CBSE Class 10 Maths (Standard) Question Paper 2026 Solved Code 30-1-3

SECTION A

1. [1 mark] For any natural number n, 6ⁿ ends with the digit:
(A) 0
(B) 6
(C) 3
(D) 2

Answer : (B) 6. Since 6ⁿ = (2 × 3)ⁿ, its only prime factors are 2 and 3 (never 5), so it can never end in 0. Checking 6¹ = 6, 6² = 36, 6³ = 216, the units digit is always 6.

Teacher's Note:
A number ends in 0 only if both 2 and 5 are its factors.
6ⁿ has only 2 and 3 as prime factors, never 5.
So the last digit of every power of 6 is always 6.

2. [1 mark] The graph of y = f(x) is given (a wavy curve lying entirely above the X-axis, rising from near the origin, dipping down but not touching the X-axis, then rising again). The number of zeroes of f(x) is:
(A) 0
(B) 1
(C) 2
(D) 4

Answer : (A) 0. The zeroes of f(x) are the x-coordinates of the points where the graph cuts or touches the x-axis. Since the curve never meets the x-axis, f(x) has 0 zeroes.

Teacher's Note:
Zeroes of a polynomial = points where its graph meets the x-axis.
A curve that stays completely above or below the x-axis has no real zeroes.
Count the number of x-axis crossings, not the humps in the curve.

3. [1 mark] If a pair of linear equations in two variables is represented by two coincident lines, then the pair of equations has:
(A) a unique solution
(B) two solutions
(C) no solution
(D) an infinite number of solutions

Answer : (D) an infinite number of solutions, since every point on one line also lies on the other.

Teacher's Note:
Coincident lines mean a1/a2 = b1/b2 = c1/c2.
This condition always gives infinitely many solutions.
Intersecting lines give a unique solution; parallel (non-coincident) lines give no solution.

4. [1 mark] The common difference of the AP: √2, 2√2, 3√2, 4√2, ..... is:
(A) √2
(B) 1
(C) 2√2
(D) −√2

Answer : (A) √2, since d = a2 − a1 = 2√2 − √2 = √2.

Teacher's Note:
Common difference d = (any term) − (previous term).
Always check using consecutive terms, not the first and last.
Here every term is a multiple of √2, so the pattern is easy to spot.

5. [1 mark] If Δ ABC and Δ DEF are similar such that 2 AB = DE and BC = 8 cm, then EF is equal to:
(A) 4 cm
(B) 8 cm
(C) 12 cm
(D) 16 cm

Answer : (D) 16 cm. Since 2AB = DE, the scale factor DE/AB = 2, so by similarity EF/BC = 2, giving EF = 2 × 8 = 16 cm.

Teacher's Note:
In similar triangles, all corresponding sides share the same ratio.
Match vertices correctly: A↔D, B↔E, C↔F, so BC corresponds to EF.
Find the scale factor from one known pair of sides first.

6. [1 mark] The mid-point of the line segment joining the points (5, − 4) and (6, 4) lies on:
(A) x-axis
(B) y-axis
(C) origin
(D) neither x-axis nor y-axis

Answer : (A) x-axis. Midpoint = ((5+6)/2, (−4+4)/2) = (11/2, 0). Since the y-coordinate is 0, this point lies on the x-axis.

Teacher's Note:
Midpoint formula: ((x1+x2)/2, (y1+y2)/2).
A point lies on the x-axis when its y-coordinate is 0.
A point lies on the y-axis when its x-coordinate is 0.

7. [1 mark] Given that sin θ = a/b, then cos θ is equal to:
(A) b/√(b²−a²)
(B) b/a
(C) √(b²−a²)/b
(D) a/√(b²−a²)

Answer : (C) √(b²−a²)/b. Taking a right triangle with perpendicular = a, hypotenuse = b, base = √(b²−a²) by Pythagoras, so cos θ = base/hypotenuse = √(b²−a²)/b.

Teacher's Note:
sin θ = perpendicular/hypotenuse; use Pythagoras to get the base.
cos θ = base/hypotenuse.
Draw a quick right triangle whenever a ratio like a/b is given.

8. [1 mark] If cos A = 1/2, then the value of sin² A + 2 cos² A is:
(A) 3/2
(B) 5/4
(C) −1
(D) 1/2

Answer : (B) 5/4. sin² A + 2 cos² A = sin²A + cos²A + cos²A = 1 + cos²A = 1 + 1/4 = 5/4.

Teacher's Note:
Always try to use sin²θ + cos²θ = 1 to simplify such expressions.
Split the extra cos²A term instead of finding sin A separately.
This saves a calculation step and avoids square-root mistakes.

9. [1 mark] The string of a flying kite is tied to a point on the ground. The length of the string between the kite and the point on the ground is 80 m. The string makes an angle of 30° with the ground. The height of the kite above the ground is:
(A) 20√3 m
(B) 40 m
(C) 40√3 m
(D) 80√3 m

Answer : (B) 40 m. Height = string length × sin 30° = 80 × (1/2) = 40 m.

Teacher's Note:
Height (opposite side) = hypotenuse × sin θ.
Remember standard values: sin 30° = 1/2.
Draw the right triangle with the string as hypotenuse before applying the ratio.

10. [1 mark] If TP and TQ are two tangents to a circle with centre O from an external point T so that ∠POQ = 120°, then ∠PTQ is equal to:
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Answer : (A) 60°. Since OPTQ is a quadrilateral with ∠OPT = ∠OQT = 90°, the angle sum gives ∠PTQ = 360° − 90° − 90° − 120° = 60°.

Teacher's Note:
Tangent is always perpendicular to the radius at the point of contact.
∠PTQ + ∠POQ = 180° is a useful shortcut for this quadrilateral.
Here 180° − 120° = 60°, matching the direct calculation.

11. [1 mark] In the given figure, PA is a tangent from an external point P to a circle with centre O (P outside, tangent PA touching the circle at A, and OB is a radius to another point B on the circle, with O joined to P). If ∠POB = 125°, then ∠APO is equal to:
(A) 25°
(B) 65°
(C) 90°
(D) 35°

Answer : (D) 35°. Since PA is a tangent at A, the radius OA is perpendicular to it, so ∠OAP = 90°. Also, A, O, B are positioned so that ∠AOP = 180° − ∠POB = 180° − 125° = 55°. Now, using the angle sum property in triangle OAP: ∠OAP + ∠APO + ∠AOP = 180°, so 90° + ∠APO + 55° = 180°, giving ∠APO = 35°.

Teacher's Note:
The tangent is perpendicular to the radius at the point of contact.
Use the angle sum property of triangle OAP once ∠OAP = 90° is known.
∠AOP here is the supplement of the given ∠POB since A, O, B are positioned with AOB forming a straight consideration through O's angle split.

12. [1 mark] Shown in the given figure is a circle with centre O. Points S and T lie on the circle with the minor sector OST making an angle of 30° at O. The area of the minor sector is 7 cm². Area of circle is:
(A) 84 π cm²
(B) 84/11 cm²
(C) 84 cm²
(D) √84/√π cm²

Answer : (C) 84 cm². Area of sector = (θ/360) × Area of circle, so 7 = (30/360) × Area, giving Area = 7 × 12 = 84 cm².

Teacher's Note:
Sector area is a fraction of the full circle's area, based on the angle at the centre.
θ/360 gives that fraction directly, no need to find the radius first.
Here 360/30 = 12, so multiply the sector area by 12.

13. [1 mark] In the given figure, O is the centre of the circle. XYZ is an arc of the circle subtending an angle of 45° at the centre. If the radius of the circle is 32 cm, then the length of the arc XYZ is:
(A) 4π cm
(B) 8π cm
(C) 64π cm
(D) 128π cm

Answer : (B) 8π cm. Arc length = (θ/360) × 2πr = (45/360) × 2π × 32 = (1/8) × 64π = 8π cm.

Teacher's Note:
Arc length formula: (θ/360) × 2πr.
Simplify the fraction θ/360 first to make the multiplication easy.
Do not confuse arc length with sector area, they use the same fraction but different base formulas.

14. [1 mark] The radius of a sphere (in cm) whose volume is 36π cm³, is:
(A) 3
(B) 3√3
(C) 3^2/3
(D) 3^1/3

Answer : (A) 3. Volume = (4/3)πr³ = 36π, so r³ = 27, giving r = 3 cm.

Teacher's Note:
Volume of a sphere = (4/3)πr³, learn this formula by heart.
Cancel π from both sides first to simplify the equation.
Take the cube root at the end to find r.

15. [1 mark] If the mean and mode of a data are 12 and 21 respectively, then its median is:
(A) 6
(B) 13.5
(C) 15
(D) 14

Answer : (C) 15. Using the empirical relation Mode = 3 Median − 2 Mean: 21 = 3(Median) − 2(12) = 3(Median) − 24, so 3(Median) = 45, Median = 15.

Teacher's Note:
Empirical relation: Mode = 3 Median − 2 Mean.
Rearrange carefully to isolate Median.
This formula is used only when direct data / class intervals are not given.

16. [1 mark] A die is thrown once. Probability of getting a number other than 3 is:
(A) 1/6
(B) 3/6
(C) 5/6
(D) 1

Answer : (C) 5/6, since 5 out of the 6 equally likely outcomes {1, 2, 4, 5, 6} are not 3.

Teacher's Note:
Total outcomes on a single die throw = 6.
P(not event) = 1 − P(event) is often the fastest approach.
Here P(3) = 1/6, so P(not 3) = 1 − 1/6 = 5/6.

17. [1 mark] The HCF of 960 and 432 is:
(A) 48
(B) 54
(C) 72
(D) 36

Answer : (A) 48. 960 = 2⁶ × 3 × 5 and 432 = 2⁴ × 3³. HCF = 2⁴ × 3 = 16 × 3 = 48.

Teacher's Note:
Write the prime factorisation of both numbers first.
HCF = product of the smallest power of each common prime factor.
Double-check by dividing both numbers by your answer, both should divide exactly.

18. [1 mark] The natural number 2 is:
(A) a prime number
(B) a composite number
(C) prime as well as composite
(D) neither prime nor composite

Answer : (A) a prime number, since 2 has exactly two factors, 1 and 2 itself. It is also the only even prime number.

Teacher's Note:
A prime number has exactly two distinct factors, 1 and itself.
2 is unique, it is the smallest and the only even prime.
1 is neither prime nor composite, don't confuse it with 2.

19. [1 mark] Assertion (A): The polynomial p(y) = y² + 4y + 3 has two zeroes.
Reason (R): A quadratic polynomial can have at most two zeroes.
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer : (B) Both A and R are true, but R is not the correct explanation of A. Factorising, p(y) = y² + 4y + 3 = (y+1)(y+3), so its zeroes are −1 and −3, i.e. it does have two zeroes, so A is true. R is also a true general statement, but it only tells us the maximum possible number of zeroes, it does not explain why this particular polynomial has exactly two.

Teacher's Note:
Factorise the quadratic to confirm the number and value of its zeroes.
"At most two zeroes" is a general fact about all quadratics, not proof for this specific one.
For assertion-reason questions, check both statements' truth first, then check the link between them.

20. [1 mark] Assertion (A): The probability that a leap year has 53 Mondays is 2/7.
Reason (R): The probability that a non-leap year has 53 Mondays is 5/7.
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer : (C) Assertion (A) is true, but Reason (R) is false. A leap year has 366 days = 52 weeks + 2 extra days. These 2 extra days can be any of the 7 pairs {(Sun,Mon), (Mon,Tue), ..., (Sat,Sun)}, and 2 of these pairs contain a Monday, so P(53 Mondays) = 2/7, confirming A. A non-leap year has 365 days = 52 weeks + 1 extra day. That 1 extra day can be any of the 7 days with equal probability, so P(53 Mondays) = 1/7, not 5/7, making R false.

Teacher's Note:
Leap year: 366 days = 52 weeks + 2 days; find pairs of extra days containing the target day.
Non-leap year: 365 days = 52 weeks + 1 day; probability of the extra day being the target day is always 1/7.
Don't confuse the leap-year probability formula with the non-leap-year case.

SECTION B

21. [2 marks] Do the points P (1, 0), Q (−5, 0) and R (−2, 5) form a triangle? If so, name the type of triangle formed.

Answer :

First, we find the lengths of all three sides using the distance formula.
PQ = √[(−5−1)² + (0−0)²] = √36 = 6
QR = √[(−2−(−5))² + (5−0)²] = √(9+25) = √34
PR = √[(−2−1)² + (5−0)²] = √(9+25) = √34

Since the sum of any two sides is greater than the third side, P, Q and R do form a triangle.
Also, QR = PR = √34, so two sides are equal.
Hence, triangle PQR is an isosceles triangle.

Teacher's Note:
Distance formula: √[(x2−x1)² + (y2−y1)²].
Three points form a triangle only if the sum of any two sides exceeds the third.
Equal side lengths tell you the triangle type, here two equal sides mean isosceles.

22. [2 marks] (A) If tan θ = 24/7, then find the value of sin θ + cos θ.

Answer (a) :

Since tan θ = 24/7 = Perpendicular/Base, we take perpendicular = 24, base = 7.
By Pythagoras, hypotenuse = √(24² + 7²) = √(576+49) = √625 = 25.
So sin θ = 24/25 and cos θ = 7/25.
Therefore, sin θ + cos θ = 24/25 + 7/25 = 31/25.

OR

(B) If cot θ = 7/8, then find the value of (1 + sin θ)(1 − sin θ) / [(1 + cos θ)(1 − cos θ)].

Answer (b) :

We simplify the expression first using the identity (1−x)(1+x) = 1 − x².
(1 + sin θ)(1 − sin θ) / [(1 + cos θ)(1 − cos θ)] = (1 − sin²θ) / (1 − cos²θ) = cos²θ / sin²θ = cot²θ.
Now substitute cot θ = 7/8: cot²θ = (7/8)² = 49/64.
So, the value of the expression = 49/64.

Teacher's Note:
When tan θ = P/B is given, use Pythagoras to find the hypotenuse quickly.
1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ are key identities here.
Simplify the trig expression to a single ratio before substituting values.

23. [2 marks] Two concentric circles are of radii 5 cm and 4 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer :

Let O be the common centre, and let AB be the chord of the larger circle (radius 5 cm) that touches the smaller circle (radius 4 cm) at point M.
Since AB is tangent to the smaller circle at M, OM ⊥ AB, and OM = 4 cm (radius of smaller circle).

In right triangle OMA, OA = 5 cm (radius of larger circle) and OM = 4 cm.
By Pythagoras, AM = √(OA² − OM²) = √(5² − 4²) = √(25−16) = √9 = 3 cm.

Since OM ⊥ AB, M is the midpoint of AB, so AB = 2 × AM = 2 × 3 = 6 cm.
Hence, the required length of the chord is 6 cm.

Teacher's Note:
The radius drawn to the point of tangency is always perpendicular to the tangent (chord here).
This perpendicular from the centre also bisects the chord.
Use the Pythagoras theorem in the right triangle formed by the two radii and half the chord.

24. [2 marks] Find a quadratic polynomial whose zeroes are (5 − 2√3) and (5 + 2√3).

Answer :

Let α = 5 − 2√3 and β = 5 + 2√3 be the zeroes of the required polynomial.

First, we find the sum of the zeroes: α + β = (5 − 2√3) + (5 + 2√3) = 10.
Next, we find the product of the zeroes: αβ = (5 − 2√3)(5 + 2√3) = 5² − (2√3)² = 25 − 12 = 13.

A quadratic polynomial with given sum and product of zeroes is x² − (sum)x + (product).
So, the required polynomial is x² − 10x + 13.

Teacher's Note:
Quadratic polynomial formula: x² − (α+β)x + αβ.
Use (a−b)(a+b) = a² − b² to quickly find the product of such conjugate surd zeroes.
Always double check the sum and product before writing the final polynomial.

25. [2 marks] (A) In Δ ABC, DE ∥ BC. If AD = x, DB = x − 2, AE = x + 2 and EC = x − 1, then find the value of x.

Answer (a) :

Since DE ∥ BC, by the Basic Proportionality Theorem, AD/DB = AE/EC.
So, x/(x−2) = (x+2)/(x−1).

Cross-multiplying: x(x−1) = (x+2)(x−2)
x² − x = x² − 4
−x = −4
x = 4.

So, the value of x = 4.

OR

(B) In the figure, Δ ABC ~ Δ XYZ, with AB = 4 cm, AC = y, BC = 6 cm in Δ ABC, and XY = x, XZ = 6 cm, YZ = 7.2 cm in Δ XYZ. Find the values of x and y.

Answer (b) :

Since Δ ABC ~ Δ XYZ, corresponding sides are proportional: AB/XY = BC/YZ = AC/XZ.
So, 4/x = 6/7.2 = y/6.

Using 4/x = 6/7.2: x = (4 × 7.2)/6 = 28.8/6 = 4.8 cm.
Using 6/7.2 = y/6: y = (6 × 6)/7.2 = 36/7.2 = 5 cm.

So, x = 4.8 cm and y = 5 cm.

Teacher's Note:
Basic Proportionality Theorem: a line parallel to one side of a triangle divides the other two sides proportionally.
In similar triangles, match up corresponding vertices carefully before writing ratios.
Cross-multiplication is the safest way to solve such proportion equations.

SECTION C

26. [3 marks] (A) If x = h + a cos θ, y = k + b sin θ, then prove that: [(x−h)/a]² + [(y−k)/b]² = 1

Answer (a) :

We are given x = h + a cos θ, so x − h = a cos θ, which gives (x−h)/a = cos θ.
We are also given y = k + b sin θ, so y − k = b sin θ, which gives (y−k)/b = sin θ.

Now, LHS = [(x−h)/a]² + [(y−k)/b]² = cos²θ + sin²θ = 1 = RHS.
Hence proved.

OR

(B) Prove that: tan A/(1 + sec A) − tan A/(1 − sec A) = 2 cosec A

Answer (b) :

We start from the left-hand side and write tan A and sec A in terms of sin A and cos A.
LHS = tan A/(1+sec A) − tan A/(1−sec A) = (sin A/cos A)/(1 + 1/cos A) − (sin A/cos A)/(1 − 1/cos A)

Simplifying each fraction: = sin A/(cos A + 1) − sin A/(cos A − 1)

Taking the common denominator (cos A + 1)(cos A − 1) = cos²A − 1 = −sin²A:
= sin A [(cos A − 1) − (cos A + 1)] / (cos²A − 1)
= sin A (−2) / (−sin²A)
= 2/sin A
= 2 cosec A = RHS.

Hence proved.

Teacher's Note:
For coordinate-form identities, isolate cos θ and sin θ first, then use sin²θ + cos²θ = 1.
For trig identity proofs, convert everything to sin and cos before combining fractions.
Use cos²A − 1 = −sin²A to simplify denominators quickly.

27. [3 marks] (A) In the given figure, Δ ABC is a right triangle in which ∠B = 90°, AB = 4 cm and BC = 3 cm (with a circle inscribed inside the triangle, touching all three sides). Find the radius of the circle inscribed in the triangle ABC.

Answer (a) :

First, we find the hypotenuse AC using Pythagoras theorem.
AC = √(AB² + BC²) = √(4² + 3²) = √25 = 5 cm.

Let r be the radius of the inscribed circle. We know area of Δ ABC can be found in two ways.
Area of Δ ABC = (1/2) × AB × BC = (1/2) × 4 × 3 = 6 cm².

Also, joining the incentre O to all three vertices splits the triangle into three smaller triangles with height r:
Area of Δ ABC = (1/2 × r × AB) + (1/2 × r × BC) + (1/2 × r × AC) = (1/2) × r × (4 + 3 + 5) = 6r.

Equating both expressions for area: 6r = 6, so r = 1.
Hence, the radius of the inscribed circle is 1 cm.

OR

(B) In the given figure, a circle touches the side QR of Δ PQR at S, and the extended sides PQ and PR at M and N respectively. Prove that: PM = (1/2)(PQ + QR + PR)

Answer (b) :

Since tangents drawn from an external point to a circle are equal in length, we can write three such pairs.
PM = PN (tangents from P)
QS = QM (tangents from Q)
RS = RN (tangents from R)

Adding PM and PN: PM + PN = PQ + QM + PR + RN (since PM = PQ + QM, and PN = PR + RN)
Since PM = PN, this gives 2PM = PQ + QM + PR + RN.

Now substitute QM = QS and RN = RS:
2PM = PQ + QS + PR + RS = PQ + (QS + RS) + PR = PQ + QR + PR (since QS + RS = QR).

Therefore, PM = (1/2)(PQ + QR + PR). Hence proved.

Teacher's Note:
For an inscribed circle, area of the triangle = r × semi-perimeter is a fast alternative method.
Tangents drawn from an external point to a circle are always equal in length, this is the key tool for tangent proofs.
Break the required side into known tangent segments to build the proof step by step.

28. [3 marks] A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surface areas are in the ratio 8 : 5, then find the ratio between the radius of their bases to their height.

Answer :

Let r and h be the common radius and height of the cylinder and the cone.
Curved surface area of cylinder = 2πrh.
Curved surface area of cone = πrl, where l = √(r² + h²) is the slant height.

We are given: 2πrh / πrl = 8/5, which simplifies to 2h/l = 8/5, i.e. 2h/√(r²+h²) = 8/5.

Cross-multiplying: 10h = 8√(r²+h²)
Squaring both sides: 100h² = 64(r² + h²) = 64r² + 64h²
100h² − 64h² = 64r²
36h² = 64r²

So, r²/h² = 36/64 = 9/16, which gives r/h = 3/4.
Hence, the required ratio of radius to height is 3 : 4.

Teacher's Note:
CSA of cylinder = 2πrh; CSA of cone = πrl, where l = √(r²+h²).
Cancel common factors (π, r) early to simplify the ratio equation.
Square both sides carefully when a square root is present, and simplify step by step.

29. [3 marks] Two different coins are tossed simultaneously. What is the probability of getting:
(i) at least one head?
(ii) at most one tail?
(iii) a head and a tail?

Answer :

When two different coins are tossed, the possible outcomes are HH, HT, TH, TT, so there are 4 equally likely outcomes in total.

(i) At least one head means HH, HT, TH, that is 3 favourable outcomes.
P(at least one head) = 3/4.

(ii) At most one tail means outcomes with 0 or 1 tail: HH, HT, TH, that is 3 favourable outcomes.
P(at most one tail) = 3/4.

(iii) A head and a tail means HT or TH, that is 2 favourable outcomes.
P(a head and a tail) = 2/4 = 1/2.

Teacher's Note:
Always list the full sample space first: HH, HT, TH, TT for two coins.
"At least one" and "at most one" are opposite-style phrases, read them carefully to count outcomes correctly.
Simplify the final probability fraction wherever possible.

30. [3 marks] Prove that √3 is an irrational number.

Answer :

Let us assume, to the contrary, that √3 is a rational number.
Then we can write √3 = p/q, where p and q are coprime integers (no common factor other than 1) and q ≠ 0.

Squaring both sides: 3 = p²/q², so p² = 3q².
This means 3 divides p², and therefore 3 divides p as well. .....(i)

So, we can write p = 3a for some integer a.
Substituting back: (3a)² = 3q², so 9a² = 3q², which gives q² = 3a².
This means 3 divides q², and therefore 3 divides q as well. .....(ii)

From (i) and (ii), both p and q are divisible by 3, which contradicts our assumption that p and q are coprime.
This contradiction shows that our original assumption was wrong.
Hence, √3 is an irrational number.

Teacher's Note:
This is a proof by contradiction, always start by assuming the opposite of what you want to prove.
Key logic: if a prime divides p², it must also divide p.
End clearly by pointing out the contradiction with the coprime assumption.

31. [3 marks] Find the ratio in which the x-axis divides the line segment joining the points (−6, 5) and (−4, −1). Also, find the point of intersection.

Answer :

Let the x-axis divide the segment joining A(−6, 5) and B(−4, −1) at point P(x, 0) in the ratio k : 1.

By the section formula, the coordinates of P are:
P = [(k×(−4) + 1×(−6))/(k+1), (k×(−1) + 1×5)/(k+1)] = (x, 0)

Since the y-coordinate of P is 0, we equate: (−k + 5)/(k+1) = 0
So, −k + 5 = 0, which gives k = 5.

Hence, the x-axis divides the segment in the ratio 5 : 1.

Now we find the x-coordinate using k = 5:
x = (5×(−4) + 1×(−6))/(5+1) = (−20 − 6)/6 = −26/6 = −13/3.

So, the point of intersection is (−13/3, 0).

Teacher's Note:
Section formula: point dividing AB in ratio k:1 is [(k x2 + x1)/(k+1), (k y2 + y1)/(k+1)].
Set the y-coordinate to 0 since we want the point on the x-axis.
Once k is known, substitute back to get the exact point of intersection.

SECTION D

32. [5 marks] (A) State and prove Basic Proportionality Theorem.

Answer (a) :

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: A triangle ABC in which a line DE is drawn parallel to side BC, intersecting AB at D and AC at E.
To prove: AD/DB = AE/EC
Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.

Proof:
Area of Δ ADE = (1/2) × AD × EN, and Area of Δ DBE = (1/2) × DB × EN.
So, ar(ADE)/ar(DBE) = AD/DB. .....(i)

Similarly, Area of Δ ADE = (1/2) × AE × DM, and Area of Δ DEC = (1/2) × EC × DM.
So, ar(ADE)/ar(DEC) = AE/EC. .....(ii)

Now, Δ DBE and Δ DEC have the same base DE and lie between the same parallels DE and BC, so they have equal areas.
ar(DBE) = ar(DEC) .....(iii)

From (i), (ii) and (iii): AD/DB = AE/EC.
Hence proved.

OR

(B) In the given figure, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, then prove that:
(i) Δ AMC ~ Δ PNR
(ii) Δ CMB ~ Δ RNQ

Answer (b) :

(i) Since Δ ABC ~ Δ PQR, we know ∠A = ∠P, and AB/PQ = AC/PR.

Since CM and RN are medians, M and N are midpoints of AB and PQ respectively, so AM = (1/2)AB and PN = (1/2)PQ.
Therefore, AB/PQ = 2AM/2PN = AM/PN.
So, AM/PN = AC/PR.

In Δ AMC and Δ PNR, we have AM/PN = AC/PR and the included ∠A = ∠P.
By SAS similarity criterion, Δ AMC ~ Δ PNR.

(ii) Since Δ ABC ~ Δ PQR, we also know ∠B = ∠Q, and AB/PQ = BC/QR.

Since M and N are midpoints, BM = (1/2)AB and QN = (1/2)PQ, so AB/PQ = 2BM/2QN = BM/QN.
So, BM/QN = BC/QR.

In Δ CMB and Δ RNQ, we have BM/QN = BC/QR and the included ∠B = ∠Q.
By SAS similarity criterion, Δ CMB ~ Δ RNQ.

Teacher's Note:
BPT proof structure: same base and height triangles have area ratios equal to their base ratio.
Triangles on the same base between the same parallels are always equal in area.
For median-similarity proofs, use "half of proportional sides remain proportional" along with SAS criterion.

33. [5 marks] The marks obtained by 80 students of class X in a mock test of Mathematics are given below in the table. Find median and the mode of the data:

Answer :

First, we convert this "more than type" cumulative frequency table into a normal frequency distribution table with class intervals.

Finding the median:
Here n = 80, so n/2 = 40.
The cumulative frequency just greater than 40 is 52, which corresponds to the class 50-60. So, 50-60 is the median class.
Here l = 50, cf (of class before median class) = 37, f = 15, h = 10.

Median = l + [(n/2 − cf)/f] × h = 50 + [(40 − 37)/15] × 10 = 50 + (3/15) × 10 = 50 + 2 = 52.

Finding the mode:
The class with the highest frequency (15) is 50-60, so 50-60 is the modal class.
Here l = 50, f1 = 15, f0 = 12, f2 = 12, h = 10.

Mode = l + [(f1 − f0)/(2f1 − f0 − f2)] × h = 50 + [(15−12)/(2×15−12−12)] × 10 = 50 + (3/6) × 10 = 50 + 5 = 55.

So, the median of the data is 52 and the mode of the data is 55.

Teacher's Note:
Convert a "more than type" table to a normal frequency table by subtracting consecutive cumulative frequencies.
Median class: the class whose cf is just greater than n/2.
Modal class: the class with the highest frequency; use the standard mode formula carefully with f1, f0, f2.

34. [5 marks] Draw the graph of the pair of linear equations x − y + 2 = 0 and 4x − y − 4 = 0. Calculate the area of the triangle formed by the lines so drawn and the x-axis.

Answer :

First, we find points to plot each line.

For x − y + 2 = 0, i.e. y = x + 2:
When x = −2, y = 0; when x = 0, y = 2; when x = 2, y = 4.
This line meets the x-axis at A(−2, 0).

For 4x − y − 4 = 0, i.e. y = 4x − 4:
When x = 1, y = 0; when x = 0, y = −4; when x = 2, y = 4.
This line meets the x-axis at C(1, 0).

Now we find the point where the two lines intersect, by solving them together.
x + 2 = 4x − 4
2 + 4 = 4x − x
6 = 3x
x = 2, and y = x + 2 = 4.
So, the lines intersect at B(2, 4).

Plotting A(−2, 0), C(1, 0) and B(2, 4) and drawing both lines and the x-axis gives a triangle ABC.

The base of this triangle lies along the x-axis, from A(−2, 0) to C(1, 0), so base AC = 1 − (−2) = 3 units.
The height of the triangle is the perpendicular distance of B from the x-axis, which is the y-coordinate of B, i.e. 4 units.

Area of triangle ABC = (1/2) × base × height = (1/2) × 3 × 4 = 6 sq. units.

Teacher's Note:
Find at least three points for each line, including where it crosses the x-axis, to plot accurately.
Solve the two equations simultaneously to get the triangle's apex (intersection point).
When the base of the triangle lies on the x-axis, the height is simply the y-coordinate of the third vertex.

35. [5 marks] (A) A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains.

Answer (a) :

Let the speed of the faster train be x km/h.
Then, the speed of the slower train = (x − 10) km/h.

Time taken by faster train for 200 km = 200/x hours.
Time taken by slower train for 200 km = 200/(x−10) hours.

Since the faster train takes one hour less than the slower train:
200/(x−10) − 200/x = 1

Taking LCM and simplifying:
[200x − 200(x−10)] / [x(x−10)] = 1
2000 / (x² − 10x) = 1
x² − 10x = 2000
x² − 10x − 2000 = 0

Solving this quadratic equation by factorisation:
(x − 50)(x + 40) = 0
So, x = 50 or x = −40.

Since speed cannot be negative, we reject x = −40.
So, x = 50.

Hence, speed of the faster train = 50 km/h, and speed of the slower train = 50 − 10 = 40 km/h.

OR

(B) The sum of the areas of two squares is 640 m². If the difference in their perimeters is 64 m, find the sides of the two squares.

Answer (b) :

Let the sides of the two squares be x m and y m, with x > y.

Sum of areas: x² + y² = 640 .....(i)
Difference of perimeters: 4x − 4y = 64, so x − y = 16, which gives y = x − 16 .....(ii)

Substituting (ii) into (i):
x² + (x − 16)² = 640
x² + x² − 32x + 256 = 640
2x² − 32x + 256 − 640 = 0
2x² − 32x − 384 = 0
x² − 16x − 192 = 0

Solving this quadratic equation by factorisation:
(x − 24)(x + 8) = 0
So, x = 24 or x = −8.

Since side length cannot be negative, we reject x = −8.
So, x = 24, and from (ii), y = 24 − 16 = 8.

Hence, the sides of the two squares are 24 m and 8 m.

Teacher's Note:
Translate the word problem into equations carefully using speed = distance/time, or area/perimeter formulas.
Form and simplify a quadratic equation, then solve by factorisation.
Always reject the negative root when the unknown represents a physical quantity like speed or length.

SECTION E

36. [4 marks] A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 cm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors as shown in figure.
Based on the above information, answer the following questions:

(i) [1 mark] What is the radius of circle?

Answer : Radius = diameter/2 = 35/2 = 17.5 cm.

(ii) [1 mark] What is the circumference of the brooch?

Answer :

Circumference = 2πr = 2 × (22/7) × (35/2) = 22 × 5 = 110 cm.

(iii) (a) [2 marks] What is the total length of silver wire required?

Answer (a) :

The wire is used to make the circle (circumference) plus 5 full diameters.
Total length of wire = Circumference + 5 × diameter = 110 + (5 × 35) = 110 + 175 = 285 cm.

OR

(iii) (b) [2 marks] What is the area of each sector of the brooch?

Answer (b) :

Since the circle is divided into 10 equal sectors, the central angle of each sector = 360°/10 = 36°.

Area of each sector = (θ/360) × πr² = (36/360) × (22/7) × (35/2) × (35/2)
= (1/10) × (22/7) × (1225/4)
= 385/4
= 96.25 cm².

Teacher's Note:
5 diameters drawn through the centre create exactly 10 equal sectors.
Total wire used = outer circle (circumference) + all the internal diameters drawn.
Sector area formula: (θ/360) × πr², with θ found by dividing 360° by the number of sectors.

37. [4 marks] In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato. The other potatoes are arranged 3 m apart in a straight line, with a total of 10 potatoes. A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.
Based on the above information, answer the following questions:

(i) [1 mark] What is the distance covered to pick up the first potato and drop it in bucket?

Answer : The first potato is 5 m away, so the competitor runs 5 m to reach it and 5 m back to the bucket. Distance = 5 + 5 = 10 m.

(ii) [1 mark] What is the distance covered to pick up the second potato and drop it in bucket?

Answer : The second potato is 5 + 3 = 8 m from the bucket, so the round trip distance = 8 + 8 = 16 m.

(iii) (a) [2 marks] What is the total distance the competitor has to run?

Answer (a) :

The distances covered for each potato form an AP: 10, 16, 22, ..... up to 10 terms, with first term a = 10 and common difference d = 6.

Using the sum formula for an AP, S_n = (n/2)[2a + (n−1)d]:
S₁₀ = (10/2)[2×10 + 9×6] = 5[20 + 54] = 5 × 74 = 370 m.

So, the total distance the competitor has to run is 370 m.

OR

(iii) (b) [2 marks] If average speed of competitor is 5 m/s, then find the average time taken by competitor to put all the potatoes in the bucket.

Answer (b) :

As found above, the distances covered form an AP with a = 10 and d = 6, and the total distance covered, S₁₀ = 370 m.

Since speed = distance/time, time = distance/speed.
Time = 370/5 = 74 seconds.

So, the average time taken is 74 seconds.

Teacher's Note:
Each "there and back" trip distance forms an arithmetic progression, since each potato is 3 m farther than the last.
Use S_n = (n/2)[2a + (n−1)d] to add up all 10 trip distances at once instead of adding term by term.
Time = total distance ÷ speed, once the total distance (from the AP sum) is known.

38. [4 marks] Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O'. Distance between the base of the tower (P) and point 'O' is 6 m. From point 'O', the angle of elevation of the top of section 'B' is 30° and the angle of elevation of the top of section 'A' (the top of the whole tower) is 60°.
Based on the above information, answer the following questions:

(i) [1 mark] Find the length of the wire from the point 'O' to the top of section 'B'.

Answer :

In right triangle OPB, ∠OPB = 90°, PO = 6 m, and the angle of elevation at O is 30°.
cos 30° = PO/OB, so √3/2 = 6/OB.
OB = 12/√3 = 4√3 m.

So, the length of the wire from O to the top of section B is 4√3 m.

(ii) [1 mark] Find the length of the wire from the point 'O' to the top of section 'A'.

Answer :

In right triangle OPA, ∠OPA = 90°, PO = 6 m, and the angle of elevation at O is 60°.
cos 60° = PO/OA, so 1/2 = 6/OA.
OA = 12 m.

So, the length of the wire from O to the top of section A is 12 m.

(iii) (a) [2 marks] Find the distance AB.

Answer (a) :

First, we find BP using triangle OPB.
tan 30° = BP/PO, so 1/√3 = BP/6, giving BP = 6/√3 = 2√3 m.

Next, we find AP using triangle OPA.
tan 60° = AP/PO, so √3 = AP/6, giving AP = 6√3 m.

Since section B is below section A on the same vertical tower, AB = AP − BP = 6√3 − 2√3 = 4√3 m.

So, the distance AB is 4√3 m.

OR

(iii) (b) [2 marks] Find the area of Δ OPB.

Answer (b) :

From triangle OPB, tan 30° = BP/PO, so 1/√3 = BP/6, giving BP = 2√3 m.

Since Δ OPB is right-angled at P, with base PO = 6 m and height BP = 2√3 m:
Area of Δ OPB = (1/2) × BP × PO = (1/2) × 2√3 × 6 = 6√3 m².

So, the area of Δ OPB is 6√3 m².

Teacher's Note:
The tower PA is vertical and perpendicular to the ground PO, so every triangle formed (OPA, OPB) is right-angled at P.
Use cos θ = adjacent/hypotenuse when the wire (hypotenuse) is wanted, and tan θ = opposite/adjacent when the height is wanted.
AB is simply the difference between the full tower height AP and the lower section height BP.

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