CBSE Class 10 Maths (Standard) Board Paper 2025 (30/1/1): Solutions

α, β are zeroes of 3x² + 6x + k and α + β + αβ = −2/3. Find k.

1. For the quadratic equation 3x² + 6x + k = 0, the sum of zeroes is α + β = -b/a = -6/3 = -2.
2. The product of zeroes is αβ = c/a = k/3.
3. Substitute these values into the given condition α + β + αβ = -2/3 to get -2 + k/3 = -2/3.
4. Solving for k gives k/3 = -2/3 + 2 = 4/3, so k = 4.

Answer: α + β = −2 and αβ = k/3, so −2 + k/3 = −2/3 gives k = 4 — option (D).

x = 1, y = 2 solves 2x − 3y + a = 0 as well as 2x + 3y − b = 0. Which relation between a and b holds?

1. Substitute x = 1 and y = 2 into the first equation 2x − 3y + a = 0 to get 2(1) − 3(2) + a = 0, which simplifies to 2 − 6 + a = 0, so a = 4.
2. Substitute x = 1 and y = 2 into the second equation 2x + 3y − b = 0 to get 2(1) + 3(2) − b = 0, which simplifies to 2 + 6 − b = 0, so b = 8.
3. Compare the values a = 4 and b = 8 to find the relation: 2 × 4 = 8, which means 2a = b.

Answer: a = 4 and b = 8, so 2a = b — option (B).

Where does the mid-point of Q(4, 6) and P(−4, 5) lie?

1. Use the mid-point formula to find the coordinates: x = (4 + (-4))/2 = 0 and y = (6 + 5)/2 = 11/2, so the mid-point is (0, 11/2).
2. Since the x-coordinate is 0, the point lies on the y-axis, which corresponds to option (B).

Answer: Mid-point is (0, 11/2), which lies on the y-axis — option (B).

θ is acute and 7 + 4 sin θ = 9. Find θ.

1. Given 7 + 4 sin θ = 9, subtract 7 from both sides to get 4 sin θ = 2.
2. Divide by 4 to find sin θ = 2/4 = 1/2.
3. Since θ is acute, the angle whose sine is 1/2 is 30°, so θ = 30°.

Answer: sin θ = 1/2, so θ = 30° — option (B).

Value of tan²θ − (1/cos θ × sec θ).

1. Identify that 1/cos θ is equal to sec θ, so the term inside the bracket is sec θ × sec θ = sec²θ.
2. Substitute this back into the expression to get tan²θ − sec²θ, and apply the trigonometric identity sec²θ − tan²θ = 1 to find that tan²θ − sec²θ = −1.

Answer: The bracket is sec²θ, and tan²θ − sec²θ = −1 — option (C).

m = HCF(98, 28) and n = LCM(98, 28). Find n − 7m.

1. Prime factorization of 98 is 2 × 7² and 28 is 2² × 7, so HCF(98, 28) = 2 × 7 = 14.
2. LCM(98, 28) = 2² × 7² = 196, so n − 7m = 196 − 7(14) = 196 − 98 = 98.

Answer: m = 14, n = 196, so n − 7m = 98 — option (C).

How are the tangents at the two ends of a diameter of a circle related?

1. The radius drawn to the point of contact is perpendicular to the tangent at that point.
2. Since the tangents at the ends of a diameter are both perpendicular to the same line, they are parallel to each other.

Answer: Both are perpendicular to the same diameter, so they are parallel — option (A).

Triangles ABC, DEF have ∠B = ∠E and ∠C = ∠F, with AB = 3 DE. How are they related?

1. Since ∠B = ∠E and ∠C = ∠F, the triangles are similar by the AA similarity criterion.
2. Given AB = 3 DE, the ratio of corresponding sides is 3 : 1, which is not 1 : 1.
3. Therefore, the triangles are similar but not congruent, so option (D) is correct.

Answer: Equal angles give similarity (AA), but sides are in ratio 3 : 1, so they are similar but not congruent — option (D).

(−1)ⁿ + (−1)⁸ = 0. What kind of number is n?

1. Evaluate the known term: (−1)⁸ = 1.
2. Substitute into the equation: (−1)ⁿ + 1 = 0, which gives (−1)ⁿ = −1.
3. Since (−1) raised to an odd power equals −1, n must be an odd number.

Answer: (−1)⁸ = 1, so (−1)ⁿ = −1, which needs n to be any odd number — option (C).

A graph shows two polynomials. How many distinct zeroes do they have together?

Uses a figure printed in CBSE's paper — what the figure shows: The graph shows an upward parabola and a downward parabola; both cross the x-axis at the same two points (one left, one right of the y-axis) and nowhere else.

1. The graph shows two parabolas that intersect the x-axis at the same two points.
2. Since both curves share these two intersection points and have no other x-intercepts, the total number of distinct zeroes is 2.

Answer: Both curves cut the x-axis at the same two points, so there are 2 distinct zeroes — option (C).

An AP has Sₘ = 2m² + 3m (sum of its first m terms). Find a₂.

1. The sum of the first m terms is Sₘ = 2m² + 3m.
2. Put m = 1: S₁ = 2(1)² + 3(1) = 5, and put m = 2: S₂ = 2(2)² + 3(2) = 8 + 6 = 14.
3. The second term is a₂ = S₂ − S₁ = 14 − 5 = 9, which is option (B).

Answer: a₂ = S₂ − S₁ = 14 − 5 = 9 — option (B).

A data set has mode 15x and mean 18x. Find its median.

1. Use the empirical relation 3 Median = Mode + 2 Mean.
2. 3 Median = 15x + 2(18x) = 51x, so Median = 51x ÷ 3 = 17x, which is option (C).

Answer: 3 Median = Mode + 2 Mean = 51x, so median = 17x — option (C).

One card is drawn from a pack of 52. Probability that it is a red face card?

1. Total number of cards in the pack is 52.
2. Red face cards are the Jack, Queen, and King of Hearts and Diamonds, totaling 6 cards.
3. Probability is the ratio of favorable outcomes to total outcomes: 6/52.
4. Simplifying the fraction gives 3/26, which corresponds to option (D).

Answer: There are 6 red face cards, so P = 6/52 = 3/26 — option (D).

Pick the rational number lying between √3 and √5 (choices include 1.4142387954012..., 2.32̄6, π and 1.857142).

1. √3 ≈ 1.732 and √5 ≈ 2.236, so the number must be between 1.732 and 2.236.
2. 1.4142387954012... is less than 1.732, so it is not between √3 and √5.
3. 2.32̄6 is greater than 2.236, so it is not between √3 and √5.
4. π is an irrational number, so it is not rational.
5. 1.857142 is a terminating decimal, hence rational, and 1.732 < 1.857142 < 2.236, so it lies between √3 and √5.

Answer: 1.857142 is a terminating decimal (rational) and lies between √3 ≈ 1.732 and √5 ≈ 2.236 — option (D).

A sector with central angle 72° has area 40π sq. units. Find the radius.

1. Use the formula for the area of a sector: Area = (θ/360) × πr².
2. Substitute the given values: 40π = (72/360) × πr².
3. Simplify the fraction and solve for r²: 40 = (1/5)r², so r² = 200.
4. Take the square root to find the radius: r = √200 = 10√2 units.

Answer: (72/360)πr² = 40π gives r² = 200, so r = 10√2 units — option (D).

PA touches a circle (centre O) at A from outside point P; AOB is a diameter and ∠POB = 115°. Find ∠APO.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure A, O, B lie on one straight line (diameter AB), PA is the tangent at A, and the 115° angle is marked between OP and OB.

1. Since PA is a tangent to the circle at A and OA is the radius, the angle between them is 90°, so ∠OAP = 90°.
2. Points A, O, and B lie on a straight line, making ∠AOB a linear pair with ∠POB, so ∠AOP = 180° − 115° = 65°.
3. In triangle AOP, the sum of angles is 180°, so ∠APO = 180° − 90° − 65° = 25°.

Answer: ∠AOP = 180° − 115° = 65° and ∠OAP = 90°, so ∠APO = 25° — option (A).

A kite is 150 m above the ground; its string makes 30° with the horizontal. Find the string length.

1. In the right-angled triangle formed by the kite, the string is the hypotenuse L, and the height 150 m is opposite to the 30° angle.
2. Using the sine ratio, sin 30° = Opposite / Hypotenuse = 150 / L.
3. Substitute the value sin 30° = 1/2 to get 1/2 = 150 / L, which gives L = 300 m.

Answer: sin 30° = 150/L, so L = 300 m — option (B).

A 20 cm wire is bent into an arc of radius 60/π cm. Find the angle at the centre.

1. Use the arc length formula: (θ/360) × 2πr = 20, where r = 60/π cm.
2. Substitute r: (θ/360) × 2π × (60/π) = 20, which simplifies to (θ/360) × 120 = 20.
3. Solve for θ: θ = (20 × 360) / 120 = 60°.

Answer: (θ/360) × 2π × (60/π) = 20 gives θ = 60° — option (B).

Assertion: choosing a number at random from 1 to 20 has probability 1. Reason: an event with probability 1 is a sure event.

1. Choosing a number from 1 to 20 is a sure event because the outcome is always within this range, so its probability is 1.
2. An event with probability 1 is defined as a sure event, so the reason correctly explains the assertion.
3. Both the assertion and the reason are true, and the reason is the correct explanation of the assertion, making option (A) the right choice.

Answer: Any number picked is always from 1 to 20, so it is a sure event; both are true and the reason explains the assertion — option (A).

Assertion: two equal hemispheres joined base to base form a sphere. Reason: a sphere of radius r has total surface area 3πr².

1. The assertion is true because joining two equal hemispheres at their circular bases results in a complete sphere.
2. The reason is false because the total surface area of a sphere with radius r is 4πr², not 3πr².
3. Therefore, the assertion is true but the reason is false, which corresponds to option (C).

Answer: The assertion is true, but a sphere's surface area is 4πr², so the reason is false — option (C).

Given y cos 0° + x cos 60° + sin 30° − cot 45° = 5, find x + 2y.

1. Substitute the standard trigonometric values: cos 0° = 1, cos 60° = 1/2, sin 30° = 1/2, and cot 45° = 1.
2. The equation becomes y(1) + x(1/2) + 1/2 − 1 = 5, which simplifies to x/2 + y − 1/2 = 5.
3. Add 1/2 to both sides to isolate the variable terms: x/2 + y = 5 + 1/2 = 11/2.
4. Multiply the entire equation by 2 to find the value of x + 2y: 2(x/2 + y) = 2(11/2), so x + 2y = 11.

Answer: x/2 + y + 1/2 − 1 = 5, so x/2 + y = 11/2 and x + 2y = 11.

Find tan²60° ÷ (sin²60° + cos²30°).

1. Substitute the standard trigonometric values: tan 60° = √3, sin 60° = √3/2, and cos 30° = √3/2.
2. Square the values to get tan²60° = 3, sin²60° = 3/4, and cos²30° = 3/4.
3. Add the terms in the denominator: 3/4 + 3/4 = 6/4 = 3/2.
4. Divide the numerator by the denominator: 3 ÷ (3/2) = 3 × (2/3) = 2.

Answer: 3 ÷ (3/4 + 3/4) = 3 ÷ 3/2 = 2.

Zeroes of p(x) = x² + (4/3)x − 4/3.

1. Factorize the quadratic expression by splitting the middle term: x² + (4/3)x − 4/3 = x² + 2x − (2/3)x − 4/3.
2. Group and factor the terms: x(x + 2) − (2/3)(x + 2) = (x + 2)(x − 2/3).
3. Set the factored form to zero to find the roots: (x + 2)(x − 2/3) = 0, which gives x = −2 or x = 2/3.

Answer: p(x) = (1/3)(3x² + 4x − 4) = (1/3)(3x − 2)(x + 2), so the zeroes are 2/3 and -2.

A circle has centre (2a, a − 7), passes through (11, −9) and has diameter 10√2. Find a.

1. The radius of the circle is half of the diameter, so radius = 10√2 ÷ 2 = 5√2.
2. Using the distance formula between the centre (2a, a − 7) and the point (11, −9), we have (2a − 11)² + (a − 7 − (−9))² = (5√2)².
3. Simplifying the terms gives (2a − 11)² + (a + 2)² = 50, which expands to 4a² − 44a + 121 + a² + 4a + 4 = 50.
4. Combining like terms and subtracting 50 from both sides results in 5a² − 40a + 75 = 0, which simplifies to a² − 8a + 15 = 0.
5. Factoring the quadratic equation gives (a − 3)(a − 5) = 0, so a = 3 or a = 5.

Answer: Radius = 5√2, so (2a − 11)² + (a + 2)² = 50, i.e. a² − 8a + 15 = 0, giving a = 3 or 5.

ΔABC ~ ΔPQR with AC = 8 cm, AB = 6 cm, BC = 4 cm and PR = 6 cm. Find PQ + QR.

1. Since ΔABC ~ ΔPQR, the ratio of corresponding sides is equal, so PR/AC = PQ/AB = QR/BC.
2. Substitute the given values to find the scale factor: PR/AC = 6/8 = 3/4.
3. Calculate PQ using the ratio: PQ/AB = 3/4, so PQ/6 = 3/4, which gives PQ = 4.5 cm.
4. Calculate QR using the ratio: QR/BC = 3/4, so QR/4 = 3/4, which gives QR = 3 cm.
5. Find the sum: PQ + QR = 4.5 + 3 = 7.5 cm.

Answer: Scale PR/AC = 6/8, so PQ = 4.5 cm, QR = 3 cm and PQ + QR = 7.5 cm.

In the figure, QR/QS = QT/PR and ∠1 = ∠2. Prove ΔPQS ~ ΔTQR.

Uses a figure printed in CBSE's paper — what the figure shows: Q, S, R lie on a line (S between Q and R); Q, P, T lie on a line (P between Q and T). PS, PR and TR are joined. ∠1 = ∠PQR at Q and ∠2 = ∠PRQ at R.

1. Given that ∠1 = ∠2, and from the figure ∠1 = ∠PQR and ∠2 = ∠PRQ, we have ∠PQR = ∠PRQ.
2. In ΔPQR, since the angles opposite to sides PQ and PR are equal, the sides are equal, so PQ = PR.
3. Substituting PR with PQ in the given ratio QR/QS = QT/PR, we get QR/QS = QT/PQ.
4. Rearranging the proportion gives QS/QR = PQ/QT.
5. In ΔPQS and ΔTQR, the ratio of corresponding sides is equal (QS/QR = PQ/QT) and the included angle ∠Q is common.
6. By the SAS similarity criterion, ΔPQS ~ ΔTQR.

Answer: ∠1 = ∠2 gives PQ = PR, so QR/QS = QT/PQ, i.e. QS/QR = PQ/QT; with common ∠Q, SAS similarity gives ΔPQS ~ ΔTQR.

P is 26 m from the centre O of a circular ground; tangents PA and PB are each 10 m. Find the radius.

1. Since PA is a tangent to the circle at A, the radius OA is perpendicular to PA, making triangle OPA a right-angled triangle.
2. Using the Pythagorean theorem in triangle OPA, OA² + PA² = OP², so OA² = 26² − 10².
3. Calculating the values gives OA² = 676 − 100 = 576, so the radius OA = √576 = 24 m.

Answer: OA ⊥ PA, so OA² = 26² − 10² = 576 and the radius is 24 m.

BCD touches a circle with centre O at C, and A is a point on the circle with B, O, A in one line. Prove ∠BAC + ∠ACD = 90°.

Uses a figure printed in CBSE's paper — what the figure shows: Line BA passes through centre O and cuts the circle at P (near B) and A; B is outside the circle on the tangent line BCD; OC is drawn to the point of contact C; AC is joined.

1. Since OA and OC are radii of the circle, OA = OC.
2. In triangle OAC, since OA = OC, the angles opposite to these sides are equal, so ∠OCA = ∠OAC.
3. Since BCD is a tangent to the circle at C and OC is the radius, OC ⊥ BCD.
4. Therefore, ∠OCD = 90°, which implies ∠OCA + ∠ACD = 90°.
5. Substituting ∠OAC for ∠OCA, we get ∠OAC + ∠ACD = 90°.
6. Since B, O, and A are collinear, ∠OAC is the same as ∠BAC, hence ∠BAC + ∠ACD = 90°.

Answer: OA = OC gives ∠OCA = ∠OAC; OC ⊥ BD gives ∠OCA + ∠ACD = 90°, so ∠BAC + ∠ACD = 90°.

A quadrilateral is drawn around a circle. Prove each pair of opposite sides makes angles at the centre adding to 180°.

1. Let ABCD touch the circle (centre O) at P, Q, R, S on AB, BC, CD, DA. Join O to A, B, C, D and to P, Q, R, S.
2. In △OAP and △OAS: OP = OS (radii), AP = AS (tangents from A) and OA is common, so they are congruent (SSS) and ∠AOP = ∠AOS.
3. In the same way ∠BOP = ∠BOQ, ∠COQ = ∠COR and ∠DOR = ∠DOS.
4. The eight angles at O add up to 360°, so 2(∠AOP + ∠BOP + ∠COR + ∠DOR) = 360°, giving ∠AOP + ∠BOP + ∠COR + ∠DOR = 180°.
5. But ∠AOP + ∠BOP = ∠AOB and ∠COR + ∠DOR = ∠COD, so ∠AOB + ∠COD = 180°.
6. The four angles ∠AOB, ∠BOC, ∠COD, ∠DOA make 360°, so ∠BOC + ∠AOD = 360° − 180° = 180° as well.

Answer: Join the centre to the vertices and contact points; equal tangents give congruent triangles, so the 8 angles at the centre pair up equally; their sum 360° gives ∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°.

Prove tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ.

1. Express tan θ and cot θ in terms of sin θ and cos θ: LHS = (sin θ / cos θ) / (1 − cos θ / sin θ) + (cos θ / sin θ) / (1 − sin θ / cos θ).
2. Simplify the denominators to get LHS = (sin θ / cos θ) / ((sin θ − cos θ) / sin θ) + (cos θ / sin θ) / ((cos θ − sin θ) / cos θ).
3. Rewrite the fractions as LHS = sin²θ / [cos θ (sin θ − cos θ)] − cos²θ / [sin θ (sin θ − cos θ)].
4. Combine the terms over the common denominator sin θ cos θ (sin θ − cos θ) to get LHS = (sin³θ − cos³θ) / [sin θ cos θ (sin θ − cos θ)].
5. Factor the numerator using the identity a³ − b³ = (a − b)(a² + ab + b²) to get LHS = [(sin θ − cos θ)(sin²θ + sin θ cos θ + cos²θ)] / [sin θ cos θ (sin θ − cos θ)].
6. Cancel the common factor (sin θ − cos θ) and use the identity sin²θ + cos²θ = 1 to simplify the numerator to 1 + sin θ cos θ.
7. Split the fraction to get LHS = 1/(sin θ cos θ) + 1, which equals 1 + sec θ cosec θ, proving the identity.

Answer: Write in sin and cos: LHS = (sin³θ − cos³θ)/[sin θ cos θ (sin θ − cos θ)] = (1 + sin θ cos θ)/(sin θ cos θ) = 1 + sec θ cosec θ.

Let s = sin A, c = cos A. Prove (s + c)/(s − c) + (s − c)/(s + c) = 2/(2s² − 1).

1. Take (s + c)(s − c) as the common denominator: LHS = [(s + c)² + (s − c)²] ÷ [(s + c)(s − c)].
2. Numerator: (s² + 2sc + c²) + (s² − 2sc + c²) = 2s² + 2c² = 2(s² + c²) = 2 × 1 = 2, since sin²A + cos²A = 1.
3. Denominator: (s + c)(s − c) = s² − c², and with c² = 1 − s² this is s² − (1 − s²) = 2s² − 1.
4. So LHS = 2/(2s² − 1) = RHS.

Answer: Combine: [(sin A + cos A)² + (sin A − cos A)²]/(sin²A − cos²A) = 2/(sin²A − (1 − sin²A)) = 2/(2 sin²A − 1).

In what ratio does the y-axis cut the segment from (5, −6) to (−1, −4)? Give the meeting point too.

1. Let the y-axis divide the segment joining A(5, -6) and B(-1, -4) in the ratio k : 1.
2. Using the section formula, the x-coordinate of the point of division is (-k + 5)/(k + 1).
3. Since the point lies on the y-axis, its x-coordinate is 0, so (-k + 5)/(k + 1) = 0.
4. Solving -k + 5 = 0 gives k = 5, so the ratio is 5 : 1.
5. The y-coordinate is (-4k - 6)/(k + 1) = (-4(5) - 6)/(5 + 1) = -26/6 = -13/3.
6. Thus, the ratio is 5 : 1 and the meeting point is (0, -13/3).

Answer: Ratio k : 1 with (−k + 5)/(k + 1) = 0 gives k = 5: ratio 5:1, point (0, -13/3).

Show that 1/√5 is irrational.

1. Assume for contradiction that 1/√5 is rational, so 1/√5 = p/q where p and q are co-prime integers and q ≠ 0.
2. Squaring both sides gives 1/5 = p²/q², which rearranges to q² = 5p².
3. Since 5 divides q², 5 must divide q; let q = 5a for some integer a.
4. Substituting q = 5a into q² = 5p² gives 25a² = 5p², so p² = 5a².
5. Since 5 divides p², 5 must divide p, meaning both p and q are divisible by 5.
6. This contradicts the assumption that p and q are co-prime, so 1/√5 is irrational.

Answer: Assume 1/√5 = p/q (co-prime). Then q² = 5p², so 5 divides q; put q = 5a to get p² = 5a², so 5 divides p too — contradiction; hence 1/√5 is irrational.

A room is a cylinder topped by a hemispherical dome; dome radius = half the cylinder height. Air volume 1408/21 m³. Find the cylinder height (π = 22/7).

1. Let the radius of the cylinder be r and its height be h. Given that the dome radius is half the cylinder height, r = h/2, which implies h = 2r.
2. The total volume of air is the sum of the cylinder volume and the hemispherical dome volume: V = πr²h + (2/3)πr³.
3. Substitute h = 2r into the volume formula: V = πr²(2r) + (2/3)πr³ = 2πr³ + (2/3)πr³ = (8/3)πr³.
4. Equate the total volume to the given value: (8/3)πr³ = 1408/21.
5. Substitute π = 22/7 and solve for r³: (8/3)(22/7)r³ = 1408/21, which simplifies to (176/21)r³ = 1408/21, so r³ = 1408/176 = 8.
6. Calculate r = ∛8 = 2 m, and then find the height h = 2r = 2(2) = 4 m.

Answer: With h = 2r: (2/3)πr³ + πr²(2r) = (8/3)πr³ = 1408/21, so r³ = 8, r = 2 m and the height is 4 m.

Two dice are rolled together. Probability that their numbers differ by 2?

1. Total number of outcomes when two dice are rolled is 6 × 6 = 36.
2. The favourable outcomes where the numbers differ by 2 are (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), and (6,4).
3. There are 8 such favourable outcomes.
4. Probability = Number of favourable outcomes / Total number of outcomes = 8/36.
5. Simplifying the fraction, 8/36 = 2/9.

Answer: 8 of 36 outcomes work: (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4); P = 8/36 = 2/9.

Vijay put money in scheme A (8% p.a.) and scheme B (9% p.a.) and got ₹1,860 interest a year. Swapping the amounts would give ₹20 more. Find each amount.

1. Let x be the amount in scheme A and y be the amount in scheme B.
2. From the given interest, 8x/100 + 9y/100 = 1860, which simplifies to 8x + 9y = 186000.
3. Swapping the amounts gives 9x/100 + 8y/100 = 1880, which simplifies to 9x + 8y = 188000.
4. Subtracting the first equation from the second gives x - y = 2000, so x = y + 2000.
5. Substituting x = y + 2000 into 8x + 9y = 186000 gives 8(y + 2000) + 9y = 186000.
6. Solving 17y + 16000 = 186000 gives 17y = 170000, so y = 10000.
7. Calculating x gives x = 10000 + 2000 = 12000, so scheme A has 12000 and scheme B has 10000.

Answer: 8x + 9y = 1,86,000 and 9x + 8y = 1,88,000 give ₹12,000 in scheme A and ₹10,000 in scheme B.

ABCD is a parallelogram, E lies on BC, and AE meets diagonal BD at F. Prove DF × EF = FB × FA.

1. Consider triangles ADF and EBF.
2. Since AE and BD are intersecting lines, angle AFD is equal to angle EFB as vertically opposite angles.
3. Since ABCD is a parallelogram, AD is parallel to BC, so angle ADF is equal to angle EBF as alternate interior angles.
4. By the AA similarity criterion, triangle ADF is similar to triangle EBF.
5. From the similarity, the ratio of corresponding sides is DF/FB = FA/EF.
6. Cross-multiplying the ratios gives DF × EF = FB × FA.

Answer: In ΔADF and ΔEBF, vertically opposite angles at F are equal and ∠ADF = ∠EBF (alternate, AD ∥ BC); so they are similar (AA), DF/FB = FA/EF, hence DF × EF = FB × FA.

In ΔABC, AD ⊥ BC and AD² = BD × DC. Prove ∠BAC = 90°.

1. AD ⊥ BC, so ∠ADB = ∠ADC = 90°.
2. AD² = BD × DC can be written as BD/AD = AD/DC.
3. In △DBA and △DAC the sides around the right angle at D are in the same ratio, so △DBA ~ △DAC (SAS similarity).
4. Corresponding angles are equal: ∠DBA = ∠DAC and ∠BAD = ∠DCA.
5. Angle sum in △ABC: ∠BAC + ∠ABC + ∠ACB = 180°, i.e. ∠BAC + ∠DAC + ∠BAD = 180°.
6. Since D lies on BC, ∠BAD + ∠DAC = ∠BAC, so 2∠BAC = 180° and ∠BAC = 90°.

Answer: AD/DC = BD/AD with ∠ADB = ∠ADC = 90°, so ΔDBA ~ ΔDAC (SAS); then ∠DBA = ∠DAC and ∠BAD = ∠DCA; adding, ∠BAC equals half of 180°, i.e. 90°.

A right triangle has perimeter 60 cm and hypotenuse 25 cm. Find its other two sides.

1. Let the two legs be x and y; since the perimeter is 60 cm and the hypotenuse is 25 cm, x + y = 60 − 25 = 35.
2. By the Pythagorean theorem, x² + y² = 25² = 625.
3. Square the sum equation to get (x + y)² = 35², which expands to x² + 2xy + y² = 1225.
4. Substitute x² + y² = 625 into the expanded equation to find 2xy = 1225 − 625 = 600, so xy = 300.
5. Form the quadratic equation t² − 35t + 300 = 0 using the sum and product of the roots.
6. Factor the quadratic as (t − 15)(t − 20) = 0, giving roots t = 15 and t = 20.
7. Thus, the other two sides of the triangle are 15 cm and 20 cm.

Answer: x + y = 35 and x² + y² = 625 give x² − 35x + 300 = 0, so the sides are 15 cm and 20 cm.

A train covers 480 km at steady speed; 8 km/h slower it would need 3 h extra. Find its speed.

1. Let the steady speed be x km/h, so the time taken is 480/x hours.
2. At the slower speed of (x − 8) km/h, the time taken is 480/(x − 8) hours.
3. Since the slower speed takes 3 hours extra, we form the equation: 480/(x − 8) − 480/x = 3.
4. Simplifying the left side by taking LCM gives 3840 / [x(x − 8)] = 3.
5. Cross-multiplying and dividing by 3 yields x² − 8x − 1280 = 0.
6. Factoring the quadratic gives (x − 40)(x + 32) = 0, so x = 40 or x = −32.
7. Rejecting the negative value, the speed is 40 km/h.

Answer: 480/(x − 8) − 480/x = 3 gives x² − 8x − 1280 = 0, so x = 40 (x = −32 rejected): speed 40 km/h.

Daily allowance classes 11–13, 13–15, ..., 23–25 have frequencies 7, 6, 9, 13, f, 5, 4. Mean is 18. Find f and the mode.

1. Identify midpoints as 12, 14, 16, 18, 20, 22, 24 and calculate Σf = 44 + f and Σfx = 752 + 20f.
2. Use the mean formula 18 = (752 + 20f) / (44 + f) to set up the equation 18(44 + f) = 752 + 20f.
3. Simplify the equation to 792 + 18f = 752 + 20f, which gives 2f = 40, so f = 20.
4. Identify the modal class as 19–21 with frequency 20, previous frequency 13, and next frequency 5.
5. Apply the mode formula: Mode = 19 + [(20 - 13) / (2×20 - 13 - 5)] × 2.
6. Calculate the mode as 19 + [7 / 22] × 2 = 19 + 14/22 ≈ 19.64.

Answer: Σf = 44 + f, Σfx = 752 + 20f; 18(44 + f) = 752 + 20f gives f = 20. Modal class 19–21: mode = 19 + 7/(40 − 13 − 5) × 2, so f = 20, mode ≈ 19.64.

Note: CBSE's marking scheme prints mode = 19.95 because it uses 3 as the class size; the classes are 2 wide, so the correct mode is 19 + 14/22 ≈ 19.64. Both agree f = 20.

Charity run: round 1 is 300 m and each next round is 50 m longer (10 rounds). Write rounds 4, 5 and 6.

1. Identify the first term a = 300 m and the common difference d = 50 m.
2. Calculate the 4th round: a + 3d = 300 + 3(50) = 300 + 150 = 450 m.
3. Calculate the 5th round: a + 4d = 300 + 4(50) = 300 + 200 = 500 m.
4. Calculate the 6th round: a + 5d = 300 + 5(50) = 300 + 250 = 550 m.

Answer: 450, 500, 550 m.

Same run (300 m, then +50 m per round). Length of the 8th round?

1. Identify the first term a = 300 m and the common difference d = 50 m for the arithmetic progression of round lengths.
2. Calculate the length of the 8th round using the formula a₈ = a + (8 - 1)d = 300 + 7 × 50 = 650 m.

Answer: a₈ = 300 + 7 × 50 = 650 m.

Same run (300 m, then +50 m per round). Total distance for all 10 rounds?

1. The distances form an arithmetic progression with first term a = 300 m and common difference d = 50 m.
2. The 10th term is a₁₀ = 300 + 9(50) = 750 m.
3. Using the sum formula S₁₀ = 10/2 × (300 + 750) = 5 × 1050 = 5250 m.

Answer: S₁₀ = 5 × (600 + 450) = 5250 m.

Same run (300 m, then +50 m per round). Total distance if a runner stops after 6 rounds?

1. The first round is 300 m, and each subsequent round increases by 50 m, forming an arithmetic progression with a = 300 and d = 50.
2. The distance of the 6th round is a₆ = 300 + (6 - 1) × 50 = 300 + 250 = 550 m.
3. The total distance for 6 rounds is S₆ = 6/2 × (300 + 550) = 3 × 850 = 2550 m.

Answer: S₆ = 3 × (600 + 250) = 2550 m.

Note: CBSE's marking scheme prints 2250 m; the correct total is 2550 m (300 + 350 + 400 + 450 + 500 + 550).

A brooch is a silver-wire circle of diameter 35 mm with 5 diameters making 10 equal sectors. Central angle of each sector?

1. The 5 diameters divide the circle into 10 equal sectors.
2. Central angle of each sector = 360° ÷ 10 = 36°.

Answer: 360° ÷ 10 = 36°.

Same brooch (diameter 35 mm, 10 equal sectors). Length of arc ACB, the arc of one sector?

Uses a figure printed in CBSE's paper — what the figure shows: A and B are the ends of one sector's arc on the circle and C is a point on that arc between them.

1. Diameter of the brooch is 35 mm, so the radius r is 35/2 = 17.5 mm.
2. The brooch is divided into 10 equal sectors, so the angle of one sector is 360°/10 = 36°.
3. Length of arc ACB = (36/360) × 2 × π × 17.5 = (1/10) × 35π = 3.5π mm.
4. Using π ≈ 22/7, Arc length = 3.5 × 22/7 = 11 mm.

Answer: Arc = (1/10) × π × 35 = 11 mm.

Same brooch (diameter 35 mm, 10 equal sectors). Area of one sector?

1. Radius of the brooch is half of the diameter, so r = 35 / 2 = 17.5 mm.
2. Area of one sector is 1/10 of the total area, calculated as (1/10) × (22/7) × 17.5 × 17.5 = 96.25 mm².

Answer: (1/10) × (22/7) × 17.5 × 17.5 = 385/4 = 96.25 mm².

Same brooch (diameter 35 mm, 5 diameters). Total silver wire used?

1. Calculate the circumference of the brooch using the formula C = πd, where d = 35 mm and π = 22/7, giving C = (22/7) × 35 = 110 mm.
2. Calculate the total length of the 5 diameters by multiplying the diameter by 5, which is 5 × 35 = 175 mm.
3. Add the circumference and the total length of the diameters to find the total silver wire used, resulting in 110 + 175 = 285 mm.

Answer: Circumference 110 mm + 5 × 35 mm = 285 mm.

Lighthouse CE; Amrita at A (foot level) sees top C at 60°, then from B, 40 m above A, sees C at 45°. With CD = h, find BD in terms of h.

Uses a figure printed in CBSE's paper — what the figure shows: ABDE is a rectangle: AB = DE = 40 m vertical, BD horizontal and equal to AE; C is the lighthouse top above D and E; ∠CBD = 45°, ∠CAE = 60°; CD = h.

1. In right triangle CBD, angle CBD is 45 degrees and CD is h.
2. Using the tangent ratio, tan 45 degrees equals CD divided by BD, which is h/BD.
3. Since tan 45 degrees is 1, we have 1 = h/BD, so BD = h m.

Answer: tan 45° = h/BD = 1, so BD = h m.

Same lighthouse set-up (CD = h, angle at B is 45°). Find BC in terms of h.

Uses a figure printed in CBSE's paper — what the figure shows: Right triangle BDC with right angle at D, ∠CBD = 45°, CD = h.

1. In right triangle BDC, sin 45° = CD/BC = h/BC.
2. Since sin 45° = 1/√2, we have 1/√2 = h/BC, so BC = √2 h m.

Answer: sin 45° = h/BC, so BC = √2 h m.

Same set-up (B is 40 m above A; elevations 60° from A, 45° from B). Find the lighthouse height CE, taking √3 = 1.73.

Uses a figure printed in CBSE's paper — what the figure shows: Same figure as 38(i): AE = BD, CE = CD + DE = h + 40.

1. Let CD = h, so CE = h + 40. In triangle BDC, tan 45° = CD/BD, so BD = h.
2. Since AE = BD, AE = h. In triangle ACE, tan 60° = CE/AE, so √3 = (h + 40)/h.
3. Rearranging gives h√3 = h + 40, so h(√3 − 1) = 40, which means h = 40/(√3 − 1).
4. Rationalizing the denominator: h = 40(√3 + 1)/(3 − 1) = 20(√3 + 1).
5. Substituting √3 = 1.73: h = 20(1.73 + 1) = 20(2.73) = 54.6 m.
6. Total height CE = h + 40 = 54.6 + 40 = 94.6 m.

Answer: AE = BD = h and tan 60° = (h + 40)/h, so h = 40/(√3 − 1) = 20(√3 + 1) = 54.6 m; CE = 94.6 m.

Same set-up (angle CAE = 60°). If AC = 100 m, find AE.

Uses a figure printed in CBSE's paper — what the figure shows: Right triangle AEC with right angle at E and ∠CAE = 60°.

1. In right △AEC (right angle at E), ∠CAE = 60° and AC = 100 m.
2. cos 60° = AE/AC, so 1/2 = AE/100.
3. AE = 100 × 1/2 = 50 m.

Answer: cos 60° = AE/AC, so AE = 50 m.