CBSE Class 10 Maths (Standard) Board Paper 2024 (30/1/1): Solutions

The zeroes of 2x² − k√2 x + 1 add up to √2. Find k.

1. Sum of zeroes of 2x² − k√2 x + 1 is −(−k√2)/2 = k√2/2.
2. Given sum = √2, so k√2/2 = √2 → k = 2.

Answer: Sum of zeroes = k√2/2 = √2, so k = 2 — option (b).

A player wins a game with probability 0.79. Probability that he loses it?

1. The sum of the probabilities of all possible outcomes in a game is 1.
2. P(lose) = 1 − P(win) = 1 − 0.79 = 0.21, so option (d) is correct.

Answer: P(lose) = 1 − 0.79 = 0.21 — option (d).

ax² + bx + c = 0 (a ≠ 0) has real and equal roots. Which relation is true?

1. Real and equal roots require the discriminant to be zero, so b² − 4ac = 0.
2. Rearranging the equation gives 4ac = b², which means ac = b²/4, matching option (c).

Answer: Equal roots need b² − 4ac = 0, so ac = b²/4 — option (c).

An AP has a = 7, last (nth) term 84 and sum of n terms 2093/2. Find n.

1. Use the sum formula Sₙ = n(a + l)/2 with a = 7 and l = 84.
2. Substitute values: 2093/2 = n(7 + 84)/2.
3. Simplify: 2093/2 = 91n/2.
4. Multiply both sides by 2: 2093 = 91n.
5. Solve for n: n = 2093/91 = 23.

Answer: Sₙ = n(7 + 84)/2 = 2093/2 gives 91n = 2093, so n = 23 — option (c).

p = 18a²b⁴ and q = 20a³b², with a, b prime. Find LCM(p, q).

1. Identify the prime factorization of the numerical coefficients: 18 = 2 × 3² and 20 = 2² × 5.
2. Find the LCM of the coefficients by taking the highest power of each prime factor: LCM(18, 20) = 2² × 3² × 5 = 180.
3. Find the LCM of the variable parts by taking the highest power of each variable: LCM(a², a³) = a³ and LCM(b⁴, b²) = b⁴.
4. Combine the results to get the final LCM: 180a³b⁴.

Answer: LCM(18, 20) = 180 and take the highest powers a³, b⁴: LCM = 180a³b⁴ — option (d).

Triangle ABC has A(5, −6), B(6, 4), C(0, 0). Find the length of median AD.

1. Find the midpoint D of side BC using the coordinates B(6, 4) and C(0, 0): D = ((6 + 0)/2, (4 + 0)/2) = (3, 2).
2. Calculate the length of median AD using the distance formula between A(5, −6) and D(3, 2): AD = √((3 − 5)² + (2 − (−6))²) = √((−2)² + 8²) = √(4 + 64) = √68 units.

Answer: D = midpoint of BC = (3, 2); AD = √(2² + 8²) = √68 units — option (a).

Given sec θ − tan θ = m, find sec θ + tan θ.

1. Recall the trigonometric identity: sec²θ − tan²θ = 1.
2. Factor the left side as (sec θ − tan θ)(sec θ + tan θ) = 1.
3. Substitute the given value sec θ − tan θ = m into the equation.
4. Solve for sec θ + tan θ to get 1/m, which corresponds to option (c).

Answer: (sec θ − tan θ)(sec θ + tan θ) = sec²θ − tan²θ = 1, so sec θ + tan θ = 1/m — option (c).

Data: 1, 4, 7, 9, 16, 21, 25. Drop the even values; probability that a value picked at random from the rest is prime?

1. Remove the even values 4 and 16 from the data set, leaving 1, 7, 9, 21, and 25.
2. Identify that only 7 is a prime number among the remaining five values, so the probability is 1/5.

Answer: Left: 1, 7, 9, 21, 25; only 7 is prime, so P = 1/5 — option (b).

For values x₁ … xₙ with frequencies f₁ … fₙ and mean x̄, find Σ fᵢ(xᵢ − x̄).

1. Expand the expression to get Σfᵢxᵢ − x̄Σfᵢ.
2. Substitute the mean formula x̄ = Σfᵢxᵢ/Σfᵢ to show the result is 0.

Answer: Σfᵢxᵢ − x̄Σfᵢ = 0, since x̄ = Σfᵢxᵢ/Σfᵢ — option (d).

Zeroes of x² + px + q are double those of 4x² − 5x − 6. Find p.

1. For the quadratic equation 4x² − 5x − 6 = 0, the sum of the zeroes is -b/a = -(-5)/4 = 5/4.
2. Since the zeroes of x² + px + q are double those of the first equation, their sum is 2 × (5/4) = 5/2.
3. For the equation x² + px + q = 0, the sum of the zeroes is -p/1 = -p.
4. Equating the sums, -p = 5/2, which gives p = -5/2.

Answer: Sum of zeroes of 4x² − 5x − 6 is 5/4, so the new sum is 5/2 = −p, giving p = −5/2 — option (a).

Points (3, −5) and (x, −5) are 15 units apart. Find the values of x.

1. Since the y-coordinates are the same, the distance is the absolute difference of the x-coordinates: |x − 3| = 15.
2. Solving gives x − 3 = 15 or x − 3 = −15, so x = 18 or x = −12.

Answer: |x − 3| = 15, so x = 18, -12 — option (b).

If cos(α + β) = 0, find cos((α + β)/2).

1. Given cos(α + β) = 0, we know that the angle whose cosine is zero is 90°.
2. Therefore, α + β = 90°.
3. Dividing both sides by 2, we get (α + β)/2 = 45°.
4. Thus, cos((α + β)/2) = cos 45° = 1/√2.

Answer: α + β = 90°, so cos 45° = 1/√2 — option (a).

A solid sphere is cut into two hemispheres. Ratio of the sphere's surface area to the total surface area of both hemispheres?

1. Surface area of the sphere = 4πr².
2. Total surface area of one hemisphere = 3πr² (curved + flat base).
3. Total surface area of both hemispheres = 2 × 3πr² = 6πr².
4. Ratio = 4πr² : 6πr² = 2 : 3, so option (c) is correct.

Answer: 4πr² : 2 × 3πr² = 2/3, i.e. 2 : 3 — option (c).

What is the name of the middle value of data arranged in order?

1. The middle value of a dataset arranged in ascending or descending order is called the median.
2. Therefore, the correct option is (b).

Answer: The median — option (b).

Volume of the biggest right circular cone cut from a solid cube of edge 2 cm?

1. The radius of the cone is half the edge of the cube, so r = 2/2 = 1 cm, and the height equals the edge, so h = 2 cm.
2. Using the volume formula V = (1/3)πr²h, we get V = (1/3)π(1)²(2) = 2π/3 cu cm.

Answer: r = 1 cm, h = 2 cm: V = (1/3)π(1)²(2) = 2π/3 cu cm — option (d).

Two dice are thrown. Probability that the total is 2, 3 or 5?

1. Total outcomes when two dice are thrown = 6 × 6 = 36.
2. Outcomes for total 2: (1,1) = 1 way; for total 3: (1,2), (2,1) = 2 ways; for total 5: (1,4), (2,3), (3,2), (4,1) = 4 ways.
3. Total favorable outcomes = 1 + 2 + 4 = 7.
4. Probability = Favorable outcomes / Total outcomes = 7/36.

Answer: Totals 2, 3, 5 come in 1 + 2 + 4 = 7 of 36 ways, so P = 7/36 — option (a).

A circle has centre (2, 0); one end of a diameter is (6, 0). Find the other end.

1. Let the other end of the diameter be (x, y).
2. Using the midpoint formula, the centre (2, 0) is the average of (6, 0) and (x, y).
3. For the x-coordinate: (6 + x)/2 = 2, so 6 + x = 4, which gives x = -2.
4. For the y-coordinate: (0 + y)/2 = 0, so y = 0.
5. Thus, the other end is (-2, 0).

Answer: The centre is the midpoint, so the other end is (-2, 0) — option (c).

The figure shows the graphs of two linear equations. What kind of pair is it?

Uses a figure printed in CBSE's paper — what the figure shows: Two straight lines with different slopes (both rising to the right, one steeper) drawn near the origin; they are not parallel, so they intersect when extended.

1. The two lines have different slopes, so they are not parallel.
2. Since they are not parallel, they intersect at exactly one point, meaning the system has a unique solution.

Answer: The two lines are not parallel, so they meet in one point: consistent with a unique solution — option (a).

Assertion: tangents at the two ends of a diameter are parallel. Reason: a diameter is the longest chord.

1. Both the Assertion and the Reason are true statements.
2. The Assertion is true because tangents at the ends of a diameter are perpendicular to the same line, making them parallel.
3. The Reason is true but does not explain why the tangents are parallel, so the correct option is (b).

Answer: Both tangents are perpendicular to the same diameter, so A is true; R is also true but does not explain A — option (b).

Assertion: a polynomial whose graph touches the x-axis at just one point cannot be quadratic. Reason: a degree-n polynomial (n > 1) has at most n zeroes.

1. The assertion is false because the polynomial y = x² touches the x-axis at exactly one point (the origin), yet it is a quadratic polynomial.
2. The reason is true because a polynomial of degree n has at most n zeroes, which is a standard property of polynomials.
3. Since the assertion is false and the reason is true, the correct option is (d).

Answer: A is false (e.g. y = x² touches the x-axis only at the origin); R is true — option (d).

Solve 7x − 2y = 5 and 8x + 7y = 15, then verify.

1. Given equations are 7x - 2y = 5 and 8x + 7y = 15.
2. Multiply the first equation by 7 and the second by 2 to eliminate y: 49x - 14y = 35 and 16x + 14y = 30.
3. Add the two new equations: 49x + 16x = 35 + 30, which simplifies to 65x = 65.
4. Solve for x: x = 65/65 = 1.
5. Substitute x = 1 into the first equation: 7(1) - 2y = 5, so 7 - 2y = 5.
6. Solve for y: -2y = 5 - 7 = -2, so y = -2/-2 = 1.
7. Verification: Substitute x = 1 and y = 1 into both original equations. For 7x - 2y: 7(1) - 2(1) = 7 - 2 = 5. For 8x + 7y: 8(1) + 7(1) = 8 + 7 = 15. Both equations are satisfied.

Answer: x = 1, y = 1 (check: 7 − 2 = 5 and 8 + 7 = 15).

A black card is missing from a 52-card pack. One card is drawn from the rest. Find P(queen of hearts).

1. Total cards in a standard pack are 52, and since one black card is missing, the total number of remaining cards is 52 - 1 = 51.
2. The queen of hearts is a red card, so it is still present in the pack, making the number of favorable outcomes 1.
3. Probability is calculated as favorable outcomes divided by total outcomes, which gives P(queen of hearts) = 1/51.

Answer: 51 cards remain and the queen of hearts is still there, so P = 1/51.

Find the value of 2√2 cos 45° sin 30° + 2√3 cos 30°.

1. Substitute the standard trigonometric values: cos 45° = 1/√2, sin 30° = 1/2, and cos 30° = √3/2.
2. Simplify the first term: 2√2 × (1/√2) × (1/2) = 2 × (1/2) = 1.
3. Simplify the second term: 2√3 × (√3/2) = (2 × 3) / 2 = 3.
4. Add the simplified terms to find the final value: 1 + 3 = 4.

Answer: 2√2 × (1/√2) × (1/2) + 2√3 × (√3/2) = 1 + 3 = 4.

With A = 60° and B = 30°, check that sin(A + B) = sin A cos B + cos A sin B.

1. Substitute A = 60° and B = 30° into the left-hand side: LHS = sin(60° + 30°) = sin 90° = 1.
2. Substitute the values into the right-hand side: RHS = sin 60° cos 30° + cos 60° sin 30°.
3. Use the standard trigonometric values: RHS = (√3/2)(√3/2) + (1/2)(1/2).
4. Simplify the terms: RHS = 3/4 + 1/4 = 1.
5. Since LHS = 1 and RHS = 1, LHS = RHS, verifying the identity.

Answer: LHS = sin 90° = 1; RHS = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1; so LHS = RHS.

In quadrilateral ABCD, diagonal BD bisects both ∠B and ∠D. Show (i) △ABD ~ △CBD and (ii) AB = BC.

Uses a figure printed in CBSE's paper — what the figure shows: Quadrilateral ABCD with diagonal BD drawn; BD bisects angle B (∠ABD = ∠CBD) and angle D (∠ADB = ∠CDB).

1. Given that BD bisects ∠B and ∠D, we have ∠ABD = ∠CBD and ∠ADB = ∠CDB.
2. In △ABD and △CBD, two pairs of corresponding angles are equal, so △ABD ~ △CBD by AA similarity criterion.
3. Since the triangles are similar and share the side BD, the ratio of corresponding sides is 1, which implies AB = BC.

Answer: ∠ABD = ∠CBD and ∠ADB = ∠CDB, so △ABD ~ △CBD by AA. They also share BD, so the ratio of similarity is BD/BD = 1, giving AB = BC.

Given that √3 is irrational, prove that 5 − 2√3 is irrational.

1. Assume 5 − 2√3 is rational, so 5 − 2√3 = a/b where a and b are integers and b ≠ 0.
2. Rearranging the equation gives 2√3 = 5 − a/b, which simplifies to √3 = (5b − a)/(2b).
3. Since a and b are integers, (5b − a)/(2b) is a rational number, implying √3 is rational.
4. This contradicts the given fact that √3 is irrational, so the assumption is false and 5 − 2√3 is irrational.

Answer: Suppose 5 − 2√3 = a/b (a, b integers, b ≠ 0). Then √3 = (5b − a)/(2b) is rational — a contradiction. So 5 − 2√3 is irrational.

Show that 5 × 11 × 17 + 3 × 11 is composite.

1. Factor out the common term 11 from the expression to get 11 × (5 × 17 + 3).
2. Simplify the expression inside the brackets to obtain 11 × (85 + 3) = 11 × 88.
3. Calculate the product to get 968, which has factors other than 1 and itself, so it is composite.

Answer: It equals 11 × (85 + 3) = 11 × 88 = 968, which has factors other than 1 and itself, so it is composite.

In what ratio does (8/5, y) divide the join of (1, 2) and (2, 3)? Also find y.

1. Let the point (8/5, y) divide the line segment joining (1, 2) and (2, 3) in the ratio k : 1.
2. Using the section formula for the x-coordinate, we have (2k + 1)/(k + 1) = 8/5.
3. Cross-multiplying gives 5(2k + 1) = 8(k + 1), which simplifies to 10k + 5 = 8k + 8.
4. Solving for k, we get 2k = 3, so k = 3/2, meaning the ratio is 3 : 2.
5. Using the section formula for the y-coordinate with k = 3/2, y = (3(3) + 2(2))/(3 + 2).
6. Calculating the value, y = (9 + 4)/5 = 13/5.

Answer: Ratio k : 1 with (2k + 1)/(k + 1) = 8/5 gives k = 3/2, i.e. 3 : 2; then y = (3×3 + 2×2)/5 = 13/5.

Rectangle ABCD: A(−1, −1), B(−1, 6), C(3, 6), D(3, −1). P, Q, R, S are midpoints of AB, BC, CD, DA. Show the diagonals of PQRS bisect each other.

1. Find the coordinates of the midpoints P, Q, R, and S using the midpoint formula: P(−1, 5/2), Q(1, 6), R(3, 5/2), S(1, −1).
2. Calculate the midpoint of diagonal PR: ((−1 + 3)/2, (5/2 + 5/2)/2) = (1, 5/2).
3. Calculate the midpoint of diagonal QS: ((1 + 1)/2, (6 + (−1))/2) = (1, 5/2).
4. Since the midpoints of both diagonals are the same, the diagonals of PQRS bisect each other.

Answer: P(−1, 5/2), Q(1, 6), R(3, 5/2), S(1, −1). Midpoint of PR = (1, 5/2) = midpoint of QS, so the diagonals bisect each other.

A workshop has 48 French, 80 Hindi and 144 English teachers. Each room seats an equal number of teachers, all of one subject. Least number of rooms?

1. To find the least number of rooms, we first determine the maximum number of teachers that can be seated in each room by finding the HCF of 48, 80, and 144.
2. Prime factorization: 48 = 2⁴ × 3, 80 = 2⁴ × 5, and 144 = 2⁴ × 3².
3. The HCF is the product of the lowest powers of common prime factors, which is 2⁴ = 16.
4. Thus, each room will seat 16 teachers of the same subject.
5. Total number of teachers = 48 + 80 + 144 = 272.
6. Least number of rooms = Total teachers ÷ Teachers per room = 272 ÷ 16 = 17.

Answer: HCF(48, 80, 144) = 16 per room, so rooms = (48 + 80 + 144)/16 = 17.

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

1. Express LHS in terms of sine and cosine: (sin θ/cos θ)/(1 − cos θ/sin θ) + (cos θ/sin θ)/(1 − sin θ/cos θ).
2. Simplify each term by combining denominators: (sin²θ)/(cos θ(sin θ − cos θ)) − (cos²θ)/(sin θ(sin θ − cos θ)).
3. Combine over the common denominator sin θ cos θ (sin θ − cos θ): (sin³θ − cos³θ) / [sin θ cos θ (sin θ − cos θ)].
4. Factor the numerator using a³ − b³ = (a − b)(a² + ab + b²): [(sin θ − cos θ)(sin²θ + sin θ cos θ + cos²θ)] / [sin θ cos θ (sin θ − cos θ)].
5. Cancel (sin θ − cos θ) and use sin²θ + cos²θ = 1 to get (1 + sin θ cos θ) / (sin θ cos θ).
6. Split the fraction into 1/(sin θ cos θ) + 1, which simplifies to cosec θ sec θ + 1.
7. Rearrange to match the RHS: 1 + sec θ cosec θ.

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.

3 years ago Rashmi's age was 3 times Nazma's; 10 years from now it will be 2 times Nazma's. Find their present ages.

1. Let the present age of Rashmi be x years and the present age of Nazma be y years.
2. From the condition 3 years ago, x − 3 = 3(y − 3), which simplifies to x − 3y = −6.
3. From the condition 10 years from now, x + 10 = 2(y + 10), which simplifies to x − 2y = 10.
4. Subtracting the first equation from the second gives (x − 2y) − (x − 3y) = 10 − (−6), so y = 16.
5. Substituting y = 16 into x − 2y = 10 gives x − 32 = 10, so x = 42.
6. Thus, Rashmi's present age is 42 years and Nazma's present age is 16 years.

Answer: x − 3 = 3(y − 3) and x + 10 = 2(y + 10) give Rashmi 42 years, Nazma 16 years.

AB is a diameter of a circle with centre O; AQ, BP (tangents at A, B) and PQ (touching the circle at R) are tangents. Prove ∠POQ = 90°.

Uses a figure printed in CBSE's paper — what the figure shows: AB is a vertical diameter; parallel tangent lines at A and B; Q on the tangent at A and P on the tangent at B; PQ is a third tangent touching the circle at a point R between them; OP and OQ are joined.

1. Join OR. Since OA and OR are radii, OA = OR. Since AQ and RQ are tangents from Q, AQ = RQ. OQ is common.
2. By SSS congruence, △AOQ ≅ △ROQ. Thus, ∠AOQ = ∠ROQ, so OQ bisects ∠AOR.
3. Similarly, OB = OR (radii) and BP = RP (tangents from P). OP is common.
4. By SSS congruence, △BOP ≅ △ROP. Thus, ∠BOP = ∠ROP, so OP bisects ∠BOR.
5. Since AB is a diameter, ∠AOB = 180°. Therefore, ∠AOR + ∠BOR = 180°.
6. Substituting the bisected angles, 2∠QOR + 2∠ROP = 180°, which simplifies to ∠QOR + ∠ROP = 90°.
7. Since ∠POQ = ∠QOR + ∠ROP, it follows that ∠POQ = 90°.

Answer: Join OR. △AOQ ≅ △ROQ and △BOP ≅ △ROP, so OQ bisects ∠AOR and OP bisects ∠BOR. Since ∠AOR + ∠BOR = 180°, ∠QOR + ∠ROP = 90°, i.e. ∠POQ = 90°.

A circle (centre O, radius 8 cm) is inscribed in quadrilateral ABCD, touching AD, AB, BC, DC at P, S, R, Q. AD ⊥ DC, BC = 30 cm, BS = 24 cm. Find DC.

Uses a figure printed in CBSE's paper — what the figure shows: P on AD, S on AB, R on BC, Q on DC are the points of contact; right angle at D.

1. Since AB and BC are tangents from point B to the circle, the lengths of the tangents are equal, so BR = BS = 24 cm.
2. Calculate the length of CR by subtracting BR from BC: CR = BC - BR = 30 - 24 = 6 cm.
3. Since BC and DC are tangents from point C to the circle, the lengths of the tangents are equal, so CQ = CR = 6 cm.
4. Given that AD is perpendicular to DC, angle D is 90 degrees. Since OP and OQ are radii perpendicular to the tangents AD and DC respectively, angles OPD and OQD are also 90 degrees.
5. In quadrilateral OPDQ, three angles are 90 degrees, making it a rectangle. Since OP and OQ are both radii of the circle, OP = OQ = 8 cm, which means OPDQ is a square.
6. Therefore, the side DQ is equal to the radius, so DQ = 8 cm.
7. Find the total length of DC by adding DQ and CQ: DC = DQ + CQ = 8 + 6 = 14 cm.

Answer: BR = BS = 24, so CR = 6 = CQ; with AD ⊥ DC, OPDQ is a square, so DQ = 8. DC = 8 + 6 = 14 cm.

A hollow cylinder of length 14 cm has outer radius 1 cm more than inner radius. Metal volume is 176 cm³. Find both radii.

1. Let the inner radius be r cm and the outer radius be R cm, so R - r = 1.
2. The volume of the metal is given by the formula V = πh(R² - r²).
3. Substitute the given values into the formula: 176 = (22/7) × 14 × (R² - r²).
4. Simplify the equation to find R² - r² = 4.
5. Factorize the difference of squares as (R + r)(R - r) = 4.
6. Substitute R - r = 1 to get R + r = 4.
7. Solve the system of equations R + r = 4 and R - r = 1 to find R = 2.5 cm and r = 1.5 cm.

Answer: (22/7) × 14 × (R² − r²) = 176 gives R + r = 4; with R − r = 1, outer radius 2.5 cm and inner radius 1.5 cm.

In a circle of radius 21 cm an arc subtends 60° at the centre. Find (i) the arc length and (ii) the area of the minor segment cut off by its chord.

1. Given radius r = 21 cm and angle θ = 60°.
2. Arc length = (θ/360) × 2πr = (60/360) × 2 × (22/7) × 21 = 22 cm.
3. Area of sector = (θ/360) × πr² = (60/360) × (22/7) × 21² = 231 cm².
4. Triangle formed by radii and chord is equilateral with side 21 cm.
5. Area of triangle = (√3/4) × 21² = 441√3/4 cm².
6. Area of segment = Area of sector − Area of triangle = 231 − 441√3/4.
7. Using √3 ≈ 1.732, Area of segment = 231 − 190.95 = 40.05 cm².

Answer: (i) Arc = 2 × (22/7) × 21 × 60/360 = 22 cm. (ii) Sector = 231 cm², triangle (equilateral) = 441√3/4 cm², so segment = 231 − 441√3/4 ≈ 40.05 cm².

Note: Uses √3 ≈ 1.732 for the decimal value.

In an AP, first term + eighth term = 32 and their product = 60. Find a and d, and the sum of the first 20 terms.

1. Let the first term be a and common difference be d. The 8th term is a + 7d.
2. Given a + (a + 7d) = 32, so 2a + 7d = 32, which implies a + 7d = 32 - a.
3. Given a(a + 7d) = 60. Substituting the expression for a + 7d gives a(32 - a) = 60.
4. Rearranging gives a² - 32a + 60 = 0, which factors to (a - 2)(a - 30) = 0, so a = 2 or a = 30.
5. If a = 2, then 2(2) + 7d = 32 gives 7d = 28, so d = 4.
6. If a = 30, then 2(30) + 7d = 32 gives 7d = -28, so d = -4.
7. Using S₂₀ = 20/2 [2a + 19d], for a=2, d=4: S₂₀ = 10[4 + 76] = 800; for a=30, d=-4: S₂₀ = 10[60 - 76] = -160.

Answer: 2a + 7d = 32 and a(a + 7d) = 60 give a = 2, d = 4 (S₂₀ = 800) or a = 30, d = −4 (S₂₀ = -160).

Note: Both APs (2, 6, 10, … and 30, 26, 22, …) satisfy the conditions; CBSE accepts both.

An AP has 40 terms. Its first 9 terms add to 153 and its last 6 terms add to 687. Find a, d and the sum of all terms.

1. Sum of first 9 terms is 9/2[2a + 8d] = 153, which simplifies to a + 4d = 17.
2. Sum of last 6 terms (T35 to T40) is 6/2[2(a + 34d) + 5d] = 687, simplifying to 2a + 73d = 229.
3. Substitute a = 17 - 4d into the second equation to get 2(17 - 4d) + 73d = 229.
4. Solving 34 - 8d + 73d = 229 gives 65d = 195, so d = 3.
5. Substituting d = 3 into a + 4d = 17 gives a = 17 - 12 = 5.
6. Sum of all 40 terms S40 = 40/2[2(5) + 39(3)] = 20[10 + 117].
7. Calculating the final sum gives S40 = 20[127] = 2540.

Answer: a + 4d = 17 and 2a + 73d = 229 give a = 5, d = 3; S₄₀ = 20(10 + 117) = 2540.

Prove: a line drawn parallel to one side of a triangle, cutting the other two sides at distinct points, divides those sides proportionally.

1. Take △ABC with DE ∥ BC, D on AB and E on AC. Join BE and CD.
2. △ADE and △BDE have the same height from E to line AB, so ar(ADE)/ar(BDE) = AD/DB.
3. △ADE and △DEC have the same height from D to line AC, so ar(ADE)/ar(DEC) = AE/EC.
4. △BDE and △DEC stand on the same base DE and lie between the same parallels DE and BC, so ar(BDE) = ar(DEC).
5. The two ratios therefore have the same numerator and equal denominators, so AD/DB = AE/EC.

Answer: Take DE ∥ BC in △ABC. Join BE, CD and draw heights. ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(DEC) = AE/EC; ar(BDE) = ar(DEC) (same base DE, same parallels), so AD/DB = AE/EC.

PA, QB, RC are perpendicular to AC with AP = x, BQ = y, CR = z (Q on both PC and AR). Prove 1/x + 1/z = 1/y.

Uses a figure printed in CBSE's paper — what the figure shows: A, B, C on a horizontal line with B between A and C; PA, QB, RC vertical; P joined to C and A joined to R, the two lines crossing at Q above B.

1. Since PA, QB, and RC are perpendicular to AC, angle PAC = angle QBC = 90° and angle RCA = angle QBA = 90°.
2. In triangles PAC and QBC, angle PCA is common and the right angles are equal, so triangle PAC is similar to triangle QBC by AA criterion.
3. From this similarity, the ratio of corresponding sides gives AP/BQ = AC/BC, which simplifies to y/x = BC/AC.
4. In triangles RCA and QBA, angle RAC is common and the right angles are equal, so triangle RCA is similar to triangle QBA by AA criterion.
5. From this similarity, the ratio of corresponding sides gives CR/BQ = AC/AB, which simplifies to y/z = AB/AC.
6. Adding the two equations yields y/x + y/z = BC/AC + AB/AC = (BC + AB)/AC.
7. Since B lies between A and C, AB + BC = AC, so the right side equals 1; dividing by y gives 1/x + 1/z = 1/y.

Answer: △PAC ~ △QBC gives y/x = BC/AC; △RCA ~ △QBA gives y/z = AB/AC. Adding, y/x + y/z = (BC + AB)/AC = 1, so 1/x + 1/z = 1/y.

A 6 m pole stands on a tower. From P on the ground the pole's top has elevation 60°; from the tower's top, P has depression 45°. Find the tower's height and its distance from P (√3 = 1.73).

1. Let the height of the tower be h and the distance from P be x.
2. From the top of the tower, the angle of depression to P is 45°, so tan 45° = h/x, which gives h = x.
3. From P, the angle of elevation to the top of the pole is 60°, so tan 60° = (h + 6)/x.
4. Substitute h = x and tan 60° = √3 to get √3 = (x + 6)/x.
5. Rearrange to find x(√3 - 1) = 6, so x = 6/(√3 - 1).
6. Rationalize to get x = 3(√3 + 1), and substitute √3 = 1.73 to calculate x = 3(2.73) = 8.19 m.
7. Since h = x, the height of the tower is 8.19 m and the distance from P is 8.19 m.

Answer: h = x (45°) and h + 6 = x√3 (60°) give h = 6/(√3 − 1) = 3(√3 + 1) = 8.19 m; P is 8.19 m from the foot.

Note: The rationalised form 3(√3 + 1) gives 8.19 m; dividing 6 by 0.73 directly gives about 8.22 m.

A floor takes 200 square tiles of side x; with side x + 1 it takes 128. Form the equation in x.

1. The area of the floor using tiles of side x is 200 × x².
2. The area of the floor using tiles of side (x + 1) is 128 × (x + 1)².
3. Since the floor area is constant, equate the two expressions: 200x² = 128(x + 1)².

Answer: Same floor area: 200x² = 128(x + 1)².

Same tiles set-up. Put the equation in standard form.

1. Divide the entire equation by 8 to simplify: 25x² = 16x² + 32x + 16.
2. Move all terms to the left side to set the equation to zero: 25x² − 16x² − 32x − 16 = 0, which gives 9x² − 32x − 16 = 0.

Answer: Divide by 8: 25x² = 16x² + 32x + 16, so 9x² − 32x − 16 = 0.

Same tiles set-up. Find the tile side x by factorising.

1. Standard form: 9x² − 32x − 16 = 0. Here 9 × (−16) = −144 = (−36) × 4 and −36 + 4 = −32.
2. Split the middle term: 9x² − 36x + 4x − 16 = 0, so 9x(x − 4) + 4(x − 4) = 0.
3. This gives (9x + 4)(x − 4) = 0, so x = 4 or x = −4/9.
4. A side cannot be negative, so x = 4 units.

Answer: (9x + 4)(x − 4) = 0; x cannot be −4/9, so x = 4 units.

Same tiles set-up. Solve 9x² − 32x − 16 = 0 with the quadratic formula.

1. In 9x² − 32x − 16 = 0, a = 9, b = −32, c = −16.
2. Discriminant = (−32)² − 4 × 9 × (−16) = 1024 + 576 = 1600, and √1600 = 40.
3. x = (32 ± 40)/18, so x = 72/18 = 4 or x = −8/18 = −4/9.
4. A side cannot be negative, so x = 4 units.

Answer: x = (32 ± √1600)/18 = (32 ± 40)/18, so x = 4 or −4/9; side x = 4 units.

BINGO data (48 balls): 0-15: 8, 15-30: 9, 30-45: 10, 45-60: 12, 60-75: 9. Give the median class.

1. Calculate n/2 = 48/2 = 24.
2. Find the cumulative frequency just greater than 24, which is 27, corresponding to the class 30–45.

Answer: n/2 = 24; cumulative frequencies 8, 17, 27, … so the median class is 30–45.

Balls are numbered 1 to 75. For the first ball drawn, find P(even number).

1. Total number of balls = 75.
2. Number of even numbers from 1 to 75 = 37.
3. P(even number) = 37/75.

Answer: There are 37 even numbers in 1–75, so P = 37/75.

Same BINGO table. Find the median.

1. Identify the median class as 30-45 with lower limit l = 30, class width h = 15, frequency f = 10, and cumulative frequency of the preceding class cf = 17.
2. Determine that half of the total observations is n/2 = 24.
3. Substitute these values into the median formula: Median = 30 + ((24 − 17)/10) × 15.
4. Calculate the result: Median = 30 + (7/10) × 15 = 30 + 10.5 = 40.5.

Answer: Median = 30 + ((24 − 17)/10) × 15 = 40.5.

Same BINGO table. Find the mode.

1. The highest frequency is 12, so the modal class is 45–60: l = 45, f₁ = 12, f₀ = 10, f₂ = 9, h = 15.
2. Mode = l + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h = 45 + (12 − 10)/(24 − 10 − 9) × 15.
3. Mode = 45 + (2/5) × 15 = 45 + 6 = 51.

Answer: Modal class 45–60: mode = 45 + (12 − 10)/(24 − 10 − 9) × 15 = 51.

Right triangle ABC (right angle at B), AB = 7 m, BC = 15 m, with incircle touching AC, BC, AB at P, Q, R; AP = x m. Find AR.

1. Since AR and AP are tangents drawn from the same external point A to the incircle, their lengths are equal.
2. Therefore, AR = AP = x m.

Answer: Tangents from A are equal, so AR = AP = x m.

Same incircle set-up (O is the centre). What kind of quadrilateral is BQOR?

1. The radii OQ and OR meet the tangents BC and AB at right angles, so ∠OQB = ∠ORB = 90°.
2. ∠QBR = ∠ABC = 90°, so BQOR has three right angles and is a rectangle.
3. Its adjacent sides OQ and OR are both radii, so they are equal; hence BQOR is a square.

Answer: Three right angles (at B, Q, R) and adjacent sides OQ = OR, so BQOR is a square.

Same incircle set-up. Express PC in terms of x and then find x.

1. Tangents from A are equal: AR = AP = x, so BR = 7 − x, and tangents from B give BQ = BR = 7 − x.
2. Tangents from C are equal: PC = CQ = BC − BQ = 15 − (7 − x) = 8 + x.
3. AC = AP + PC = x + 8 + x = 8 + 2x.
4. By Pythagoras, AC = √(7² + 15²) = √(49 + 225) = √274.
5. So 8 + 2x = √274, giving x = (√274 − 8)/2 ≈ (16.55 − 8)/2 ≈ 4.28 m.

Answer: BR = 7 − x = BQ, so PC = CQ = 15 − (7 − x) = 8 + x. AC = 8 + 2x = √(49 + 225) = √274, so x = (√274 − 8)/2 ≈ 4.28 m.

Same incircle set-up. Find x and then the radius r.

1. AR = AP = x, so BR = BQ = 7 − x and PC = CQ = 15 − (7 − x) = 8 + x; hence AC = x + 8 + x = 8 + 2x.
2. By Pythagoras, AC = √(49 + 225) = √274, so x = (√274 − 8)/2 ≈ 4.28 m.
3. BQOR is a square, so r = OR = BR = 7 − x.
4. r = 7 − (√274 − 8)/2 = (22 − √274)/2 ≈ (22 − 16.55)/2 ≈ 2.72 m.

Answer: x = (√274 − 8)/2 ≈ 4.28 m; r = BR = 7 − x = (22 − √274)/2 ≈ 2.72 m.

Note: CBSE's marking scheme writes r = 11 − √274/2, which is the same value.