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

CBSE Class 10 Mathematics Standard: 2026 board paper, set 30/4/1

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 1171-1_30-4-1 (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
C	26-31	Short Answer
D	32-35	Long Answer
E	36-38	Case study based

Instructions in brief: All questions compulsory; no overall choice, internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 two-mark questions in Section E; calculator not allowed; take pi = 22/7 where required.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQQuadratic Equations
If the quadratic equation $9 x^{2} + 8 k x + 16 = 0$ has real and equal roots, then the value of k is
1. (A)3
2. (B)–3
3. (C)–4
4. (D) $\frac{3}{2}$
Try it first, then show our solutionSolution
1. Equal roots $\Rightarrow D = b^{2} - 4 a c = 0$ : $(8 k)^{2} - 4 (9) (16) = 0 \Rightarrow 64 k^{2} = 576 \Rightarrow k^{2} = 9$ .
2. So $k = \pm 3$ ; both 3 (A) and –3 (B) are listed, so either option is correct.
  1 mark
Answer: k = ±3; option (A) 3 (or (B) –3)
Where the marks are
- Uses $D = 0$ to get $k = \pm 3$ and chooses (A) or (B) (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQTriangles
It is given that $△ A B C \sim △ E D F$ . Which of the following is not true ?
1. (A) $\frac{Perimeter of △ A B C}{Perimeter of △ E D F} = \frac{A B}{E D}$
2. (B) $\frac{A B}{E D} = \frac{A C}{E F}$
3. (C) $∠ A = ∠ D, ∠ C = ∠ F$
4. (D) $\frac{A B + B C}{A C} = \frac{D E + D F}{E F}$
Try it first, then show our solutionSolution
1. $△ A B C \sim △ E D F$ matches A↔E, B↔D, C↔F, so $∠ A = ∠ E$ and $∠ B = ∠ D$ , not $∠ A = ∠ D$ .
2. Options (A), (B) and (D) follow from proportional corresponding sides, so (C) is the statement that is not true.
  1 mark
Answer: Option (C)
Where the marks are
- Identifies (C) as the false statement (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQArithmetic Progressions
The $n^{t h}$ term of an A.P. is $2 n + 1$ . Its common difference is
1. (A) $2$
2. (B) $2 n$
3. (C)1
4. (D) $2 + 1$
Try it first, then show our solutionSolution
1. $d = a_{n + 1} - a_{n} = [2 (n + 1) + 1] - [2 n + 1] = 2$ .
  1 mark
Answer: $2$ , option (A)
Where the marks are
- Finds $d = 2$ (1 mark)
Q4 · 1 mark · MCQCircles
If PQ and PR are tangents to the circle with centre O and radius 4 cm such that $∠ QP R = 9 0^{\circ}$ , then the length OP is
1. (A)4 cm
2. (B) $42$ cm
3. (C)8 cm
4. (D) $22$ cm
Figure in the paper: Circle with centre O; external point P; tangents PQ and PR touch the circle at Q and R; OQ, OR and OP are joined; angle QPR = 90°. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O QP = ∠ O R P = 9 0^{\circ}$ [tangent ⟂ radius] and $∠ QP R = 9 0^{\circ}$ , with $O Q = O R$ , so OQPR is a square of side 4 cm.
2. OP is its diagonal: $O P = 4^{2} + 4^{2} = 42$ cm.
  1 mark
Answer: $42$ cm, option (B)
Where the marks are
- Recognises square OQPR and gets $O P = 42$ cm (1 mark)
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Q5 · 1 mark · MCQPolynomials
Observe the graph of polynomial p(x). Number of zeroes of p(x) is
1. (A)5
2. (B)4
3. (C)6
4. (D)3
Figure in the paper: Graph of y = p(x) on XY axes: a cubic-like curve that rises to a maximum on the left of the y-axis, falls crossing the x-axis, reaches a minimum below the x-axis just right of the origin, then rises crossing the x-axis again; it cuts the x-axis at 3 points. See it in the official paper.
Try it first, then show our solutionSolution
1. Number of zeroes = number of points where the graph meets the x-axis. The curve cuts the x-axis at 3 points, so p(x) has 3 zeroes.
  1 mark
Answer: 3, option (D)
Where the marks are
- Counts 3 intersections with x-axis (1 mark)
Q6 · 1 mark · MCQTriangles
In the given figure, DE || BC. If $\frac{A D}{D B} = \frac{1}{3}$ and AC = 6 cm, then length AE is
1. (A)1.5 cm
2. (B)1 cm
3. (C)2 cm
4. (D)3 cm
Figure in the paper: Triangle ABC with D on AB and E on AC such that DE is parallel to BC. See it in the official paper.
Try it first, then show our solutionSolution
1. By BPT, $\frac{A E}{E C} = \frac{A D}{D B} = \frac{1}{3}$ , so $A E = \frac{1}{4} A C = \frac{1}{4} \times 6 = 1.5$ cm.
  1 mark
Answer: 1.5 cm, option (A)
Where the marks are
- Applies BPT to get AE = 1.5 cm (1 mark)
Q7 · 1 mark · MCQCoordinate Geometry
In the given figure, a circle is centred at (1, 2). The diameter of the circle is
1. (A)4
2. (B) $22$
3. (C) $5$
4. (D) $25$
Figure in the paper: Circle with centre (1, 2) passing through the origin O; a diameter is drawn from a point on the y-axis to a point on the x-axis, passing through the centre; the circle touches/passes through O. See it in the official paper.
Try it first, then show our solutionSolution
1. The circle passes through O(0,0), so radius $= (1 - 0)^{2} + (2 - 0)^{2} = 5$ .
2. Diameter $= 25$ .
  1 mark
Answer: $25$ , option (D)
Where the marks are
- Radius $5$ by distance formula, diameter $25$ (1 mark)
Q8 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of k for which the system of linear equations $\frac{x}{2} + \frac{y}{3} = 5$ and $2 x + k y = 7$ is inconsistent, is
1. (A) $\frac{3}{4}$
2. (B) $\frac{4}{3}$
3. (C) $\frac{1}{3}$
4. (D)3
Try it first, then show our solutionSolution
1. Inconsistent: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , i.e. $\frac{1/2}{2} = \frac{1/3}{k}$ .
2. $\frac{1}{4} = \frac{1}{3 k} \Rightarrow k = \frac{4}{3}$ ; and $\frac{1}{4} \neq = \frac{5}{7}$ , so the system is inconsistent.
  1 mark
Answer: $k = \frac{4}{3}$ , option (B)
Where the marks are
- Uses ratio condition to get $k = 4/3$ (1 mark)
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Q9 · 1 mark · MCQAreas Related to Circles
A circle is divided into 16 identical sectors. If radius of the circle is 7 cm, area of each sector is
1. (A) $\frac{77}{4}$ cm $^{2}$
2. (B)77 cm $^{2}$
3. (C)154 cm $^{2}$
4. (D) $\frac{77}{8}$ cm $^{2}$
Try it first, then show our solutionSolution
1. Area of circle $= \frac{22}{7} \times 7^{2} = 154$ cm²; each sector $= \frac{154}{16} = \frac{77}{8}$ cm².
  1 mark
Answer: $\frac{77}{8}$ cm², option (D)
Where the marks are
- Divides circle area 154 by 16 (1 mark)
Q10 · 1 mark · MCQProbability
Two different dice are rolled together. The probability that both the obtained numbers are less than 4, is
1. (A) $\frac{2}{9}$
2. (B) $\frac{7}{36}$
3. (C) $\frac{1}{4}$
4. (D) $\frac{2}{3}$
Try it first, then show our solutionSolution
1. Total outcomes = 36. Each die must show 1, 2 or 3: favourable = $3 \times 3 = 9$ .
2. $P = \frac{9}{36} = \frac{1}{4}$ .
  1 mark
Answer: $\frac{1}{4}$ , option (C)
Where the marks are
- 9 favourable out of 36 gives 1/4 (1 mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
When $sin A = \frac{1}{3}$ , the value of $cot A$ is
1. (A) $\frac{2 2}{3}$
2. (B) $22$
3. (C) $\frac{1}{2 2}$
4. (D)3
Try it first, then show our solutionSolution
1. $cos A = 1 - \frac{1}{9} = \frac{2 2}{3}$ .
2. $cot A = \frac{c o s A}{s i n A} = \frac{2 2 /3}{1/3} = 22$ .
  1 mark
Answer: $22$ , option (B)
Where the marks are
- Finds $cos A$ and $cot A = 22$ (1 mark)
Q12 · 1 mark · MCQIntroduction to Trigonometry
$\frac{1 + t a n ^{2} A}{1 + c o t ^{2} A}$ equals to :
1. (A) $tan^{2} A$
2. (B)–1
3. (C) $- tan^{2} A$
4. (D) $cot^{2} A$
Try it first, then show our solutionSolution
1. $\frac{1 + t a n ^{2} A}{1 + c o t ^{2} A} = \frac{s e c ^{2} A}{c s c ^{2} A} = \frac{s i n ^{2} A}{c o s ^{2} A} = tan^{2} A$ .
  1 mark
Answer: $tan^{2} A$ , option (A)
Where the marks are
- Uses identities to get $tan^{2} A$ (1 mark)
Q13 · 1 mark · MCQCompetency-basedSurface Areas and Volumes
Three tennis balls are just packed in a cylindrical jar. If radius of each ball is r, volume of air inside the jar is
1. (A) $2 π r^{3}$
2. (B) $3 π r^{3}$
3. (C) $5 π r^{3}$
4. (D) $4 π r^{3}$
Figure in the paper: A tall cylindrical jar containing three tennis balls stacked vertically, each touching the jar's sides, top and bottom. See it in the official paper.
Try it first, then show our solutionSolution
1. Cylinder: radius r, height 6r, volume $= π r^{2} (6 r) = 6 π r^{3}$ . Three balls: $3 \times \frac{4}{3} π r^{3} = 4 π r^{3}$ .
2. Air $= 6 π r^{3} - 4 π r^{3} = 2 π r^{3}$ .
  1 mark
Answer: $2 π r^{3}$ , option (A)
Where the marks are
- Cylinder volume minus ball volumes = $2 π r^{3}$ (1 mark)
Q14 · 1 mark · MCQCircles
PQ is tangent to a circle with centre O. If $∠ P O R = 6 5^{\circ}$ , then $m ∠ P T R$ is
1. (A)65°
2. (B)58.5°
3. (C)57.5°
4. (D)45°
Figure in the paper: Circle with centre O; tangent line PQ touches circle at P; S is diametrically opposite P (SP is a diameter); R is a point on the circle; angle POR = 65°; line SR extended meets the tangent at T (T lies between P and Q); angle PTR is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ P S R = \frac{1}{2} ∠ P O R = 32. 5^{\circ}$ [angle at centre is twice angle at circumference]. $∠ S P T = 9 0^{\circ}$ [tangent ⟂ radius].
2. In $△ S P T$ : $∠ P T R = 18 0^{\circ} - 9 0^{\circ} - 32. 5^{\circ} = 57. 5^{\circ}$ .
  1 mark
Answer: 57.5°, option (C)
Where the marks are
- Gets $∠ P S R = 32. 5^{\circ}$ and angle sum gives 57.5° (1 mark)
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Q15 · 1 mark · MCQCoordinate Geometry
The distance between the points $(a cos θ + b sin θ, 0)$ and $(0, a sin θ - b cos θ)$ is
1. (A) $a^{2} + b^{2}$
2. (B) $a^{2} - b^{2}$
3. (C) $a^{2} - b^{2}$
4. (D) $a^{2} + b^{2}$
Try it first, then show our solutionSolution
1. $d^{2} = (a cos θ + b sin θ)^{2} + (a sin θ - b cos θ)^{2} = a^{2} (cos^{2} θ + sin^{2} θ) + b^{2} (sin^{2} θ + cos^{2} θ) = a^{2} + b^{2}$ (cross terms cancel).
2. $d = a^{2} + b^{2}$ .
  1 mark
Answer: $a^{2} + b^{2}$ , option (A)
Where the marks are
- Distance formula with cancellation gives $a^{2} + b^{2}$ (1 mark)
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Q16 · 1 mark · MCQCompetency-basedSurface Areas and Volumes
An ice-cream cone of radius r and height h is completely filled by two spherical scoopes of ice-cream. If radius of each spherical scoop is $\frac{r}{2}$ , then h : 2r equals
1. (A)1 : 8
2. (B)1 : 2
3. (C)1 : 1
4. (D)2 : 1
Try it first, then show our solutionSolution
1. $\frac{1}{3} π r^{2} h = 2 \times \frac{4}{3} π (\frac{r}{2})^{3} = \frac{1}{3} π r^{3} \Rightarrow h = r$ .
2. $h : 2 r = r : 2 r = 1 : 2$ .
  1 mark
Answer: 1 : 2, option (B)
Where the marks are
- Equates volumes to get h = r, ratio 1 : 2 (1 mark)
Q17 · 1 mark · MCQAreas Related to Circles
Arc PQ subtends an angle $θ$ at the centre of the circle with radius 6.3 cm. If $P Q$ = 11 cm, then the value of $θ$ is
1. (A)10°
2. (B)60°
3. (C)45°
4. (D)100°
Try it first, then show our solutionSolution
1. $\frac{θ}{360} \times 2 \times \frac{22}{7} \times 6.3 = 11 \Rightarrow \frac{θ}{360} \times 39.6 = 11$ .
2. $θ = \frac{11 \times 360}{39.6} = 10 0^{\circ}$ .
  1 mark
Answer: 100°, option (D)
Where the marks are
- Arc length formula gives θ = 100° (1 mark)
Q18 · 1 mark · MCQStatistics
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is
1. (A)34
2. (B)43
3. (C)38.5
4. (D)41.5
Try it first, then show our solutionSolution
1. Empirical relation: Mode = 3 Median − 2 Mean = 3(40) − 2(43) = 120 − 86 = 34.
  1 mark
Answer: 34, option (A)
Where the marks are
- Uses 3 Median − 2 Mean = 34 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A) : If probability of happening of an event is 0.2p, p > 0, then p can’t be more than 5.
Reason (R) : $P (\overline{E}) = 1 - P (E)$ for an event E.
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 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. A: $0 \leq 0.2 p \leq 1 \Rightarrow p \leq 5$ , so A is true.
2. R is a true fact, but A follows from $0 \leq P (E) \leq 1$ , not from the complement rule; so R does not explain A. Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Both true, R not the explanation (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A) : $(3 + 5)$ is an irrational number.
Reason (R) : Sum of the any two irrational numbers is always irrational.
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 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. A is true: $3 + 5$ is irrational (if it were rational r, squaring gives $15$ rational, a contradiction).
2. R is false: e.g. $2 + (- 2) = 0$ is rational. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- A true, R false via counterexample (1 mark)
Open Q20 on its own page

Section B

Very Short Answer

Q21 · 2 marks · Very short answerQuadratic Equations
Verify that roots of the quadratic equation $(p - q) x^{2} + (q - r) x + (r - p) = 0$ are equal when $q + r = 2 p$ .
Try it first, then show our solutionSolution
1. Compare with $a x^{2} + b x + c = 0$ : $a = p - q$ , $b = q - r$ , $c = r - p$ . The discriminant is $D = b^{2} - 4 a c = (q - r)^{2} - 4 (p - q) (r - p)$ .
  ½ mark
2. Expand: $D = q^{2} - 2 q r + r^{2} - 4 (p r - p^{2} - q r + pq) = q^{2} + 2 q r + r^{2} - 4 p (q + r) + 4 p^{2} = (q + r - 2 p)^{2}$ .
  ½ mark
3. Put $q + r = 2 p$ : $D = (2 p - 2 p)^{2} = 0$ .
  ½ mark
4. Since $D = 0$ , the roots of the given equation are real and equal. Hence verified.
  ½ mark
Answer: $D = (q + r - 2 p)^{2} = 0$ when $q + r = 2 p$ , so the roots are equal.
Where the marks are
- Writes $D = (q - r)^{2} - 4 (p - q) (r - p)$ (½ mark)
- Simplifies to $(q + r - 2 p)^{2}$ (½ mark)
- Substitutes $q + r = 2 p$ to get $D = 0$ (½ mark)
- Concludes that the roots are equal (½ mark)
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Q22 · 2 marks · Very short answerTriangles
D is a point on the side BC of $△ A B C$ such that $∠ C A B = ∠ C D A$ . Show that $C A^{2} = C B \times C D$ .
Figure in the paper: Triangle ABC with D on side BC; AD is joined; angle CAB (at A) and angle CDA (at D) are marked equal. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A D C$ and $△ B A C$ : $∠ C D A = ∠ C A B$ (given) and $∠ A C D = ∠ B C A$ (common angle). So $△ A D C \sim △ B A C$ [AA similarity].
  1 mark
2. Corresponding sides are proportional: $\frac{D C}{A C} = \frac{A C}{B C}$ .
  ½ mark
3. Cross-multiplying, $A C^{2} = B C \times D C$ , i.e. $C A^{2} = C B \times C D$ . Hence shown.
  ½ mark
Answer: $△ A D C \sim △ B A C$ (AA), so $C A^{2} = C B \times C D$ .
Where the marks are
- Proves $△ A D C \sim △ B A C$ by AA with reasons (1 mark)
- Writes the correct ratio $D C / A C = A C / B C$ (½ mark)
- Concludes $C A^{2} = C B \times C D$ (½ mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerIntroduction to Trigonometry
Prove that : $\frac{1 + s i n A}{1 - s i n A} = sec A + tan A$
Try it first, then show our solutionSolution
1. LHS $= \frac{1 + sin A}{1 - sin A} \times \frac{1 + sin A}{1 + sin A}$ (rationalising).
  ½ mark
2. $= \frac{( 1 + sin A ) ^{2}}{1 - sin ^{2} A} = \frac{1 + sin A}{1 - sin ^{2} A}$
  ½ mark
3. $= \frac{1 + sin A}{cos A}$ [since $1 - sin^{2} A = cos^{2} A$ ]
  ½ mark
4. $= \frac{1}{cos A} + \frac{sin A}{cos A} = sec A + tan A =$ RHS. Hence proved.
  ½ mark
Answer: LHS $= \frac{1 + sin A}{cos A} = sec A + tan A =$ RHS
Where the marks are
- Multiplies numerator and denominator by $1 + sin A$ (½ mark)
- Gets $\frac{1 + s i n A}{1 - s i n ^{2} A}$ (½ mark)
- Uses $1 - sin^{2} A = cos^{2} A$ (½ mark)
- Splits into $sec A + tan A$ (½ mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $\frac{3 c o s ^{2} 3 0 ^{\circ} - 6 c s c ^{2} 3 0 ^{\circ}}{t a n ^{2} 6 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. Use $cos 3 0^{\circ} = \frac{3}{2}$ , $csc 3 0^{\circ} = 2$ , $tan 6 0^{\circ} = 3$ : expression $= \frac{3 ( \frac{3}{2} ) ^{2} - 6 ( 2 ) ^{2}}{( 3 ) ^{2}} = \frac{\frac{9}{4} - 24}{3}$
  1½ marks
2. $= \frac{- \frac{87}{4}}{3} = - \frac{87}{12} = - \frac{29}{4}$
  ½ mark
Answer: $- \frac{29}{4}$
Where the marks are
- Substitutes correct values of $cos 3 0^{\circ}$ , $csc 3 0^{\circ}$ , $tan 6 0^{\circ}$ (1½ marks)
- Simplifies to $- \frac{29}{4}$ (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerReal Numbers
Prove that $2 + 35$ is an irrational number given that $5$ is irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $2 + 35$ is rational. Then $2 + 35 = \frac{p}{q}$ , where $p, q$ are integers and $q \neq = 0$ .
  ½ mark
2. Rearranging: $35 = \frac{p}{q} - 2$ , so $5 = \frac{p - 2 q}{3 q}$ .
  ½ mark
3. Since $p, q$ are integers, $\frac{p - 2 q}{3 q}$ is rational, so $5$ would be rational. This contradicts the given fact that $5$ is irrational. So the assumption is wrong, and $2 + 35$ is irrational.
  1 mark
Answer: $2 + 35$ is irrational (proved by contradiction).
Where the marks are
- Assumes $2 + 35 = \frac{p}{q}$ , $q \neq = 0$ (½ mark)
- Expresses $5 = \frac{p - 2 q}{3 q}$ (½ mark)
- States the contradiction and concludes (1 mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerReal Numbers
If the HCF of 210 and 55 is expressed as $210 \times 5 + 55 m$ , then find the value of m.
Try it first, then show our solutionSolution
1. $210 = 2 \times 3 \times 5 \times 7$ and $55 = 5 \times 11$ , so HCF $(210, 55) = 5$ .
  1 mark
2. So $5 = 210 \times 5 + 55 m$ , i.e. $5 = 1050 + 55 m$ .
  ½ mark
3. $55 m = 5 - 1050 = - 1045 \Rightarrow m = - 19$ .
  ½ mark
Answer: $m = - 19$
Where the marks are
- Finds HCF = 5 by prime factorisation (1 mark)
- Forms $5 = 210 \times 5 + 55 m$ (½ mark)
- Gets $m = - 19$ (½ mark)
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Q25 · 2 marks · Very short answerPolynomials
$α$ and $β$ are the zeroes of the polynomial $5 x^{2} - 16 x - 10$ . Find the value of $\frac{α}{β} + \frac{β}{α}$ .
Try it first, then show our solutionSolution
1. For $5 x^{2} - 16 x - 10$ : $α + β = - \frac{b}{a} = \frac{16}{5}$ and $α β = \frac{c}{a} = \frac{- 10}{5} = - 2$ .
  1 mark
2. $\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β} = \frac{( α + β ) ^{2} - 2 α β}{α β} = \frac{\frac{256}{25} + 4}{- 2}$
  ½ mark
3. $= \frac{\frac{356}{25}}{- 2} = - \frac{178}{25}$
  ½ mark
Answer: $- \frac{178}{25}$
Where the marks are
- Writes $α + β = \frac{16}{5}$ and $α β = - 2$ (1 mark)
- Uses $\frac{( α + β ) ^{2} - 2 α β}{α β}$ with substitution (½ mark)
- Gets $- \frac{178}{25}$ (½ mark)
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Section C

Short Answer

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
In a class test, Veer scored 6 more than twice as many marks as Kevin scored. If one of them had scored 4 more marks, their total score would have been 40. Find the marks obtained by Veer and Kevin.
Try it first, then show our solutionSolution
1. Let Veer's marks be $V$ and Kevin's marks be $K$ . Veer scored 6 more than twice Kevin's marks: $V = 2 K + 6$ ...(i)
  1 mark
2. If one of them had scored 4 more, the total would be 40: $V + K + 4 = 40$ , i.e. $V + K = 36$ ...(ii)
  1 mark
3. Substituting (i) in (ii): $2 K + 6 + K = 36 \Rightarrow 3 K = 30 \Rightarrow K = 10$ . Then $V = 2 (10) + 6 = 26$ . So Veer obtained 26 marks and Kevin obtained 10 marks.
  1 mark
Answer: Veer = 26 marks, Kevin = 10 marks
Where the marks are
- Forms $V = 2 K + 6$ (1 mark)
- Forms $V + K + 4 = 40$ (1 mark)
- Solves to get $V = 26$ and $K = 10$ (½ + ½ marks)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve the linear equations $3 x + y = 14$ and $y = 2$ graphically.
Try it first, then show our solutionSolution
1. For $3 x + y = 14$ , i.e. $y = 14 - 3 x$ : table of values $x = 3, y = 5$ ; $x = 4, y = 2$ ; $x = 5, y = - 1$ . Plot $(3, 5)$ , $(4, 2)$ , $(5, - 1)$ and join them to get a straight line.
  1½ marks
2. $y = 2$ is a straight line parallel to the x-axis through $(0, 2)$ . Points: $(0, 2)$ , $(2, 2)$ , $(4, 2)$ . Draw it on the same axes.
  ½ mark
3. The two lines intersect at $(4, 2)$ . Hence the solution is $x = 4$ , $y = 2$ .
  1 mark
Answer: $x = 4$ , $y = 2$
Where the marks are
- Correct table and graph of $3 x + y = 14$ (1½ marks)
- Correct graph of $y = 2$ (½ mark)
- Reads the intersection point: $x = 4, y = 2$ (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerAreas Related to Circles
Chord AB of a circle with centre O and radius 21 mm subtends an angle of $12 0^{\circ}$ at the centre. Find the perimeters of the shaded region. (Use $3$ = 1.73)
Figure in the paper: Circle with centre O; radii OA and OB with angle AOB = 120°; chord AB drawn; the minor segment between chord AB and the arc AB is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. The perimeter of the shaded segment = length of chord AB + length of minor arc AB. Draw $O D ⊥ A B$ . The perpendicular from the centre bisects the chord and $∠ A O B$ , so $∠ A O D = 6 0^{\circ}$ . In right $△ O D A$ : $sin 6 0^{\circ} = \frac{A D}{O A} \Rightarrow \frac{3}{2} = \frac{A D}{21} \Rightarrow A D = \frac{21 3}{2}$ mm.
  1 mark
2. $A B = 2 A D = 213 = 21 \times 1.73 = 36.33$ mm.
  ½ mark
3. Length of minor arc AB $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{120}{360} \times 2 \times \frac{22}{7} \times 21 = 44$ mm.
  1 mark
4. Perimeter of the shaded region $= 44 + 36.33 = 80.33$ mm.
  ½ mark
Answer: 80.33 mm
Where the marks are
- Draws the perpendicular from O and finds $A D = \frac{21 3}{2}$ mm using $sin 6 0^{\circ}$ (1 mark)
- Gets chord $A B = 213$ mm (½ mark)
- Finds arc length = 44 mm (1 mark)
- Adds to get perimeter 80.33 mm (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{s i n θ - c o s θ + 1}{s i n θ + c o s θ - 1} = \frac{1}{s e c θ - t a n θ}$
Try it first, then show our solutionSolution
1. Divide the numerator and denominator of the LHS by $cos θ$ : LHS $= \frac{tan θ - 1 + sec θ}{tan θ + 1 - sec θ}$ .
  1 mark
2. Write $1 = sec^{2} θ - tan^{2} θ$ in the numerator: $= \frac{( sec θ + tan θ ) - ( sec ^{2} θ - tan ^{2} θ )}{tan θ + 1 - sec θ}$
  ½ mark
3. $= \frac{( sec θ + tan θ ) ( 1 - sec θ + tan θ )}{tan θ + 1 - sec θ} = sec θ + tan θ$
  ½ mark
4. Multiply and divide by $(sec θ - tan θ)$ : $= \frac{( sec θ + tan θ ) ( sec θ - tan θ )}{sec θ - tan θ} = \frac{sec ^{2} θ - tan ^{2} θ}{sec θ - tan θ}$
  ½ mark
5. $= \frac{1}{sec θ - tan θ} =$ RHS [since $sec^{2} θ - tan^{2} θ = 1$ ]. Hence proved.
  ½ mark
Answer: LHS $= sec θ + tan θ = \frac{1}{sec θ - tan θ} =$ RHS
Where the marks are
- Divides by $cos θ$ to get $\frac{t a n θ - 1 + s e c θ}{t a n θ + 1 - s e c θ}$ (1 mark)
- Replaces 1 by $sec^{2} θ - tan^{2} θ$ (½ mark)
- Factorises and cancels to get $sec θ + tan θ$ (½ mark)
- Multiplies and divides by $sec θ - tan θ$ (½ mark)
- Uses the identity to reach RHS (½ mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCompetency-basedSurface Areas and Volumes
To protect plants from heat, a shed of iron rods covered with green cloth is made. The lower part of the shed is a cuboid mounted by semi-cylinder as shown in the figure. Find the area of the cloth required to make this shed, if dimensions of the cuboid are 14 m × 25 m × 16 m.
Figure in the paper: A shed: cuboid with a semi-cylindrical roof on top. Front face width 14 m (semicircle of diameter 14 m on top), length 25 m, height of cuboid 16 m. See it in the official paper.
Try it first, then show our solutionSolution
1. Cloth for the four walls of the cuboid (no floor, no top) $= 2 (l + b) h = 2 (14 \times 16 + 25 \times 16) = 2 (224 + 400) = 1248 m^{2}$ .
  1 mark
2. Radius of semi-cylinder $r = \frac{14}{2} = 7$ m, length $h = 25$ m. Cloth for the semi-cylindrical part = curved surface of half-cylinder + two semicircular ends $= π r h + 2 \times \frac{1}{2} π r^{2} = \frac{22}{7} \times 7 \times 25 + \frac{22}{7} \times 7^{2}$
  1 mark
3. $= 550 + 154 = 704 m^{2}$
  ½ mark
4. Total cloth required $= 1248 + 704 = 1952 m^{2}$ .
  ½ mark
Answer: 1952 m²
Where the marks are
- Area of four walls = 1248 m² (1 mark)
- Correct expression $π r h + π r^{2}$ for semi-cylinder with r = 7 m (1 mark)
- Evaluates to 704 m² (½ mark)
- Total cloth = 1952 m² (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerSurface Areas and Volumes
The internal and external radii of a hollow hemisphere are $52$ cm and 10 cm respectively. A cone of height $57$ cm and radius $52$ cm is surmounted on the hemisphere as shown in the figure. Find the total surface area of the object in terms of $π$ . (Use $2$ = 1.4)
Figure in the paper: A hollow hemisphere (shaded flat annular rim between internal and external radii) with a cone placed on its top, the cone's base lying inside the hemisphere's opening. See it in the official paper.
Try it first, then show our solutionSolution
1. Let internal radius $r_{1} = 52$ cm (= radius of cone) and external radius $r_{2} = 10$ cm. Slant height of the cone $l = r_{1}^{2} + h^{2} = (52)^{2} + (57)^{2} = 50 + 175 = 225 = 15$ cm.
  1 mark
2. TSA of object = outer curved surface of hemisphere + curved surface of cone + area of the annular rim $= 2 π r_{2}^{2} + π r_{1} l + π (r_{2}^{2} - r_{1}^{2})$ $= π (2 \times 1 0^{2} + 52 \times 15 + 1 0^{2} - (52)^{2})$
  1 mark
3. $= π (200 + 75 \times 1.4 + 100 - 50) = π (200 + 105 + 50) = 355 π cm^{2}$
  1 mark
Answer: $355 π$ cm²
Where the marks are
- Finds slant height $l = 15$ cm (1 mark)
- Writes TSA $= 2 π r_{2}^{2} + π r_{1} l + π (r_{2}^{2} - r_{1}^{2})$ with substitution (1 mark)
- Evaluates to $355 π$ cm² (1 mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerProbability
A bag contains 30 balls out of which ‘m’ number of balls are blue in colour.
1. (i)
  Find the probability that a ball drawn at random from the bag is not blue.
  0 marks
2. (ii)
  If 6 more blue balls are added in the bag, then the probability of drawing a blue ball will be $\frac{5}{4}$ times the probability of drawing a blue ball in the first case. Find the value of m.
  0 marks
Try it first, then show our solutionSolution
1. (i) Number of balls that are not blue $= 30 - m$ . P(not blue) $= \frac{30 - m}{30}$ (or $1 - \frac{m}{30}$ ).
  1 mark
2. (ii) After adding 6 blue balls: total balls $= 36$ , blue balls $= m + 6$ . So P(blue) $= \frac{m + 6}{36}$ .
  1 mark
3. Given: $\frac{m + 6}{36} = \frac{5}{4} \times \frac{m}{30} = \frac{m}{24}$ .
  ½ mark
4. $24 (m + 6) = 36 m \Rightarrow 24 m + 144 = 36 m \Rightarrow 12 m = 144 \Rightarrow m = 12$ .
  ½ mark
Answer: (i) $\frac{30 - m}{30}$ ; (ii) $m = 12$
Where the marks are
- P(not blue) $= \frac{30 - m}{30}$ (1 mark)
- New P(blue) $= \frac{m + 6}{36}$ (1 mark)
- Forms equation $\frac{m + 6}{36} = \frac{5}{4} \times \frac{m}{30}$ (½ mark)
- Solves to get $m = 12$ (½ mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerReal Numbers
Find the greatest number less than 10,000 which is exactly divisible by 48, 60 and 65.
Try it first, then show our solutionSolution
1. Prime factorisation: $48 = 2^{4} \times 3$ , $60 = 2^{2} \times 3 \times 5$ , $65 = 5 \times 13$ .
  1½ marks
2. LCM $(48, 60, 65) = 2^{4} \times 3 \times 5 \times 13 = 3120$ .
  1 mark
3. Any number divisible by all three is a multiple of 3120: $3120 \times 3 = 9360 < 10000$ , but $3120 \times 4 = 12480 > 10000$ . So the required number is $9360$ .
  ½ mark
Answer: 9360
Where the marks are
- Correct prime factorisation of 48, 60 and 65 (1½ marks)
- LCM = 3120 (1 mark)
- Greatest multiple below 10,000 = 9360 (½ mark)
Open Q31 on its own page

Section D

Long Answer

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
A person on tour has ₹ 5,400 for his expenses. If he extends his tour by 5 days, he has to cut down his daily expenses by ₹ 180. Find the original duration of the tour and daily expense.
Try it first, then show our solutionSolution
1. Let the original duration be $x$ days. Daily expense $= ₹ \frac{5400}{x}$ ; if extended to $x + 5$ days, daily expense $= ₹ \frac{5400}{x + 5}$ . Given: $\frac{5400}{x} - \frac{5400}{x + 5} = 180$ .
  2 marks
2. $\frac{5400 \times 5}{x ( x + 5 )} = 180 \Rightarrow x (x + 5) = 150 \Rightarrow x^{2} + 5 x - 150 = 0$ .
  1 mark
3. $x^{2} + 15 x - 10 x - 150 = 0 \Rightarrow (x + 15) (x - 10) = 0 \Rightarrow x = - 15$ or $x = 10$ .
  1 mark
4. Number of days cannot be negative, so $x = 10$ .
  ½ mark
5. Original duration = 10 days and daily expense $= \frac{5400}{10} = ₹540$ .
  ½ mark
Answer: Original duration = 10 days; daily expense = ₹540
Where the marks are
- Forms $\frac{5400}{x} - \frac{5400}{x + 5} = 180$ (2 marks)
- Reduces to $x^{2} + 5 x - 150 = 0$ (1 mark)
- Factorises to $(x + 15) (x - 10) = 0$ (1 mark)
- Rejects negative root, $x = 10$ (½ mark)
- States daily expense ₹540 (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerCompetency-basedQuadratic Equations
The total cost of certain piece of cloth was ₹ 2,100. During special sale time, the shopkeeper offered 2 m extra cloth for free thus reducing the price of cloth per metre by ₹ 120. What was the original per metre price of cloth and its length ?
Try it first, then show our solutionSolution
1. Let the original length be $x$ m. Original price per metre $= ₹ \frac{2100}{x}$ ; with 2 m extra, price per metre $= ₹ \frac{2100}{x + 2}$ . Given: $\frac{2100}{x} - \frac{2100}{x + 2} = 120$ .
  2 marks
2. $\frac{2100 \times 2}{x ( x + 2 )} = 120 \Rightarrow x (x + 2) = 35 \Rightarrow x^{2} + 2 x - 35 = 0$ .
  1 mark
3. $x^{2} + 7 x - 5 x - 35 = 0 \Rightarrow (x + 7) (x - 5) = 0 \Rightarrow x = - 7$ or $x = 5$ .
  1 mark
4. Length cannot be negative, so $x = 5$ .
  ½ mark
5. Original length = 5 m and original price per metre $= \frac{2100}{5} = ₹420$ .
  ½ mark
Answer: Original price = ₹420 per metre; original length = 5 m
Where the marks are
- Forms $\frac{2100}{x} - \frac{2100}{x + 2} = 120$ (2 marks)
- Reduces to $x^{2} + 2 x - 35 = 0$ (1 mark)
- Factorises to $(x + 7) (x - 5) = 0$ (1 mark)
- Rejects negative root, $x = 5$ (½ mark)
- States price ₹420 per metre (½ mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerCircles
In the given figure, TP and TQ are tangents to a circle with centre M, touching another circle with centre N at A and B respectively. It is given that MQ = 13 cm, NB = 8 cm, BQ = 35 cm and TP = 80 cm.
1. (i)
  Name the quadrilateral MQBN.
  1 mark
2. (ii)
  Is MN parallel to PA ? Justify your answer.
  1 mark
3. (iii)
  Find length TB.
  1 mark
4. (iv)
  Find length MN.
  2 marks
Figure in the paper: A larger circle with centre M and a smaller circle with centre N. Two lines from T: tangent line TQ along the bottom touching the big circle at Q and small circle at B; and the other tangent line T-A-P touching the small circle at A and the big circle at P. MQ = 13 (perpendicular to TQ), NB = 8 (perpendicular to TQ), QB = 35 cm, MP and NA are radii perpendicular to the upper tangent. M and N are joined; a dashed horizontal line from M's level toward N is drawn with right angle marks. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $M Q ⊥ QB$ and $N B ⊥ QB$ [radius ⟂ tangent], so $M Q ∥ N B$ . Since $M Q = 13 \neq = N B = 8$ , MN is not parallel to QB. Exactly one pair of opposite sides is parallel, so MQBN is a trapezium (a right trapezium).
  1 mark
2. (ii) No. $M P ⊥ P A$ and $N A ⊥ P A$ [radius ⟂ tangent], so $M P ∥ N A$ . If MN were parallel to PA, MPAN would be a rectangle and $M P = N A$ ; but $M P = 13$ cm $\neq = N A = 8$ cm. Hence MN is not parallel to PA.
  1 mark
3. (iii) $T Q = T P = 80$ cm [tangents from an external point are equal]. So $T B = T Q - B Q = 80 - 35 = 45$ cm.
  1 mark
4. (iv) Draw $N S ⊥ M Q$ meeting MQ at S. NSQB is a rectangle, so $N S = B Q = 35$ cm and $S Q = N B = 8$ cm, giving $M S = 13 - 8 = 5$ cm. In right $△ M S N$ : $M N^{2} = N S^{2} + M S^{2} = 3 5^{2} + 5^{2}$
  1 mark
5. $= 1225 + 25 = 1250$
  ½ mark
6. $M N = 1250 = 252$ cm.
  ½ mark
Answer: (i) Trapezium; (ii) No, since MP ≠ NA; (iii) TB = 45 cm; (iv) MN = $252$ cm
Where the marks are
- (i) Identifies MQBN as a trapezium with reason (1 mark)
- (ii) Says No, justified by $M P \neq = N A$ (1 mark)
- (iii) Uses $T Q = T P = 80$ cm to get TB = 45 cm (1 mark)
- (iv) Applies Pythagoras with $N S = 35$ , $M S = 5$ (1 mark)
- (iv) $M N^{2} = 1250$ (½ mark)
- (iv) $M N = 252$ cm (½ mark)
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Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A kite is flying at a height of 60 m above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as $3 0^{\circ}$ . From the bottom of the same building, the angle of elevation of kite is $4 5^{\circ}$ . Find the length of the string and height of roof from the ground. (Use $3$ = 1.73)
Try it first, then show our solutionSolution
1. Figure: let K be the kite and G the point on the ground directly below it, so KG = 60 m. TR is the building, with T its bottom and R the roof where Ravi stands. Draw RS ⟂ KG with S on KG, so SG = TR and SR = GT. The angle of elevation of K from R is $∠ K R S = 3 0^{\circ}$ , and from T it is $∠ K T G = 4 5^{\circ}$ .
  1 mark
2. In right $△ K GT$ : $tan 4 5^{\circ} = \frac{K G}{GT} \Rightarrow 1 = \frac{60}{GT} \Rightarrow GT = 60$ m, so $S R = 60$ m.
  1 mark
3. In right $△ K S R$ : $tan 3 0^{\circ} = \frac{K S}{S R} \Rightarrow \frac{1}{3} = \frac{K S}{60}$
  1 mark
4. $K S = \frac{60}{3} = 203 = 20 \times 1.73 = 34.6$ m
  ½ mark
5. Height of roof $T R = S G = K G - K S = 60 - 34.6 = 25.4$ m.
  ½ mark
6. Length of string KR: $sin 3 0^{\circ} = \frac{K S}{K R} \Rightarrow \frac{1}{2} = \frac{20 3}{K R}$
  ½ mark
7. $K R = 403 = 40 \times 1.73 = 69.2$ m. So the string is 69.2 m long and the roof is 25.4 m above the ground.
  ½ mark
Answer: Length of string = 69.2 m; height of roof = 25.4 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 4 5^{\circ}$ to get GT = 60 m (1 mark)
- Uses $tan 3 0^{\circ} = \frac{K S}{S R}$ (1 mark)
- KS = $203$ = 34.6 m (½ mark)
- Height of roof = 25.4 m (½ mark)
- Uses $sin 3 0^{\circ} = \frac{K S}{K R}$ (½ mark)
- String length = 69.2 m (½ mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerStatistics
The median of the following data is 50 and sum of all frequencies is 90 :
Class : 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90
Frequency : p 15 25 20 q 8 10
Find the values of p and q.
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Class $f$ c.f.
  20–30 $p$ $p$
  30–40 15 $p + 15$
  40–50 25 $p + 40$
  50–60 20 $p + 60$
  60–70 $q$ $p + q + 60$
  70–80 8 $p + q + 68$
  80–90 10 $p + q + 78$
  1½ marks
2. Median = 50, which lies in the class 50–60. So the median class is 50–60, with $l = 50$ , $h = 10$ , $f = 20$ , $c f = p + 40$ and $\frac{N}{2} = \frac{90}{2} = 45$ .
  ½ mark
3. Median $= l + (\frac{\frac{N}{2} - c f}{f}) \times h$
  $\Rightarrow 50 = 50 + \frac{45 - ( p + 40 )}{20} \times 10$
  1 mark
4. $\Rightarrow \frac{5 - p}{2} = 0 \Rightarrow p = 5$
  ½ mark
5. The sum of all frequencies is 90, so $p + q + 78 = 90$ .
  $\Rightarrow 5 + q = 12$
  1 mark
6. $∴ q = 7$
  Hence $p = 5$ and $q = 7$ .
  ½ mark
Answer: p = 5, q = 7
Where the marks are
- Correct cumulative frequency table in terms of p and q (1½ marks)
- Identifies median class 50–60 (½ mark)
- Substitutes correctly in the median formula (1 mark)
- Gets p = 5 (½ mark)
- Uses total frequency: p + q + 78 = 90 (1 mark)
- Gets q = 7 (½ mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerStatistics
Find mean and mode of the following distribution :
Class : 0 – 15 15 – 30 30 – 45 45 – 60 60 – 75 75 – 90 90 – 105
Frequency : 4 8 11 14 10 7 6
Try it first, then show our solutionSolution
1. Take assumed mean $a = 52.5$ and $h = 15$ , with $u_{i} = \frac{x _{i} - 52.5}{15}$ .
  Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
  0–15 4 7.5 −3 −12
  15–30 8 22.5 −2 −16
  30–45 11 37.5 −1 −11
  45–60 14 52.5 0 0
  60–75 10 67.5 1 10
  75–90 7 82.5 2 14
  90–105 6 97.5 3 18
  Total 60 3
  2 marks
2. Mean $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 52.5 + 15 \times \frac{3}{60}$
  ½ mark
3. $= 52.5 + 0.75 = 53.25$
  ½ mark
4. The highest frequency is 14, so the modal class is 45–60.
  Here $l = 45$ , $f_{1} = 14$ , $f_{0} = 11$ , $f_{2} = 10$ and $h = 15$ .
  ½ mark
5. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 45 + \frac{14 - 11}{28 - 11 - 10} \times 15$
  1 mark
6. $= 45 + \frac{45}{7} = 45 + 6.43 \approx 51.43$
  Mean = 53.25 and Mode ≈ 51.43.
  ½ mark
Answer: Mean = 53.25; Mode ≈ 51.43
Where the marks are
- Correct table with class marks, u_i and f_iu_i, giving Σf = 60 and Σf_iu_i = 3 (2 marks)
- Correct substitution in the step-deviation formula (½ mark)
- Mean = 53.25 (½ mark)
- Identifies modal class 45–60 (½ mark)
- Correct substitution in the mode formula (1 mark)
- Mode ≈ 51.4 (½ mark)
Open Q35(b) on its own page

Class :	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90
Frequency :	p	15	25	20	q	8	10

Class	$f$	c.f.
20–30	$p$	$p$
30–40	15	$p + 15$
40–50	25	$p + 40$
50–60	20	$p + 60$
60–70	$q$	$p + q + 60$
70–80	8	$p + q + 68$
80–90	10	$p + q + 78$

Class :	0 – 15	15 – 30	30 – 45	45 – 60	60 – 75	75 – 90	90 – 105
Frequency :	4	8	11	14	10	7	6

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
0–15	4	7.5	−3	−12
15–30	8	22.5	−2	−16
30–45	11	37.5	−1	−11
45–60	14	52.5	0	0
60–75	10	67.5	1	10
75–90	7	82.5	2	14
90–105	6	97.5	3	18
Total	60			3

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedTriangles
Carom board is a very popular game. The board is a square of side length 65 cm. It has circular pockets in each corner.
Ansh strikes a disc, kept at position P with a striker. The disc, hits the boundary of the board at R and goes straight to pocket at corner C. It is given that PS = 9 cm, PQ = 35 cm, BR = $x$ , $∠ P R Q = α$ and $∠ C R B = θ$ .
Based on the above information, answer the following questions :
1. (i)
  Using law of reflection i.e. $∠ P R T = ∠ C R T$ , prove that $θ = α$ .
  1 mark
2. (ii)
  Prove that $△ P QR \sim △ C B R$ given that PQ is perpendicular to AB.
  1 mark
3. (iii)(a)
  Find the value of $x$ using similarity of triangles.
  2 marks
4. OR
  (iii)(b)
  If $\frac{Area △ P QR}{Area △ C B R} = \frac{P Q ^{2}}{C B ^{2}}$ , then find the value of $x$ .
  2 marks
Figure in the paper: Photo of a carom board plus a diagram: square board with corners C (top-left), B (top-right), A (bottom-right), D (bottom-left); side CB = 65 cm. Pockets (circles) at each corner. R lies on side AB at distance x from B. Q is on AB below R, with PQ horizontal = 35 cm, PQ perpendicular to AB. P is a point inside; PS = 9 cm is the perpendicular from P to side DA (bottom side) with S on DA. T is on a dashed horizontal line through R extended into the board (normal at R). Line CR is drawn from C to R; line PR from P to R. Angle PRQ = α, angle CRB = θ. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) TR is the normal at R, so TR ⟂ AB. Hence $∠ T R Q = ∠ T R B = 9 0^{\circ}$ .
  $∴ α + ∠ P R T = 9 0^{\circ} = θ + ∠ C R T$
  Since $∠ P R T = ∠ C R T$ [law of reflection], we get $α = θ$ .
  1 mark
2. (ii) In $△ P QR$ and $△ C B R$ :
  $∠ P R Q = ∠ C R B$ [ $α = θ$ , from (i)]
  $∠ P QR = ∠ C B R = 9 0^{\circ}$ [PQ ⟂ AB, and the board is a square]
  $∴ △ P QR \sim △ C B R$ [AA similarity].
  1 mark
3. (iii)(a) Since $△ P QR \sim △ C B R$ , $\frac{P Q}{C B} = \frac{QR}{B R}$ .
  ½ mark
4. We have $Q A = P S = 9$ cm, so $QR = A B - Q A - B R = 65 - 9 - x = 56 - x$ .
  $\Rightarrow \frac{35}{65} = \frac{56 - x}{x}$
  1 mark
5. $\Rightarrow 35 x = 65 (56 - x) \Rightarrow 100 x = 3640 \Rightarrow x = 36.4$ cm
  ½ mark
6. (iii)(b) $\frac{ar ( △ P QR )}{ar ( △ C B R )} = \frac{\frac{1}{2} \times P Q \times QR}{\frac{1}{2} \times C B \times B R} = \frac{P Q ^{2}}{C B ^{2}}$
  $\Rightarrow \frac{\frac{1}{2} \times 35 \times ( 65 - 9 - x )}{\frac{1}{2} \times 65 \times x} = \frac{35 \times 35}{65 \times 65}$
  1 mark
7. $\Rightarrow \frac{56 - x}{x} = \frac{35}{65}$
  ½ mark
8. $\Rightarrow 100 x = 3640 \Rightarrow x = 36.4$ cm
  ½ mark
Answer: (i) θ = α proved; (ii) ΔPQR ~ ΔCBR by AA; (iii)(a) x = 36.4 cm; (iii)(b) x = 36.4 cm
Where the marks are
- (i) Uses TR ⟂ AB to write α + ∠PRT = θ + ∠CRT and concludes α = θ (1 mark)
- (ii) Equal angles α = θ and two right angles, giving similarity by AA (1 mark)
- (iii)(a) Writes PQ/CB = QR/BR (½ mark)
- (iii)(a) Substitutes QR = 56 − x to get 35/65 = (56 − x)/x (1 mark)
- (iii)(a) x = 36.4 cm (½ mark)
- (iii)(b) Writes the area ratio using ½ × base × height and equates it to 35²/65² (1 mark)
- (iii)(b) Simplifies to (56 − x)/x = 35/65 (½ mark)
- (iii)(b) x = 36.4 cm (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
‘Kolam’ is a decorative art which is made with rice flour in South Indian States. It is drawn on grid pattern of dots. One such art work is shown below.
Observe the given figure carefully. There are 4 dots in first square, 8 dots in second square, 12 dots in third square and so on.
Based on the above, answer the following questions :
1. (i)
  Show that number of dots given above form an A.P. Write the first term and common difference.
  1 mark
2. (ii)
  Write $n^{t h}$ term of the A.P. formed.
  1 mark
3. (iii)(a)
  The pattern is expanded on a large ground. If total 220 dots are used, then find the number of squares formed.
  2 marks
4. OR
  (iii)(b)
  Is it possible to complete n number of squares using 100 dots ? If yes, then find the value of n.
  2 marks
Figure in the paper: A Kolam pattern drawn on a diamond-shaped grid of dots: concentric square rings of dots (rotated 45°) around a central black filled square, with looping curves around the dots. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The numbers of dots are 4, 8, 12, …
  Here $8 - 4 = 12 - 8 = 4$ . The difference is constant, so the numbers form an A.P.
  ½ mark
2. First term $a = 4$ and common difference $d = 4$ .
  ½ mark
3. (ii) $a_{n} = a + (n - 1) d = 4 + (n - 1) 4 = 4 n$
  1 mark
4. (iii)(a) Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with $S_{n} = 220$ :
  $220 = \frac{n}{2} [8 + 4 (n - 1)]$
  ½ mark
5. $\Rightarrow 220 = 2 n (n + 1) \Rightarrow n^{2} + n - 110 = 0$
  ½ mark
6. $\Rightarrow (n + 11) (n - 10) = 0 \Rightarrow n = 10$ or $n = - 11$
  $n$ cannot be negative, so $n = 10$ . Hence 10 squares are formed.
  1 mark
7. (iii)(b) Take $S_{n} = 100$ :
  $100 = \frac{n}{2} [8 + 4 (n - 1)] = 2 n (n + 1)$
  ½ mark
8. $\Rightarrow n^{2} + n - 50 = 0$
  ½ mark
9. $n = \frac{- 1 \pm 1 + 200}{2} = \frac{- 1 \pm 201}{2}$
  $201$ is not an integer, so $n$ is not a natural number. Therefore it is not possible to complete squares using exactly 100 dots.
  1 mark
Answer: (i) AP with a = 4, d = 4; (ii) aₙ = 4n; (iii)(a) 10 squares; (iii)(b) Not possible, since n = (−1 ± √201)/2 is not a natural number
Where the marks are
- (i) Shows the common difference 4 is constant, so the dots form an A.P. (½ mark)
- (i) States a = 4, d = 4 (½ mark)
- (ii) aₙ = 4n (1 mark)
- (iii)(a) Correct substitution in the Sₙ formula with Sₙ = 220 (½ mark)
- (iii)(a) Forms n² + n − 110 = 0 (½ mark)
- (iii)(a) Solves to get n = 10 squares, rejecting −11 (1 mark)
- (iii)(b) Correct substitution with Sₙ = 100 (½ mark)
- (iii)(b) Forms n² + n − 50 = 0 (½ mark)
- (iii)(b) Shows n is not natural, so it is not possible (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Observe the map of Jaipur city placed on a Cartesian plane. Taking Rambagh Palace as origin, the location of some places are given below :
Point A : (–4, 2) Rajasthan High Court
Point B : (4, –4) Birla Mandir
Point C : (4, 3) Heera Bagh
Point D : (–5, –2) Amar Jawan Jyoti
Based on the above, answer the following questions :
1. (i)
  Advocate Rehana stays at Heera Bagh. How much distance she has to cover daily to go to the court and coming back home ?
  1 mark
2. (ii)
  There is a crossing on X-axis which divides AD in a certain ratio. Find the ratio.
  1 mark
3. (iii)(a)
  Is Birla Mandir equidistant from Heera Bagh and Amar Jawan Jyoti ? Justify your answer.
  2 marks
4. OR
  (iii)(b)
  Using section formula, show that points A, O and B are not collinear.
  2 marks
Figure in the paper: Street map of Jaipur overlaid with X and Y axes through origin O (Rambagh Palace). Marked points: A (Rajasthan High Court) upper left, C (Heerabagh) upper right on positive y side, D (Amar Jawan Jyoti) lower left, B (Birla Mandir, Jaipur) lower right. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Rehana goes from C(4, 3) to A(−4, 2) and back, so distance = $2 \times A C$ .
  $A C = (- 4 - 4)^{2} + (2 - 3)^{2}$
  ½ mark
2. $A C = 64 + 1 = 65$
  Distance covered daily $= 265$ units.
  ½ mark
3. (ii) Let the point P(x, 0) on the X-axis divide AD in the ratio $k : 1$ , with A(−4, 2) and D(−5, −2).
  By the section formula, the y-coordinate gives $0 = \frac{k ( - 2 ) + 1 ( 2 )}{k + 1}$ .
  ½ mark
4. $\Rightarrow - 2 k + 2 = 0 \Rightarrow k = 1$
  The required ratio is 1 : 1.
  ½ mark
5. (iii)(a) Distance of Birla Mandir B(4, −4) from Heera Bagh C(4, 3):
  $B C = (4 - 4)^{2} + (- 4 - 3)^{2} = 49 = 7$ units
  1 mark
6. Distance of B from Amar Jawan Jyoti D(−5, −2):
  $B D = (4 + 5)^{2} + (- 4 + 2)^{2} = 81 + 4 = 85$ units
  Since $B C \neq = B D$ , Birla Mandir is not equidistant from Heera Bagh and Amar Jawan Jyoti.
  1 mark
7. (iii)(b) Suppose A(−4, 2), O(0, 0) and B(4, −4) are collinear, and O divides AB in the ratio $k : 1$ .
  ½ mark
8. x-coordinate: $0 = \frac{4 k - 4}{k + 1} \Rightarrow k = 1$
  ½ mark
9. y-coordinate: $0 = \frac{- 4 k + 2}{k + 1}$
  ½ mark
10. $\Rightarrow k = \frac{1}{2}$
  The two values of $k$ are different, so no single ratio works. Hence A, O and B are not collinear.
  ½ mark
Answer: (i) 2√65 units; (ii) 1 : 1; (iii)(a) No, BC = 7 units but BD = √85 units; (iii)(b) k = 1 from x and k = 1/2 from y, so A, O, B are not collinear
Where the marks are
- (i) Uses distance = 2AC with correct substitution in the distance formula (½ mark)
- (i) 2√65 units (½ mark)
- (ii) Uses the section formula with y-coordinate = 0 (½ mark)
- (ii) Ratio 1 : 1 (½ mark)
- (iii)(a) BC = 7 units (1 mark)
- (iii)(a) BD = √85 units and concludes not equidistant (1 mark)
- (iii)(b) Assumes O divides AB in the ratio k : 1 (½ mark)
- (iii)(b) k = 1 from the x-coordinate (½ mark)
- (iii)(b) Sets up the y-coordinate equation (½ mark)
- (iii)(b) k = 1/2, values differ, so not collinear (½ mark)
Open Q38 on its own page

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.