CBSE Class 10 Maths (Standard) Board Paper 2023 (30/1/1): Solutions

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

Uses a figure printed in CBSE's paper — what the figure shows: The graph is a downward-opening parabola whose vertex lies on the positive x-axis: it touches the x-axis at exactly one point and does not cross it.

1. The graph of y = p(x) touches the x-axis at exactly one point and does not cross it.
2. Since the curve intersects the x-axis at only one point, p(x) has 1 zero, which corresponds to option (B).

Answer: The curve only touches the x-axis at one point, so p(x) has 1 zero — option (B).

For which k do kx = y + 2 and 6x = 2y + 3 have infinitely many solutions?

1. Write the equations as kx − y − 2 = 0 and 6x − 2y − 3 = 0.
2. Infinitely many solutions need k/6 = (−1)/(−2) = (−2)/(−3), that is k/6 = 1/2 = 2/3. Since 1/2 ≠ 2/3, no value of k works — option (B).

Answer: Equal ratios need k/6 = 1/2 = 2/3, but 1/2 ≠ 2/3, so no such k exists — option (B).

p − 1, p + 1, 2p + 3 are consecutive AP terms. Find p.

1. For three terms to be in AP, the middle term must be the average of the other two, so 2(p + 1) = (p − 1) + (2p + 3).
2. Simplifying the equation gives 2p + 2 = 3p + 2, which leads to p = 0.

Answer: 2(p + 1) = (p − 1) + (2p + 3) gives 2p + 2 = 3p + 2, so p = 0 — option (C).

In what ratio does the x-axis cut the segment from A(3, 6) to B(−12, −3)?

1. The x-axis has the equation y = 0, so the point of division must have a y-coordinate of 0.
2. Using the section formula for the y-coordinate, 0 = (m(-3) + n(6)) / (m + n), which simplifies to 3m = 6n, giving the ratio m : n = 2 : 1.

Answer: On the x-axis y = 0: ratio = 6 : 3 = 2 : 1 — option (D).

PQ touches the circle (centre O) at B; chord AB has ∠AOB = 95°. Find ∠ABQ.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, OA and OB are radii with ∠AOB = 95°; tangent line PQ touches the circle at B, with Q on the same side as A.

1. Since OA and OB are radii of the circle, triangle OAB is isosceles, so ∠OAB = ∠OBA.
2. Using the angle sum property in triangle OAB: ∠OBA = (180° − 95°)/2 = 42.5°.
3. The radius OB is perpendicular to the tangent PQ at point B, so ∠OBQ = 90°.
4. Since Q is on the same side as A, ∠ABQ = ∠OBQ − ∠OBA = 90° − 42.5° = 47.5°.

Answer: ∠OBA = (180° − 95°)/2 = 42.5° and ∠OBQ = 90°, so ∠ABQ = 47.5° — option (A).

Given 2 tan A = 3, evaluate (4 sin A + 3 cos A)/(4 sin A − 3 cos A).

1. Divide the numerator and denominator by cos A to express the terms in tan A: (4 sin A + 3 cos A)/(4 sin A − 3 cos A) = (4 tan A + 3)/(4 tan A − 3).
2. Substitute the given value 2 tan A = 3, which implies 4 tan A = 6, into the expression: (6 + 3)/(6 − 3) = 9/3 = 3.

Answer: Divide by cos A: (4 tan A + 3)/(4 tan A − 3) = (6 + 3)/(6 − 3) = 3 — option (C).

α, β are zeroes of x² + x − 1. Find 1/α + 1/β.

1. Use the sum and product of zeroes for the equation x² + x − 1 = 0, where α + β = −1 and αβ = −1.
2. Calculate 1/α + 1/β as (α + β)/(αβ) = (−1)/(−1) = 1.

Answer: (α + β)/(αβ) = (−1)/(−1) = 1 — option (A).

Smallest positive k for which 2x² + kx − 4 = 0 has rational roots.

1. For the quadratic equation 2x² + kx − 4 = 0 to have rational roots, its discriminant D = b² − 4ac must be a perfect square.
2. Substituting a = 2, b = k, and c = −4, we get D = k² − 4(2)(−4) = k² + 32.
3. Testing small positive integers for k: if k = 1, D = 1 + 32 = 33 (not a perfect square); if k = 2, D = 4 + 32 = 36 (which is 6², a perfect square).
4. Thus, the smallest positive value of k is 2, corresponding to option (B).

Answer: Need k² + 32 to be a perfect square; k = 1 gives 33 (no), k = 2 gives 36 (yes), so k = 2 — option (B).

Evaluate (3/4)tan²30° − sec²45° + sin²60°.

1. Substitute the standard trigonometric values: tan 30° = 1/√3, sec 45° = √2, and sin 60° = √3/2.
2. Square the values: tan²30° = (1/√3)² = 1/3, sec²45° = (√2)² = 2, and sin²60° = (√3/2)² = 3/4.
3. Substitute these into the expression: (3/4)(1/3) − 2 + 3/4.
4. Simplify the terms: 1/4 − 2 + 3/4 = 1 − 2 = −1.

Answer: (3/4)(1/3) − 2 + 3/4 = 1/4 − 2 + 3/4 = -1 — option (A).

A cylinder 5 cm tall has curved surface area 94.2 cm². Find its radius (π = 3.14).

1. Use the formula for the curved surface area of a cylinder, CSA = 2πrh, and substitute the given values: 94.2 = 2 × 3.14 × r × 5.
2. Simplify the equation to 94.2 = 31.4r, then solve for r to get r = 94.2 / 31.4 = 3 cm.

Answer: 2πrh = 94.2 gives r = 94.2/(2 × 3.14 × 5) = 3 cm — option (B).

Marks of 80 students as a 'less than' table: <10: 3, <20: 12, <30: 27, <40: 57, <50: 75, <60: 80. Find the modal class.

1. Convert the 'less than' cumulative frequencies into class frequencies by subtracting consecutive values: 12 − 3 = 9, 27 − 12 = 15, 57 − 27 = 30, 75 − 57 = 18, and 80 − 75 = 5.
2. Identify the class with the highest frequency, which is 30 for the interval 30-40, making it the modal class.

Answer: Class frequencies are 3, 9, 15, 30, 18, 5; the highest is 30, so the modal class is 30-40 — option (C).

Curved surface area of a cone with h = 24 cm and r = 7 cm.

1. Slant height l = √(h² + r²) = √(24² + 7²) = √(576 + 49) = √625 = 25 cm.
2. Curved surface area = πrl = (22/7) × 7 × 25 = 22 × 25 = 550 cm² — option (C).

Answer: l = √(24² + 7²) = 25 cm, so πrl = (22/7) × 7 × 25 = 550 cm² — option (C).

Distance between (0, 2√5) and (−2√5, 0).

1. Use the distance formula d = √((x₂ - x₁)² + (y₂ - y₁)²) with points (0, 2√5) and (−2√5, 0).
2. Substitute the coordinates to get d = √((−2√5 - 0)² + (0 - 2√5)²) = √((−2√5)² + (−2√5)²).
3. Simplify the squares inside the radical: (−2√5)² = 4 × 5 = 20, so d = √(20 + 20) = √40 = 2√10 units.

Answer: √(20 + 20) = √40 = 2√10 units — option (A).

Which quadratic has zeroes −2/3 and 2/3?

1. Sum of zeroes = −2/3 + 2/3 = 0 and product = (−2/3) × (2/3) = −4/9, so the polynomial is k(x² − 4/9) for any non-zero k.
2. Taking k = 9 gives 9x² − 4, and any multiple such as 5(9x² − 4) = 45x² − 20 has the same zeroes — option (D).

Answer: Sum 0, product −4/9, so any multiple of 9x² − 4 works: 5(9x² − 4) — option (D).

Every observation is increased by 3. What happens to the mean?

1. Let the original observations be x1, x2, ..., xn with mean x̄ = (x1 + x2 + ... + xn)/n.
2. If every observation is increased by 3, the new observations are (x1+3), (x2+3), ..., (xn+3).
3. The new mean is [(x1+3) + (x2+3) + ... + (xn+3)]/n = [(x1 + x2 + ... + xn) + 3n]/n = x̄ + 3.
4. Thus, the mean increases by 3, so option (B) is correct.

Answer: The mean also goes up by 3 — option (B).

p = P(event happens), q = P(it does not happen). How are p and q related?

1. The event happening and the event not happening are complementary events, meaning their probabilities sum to 1.
2. Therefore, p + q = 1, which corresponds to option (A).

Answer: Complementary events: p + q = 1 — option (A).

6000 lottery tickets are sold; a girl's chance of first prize is 0.08. How many tickets did she buy?

1. The probability of winning is calculated as the number of tickets bought divided by the total number of tickets sold.
2. Let the number of tickets bought be x, so x / 6000 = 0.08.
3. Solving for x gives x = 0.08 × 6000 = 480 tickets.

Answer: 0.08 × 6000 = 480 tickets — option (C).

Of 20 people, 5 cannot swim. Probability that a randomly chosen person can swim.

1. Total number of people = 20, and number of people who cannot swim = 5.
2. Number of people who can swim = 20 − 5 = 15.
3. Probability that a randomly chosen person can swim = 15/20 = 3/4, so option (A) is correct.

Answer: 15 of 20 can swim: P = 3/4 — option (A).

A: P(0, 2) is where the line 3x + 2y = 4 meets the y-axis. R: P(0, 2) is 2 units from the x-axis.

1. Substitute x = 0 into the equation 3x + 2y = 4 to find the y-intercept, which gives 2y = 4, so y = 2, making Assertion A true.
2. The distance of point P(0, 2) from the x-axis is its y-coordinate, which is 2, making Reason R true.
3. Since Reason R describes a property of the point but does not derive the coordinates from the line equation, it is not the correct explanation for Assertion A, so the answer is option (B).

Answer: A is true (3·0 + 2·2 = 4) and R is true, but R does not explain A — option (B).

Right triangle ABC with AB = 2 cm, BC = 3 cm, ∠B = 90°. A: its perimeter is rational. R: a sum of squares of two rationals is rational.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, triangle ABC is right-angled at B with AB = 2 cm and BC = 3 cm.

1. Since ∠B = 90°, apply the Pythagorean theorem to find AC: AC² = AB² + BC² = 2² + 3² = 4 + 9 = 13, so AC = √13.
2. Calculate the perimeter of triangle ABC: Perimeter = AB + BC + AC = 2 + 3 + √13 = 5 + √13.
3. Since √13 is an irrational number, the sum 5 + √13 is also irrational, making Assertion A false.
4. Reason R states that the sum of squares of two rationals is rational, which is a true mathematical statement (e.g., 1² + 2² = 5), so R is true.
5. Thus, Assertion A is false and Reason R is true, which corresponds to option (D).

Answer: AC = √13, so the perimeter 5 + √13 is irrational: A is false, R is true — option (D).

Solve x = 3 and y = −4 by drawing their graphs.

1. Draw the line x = 3 as a vertical line passing through the point (3, 0) on the x-axis.
2. Draw the line y = −4 as a horizontal line passing through the point (0, −4) on the y-axis.
3. Observe that the two lines intersect at the point (3, −4).
4. Therefore, the solution of the system is x = 3 and y = −4.

Answer: x = 3 is a vertical line and y = −4 a horizontal line; they meet at (3, −4), so x = 3, y = −4.

Graphically decide whether x = 0, y = −7 is a consistent system.

1. Plot the line x = 0, which is the y-axis, and the line y = -7, which is a horizontal line parallel to the x-axis.
2. Observe that the two lines intersect at the point (0, -7).
3. Since the lines intersect at a unique point, the system of equations is consistent.

Answer: x = 0 (the y-axis) and the line y = −7 cross at (0, −7), so the system is consistent.

In triangle ABC, X on AB and Z on AC with XZ ∥ BC; AM (M on BC) cuts XZ at Y. Given ZC = 2, AZ = 3, MC = 5, BM = 3 (all cm), find XY.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, X lies on AB and Z on AC with XZ ∥ BC; segment AM goes from A to M on BC and crosses XZ at Y.

1. Since XZ is parallel to BC, by the Basic Proportionality Theorem, AX/XB = AZ/ZC = 3/2.
2. Therefore, AX/AB = AX/(AX + XB) = 3/(3 + 2) = 3/5.
3. In triangles AXY and ABM, angle A is common and angle AXY equals angle ABM (corresponding angles), so triangle AXY is similar to triangle ABM.
4. Thus, XY/BM = AX/AB, which gives XY/3 = 3/5, so XY = 1.8 cm.

Answer: AX/XB = AZ/ZC = 3/2, so AX/AB = 3/5; triangles AXY and ABM are similar, so XY = (3/5) × 3 = 1.8 cm.

Given sin θ + cos θ = √3, find sin θ · cos θ.

1. Square both sides of the given equation: (sin θ + cos θ)² = (√3)².
2. Expand the left side and simplify the right side: sin²θ + 2 sin θ cos θ + cos²θ = 3.
3. Use the identity sin²θ + cos²θ = 1 to substitute: 1 + 2 sin θ cos θ = 3.
4. Solve for sin θ cos θ: 2 sin θ cos θ = 2, so sin θ cos θ = 1.

Answer: Squaring: 1 + 2 sin θ cos θ = 3, so sin θ cos θ = 1.

Note: No real angle actually satisfies sin θ + cos θ = √3, because sin θ + cos θ is never more than √2 (and sin θ cos θ is never more than 1/2). CBSE's marking scheme still accepts the algebraic answer 1.

sin α = 1/√2 and cot β = √3. Find cosec α + cosec β.

1. Given sin α = 1/√2, we find cosec α as the reciprocal of sin α, so cosec α = √2.
2. Given cot β = √3, we use the identity cosec²β = 1 + cot²β to find cosec β.
3. Substituting the value, cosec²β = 1 + (√3)² = 1 + 3 = 4, so cosec β = √4 = 2.
4. Adding the values, cosec α + cosec β = √2 + 2.

Answer: cosec α = √2 and cosec β = √(1 + cot²β) = 2, so the sum is √2 + 2.

Largest number that leaves remainder 1 on dividing 85 and remainder 2 on dividing 72.

1. Subtract the respective remainders from the numbers to find the values divisible by the required number: 85 − 1 = 84 and 72 − 2 = 70.
2. Calculate the HCF of 84 and 70 to find the largest such number: HCF(84, 70) = 14.

Answer: HCF(85 − 1, 72 − 2) = HCF(84, 70) = 14.

A bag has 4 red, 3 blue, 2 yellow balls; one is drawn at random. Find P(red) and P(yellow).

1. Total number of balls = 4 + 3 + 2 = 9.
2. Number of red balls = 4, so P(red) = 4/9.
3. Number of yellow balls = 2, so P(yellow) = 2/9.

Answer: Total 9 balls, so P(red), P(yellow) = 4/9 and 2/9.

Two numbers: half their difference is 2, and the larger plus twice the smaller is 13. Find them.

1. Let the larger number be x and the smaller number be y.
2. Half their difference is 2, so x - y = 4.
3. The larger plus twice the smaller is 13, so x + 2y = 13.
4. Subtract the first equation from the second: (x + 2y) - (x - y) = 13 - 4.
5. Simplify to get 3y = 9, so y = 3.
6. Substitute y = 3 into x - y = 4 to get x - 3 = 4, so x = 7.
7. The numbers are 7 and 3.

Answer: x − y = 4 and x + 2y = 13 give 3y = 9: the numbers are 7 and 3.

Show that √5 is irrational.

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

Answer: Assume √5 = p/q in lowest terms. Then p² = 5q², so 5 divides p; write p = 5a, giving q² = 5a², so 5 divides q too. This contradicts p, q being co-prime, hence √5 is irrational.

(−5, 3) and (5, 3) are vertices of an equilateral triangle that contains the origin. Find the third vertex (√3 = 1.7).

1. Let the third vertex be C(x, y). Since the triangle is equilateral, the perpendicular bisector of the side joining (-5, 3) and (5, 3) passes through C.
2. The midpoint of the side is (0, 3) and the side is horizontal, so the perpendicular bisector is the line x = 0. Thus, the x-coordinate of the third vertex is 0.
3. The length of the side is 5 - (-5) = 10. The height of an equilateral triangle with side 10 is (√3/2) × 10 = 5√3.
4. Using the given value √3 = 1.7, the height is 5 × 1.7 = 8.5.
5. The y-coordinate of the third vertex is 3 ± 8.5, giving possible values of 11.5 or -5.5.
6. For the triangle to contain the origin (0, 0), the third vertex must be below the side y = 3. Therefore, the y-coordinate is -5.5.
7. The third vertex is (0, -5.5).

Answer: Side = 10, so the third vertex is on x = 0 at height 5√3 = 8.5 from y = 3; to contain the origin it must be below: (0, -5.5).

TP and TQ are tangents from external point T to a circle with centre O. Show ∠PTQ = 2∠OPQ.

1. Let ∠PTQ = θ. Tangents from an external point are equal, so TP = TQ and ∠TPQ = ∠TQP.
2. Angle sum in triangle TPQ: 2∠TPQ = 180° − θ, so ∠TPQ = 90° − θ/2.
3. The radius OP is perpendicular to the tangent TP, so ∠OPT = 90°, and ∠OPQ = ∠OPT − ∠TPQ.
4. So ∠OPQ = 90° − (90° − θ/2) = θ/2, which means ∠PTQ = 2∠OPQ.

Answer: TP = TQ, so ∠TPQ = ∠TQP = 90° − ∠PTQ/2. Since ∠OPT = 90°, ∠OPQ = 90° − ∠TPQ = ∠PTQ/2, i.e. ∠PTQ = 2∠OPQ.

A circle (centre O, radius r) is inscribed in quadrilateral ABCD with ∠B = 90°. AD = 17 cm, AB = 20 cm, DS = 3 cm. Find r.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the circle touches AD at R, DC at S, CB at P and AB at Q; OP and OQ are radii, and ∠B = 90°.

1. Since tangents from an external point are equal, DR = DS = 3 cm.
2. Calculate AR as AD - DR = 17 - 3 = 14 cm.
3. Since tangents from A are equal, AQ = AR = 14 cm.
4. Calculate QB as AB - AQ = 20 - 14 = 6 cm.
5. In quadrilateral OPBQ, angles at P, Q, and B are 90 degrees because radii are perpendicular to tangents and angle B is given as 90 degrees.
6. Since OP = OQ = r and adjacent sides are equal with all angles 90 degrees, OPBQ is a square.
7. Therefore, the radius r is equal to side QB, so r = 6 cm.

Answer: DR = DS = 3, so AR = 14 = AQ and QB = 20 − 14 = 6; OPBQ is a square, so r = 6 cm.

Prove (tan θ + sec θ − 1)/(tan θ − sec θ + 1) = (1 + sin θ)/cos θ.

1. Replace 1 in the numerator with sec²θ − tan²θ to get (tan θ + sec θ − (sec²θ − tan²θ)) / (tan θ − sec θ + 1).
2. Simplify the numerator by factoring: tan θ + sec θ − sec²θ + tan²θ = (tan θ + sec θ)(1 − sec θ + tan θ).
3. Observe that the denominator is tan θ − sec θ + 1, which is identical to the factor (1 − sec θ + tan θ) in the numerator.
4. Cancel the common factor (1 − sec θ + tan θ) from the numerator and denominator, leaving tan θ + sec θ.
5. Express tan θ + sec θ in terms of sine and cosine: sin θ/cos θ + 1/cos θ = (sin θ + 1)/cos θ.
6. Rearrange the numerator to match the required form: (1 + sin θ)/cos θ.

Answer: Replace 1 in the numerator by sec²θ − tan²θ: numerator = (tan θ + sec θ)(1 − sec θ + tan θ), which cancels with the denominator to leave tan θ + sec θ = (1 + sin θ)/cos θ.

A room is a cylinder topped by a hemispherical dome; the dome radius is half the cylinder height. The room holds 1408/21 m³ of air. Find its total height (π = 22/7).

1. Let the dome radius be r; then the cylinder height is h = 2r.
2. Volume = πr²h + (2/3)πr³ = 2πr³ + (2/3)πr³ = (8/3)πr³.
3. So (8/3) × (22/7) × r³ = 1408/21, i.e. (176/21)r³ = 1408/21, giving r³ = 1408/176 = 8 and r = 2 m.
4. Total height = h + r = 4 + 2 = 6 m.

Answer: With h = 2r: πr²(2r) + (2/3)πr³ = (8/3)πr³ = 1408/21, so r³ = 8, r = 2 m, h = 4 m; total height = 6 m.

A cone (r = 3 cm, h = 12 cm) is filled with ice-cream except its bottom 1/6 of volume, with a hemisphere of ice-cream on top. Find the ice-cream volume (π = 3.14).

1. Volume of the cone is calculated as (1/3) × π × r² × h = (1/3) × π × 3² × 12 = 36π cm³.
2. The bottom 1/6 of the cone is empty, so the filled part is 5/6 of the cone's volume, which is (5/6) × 36π = 30π cm³.
3. Volume of the hemisphere on top is (2/3) × π × r³ = (2/3) × π × 3³ = 18π cm³.
4. Total ice-cream volume is the sum of the filled cone part and the hemisphere: 30π + 18π = 48π cm³.
5. Substituting π = 3.14, the total volume is 48 × 3.14 = 150.72 cm³.

Answer: Cone = 36π, filled part = 30π; hemisphere = 18π; total 48π = 150.72 cm³.

Prove: a line parallel to one side of a triangle, cutting the other two sides at distinct points, divides them in the same ratio.

1. Take triangle ABC with DE ∥ BC, D on AB and E on AC. To prove: AD/DB = AE/EC.
2. Join BE and CD. Draw EN ⊥ AB and DM ⊥ AC.
3. Triangles ADE and BDE have bases AD and DB on AB and the same height EN, so ar(ADE)/ar(BDE) = AD/DB.
4. Triangles ADE and CDE have bases AE and EC on AC and the same height DM, so ar(ADE)/ar(CDE) = AE/EC.
5. Triangles BDE and CDE stand on the same base DE between the same parallels DE and BC, so ar(BDE) = ar(CDE).
6. Hence the two ratios are equal: AD/DB = AE/EC.

Answer: Take DE ∥ BC in triangle ABC (D on AB, E on AC); join BE, CD and draw heights. ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(CDE) = AE/EC. Triangles BDE and CDE share base DE between the same parallels, so their areas are equal; hence AD/DB = AE/EC.

A 24 m tower's top is seen at 60° elevation from the foot of a second tower; the second tower's top is seen at 30° from the foot of the first. Find the distance between them, the second tower's height, and the length of a wire joining the two tops.

1. Let the distance between the two towers be x m.
2. In the right-angled triangle formed by the first tower, tan 60° = 24/x.
3. Solving for x, x = 24/tan 60° = 24/√3 = 8√3 m.
4. For the second tower, tan 30° = h/x, where h is its height.
5. Substituting values, h = 8√3 × (1/√3) = 8 m.
6. The horizontal distance is 8√3 m and the vertical difference is 24 − 8 = 16 m.
7. The wire length is √((8√3)² + 16²) = √(192 + 256) = √448 = 8√7 m.

Answer: Distance x = 24/tan 60° = 8√3 m; other tower's height = x tan 30° = 8 m; wire = √448 = 8√7 m, since √(16² + (8√3)²) = √448.

A spherical balloon of radius r subtends 60° at an observer's eye, and its centre is at 45° elevation. Show the centre's height is √2 r.

1. Let O be the centre of the balloon, B be the observer's eye, and A be the point on the ground directly below O.
2. The balloon subtends 60° at B, so the angle between the line of sight to the centre BO and the tangent to the sphere is 30°.
3. In the right-angled triangle formed by the radius r, the line of sight OB, and the tangent, we have sin 30° = r/OB.
4. Substituting sin 30° = 1/2 gives 1/2 = r/OB, so OB = 2r.
5. The centre is at 45° elevation, so in the right-angled triangle OAB, sin 45° = OA/OB.
6. Substituting sin 45° = 1/√2 and OB = 2r gives 1/√2 = OA/2r, so OA = 2r/√2 = √2 r.

Answer: The tangent from the eye B makes 30° with BO, so sin 30° = r/OB gives OB = 2r. Height OA = OB sin 45° = 2r/√2 = √2 r.

In a circle of radius 14 cm a chord subtends 60° at the centre. Find the areas of the minor and the major segments.

1. Area of the minor sector = (60/360) × (22/7) × 14 × 14 = 308/3 cm² ≈ 102.67 cm².
2. The radii and chord form an equilateral triangle of side 14 cm (two equal radii with a 60° angle), with area (√3/4) × 14² = 49√3 cm² ≈ 84.77 cm² (√3 ≈ 1.73).
3. Minor segment = sector − triangle = 308/3 − 49√3 cm² ≈ 102.67 − 84.77 = 17.90 cm².
4. Area of the circle = (22/7) × 14 × 14 = 616 cm².
5. Major segment = 616 − (308/3 − 49√3) = 1540/3 + 49√3 cm² ≈ 513.33 + 84.77 = 598.10 cm².

Answer: Minor segment = sector − equilateral triangle; major segment = 616 - minor; the two areas are 308/3 - 49√3 cm² and 1540/3 + 49√3 cm² (≈ 17.9 cm² and 598.1 cm² with √3 ≈ 1.73).

In an AP, a₁₁ : a₁₇ = 3 : 4. Find a₅ : a₂₁ and S₅ : S₂₁.

1. Let the first term be a and common difference be d. Given a₁₁ : a₁₇ = 3 : 4, so (a + 10d) / (a + 16d) = 3/4.
2. Cross-multiplying gives 4(a + 10d) = 3(a + 16d), which simplifies to 4a + 40d = 3a + 48d.
3. Solving for a, we get a = 8d.
4. Calculate a₅ = a + 4d = 12d and a₂₁ = a + 20d = 28d, so a₅ : a₂₁ = 12d : 28d = 3 : 7.
5. Using Sₙ = n/2 [2a + (n-1)d], S₅ = 5/2 [2(8d) + 4d] = 5/2 [20d] = 50d.
6. Similarly, S₂₁ = 21/2 [2(8d) + 20d] = 21/2 [36d] = 378d.
7. Thus, S₅ : S₂₁ = 50d : 378d = 25 : 189.

Answer: 4(a + 10d) = 3(a + 16d) gives a = 8d. a₅ : a₂₁ = 12d : 28d and S₅ : S₂₁ = 5(20d) : 21(36d), i.e. 3:7 and 25:189.

250 logs sit in rows: 22 at the base, then 21, then 20, each row one fewer. How many rows, and how many logs in the top row?

1. Let n be the number of rows. The logs form an arithmetic progression with first term a = 22 and common difference d = -1.
2. Using the sum formula S = n/2 [2a + (n - 1)d], substitute the values: 250 = n/2 [2(22) + (n - 1)(-1)].
3. Simplify the equation: 250 = n/2 [44 - n + 1] = n/2 [45 - n].
4. Multiply by 2 and rearrange to form a quadratic equation: 500 = 45n - n², so n² - 45n + 500 = 0.
5. Factorize the quadratic: (n - 20)(n - 25) = 0, giving possible values n = 20 or n = 25.
6. Reject n = 25 because it results in a negative number of logs in the top row (22 - 24 = -2), so n = 20.
7. Calculate the logs in the top row for n = 20: 22 + (20 - 1)(-1) = 22 - 19 = 3.

Answer: n/2[44 − (n − 1)] = 250 gives n² − 45n + 500 = 0, so n = 20 or 25; n = 25 would make the top row negative, so n = 20, top row 3 logs.

An 18 cm × 12 cm photo has both length and width increased by x so that its area doubles. Write the equation.

1. The new length is (18 + x) cm and the new width is (12 + x) cm.
2. Since the area doubles, the new area equals 2 × 18 × 12, which is 432 cm².
3. Equating the new area to the doubled original area gives the equation (18 + x)(12 + x) = 432.

Answer: (18 + x)(12 + x) = 2 × 18 × 12 = 432.

Write that equation as a quadratic in standard form.

1. Expand the left side: (18 + x)(12 + x) = x² + 30x + 216, so x² + 30x + 216 = 432.
2. Subtract 432 from both sides: x² + 30x − 216 = 0.

Answer: x² + 30x − 216 = 0.

Find the dimensions of the enlarged photo.

1. Solve x² + 30x − 216 = 0: split 30x as 36x − 6x, giving (x + 36)(x − 6) = 0.
2. So x = 6 or x = −36; an increase in length cannot be negative, so x = 6 cm.
3. New length = 18 + 6 = 24 cm and new width = 12 + 6 = 18 cm, so the enlarged photo is 24 cm × 18 cm (check: 24 × 18 = 432 cm²).

Answer: (x + 36)(x − 6) = 0, x = 6 (x cannot be negative), so the new photo is 24 cm × 18 cm.

Could a rational x make the new area exactly 220 cm²?

1. Set up the equation for the new area: (18 + x)(12 + x) = 220.
2. Expand and simplify to form the quadratic equation: x² + 30x + 216 = 220, which becomes x² + 30x − 4 = 0.
3. Calculate the discriminant D = b² − 4ac = 30² − 4(1)(−4) = 900 + 16 = 916.
4. Since 916 is not a perfect square, the roots are irrational, so no rational value of x exists.

Answer: (18 + x)(12 + x) = 220 gives x² + 30x − 4 = 0 with D = 916, not a perfect square, so no rational x works.

2018 monsoon rainfall classes of width 200 mm from 200 to 1800 mm have sub-division counts 2, 4, 7, 4, 2, 3, 1, 1 (in order). Give the modal class.

1. The class intervals are 200-400, 400-600, 600-800, 800-1000, 1000-1200, 1200-1400, 1400-1600, and 1600-1800 mm.
2. The frequencies are 2, 4, 7, 4, 2, 3, 1, and 1 respectively.
3. The highest frequency is 7, which corresponds to the class interval 600-800 mm.
4. Therefore, the modal class is 600-800 mm.

Answer: Highest frequency 7, so the modal class is 600-800 mm.

Find the median rainfall from the same table.

1. Total n = 2 + 4 + 7 + 4 + 2 + 3 + 1 + 1 = 24, so n/2 = 12. Cumulative frequencies are 2, 6, 13, …, so the median class is 600-800.
2. For this class l = 600, cf before it = 6, f = 7, h = 200.
3. Median = l + ((n/2 − cf)/f) × h = 600 + (6/7) × 200 = 600 + 1200/7 = 5400/7 ≈ 771.43 mm.

Answer: n = 24, n/2 = 12; median class 600-800 (cf before 6, f = 7, h = 200): 600 + 200 × 6/7 = 5400/7 ≈ 771.43 mm.

Find the mean rainfall from the same table.

1. Class mid-points are 300, 500, 700, 900, 1100, 1300, 1500, 1700 with frequencies 2, 4, 7, 4, 2, 3, 1, 1 (total 24).
2. Σfx = 600 + 2000 + 4900 + 3600 + 2200 + 3900 + 1500 + 1700 = 20400.
3. Mean = 20400 ÷ 24 = 850 mm.

Answer: Σfx = 20400 over 24 sub-divisions, so mean = 850 mm.

Sub-divisions with 1000 mm or more count as good rainfall. How many are there?

1. Classes from 1000 mm upward are 1000-1200, 1200-1400, 1400-1600 and 1600-1800, with counts 2, 3, 1 and 1.
2. Sub-divisions with good rainfall = 2 + 3 + 1 + 1 = 7.

Answer: 2 + 3 + 1 + 1 = 7 sub-divisions.

AB touches a circle (centre O, radius 75 cm) at A, and ∠ABO = 30°. Find AB.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, OA is the radius to the point of contact A; OB cuts the circle at Q; P is on AB with PQ ∥ OA.

1. Since AB is a tangent to the circle at A, the radius OA is perpendicular to AB, so ∠OAB = 90°.
2. In right-angled triangle OAB, tan(∠ABO) = OA/AB, which gives tan 30° = 75/AB.
3. Substituting tan 30° = 1/√3, we get 1/√3 = 75/AB, so AB = 75√3 cm.

Answer: OA ⊥ AB, so tan 30° = 75/AB and AB = 75√3 cm.

Same set-up: find OB.

Uses a figure printed in CBSE's paper — what the figure shows: Same figure as 38(a).

1. In the right-angled triangle, the ratio of the opposite side to the hypotenuse is the sine of the angle.
2. Using the given values, sin 30° = 75/OB.
3. Substitute the value of sin 30° as 1/2 to get 1/2 = 75/OB.
4. Solve for OB by cross-multiplying: OB = 75 × 2 = 150 cm.

Answer: sin 30° = 75/OB, so OB = 150 cm.

Same set-up, with Q where OB meets the circle and PQ ∥ OA (P on AB). Find AP.

Uses a figure printed in CBSE's paper — what the figure shows: Same figure as 38(a): Q lies on OB and on the circle; P lies on AB with PQ ∥ OA.

1. Since Q lies on OB and the circle, OQ is a radius, so OQ = 75 cm.
2. Given OB = 150 cm, calculate QB = OB - OQ = 150 - 75 = 75 cm.
3. Since OQ = QB, Q is the mid-point of OB.
4. In triangle OAB, PQ is parallel to OA and Q is the mid-point of OB, so by the Midpoint Theorem, P is the mid-point of AB.
5. Calculate AP = AB / 2 = (75√3) / 2 = 37.5√3 cm.

Answer: QB = 150 − 75 = 75 cm, so Q is the mid-point of OB; PQ ∥ OA makes P the mid-point of AB, so AP = 37.5√3 cm (= 75√3/2 cm).

Same set-up: find PQ.

Uses a figure printed in CBSE's paper — what the figure shows: Same figure as 38(a).

1. Q lies on the circle, so OQ = 75 cm and BQ = OB − OQ = 150 − 75 = 75 cm.
2. PQ ∥ OA, so in triangles BQP and BOA, ∠B is common and ∠BQP = ∠BOA (corresponding angles); hence the triangles are similar (AA).
3. So PQ/OA = BQ/BO = 75/150 = 1/2, giving PQ = 75/2 = 37.5 cm.

Answer: Triangles BQP and BOA are similar with BQ/BO = 75/150 = 1/2, so PQ = 75/2 = 37.5 cm.