MathsCBSE Class 10 · Standard · Board exam 2025 (session 2024-25)

CBSE Class 10 Mathematics Standard: 2025 board paper, set 30/4/1

The question paper set in the CBSE Class 10 board exam of 2025, 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/2025/X/041_Mathematics_Standard.zip (a zip; this paper is the file 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; internal choice in 22 and 24
C	26-31	short answer; internal choice in 26 and 29
D	32-35	long answer; internal choice in 32 and 33
E	36-38	case-study based; internal choice in 2-mark sub-part of each case

Instructions in brief: No overall choice; internal choice given in 2 questions of Section B, 2 of Section C, 2 of Section D and in the 2-mark sub-part of 3 case-study questions in Section E. Take π = 22/7 where required if not stated; calculators not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If $x = a b^{3}$ and $y = a^{3} b$ , where $a$ and $b$ are prime numbers, then $[HCF (x, y) - LCM (x, y)]$ is equal to :
1. (A) $1 - a^{3} b^{3}$
2. (B) $ab (1 - ab)$
3. (C) $ab - a^{4} b^{4}$
4. (D) $ab (1 - ab) (1 + ab)$
Try it first, then show our solutionSolution
1. HCF $(x, y) = ab$ (lowest powers), LCM $(x, y) = a^{3} b^{3}$ (highest powers).
2. HCF − LCM $= ab - a^{3} b^{3} = ab (1 - a^{2} b^{2}) = ab (1 - ab) (1 + ab)$ , option (D).
  1 mark
Answer: $ab (1 - ab) (1 + ab)$ , option (D)
Where the marks are
- Finds HCF $= ab$ , LCM $= a^{3} b^{3}$ and factorises the difference to get (D) (1 mark)
Q2 · 1 mark · MCQReal Numbers
$(1 + 3)^{2} - (1 - 3)^{2}$ is :
1. (A)a positive rational number.
2. (B)a negative integer.
3. (C)a positive irrational number.
4. (D)a negative irrational number.
Try it first, then show our solutionSolution
1. Using $(p + q)^{2} - (p - q)^{2} = 4 pq$ : value $= 43$ .
2. $43 > 0$ and is irrational, so option (C).
  1 mark
Answer: $43$ , option (C)
Where the marks are
- Simplifies to $43$ and identifies it as positive irrational (1 mark)
Q3 · 1 mark · MCQQuadratic Equations
The value of ' $a$ ' for which $a x^{2} + x + a = 0$ has equal and positive roots is :
1. (A) $2$
2. (B) $- 2$
3. (C) $\frac{1}{2}$
4. (D) $- \frac{1}{2}$
Try it first, then show our solutionSolution
1. Equal roots: $D = 1 - 4 a^{2} = 0 \Rightarrow a = \pm \frac{1}{2}$ .
2. Root $= - \frac{1}{2 a}$ is positive only if $a < 0$ , so $a = - \frac{1}{2}$ (root = 1), option (D).
  1 mark
Answer: $a = - \frac{1}{2}$ , option (D)
Where the marks are
- Uses $D = 0$ and positivity of root to get $a = - \frac{1}{2}$ (1 mark)
Q4 · 1 mark · MCQCoordinate Geometry
The distance of a point A from $x$ -axis is 3 units. Which of the following cannot be coordinates of the point A ?
1. (A) $(1, 3)$
2. (B) $(- 3, - 3)$
3. (C) $(- 3, 3)$
4. (D) $(3, 1)$
Try it first, then show our solutionSolution
1. Distance from $x$ -axis $= ∣ y ∣$ , so $∣ y ∣ = 3$ .
2. $(3, 1)$ has $∣ y ∣ = 1 \neq = 3$ , so it cannot be A; option (D).
  1 mark
Answer: $(3, 1)$ , option (D)
Where the marks are
- Uses distance from $x$ -axis $= ∣ y ∣$ to reject $(3, 1)$ (1 mark)
Q5 · 1 mark · MCQProbability
The number of red balls in a bag is 10 more than the number of black balls. If the probability of drawing a red ball at random from this bag is $\frac{3}{5}$ , then the total number of balls in the bag is :
1. (A) $50$
2. (B) $60$
3. (C) $80$
4. (D) $40$
Try it first, then show our solutionSolution
1. Black $= b$ , red $= b + 10$ , total $= 2 b + 10$ ; $\frac{b + 10}{2 b + 10} = \frac{3}{5} \Rightarrow 5 b + 50 = 6 b + 30 \Rightarrow b = 20$ .
2. Total $= 2 (20) + 10 = 50$ , option (A).
  1 mark
Answer: 50, option (A)
Where the marks are
- Forms equation and gets total 50 (1 mark)
Q6 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of ' $p$ ' for which the equations $p x + 3 y = p - 3$ , $12 x + p y = p$ has infinitely many solutions is :
1. (A) $- 6$ only
2. (B) $6$ only
3. (C) $\pm 6$
4. (D)Any real number except $\pm 6$
Try it first, then show our solutionSolution
1. Write as $p x + 3 y - (p - 3) = 0$ , $12 x + p y - p = 0$ . Need $\frac{p}{12} = \frac{3}{p} = \frac{p - 3}{p}$ .
2. $p^{2} = 36 \Rightarrow p = \pm 6$ ; $\frac{3}{p} = \frac{p - 3}{p} \Rightarrow p = 6$ . So $p = 6$ only, option (B).
  1 mark
Answer: $p = 6$ only, option (B)
Where the marks are
- Applies $a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}$ and gets $p = 6$ (1 mark)
Open Q6 on its own page
Q7 · 1 mark · MCQTriangles
$Δ$ ABC and $Δ$ PQR are shown in the adjoining figures. The measure of $∠ C$ is :
1. (A) $14 0^{\circ}$
2. (B) $8 0^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $4 0^{\circ}$
Figure in the paper: Triangle ABC: AB = 3.8 cm, BC = 6 cm, AC = $33$ cm, angle at C marked. Triangle PQR: PR = $63$ cm, RQ = 7.6 cm, PQ = 12 cm, angle at R = 80 $^{\circ}$ , angle at Q = 60 $^{\circ}$ . See it in the official paper.
Try it first, then show our solutionSolution
1. $\frac{A B}{R Q} = \frac{3.8}{7.6} = \frac{1}{2}, \frac{B C}{QP} = \frac{6}{12} = \frac{1}{2}, \frac{C A}{P R} = \frac{3 3}{6 3} = \frac{1}{2}$ , so $△ A B C \sim △ R QP$ (SSS).
2. $∠ C = ∠ P = 18 0^{\circ} - 8 0^{\circ} - 6 0^{\circ} = 4 0^{\circ}$ , option (D).
  1 mark
Answer: $4 0^{\circ}$ , option (D)
Where the marks are
- Establishes SSS similarity with correct correspondence and finds $∠ C = ∠ P = 4 0^{\circ}$ (1 mark)
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Q8 · 1 mark · MCQIntroduction to Trigonometry
$tan 2 A = 3 tan A$ is true, when the measure of $∠ A$ is :
1. (A) $9 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $3 0^{\circ}$
Try it first, then show our solutionSolution
1. For $A = 3 0^{\circ}$ : $tan 6 0^{\circ} = 3$ and $3 tan 3 0^{\circ} = \frac{3}{3} = 3$ . Equal.
  1 mark
2. Others fail ( $tan 9 0^{\circ}$ undefined, $tan 12 0^{\circ} < 0$ , $tan 9 0^{\circ}$ for 45°). Option (D).
Answer: $3 0^{\circ}$ , option (D)
Where the marks are
- Verifies $tan 6 0^{\circ} = 3 tan 3 0^{\circ}$ (1 mark)
Open Q8 on its own page
Q9 · 1 mark · MCQIntroduction to Trigonometry
Which of the following statements is true ?
1. (A) $sin 2 0^{\circ} > sin 7 0^{\circ}$
2. (B) $sin 2 0^{\circ} > cos 2 0^{\circ}$
3. (C) $cos 2 0^{\circ} > cos 7 0^{\circ}$
4. (D) $tan 2 0^{\circ} > tan 7 0^{\circ}$
Try it first, then show our solutionSolution
1. From 0° to 90°, cos decreases while sin and tan increase.
2. Hence $cos 2 0^{\circ} > cos 7 0^{\circ}$ ; (A),(B),(D) are false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Uses decreasing nature of cosine in $0^{\circ}$ – $9 0^{\circ}$ (1 mark)
Q10 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A 30 m long rope is tightly stretched and tied from the top of pole to the ground. If the rope makes an angle of $6 0^{\circ}$ with the ground, the height of the pole is :
1. (A) $103$ m
2. (B) $303$ m
3. (C) $15$ m
4. (D) $153$ m
Try it first, then show our solutionSolution
1. Rope is hypotenuse: $h = 30 sin 6 0^{\circ} = 30 \cdot \frac{3}{2} = 153$ m, option (D).
  1 mark
Answer: $153$ m, option (D)
Where the marks are
- Uses $sin 6 0^{\circ} = h /30$ to get $153$ m (1 mark)
Q11 · 1 mark · MCQCompetency-basedSurface Areas and Volumes
On the top face of the wooden cube of side 7 cm, hemispherical depressions of radius 0.35 cm are to be formed by taking out the wood. The maximum number of depressions that can be formed is :
1. (A) $400$
2. (B) $100$
3. (C) $20$
4. (D) $10$
Try it first, then show our solutionSolution
1. Each depression needs diameter $0.7$ cm; along a side: $7/0.7 = 10$ .
2. Maximum $= 10 \times 10 = 100$ , option (B).
  1 mark
Answer: 100, option (B)
Where the marks are
- Fits 10 per row, 10 rows → 100 (1 mark)
Q12 · 1 mark · MCQStatistics
The cumulative frequency for calculating median is obtained by adding the frequencies of all the :
1. (A)classes up to the median class
2. (B)classes following the median class
3. (C)classes preceding the median class
4. (D)all classes
Try it first, then show our solutionSolution
1. In median $= l + \frac{n /2 - c f}{f} \times h$ , $c f$ is the cumulative frequency of the class preceding the median class, i.e. sum of frequencies of all classes before it. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Identifies $c f$ as sum over classes preceding the median class (1 mark)
Q13 · 1 mark · MCQStatistics
If mean and median of given set of observations are 10 and 11 respectively, then the value of mode is :
1. (A) $10.5$
2. (B) $8$
3. (C) $13$
4. (D) $21$
Try it first, then show our solutionSolution
1. Mode $= 3$ Median $- 2$ Mean $= 33 - 20 = 13$ , option (C).
  1 mark
Answer: 13, option (C)
Where the marks are
- Applies empirical relation to get 13 (1 mark)
Q14 · 1 mark · MCQCircles
In the adjoining figure, AB is the chord of the larger circle touching the smaller circle. The centre of both the circles is O. If $A B = 2 r$ and $O P = r$ , then the radius of larger circle is :
1. (A) $2 r$
2. (B) $3 r$
3. (C) $22 r$
4. (D) $2 r$
Figure in the paper: Two concentric circles with centre O. AB is a chord of the larger circle and is tangent to the smaller circle at point P, where OP = r is the radius of the smaller circle. A and B lie on the larger circle. See it in the official paper.
Try it first, then show our solutionSolution
1. $O P ⊥ A B$ [tangent ⟂ radius], and perpendicular from centre bisects chord, so $A P = r$ .
2. $O A = r^{2} + r^{2} = 2 r$ , option (D).
  1 mark
Answer: $2 r$ , option (D)
Where the marks are
- Uses $O P ⊥ A B$ , $A P = r$ and Pythagoras (1 mark)
Q15 · 1 mark · MCQCircles
A parallelogram having one of its sides 5 cm circumscribes a circle. The perimeter of parallelogram is :
1. (A) $20$ cm
2. (B)less than $20$ cm
3. (C)more than $20$ cm but less than $40$ cm
4. (D) $40$ cm
Try it first, then show our solutionSolution
1. A parallelogram circumscribing a circle is a rhombus (since $A B + C D = A D + B C$ ), so all sides are 5 cm.
2. Perimeter $= 4 \times 5 = 20$ cm, option (A).
  1 mark
Answer: 20 cm, option (A)
Where the marks are
- Recognises rhombus and gets 20 cm (1 mark)
Q16 · 1 mark · MCQTriangles
E and F are points on the sides AB and AC respectively of a $Δ$ ABC such that $\frac{A E}{E B} = \frac{A F}{F C} = \frac{1}{2}$ . Which of the following relation is true ?
1. (A) $E F = 2 B C$
2. (B) $B C = 2 E F$
3. (C) $E F = 3 B C$
4. (D) $B C = 3 E F$
Try it first, then show our solutionSolution
1. By converse of BPT, $E F ∥ B C$ , so $△ A E F \sim △ A B C$ with $\frac{A E}{A B} = \frac{1}{3}$ .
2. $\frac{E F}{B C} = \frac{1}{3} \Rightarrow B C = 3 E F$ , option (D).
  1 mark
Answer: $B C = 3 E F$ , option (D)
Where the marks are
- Uses similarity with ratio $1 : 3$ (1 mark)
Q17 · 1 mark · MCQPolynomials
Which of the following statements is true for a polynomial $p (x)$ of degree 3?
1. (A) $p (x)$ has at most two distinct zeroes.
2. (B) $p (x)$ has at least two distinct zeroes.
3. (C) $p (x)$ has exactly three distinct zeroes.
4. (D) $p (x)$ has at most three distinct zeroes.
Try it first, then show our solutionSolution
1. A polynomial of degree $n$ has at most $n$ zeroes; e.g. $x^{3}$ has only one distinct zero, so (B),(C) fail, and $x (x - 1) (x - 2)$ has three, so (A) fails. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- States degree-3 polynomial has at most 3 zeroes (1 mark)
Q18 · 1 mark · MCQProbability
A pair of dice is thrown. The probability that sum of numbers appearing on top faces is at most 10 is :
1. (A) $\frac{1}{11}$
2. (B) $\frac{10}{11}$
3. (C) $\frac{5}{6}$
4. (D) $\frac{11}{12}$
Try it first, then show our solutionSolution
1. Sum > 10: (5,6),(6,5),(6,6) — 3 outcomes of 36.
2. P(sum ≤ 10) $= 1 - \frac{3}{36} = \frac{33}{36} = \frac{11}{12}$ , option (D).
  1 mark
Answer: $\frac{11}{12}$ , option (D)
Where the marks are
- Uses complement and gets $\frac{11}{12}$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Directions: In question 19, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion (A): $4^{n}$ ends with digit 0 for some natural number $n$ .
Reason (R): For a number ' $x$ ' having 2 and 5 as its prime factors, $x^{n}$ always ends with digit 0 for every natural number $n$ .
1. (A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of 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. $4^{n} = 2^{2 n}$ has no prime factor 5, so it never ends in 0: A is false.
2. If 2 and 5 divide $x$ , then 10 divides $x^{n}$ , so $x^{n}$ ends in 0: R is true. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- A false (no factor 5), R true (factor 10) (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedCircles
Directions: In question 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion (A): Tangents drawn at the end points of a diameter of a circle are always parallel to each other.
Reason (R): The lengths of tangents drawn to a circle from a point outside the circle are always equal.
1. (A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of 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. Both tangents are ⟂ to the same diameter [tangent ⟂ radius], so they are parallel: A true.
2. R is a true theorem, but it does not explain A (A follows from tangent ⟂ radius). Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Both true, R not the explanation (1 mark)
Open Q20 on its own page

Section B

very short answer; internal choice in 22 and 24

Q21 · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve the following system of equations algebraically : $30 x + 44 y = 10$ ; $40 x + 55 y = 13$
Try it first, then show our solutionSolution
1. Multiply the first equation by 4 and the second by 3 so that the coefficients of $x$ are equal:
  $120 x + 176 y = 40$ ...(i)
  $120 x + 165 y = 39$ ...(ii)
  1 mark
2. Subtract (ii) from (i): $11 y = 1 \Rightarrow y = \frac{1}{11}$
  ½ mark
3. Substitute in $30 x + 44 y = 10$ : $30 x + 4 = 10 \Rightarrow 30 x = 6 \Rightarrow x = \frac{1}{5}$
  ½ mark
Answer: $x = \frac{1}{5}, y = \frac{1}{11}$
Where the marks are
- Makes the coefficients of one variable equal, e.g. $120 x + 176 y = 40$ and $120 x + 165 y = 39$ (1 mark)
- Gets $y = \frac{1}{11}$ (½ mark)
- Gets $x = \frac{1}{5}$ (½ mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(A) · 2 marks · Very short answerCompetency-basedSome Applications of Trigonometry
A 1.5 m tall boy is walking away from the base of a lamp post which is 12 m high, at the speed of 2.5 m/sec. Find the length of his shadow after 3 seconds.
Try it first, then show our solutionSolution
1. Figure: lamp post AB = 12 m with B on the ground; boy CD = 1.5 m standing at D; the line from the lamp top A through the boy's head C meets the ground at E, so DE = $x$ m is the shadow. B, D, E lie on the ground line.
  ½ mark
2. Distance walked in 3 s $= 2.5 \times 3 = 7.5$ m, so BD = 7.5 m and BE = $(7.5 + x)$ m.
  ½ mark
3. In $Δ A B E$ and $Δ C D E$ : $∠ B = ∠ D = 9 0^{\circ}$ and $∠ E$ is common, so $Δ A B E \sim Δ C D E$ (AA). Hence $\frac{C D}{A B} = \frac{D E}{B E}$ .
  ½ mark
4. $\frac{1.5}{12} = \frac{x}{7.5 + x} \Rightarrow 12 x = 11.25 + 1.5 x \Rightarrow 10.5 x = 11.25 \Rightarrow x = \frac{15}{14} \approx 1.07$ m.
  Length of shadow ≈ 1.07 m.
  ½ mark
Answer: Shadow $= \frac{15}{14}$ m ≈ 1.07 m
Where the marks are
- Correct labelled figure (½ mark)
- Distance covered in 3 s = 7.5 m (½ mark)
- Uses similarity $Δ A B E \sim Δ C D E$ to write $\frac{C D}{A B} = \frac{D E}{B E}$ (½ mark)
- Solves to get shadow ≈ 1.07 m (½ mark)
Open Q22(A) on its own page
OR
Q22(B) · 2 marks · Very short answerTriangles
In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that $Δ$ ABE $\sim$ $Δ$ CFB
Try it first, then show our solutionSolution
1. Figure: parallelogram ABCD with AD produced beyond D to E; BE cuts CD at F.
  ½ mark
2. In $Δ A B E$ and $Δ C F B$ :
  $∠ A = ∠ C$ [opposite angles of a parallelogram]
  $∠ A E B = ∠ C B F$ [alternate angles, since $A E ∥ B C$ and BE is a transversal]
  1 mark
3. Therefore $Δ A B E \sim Δ C F B$ [AA similarity criterion].
  ½ mark
Answer: $Δ A B E \sim Δ C F B$ by AA similarity
Where the marks are
- Correct figure (½ mark)
- States the two pairs of equal angles with reasons (1 mark)
- Concludes similarity by AA (½ mark)
Open Q22(B) on its own page
Q23 · 2 marks · Very short answerCoordinate Geometry
Find the coordinates of the point C which lies on the line AB produced such that $A C = 2 B C$ , where coordinates of points A and B are $(- 1, 7)$ and $(4, - 3)$ respectively.
Try it first, then show our solutionSolution
1. Let C be $(x, y)$ . C lies on AB produced with AC = 2BC, so AB = BC, i.e. B is the mid-point of AC.
  ½ mark
2. $\frac{- 1 + x}{2} = 4 \Rightarrow x = 9$
  ½ mark
3. $\frac{7 + y}{2} = - 3 \Rightarrow y = - 13$
  ½ mark
4. Coordinates of C are $(9, - 13)$ .
  ½ mark
Answer: C(9, −13)
Where the marks are
- Recognises B is the mid-point of AC (½ mark)
- Finds $x = 9$ (½ mark)
- Finds $y = - 13$ (½ mark)
- States C(9, −13) (½ mark)
Open Q23 on its own page
Answer one of these 2 alternatives.
Q24(A) · 2 marks · Very short answerIntroduction to Trigonometry
Find the value of $x$ for which $(sin A + cosec A)^{2} + (cos A + sec A)^{2} = x + tan^{2} A + cot^{2} A$
Try it first, then show our solutionSolution
1. Expand: $sin^{2} A + cosec^{2} A + 2 sin A cosec A + cos^{2} A + sec^{2} A + 2 cos A sec A$
  $= sin^{2} A + cosec^{2} A + 2 + cos^{2} A + sec^{2} A + 2$
  ½ mark
2. Use $sin^{2} A + cos^{2} A = 1$ , $cosec^{2} A = 1 + cot^{2} A$ , $sec^{2} A = 1 + tan^{2} A$ :
  LHS $= 1 + 4 + (1 + cot^{2} A) + (1 + tan^{2} A) = 7 + tan^{2} A + cot^{2} A$
  1 mark
3. Comparing with $x + tan^{2} A + cot^{2} A$ : $x = 7$ .
  ½ mark
Answer: $x = 7$
Where the marks are
- Correct expansion using $sin A cosec A = 1$ , $cos A sec A = 1$ (½ mark)
- Applies the three identities to simplify (1 mark)
- $x = 7$ (½ mark)
Open Q24(A) on its own page
OR
Q24(B) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate the following : $\frac{3 s i n 3 0 ^{\circ} - 4 s i n ^{3} 3 0 ^{\circ}}{2 s i n ^{2} 5 0 ^{\circ} + 2 c o s ^{2} 5 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. $= \frac{3 \times \frac{1}{2} - 4 \times \frac{1}{8}}{2 ( sin ^{2} 5 0 ^{\circ} + cos ^{2} 5 0 ^{\circ} )}$ [ $sin 3 0^{\circ} = \frac{1}{2}$ ]
  1 mark
2. $= \frac{\frac{3}{2} - \frac{1}{2}}{2 \times 1}$ [ $sin^{2} θ + cos^{2} θ = 1$ ]
  ½ mark
3. $= \frac{1}{2}$
  ½ mark
Answer: $\frac{1}{2}$
Where the marks are
- Substitutes $sin 3 0^{\circ} = \frac{1}{2}$ and takes 2 common in denominator (1 mark)
- Uses $sin^{2} 5 0^{\circ} + cos^{2} 5 0^{\circ} = 1$ (½ mark)
- Final value $\frac{1}{2}$ (½ mark)
Open Q24(B) on its own page
Q25 · 2 marks · Very short answerCompetency-basedProbability
Two friends Anil and Ashraf were born in the December month in the year 2010. Find the probability that :
1. (i)
  they share same date of birth.
  0 marks
2. (ii)
  they have different dates of birth.
  0 marks
Try it first, then show our solutionSolution
1. December has 31 days. Whatever Anil's birth date, Ashraf's date has 31 equally likely possibilities.
  (i) Favourable: only 1 (the same date). P(same date) $= \frac{1}{31}$
  1 mark
2. (ii) P(different dates) $= 1 - \frac{1}{31} = \frac{30}{31}$
  1 mark
Answer: (i) $\frac{1}{31}$ ; (ii) $\frac{30}{31}$
Where the marks are
- P(same date of birth) = 1/31 (1 mark)
- P(different dates of birth) = 30/31 (1 mark)
Open Q25 on its own page

Section C

short answer; internal choice in 26 and 29

Answer one of these 2 alternatives.
Q26(A) · 3 marks · Short answerReal Numbers
Prove that $2$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $2$ is rational. Then $2 = \frac{p}{q}$ , where $p$ and $q$ are co-prime integers and $q \neq = 0$ .
  ½ mark
2. Then $2 q = p$ , so $2 q^{2} = p^{2}$ . Hence 2 divides $p^{2}$ , so 2 divides $p$ [if a prime divides $a^{2}$ , it divides $a$ ]. Let $p = 2 m$ for some integer $m$ .
  1 mark
3. Then $2 q^{2} = 4 m^{2}$ , so $q^{2} = 2 m^{2}$ . Hence 2 divides $q^{2}$ , so 2 divides $q$ . Thus 2 is a common factor of $p$ and $q$ , which contradicts that $p$ and $q$ are co-prime.
  1 mark
4. So our assumption is wrong, and $2$ is irrational.
  ½ mark
Answer: √2 is irrational (proved by contradiction)
Where the marks are
- Assumes $2 = p / q$ with p, q co-prime and q ≠ 0 (½ mark)
- Shows 2 divides p (1 mark)
- Shows 2 divides q, which contradicts co-primeness (1 mark)
- Concludes √2 is irrational (½ mark)
Open Q26(A) on its own page
OR
Q26(B) · 3 marks · Short answerReal Numbers
Let $x$ and $y$ be two distinct prime numbers and $p = x^{2} y^{3}$ , $q = x y^{4}$ , $r = x^{5} y^{2}$ . Find the HCF and LCM of $p$ , $q$ and $r$ . Further check if $HCF (p, q, r) \times LCM (p, q, r) = p \times q \times r$ or not.
Try it first, then show our solutionSolution
1. HCF = product of smallest powers of common primes: $HCF (p, q, r) = x^{1} y^{2} = x y^{2}$
  1 mark
2. LCM = product of greatest powers: $LCM (p, q, r) = x^{5} y^{4}$
  1 mark
3. $HCF \times LCM = x y^{2} \times x^{5} y^{4} = x^{6} y^{6}$ , while $p \times q \times r = x^{2 + 1 + 5} y^{3 + 4 + 2} = x^{8} y^{9}$ .
  Since $x^{6} y^{6} \neq = x^{8} y^{9}$ , $HCF (p, q, r) \times LCM (p, q, r) \neq = p \times q \times r$ .
  1 mark
Answer: HCF $= x y^{2}$ , LCM $= x^{5} y^{4}$ ; HCF × LCM $= x^{6} y^{6} \neq = x^{8} y^{9} = pq r$
Where the marks are
- HCF $= x y^{2}$ (1 mark)
- LCM $= x^{5} y^{4}$ (1 mark)
- Compares $x^{6} y^{6}$ with $x^{8} y^{9}$ and concludes they are not equal (1 mark)
Open Q26(B) on its own page
Q27 · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
The monthly incomes of two persons are in the ratio 9 : 7 and their monthly expenditures are in the ratio 4 : 3. If each saved ₹ 5,000, express the given situation algebraically as a system of linear equations in two variables. Hence, find their respective monthly incomes.
Try it first, then show our solutionSolution
1. Let the incomes be ₹ $9 x$ and ₹ $7 x$ and expenditures ₹ $4 y$ and ₹ $3 y$ . Saving = income − expenditure:
  $9 x - 4 y = 5000$ ...(i)
  1 mark
2. $7 x - 3 y = 5000$ ...(ii)
  1 mark
3. (i)×3: $27 x - 12 y = 15000$ ; (ii)×4: $28 x - 12 y = 20000$ . Subtracting: $x = 5000$ (and $y = 10000$ ).
  ½ mark
4. Monthly incomes: $9 \times 5000 =$ ₹45000 and $7 \times 5000 =$ ₹35000.
  ½ mark
Answer: ₹45,000 and ₹35,000
Where the marks are
- Forms $9 x - 4 y = 5000$ (1 mark)
- Forms $7 x - 3 y = 5000$ (1 mark)
- Solves to get $x = 5000$ (½ mark)
- States incomes ₹45000 and ₹35000 (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerCoordinate Geometry
P $(x, y)$ , Q $(- 2, - 3)$ and R $(2, 3)$ are the vertices of a right triangle PQR right angled at P. Find the relationship between $x$ and $y$ . Hence, find all possible values of $x$ for which $y = 2$ .
Try it first, then show our solutionSolution
1. Since $∠ P = 9 0^{\circ}$ , by Pythagoras theorem $P Q^{2} + P R^{2} = Q R^{2}$ :
  $(x + 2)^{2} + (y + 3)^{2} + (x - 2)^{2} + (y - 3)^{2} = 4^{2} + 6^{2} = 52$
  1 mark
2. $2 x^{2} + 2 y^{2} + 26 = 52 \Rightarrow x^{2} + y^{2} = 13$ (required relation)
  1 mark
3. For $y = 2$ : $x^{2} = 13 - 4 = 9 \Rightarrow x = \pm 3$
  1 mark
Answer: $x^{2} + y^{2} = 13$ ; $x = 3$ or $x = - 3$
Where the marks are
- Applies Pythagoras with distance formula (1 mark)
- Simplifies to $x^{2} + y^{2} = 13$ (1 mark)
- Finds $x = \pm 3$ (1 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{c o s A + s i n A - 1}{c o s A - s i n A + 1} = cosec A - cot A$
Try it first, then show our solutionSolution
1. Divide numerator and denominator by $sin A$ :
  LHS $= \frac{cot A + 1 - cosec A}{cot A - 1 + cosec A}$
  1 mark
2. Replace 1 in the numerator by $cosec^{2} A - cot^{2} A$ :
  $= \frac{( cot A - cosec A ) + ( cosec ^{2} A - cot ^{2} A )}{cot A - 1 + cosec A}$
  1 mark
3. $= \frac{( cosec A - cot A ) ( - 1 + cosec A + cot A )}{cot A - 1 + cosec A}$
  ½ mark
4. $= cosec A - cot A =$ RHS. Hence proved.
  ½ mark
Answer: LHS = cosec A − cot A = RHS (proved)
Where the marks are
- Divides by sin A to get expression in cot A and cosec A (1 mark)
- Uses $1 = cosec^{2} A - cot^{2} A$ (1 mark)
- Factorises numerator (½ mark)
- Cancels to get RHS (½ mark)
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerIntroduction to Trigonometry
If $cot θ + cos θ = p$ and $cot θ - cos θ = q$ , prove that $p^{2} - q^{2} = 4 pq$
Try it first, then show our solutionSolution
1. LHS $= p^{2} - q^{2} = (p + q) (p - q) = [(cot θ + cos θ) + (cot θ - cos θ)] [(cot θ + cos θ) - (cot θ - cos θ)]$
  1 mark
2. $= 2 cot θ \times 2 cos θ = 4 cot θ cos θ$
  ½ mark
3. RHS $= 4 (cot θ + cos θ) (cot θ - cos θ) = 4 cot^{2} θ - cos^{2} θ$
  ½ mark
4. $= 4 \frac{cos ^{2} θ}{sin ^{2} θ} - cos^{2} θ = 4 cos^{2} θ (cosec^{2} θ - 1)$
  ½ mark
5. $= 4 cos^{2} θ cot^{2} θ = 4 cot θ cos θ$ . Hence LHS = RHS.
  ½ mark
Answer: $p^{2} - q^{2} = 4 cot θ cos θ = 4 pq$ (proved)
Where the marks are
- Uses $p^{2} - q^{2} = (p + q) (p - q)$ (1 mark)
- LHS $= 4 cot θ cos θ$ (½ mark)
- $pq = cot^{2} θ - cos^{2} θ$ (½ mark)
- Takes $cos^{2} θ$ common to get $cosec^{2} θ - 1$ (½ mark)
- Uses $cosec^{2} θ - 1 = cot^{2} θ$ to get RHS $= 4 cot θ cos θ$ (½ mark)
Open Q29(B) on its own page
Q30 · 3 marks · Short answerQuadratic Equations
$α$ and $β$ are zeroes of a quadratic polynomial $p x^{2} + q x + 1$ . Form a quadratic polynomial whose zeroes are $\frac{2}{α}$ and $\frac{2}{β}$ .
Try it first, then show our solutionSolution
1. For $p x^{2} + q x + 1$ : $α + β = - \frac{q}{p}$ , $α β = \frac{1}{p}$
  ½ mark
2. Sum of new zeroes $= \frac{2}{α} + \frac{2}{β} = \frac{2 ( α + β )}{α β} = 2 \times (- \frac{q}{p}) \times p = - 2 q$
  1 mark
3. Product of new zeroes $= \frac{2}{α} \times \frac{2}{β} = \frac{4}{α β} = 4 p$
  1 mark
4. Required polynomial $= x^{2} - (sum) x + product = x^{2} + 2 q x + 4 p$ (or any non-zero multiple $k (x^{2} + 2 q x + 4 p)$ ).
  ½ mark
Answer: $x^{2} + 2 q x + 4 p$
Where the marks are
- Writes sum and product of zeroes $- \frac{q}{p}$ and $\frac{1}{p}$ (½ mark)
- Sum of new zeroes $= - 2 q$ (1 mark)
- Product of new zeroes $= 4 p$ (1 mark)
- Forms $x^{2} + 2 q x + 4 p$ (½ mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCircles
Rectangle ABCD circumscribes the circle of radius 10 cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.
Try it first, then show our solutionSolution
1. Figure: rectangle ABCD with a circle (centre O, radius 10 cm) touching AB at P, BC at Q, CD at R and DA at S.
  ½ mark
2. Tangents drawn from an external point are equal: $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ .
  1 mark
3. Adding: $A P + B P + C R + D R = A S + B Q + C Q + D S \Rightarrow A B + C D = A D + B C$ ...(i)
  ½ mark
4. In rectangle ABCD, $A B = C D$ and $B C = A D$ , so (i) gives $2 A B = 2 A D \Rightarrow A B = A D$ . A rectangle with adjacent sides equal is a square, so ABCD is a square.
  ½ mark
5. Side of square = distance between opposite tangent points = diameter $= 2 \times 10 = 20$ cm. Perimeter $= 4 \times 20 = 80$ cm.
  ½ mark
Answer: ABCD is a square; perimeter = 80 cm
Where the marks are
- Correct figure (½ mark)
- Equal tangent lengths from each vertex (1 mark)
- Adds to get $A B + C D = A D + B C$ (½ mark)
- Uses rectangle property to get $A B = A D$ , hence square (½ mark)
- Side = 20 cm, perimeter = 80 cm (½ mark)
Open Q31 on its own page

Section D

long answer; internal choice in 32 and 33

Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerQuadratic Equations
The sides of a right triangle are such that the longest side is 4 m more than the shortest side and the third side is 2 m less than the longest side. Find the length of each side of the triangle. Also, find the difference between the numerical values of the area and the perimeter of the given triangle.
Try it first, then show our solutionSolution
1. Let the shortest side be $x$ m. Then longest side (hypotenuse) $= (x + 4)$ m and third side $= (x + 4) - 2 = (x + 2)$ m.
  1 mark
2. By Pythagoras theorem: $(x + 4)^{2} = x^{2} + (x + 2)^{2}$
  $x^{2} + 8 x + 16 = 2 x^{2} + 4 x + 4 \Rightarrow x^{2} - 4 x - 12 = 0$
  1 mark
3. $(x - 6) (x + 2) = 0 \Rightarrow x = 6$ (since $x = - 2$ is rejected, a length cannot be negative)
  1 mark
4. Sides: 6 m, 8 m and 10 m.
  ½ mark
5. Area $= \frac{1}{2} \times 6 \times 8 = 24$ m²
  ½ mark
6. Perimeter $= 6 + 8 + 10 = 24$ m
  ½ mark
7. Difference between numerical values $= 24 - 24 = 0$
  ½ mark
Answer: Sides 6 m, 8 m, 10 m; difference = 0
Where the marks are
- Expresses sides as $x$ , $x + 2$ , $x + 4$ (1 mark)
- Pythagoras equation leading to $x^{2} - 4 x - 12 = 0$ (1 mark)
- Solves to get $x = 6$ (1 mark)
- Sides 6, 8, 10 m (½ mark)
- Area 24 m² (½ mark)
- Perimeter 24 m (½ mark)
- Difference 0 (½ mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerQuadratic Equations
Express the equation $\frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3}$ ; $(x \neq = 3, 5)$ as a quadratic equation in standard form. Hence, find the roots of the equation so formed.
Try it first, then show our solutionSolution
1. $\frac{( x - 2 ) ( x - 5 ) + ( x - 4 ) ( x - 3 )}{( x - 3 ) ( x - 5 )} = \frac{10}{3}$
  $\Rightarrow \frac{( x ^{2} - 7 x + 10 ) + ( x ^{2} - 7 x + 12 )}{x ^{2} - 8 x + 15} = \frac{10}{3}$
  1½ marks
2. $3 (2 x^{2} - 14 x + 22) = 10 (x^{2} - 8 x + 15)$
  $6 x^{2} - 42 x + 66 = 10 x^{2} - 80 x + 150 \Rightarrow 4 x^{2} - 38 x + 84 = 0$
  $\Rightarrow 2 x^{2} - 19 x + 42 = 0$ (standard form)
  1½ marks
3. Splitting the middle term: $2 x^{2} - 12 x - 7 x + 42 = 0 \Rightarrow 2 x (x - 6) - 7 (x - 6) = 0 \Rightarrow (x - 6) (2 x - 7) = 0$
  1 mark
4. $x = 6$ or $x = \frac{7}{2}$ (both allowed since $x \neq = 3, 5$ )
  1 mark
Answer: $2 x^{2} - 19 x + 42 = 0$ ; roots $x = 6, \frac{7}{2}$
Where the marks are
- Combines LHS into a single fraction (1½ marks)
- Simplifies to $2 x^{2} - 19 x + 42 = 0$ (1½ marks)
- Factorises as $(x - 6) (2 x - 7) = 0$ (1 mark)
- Roots $6$ and $\frac{7}{2}$ (1 mark)
Open Q32(B) on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerTriangles
The corresponding sides of $Δ$ ABC and $Δ$ PQR are in the ratio $3 : 5$ . $A D ⊥ B C$ and $P S ⊥ QR$ as shown in the following figures :
1. (i)
  Prove that $Δ$ ADC $\sim$ $Δ$ PSR
  0 marks
2. (ii)
  If AD = 4 cm, find the length of PS.
  0 marks
3. (iii)
  Using (ii) find ar( $Δ$ ABC) : ar( $Δ$ PQR)
  0 marks
Figure in the paper: Triangle ABC with altitude AD perpendicular to BC (D on BC, right angle marked at D). Triangle PQR with altitude PS perpendicular to QR (S on QR, right angle marked at S). See it in the official paper.
Try it first, then show our solutionSolution
1. Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R} = \frac{3}{5}$ , so $Δ A B C \sim Δ P QR$ (SSS similarity), hence $∠ C = ∠ R$ (corresponding angles).
  ½ mark
2. (i) In $Δ A D C$ and $Δ P S R$ :
  $∠ A D C = ∠ P S R = 9 0^{\circ}$ [AD ⟂ BC, PS ⟂ QR]
  $∠ C = ∠ R$ [proved above]
  1 mark
3. $∴ Δ A D C \sim Δ P S R$ [AA similarity]
  ½ mark
4. (ii) Corresponding sides of similar triangles are proportional: $\frac{A D}{P S} = \frac{A C}{P R} = \frac{3}{5}$
  ½ mark
5. $\frac{4}{P S} = \frac{3}{5}$
  ½ mark
6. $P S = \frac{20}{3}$ cm
  ½ mark
7. (iii) $\frac{a r ( Δ A B C )}{a r ( Δ P QR )} = \frac{\frac{1}{2} \times B C \times A D}{\frac{1}{2} \times QR \times P S} = \frac{B C}{QR} \times \frac{A D}{P S}$
  1 mark
8. $= \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$ , so $a r (Δ A B C) : a r (Δ P QR) = 9 : 25$
  ½ mark
Answer: (i) Proved by AA; (ii) PS = 20/3 cm; (iii) 9 : 25
Where the marks are
- $Δ A B C \sim Δ P QR$ so $∠ C = ∠ R$ (½ mark)
- Two pairs of equal angles in $Δ A D C$ , $Δ P S R$ (1 mark)
- Concludes $Δ A D C \sim Δ P S R$ (½ mark)
- Writes $\frac{A D}{P S} = \frac{A C}{P R} = \frac{3}{5}$ (½ mark)
- Substitutes AD = 4 (½ mark)
- PS = 20/3 cm (½ mark)
- Area ratio as ratio of $\frac{1}{2} \times$ base × height (1 mark)
- Ratio 9 : 25 (½ mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerTriangles
State basic proportionality theorem. Use it to prove the following : If three parallel lines $l$ , $m$ , $n$ are intersected by transversals $q$ and $s$ as shown in the adjoining figure, then $\frac{A B}{B C} = \frac{D E}{E F}$ .
Figure in the paper: Three parallel horizontal lines l (top), m (middle), n (bottom). Transversal q meets l at A, m at B, n at C. Transversal s meets l at D, m at E, n at F. Arrows mark the lines as parallel. See it in the official paper.
Try it first, then show our solutionSolution
1. Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
  1 mark
2. Construction: Join AF; let it cut line $m$ at G.
  1 mark
3. In $Δ A C F$ , $B G ∥ C F$ (parts of $m ∥ n$ ). By BPT: $\frac{A B}{B C} = \frac{A G}{GF}$ ...(i)
  1 mark
4. In $Δ F D A$ , $GE ∥ A D$ (parts of $m ∥ l$ ). By BPT: $\frac{F E}{E D} = \frac{F G}{G A}$ , i.e. $\frac{D E}{E F} = \frac{A G}{GF}$ ...(ii)
  1 mark
5. From (i) and (ii): $\frac{A B}{B C} = \frac{D E}{E F}$ . Hence proved.
  1 mark
Answer: BPT stated; $\frac{A B}{B C} = \frac{D E}{E F}$ proved using diagonal AF
Where the marks are
- Correct statement of BPT (1 mark)
- Construction: join AF meeting m at G (1 mark)
- BPT in $Δ A C F$ : $\frac{A B}{B C} = \frac{A G}{GF}$ (1 mark)
- BPT in $Δ F D A$ : $\frac{D E}{E F} = \frac{A G}{GF}$ (1 mark)
- Combines to conclude (1 mark)
Open Q33(B) on its own page
Q34 · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A wooden cubical die is formed by forming hemispherical depressions on each face of the cube such that face 1 has one depression, face 2 has two depressions and so on. The sum of number of hemispherical depressions on opposite faces is always 7. If the edge of the cubical die measures 5 cm and each hemispherical depression is of diameter 1.4 cm, find the total surface area of the die so formed.
Try it first, then show our solutionSolution
1. Number of depressions $= 1 + 2 + 3 + 4 + 5 + 6 = 21$ . Radius of each $r = 0.7$ cm.
  1 mark
2. TSA of die = TSA of cube + CSA of 21 hemispheres − area of 21 circular openings.
  TSA of cube $= 6 \times 5^{2} = 150$ cm²
  1 mark
3. CSA of 21 hemispheres $= 21 \times 2 π r^{2} = 21 \times 2 \times \frac{22}{7} \times (0.7)^{2} = 64.68$ cm²
  1 mark
4. Area of 21 circular bases removed $= 21 \times π r^{2} = 21 \times \frac{22}{7} \times (0.7)^{2} = 32.34$ cm²
  1 mark
5. TSA $= 150 + 64.68 - 32.34 = 182.34$ cm²
  1 mark
Answer: 182.34 cm²
Where the marks are
- Total depressions = 21 (1 mark)
- TSA of cube = 150 cm² (1 mark)
- CSA of 21 hemispheres = 64.68 cm² (1 mark)
- Base area of 21 depressions = 32.34 cm² (1 mark)
- Total surface area 182.34 cm² (1 mark)
Open Q34 on its own page
Q35 · 5 marks · Long answerStatistics
The following table shows the number of patients of different age group who were discharged from the hospital in a particular month :
Age (in years) Number of Patients Discharged
5-15 6
15-25 11
25-35 21
35-45 23
45-55 14
55-65 5
Total 80
Find the 'mean' and the 'mode' of the above data.
Try it first, then show our solutionSolution
1. Find the class marks $x_{i} = \frac{lower limit + upper limit}{2}$ and the products $f_{i} x_{i}$ :
  Age (years) $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  5-15 6 10 60
  15-25 11 20 220
  25-35 21 30 630
  35-45 23 40 920
  45-55 14 50 700
  55-65 5 60 300
  Total $\sum f_{i} = 80$ $\sum f_{i} x_{i} = 2830$
  1½ marks
2. Mean (direct method): $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2830}{80}$
  1 mark
3. $\overset{x}{ˉ} = \frac{283}{8} = 35.375 \approx 35.38$ years
  ½ mark
4. The highest frequency is 23, so the modal class is 35-45.
  Hence $l = 35$ , $f_{1} = 23$ , $f_{0} = 21$ , $f_{2} = 14$ and $h = 10$ .
  ½ mark
5. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 35 + \frac{23 - 21}{46 - 21 - 14} \times 10$
  1 mark
6. $= 35 + \frac{2}{11} \times 10 = 35 + \frac{20}{11} = \frac{405}{11} \approx 36.82$ years
  ½ mark
Answer: Mean = 283/8 ≈ 35.38 years; Mode = 405/11 ≈ 36.82 years
Where the marks are
- Correct table of class marks $x_{i}$ and $f_{i} x_{i}$ with $\sum f_{i} x_{i} = 2830$ (1½ marks)
- Uses the mean formula $\frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2830}{80}$ (1 mark)
- Mean ≈ 35.38 years (½ mark)
- Identifies the modal class as 35-45 (½ mark)
- Correct substitution in the mode formula (1 mark)
- Mode ≈ 36.82 years (½ mark)
Open Q35 on its own page

Age (in years)	Number of Patients Discharged
5-15	6
15-25	11
25-35	21
35-45	23
45-55	14
55-65	5
Total	80

Age (years)	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
Total	$\sum f_{i} = 80$		$\sum f_{i} x_{i} = 2830$

Section E

case-study based; internal choice in 2-mark sub-part of each case

Q36 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
The Olympic symbol comprising five interlocking rings represents the union of the five continents of the world and the meeting of athletes from all over the world at the Olympic games. In order to spread awareness about Olympic games, students of Class-X took part in various activities organised by the school. One such group of students made 5 circular rings in the school lawn with the help of ropes. Each circular ring required 44 m of rope.
Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that $Δ$ OAB is an equilateral triangle and all unshaded regions are congruent.
Based on above information, answer the following questions :
1. (i)
  Find the radius of each circular ring.
  1 mark
2. (ii)
  What is the measure of $∠$ AOB ?
  1 mark
3. (iii)(a)
  Find the area of shaded region $R_{1}$ .
  2 marks
4. OR
  (iii)(b)
  Find the length of rope around the unshaded regions.
  2 marks
Figure in the paper: Left: five interlocking rings (Olympic symbol) in two rows, three on top and two below. Right: a rangoli of five overlapping shaded circles with a central equilateral triangle OAB formed by the centres O, A, B; the unshaded congruent regions lie between the circles near the triangle. Region R1 is marked in the upper-left circle, with radius OR1 and points A, B, O labelled. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The rope used for one ring is its circumference: $2 π r = 44$ , so $2 \times \frac{22}{7} \times r = 44$ .
  ½ mark
2. (i) Therefore $r = \frac{44 \times 7}{44} = 7$ m.
  ½ mark
3. (ii) $Δ O A B$ is equilateral, so each of its angles is $6 0^{\circ}$ . Hence $∠ A O B = 6 0^{\circ}$ .
  1 mark
4. (iii)(a) Region $R_{1}$ is the circle with centre O minus two congruent segments, each cut off by a chord subtending $6 0^{\circ}$ at O.
  Area of one segment $= \frac{60}{360} \times \frac{22}{7} \times 7^{2} - \frac{3}{4} \times 7^{2} = \frac{77}{3} - \frac{49 3}{4}$
  Area of $R_{1} = \frac{22}{7} \times 7 \times 7 - 2 (\frac{77}{3} - \frac{49 3}{4})$
  1 mark
5. (iii)(a) $= 154 - \frac{154}{3} + \frac{49 3}{2} = (\frac{308}{3} + \frac{49 3}{2})$ m² $\approx 145.05$ m² (using $3 = 1.73$ )
  1 mark
6. (iii)(b) There are 4 congruent unshaded regions. Each is bounded by two equal arcs, each subtending $6 0^{\circ}$ at a centre, so the rope needed is $8 \times$ (length of one arc).
  ½ mark
7. (iii)(b) Length of one arc $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{22}{3}$ m, so the total length $= 8 \times \frac{22}{3}$ .
  1 mark
8. (iii)(b) $= \frac{176}{3}$ m $\approx 58.67$ m
  ½ mark
Answer: (i) 7 m; (ii) 60°; (iii)(a) (308/3 + 49√3/2) m² ≈ 145.05 m²; (iii)(b) 176/3 m ≈ 58.67 m
Where the marks are
- Writes $2 π r = 44$ (½ mark)
- Gets $r = 7$ m (½ mark)
- States $∠ A O B = 6 0^{\circ}$ because the triangle is equilateral (1 mark)
- (iii)(a) Writes area = circle − 2 segments with correct substitution (1 mark)
- (iii)(a) Area $= \frac{308}{3} + \frac{49 3}{2} \approx 145.05$ m² (1 mark)
- (iii)(b) Recognises that the rope length is 8 arcs of $6 0^{\circ}$ (½ mark)
- (iii)(b) Correct arc-length substitution (1 mark)
- (iii)(b) Length $= \frac{176}{3} \approx 58.67$ m (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Cable cars at hill stations are one of the major tourist attractions. On a hill station, the length of cable car ride from base point to top most point on the hill is 5000 m. Poles are installed at equal intervals on the way to provide support to the cables on which car moves.
The distance of first pole from base point is 200 m and subsequent poles are installed at equal interval of 150 m. Further, the distance of last pole from the top is 300 m.
Based on above information, answer the following questions using Arithmetic Progression :
1. (i)
  Find the distance of $1 0^{th}$ pole from the base.
  1 mark
2. (ii)
  Find the distance between $1 5^{th}$ pole and $2 5^{th}$ pole.
  1 mark
3. (iii)(a)
  Find the time taken by cable car to reach $1 5^{th}$ pole from the top if it is moving at the speed of 5 m/sec and coming from top.
  2 marks
4. OR
  (iii)(b)
  Find the total number of poles installed along the entire journey.
  2 marks
Figure in the paper: Illustration of cable cars hanging from a cable supported by several poles on a hill slope; the cars travel along the cable between the base and the top station. See it in the official paper.
Try it first, then show our solutionSolution
1. The distances of the poles from the base form an AP 200, 350, 500, … with $a = 200$ and $d = 150$ .
  (i) $a_{10} = a + 9 d = 200 + 9 \times 150 = 1550$ m
  1 mark
2. (ii) $a_{25} - a_{15} = (a + 24 d) - (a + 14 d) = 10 d = 10 \times 150 = 1500$ m
  1 mark
3. (iii)(a) Counting from the top, the poles are at distances 300, 450, 600, … from the top (an AP with $a = 300$ , $d = 150$ ).
  Distance of the 15th pole from the top $= 300 + 14 \times 150 = 2400$ m
  1 mark
4. (iii)(a) Time $= \frac{distance}{speed} = \frac{2400}{5} = 480$ s, i.e. 8 minutes.
  1 mark
5. (iii)(b) Distance of the last pole from the base $= 5000 - 300 = 4700$ m, so $a_{n} = 4700$ .
  ½ mark
6. (iii)(b) $200 + (n - 1) \times 150 = 4700 \Rightarrow (n - 1) \times 150 = 4500 \Rightarrow n - 1 = 30$
  1 mark
7. (iii)(b) $n = 31$ , so 31 poles are installed.
  ½ mark
Answer: (i) 1550 m; (ii) 1500 m; (iii)(a) 480 s (8 minutes); (iii)(b) 31 poles
Where the marks are
- Finds $a_{10} = 200 + 9 \times 150 = 1550$ m (1 mark)
- Finds $a_{25} - a_{15} = 10 d = 1500$ m (1 mark)
- (iii)(a) Distance of the 15th pole from the top $= 300 + 14 \times 150 = 2400$ m (1 mark)
- (iii)(a) Time $= \frac{2400}{5} = 480$ s (1 mark)
- (iii)(b) Last pole is 4700 m from the base (½ mark)
- (iii)(b) Forms $200 + (n - 1) 150 = 4700$ (1 mark)
- (iii)(b) Gets $n = 31$ (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident. The ambulance was travelling at the speed of 60 km/h. The drone stopped at a point Q, 100 m vertically above the point P. The angle of depression of the ambulance was found to be $3 0^{\circ}$ at a particular instant.
Based on above information, answer the following questions :
1. (i)
  Represent the above situation with the help of a diagram.
  1 mark
2. (ii)
  Find the distance between the ambulance and the site of accident (P) at the particular instant. (Use $3 = 1.73$ )
  1 mark
3. (iii)(a)
  Find the time (in seconds) in which the angle of depression changes from $3 0^{\circ}$ to $4 5^{\circ}$ .
  2 marks
4. OR
  (iii)(b)
  How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is $6 0^{\circ}$ from the drone ?
  2 marks
Figure in the paper: Drone at point Q, vertically 100 m above accident point P on the ground. An ambulance on the straight highway is shown to the right, with a line of sight from the drone to the ambulance making the angle of depression at Q. A police car is also shown on the highway between P and the ambulance. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Diagram: draw a vertical segment QP of length 100 m, with Q the drone and P the accident site on the ground. Draw a horizontal ground line from P with the ambulance at R on it, so that $∠ QP R = 9 0^{\circ}$ and $P R = d$ . Draw a horizontal ray QS through Q and join QR. Mark $∠ S QR = 3 0^{\circ}$ (angle of depression), which equals $∠ QR P = 3 0^{\circ}$ [alternate angles].
  1 mark
2. (ii) In right $Δ QP R$ : $tan 3 0^{\circ} = \frac{QP}{P R} \Rightarrow \frac{100}{d} = \frac{1}{3}$
  ½ mark
3. (ii) $d = 1003 = 100 \times 1.73 = 173$ m
  ½ mark
4. (iii)(a) Figure: mark a point M on PR where the angle of depression is $4 5^{\circ}$ , so $∠ QM P = 4 5^{\circ}$ . Let $R M = x$ , so that $P M = 173 - x$ .
  ½ mark
5. (iii)(a) In right $Δ QP M$ : $tan 4 5^{\circ} = \frac{100}{173 - x} = 1$
  ½ mark
6. (iii)(a) $173 - x = 100 \Rightarrow x = 73$ m
  ½ mark
7. (iii)(a) Speed $= 60 \times \frac{5}{18} = \frac{50}{3}$ m/s, so time $= \frac{73}{50/3} = \frac{219}{50} \approx 4.4$ s.
  ½ mark
8. (iii)(b) Figure: mark a point T on the highway with $∠ QT P = 6 0^{\circ}$ (angle of depression $6 0^{\circ}$ ), and let $P T = y$ .
  ½ mark
9. (iii)(b) In right $Δ QP T$ : $tan 6 0^{\circ} = \frac{100}{y} = 3$
  ½ mark
10. (iii)(b) $y = \frac{100}{3} = \frac{100 3}{3} = \frac{173}{3}$ m $\approx 57.67$ m
  ½ mark
11. (iii)(b) Time $= \frac{173/3}{50/3} = \frac{173}{50} \approx 3.5$ s (exactly $23$ s).
  ½ mark
Answer: (ii) 173 m; (iii)(a) ≈ 4.4 s (219/50 s); (iii)(b) 2√3 ≈ 3.5 s
Where the marks are
- (i) Correct labelled figure showing QP = 100 m, a right angle at P and the $3 0^{\circ}$ angle of depression (1 mark)
- (ii) Uses $tan 3 0^{\circ} = \frac{100}{d}$ (½ mark)
- (ii) Gets $d = 173$ m (½ mark)
- (iii)(a) Correct figure with the $4 5^{\circ}$ position marked (½ mark)
- (iii)(a) Forms $\frac{100}{173 - x} = 1$ (½ mark)
- (iii)(a) Gets $x = 73$ m (½ mark)
- (iii)(a) Converts the speed and finds time ≈ 4.4 s (½ mark)
- (iii)(b) Correct figure with the $6 0^{\circ}$ position (½ mark)
- (iii)(b) Forms $\frac{100}{y} = 3$ (½ mark)
- (iii)(b) Gets $y = \frac{173}{3}$ m (½ mark)
- (iii)(b) Time $= 23 \approx 3.5$ s (½ 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.