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

CBSE Class 10 Mathematics Basic: 2026 board paper, set 430/4/1

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 1158-1_430-4-1 Mathematics (Basic).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 28 and 31
D	32-35	Long Answer; internal choice in 33 and 35
E	36-38	Case study based; internal choice in each case study

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (in the case studies, the choice is within the 2-mark subparts). Use of calculator is not allowed; take π = 22/7 where required if not stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If HCF $(850, 325)$ is $25$ , then LCM $(850, 325)$ is :
1. (A) $442$
2. (B) $11050$
3. (C) $8450$
4. (D) $2210$
Try it first, then show our solutionSolution
1. For two numbers, $HCF \times LCM =$ product of the numbers.
2. $LCM = \frac{850 \times 325}{25} = 34 \times 325 = 11050$ , so option (B).
  1 mark
Answer: LCM = 11050, option (B)
Where the marks are
- Uses HCF × LCM = product of the numbers to get 11050 (1 mark)
Q2 · 1 mark · MCQQuadratic Equations
The value of k for which the equation $2 k x^{2} -6 x + 3 = 0$ has real and equal roots, is :
1. (A) $\frac{3}{2}$
2. (B) $\frac{1}{2}$
3. (C) $- \frac{3}{2}$
4. (D) $2$
Try it first, then show our solutionSolution
1. For real and equal roots, $D = b^{2} - 4 a c = 0$ .
2. $(- 6)^{2} - 4 (2 k) (3) = 0 \Rightarrow 36 - 24 k = 0 \Rightarrow k = \frac{3}{2}$ , so option (A).
  1 mark
Answer: k = 3/2, option (A)
Where the marks are
- Sets the discriminant to zero and gets $k = \frac{3}{2}$ (1 mark)
Q3 · 1 mark · MCQSurface Areas and Volumes
Two wooden solid hemispheres of same radii $r$ , are joined at a point, as shown in the figure. The total surface area of the object is :
1. (A) $4 π r^{2}$
2. (B) $6 π r^{2}$
3. (C) $3 π r^{2}$
4. (D) $5 π r^{2}$
Figure in the paper: Two solid hemispheres of equal radius r joined at their flat faces' rims to a common point, one hemisphere above the other, with the curved surfaces facing outward (a bowl-shaped top and bottom, meeting at a point). See it in the official paper.
Try it first, then show our solutionSolution
1. The hemispheres touch only at a point, so no surface is hidden. Each hemisphere shows its curved surface $2 π r^{2}$ and its flat face $π r^{2}$ , a total of $3 π r^{2}$ .
2. Total surface area $= 2 \times 3 π r^{2} = 6 π r^{2}$ , so option (B).
  1 mark
Answer: $6 π r^{2}$ , option (B)
Where the marks are
- Adds the full TSA of both hemispheres to get $6 π r^{2}$ (1 mark)
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Q4 · 1 mark · MCQCircles
PT is tangent to the circle with centre O and radius $5$ cm. OP intersects the circle at Q. If $P Q = x$ , then $P T^{2}$ equals :
1. (A) $x^{2} + 5 x$
2. (B) $x^{2} + 10 x + 50$
3. (C) $x^{2} + 10 x$
4. (D) $x^{2} -25$
Figure in the paper: Circle with centre O; T is a point on the circle where tangent PT touches it; P lies outside on the extension of OQ; Q is the point where OP meets the circle, so OQ = 5 cm and P is beyond Q. See it in the official paper.
Try it first, then show our solutionSolution
1. $O T ⊥ P T$ [tangent ⟂ radius] and $O P = O Q + QP = 5 + x$ .
2. $P T^{2} = O P^{2} - O T^{2} = (5 + x)^{2} - 25 = x^{2} + 10 x$ , so option (C).
  1 mark
Answer: $x^{2} + 10 x$ , option (C)
Where the marks are
- Applies Pythagoras in triangle OTP with $O P = 5 + x$ (1 mark)
Q5 · 1 mark · MCQReal Numbers
Probability of getting an irrational number at random from the numbers $3, 4, 39, 38, 5, 0, 4^{\frac{3}{2}}$ is :
1. (A) $0$
2. (B) $\frac{4}{7}$
3. (C) $\frac{3}{7}$
4. (D) $1$
Try it first, then show our solutionSolution
1. $4 = 2$ , $39 = 9$ , $0$ and $4^{3/2} = 8$ are rational. $3$ , $38 = 62$ and $5$ are irrational.
2. There are 3 irrational numbers out of 7, so $P = \frac{3}{7}$ , option (C).
  1 mark
Answer: 3/7, option (C)
Where the marks are
- Identifies the 3 irrational numbers out of 7 (1 mark)
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Q6 · 1 mark · MCQPolynomials
The graph of polynomial $p (x) = k$ is shown here. Number of zeroes of polynomial $p (x)$ is :
1. (A) $0$
2. (B) $1$
3. (C) $2$
4. (D)infinitely many
Figure in the paper: Cartesian graph with axes XX′ (horizontal) and YY′ (vertical) meeting at origin 0; a horizontal line labelled p(x) = k drawn parallel to the x-axis, above it, not meeting the x-axis. See it in the official paper.
Try it first, then show our solutionSolution
1. Zeroes are the points where the graph meets the x-axis.
2. The line $y = k$ is parallel to the x-axis and lies above it, so it never meets the x-axis. There are 0 zeroes, option (A).
  1 mark
Answer: 0, option (A)
Where the marks are
- Notes that the graph does not cut the x-axis (1 mark)
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Q7 · 1 mark · MCQArithmetic Progressions
The sum of first $n$ terms of an A.P. is $502$ . If the first and the last terms are $2$ and $192$ respectively, then the value of $n$ is :
1. (A) $10$
2. (B) $5$
3. (C) $15$
4. (D) $20$
Try it first, then show our solutionSolution
1. Use $S_{n} = \frac{n}{2} (a + l)$ .
2. $502 = \frac{n}{2} (2 + 192) = 10 n 2 \Rightarrow n = 5$ , so option (B).
  1 mark
Answer: n = 5, option (B)
Where the marks are
- Uses $S_{n} = \frac{n}{2} (a + l)$ to get $n = 5$ (1 mark)
Q8 · 1 mark · MCQStatistics
While calculating mean of a grouped frequency distribution using step deviation method $(u = \frac{x - a}{h})$ it was found that $\overset{x}{ˉ} = 62$ , $a = 47.5$ , $h = 5$ . The value of $\overset{u}{ˉ}$ is :
1. (A) $3$
2. (B) $14.5$
3. (C) $2.9$
4. (D) $3.1$
Try it first, then show our solutionSolution
1. In the step deviation method, $\overset{x}{ˉ} = a + h \overset{u}{ˉ}$ .
2. $62 = 47.5 + 5 \overset{u}{ˉ} \Rightarrow \overset{u}{ˉ} = \frac{14.5}{5} = 2.9$ , so option (C).
  1 mark
Answer: 2.9, option (C)
Where the marks are
- Uses $\overset{x}{ˉ} = a + h \overset{u}{ˉ}$ to get 2.9 (1 mark)
Q9 · 1 mark · MCQProbability
A card is drawn from a well shuffled deck of 52 cards. The probability of getting an ace or a ten is :
1. (A) $\frac{1}{13}$
2. (B) $\frac{2}{13}$
3. (C) $\frac{1}{26}$
4. (D) $\frac{4}{13}$
Try it first, then show our solutionSolution
1. A deck has 4 aces and 4 tens, giving 8 favourable outcomes.
2. $P = \frac{8}{52} = \frac{2}{13}$ , so option (B).
  1 mark
Answer: 2/13, option (B)
Where the marks are
- Counts 8 favourable cards and gets 2/13 (1 mark)
Q10 · 1 mark · MCQCoordinate Geometry
The distance between the points $(-4, 2)$ and $(1, 0)$ is :
1. (A) $13$ units
2. (B) $3$ units
3. (C) $9$ units
4. (D) $29$ units
Try it first, then show our solutionSolution
1. Distance $= (1 + 4)^{2} + (0 - 2)^{2} = 25 + 4 = 29$ units, so option (D).
  1 mark
Answer: $29$ units, option (D)
Where the marks are
- Applies the distance formula to get $29$ (1 mark)
Q11 · 1 mark · MCQSurface Areas and Volumes
A cone with slant height $10$ cm and radius $6$ cm is surmounted on a hemisphere of same radius. The height of the toy is :
1. (A) $12$ cm
2. (B) $16$ cm
3. (C) $14$ cm
4. (D) $10$ cm
Figure in the paper: A cone with its circular base on top of a hemisphere of the same radius; the flat face of the hemisphere is at the bottom and the cone's base circle coincides with the hemisphere's flat circular face. See it in the official paper.
Try it first, then show our solutionSolution
1. Height of the cone $= 1 0^{2} - 6^{2} = 8$ cm.
2. Height of the toy $= 8 + 6 = 14$ cm (cone height + hemisphere radius), so option (C).
  1 mark
Answer: 14 cm, option (C)
Where the marks are
- Finds the cone height 8 cm and adds the radius 6 cm (1 mark)
Q12 · 1 mark · MCQCircles
In the given figure, PQ and PR are two tangents drawn to a circle with centre O. If $∠ O R Q = 2 5^{\circ}$ , then the measure of $∠ P QR$ is :
1. (A) $6 5^{\circ}$
2. (B) $2 5^{\circ}$
3. (C) $5 0^{\circ}$
4. (D) $7 5^{\circ}$
Figure in the paper: Circle with centre O; external point P with tangents PQ and PR touching the circle at Q and R; chord QR is drawn; OR is joined; ∠ORQ = 25° is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. $O Q = O R$ (radii), so $∠ O QR = 2 5^{\circ}$ .
2. $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ P QR = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ , option (A).
  1 mark
Answer: 65°, option (A)
Where the marks are
- Uses the isosceles triangle OQR and tangent ⟂ radius to get 65° (1 mark)
Q13 · 1 mark · MCQIntroduction to Trigonometry
The value of $sin 9 0^{\circ} cos 9 0^{\circ} - sin^{2} 6 0^{\circ}$ is :
1. (A) $\frac{1}{4}$
2. (B) $\frac{3}{4}$
3. (C) $- \frac{3}{4}$
4. (D) $\frac{5}{4}$
Try it first, then show our solutionSolution
1. $sin 9 0^{\circ} cos 9 0^{\circ} - sin^{2} 6 0^{\circ} = 1 \times 0 - \frac{3}{4} = - \frac{3}{4}$ , so option (C).
  1 mark
Answer: −3/4, option (C)
Where the marks are
- Substitutes the standard values to get −3/4 (1 mark)
Q14 · 1 mark · MCQArithmetic Progressions
$n^{th}$ term of the A.P. $- \frac{3}{2}, \frac{3}{2}, \frac{9}{2}, \dots$ is :
1. (A) $\frac{3 n}{2} -3$
2. (B) $3 n - \frac{9}{2}$
3. (C) $\frac{3 n -9}{2}$
4. (D) $3 n + \frac{3}{2}$
Try it first, then show our solutionSolution
1. Here $a = - \frac{3}{2}$ and $d = 3$ .
2. $a_{n} = - \frac{3}{2} + (n - 1) 3 = 3 n - \frac{9}{2}$ , so option (B).
  1 mark
Answer: $3 n - \frac{9}{2}$ , option (B)
Where the marks are
- Applies $a_{n} = a + (n - 1) d$ correctly (1 mark)
Q15 · 1 mark · MCQTriangles
In the given figure, $P Q ∥ B C$ . If $A P : A B = 3 : 7$ then, $A Q : QC$ equals :
1. (A) $3 : 7$
2. (B) $3 : 10$
3. (C) $7 : 3$
4. (D) $3 : 4$
Figure in the paper: Triangle ABC with apex A at top; P lies on AB and Q lies on AC; PQ is drawn parallel to base BC (arrows marked on PQ and BC). See it in the official paper.
Try it first, then show our solutionSolution
1. $A P : P B = 3 : (7 - 3) = 3 : 4$ .
2. By BPT, $A Q : QC = A P : P B = 3 : 4$ , so option (D).
  1 mark
Answer: 3 : 4, option (D)
Where the marks are
- Finds AP : PB = 3 : 4 and applies BPT (1 mark)
Q16 · 1 mark · MCQIntroduction to Trigonometry
If $sin θ = \frac{1}{7}$ , then $tan θ$ is :
1. (A) $\frac{1}{4 3}$
2. (B) $\frac{1}{2 3}$
3. (C) $\frac{4 3}{7}$
4. (D) $\frac{6}{7}$
Try it first, then show our solutionSolution
1. $cos θ = 1 - \frac{1}{49} = \frac{48}{7} = \frac{4 3}{7}$ .
2. $tan θ = \frac{sin θ}{cos θ} = \frac{1}{4 3}$ , so option (A).
  1 mark
Answer: $\frac{1}{4 3}$ , option (A)
Where the marks are
- Finds $cos θ$ and then $tan θ$ (1 mark)
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Q17 · 1 mark · MCQReal Numbers
$7 \times 29 \times 23 + 1$ is :
1. (A)a prime number.
2. (B)divisible by $23$ .
3. (C)an odd number.
4. (D)a composite number.
Try it first, then show our solutionSolution
1. $7 \times 29 \times 23 = 4669$ , which is odd, so $4669 + 1 = 4670$ is even.
2. $4670 = 2 \times 5 \times 467$ has factors other than 1 and itself, so it is composite, option (D).
  1 mark
Answer: a composite number, option (D)
Where the marks are
- Shows that the number is even, hence composite (1 mark)
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Q18 · 1 mark · MCQCircles
Chord AB subtends an angle of $4 0^{\circ}$ at the centre of the circle of radius $9$ cm. The length of arc AB is :
1. (A) $\frac{22}{7}$ cm
2. (B) $\frac{198}{7}$ cm
3. (C) $\frac{44}{7}$ cm
4. (D) $\frac{54}{7}$ cm
Try it first, then show our solutionSolution
1. Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{40}{360} \times 2 \times \frac{22}{7} \times 9 = \frac{44}{7}$ cm, so option (C).
  1 mark
Answer: 44/7 cm, option (C)
Where the marks are
- Uses the arc length formula to get 44/7 cm (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A): The probability of an event can not be $\frac{1}{0.9}$ .
Reason (R): $0 \leq P (E) \leq 1$ for an event E.
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 correct explanation for Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $\frac{1}{0.9} \approx 1.11 > 1$ .
2. Since $0 \leq P (E) \leq 1$ , a probability cannot exceed 1. Both A and R are true, and R explains A, so option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises $\frac{1}{0.9} > 1$ , which violates $0 \leq P (E) \leq 1$ (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A): $4^{n}$ can not end with the digit zero.
Reason (R): Prime factorisation of $4^{n}$ is unique.
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 correct explanation for 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 factor 5, so it cannot end in 0. A is true.
2. R is true by the Fundamental Theorem of Arithmetic, but uniqueness alone does not explain A; the absence of the factor 5 does. So option (B).
  1 mark
Answer: Option (B)
Where the marks are
- States both A and R are true, but R does not explain A (1 mark)
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Section B

Very Short Answer; internal choice in 22 and 24

Q21 · 2 marks · Very short answerIntroduction to Trigonometry
If $sin (A - B) = \frac{1}{2}$ and $tan (A + B) = 3$ , $0^{\circ} \leq A + B < 9 0^{\circ}$ , $A > B$ then find the values of A and B.
Try it first, then show our solutionSolution
1. $sin (A - B) = \frac{1}{2} = sin 3 0^{\circ} \Rightarrow A - B = 3 0^{\circ}$ ...(i)
  ½ mark
2. $tan (A + B) = 3 = tan 6 0^{\circ} \Rightarrow A + B = 6 0^{\circ}$ ...(ii)
  ½ mark
3. Adding (i) and (ii): $2 A = 9 0^{\circ} \Rightarrow A = 4 5^{\circ}$ . Substituting in (ii): $B = 6 0^{\circ} - 4 5^{\circ} = 1 5^{\circ}$ .
  1 mark
Answer: $A = 4 5^{\circ}$ , $B = 1 5^{\circ}$
Where the marks are
- Obtains $A - B = 3 0^{\circ}$ (½ mark)
- Obtains $A + B = 6 0^{\circ}$ (½ mark)
- Solves to get $A = 4 5^{\circ}$ , $B = 1 5^{\circ}$ (1 mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerPolynomials
If $α, β$ are the zeroes of polynomial $p (x) = 6 x^{2} -5 x -3$ , then find the value of $\frac{1}{α} + \frac{1}{β}$ .
Try it first, then show our solutionSolution
1. For $6 x^{2} - 5 x - 3$ : $a = 6, b = - 5, c = - 3$ . So $α + β = - \frac{b}{a} = \frac{5}{6}$ and $α β = \frac{c}{a} = \frac{- 3}{6} = - \frac{1}{2}$ .
  1 mark
2. $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{5/6}{- 1/2} = - \frac{5}{3}$
  1 mark
Answer: $\frac{1}{α} + \frac{1}{β} = - \frac{5}{3}$
Where the marks are
- Writes sum $= \frac{5}{6}$ and product $= - \frac{3}{6}$ of zeroes (1 mark)
- Uses $\frac{α + β}{α β}$ to get $- \frac{5}{3}$ (1 mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerPolynomials
One zero of the polynomial $4 x^{2} -12 x + (2 k + 1)$ is five times the other. Find the value of k.
Try it first, then show our solutionSolution
1. Let the zeroes be $α$ and $5 α$ .
  ½ mark
2. Sum of zeroes: $α + 5 α = \frac{12}{4} = 3 \Rightarrow 6 α = 3 \Rightarrow α = \frac{1}{2}$
  ½ mark
3. Product of zeroes: $α \times 5 α = \frac{2 k + 1}{4} \Rightarrow 5 \cdot \frac{1}{4} = \frac{2 k + 1}{4}$
  ½ mark
4. $\Rightarrow 2 k + 1 = 5 \Rightarrow k = 2$
  ½ mark
Answer: $k = 2$
Where the marks are
- Takes zeroes as $α$ and $5 α$ (½ mark)
- Uses sum of zeroes to get $α = \frac{1}{2}$ (½ mark)
- Writes product $5 α^{2} = \frac{2 k + 1}{4}$ (½ mark)
- Obtains $k = 2$ (½ mark)
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Q23 · 2 marks · Very short answerCoordinate Geometry
A $(-2, 3)$ and B $(4, 1)$ are end points of the diameter of a semi-circle. The semi-circle intersects y-axis at point P. Find the co-ordinates of the point P.
Try it first, then show our solutionSolution
1. Centre O is the mid-point of diameter AB: $O = (\frac{- 2 + 4}{2}, \frac{3 + 1}{2}) = (1, 2)$ .
  ½ mark
2. Let $P = (0, y)$ (it lies on the y-axis). Since P is on the circle, $O P = O B$ (radii): $O P^{2} = O B^{2} \Rightarrow (0 - 1)^{2} + (y - 2)^{2} = (4 - 1)^{2} + (1 - 2)^{2}$
3. $1 + (y - 2)^{2} = 10 \Rightarrow (y - 2)^{2} = 9 \Rightarrow y - 2 = \pm 3 \Rightarrow y = 5$ or $y = - 1$
  1 mark
4. Hence P is $(0, 5)$ or $(0, - 1)$ (depending on which side of AB the semi-circle lies; the figure shows it above AB, giving $(0, 5)$ ).
  ½ mark
Answer: P(0, 5) or P(0, −1)
Where the marks are
- Finds centre $(1, 2)$ using mid-point formula (½ mark)
- Sets $O P^{2} = O B^{2}$ and solves to get $y = 5$ or $- 1$ (1 mark)
- States coordinates of P (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerCircles
In the given figure, AP, AQ and BC are tangents to the circle with centre O. If $A B = 6$ cm, $A C = 7$ cm and $B C = 5$ cm, then what is the length of AP ?
Figure in the paper: A circle with centre O inside a triangle-like figure: apex A at top with tangent lines AP and AQ touching the circle at P (left) and Q (right); a third tangent BC touches the circle at R, with B on AP and C on AQ. See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents from an external point are equal: $B P = B R$ , $C Q = C R$ , $A P = A Q$ . So $A P = A B + B P = A B + B R$ and $A Q = A C + C Q = A C + C R$ .
  ½ mark
2. Adding: $A P + A Q = A B + A C + (B R + C R) = A B + A C + B C$
  ½ mark
3. Since $A P = A Q$ : $2 A P = 6 + 7 + 5 = 18$
  ½ mark
4. $\Rightarrow A P = 9$ cm
  ½ mark
Answer: AP = 9 cm
Where the marks are
- Writes $A P = A B + B R$ and $A Q = A C + C R$ using equal tangents (½ mark)
- Adds to get $A P + A Q = A B + B C + A C$ (½ mark)
- $2 A P = 18$ (½ mark)
- $A P = 9$ cm (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerCircles
In the given figure, TP is tangent to a circle with centre O. Diameter BA when produced meets the tangent at T. If $∠ A B P = 3 5^{\circ}$ , then find the measure of $∠ P T A$ .
Figure in the paper: Circle with centre O; diameter BA extended beyond A to meet tangent TP at T; P is a point on the circle; chords BP and AP are drawn; ∠ABP = 35° is marked at B. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ O B P$ , $O B = O P$ (radii), so $∠ O P B = ∠ O B P = 3 5^{\circ}$ . Hence $∠ B O P = 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ and $∠ P O T = 18 0^{\circ} - 11 0^{\circ} = 7 0^{\circ}$ [linear pair].
  1 mark
2. $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius]. In $△ O P T$ : $∠ P T O = 18 0^{\circ} - 9 0^{\circ} - 7 0^{\circ} = 2 0^{\circ}$ , i.e. $∠ P T A = 2 0^{\circ}$ .
  1 mark
Answer: $∠ P T A = 2 0^{\circ}$
Where the marks are
- Uses $O B = O P$ to get $∠ B O P = 11 0^{\circ}$ (or $∠ P O A = 7 0^{\circ}$ ) (1 mark)
- Uses $∠ O P T = 9 0^{\circ}$ and angle sum to get $∠ P T A = 2 0^{\circ}$ (1 mark)
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Q25 · 2 marks · Very short answerTriangles
In the given figure, $QR ∥ C B$ and $R P ∥ A C$ . If $B R = 10$ cm, $Q A = 12$ cm, $B P = 12$ cm and $P C = 18$ cm, then find the lengths of AR and QC.
Figure in the paper: Triangle ABC with apex A; R lies on AB, Q lies on AC, P lies on BC; QR is parallel to CB (horizontal); RP is parallel to AC. Given BR = 10 cm, QA = 12 cm, BP = 12 cm, PC = 18 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ , $R P ∥ A C$ , so by BPT: $\frac{B P}{P C} = \frac{B R}{R A} \Rightarrow \frac{12}{18} = \frac{10}{A R} \Rightarrow A R = 15$ cm
  1 mark
2. In $△ A B C$ , $QR ∥ C B$ , so by BPT: $\frac{A R}{R B} = \frac{A Q}{QC} \Rightarrow \frac{15}{10} = \frac{12}{QC} \Rightarrow QC = 8$ cm
  1 mark
Answer: AR = 15 cm, QC = 8 cm
Where the marks are
- Applies BPT with $R P ∥ A C$ to get $A R = 15$ cm (1 mark)
- Applies BPT with $QR ∥ C B$ to get $QC = 8$ cm (1 mark)
Open Q25 on its own page

Section C

Short Answer; internal choice in 28 and 31

Q26 · 3 marks · Short answerCircles
Prove that a parallelogram circumscribing a circle, is a rhombus.
Try it first, then show our solutionSolution
1. Figure: parallelogram ABCD with a circle inside touching AB at P, BC at Q, CD at R and DA at S. Given: ABCD is a parallelogram circumscribing the circle. To prove: ABCD is a rhombus.
  ½ mark
2. Tangents 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: $A P + B P + C R + D R = A S + B Q + C Q + D S$ , i.e. $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$ , so $A B + C D = A D + B C$ .
  1 mark
4. In a parallelogram $A B = C D$ and $A D = B C$ , so $2 A B = 2 B C \Rightarrow A B = B C$ . Adjacent sides are equal, so all four sides are equal; hence ABCD is a rhombus.
  ½ mark
Answer: AB = BC = CD = DA, so ABCD is a rhombus.
Where the marks are
- Correct labelled figure (½ mark)
- Writes the four pairs of equal tangents (1 mark)
- Adds to get $A B + C D = A D + B C$ (1 mark)
- Uses opposite sides equal to conclude $A B = B C$ , rhombus (½ mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerAreas Related to Circles
A chord of a circle of radius $14$ cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major segment.
Try it first, then show our solutionSolution
1. (i) Area of minor segment = area of sector − area of right triangle $= \frac{90}{360} \times \frac{22}{7} \times 14 \times 14 - \frac{1}{2} \times 14 \times 14$
  1 mark
2. $= 154 - 98 = 56 cm^{2}$
  1 mark
3. (ii) Area of major segment = area of circle − minor segment $= \frac{22}{7} \times 14 \times 14 - 56$
  ½ mark
4. $= 616 - 56 = 560 cm^{2}$
  ½ mark
Answer: (i) 56 cm²; (ii) 560 cm²
Where the marks are
- Correct expression sector − triangle for minor segment (1 mark)
- Minor segment $= 56$ cm² (1 mark)
- Major segment = circle − minor segment (½ mark)
- Major segment $= 560$ cm² (½ mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{s e c A - t a n A}{s e c A + t a n A}$
Try it first, then show our solutionSolution
1. LHS $= \frac{\frac{c o s A}{s i n A} - c o s A}{\frac{c o s A}{s i n A} + c o s A} = \frac{c o s A - s i n A c o s A}{c o s A + s i n A c o s A}$ (multiplying numerator and denominator by $sin A$ )
  1 mark
2. $= \frac{c o s A ( 1 - s i n A )}{c o s A ( 1 + s i n A )} = \frac{1 - s i n A}{1 + s i n A}$
  ½ mark
3. Dividing numerator and denominator by $cos A$ : $= \frac{\frac{1}{c o s A} - \frac{s i n A}{c o s A}}{\frac{1}{c o s A} + \frac{s i n A}{c o s A}}$
  1 mark
4. $= \frac{s e c A - t a n A}{s e c A + t a n A} =$ RHS. Hence proved.
  ½ mark
Answer: LHS = $\frac{1 - s i n A}{1 + s i n A}$ = RHS; identity proved.
Where the marks are
- Writes cot A as $\frac{c o s A}{s i n A}$ and simplifies (1 mark)
- Reduces to $\frac{1 - s i n A}{1 + s i n A}$ (½ mark)
- Divides by $cos A$ (1 mark)
- Obtains $\frac{s e c A - t a n A}{s e c A + t a n A}$ (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $(cosec A - sin A) (sec A - cos A) = \frac{1}{t a n A + c o t A}$
Try it first, then show our solutionSolution
1. LHS $= (\frac{1}{s i n A} - sin A) (\frac{1}{c o s A} - cos A) = \frac{1 - s i n ^{2} A}{s i n A} \times \frac{1 - c o s ^{2} A}{c o s A}$
  1 mark
2. $= \frac{c o s ^{2} A}{s i n A} \times \frac{s i n ^{2} A}{c o s A} = sin A cos A$ [using $sin^{2} A + cos^{2} A = 1$ ]
  1 mark
3. $= \frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A}$ ; dividing numerator and denominator by $sin A cos A$ : $= \frac{1}{\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}} = \frac{1}{t a n A + c o t A} =$ RHS. Hence proved.
  1 mark
Answer: Both sides equal $sin A cos A$ ; identity proved.
Where the marks are
- Writes cosec A and sec A in terms of sin and cos and simplifies brackets (1 mark)
- Reduces LHS to $sin A cos A$ (1 mark)
- Shows $sin A cos A = \frac{1}{t a n A + c o t A}$ (1 mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerReal Numbers
Prove that $5$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $5$ is rational.
  Then $5 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are co-prime.
  ½ mark
2. Squaring gives $5 q^{2} = p^{2}$ . So $p^{2}$ is divisible by 5.
  Since 5 is prime, $p$ is divisible by 5 [if a prime divides $p^{2}$ , it divides $p$ ].
  So $p = 5 a$ for some integer $a$ .
  1 mark
3. Substitute: $25 a^{2} = 5 q^{2} \Rightarrow q^{2} = 5 a^{2}$ . So $q^{2}$ is divisible by 5, and hence $q$ is divisible by 5.
  ½ mark
4. Now 5 is a common factor of $p$ and $q$ . This contradicts the assumption that $p$ and $q$ are co-prime.
  Hence our assumption is wrong, and $5$ is irrational.
  1 mark
Answer: $5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 = p / q$ with $p, q$ co-prime (½ mark)
- Shows 5 divides $p$ , writes $p = 5 a$ (1 mark)
- Shows 5 divides $q$ (½ mark)
- States the contradiction and concludes (1 mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerProbability
Two dice are rolled together. Find the probability that :
1. (i)
  the sum of the numbers obtained is $10$ .
  0 marks
2. (ii)
  the product of the numbers obtained is $6$ .
  0 marks
Try it first, then show our solutionSolution
1. Total possible outcomes when two dice are rolled $= 6 \times 6 = 36$ .
2. (i) Sum 10: favourable outcomes $(4, 6), (5, 5), (6, 4)$ , i.e. 3. $P = \frac{3}{36} = \frac{1}{12}$
  1½ marks
3. (ii) Product 6: favourable outcomes $(1, 6), (6, 1), (2, 3), (3, 2)$ , i.e. 4. $P = \frac{4}{36} = \frac{1}{9}$
  1½ marks
Answer: (i) $\frac{1}{12}$ ; (ii) $\frac{1}{9}$
Where the marks are
- Lists 3 favourable outcomes for sum 10 and gets $\frac{3}{36} = \frac{1}{12}$ (1½ marks)
- Lists 4 favourable outcomes for product 6 and gets $\frac{4}{36} = \frac{1}{9}$ (1½ marks)
Open Q30 on its own page
Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerCoordinate Geometry
The vertices of a rhombus ABCD are $A (-3, -4)$ , $B (5, -3)$ , $C (1, 4)$ and $D (-7, 3)$ . Find the length of both the diagonals. Hence, find area of the rhombus ABCD.
Try it first, then show our solutionSolution
1. $A C = (1 + 3)^{2} + (4 + 4)^{2} = 16 + 64 = 80 = 45$ units
  1 mark
2. $B D = (- 7 - 5)^{2} + (3 + 3)^{2} = 144 + 36 = 180 = 65$ units
  1 mark
3. Area of rhombus $= \frac{1}{2} \times d_{1} \times d_{2} = \frac{1}{2} \times 45 \times 65 = \frac{1}{2} \times 120 = 60$ square units
  1 mark
Answer: AC = $45$ units, BD = $65$ units, Area = 60 sq. units
Where the marks are
- $A C = 45$ (1 mark)
- $B D = 65$ (1 mark)
- Area $= \frac{1}{2} d_{1} d_{2} = 60$ sq units (1 mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerCoordinate Geometry
The line segment joining the points $A (-5, 1)$ and $B (7, 6)$ is trisected at the points P and Q such that P is nearer to A. If P lies on the line $x + y = k$ , then find the value of $k$ .
Try it first, then show our solutionSolution
1. P is the point of trisection nearer to A, so $A P : P B = 1 : 2$ .
  ½ mark
2. By section formula, $P = (\frac{1 \times 7 + 2 \times ( - 5 )}{1 + 2}, \frac{1 \times 6 + 2 \times 1}{1 + 2}) = (\frac{- 3}{3}, \frac{8}{3}) = (- 1, \frac{8}{3})$
  1½ marks
3. P lies on $x + y = k$ : $k = - 1 + \frac{8}{3} = \frac{5}{3}$
  1 mark
Answer: $k = \frac{5}{3}$
Where the marks are
- Identifies ratio $A P : P B = 1 : 2$ (½ mark)
- Finds $P (- 1, \frac{8}{3})$ by section formula (1½ marks)
- Substitutes in $x + y = k$ to get $k = \frac{5}{3}$ (1 mark)
Open Q31(b) on its own page

Section D

Long Answer; internal choice in 33 and 35

Q32 · 5 marks · Long answerArithmetic Progressions
The third and ninth term of an A.P. are $4$ and $-8$ respectively.
1. (i)
  Which term of the A.P. is zero ?
  0 marks
2. (ii)
  Find the value of $n$ if $S_{n} = -36$ .
  0 marks
Try it first, then show our solutionSolution
1. (i) $a_{3} = a + 2 d = 4$ ...(1)
  ½ mark
2. $a_{9} = a + 8 d = - 8$ ...(2)
  ½ mark
3. Subtracting (1) from (2): $6 d = - 12 \Rightarrow d = - 2$ ; then $a = 4 - 2 (- 2) = 8$ .
  1 mark
4. $a_{n} = 0 \Rightarrow 8 + (n - 1) (- 2) = 0 \Rightarrow n - 1 = 4 \Rightarrow n = 5$ . So the 5th term is zero.
  1 mark
5. (ii) $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = - 36 \Rightarrow \frac{n}{2} [16 + (n - 1) (- 2)] = - 36$
  1 mark
6. $\Rightarrow \frac{n}{2} (18 - 2 n) = - 36 \Rightarrow 9 n - n^{2} = - 36 \Rightarrow n^{2} - 9 n - 36 = 0$
  ½ mark
7. $\Rightarrow (n - 12) (n + 3) = 0 \Rightarrow n = 12$ or $n = - 3$ . Since n is a natural number, reject $n = - 3$ ; hence $n = 12$ .
  ½ mark
Answer: (i) 5th term; (ii) n = 12
Where the marks are
- Writes $a + 2 d = 4$ (½ mark)
- Writes $a + 8 d = - 8$ (½ mark)
- Solves to get $d = - 2$ , $a = 8$ (½ + ½ marks)
- Gets $n = 5$ from $a_{n} = 0$ (1 mark)
- Correct $S_{n}$ equation (1 mark)
- Forms $n^{2} - 9 n - 36 = 0$ (½ mark)
- Factorises and takes $n = 12$ , rejecting $- 3$ (½ mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerTriangles
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ . Take a line parallel to $B C$ that meets $A B$ at $D$ and $A C$ at $E$ . Join $B E$ and $C D$ . Draw $E N ⊥ A B$ (N on AB) and $D M ⊥ A C$ (M on AC).
  ½ mark
2. Given: In $△ A B C$ , $D E ∥ B C$ , with $D E$ meeting $A B$ at $D$ and $A C$ at $E$ .
  ½ mark
3. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  ½ mark
4. Construction: Join $B E$ and $C D$ . Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  ½ mark
5. Proof: Use area $= \frac{1}{2} \times$ base $\times$ height.
  In $△ A D E$ and $△ B D E$ , the height from $E$ is $E N$ for both:
  $\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} \cdot A D \cdot E N}{\frac{1}{2} \cdot D B \cdot E N} = \frac{A D}{D B}$ ... (1)
  1 mark
6. In $△ A D E$ and $△ D E C$ , the height from $D$ is $D M$ for both:
  $\frac{a r ( A D E )}{a r ( D E C )} = \frac{\frac{1}{2} \cdot A E \cdot D M}{\frac{1}{2} \cdot E C \cdot D M} = \frac{A E}{E C}$ ... (2)
  1 mark
7. $△ B D E$ and $△ D E C$ have the same base $D E$ and lie between the same parallels $D E$ and $B C$ . So $a r (B D E) = a r (D E C)$ ... (3)
  From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: Proved: if DE ∥ BC in △ABC, then AD/DB = AE/EC (Basic Proportionality Theorem).
Where the marks are
- Draws a correct labelled figure (½ mark)
- States what is given (½ mark)
- States what is to be proved (½ mark)
- States the construction: join BE and CD, draw EN ⊥ AB and DM ⊥ AC (½ mark)
- Expresses $a r (A D E) / a r (B D E)$ as $A D / D B$ (1 mark)
- Expresses $a r (A D E) / a r (D E C)$ as $A E / E C$ (1 mark)
- Uses equal areas of triangles on the same base between the same parallels to conclude (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerTriangles
$A D$ and $P S$ are respectively, the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ , then prove that
1. (i)
  $△ A D C \sim △ P S R$
  0 marks
2. (ii)
  $\frac{A D}{P S} = \frac{B C}{QR}$
  0 marks
Figure in the paper: Two triangles side by side: triangle ABC with median AD (D on BC, B left, C right, A apex); triangle PQR with median PS (S on QR, Q left, R right, P apex). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Since $△ A B C \sim △ P QR$ , corresponding sides are proportional: $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{B C}{QR}$ . As D and S are mid-points of BC and QR, $B C = 2 D C$ and $QR = 2 S R$ , so $\frac{B C}{QR} = \frac{2 D C}{2 S R} = \frac{D C}{S R}$ .
  2 marks
2. Hence $\frac{A C}{P R} = \frac{D C}{S R}$ .
  ½ mark
3. Also $∠ C = ∠ R$ (corresponding angles of similar triangles). So in $△ A D C$ and $△ P S R$ , two sides are proportional and the included angle is equal; $∴ △ A D C \sim △ P S R$ [SAS similarity].
  ½ mark
4. (ii) Since $△ A D C \sim △ P S R$ , $\frac{A D}{P S} = \frac{D C}{S R}$ .
  1 mark
5. $\frac{A D}{P S} = \frac{B C /2}{QR /2} = \frac{B C}{QR}$ . Hence proved.
  1 mark
Answer: (i) $△ A D C \sim △ P S R$ by SAS; (ii) $\frac{A D}{P S} = \frac{B C}{QR}$
Where the marks are
- Writes ratio of sides from similarity and $\frac{B C}{QR} = \frac{2 D C}{2 S R}$ (2 marks)
- Gets $\frac{A C}{P R} = \frac{D C}{S R}$ (½ mark)
- Uses $∠ C = ∠ R$ and SAS similarity (½ mark)
- Writes $\frac{A D}{P S} = \frac{D C}{S R}$ (1 mark)
- Replaces DC, SR by halves of BC, QR to conclude (1 mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
The angle of elevation of the top of a building from the foot of the tower is $3 0^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $6 0^{\circ}$ . If the tower is $60$ m high, find the height of the building and distance between the building and the tower. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: let CD = 60 m be the tower (foot C) and AB = h m be the building (foot A), both vertical on level ground, with CA = x m. At A, the angle of elevation of D is $6 0^{\circ}$ ( $∠ D A C = 6 0^{\circ}$ ); at C, the angle of elevation of B is $3 0^{\circ}$ ( $∠ B C A = 3 0^{\circ}$ ).
  1 mark
2. In right $△ D C A$ : $tan 6 0^{\circ} = \frac{D C}{C A} = \frac{60}{x}$
  1 mark
3. $\Rightarrow 3 = \frac{60}{x} \Rightarrow x = \frac{60}{3} = 203 = 20 \times 1.73 = 34.6$ m
  1 mark
4. In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{C A} = \frac{h}{x}$
  1 mark
5. $\Rightarrow \frac{1}{3} = \frac{h}{20 3} \Rightarrow h = 20$ m
  1 mark
Answer: Height of building = 20 m; distance between building and tower = $203$ m = 34.6 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ} = \frac{60}{x}$ (1 mark)
- Gets $x = 203 = 34.6$ m (1 mark)
- Uses $tan 3 0^{\circ} = \frac{h}{x}$ (1 mark)
- Gets $h = 20$ m (1 mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables
The difference between two numbers is $12$ . The greater number is $6$ less than twice the smaller one.
1. (i)
  Representing the above situation, frame two linear equations in two variables.
  0 marks
2. (ii)
  Show that the equations have unique solution.
  0 marks
3. (iii)
  Solve the equations and hence find the numbers.
  0 marks
Try it first, then show our solutionSolution
1. (i) Let the greater number be x and the smaller be y. Difference is 12: $x - y = 12$ ...(1)
  1 mark
2. Greater is 6 less than twice the smaller: $x = 2 y - 6$ , i.e. $x - 2 y = - 6$ ...(2)
  1 mark
3. (ii) $\frac{a _{1}}{a _{2}} = \frac{1}{1} = 1$ , $\frac{b _{1}}{b _{2}} = \frac{- 1}{- 2} = \frac{1}{2}$ . Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect, so the pair has a unique solution.
  1 mark
4. (iii) Subtracting (2) from (1): $(x - y) - (x - 2 y) = 12 - (- 6) \Rightarrow y = 18$
  1 mark
5. From (1): $x = 12 + 18 = 30$ . The numbers are 30 and 18.
  1 mark
Answer: (i) $x - y = 12$ , $x - 2 y = - 6$ ; (ii) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , unique solution; (iii) numbers are 30 and 18
Where the marks are
- Forms $x - y = 12$ (1 mark)
- Forms $x - 2 y = - 6$ (1 mark)
- Compares ratios $\frac{1}{1} \neq = \frac{- 1}{- 2}$ to show unique solution (1 mark)
- Finds $y = 18$ (1 mark)
- Finds $x = 30$ (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerPair of Linear Equations in Two Variables
Solve the following equations graphically : $x + y = 7$ and $2 x -5 y = 7$
Try it first, then show our solutionSolution
1. For $x + y = 7$ , i.e. $y = 7 - x$ : table of values $x = 0, 1, 7$ gives $y = 7, 6, 0$ ; points $(0, 7), (1, 6), (7, 0)$ . Plot them and join to draw the line.
  2 marks
2. For $2 x - 5 y = 7$ , i.e. $y = \frac{2 x - 7}{5}$ : table of values $x = 1, 3.5, 6$ gives $y = - 1, 0, 1$ ; points $(1, - 1), (3.5, 0), (6, 1)$ . Plot them and join to draw the second line.
  2 marks
3. The two lines intersect at $(6, 1)$ . Check: $6 + 1 = 7$ and $2 (6) - 5 (1) = 7$ . Hence the solution is $x = 6, y = 1$ .
  1 mark
Answer: $x = 6$ , $y = 1$ (lines meet at (6, 1))
Where the marks are
- Correct points and line for $x + y = 7$ (2 marks)
- Correct points and line for $2 x - 5 y = 7$ (2 marks)
- Reads intersection: $x = 6$ , $y = 1$ (1 mark)
Open Q35(b) on its own page

Section E

Case study based; internal choice in each case study

Q36 · 4 marks · Case-basedCompetency-basedQuadratic Equations
Observe the figure given above. It shows six identical rectangular enclosures made by using fencing wire mesh. These enclosures are used to protect baby animals in a zoo. Dimensions of each enclosure is $x$ feet $\times$ $y$ feet. The total length of fencing required is $152$ feet and area of each enclosure is $80$ square feet.
Based on the above, answer the following questions :
1. (i)
  Write an expression for length of fencing required in terms of $x$ and $y$ .
  1 mark
2. (ii)
  Write the area of each enclosure in terms of $x$ .
  1 mark
3. (iii)
  (a) Write the above equation in quadratic equation form and thus find the dimensions of each enclosure using factorisation method.
  2 marks
4. OR
  (iii)
  (b) Using above equation in quadratic form, solve the equation and find the dimensions of each enclosure using quadratic formula.
  2 marks
Figure in the paper: Left: photo of an animal behind wire mesh. Right: a rectangle of width x (top, marked with arrows) and height y (right side), divided into six identical rectangular enclosures arranged in 2 columns and 3 rows, with fencing lines and tick marks on the sides. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The six enclosures sit in 2 columns and 3 rows. There are 4 horizontal fence lines, each of length $2 x$ , which give $8 x$ . There are 3 vertical fence lines, each of length $3 y$ , which give $9 y$ .
  Length of fencing: $8 x + 9 y = 152$
  1 mark
2. (ii) From (i), $y = \frac{152 - 8 x}{9}$ .
  Area of each enclosure $= x y = \frac{x ( 152 - 8 x )}{9}$ , so $\frac{x ( 152 - 8 x )}{9} = 80$ .
  1 mark
3. (iii)(a) Rearrange into quadratic form:
  $152 x - 8 x^{2} = 720 \Rightarrow 8 x^{2} - 152 x + 720 = 0 \Rightarrow x^{2} - 19 x + 90 = 0$
  ½ mark
4. Factorise by splitting the middle term:
  $x^{2} - 10 x - 9 x + 90 = 0 \Rightarrow (x - 10) (x - 9) = 0$
  ½ mark
5. So $x = 10$ or $x = 9$ .
  ½ mark
6. Find $y$ from $y = \frac{80}{x}$ :
  When $x = 10$ , $y = 8$ .
  When $x = 9$ , $y = \frac{80}{9}$ .
  Dimensions of each enclosure: $10$ ft $\times$ $8$ ft, or $9$ ft $\times \frac{80}{9}$ ft.
  ½ mark
7. (iii)(b) The quadratic is $x^{2} - 19 x + 90 = 0$ , with $a = 1$ , $b = - 19$ , $c = 90$ .
  Discriminant: $D = b^{2} - 4 a c = 361 - 360 = 1$
  ½ mark
8. By the quadratic formula:
  $x = \frac{- b \pm D}{2 a} = \frac{19 \pm 1}{2} = \frac{19 \pm 1}{2}$
  ½ mark
9. So $x = 10$ or $x = 9$ .
  ½ mark
10. Then $y = \frac{80}{x}$ gives $y = 8$ or $y = \frac{80}{9}$ .
  Dimensions: $10$ ft $\times$ $8$ ft, or $9$ ft $\times \frac{80}{9}$ ft.
  ½ mark
Answer: (i) $8 x + 9 y = 152$ ; (ii) $\frac{x ( 152 - 8 x )}{9} = 80$ ; (iii) $x^{2} - 19 x + 90 = 0$ , so $x = 10$ , $y = 8$ (or $x = 9$ , $y = \frac{80}{9}$ ): each enclosure is 10 ft × 8 ft (or 9 ft × 80/9 ft)
Where the marks are
- Writes the fencing equation $8 x + 9 y = 152$ (1 mark)
- Writes the area as $\frac{x ( 152 - 8 x )}{9} = 80$ (1 mark)
- Forms $x^{2} - 19 x + 90 = 0$ (½ mark)
- Factorises as $(x - 10) (x - 9) = 0$ in (a), or substitutes correctly into the quadratic formula in (b) (½ mark)
- Gets $x = 10$ or $9$ (½ mark)
- Gets $y = 8$ or $\frac{80}{9}$ (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedStatistics
The Air Quality Index (AQI) is a scale from 0 to 500 that indicates air quality, with higher numbers signifying more pollution and greater health concerns. Mansi collected the daily data of AQI of her city for a month and presented it as given below :
AQI Range 1 – 100 101 – 200 201 – 300 301 – 400 401 – 500
Number of Days 3 9 12 4 2
Based on the above, answer the following questions :
1. (i)
  Convert the data to continuous frequency distribution.
  1 mark
2. (ii)
  What is the quality of air in most of the days of the month ?
  1 mark
3. (iii)
  (a) Using table formed in part (i), find mode of the data.
  2 marks
4. OR
  (iii)
  (b) Using table formed in part (i), find median of the data.
  2 marks
Figure in the paper: Table titled 'Central Pollution Control Board's Air Quality Standards': AQI 0-50 Good; 51-100 Satisfactory; 101-200 Moderate; 201-300 Poor; 301-400 Very Poor; 401-500 Severe. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The gap between consecutive classes is $101 - 100 = 1$ . Subtract $0.5$ from each lower limit and add $0.5$ to each upper limit.
  AQI (class) Number of days
  0.5 – 100.5 3
  100.5 – 200.5 9
  200.5 – 300.5 12
  300.5 – 400.5 4
  400.5 – 500.5 2
  1 mark
2. (ii) The highest number of days (12) falls in the range 201–300. By the CPCB scale this range is 'Poor', so the air quality was Poor on most days.
  1 mark
3. (iii)(a) The modal class is $200.5 - 300.5$ , since its frequency 12 is the highest.
  Here $l = 200.5$ , $f_{1} = 12$ , $f_{0} = 9$ , $f_{2} = 4$ and $h = 100$ .
  Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 200.5 + (\frac{12 - 9}{24 - 9 - 4}) \times 100 = 200.5 + \frac{300}{11}$
  1½ marks
4. $= 200.5 + 27.27 = 227.77$ (approx.)
  ½ mark
5. (iii)(b) Cumulative frequency table:
  Class f cf
  0.5 – 100.5 3 3
  100.5 – 200.5 9 12
  200.5 – 300.5 12 24
  300.5 – 400.5 4 28
  400.5 – 500.5 2 30
  $N = 30$ , so $\frac{N}{2} = 15$ . The cf first exceeds 15 in the class $200.5 - 300.5$ , so this is the median class.
  ½ mark
6. Here $l = 200.5$ , $c f = 12$ , $f = 12$ and $h = 100$ .
  Median $= l + (\frac{\frac{N}{2} - c f}{f}) \times h = 200.5 + (\frac{15 - 12}{12}) \times 100$
  1 mark
7. $= 200.5 + 25 = 225.5$
  ½ mark
Answer: (i) Classes 0.5–100.5, 100.5–200.5, 200.5–300.5, 300.5–400.5, 400.5–500.5 with frequencies 3, 9, 12, 4, 2; (ii) Poor; (iii)(a) Mode ≈ 227.77; (iii)(b) Median = 225.5
Where the marks are
- Writes the correct continuous class intervals with their frequencies (1 mark)
- States that the air quality was Poor on most days (1 mark)
- Identifies the modal class and substitutes correctly into the mode formula (1½ marks)
- Gets mode ≈ 227.77 (½ mark)
- Makes the correct cumulative frequency table and median class (½ mark)
- Substitutes correctly into the median formula (1 mark)
- Gets median = 225.5 (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
'Gilli Danda' is a very popular traditional game of India which is played with two wooden sticks – the larger one is called 'Danda' and smaller one 'Gilli'.
'Danda' – It is cylindrical in shape with diameter $4$ cm and length $42$ cm.
Gilli – It is cylindrical in middle with identical conical ends of same radius $1.5$ cm and length $2.8$ cm. The length of cylindrical part is $7$ cm.
Based on the above, answer the following questions :
1. (i)
  Find the volume of wood used in making both the conical parts of Gilli.
  1 mark
2. (ii)
  Find the volume of wood used in making cylindrical part of Gilli.
  1 mark
3. (iii)
  (a) A cylindrical log of wood of radius $1.5$ cm and length $14$ cm is used to make Gilli. Find the volume of the wood scrapped.
  2 marks
4. OR
  (iii)
  (b) Find the total surface area of 'Danda'.
  2 marks
Figure in the paper: Left: a long cylindrical stick (Danda) and a short stick (Gilli). Right: Danda is a long cylinder with diameter 4 cm and length 42 cm. Gilli is a cylinder of length 7 cm with a cone attached at each end; the cones have radius 1.5 cm and slant/height dimension 2.8 cm marked. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Each cone has $r = 1.5$ cm and $h = 2.8$ cm.
  Volume of both cones $= 2 \times \frac{1}{3} π r^{2} h = 2 \times \frac{1}{3} \times \frac{22}{7} \times 1.5 \times 1.5 \times 2.8$
  ½ mark
2. $= \frac{2}{3} \times 19.8 = 13.2$ cm³
  ½ mark
3. (ii) The cylindrical part has $r = 1.5$ cm and $h = 7$ cm.
  Volume $= π r^{2} h = \frac{22}{7} \times 1.5 \times 1.5 \times 7$
  ½ mark
4. $= 49.5$ cm³
  ½ mark
5. (iii)(a) Volume of the log $= π r^{2} h = \frac{22}{7} \times 1.5 \times 1.5 \times 14 = 99$ cm³
  1 mark
6. Volume of the Gilli $= 13.2 + 49.5 = 62.7$ cm³
  ½ mark
7. Volume of wood scrapped $= 99 - 62.7 = 36.3$ cm³
  ½ mark
8. (iii)(b) The Danda has $r = \frac{4}{2} = 2$ cm and $h = 42$ cm.
  TSA of a cylinder $= 2 π r^{2} + 2 π r h = 2 \times \frac{22}{7} \times 2 \times 2 + 2 \times \frac{22}{7} \times 2 \times 42$
  1 mark
9. $= \frac{88}{7} \times (2 + 42) = \frac{3872}{7}$ cm² $\approx 553.14$ cm²
  1 mark
Answer: (i) 13.2 cm³; (ii) 49.5 cm³; (iii)(a) 36.3 cm³; (iii)(b) $\frac{3872}{7}$ cm² ≈ 553.14 cm²
Where the marks are
- Applies $2 \times \frac{1}{3} π r^{2} h$ with $r = 1.5$ , $h = 2.8$ (½ mark) and gets 13.2 cm³ (½ mark)
- Applies $π r^{2} h$ with $h = 7$ (½ mark) and gets 49.5 cm³ (½ mark)
- Finds the volume of the log as 99 cm³ (1 mark)
- Finds the volume of the Gilli as 62.7 cm³ (½ mark)
- Finds the wood scrapped as 36.3 cm³ (½ mark)
- Uses TSA $= 2 π r^{2} + 2 π r h$ with $r = 2$ , $h = 42$ (1 mark)
- Gets $\frac{3872}{7}$ cm² ≈ 553.14 cm² (1 mark)
Open Q38 on its own page

AQI Range	1 – 100	101 – 200	201 – 300	301 – 400	401 – 500
Number of Days	3	9	12	4	2

AQI (class)	Number of days
0.5 – 100.5	3
100.5 – 200.5	9
200.5 – 300.5	12
300.5 – 400.5	4
400.5 – 500.5	2

Class	f	cf
0.5 – 100.5	3	3
100.5 – 200.5	9	12
200.5 – 300.5	12	24
300.5 – 400.5	4	28
400.5 – 500.5	2	30

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