MathsCBSE Class 10 · Standard · Board exam 2026 (session 2025-26)

CBSE Class 10 Mathematics Standard: 2026 board paper, set 30/5/1

The question paper set in the CBSE Class 10 board exam of 2026, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2026/X/Mathematics_Standard.zip (a zip; this paper is the file 1172-1_30-5-1 (Mathematics Standard).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

Practise a paper like this, timedMy board paper sets a full paper in this exact format from real board questions, with a timer.Get your written answers markedThe Answer Coach marks a photo of your handwritten answer step by step, like the examiner.

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 22 and 24
C	26-31	Short answer; internal choice in 27 and 29
D	32-35	Long answer; internal choice in 33 and 34
E	36-38	Case study based; internal choice in the 2-mark subparts of each case

Instructions in brief: All 38 questions are compulsory; no overall choice, but internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 two-mark subparts in Section E; use of calculator is not allowed and π = 22/7 unless stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQQuadratic Equations
The value of $k$ for which the equation $k x^{2} - 6 x - 4 = 0$ has real and equal roots, is
1. (A) $\frac{9}{4}$
2. (B) $- 4$
3. (C) $- \frac{9}{4}$
4. (D) $- 2$
Try it first, then show our solutionSolution
1. For real and equal roots, $D = b^{2} - 4 a c = 0$ : $(- 6)^{2} - 4 (k) (- 4) = 36 + 16 k = 0$ .
2. $k = - \frac{36}{16} = - \frac{9}{4}$ , so option (C).
  1 mark
Answer: $k = - \frac{9}{4}$ , option (C)
Where the marks are
- Uses $D = 0$ and gets $k = - \frac{9}{4}$ (1 mark)
Q2 · 1 mark · MCQCoordinate Geometry
The line segment joining the points P $(- 4, - 2)$ and Q $(10, 4)$ is divided by y-axis in the ratio
1. (A) $2 : 5$
2. (B) $1 : 2$
3. (C) $2 : 1$
4. (D) $5 : 2$
Try it first, then show our solutionSolution
1. On the y-axis $x = 0$ . Let the ratio be $k : 1$ : $\frac{10 k - 4}{k + 1} = 0 \Rightarrow k = \frac{2}{5}$ .
2. Ratio $= 2 : 5$ , option (A).
  1 mark
Answer: $2 : 5$ , option (A)
Where the marks are
- Puts x-coordinate of dividing point = 0 and gets 2 : 5 (1 mark)
Q3 · 1 mark · MCQPolynomials
If the zeroes of a polynomial $p (x)$ are $- 3$ and $8$ , then $p (x)$ equals
1. (A) $x^{2} + 5 x - 4$
2. (B) $(x + 3) (- x + 8)$
3. (C) $a (x^{2} + 5 x - 24)$
4. (D) $x^{2} - 24$
Try it first, then show our solutionSolution
1. $(x + 3) (- x + 8)$ becomes 0 at $x = - 3$ and $x = 8$ , so both are zeroes.
2. Check others: (A) at $x = 8$ gives $100 \neq = 0$ ; (C) $x^{2} + 5 x - 24 = (x + 8) (x - 3)$ has zeroes $- 8, 3$ ; (D) zeroes $\pm 24$ . Hence option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Identifies $(x + 3) (- x + 8)$ as having zeroes $- 3$ and $8$ (1 mark)
Open Q3 on its own page
Q4 · 1 mark · MCQCircles
In the given figure, PQ is tangent to the circle with centre O. S is a point on the circle such that $∠ S QT = 5 5^{\circ}$ . The $m ∠ QP S$ is
1. (A) $5 5^{\circ}$
2. (B) $2 0^{\circ}$
3. (C) $3 5^{\circ}$
4. (D) $7 0^{\circ}$
Figure in the paper: A circle with centre O. P is outside the circle on the left; PQ is a tangent touching the circle at Q (upper left point of circle). PS is a secant/line from P through O to S, the point on the right of the circle diametrically opposite to the left end, so PS passes through the centre O. QT is a ray extended from Q beyond the tangent line to the upper right, with $∠ S QT = 5 5^{\circ}$ marked between chord QS and QT. Chord QS is drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. $O Q ⊥ P T$ [tangent ⟂ radius], so $∠ O QS = 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ . Since $O Q = O S$ , $∠ O S Q = 3 5^{\circ}$ .
2. In $△ P QS$ : $∠ P QS = 18 0^{\circ} - 5 5^{\circ} = 12 5^{\circ}$ , so $∠ QP S = 18 0^{\circ} - 12 5^{\circ} - 3 5^{\circ} = 2 0^{\circ}$ . Option (B).
  1 mark
Answer: $2 0^{\circ}$ , option (B)
Where the marks are
- Uses tangent ⟂ radius and isosceles triangle to get $∠ QP S = 2 0^{\circ}$ (1 mark)
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Q5 · 1 mark · MCQTriangles
Devansh proved that $△ A B C \sim △ P QR$ using SAS similarity criteria. If he found $∠ C = ∠ R$ , then which of the following was proved true?
1. (A) $\frac{A C}{A B} = \frac{P R}{P Q}$
2. (B) $\frac{B C}{A C} = \frac{P R}{QR}$
3. (C) $\frac{A C}{B C} = \frac{P R}{P Q}$
4. (D) $\frac{A C}{B C} = \frac{P R}{QR}$
Try it first, then show our solutionSolution
1. In SAS similarity, the equal angles must be included between the proportional sides. $∠ C$ lies between AC and BC; $∠ R$ lies between PR and QR.
2. So $\frac{A C}{B C} = \frac{P R}{QR}$ , option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Identifies sides enclosing $∠ C$ and $∠ R$ (1 mark)
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Q6 · 1 mark · MCQStatistics
While calculating mean of a grouped frequency distribution, step deviation method was used $(\frac{x - a}{h} = u)$ . It was found that $\overset{x}{ˉ} = 64$ , $h = 5$ and $a = 62.5$ . The value of $\overset{u}{ˉ}$ is
1. (A) $0.5$
2. (B) $1.5$
3. (C) $0.3$
4. (D) $7.5$
Try it first, then show our solutionSolution
1. $\overset{x}{ˉ} = a + h \overset{u}{ˉ} \Rightarrow 64 = 62.5 + 5 \overset{u}{ˉ}$ .
2. $\overset{u}{ˉ} = \frac{1.5}{5} = 0.3$ , option (C).
  1 mark
Answer: $\overset{u}{ˉ} = 0.3$ , option (C)
Where the marks are
- Uses $\overset{x}{ˉ} = a + h \overset{u}{ˉ}$ to get 0.3 (1 mark)
Q7 · 1 mark · MCQSurface Areas and Volumes
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use $π = 3.14$ )
1. (A) $3142$ cm $^{2}$
2. (B) $314$ cm $^{2}$
3. (C) $\frac{3140}{3}$ cm $^{2}$
4. (D) $31402$ cm $^{2}$
Figure in the paper: A solid hemisphere shown with its flat circular face at the top. A cone is inscribed inside it: its apex touches the curved surface at the bottom, and its circular base lies on the flat top face. A vertical axis line joins the centre of the top face to the bottom apex. See it in the official paper.
Try it first, then show our solutionSolution
1. Maximum cone: $r = 10$ cm, $h = 10$ cm, so $l = 1 0^{2} + 1 0^{2} = 102$ cm.
2. CSA $= π r l = 3.14 \times 10 \times 102 = 3142$ cm $^{2}$ , option (A).
  1 mark
Answer: $3142$ cm $^{2}$ , option (A)
Where the marks are
- Takes $r = h = 10$ , $l = 102$ and computes $π r l$ (1 mark)
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Q8 · 1 mark · MCQArithmetic Progressions
If $a_{n}$ represents $n^{t h}$ term of the A.P. $- \frac{15}{4}, - \frac{10}{4}, - \frac{5}{4}, \dots\dots$ then value of $a_{16} - a_{12}$ is
1. (A) $4$
2. (B) $\frac{5}{4}$
3. (C) $5$
4. (D) $\frac{25}{4}$
Try it first, then show our solutionSolution
1. $d = - \frac{10}{4} + \frac{15}{4} = \frac{5}{4}$ .
2. $a_{16} - a_{12} = 4 d = 4 \times \frac{5}{4} = 5$ , option (C).
  1 mark
Answer: 5, option (C)
Where the marks are
- Finds $d = \frac{5}{4}$ and uses $a_{16} - a_{12} = 4 d$ (1 mark)
Q9 · 1 mark · MCQQuadratic Equations
The value of $p$ for which roots of the quadratic equation $x^{2} - p x + 6 = 0$ are rational, is
1. (A) $1$
2. (B) $- 5$
3. (C) $25$
4. (D) $5$
Try it first, then show our solutionSolution
1. Roots are rational when $D = p^{2} - 24$ is a perfect square.
2. $p = - 5$ : $D = 25 - 24 = 1 = 1^{2}$ (roots 2... actually $- 2, - 3$ ). Others: $p = 1$ gives $- 23$ , $p = 25$ gives $601$ , $p = 5$ gives $- 19$ . Option (B).
  1 mark
Answer: $p = - 5$ , option (B)
Where the marks are
- Checks $p^{2} - 24$ is a perfect square for $p = - 5$ (1 mark)
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Q10 · 1 mark · MCQProbability
Which of the following can not be the probability of an event?
1. (A) $\frac{39}{100}$
2. (B) $\frac{0.001}{20}$
3. (C) $\frac{10}{0.2}$
4. (D) $10%$
Try it first, then show our solutionSolution
1. Probability must satisfy $0 \leq P \leq 1$ .
2. $\frac{10}{0.2} = 50 > 1$ , so it cannot be a probability. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- States $0 \leq P \leq 1$ and identifies $\frac{10}{0.2} = 50$ (1 mark)
Q11 · 1 mark · MCQProbability
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is either a ten or a king is
1. (A) $\frac{1}{26}$
2. (B) $\frac{2}{13}$
3. (C) $\frac{1}{13}$
4. (D) $\frac{8}{26}$
Try it first, then show our solutionSolution
1. Favourable cards: 4 tens + 4 kings = 8.
2. $P = \frac{8}{52} = \frac{2}{13}$ , option (B).
  1 mark
Answer: $\frac{2}{13}$ , option (B)
Where the marks are
- Counts 8 favourable cards and gets $\frac{2}{13}$ (1 mark)
Q12 · 1 mark · MCQCompetency-basedSurface Areas and Volumes
A camping tent in hemispherical shape of radius 1.4 m, has a door opening of area $0.50$ m $^{2}$ . Outer surface area of the tent is
1. (A) $11.78$ m $^{2}$
2. (B) $12.32$ m $^{2}$
3. (C) $11.82$ m $^{2}$
4. (D) $12.86$ m $^{2}$
Try it first, then show our solutionSolution
1. CSA of hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 1.4 \times 1.4 = 12.32$ m $^{2}$ .
2. Outer surface area $= 12.32 - 0.50 = 11.82$ m $^{2}$ , option (C).
  1 mark
Answer: 11.82 m $^{2}$ , option (C)
Where the marks are
- Computes $2 π r^{2} = 12.32$ and subtracts door area (1 mark)
Q13 · 1 mark · MCQAreas Related to Circles
An arc of length $2.2$ cm subtends an angle $θ$ at the centre of the circle with radius $2.8$ cm. The value of $θ$ is
1. (A) $5 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $3 0^{\circ}$
Try it first, then show our solutionSolution
1. $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r \Rightarrow 2.2 = \frac{θ}{360} \times 2 \times \frac{22}{7} \times 2.8 = \frac{θ}{360} \times 17.6$ .
2. $θ = \frac{2.2 \times 360}{17.6} = 4 5^{\circ}$ , option (C).
  1 mark
Answer: $4 5^{\circ}$ , option (C)
Where the marks are
- Uses arc length formula to get $4 5^{\circ}$ (1 mark)
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Q14 · 1 mark · MCQIntroduction to Trigonometry
For an acute angle $θ$ , if $sin θ = \frac{1}{9}$ , then value of $\frac{9 cosec θ + 1}{9 cosec θ - 1}$ is
1. (A) $0$
2. (B) $\frac{80}{81}$
3. (C) $1$
4. (D) $\frac{82}{80}$
Try it first, then show our solutionSolution
1. $cosec θ = 9$ , so $9 cosec θ = 81$ .
2. Value $= \frac{81 + 1}{81 - 1} = \frac{82}{80}$ , option (D).
  1 mark
Answer: $\frac{82}{80}$ , option (D)
Where the marks are
- Uses $cosec θ = 9$ and substitutes (1 mark)
Q15 · 1 mark · MCQTriangles
In the given figure, PQ $∥$ YZ such that XP : PY = $2 : 3$ . If PQ = 5 cm, then YZ equals
1. (A) $12.5$ cm
2. (B) $10$ cm
3. (C) $15$ cm
4. (D) $7.5$ cm
Figure in the paper: Triangle XYZ with apex X at top. P lies on side XY and Q lies on side XZ. PQ is drawn parallel to the base YZ (arrow marks on PQ and YZ show parallel). Points labelled X, P, Y, Q, Z. See it in the official paper.
Try it first, then show our solutionSolution
1. $P Q ∥ Y Z \Rightarrow △ X P Q \sim △ X Y Z$ (AA), so $\frac{P Q}{Y Z} = \frac{X P}{X Y} = \frac{2}{5}$ .
2. $Y Z = \frac{5 \times 5}{2} = 12.5$ cm, option (A).
  1 mark
Answer: 12.5 cm, option (A)
Where the marks are
- Uses similarity with ratio $X P : X Y = 2 : 5$ to get 12.5 cm (1 mark)
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Q16 · 1 mark · MCQCompetency-basedProbability
Meena calculates that the probability of her winning the first prize in a lottery is $0.08$ . If total 800 tickets were sold, the number of tickets bought by her, is
1. (A) $64$
2. (B) $640$
3. (C) $100$
4. (D) $10$
Try it first, then show our solutionSolution
1. $P = \frac{her tickets}{800} = 0.08$ .
2. Her tickets $= 0.08 \times 800 = 64$ , option (A).
  1 mark
Answer: 64, option (A)
Where the marks are
- Computes $0.08 \times 800 = 64$ (1 mark)
Q17 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A wire is attached from a point A on the ground to the top of a pole BC, making an angle of elevation as $6 0^{\circ}$ . If AB = $53$ m, then length of the wire is
1. (A) $10$ m
2. (B) $103$ m
3. (C) $15$ m
4. (D) $\frac{5}{2} 3$ m
Figure in the paper: Right triangle ABC with right angle at B. BC is the vertical pole, AB is the horizontal ground of length $53$ m, and AC is the wire from A to the top C. The angle at A between ground and wire is $6 0^{\circ}$ . See it in the official paper.
Try it first, then show our solutionSolution
1. $cos 6 0^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{2} = \frac{5 3}{A C}$ .
2. $A C = 103$ m, option (B).
  1 mark
Answer: $103$ m, option (B)
Where the marks are
- Uses $cos 6 0^{\circ} = A B / A C$ to get $103$ m (1 mark)
Q18 · 1 mark · MCQIntroduction to Trigonometry
Simplest form of $\frac{s e c A}{s e c ^{2} A - 1}$ is
1. (A) $sin A$
2. (B) $tan A$
3. (C) $cosec A$
4. (D) $cos A$
Try it first, then show our solutionSolution
1. $sec^{2} A - 1 = tan A$ , so expression $= \frac{s e c A}{t a n A} = \frac{1/ c o s A}{s i n A / c o s A} = \frac{1}{s i n A}$ .
2. $= cosec A$ , option (C).
  1 mark
Answer: cosec A, option (C)
Where the marks are
- Uses $sec^{2} A - 1 = tan^{2} A$ and simplifies to cosec A (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedPair of Linear Equations in Two Variables
Assertion (A): The system of linear equations $3 x - 5 y + 7 = 0$ and $- 6 x + 10 y + 14 = 0$ is inconsistent.
Reason (R): When two linear equations don't have unique solution, they always represent parallel lines.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $\frac{a _{1}}{a _{2}} = \frac{3}{- 6} = - \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 5}{10} = - \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{7}{14} = \frac{1}{2}$ . Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , lines are parallel: no solution, so A is true.
2. R is false: no unique solution may also mean coincident lines (infinitely many solutions). Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows A true by ratio test and R false due to coincident case (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A): H.C.F. $(36 m^{2}, 18 m) = 18 m$ , where $m$ is a prime number.
Reason (R): H.C.F. of two numbers is always less than or equal to the smaller number.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $36 m^{2} = 2^{2} \cdot 3^{2} \cdot m^{2}$ , $18 m = 2 \cdot 3^{2} \cdot m$ ; HCF $= 2 \cdot 3^{2} \cdot m = 18 m$ . A is true.
2. R is true (HCF divides both numbers, so it cannot exceed the smaller), but A is established by prime factorisation, not by R. Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Verifies HCF = 18m and recognises R is true but not the explanation (1 mark)
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Section B

Very short answer; internal choice in 22 and 24

Q21 · 2 marks · Very short answerReal Numbers
Prove that $2 - 53$ is an irrational number given that $3$ is irrational.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $2 - 53$ is rational. Then $2 - 53 = \frac{a}{b}$ , where $a, b$ are integers and $b \neq = 0$ .
  ½ mark
2. Rearranging: $53 = 2 - \frac{a}{b} = \frac{2 b - a}{b}$ , so $3 = \frac{2 b - a}{5 b}$ .
  ½ mark
3. Since $a, b$ are integers and $b \neq = 0$ , $\frac{2 b - a}{5 b}$ is rational. This would make $3$ rational, which contradicts the given fact that $3$ is irrational. So our assumption is wrong, and $2 - 53$ is irrational.
  1 mark
Answer: $2 - 53$ is irrational (proved by contradiction).
Where the marks are
- Assumes $2 - 53 = \frac{a}{b}$ with integers $a, b$ and $b \neq = 0$ (½ mark)
- Expresses $3 = \frac{2 b - a}{5 b}$ (½ mark)
- States the contradiction (rational = irrational) and concludes (1 mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerCoordinate Geometry
Vertices of a right triangle ABC with $∠ B = 9 0^{\circ}$ are A(3, 4), B(1, 1) and C(–8, 7). Find the value of $tan A$ .
Try it first, then show our solutionSolution
1. Since $∠ B = 9 0^{\circ}$ , for angle A the opposite side is BC and the adjacent side is AB, so $tan A = \frac{B C}{A B}$ .
  $B C = (- 8 - 1)^{2} + (7 - 1)^{2} = 81 + 36 = 117 = 313$
  ½ mark
2. $A B = (3 - 1)^{2} + (4 - 1)^{2} = 4 + 9 = 13$
  ½ mark
3. $tan A = \frac{B C}{A B} = \frac{3 13}{13} = 3$
  1 mark
Answer: $tan A = 3$
Where the marks are
- Finds $B C = 313$ using the distance formula (½ mark)
- Finds $A B = 13$ (½ mark)
- Uses $tan A = \frac{B C}{A B}$ to get 3 (1 mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerCoordinate Geometry
Using distance formula, prove that the points A(2, 3), B(–7, 0) and C(–1, 2) are collinear.
Try it first, then show our solutionSolution
1. $A B = (- 7 - 2)^{2} + (0 - 3)^{2} = 81 + 9 = 90 = 310$
  ½ mark
2. $B C = (- 1 + 7)^{2} + (2 - 0)^{2} = 36 + 4 = 40 = 210$
  ½ mark
3. $A C = (- 1 - 2)^{2} + (2 - 3)^{2} = 9 + 1 = 10$
  ½ mark
4. $A C + B C = 10 + 210 = 310 = A B$ . The sum of two of the distances equals the third, so A, B, C are collinear (C lies between A and B).
  ½ mark
Answer: $A C + B C = A B = 310$ , so A, B, C are collinear.
Where the marks are
- $A B = 310$ (½ mark)
- $B C = 210$ (½ mark)
- $A C = 10$ (½ mark)
- Shows $A C + B C = A B$ and concludes collinear (½ mark)
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Q23 · 2 marks · Very short answerTriangles
In the given figure, AB $∥$ DE and AC $∥$ DF. Show that $△ A B C \sim △ D E F$ . If BC = 10 cm, EB = CF = 5 cm and AB = 7 cm, then find the length DE.
Figure in the paper: Triangle DEF with D at top, E at bottom left, F at bottom right. Points B and C lie on segment EF in that order (E, B, C, F), and A lies inside the triangle, joined to B and C. Segment AB is parallel to DE and segment AC is parallel to DF, both shown with arrow marks. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ and $△ D E F$ :
  $A B ∥ D E \Rightarrow ∠ A B C = ∠ D E F$ [corresponding angles]
  $A C ∥ D F \Rightarrow ∠ A C B = ∠ D F E$ [corresponding angles]
  $∴ △ A B C \sim △ D E F$ [AA similarity criterion]
  1 mark
2. $E F = E B + B C + C F = 5 + 10 + 5 = 20$ cm. Corresponding sides of similar triangles are proportional: $\frac{B C}{E F} = \frac{A B}{D E} \Rightarrow \frac{10}{20} = \frac{7}{D E}$
  ½ mark
3. $D E = \frac{7 \times 20}{10} = 14$ cm
  ½ mark
Answer: $△ A B C \sim △ D E F$ (AA); DE = 14 cm
Where the marks are
- Shows two pairs of equal angles from the parallel lines and concludes similarity by AA (1 mark)
- Writes $\frac{B C}{E F} = \frac{A B}{D E}$ with $E F = 20$ cm (½ mark)
- Gets DE = 14 cm (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $\frac{s i n ^{3} 6 0 ^{\circ} - t a n 3 0 ^{\circ}}{c o s ^{2} 4 5 ^{\circ}}$
Try it first, then show our solutionSolution
1. Substitute $sin 6 0^{\circ} = \frac{3}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ , $cos 4 5^{\circ} = \frac{1}{2}$ :
  $\frac{( \frac{3}{2} ) ^{3} - \frac{1}{3}}{( \frac{1}{2} ) ^{2}} = \frac{\frac{3 3}{8} - \frac{3}{3}}{\frac{1}{2}} = \frac{\frac{9 3 - 8 3}{24}}{\frac{1}{2}}$
  1½ marks
2. $= \frac{3}{24} \times 2 = \frac{3}{12} (= \frac{1}{4 3})$
  ½ mark
Answer: $\frac{3}{12}$ (or $\frac{1}{4 3}$ )
Where the marks are
- Correct substitution of the standard values and simplification of the numerator (1½ marks)
- Final value $\frac{3}{12}$ (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerIntroduction to Trigonometry
For acute angles A and B and A + 2B and 2A + B are acute if $tan (A + 2 B) = 3$ and $sin (2 A + B) = \frac{1}{2}$ , then find the measures of angles A and B.
Try it first, then show our solutionSolution
1. $tan (A + 2 B) = 3 = tan 6 0^{\circ} \Rightarrow A + 2 B = 6 0^{\circ}$ ...(1)
  ½ mark
2. $sin (2 A + B) = \frac{1}{2} = sin 4 5^{\circ} \Rightarrow 2 A + B = 4 5^{\circ}$ ...(2)
  ½ mark
3. Adding (1) and (2): $3 A + 3 B = 10 5^{\circ} \Rightarrow A + B = 3 5^{\circ}$ ...(3). Subtracting (3) from (1): $B = 2 5^{\circ}$ .
  ½ mark
4. From (3): $A = 3 5^{\circ} - 2 5^{\circ} = 1 0^{\circ}$ .
  ½ mark
Answer: $A = 1 0^{\circ}$ , $B = 2 5^{\circ}$
Where the marks are
- $A + 2 B = 6 0^{\circ}$ (½ mark)
- $2 A + B = 4 5^{\circ}$ (½ mark)
- $B = 2 5^{\circ}$ (½ mark)
- $A = 1 0^{\circ}$ (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerProbability
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $\frac{3}{5}$ , then find the number of yellow balls.
Try it first, then show our solutionSolution
1. Each ball is either yellow or green, so $P (yellow) = 1 - P (green) = 1 - \frac{3}{5} = \frac{2}{5}$ , i.e. $\frac{number of yellow balls}{25} = \frac{2}{5}$ .
  1 mark
2. Number of yellow balls $= 25 \times \frac{2}{5} = 10$ .
  1 mark
Answer: 10 yellow balls
Where the marks are
- Uses $P (yellow) = 1 - P (green) = \frac{2}{5}$ (1 mark)
- Number of yellow balls = 10 (1 mark)
Open Q25 on its own page

Section C

Short answer; internal choice in 27 and 29

Q26 · 3 marks · Short answerAreas Related to Circles
A circle of diameter 20 cm is equally divided into five sectors. Find the area and perimeter of one of the sectors.
Try it first, then show our solutionSolution
1. Radius $r = \frac{20}{2} = 10$ cm. Central angle of each sector $θ = \frac{36 0 ^{\circ}}{5} = 7 2^{\circ}$ .
  ½ mark
2. Area of one sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{72}{360} \times \frac{22}{7} \times 10 \times 10$
  ½ mark
3. $= \frac{440}{7}$ cm² $\approx 62.86$ cm²
  ½ mark
4. Perimeter of sector $=$ arc length $+ 2 r = \frac{72}{360} \times 2 \times \frac{22}{7} \times 10 + 10 + 10 = \frac{88}{7} + 20$
  1 mark
5. $= \frac{228}{7}$ cm $\approx 32.57$ cm
  ½ mark
Answer: Area $= \frac{440}{7}$ cm² ≈ 62.86 cm²; Perimeter $= \frac{228}{7}$ cm ≈ 32.57 cm
Where the marks are
- Radius 10 cm and central angle $7 2^{\circ}$ (½ mark)
- Correct area formula with substitution (½ mark)
- Area $= \frac{440}{7}$ cm² (½ mark)
- Perimeter = arc length + 2r with substitution (1 mark)
- Perimeter $= \frac{228}{7}$ cm (½ mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerArithmetic Progressions
In an A.P., 15th term exceeds the 8th term by 21. If sum of first 10 terms is 55, then form the A.P.
Try it first, then show our solutionSolution
1. Let first term $= a$ , common difference $= d$ . Given $a_{15} = a_{8} + 21$ : $a + 14 d = a + 7 d + 21$
  1 mark
2. $7 d = 21 \Rightarrow d = 3$
  ½ mark
3. $S_{10} = \frac{10}{2} [2 a + (10 - 1) d] \Rightarrow 55 = 5 [2 a + 27]$
  ½ mark
4. $11 = 2 a + 27 \Rightarrow 2 a = - 16 \Rightarrow a = - 8$
  ½ mark
5. The A.P. is $- 8, - 5, - 2, 1, \dots$
  ½ mark
Answer: A.P.: $- 8, - 5, - 2, 1, \dots$ ( $a = - 8$ , $d = 3$ )
Where the marks are
- Forms $a + 14 d = a + 7 d + 21$ (1 mark)
- $d = 3$ (½ mark)
- Uses $S_{10} = 55$ correctly (½ mark)
- $a = - 8$ (½ mark)
- Writes the A.P. (½ mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerArithmetic Progressions
The sum of first n terms of an A.P. is $2 n^{2} + 13 n$ . Find its $n^{t h}$ term and hence $1 0^{t h}$ term.
Try it first, then show our solutionSolution
1. $S_{n} = 2 n^{2} + 13 n$ . $S_{1} = a_{1} = 2 + 13 = 15$
  ½ mark
2. $S_{2} = a_{1} + a_{2} = 8 + 26 = 34 \Rightarrow a_{2} = 34 - 15 = 19$
  ½ mark
3. $d = a_{2} - a_{1} = 19 - 15 = 4$
  ½ mark
4. $a_{n} = a + (n - 1) d = 15 + (n - 1) \times 4 = 4 n + 11$
  1 mark
5. $a_{10} = 4 \times 10 + 11 = 51$
  ½ mark
Answer: $a_{n} = 4 n + 11$ ; $a_{10} = 51$
Where the marks are
- $a_{1} = 15$ (½ mark)
- $a_{2} = 19$ (½ mark)
- $d = 4$ (½ mark)
- $a_{n} = 4 n + 11$ (1 mark)
- $a_{10} = 51$ (½ mark)
Open Q27(b) on its own page
Q28 · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + tan θ + cot θ$ .
Try it first, then show our solutionSolution
1. LHS $= \frac{\frac{s i n θ}{c o s θ}}{1 - \frac{c o s θ}{s i n θ}} + \frac{\frac{c o s θ}{s i n θ}}{1 - \frac{s i n θ}{c o s θ}}$
  ½ mark
2. $= \frac{sin ^{2} θ}{cos θ ( sin θ - cos θ )} + \frac{cos ^{2} θ}{sin θ ( cos θ - sin θ )} = \frac{sin ^{2} θ}{cos θ ( sin θ - cos θ )} - \frac{cos ^{2} θ}{sin θ ( sin θ - cos θ )}$
  ½ mark
3. $= \frac{sin ^{3} θ - cos ^{3} θ}{sin θ cos θ ( sin θ - cos θ )}$
  ½ mark
4. Using $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ : $= \frac{( sin θ - cos θ ) ( sin ^{2} θ + cos ^{2} θ + sin θ cos θ )}{sin θ cos θ ( sin θ - cos θ )}$
  ½ mark
5. $= \frac{sin ^{2} θ}{sin θ cos θ} + \frac{cos ^{2} θ}{sin θ cos θ} + \frac{sin θ cos θ}{sin θ cos θ}$
  ½ mark
6. $= tan θ + cot θ + 1 =$ RHS. Hence proved.
  ½ mark
Answer: LHS simplifies to $1 + tan θ + cot θ$ = RHS. Hence proved.
Where the marks are
- Writes tan and cot in terms of sin and cos (½ mark)
- Simplifies each fraction (½ mark)
- Combines to $\frac{s i n ^{3} θ - c o s ^{3} θ}{s i n θ c o s θ ( s i n θ - c o s θ )}$ (½ mark)
- Factorises using $a^{3} - b^{3}$ (½ mark)
- Splits into three fractions (½ mark)
- Reaches RHS (½ mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCoordinate Geometry
A circle centred at (2, 1) passes through the points A(5, 6) and B(–3, K). Find the value(s) of K. Hence find length of chord AB.
Try it first, then show our solutionSolution
1. Let O(2, 1) be the centre. Radii are equal, so $O A = O B$ : $(5 - 2)^{2} + (6 - 1)^{2} = (- 3 - 2)^{2} + (K - 1)^{2} \Rightarrow 9 + 25 = 25 + (K - 1)^{2}$
  1 mark
2. $(K - 1)^{2} = 9 \Rightarrow K - 1 = \pm 3 \Rightarrow K = 4$ or $K = - 2$
  1 mark
3. For $K = - 2$ , B(–3, –2): $A B = (- 3 - 5)^{2} + (- 2 - 6)^{2} = 64 + 64 = 128 = 82$ units
  ½ mark
4. For $K = 4$ , B(–3, 4): $A B = (- 3 - 5)^{2} + (4 - 6)^{2} = 64 + 4 = 68 = 217$ units
  ½ mark
Answer: $K = 4$ or $K = - 2$ ; AB $= 217$ units (K = 4) or $82$ units (K = –2)
Where the marks are
- Uses OA = OB with the distance formula (1 mark)
- Finds $K = - 2, 4$ (1 mark)
- AB $= 82$ for $K = - 2$ (½ mark)
- AB $= 217$ for $K = 4$ (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCoordinate Geometry
Prove that the point P dividing the line segment joining the points A(–1, 7) and B(4, –3) in the ratio 3 : 2, lies on the line $x - 3 y = - 1$ . Also find length of PA and PB.
Try it first, then show our solutionSolution
1. By the section formula with $m : n = 3 : 2$ , A(–1, 7), B(4, –3):
  $P = (\frac{3 ( 4 ) + 2 ( - 1 )}{5}, \frac{3 ( - 3 ) + 2 ( 7 )}{5}) = (\frac{10}{5}, \frac{5}{5}) = (2, 1)$
  1 mark
2. Put $x = 2, y = 1$ in $x - 3 y$ : $2 - 3 (1) = - 1 =$ RHS. So P lies on the line $x - 3 y = - 1$ .
  1 mark
3. $P A = (2 + 1)^{2} + (1 - 7)^{2} = 9 + 36 = 45 = 35$ units
  ½ mark
4. $P B = (2 - 4)^{2} + (1 + 3)^{2} = 4 + 16 = 20 = 25$ units
  ½ mark
Answer: P(2, 1) lies on $x - 3 y = - 1$ ; PA $= 35$ units, PB $= 25$ units
Where the marks are
- Finds P = (2, 1) by section formula (1 mark)
- Verifies P satisfies $x - 3 y = - 1$ (1 mark)
- PA $= 35$ (½ mark)
- PB $= 25$ (½ mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve the system of linear equations: $x = 4$ and $3 x - 2 y = 6$ graphically.
Try it first, then show our solutionSolution
1. For $3 x - 2 y = 6$ , i.e. $y = \frac{3 x - 6}{2}$ , table of values:
  x 0 2 4
  y –3 0 3
  Plot (0, –3), (2, 0), (4, 3) on graph paper and join them to get a straight line.
  1½ marks
2. $x = 4$ is a straight line parallel to the y-axis, at a distance of 4 units to the right of it, passing through (4, 0), (4, 3), (4, 5). Draw this line.
  ½ mark
3. The two lines intersect at the point (4, 3). Hence the solution is $x = 4$ , $y = 3$ . (Check: $3 (4) - 2 (3) = 6$ ✓)
  1 mark
Answer: $x = 4$ , $y = 3$ (lines meet at (4, 3))
Where the marks are
- Correct table and graph of $3 x - 2 y = 6$ (1½ marks)
- Correct graph of $x = 4$ (½ mark)
- Reads intersection: $x = 4$ (½ mark), $y = 3$ (½ mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCompetency-basedReal Numbers
The dimensions of a window are 156 cm $\times$ 216 cm. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.
Try it first, then show our solutionSolution
1. $156 = 2 \times 2 \times 3 \times 13 = 2^{2} \times 3 \times 13$
  ½ mark
2. $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^{3} \times 3^{3}$
  ½ mark
3. The largest square that fits exactly along both sides has side = HCF(156, 216) $= 2^{2} \times 3 = 12$ cm.
  1 mark
4. Number of squares $= \frac{156 \times 216}{12 \times 12} = 13 \times 18 = 234$
  1 mark
Answer: Side of square = 12 cm; number of squares = 234
Where the marks are
- Prime factorisation of 156 (½ mark)
- Prime factorisation of 216 (½ mark)
- Side = HCF = 12 cm (1 mark)
- Number of squares = 234 (1 mark)
Open Q31 on its own page

x	0	2	4
y	–3	0	3

Section D

Long answer; internal choice in 33 and 34

Q32 · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables
Venkat can row a boat in still water at the speed of 12 km/h. He ferries tourists 15 km upstream and 18 km downstream in 3 hours. Find the speed of the stream.
Try it first, then show our solutionSolution
1. Let the speed of the stream be $x$ km/h. Upstream speed $= (12 - x)$ km/h, downstream speed $= (12 + x)$ km/h. Time = distance ÷ speed, so
  $\frac{15}{12 - x} + \frac{18}{12 + x} = 3$
  2 marks
2. $15 (12 + x) + 18 (12 - x) = 3 (144 - x^{2}) \Rightarrow 396 - 3 x = 432 - 3 x^{2} \Rightarrow 3 x^{2} - 3 x - 36 = 0 \Rightarrow x^{2} - x - 12 = 0$
  1 mark
3. $x^{2} - 4 x + 3 x - 12 = 0 \Rightarrow (x - 4) (x + 3) = 0 \Rightarrow x = 4$ or $x = - 3$
  1 mark
4. Speed cannot be negative, so $x = - 3$ is rejected. Hence the speed of the stream is 4 km/h.
  1 mark
Answer: Speed of the stream = 4 km/h
Where the marks are
- Forms the equation $\frac{15}{12 - x} + \frac{18}{12 + x} = 3$ (2 marks)
- Reduces to $x^{2} - x - 12 = 0$ (1 mark)
- Factorises to get $x = 4, - 3$ (1 mark)
- Rejects $- 3$ ; speed = 4 km/h (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerTriangles
D is the mid-point of side BC of $△ A B C$ . CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. Prove that:
(i) OBGC is a parallelogram.
(ii) EF $∥$ BC
(iii) $△ A E F \sim △ A B C$
Figure in the paper: Triangle ABC with A at top, B bottom left, C bottom right. D is the midpoint of BC (tick marks on BD and DC). AD is drawn and extended below BC to G, with OD = DG (tick marks). E lies on AB and F lies on AC; BF and CE are drawn intersecting at O on AD. EF is drawn as a horizontal segment near the top, with E and F marked. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In quadrilateral OBGC, $B D = D C$ [D is mid-point of BC] and $O D = D G$ [given]. So the diagonals BC and OG bisect each other at D. Hence OBGC is a parallelogram.
  1 mark
2. (ii) Since OBGC is a parallelogram, $O C ∥ B G$ and $O B ∥ C G$ . E lies on line CO, so $O E ∥ B G$ ; F lies on line BO, so $O F ∥ GC$ .
  ½ mark
3. In $△ A B G$ , $O E ∥ B G$ with E on AB and O on AG, so by the Basic Proportionality Theorem $\frac{A E}{E B} = \frac{A O}{O G}$ ...(1)
  1 mark
4. Similarly in $△ A GC$ , $O F ∥ GC$ , so $\frac{A F}{F C} = \frac{A O}{O G}$ ...(2) [BPT]
  ½ mark
5. From (1) and (2), $\frac{A E}{E B} = \frac{A F}{F C}$ . So in $△ A B C$ , EF divides AB and AC in the same ratio, hence $E F ∥ B C$ [converse of BPT].
  1 mark
6. (iii) In $△ A E F$ and $△ A B C$ : $∠ A E F = ∠ A B C$ [corresponding angles, $E F ∥ B C$ ] and $∠ E A F = ∠ B A C$ [common]. Hence $△ A E F \sim △ A B C$ [AA similarity].
  1 mark
Answer: (i) OBGC is a parallelogram (diagonals bisect each other); (ii) EF ∥ BC (BPT and its converse); (iii) △AEF ~ △ABC (AA)
Where the marks are
- (i) Diagonals OG and BC bisect each other ⇒ parallelogram (1 mark)
- (ii) Deduces $O E ∥ B G$ from the parallelogram (½ mark)
- Applies BPT in △ABG: $\frac{A E}{E B} = \frac{A O}{O G}$ (1 mark)
- Applies BPT in △AGC: $\frac{A F}{F C} = \frac{A O}{O G}$ (½ mark)
- Equates ratios and uses converse of BPT to get EF ∥ BC (1 mark)
- (iii) Equal corresponding angle + common angle ⇒ AA similarity (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerTriangles
Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that:
(i) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP
Figure in the paper: Parallelogram ABCD with A bottom left, B bottom right, C upper right, D upper left. Q is the midpoint of side DC (tick marks). Diagonal BD is drawn, and line AR passes through Q and meets BD at P; R lies on the extension of BC beyond C. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $D C ∥ A B$ [opposite sides of parallelogram], so $QC ∥ A B$ . In $△ R QC$ and $△ R A B$ : $∠ R$ is common and $∠ R QC = ∠ R A B$ [corresponding angles]. So $△ R QC \sim △ R A B$ (AA) and $\frac{QR}{A R} = \frac{QC}{A B} = \frac{\frac{1}{2} D C}{D C} = \frac{1}{2}$ [Q mid-point of DC, AB = DC].
  1 mark
2. $A R = 2 QR \Rightarrow A Q + QR = 2 QR \Rightarrow A Q = QR$ . Hence proved.
  1 mark
3. (ii) In $△ P Q D$ and $△ P A B$ : $∠ QP D = ∠ A P B$ [vertically opposite] and $∠ P D Q = ∠ P B A$ [alternate angles, $D Q ∥ A B$ ]. So $△ P Q D \sim △ P A B$ (AA) and $\frac{QP}{A P} = \frac{D Q}{B A} = \frac{\frac{1}{2} A B}{A B} = \frac{1}{2}$ .
  1 mark
4. Hence $A P = 2 P Q$ , i.e. $P Q = \frac{1}{2} A P$ .
  ½ mark
5. (iii) From (i), $A Q = QR \Rightarrow A P + P Q = P R - P Q$ .
  ½ mark
6. Substituting $P Q = \frac{1}{2} A P$ : $A P + \frac{1}{2} A P = P R - \frac{1}{2} A P \Rightarrow P R = 2 A P$ . Hence proved.
  1 mark
Answer: (i) AQ = QR; (ii) AP = 2PQ; (iii) PR = 2AP — all proved using similar triangles
Where the marks are
- (i) △RQC ~ △RAB giving $\frac{QR}{A R} = \frac{1}{2}$ (1 mark)
- Concludes AQ = QR (1 mark)
- (ii) △PQD ~ △PAB giving $\frac{QP}{A P} = \frac{1}{2}$ (1 mark)
- AP = 2PQ (½ mark)
- (iii) Writes $A P + P Q = P R - P Q$ (½ mark)
- Substitutes and gets PR = 2AP (1 mark)
Open Q33(b) on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerStatistics
The mean of the following frequency distribution is 28. If sum of all frequencies is 100, then find the values of p and q:
Class Interval 0–10 10–20 20–30 30–40 40–50 50–60
Frequency 12 p 27 20 q 6
Try it first, then show our solutionSolution
1. Table of class marks and products:
  Class $x_{i}$ $f_{i}$ $f_{i} x_{i}$
  0–10 5 12 60
  10–20 15 p 15p
  20–30 25 27 675
  30–40 35 20 700
  40–50 45 q 45q
  50–60 55 6 330
  Total $65 + p + q$ $1765 + 15 p + 45 q$
  2 marks
2. $\sum f_{i} = 100 \Rightarrow 65 + p + q = 100 \Rightarrow p + q = 35$ ...(1)
  ½ mark
3. Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} \Rightarrow 28 = \frac{1765 + 15 p + 45 q}{100}$
  1 mark
4. $2800 = 1765 + 15 p + 45 q \Rightarrow 15 p + 45 q = 1035 \Rightarrow p + 3 q = 69$ ...(2)
  ½ mark
5. (2) – (1): $2 q = 34 \Rightarrow q = 17$ ; from (1), $p = 35 - 17 = 18$ .
  1 mark
Answer: $p = 18$ , $q = 17$
Where the marks are
- Correct table with $x_{i}$ , $f_{i} x_{i}$ and totals (2 marks)
- $p + q = 35$ (½ mark)
- Sets mean $= \frac{1765 + 15 p + 45 q}{100} = 28$ (1 mark)
- $p + 3 q = 69$ (½ mark)
- $p = 18$ (½ mark), $q = 17$ (½ mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerStatistics
Find median and mode of the following distribution:
Class Interval 0–15 15–30 30–45 45–60 60–75 75–90 90–105
Frequency 15 10 12 9 8 10 6
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Class $f$ $c f$
  0–15 15 15
  15–30 10 25
  30–45 12 37
  45–60 9 46
  60–75 8 54
  75–90 10 64
  90–105 6 70
  $n = \sum f = 70$
  1 mark
2. $\frac{n}{2} = \frac{70}{2} = 35$ . The first $c f$ greater than 35 is 37, so the median class is 30–45.
  Here $l = 30$ , $c f = 25$ , $f = 12$ , $h = 15$ .
  ½ mark
3. Median $= l + (\frac{\frac{n}{2} - c f}{f}) \times h = 30 + \frac{35 - 25}{12} \times 15$
  1 mark
4. $= 30 + 12.5 = 42.5$
  ½ mark
5. The highest frequency is 15, so the modal class is 0–15.
  Here $l = 0$ , $f_{1} = 15$ , $f_{0} = 0$ (no class before it), $f_{2} = 10$ , $h = 15$ .
  ½ mark
6. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 0 + \frac{15 - 0}{30 - 0 - 10} \times 15$
  1 mark
7. $= \frac{15}{20} \times 15 = \frac{45}{4} = 11.25$
  Hence, median = 42.5 and mode = 11.25.
  ½ mark
Answer: Median = 42.5; Mode = 11.25
Where the marks are
- Correct cumulative frequency table (1 mark)
- Finds $n /2 = 35$ and identifies median class 30–45 (½ mark)
- Substitutes correctly in the median formula (1 mark)
- Median = 42.5 (½ mark)
- Identifies modal class 0–15 (½ mark)
- Substitutes correctly in the mode formula with $f_{0} = 0$ (1 mark)
- Mode = 11.25 (½ mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerCircles
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.
Figure in the paper: A circle with centre O. P lies outside the circle on the right. Tangents PQ (touching at Q, upper point) and PR (touching at R, lower point) are drawn from P. A third tangent AB touches the circle at C, where C lies on segment OP; A lies on PQ and B lies on PR, so AB is a tangent segment crossing OP at C. See it in the official paper.
Try it first, then show our solutionSolution
1. In right $△ O QP$ , $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius].
  So $P Q = O P^{2} - O Q^{2} = 169 - 25 = 144 = 12$ cm.
  1 mark
2. $P C = O P - O C = 13 - 5 = 8$ cm.
  Let $A C = A Q = x$ [tangents from an external point A are equal]. Then $P A = P Q - A Q = 12 - x$ .
  1 mark
3. $O C ⊥ A B$ [tangent ⟂ radius] and C lies on OP, so $∠ A C P = 9 0^{\circ}$ .
  In right $△ A C P$ : $P A^{2} = A C^{2} + P C^{2} \Rightarrow (12 - x)^{2} = x^{2} + 8^{2}$
  1 mark
4. $144 - 24 x + x^{2} = x^{2} + 64 \Rightarrow 24 x = 80 \Rightarrow x = \frac{10}{3}$ cm
  1 mark
5. In the same way, $B C = B R$ and $P B = 12 - B C$ ; with $P C = 8$ and $∠ B C P = 9 0^{\circ}$ this gives $B C = \frac{10}{3}$ cm.
  So $A B = A C + C B = 2 \times \frac{10}{3} = \frac{20}{3}$ cm $\approx 6.67$ cm.
  ½ mark
6. $P A = 12 - \frac{10}{3} = \frac{26}{3}$ cm $\approx 8.67$ cm.
  ½ mark
Answer: AB = 20/3 cm (≈ 6.67 cm); PA = 26/3 cm (≈ 8.67 cm)
Where the marks are
- Finds $P Q = 12$ cm using tangent ⟂ radius and Pythagoras (1 mark)
- $P C = 8$ cm, and $P A = 12 - x$ with $A C = A Q = x$ (1 mark)
- Uses $A B ⊥ O P$ to write $(12 - x)^{2} = x^{2} + 8^{2}$ (1 mark)
- Solves to get $x = 10/3$ cm (1 mark)
- $A B = 2 A C = 20/3$ cm (½ mark)
- $P A = 26/3$ cm (½ mark)
Open Q35 on its own page

Class Interval	0–10	10–20	20–30	30–40	40–50	50–60
Frequency	12	p	27	20	q	6

Class	$x_{i}$	$f_{i}$	$f_{i} x_{i}$
0–10	5	12	60
10–20	15	p	15p
20–30	25	27	675
30–40	35	20	700
40–50	45	q	45q
50–60	55	6	330
Total		$65 + p + q$	$1765 + 15 p + 45 q$

Class Interval	0–15	15–30	30–45	45–60	60–75	75–90	90–105
Frequency	15	10	12	9	8	10	6

Class	$f$	$c f$
0–15	15	15
15–30	10	25
30–45	12	37
45–60	9	46
60–75	8	54
75–90	10	64
90–105	6	70

Section E

Case study based; internal choice in the 2-mark subparts of each case

Q36 · 4 marks · Case-basedCompetency-basedPolynomials
An arch of a railway bridge, built on Chenab riverbed, is shown in the diagram. It is a parabolic arch connecting two hills at P and Q. If the parabolic curve is represented by the polynomial $p (x) = - 0.0025 x^{2} - 0.025 x + 136$ .
Observe the diagram and based on above information, answer the following questions:
1. (i)
  Write the co-ordinates of point A.
  1 mark
2. (ii)
  Find the span of the arch.
  1 mark
3. (iii)(a)
  Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials.
  2 marks
4. OR
  (iii)(b)
  Find the values of $p (x)$ at $x = 100$ and $x = - 100$ . Are they same?
  2 marks
Figure in the paper: Left: photograph of a steel arch railway bridge over a gorge. Right: schematic with axes X' (left) and X (right), origin O, and Y axis upward. A parabolic arch rises from Q at (–238.5, 0) on the left to P at (228.5, 0) on the right, with its peak A on the Y-axis at the top, under a horizontal 'Railway Track' line. Vertical support lines hang from the arch to the track. A dashed 'Riverbed' line lies below the X-axis. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) A lies on the y-axis, so $x = 0$ there.
  $p (0) = - 0.0025 (0)^{2} - 0.025 (0) + 136 = 136$
  So A = $(0, 136)$ .
  1 mark
2. (ii) The arch meets the x-axis at Q(–238.5, 0) and P(228.5, 0).
  Span $= QP = 228.5 - (- 238.5) = 467$ units.
  1 mark
3. (iii)(a) The curve cuts the x-axis at P and Q, so the zeroes of $p (x)$ are $228.5$ and $- 238.5$ .
  1 mark
4. (iii)(a) Sum of zeroes $= 228.5 + (- 238.5) = - 10$ .
  $- \frac{coefficient of x}{coefficient of x ^{2}} = - \frac{- 0.025}{- 0.0025} = - 10$
  Both are equal, so the relationship is verified.
  1 mark
5. (iii)(b) $p (100) = - 0.0025 (100)^{2} - 0.025 (100) + 136 = - 25 - 2.5 + 136 = 108.5$
  1 mark
6. (iii)(b) $p (- 100) = - 0.0025 (100)^{2} - 0.025 (- 100) + 136 = - 25 + 2.5 + 136 = 113.5$
  ½ mark
7. (iii)(b) $108.5 \neq = 113.5$ , so $p (100) \neq = p (- 100)$ . The values are not the same.
  ½ mark
Answer: (i) A(0, 136); (ii) 467 units; (iii)(a) zeroes 228.5 and −238.5, sum = −10 = −b/a, verified; (iii)(b) p(100) = 108.5, p(−100) = 113.5, not same
Where the marks are
- (i) A = (0, 136) (1 mark)
- (ii) Span = 238.5 + 228.5 = 467 units (1 mark)
- (iii)(a) Zeroes 228.5 and −238.5 read from the diagram (1 mark)
- (iii)(a) Sum = −10 = −(coefficient of x)/(coefficient of x²), verified (1 mark)
- (iii)(b) p(100) = 108.5 (1 mark)
- (iii)(b) p(−100) = 113.5 (½ mark)
- (iii)(b) Concludes the values are not the same (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter 7 cm is latched with a very thin rod. (Ignore the rod while making calculations.) Dimensions of the cuboid are 24 cm $\times$ 12 cm $\times$ 17 cm.
1. (i)
  Find the surface area of the bulb.
  1 mark
2. (ii)
  What could be the maximum diameter of the bulb if at least 1 cm space is left from each side?
  1 mark
3. (iii)(a)
  Find the area of the fabric used if there is a fold of 2 cm on top and bottom edges.
  2 marks
4. OR
  (iii)(b)
  Find the space available inside the lamp.
  2 marks
Figure in the paper: Left: photograph of a fabric wall lamp with a box-shaped shade and a base fixture on a wall. Right: a cuboid drawn in perspective, labelled 24 cm (top edge length), 12 cm (depth) and 17 cm (height). Open top and bottom; a circle (the spherical bulb) is drawn at the centre of the front face with a dashed vertical line through it. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Radius of bulb $r = \frac{7}{2}$ cm.
  Surface area $= 4 π r^{2} = 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 154$ cm².
  1 mark
2. (ii) The bulb must fit within the smallest dimension of the lamp, 12 cm, with at least 1 cm left on each side.
  Maximum diameter $= 12 - 2 \times 1 = 10$ cm.
  1 mark
3. (iii)(a) The lamp is open at top and bottom, so only the four side faces use fabric. A 2 cm fold at the top and at the bottom increases the height to $17 + 2 + 2 = 21$ cm.
  The effective dimensions are $24 cm \times 12 cm \times 21 cm$ .
  1 mark
4. (iii)(a) Fabric area = lateral surface area $= 2 (l + b) h = 2 \times (24 + 12) \times 21 = 1512$ cm².
  1 mark
5. (iii)(b) Space available = volume of cuboid − volume of bulb
  $= 24 \times 12 \times 17 - \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$
  1 mark
6. (iii)(b) $= 4896 - \frac{539}{3} = \frac{14688 - 539}{3} = \frac{14149}{3}$ cm³ $\approx 4716.33$ cm³
  1 mark
Answer: (i) 154 cm²; (ii) 10 cm; (iii)(a) 1512 cm²; (iii)(b) 14149/3 cm³ ≈ 4716.33 cm³
Where the marks are
- (i) Surface area of bulb = 4πr² = 154 cm² (1 mark)
- (ii) Maximum diameter = 12 − 2 = 10 cm (1 mark)
- (iii)(a) Height with folds = 21 cm, so dimensions 24 × 12 × 21 (1 mark)
- (iii)(a) Fabric area = 2 × 21 × (24 + 12) = 1512 cm² (1 mark)
- (iii)(b) Writes volume of cuboid minus volume of sphere with correct values (1 mark)
- (iii)(b) Space = 14149/3 cm³ ≈ 4716.3 cm³ (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is 54 metres away from the water tank.
From a window (W) of the building, the angle of elevation of top of the tank is $4 5^{\circ}$ and angle of depression of its foot is $3 0^{\circ}$ .
1. (i)
  Write a relation between d (the height of window) and y.
  1 mark
2. (ii)
  Determine the value of h.
  1 mark
3. (iii)(a)
  Determine height of the water tank.
  2 marks
4. OR
  (iii)(b)
  Find the value of $x$ and height of the window above ground level.
  2 marks
Figure in the paper: Right-angled figure: AB is the vertical water tank on the left, with B at top and A at the base; CD is the vertical building on the right, with C at base and D at top; AC = 54 metres is the horizontal ground distance. A window W is on the building's face. From W, a horizontal line WX meets the tank at X. The line WB (length x) makes $4 5^{\circ}$ with WX (angle of elevation to the top B), and the line WA (length y) makes $3 0^{\circ}$ with WX (angle of depression to the foot A). Labels: h = XB (tank part above X), d = AX (height of window level above the foot A). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $∠ X W A = ∠ W A C = 3 0^{\circ}$ (alternate angles, $X W ∥ A C$ ). In right $△ A X W$ , AX = d and AW = y.
  $sin 3 0^{\circ} = \frac{A X}{A W} \Rightarrow \frac{1}{2} = \frac{d}{y} \Rightarrow y = 2 d$
  1 mark
2. (ii) $W X = A C = 54$ m. In right $△ B X W$ :
  $tan 4 5^{\circ} = \frac{B X}{W X} \Rightarrow 1 = \frac{h}{54} \Rightarrow h = 54$ m
  1 mark
3. (iii)(a) In right $△ A X W$ :
  $tan 3 0^{\circ} = \frac{A X}{W X} \Rightarrow \frac{1}{3} = \frac{d}{54} \Rightarrow d = \frac{54}{3} = 183$ m
  1 mark
4. (iii)(a) Height of the tank $A B = h + d = (54 + 183)$ m $\approx 85.18$ m
  1 mark
5. (iii)(b) The height of the window above the ground is $W C$ . Since $∠ W A C = 3 0^{\circ}$ (alternate angles), in right $△ W C A$ :
  $tan 3 0^{\circ} = \frac{W C}{A C} \Rightarrow \frac{1}{3} = \frac{W C}{54} \Rightarrow W C = 183$ m $\approx 31.18$ m
  1 mark
6. (iii)(b) In right $△ B X W$ :
  $sin 4 5^{\circ} = \frac{h}{x} \Rightarrow \frac{1}{2} = \frac{54}{x} \Rightarrow x = 542$ m $\approx 76.37$ m
  1 mark
Answer: (i) y = 2d; (ii) h = 54 m; (iii)(a) (54 + 18√3) m; (iii)(b) x = 54√2 m, height of window = 18√3 m
Where the marks are
- (i) sin 30° = d/y gives y = 2d (1 mark)
- (ii) tan 45° = h/54 gives h = 54 m (1 mark)
- (iii)(a) tan 30° = d/54 gives d = 18√3 m (1 mark)
- (iii)(a) Height of tank = (54 + 18√3) m (1 mark)
- (iii)(b) Height of window WC = 18√3 m using tan 30° (1 mark)
- (iii)(b) x = 54√2 m using sin 45° (or cos 45°) (1 mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.