24/7, find sin θ + cos θ.

1. Given tan θ = 24/7, we identify the opposite side as 24 and the adjacent side as 7.
2. Using the Pythagorean theorem, the hypotenuse is √(24² + 7²) = √(576 + 49) = √625 = 25.
3. Calculate sin θ = opposite/hypotenuse = 24/25 and cos θ = adjacent/hypotenuse = 7/25.
4. Add the values to find sin θ + cos θ = 24/25 + 7/25 = 31/25.

Answer: Hypotenuse 25, sin θ = 24/25, cos θ = 7/25, so the sum is 31/25.

Given cot θ = 7/8, evaluate (1 + sin θ)(1 − sin θ) ÷ [(1 + cos θ)(1 − cos θ)].

1. Simplify the numerator using the identity (1 + sin θ)(1 − sin θ) = 1 − sin²θ = cos²θ.
2. Simplify the denominator using the identity (1 + cos θ)(1 − cos θ) = 1 − cos²θ = sin²θ.
3. Substitute these into the expression to get cos²θ ÷ sin²θ, which equals cot²θ.
4. Square the given value cot θ = 7/8 to find cot²θ = (7/8)² = 49/64.

Answer: The expression = cos²θ/sin²θ = cot²θ = 49/64.

Concentric circles have radii 5 cm and 4 cm. Find the length of a chord of the bigger circle that is tangent to the smaller one.

1. Let O be the common center, and let AB be the chord of the larger circle tangent to the smaller circle at point T.
2. Since AB is tangent to the smaller circle at T, the radius OT is perpendicular to AB, so angle OTA is 90 degrees.
3. Join OA to form the right-angled triangle OTA, where OA is the radius of the larger circle (5 cm) and OT is the radius of the smaller circle (4 cm).
4. Apply the Pythagorean theorem in triangle OTA: AT² + OT² = OA², which gives AT² + 4² = 5².
5. Solve for AT: AT² = 25 − 16 = 9, so AT = 3 cm.
6. Since the radius perpendicular to a chord bisects the chord, the total length of chord AB is 2 × AT = 2 × 3 = 6 cm.

Answer: Half-chord = √(5² − 4²) = 3 cm, so chord = 6 cm.

Show that √3 is irrational.

1. Assume for contradiction that √3 is rational, so √3 = a/b where a and b are coprime integers and b ≠ 0.
2. Squaring both sides gives 3 = a²/b², which implies a² = 3b².
3. Since 3 divides a², 3 must divide a; let a = 3c for some integer c.
4. Substituting a = 3c into a² = 3b² gives (3c)² = 3b², so 9c² = 3b².
5. Dividing by 3 yields b² = 3c², which means 3 divides b² and thus 3 divides b.
6. Since 3 divides both a and b, they share a common factor, contradicting the assumption that a and b are coprime.
7. Therefore, the assumption is false and √3 is irrational.

Answer: Assume √3 = a/b with a, b coprime and b ≠ 0. Then a² = 3b², so 3 divides a; put a = 3c, giving b² = 3c², so 3 divides b too. This contradicts a, b being coprime, hence √3 is irrational.

In what ratio does the x-axis divide the segment from (−6, 5) to (−4, −1)? Find the point where they meet.

1. Let the x-axis divide the segment joining A(-6, 5) and B(-4, -1) in the ratio k : 1.
2. Using the section formula, the y-coordinate of the point of division is given by (k(-1) + 1(5)) / (k + 1).
3. Since the point lies on the x-axis, its y-coordinate is 0, so (5 - k) / (k + 1) = 0.
4. Solving for k gives 5 - k = 0, which implies k = 5.
5. Thus, the required ratio is 5 : 1.
6. Substitute k = 5 into the x-coordinate formula: (5(-4) + 1(-6)) / (5 + 1) = (-20 - 6) / 6 = -26 / 6 = -13/3.
7. Therefore, the point of intersection is (-13/3, 0).

Answer: Taking ratio k : 1, the y-coordinate (−k + 5)/(k + 1) = 0 gives k = 5. Ratio 5 : 1, and the point is (-13/3, 0).

Given x = h + a cos θ and y = k + b sin θ, show that ((x − h)/a)² + ((y − k)/b)² = 1.

1. From the given equation x = h + a cos θ, subtract h from both sides to get x − h = a cos θ.
2. Divide both sides by a to obtain (x − h)/a = cos θ.
3. From the given equation y = k + b sin θ, subtract k from both sides to get y − k = b sin θ.
4. Divide both sides by b to obtain (y − k)/b = sin θ.
5. Substitute these values into the left-hand side: ((x − h)/a)² + ((y − k)/b)² = (cos θ)² + (sin θ)².
6. Simplify the expression to cos²θ + sin²θ, which equals 1 by the Pythagorean identity.

Answer: (x − h)/a = cos θ and (y − k)/b = sin θ, so the left side = cos²θ + sin²θ = 1.

Show that tan A/(1 + sec A) − tan A/(1 − sec A) = 2 cosec A.

1. Write tan A = sin A/cos A and sec A = 1/cos A, then multiply the top and bottom of each fraction by cos A.
2. This turns the left side into sin A/(cos A + 1) − sin A/(cos A − 1).
3. Over the common denominator (cos A + 1)(cos A − 1) = cos²A − 1, the numerator is sin A[(cos A − 1) − (cos A + 1)] = −2 sin A.
4. Since cos²A − 1 = −sin²A, the left side = −2 sin A/(−sin²A) = 2/sin A.
5. 2/sin A = 2 cosec A, which is the right side.

Answer: Writing in sin and cos: sin A/(cos A + 1) − sin A/(cos A − 1) = sin A × (−2)/(cos²A − 1) = 2 sin A/sin²A = 2 cosec A.

ΔABC is right-angled at B with AB = 4 cm and BC = 3 cm. Find the radius of its incircle.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, a circle is inscribed in ΔABC, right-angled at B.

1. In ΔABC, ∠B = 90°, AB = 4 cm, BC = 3 cm.
2. By Pythagoras theorem, AC² = AB² + BC² = 4² + 3² = 16 + 9 = 25, so AC = 5 cm.
3. Let r be the inradius. The tangent lengths from a vertex are equal, so the two segments on AB are r and (4 − r), on BC are r and (3 − r), and on AC are (4 − r) and (3 − r).
4. Since the two tangent segments from A and C meet at AC, (4 − r) + (3 − r) = AC = 5.
5. Solving: 7 − 2r = 5 → 2r = 2 → r = 1 cm.

Answer: AC = 5 cm; with tangent lengths, (4 − r) + (3 − r) = 5, so r = 1 cm.

A circle touches side QR of ΔPQR at S and sides PQ, PR produced at M, N. Show PM = ½(PQ + QR + PR).

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the circle lies below QR, touching QR at S and the extensions of PQ beyond Q at M and PR beyond R at N.

1. Since tangents drawn from an external point to a circle are equal, we have PM = PN, QM = QS, and RN = RS.
2. Expressing PM and PN in terms of the triangle sides and tangent segments gives PM = PQ + QM and PN = PR + RN.
3. Substituting the equal tangent lengths QM = QS and RN = RS into these expressions yields PM = PQ + QS and PN = PR + RS.
4. Adding these two equations results in PM + PN = PQ + QS + PR + RS.
5. Using the relation PM = PN, the left side becomes 2PM, and since QS + RS equals the side QR, the right side becomes PQ + QR + PR.
6. Thus, 2PM = PQ + QR + PR, which implies PM = ½(PQ + QR + PR).

Answer: Tangents from one point are equal: PM = PN, QM = QS, RN = RS. So 2PM = PM + PN = PQ + QS + PR + RS = PQ + QR + PR, giving PM = ½(PQ + QR + PR).

A solid cylinder with a hemisphere at each end is 20 cm tall overall; the cylinder's diameter is 7 cm. Find its volume (π = 22/7).

1. Given diameter = 7 cm, so radius r = 7/2 = 3.5 cm.
2. Total height is 20 cm, which includes the cylinder height plus two hemispheres (each of height r). So cylinder height h = 20 − 2r = 20 − 7 = 13 cm.
3. Volume of cylinder = πr²h = (22/7) × (3.5)² × 13 = (22/7) × 12.25 × 13 = 500.5 cm³.
4. Volume of two hemispheres = volume of one full sphere = (4/3)πr³ = (4/3) × (22/7) × (3.5)³ = (4/3) × (22/7) × 42.875 = 179.666... cm³.
5. Total volume = 500.5 + 179.666... = 680.166... cm³ ≈ 680.17 cm³.

Answer: r = 3.5 cm, cylinder height = 20 − 7 = 13 cm. Volume = πr²(13) + (4/3)πr³ = 4081/6 cm³ ≈ 680.17 cm³.

Note: CBSE's marking scheme prints 680.1 cm³ (approx.); the exact value is 4081/6 ≈ 680.17 cm³.

Two different-coloured dice are rolled together. List the outcomes and find P(same number on both) and P(different numbers).

1. List all possible outcomes as ordered pairs (Die 1, Die 2) from (1,1) to (6,6), giving a total of 36 outcomes.
2. Identify the outcomes where both dice show the same number: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6), which are 6 favorable outcomes.
3. Calculate P(same number) = 6/36 = 1/6.
4. Calculate P(different numbers) = 1 - P(same number) = 1 - 1/6 = 5/6.

Answer: 36 outcomes (1,1) to (6,6); 6 doubles, so the probabilities are 1/6 and 5/6.

Using graphs, find the vertices of the triangle formed by 2x − 3y + 6 = 0, 2x + 3y − 18 = 0 and x = 0, and its area.

1. Plot each line from two of its points: 2x − 3y + 6 = 0 passes through (0, 2) and (3, 4); 2x + 3y − 18 = 0 passes through (0, 6) and (9, 0); x = 0 is the y-axis.
2. Where the first line meets x = 0: −3y + 6 = 0, so y = 2, giving the vertex (0, 2).
3. Where the second line meets x = 0: 3y − 18 = 0, so y = 6, giving the vertex (0, 6).
4. The two slanted lines cross where adding them gives 4x − 12 = 0, so x = 3; then 6 − 3y + 6 = 0 gives y = 4, the vertex (3, 4).
5. The graph shows these three vertices: (0, 2), (3, 4) and (0, 6).
6. Base along the y-axis = 6 − 2 = 4 units; height = distance of (3, 4) from the y-axis = 3 units.
7. Area = ½ × 4 × 3 = 6 square units.

Answer: Vertices (0, 2), (3, 4), (0, 6). Base on the y-axis = 4, height = 3, area = 6 square units.

Over 200 km, the faster of two trains needs 1 hour less; the slower is 10 km/h slower. Find both speeds.

1. Let the speed of the faster train be x km/h, so the speed of the slower train is (x - 10) km/h.
2. The time taken by the faster train is 200/x hours and by the slower train is 200/(x - 10) hours.
3. Since the faster train takes 1 hour less, we have 200/(x - 10) - 200/x = 1.
4. Simplifying the equation gives 200x - 200(x - 10) = x(x - 10), which leads to 2000 = x² - 10x.
5. Rearranging yields the quadratic equation x² - 10x - 2000 = 0.
6. Factoring the equation gives (x - 50)(x + 40) = 0, so x = 50 or x = -40.
7. Since speed cannot be negative, x = 50 km/h, making the slower train's speed 50 - 10 = 40 km/h.

Answer: 200/(x − 10) − 200/x = 1 gives x² − 10x − 2000 = 0, so 50 km/h and 40 km/h.

Two squares have total area 640 m² and perimeters differing by 64 m. Find their sides.

1. Let the side of the larger square be x m. Since the perimeters differ by 64 m, the difference in their sides is 64 / 4 = 16 m, so the side of the smaller square is (x - 16) m.
2. The total area of the two squares is 640 m², so we write the equation: x² + (x - 16)² = 640.
3. Expand and simplify the equation: x² + x² - 32x + 256 = 640, which becomes 2x² - 32x - 384 = 0.
4. Divide the entire equation by 2 to get the standard quadratic form: x² - 16x - 192 = 0.
5. Factorize the quadratic equation: (x - 24)(x + 8) = 0, which gives x = 24 or x = -8.
6. Since the side length cannot be negative, we take x = 24 m. The side of the smaller square is 24 - 16 = 8 m.
7. Therefore, the sides of the two squares are 24 m and 8 m.

Answer: Sides x and x − 16: x² + (x − 16)² = 640 gives x² − 16x − 192 = 0, so x = 24. Sides 24 m and 8 m.

State the Basic Proportionality (Thales) Theorem and prove it.

1. Statement: if a line parallel to one side of a triangle cuts the other two sides at distinct points, it divides those two sides in the same ratio.
2. Take ΔABC with DE ∥ BC, D on AB and E on AC. To prove AD/DB = AE/EC.
3. Join BE and CD. Draw EN ⊥ AB and DM ⊥ AC.
4. ΔADE and ΔBDE have bases AD and DB on AB and the same height EN, so ar(ADE)/ar(BDE) = (½ × AD × EN)/(½ × DB × EN) = AD/DB.
5. ΔADE and ΔDEC have bases AE and EC on AC and the same height DM, so ar(ADE)/ar(DEC) = AE/EC.
6. ΔBDE and ΔDEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
7. So the two ratios have equal denominators, giving AD/DB = AE/EC.

Answer: Statement: a line parallel to one side of a triangle, cutting the other two sides at distinct points, divides those sides in the same ratio. Proof: in ΔABC with DE ∥ BC, join BE and CD, draw heights; ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(DEC) = AE/EC; ar(BDE) = ar(DEC) (same base DE, between the same parallels), so AD/DB = AE/EC.

ΔABC ~ ΔPQR and CM, RN are their medians. Prove (i) ΔAMC ~ ΔPNR and (ii) ΔCMB ~ ΔRNQ.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, M is the mid-point of AB in ΔABC and N is the mid-point of PQ in ΔPQR; CM and RN are drawn.

1. Given ΔABC ~ ΔPQR, corresponding angles are equal, so ∠A = ∠P and ∠B = ∠Q.
2. From similarity, the ratio of corresponding sides is equal, so AB/PQ = AC/PR = BC/QR.
3. M is the midpoint of AB, so AB = 2AM, and N is the midpoint of PQ, so PQ = 2PN.
4. Substituting these into the ratio gives 2AM/2PN = AC/PR, which simplifies to AM/PN = AC/PR.
5. In ΔAMC and ΔPNR, AM/PN = AC/PR and the included angle ∠A = ∠P, so ΔAMC ~ ΔPNR by SAS.
6. Similarly, MB = AB/2 and NQ = PQ/2, so MB/NQ = AB/PQ = BC/QR.
7. In ΔCMB and ΔRNQ, MB/NQ = BC/QR and the included angle ∠B = ∠Q, so ΔCMB ~ ΔRNQ by SAS.

Answer: (i) ∠A = ∠P and AB/PQ = AC/PR; since AB = 2AM, PQ = 2PN, AM/PN = AC/PR, so ΔAMC ~ ΔPNR (SAS). (ii) ∠B = ∠Q and AB/PQ = BC/QR gives MB/NQ = BC/QR, so ΔCMB ~ ΔRNQ (SAS).

Classes 0-10, 10-20, ..., 60-70 have frequencies 1, x, 5, 7, y, 3, 1. Total frequency is 25 and mean is 35. Find x and y.

1. The class midpoints are 5, 15, 25, 35, 45, 55, and 65.
2. From the total frequency, 1 + x + 5 + 7 + y + 3 + 1 = 25, which simplifies to x + y = 8.
3. The sum of midpoint times frequency is 5(1) + 15(x) + 25(5) + 35(7) + 45(y) + 55(3) + 65(1) = 605 + 15x + 45y.
4. Using the mean formula, (605 + 15x + 45y) / 25 = 35, which gives 605 + 15x + 45y = 875.
5. Simplifying the equation yields 15x + 45y = 270, or x + 3y = 18.
6. Subtracting x + y = 8 from x + 3y = 18 gives 2y = 10, so y = 5.
7. Substituting y = 5 into x + y = 8 gives x = 3.

Answer: x + y = 8 and 605 + 15x + 45y = 875, i.e. x + 3y = 18. So x = 3, y = 5.

Potato race: bucket 5 m from the first potato, 10 potatoes 3 m apart in a line; each potato is fetched back to the bucket one at a time. Distance run for the first potato?

1. The first potato is located 5 m from the bucket.
2. The distance run to fetch it is 5 m to the potato and 5 m back to the bucket, totaling 10 m.

Answer: 5 m there and 5 m back = 10 m.

Same potato race (bucket 5 m from first potato, gaps 3 m): distance run for the second potato?

1. Distance to the second potato is the bucket distance plus one gap: 5 + 3 = 8 m.
2. Total distance run is twice the one-way distance: 2 × 8 = 16 m.

Answer: 8 m each way = 16 m.

Same potato race with 10 potatoes: total distance the competitor runs.

1. The k-th potato is 5 + 3(k − 1) m from the bucket and is run to and back, so the runs are 10 m, 16 m, 22 m, ...
2. These form an AP with a = 10 and d = 6, and there are n = 10 terms.
3. S₁₀ = (10/2) × [2 × 10 + 9 × 6] = 5 × (20 + 54) = 5 × 74 = 370 m.

Answer: Distances 10, 16, 22, ... form an AP (a = 10, d = 6); S10 = 5 × (20 + 54) = 370 m.

Same potato race: at an average speed of 5 m/s, how long to collect all 10 potatoes?

1. The runs for the potatoes are 10 m, 16 m, 22 m, ..., an AP with a = 10 and d = 6.
2. Total distance for 10 potatoes: S₁₀ = 5 × [2 × 10 + 9 × 6] = 5 × 74 = 370 m.
3. Time = distance ÷ speed = 370 ÷ 5 = 74 s.

Answer: Total run = 370 m (AP, a = 10, d = 6), so time = 370 ÷ 5 = 74 s.

Tower PA stands at P with B on it below A; point O is 6 m from P on level ground. Elevation of B from O is 30°, of A is 60°. Length of wire OB?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the vertical tower stands at P with B (lower section top) and A (top) on it; wires run from O to B and to A; OP = 6 m, ∠BOP = 30°, ∠AOP = 60°.

1. In right triangle OPB, angle OPB is 90° and angle BOP is 30°.
2. Using the cosine ratio, cos 30° = OP/OB, so OB = 6/cos 30°.
3. Substituting cos 30° = √3/2, we get OB = 6/(√3/2) = 12/√3 = 4√3 m ≈ 6.93 m.

Answer: OB = 6/cos 30° = 12/√3 = 4√3 m ≈ 6.93 m.

Same tower (OP = 6 m, elevation of A from O is 60°): length of wire OA?

1. In right-angled triangle OPA, angle OPA is 90° and angle POA is 60°.
2. Using the cosine ratio, cos 60° = adjacent/hypotenuse = OP/OA.
3. Substituting the values, 1/2 = 6/OA, which gives OA = 12 m.

Answer: OA = 6/cos 60° = 12 m.

Same tower (OP = 6 m, elevations 30° to B and 60° to A): find AB.

1. In right ΔOPB, tan 30° = BP/OP, so BP = 6 × (1/√3) = 2√3 m.
2. In right ΔOPA, tan 60° = AP/OP, so AP = 6 × √3 = 6√3 m.
3. B lies between P and A, so AB = AP − BP = 6√3 − 2√3 = 4√3 m ≈ 6.93 m.

Answer: BP = 6 tan 30° = 2√3 m, AP = 6 tan 60° = 6√3 m, so AB = 4√3 m ≈ 6.93 m.

Same tower (OP = 6 m, elevation of B is 30°): find the area of ΔOPB.

1. ΔOPB is right-angled at P, with ∠BOP = 30° and OP = 6 m.
2. tan 30° = BP/OP, so BP = 6/√3 = 2√3 m.
3. Area = ½ × OP × BP = ½ × 6 × 2√3 = 6√3 m² ≈ 10.39 m².

Answer: BP = 2√3 m, area = ½ × 6 × 2√3 = 6√3 m² ≈ 10.39 m².

A silver-wire brooch is a circle of diameter 35 cm with 5 diameters making 10 equal sectors. Radius of the circle?

1. Given diameter of the circle = 35 cm.
2. Radius = Diameter / 2 = 35 / 2 = 17.5 cm.

Answer: Radius = 35/2 = 17.5 cm.

Same brooch (diameter 35 cm): its circumference?

1. Given diameter d = 35 cm, so radius r = 35/2 cm.
2. Circumference C = 2πr = 2 × (22/7) × (35/2) = (22/7) × 35 = 110 cm.

Answer: Circumference = (22/7) × 35 = 110 cm.

Same brooch (circle of diameter 35 cm plus 5 diameters): total wire needed?

1. Calculate the circumference of the brooch using the formula C = πd, which is 22/7 × 35 = 110 cm.
2. Calculate the total length of the 5 diameters by multiplying the diameter by 5, which is 5 × 35 = 175 cm.
3. Add the circumference and the total length of the diameters to find the total wire needed, which is 110 + 175 = 285 cm.

Answer: 110 + 5 × 35 = 285 cm.

Same brooch (diameter 35 cm, 10 equal sectors): area of one sector?

1. Radius r = 35/2 = 17.5 cm.
2. Angle of one sector = 360°/10 = 36°.
3. Area = (36/360) × (22/7) × 17.5² = 96.25 cm².

Answer: Angle 36°, area = (36/360) × (22/7) × 17.5² = 385/4 = 96.25 cm².