MathsCBSE Class 10 · Standard · Board exam 2023 (session 2022-23)

CBSE Class 10 Mathematics Standard: 2023 board paper, set 30/6/1

The question paper set in the CBSE Class 10 board exam of 2023, 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/2023/X/MATHEMATICS_STANDARD.zip (a zip; this paper is the file 30_6_1_Maths 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 (VSA)
C	26-31	Short Answer (SA)
D	32-35	Long Answer (LA)
E	36-38	Case study / passage based, internal choice in 2-mark subparts

Instructions in brief: All 38 questions compulsory; no overall choice; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 subparts in Section E; calculator not allowed; take π = 22/7 if not stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If $p^{2} = \frac{32}{50}$ , then p is a/an
1. (A)whole number
2. (B)integer
3. (C)rational number
4. (D)irrational number
Try it first, then show our solutionSolution
1. $p^{2} = \frac{32}{50} = \frac{16}{25}$ , so $p = \pm \frac{4}{5}$ .
2. $\pm \frac{4}{5}$ is of the form $\frac{p}{q}$ with integers $p, q$ and $q \neq = 0$ , so p is a rational number. It is not an integer. Option (C).
  1 mark
Answer: p = ±4/5, a rational number, option (C)
Where the marks are
- Simplifies to $p = \pm \frac{4}{5}$ and identifies p as rational (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQCoordinate Geometry
The distance of the point $(- 6, 8)$ from x-axis is
1. (A)6 units
2. (B) $- 6$ units
3. (C)8 units
4. (D)10 units
Try it first, then show our solutionSolution
1. The distance of a point $(x, y)$ from the x-axis is $∣ y ∣$ . Here it is $∣8∣ = 8$ units. Option (C).
  1 mark
Answer: 8 units, option (C)
Where the marks are
- Uses distance from x-axis = |y| = 8 units (1 mark)
Q3 · 1 mark · MCQPolynomials
The number of quadratic polynomials having zeroes $- 5$ and $- 3$ is
1. (A)1
2. (B)2
3. (C)3
4. (D)more than 3
Try it first, then show our solutionSolution
1. Every polynomial $k (x + 5) (x + 3) = k (x^{2} + 8 x + 15)$ with $k \neq = 0$ has zeroes $- 5$ and $- 3$ .
2. Since k can be any non-zero real number, there are infinitely many such polynomials, i.e. more than 3. Option (D).
  1 mark
Answer: More than 3, option (D)
Where the marks are
- Recognises $k (x^{2} + 8 x + 15)$ works for every $k \neq = 0$ , so there are infinitely many (1 mark)
Open Q3 on its own page
Q4 · 1 mark · MCQPair of Linear Equations in Two Variables
The point of intersection of the line represented by $3 x - y = 3$ and y-axis is given by
1. (A) $(0, - 3)$
2. (B) $(0, 3)$
3. (C) $(2, 0)$
4. (D) $(- 2, 0)$
Try it first, then show our solutionSolution
1. On the y-axis $x = 0$ , so $3 (0) - y = 3$ , which gives $y = - 3$ .
2. The point of intersection is $(0, - 3)$ . Option (A).
  1 mark
Answer: (0, −3), option (A)
Where the marks are
- Puts x = 0 and gets y = −3 (1 mark)
Q5 · 1 mark · MCQCircles
The circumferences of two circles are in the ratio $4 : 5$ . What is the ratio of their radii?
1. (A) $16 : 25$
2. (B) $25 : 16$
3. (C) $2 : 5$
4. (D) $4 : 5$
Try it first, then show our solutionSolution
1. $\frac{2 π r _{1}}{2 π r _{2}} = \frac{4}{5}$ , so $r_{1} : r_{2} = 4 : 5$ . Option (D).
  1 mark
Answer: 4 : 5, option (D)
Where the marks are
- Uses $C = 2 π r$ , so the ratio of radii equals the ratio of circumferences, 4 : 5 (1 mark)
Q6 · 1 mark · MCQPolynomials
If $α$ and $β$ are the zeroes of the polynomial $x^{2} - 1$ , then the value of $(α + β)$ is
1. (A)2
2. (B)1
3. (C) $- 1$
4. (D)0
Try it first, then show our solutionSolution
1. Sum of zeroes $= - \frac{b}{a}$ . For $x^{2} + 0 x - 1$ , this gives $α + β = - \frac{0}{1} = 0$ (the zeroes are $1$ and $- 1$ ). Option (D).
  1 mark
Answer: 0, option (D)
Where the marks are
- Applies sum of zeroes = −b/a = 0 (1 mark)
Q7 · 1 mark · MCQIntroduction to Trigonometry
$\frac{c o s ^{2} θ}{s i n ^{2} θ} - \frac{1}{s i n ^{2} θ}$ , in simplified form, is:
1. (A) $tan^{2} θ$
2. (B) $sec^{2} θ$
3. (C)1
4. (D) $- 1$
Try it first, then show our solutionSolution
1. $\frac{c o s ^{2} θ - 1}{s i n ^{2} θ} = \frac{- s i n ^{2} θ}{s i n ^{2} θ} = - 1$ [using $sin^{2} θ + cos^{2} θ = 1$ ]. Option (D).
  1 mark
Answer: −1, option (D)
Where the marks are
- Combines the fractions and uses $cos^{2} θ - 1 = - sin^{2} θ$ to get −1 (1 mark)
Q8 · 1 mark · MCQTriangles
If $△ P QR \sim △ A B C$ ; PQ = 6 cm, AB = 8 cm and the perimeter of $△ A B C$ is 36 cm, then the perimeter of $△ P QR$ is
1. (A)20.25 cm
2. (B)27 cm
3. (C)48 cm
4. (D)64 cm
Try it first, then show our solutionSolution
1. For similar triangles, the ratio of perimeters equals the ratio of corresponding sides: $\frac{perimeter of P QR}{36} = \frac{6}{8}$ .
2. Perimeter of $△ P QR = \frac{6}{8} \times 36 = 27$ cm. Option (B).
  1 mark
Answer: 27 cm, option (B)
Where the marks are
- Uses ratio of perimeters = ratio of corresponding sides and gets 27 cm (1 mark)
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Q9 · 1 mark · MCQQuadratic Equations
If the quadratic equation $a x^{2} + b x + c = 0$ has two real and equal roots, then 'c' is equal to
1. (A) $\frac{- b}{2 a}$
2. (B) $\frac{b}{2 a}$
3. (C) $\frac{- b ^{2}}{4 a}$
4. (D) $\frac{b ^{2}}{4 a}$
Try it first, then show our solutionSolution
1. For equal roots, $D = b^{2} - 4 a c = 0$ .
2. Hence $c = \frac{b ^{2}}{4 a}$ . Option (D).
  1 mark
Answer: c = b²/4a, option (D)
Where the marks are
- Sets $b^{2} - 4 a c = 0$ and gets $c = \frac{b ^{2}}{4 a}$ (1 mark)
Q10 · 1 mark · MCQTriangles
In the given figure, DE $∥$ BC. If AD = 3 cm, AB = 7 cm and EC = 3 cm, then the length of AE is
1. (A)2 cm
2. (B)2.25 cm
3. (C)3.5 cm
4. (D)4 cm
Figure in the paper: Right-angled triangle ABC with right angle at C. A is at the top, B bottom-left, C bottom-right. D lies on AB, E lies on AC, with DE parallel to BC (horizontal). AB = 7 cm (marked along the hypotenuse), AD = 3 cm, EC = 3 cm marked on the vertical side AC between E and C. See it in the official paper.
Try it first, then show our solutionSolution
1. $D B = A B - A D = 7 - 3 = 4$ cm.
2. By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ , so $\frac{3}{4} = \frac{A E}{3}$ , which gives $A E = 2.25$ cm. Option (B).
  1 mark
Answer: 2.25 cm, option (B)
Where the marks are
- Finds DB = 4 cm and applies BPT to get AE = 2.25 cm (1 mark)
Q11 · 1 mark · MCQProbability
A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball?
1. (A) $\frac{1}{4}$
2. (B) $\frac{2}{5}$
3. (C) $\frac{7}{20}$
4. (D) $\frac{13}{20}$
Try it first, then show our solutionSolution
1. Total balls $= 5 + 8 + 7 = 20$ . Neither blue nor pink means yellow, and there are 7 yellow balls.
2. $P = \frac{7}{20}$ . Option (C).
  1 mark
Answer: 7/20, option (C)
Where the marks are
- Identifies the favourable outcomes as the 7 yellow balls out of 20 (1 mark)
Q12 · 1 mark · MCQSurface Areas and Volumes
The volume of a right circular cone whose area of the base is $156 c m^{2}$ and the vertical height is 8 cm, is
1. (A) $2496 c m^{3}$
2. (B) $1248 c m^{3}$
3. (C) $1664 c m^{3}$
4. (D) $416 c m^{3}$
Try it first, then show our solutionSolution
1. $V = \frac{1}{3} \times base area \times h = \frac{1}{3} \times 156 \times 8 = 416 c m^{3}$ . Option (D).
  1 mark
Answer: 416 cm³, option (D)
Where the marks are
- Uses $V = \frac{1}{3} \times$ base area $\times$ h to get 416 cm³ (1 mark)
Q13 · 1 mark · MCQCompetency-basedPair of Linear Equations in Two Variables
3 chairs and 1 table cost ₹ 900; whereas 5 chairs and 3 tables cost ₹ 2,100. If the cost of 1 chair is ₹ x and the cost of 1 table is ₹ y, then the situation can be represented algebraically as
1. (A) $3 x + y = 900, 3 x + 5 y = 2100$
2. (B) $x + 3 y = 900, 3 x + 5 y = 2100$
3. (C) $3 x + y = 900, 5 x + 3 y = 2100$
4. (D) $x + 3 y = 900, 5 x + 3 y = 2100$
Try it first, then show our solutionSolution
1. 3 chairs and 1 table give $3 x + y = 900$ ; 5 chairs and 3 tables give $5 x + 3 y = 2100$ . Option (C).
  1 mark
Answer: 3x + y = 900, 5x + 3y = 2100, option (C)
Where the marks are
- Forms both equations correctly (1 mark)
Q14 · 1 mark · MCQCircles
In the given figure, PA and PB are tangents from external point P to a circle with centre C and Q is any point on the circle. Then the measure of $∠ A QB$ is
1. (A) $62 \frac{1}{2}^{\circ}$
2. (B) $12 5^{\circ}$
3. (C) $5 5^{\circ}$
4. (D) $9 0^{\circ}$
Figure in the paper: Circle with centre C. External point P on the left with tangents PA (touching at top point A) and PB (touching at bottom point B). Angle APB = 55° marked at P. Point Q is on the right side of the circle; chords QA and QB are drawn and angle AQB is marked at Q. Radii CA and CB are drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ A C B = 18 0^{\circ} - 5 5^{\circ} = 12 5^{\circ}$ [ $∠ P A C = ∠ P B C = 9 0^{\circ}$ , tangent ⟂ radius].
2. $∠ A QB = \frac{1}{2} ∠ A C B = 62 \frac{1}{2}^{\circ}$ [angle at the centre is twice the angle at the circumference]. Option (A).
  1 mark
Answer: 62½°, option (A)
Where the marks are
- Finds ∠ACB = 125° and halves it to get ∠AQB = 62½° (1 mark)
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Q15 · 1 mark · MCQProbability
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability of getting a face card is
1. (A) $\frac{1}{2}$
2. (B) $\frac{3}{13}$
3. (C) $\frac{4}{13}$
4. (D) $\frac{1}{13}$
Try it first, then show our solutionSolution
1. There are 12 face cards (J, Q, K of each of 4 suits). $P = \frac{12}{52} = \frac{3}{13}$ . Option (B).
  1 mark
Answer: 3/13, option (B)
Where the marks are
- Counts 12 face cards and gets 3/13 (1 mark)
Q16 · 1 mark · MCQIntroduction to Trigonometry
If $θ$ is an acute angle of a right angled triangle, then which of the following equation is not true?
1. (A) $sin θ cot θ = cos θ$
2. (B) $cos θ tan θ = sin θ$
3. (C) $cosec^{2} θ - cot^{2} θ = 1$
4. (D) $tan^{2} θ - sec^{2} θ = 1$
Try it first, then show our solutionSolution
1. The correct identity is $sec^{2} θ - tan^{2} θ = 1$ .
2. So $tan^{2} θ - sec^{2} θ = - 1$ , not 1. Hence (D) is not true; (A), (B) and (C) are true.
  1 mark
Answer: Option (D)
Where the marks are
- Uses $sec^{2} θ - tan^{2} θ = 1$ to identify (D) as false (1 mark)
Q17 · 1 mark · MCQPolynomials
If the zeroes of the quadratic polynomial $x^{2} + (a + 1) x + b$ are 2 and $- 3$ , then
1. (A) $a = - 7, b = - 1$
2. (B) $a = 5, b = - 1$
3. (C) $a = 2, b = - 6$
4. (D) $a = 0, b = - 6$
Try it first, then show our solutionSolution
1. Sum of zeroes: $2 + (- 3) = - 1 = - (a + 1)$ , so $a = 0$ .
2. Product of zeroes: $b = 2 \times (- 3) = - 6$ . Option (D).
  1 mark
Answer: a = 0, b = −6, option (D)
Where the marks are
- Uses sum and product of zeroes to get a = 0 and b = −6 (1 mark)
Q18 · 1 mark · MCQArithmetic Progressions
If the sum of the first n terms of an A.P. be $3 n^{2} + n$ and its common difference is 6, then its first term is
1. (A)2
2. (B)3
3. (C)1
4. (D)4
Try it first, then show our solutionSolution
1. $a_{1} = S_{1} = 3 (1)^{2} + 1 = 4$ . Check: $S_{2} = 14$ , so $a_{2} = 10$ and $d = 6$ ✓. Option (D).
  1 mark
Answer: 4, option (D)
Where the marks are
- Puts n = 1 to get $a = S_{1} = 4$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedQuadratic Equations
Assertion (A): If $5 + 7$ is a root of a quadratic equation with rational co-efficients, then its other root is $5 - 7$ .
Reason (R): Surd roots of a quadratic equation with rational co-efficients occur in conjugate pairs.
Choose the correct option out of the following:
1. (A)Both Assertion (A) and Reason (R) are true; and Reason (R) is the 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. R is true: irrational (surd) roots of a quadratic with rational coefficients occur in conjugate pairs $p \pm q$ .
2. Hence the other root is $5 - 7$ , so A is true and R correctly explains it. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both statements true and R explains A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Assertion (A): For $0 < θ \leq 9 0^{\circ}$ , $cosec θ - cot θ$ and $cosec θ + cot θ$ are reciprocal of each other.
Reason (R): $cosec^{2} θ - cot^{2} θ = 1$
Choose the correct option out of the following:
1. (A)Both Assertion (A) and Reason (R) are true; and Reason (R) is the 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. R is the standard identity $cosec^{2} θ - cot^{2} θ = 1$ , so R is true.
2. Factorising R: $(cosec θ - cot θ) (cosec θ + cot θ) = 1$ , so the two expressions are reciprocals. A follows from R. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Factorises the identity to show the product is 1 (1 mark)
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Section B

Very Short Answer (VSA)

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerReal Numbers
Show that $6^{n}$ can not end with digit 0 for any natural number 'n'.
Try it first, then show our solutionSolution
1. Any number ending in the digit 0 is divisible by both 2 and 5, so its prime factorisation must contain the prime 5.
  1 mark
2. Here $6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}$ . By the Fundamental Theorem of Arithmetic this factorisation is unique, so the only primes in $6^{n}$ are 2 and 3. The prime 5 never appears. Hence $6^{n}$ cannot end with the digit 0 for any natural number $n$ .
  1 mark
Answer: $6^{n} = 2^{n} 3^{n}$ has no factor 5, so it can never end in 0.
Where the marks are
- States that ending in 0 needs 5 in the prime factorisation (1 mark)
- Writes $6^{n} = 2^{n} 3^{n}$ and uses the uniqueness of factorisation to conclude (1 mark)
Open Q21(A) on its own page
OR
Q21(B) · 2 marks · Very short answerReal Numbers
Find the HCF and LCM of 72 and 120.
Try it first, then show our solutionSolution
1. Prime factorisation: $72 = 2^{3} \times 3^{2}$ and $120 = 2^{3} \times 3 \times 5$ .
2. HCF = product of the smallest powers of the common primes $= 2^{3} \times 3 = 24$ .
  1 mark
3. LCM = product of the greatest powers of all the primes $= 2^{3} \times 3^{2} \times 5 = 360$ .
  1 mark
Answer: HCF = 24, LCM = 360
Where the marks are
- Correct HCF 24 from the factorisations (1 mark)
- Correct LCM 360 (1 mark)
Open Q21(B) on its own page
Q22 · 2 marks · Very short answerCoordinate Geometry
A line intersects y-axis and x-axis at point P and Q, respectively. If R(2, 5) is the mid-point of line segment PQ, then find the coordinates of P and Q.
Try it first, then show our solutionSolution
1. P lies on the y-axis and Q lies on the x-axis, so let $P (0, y)$ and $Q (x, 0)$ .
  ½ mark
2. R(2, 5) is the mid-point of PQ, so by the mid-point formula $\frac{0 + x}{2} = 2 \Rightarrow x = 4$ .
  ½ mark
3. Similarly $\frac{y + 0}{2} = 5 \Rightarrow y = 10$ .
  ½ mark
4. Hence $P (0, 10)$ and $Q (4, 0)$ .
  ½ mark
Answer: P(0, 10), Q(4, 0)
Where the marks are
- Takes P(0, y) and Q(x, 0) (½ mark)
- Applies the mid-point formula to get x = 4 (½ mark)
- Gets y = 10 (½ mark)
- States P(0, 10) and Q(4, 0) (½ mark)
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Q23 · 2 marks · Very short answerCompetency-basedSome Applications of Trigonometry
Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation $θ$ of the sun is such that $tan θ = \frac{6}{7}$ .
Try it first, then show our solutionSolution
1. Let AB = 18 m be the pole and AP its shadow, with $∠ A P B = θ$ and $∠ A = 9 0^{\circ}$ . In right $△ P A B$ , $tan θ = \frac{A B}{A P} = \frac{18}{A P}$ .
  1 mark
2. Substituting $tan θ = \frac{6}{7}$ : $\frac{6}{7} = \frac{18}{A P}$ .
  ½ mark
3. $A P = \frac{18 \times 7}{6} = 21$ m, so the shadow is 21 m long.
  ½ mark
Answer: Length of shadow = 21 m
Where the marks are
- Sets up $tan θ = \frac{18}{A P}$ in the right triangle (1 mark)
- Substitutes $\frac{6}{7}$ (½ mark)
- Gets AP = 21 m (½ mark)
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Q24 · 2 marks · Very short answerCircles
In the given figure, PA is a tangent to the circle drawn from the external point P and PBC is the secant to the circle with BC as diameter. If $∠ A O C = 13 0^{\circ}$ , then find the measure of $∠ A P B$ , where O is the centre of the circle.
Figure in the paper: Circle with centre O and diameter BC horizontal (B left, C right, O the centre). External point P on the left, on the line BC extended. PA is a tangent touching the circle at A (upper left). OA is drawn, and the angle AOC = 130° is marked at O. See it in the official paper.
Try it first, then show our solutionSolution
1. BOC is a straight line, so $∠ A O B = 18 0^{\circ} - 13 0^{\circ} = 5 0^{\circ}$ [linear pair].
  ½ mark
2. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  ½ mark
3. In $△ O A P$ : $∠ A P B = 18 0^{\circ} - (5 0^{\circ} + 9 0^{\circ}) = 4 0^{\circ}$ [angle sum property].
  1 mark
Answer: $∠ A P B = 4 0^{\circ}$
Where the marks are
- Finds $∠ A O B = 5 0^{\circ}$ (½ mark)
- Uses $∠ O A P = 9 0^{\circ}$ (½ mark)
- Uses the angle sum of $△ O A P$ to get $4 0^{\circ}$ (1 mark)
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Answer one of these 2 alternatives.
Q25(A) · 2 marks · Very short answerTriangles
In the given figure, ABC is a triangle in which DE $∥$ BC. If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, then find the value of x.
Figure in the paper: Triangle ABC with apex A at top, B bottom-left, C bottom-right. D on AB and E on AC with DE parallel to BC. Labels: AD = x, DB = x − 2, AE = x + 2, EC = x − 1. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ , DE ∥ BC, so by the Basic Proportionality Theorem $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{x}{x - 2} = \frac{x + 2}{x - 1}$ .
  1 mark
2. Cross-multiplying: $x (x - 1) = (x + 2) (x - 2)$ , so $x^{2} - x = x^{2} - 4$ , which gives $x = 4$ .
  1 mark
Answer: x = 4
Where the marks are
- Applies BPT to form $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$ (1 mark)
- Solves to get x = 4 (1 mark)
Open Q25(A) on its own page
OR
Q25(B) · 2 marks · Very short answerTriangles
Diagonals AC and BD of trapezium ABCD with AB $∥$ DC intersect each other at point O. Show that $\frac{O A}{O C} = \frac{O B}{O D}$ .
Figure in the paper: Trapezium ABCD with AB at the bottom and DC at the top, AB parallel to DC. Diagonals AC and BD intersect at O. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A O B$ and $△ C O D$ : $∠ O A B = ∠ O C D$ [alternate angles, AB ∥ DC, AC transversal] and $∠ O B A = ∠ O D C$ [alternate angles, BD transversal]. Hence $△ A O B \sim △ C O D$ [AA similarity].
  1½ marks
2. Corresponding sides of similar triangles are proportional, so $\frac{O A}{O C} = \frac{O B}{O D}$ .
  ½ mark
Answer: $△ A O B \sim △ C O D$ (AA), hence $\frac{O A}{O C} = \frac{O B}{O D}$ .
Where the marks are
- Shows the two pairs of equal alternate angles and concludes $△ A O B \sim △ C O D$ (1½ marks)
- Writes the ratio of corresponding sides (½ mark)
Open Q25(B) on its own page

Section C

Short Answer (SA)

Q26 · 3 marks · Short answerCoordinate Geometry
Find the ratio in which the line segment joining the points A(6, 3) and B(-2, -5) is divided by x-axis.
Try it first, then show our solutionSolution
1. Let the x-axis cut AB at $P (x, 0)$ , dividing AB in the ratio $k : 1$ .
  ½ mark
2. By the section formula, the y-coordinate of P is $\frac{k ( - 5 ) + 1 ( 3 )}{k + 1}$ . Since P lies on the x-axis, $\frac{- 5 k + 3}{k + 1} = 0 \Rightarrow - 5 k + 3 = 0 \Rightarrow k = \frac{3}{5}$ .
  2 marks
3. Hence the required ratio is $3 : 5$ (internally).
  ½ mark
Answer: 3 : 5
Where the marks are
- Takes the point (x, 0) dividing AB in k : 1 (½ mark)
- Uses the section formula for the y-coordinate and gets k = 3/5 (2 marks)
- States the ratio 3 : 5 (½ mark)
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Answer one of these 2 alternatives.
Q27(A) · 3 marks · Short answerReal Numbers
Find the HCF and LCM of 26, 65 and 117, using prime factorisation.
Try it first, then show our solutionSolution
1. $26 = 2 \times 13$ , $65 = 5 \times 13$ , $117 = 3^{2} \times 13$ .
  1 mark
2. The only prime common to all three is 13, so HCF $= 13$ .
  1 mark
3. LCM $= 2 \times 3^{2} \times 5 \times 13 = 1170$ .
  1 mark
Answer: HCF = 13, LCM = 1170
Where the marks are
- Correct prime factorisation of all three numbers (1 mark)
- HCF = 13 (1 mark)
- LCM = 1170 (1 mark)
Open Q27(A) on its own page
OR
Q27(B) · 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 Q27(B) on its own page
Q28 · 3 marks · Short answerTriangles
In the given figure, E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD $⊥$ BC and EF $⊥$ AC, then prove that $△ A B D \sim △ E C F$ .
Figure in the paper: Isosceles triangle ABC with apex A at top, B and C on the horizontal base with C on the right. AD is drawn from A perpendicular to BC, with D on BC (right angle marked). E lies on line CB extended beyond B to the left. EF is drawn from E to F on side AC, with right angle marked at F. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ , AB = AC, so $∠ B = ∠ C$ [angles opposite equal sides].
  1 mark
2. In $△ A B D$ and $△ E C F$ : $∠ A D B = ∠ E F C = 9 0^{\circ}$ [AD ⟂ BC, EF ⟂ AC] and $∠ A B D = ∠ E C F$ [from the first step, since $∠ E C F$ is the same angle as $∠ C$ ].
  1 mark
3. Therefore $△ A B D \sim △ E C F$ [AA similarity criterion].
  1 mark
Answer: $△ A B D \sim △ E C F$ by AA similarity.
Where the marks are
- Uses AB = AC to get $∠ B = ∠ C$ (1 mark)
- Identifies the pair of right angles and the pair of equal base angles (1 mark)
- Concludes similarity by AA (1 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerQuadratic Equations
The sum of two numbers is 15. If the sum of their reciprocals is $\frac{3}{10}$ , find the two numbers.
Try it first, then show our solutionSolution
1. Let one number be $x$ ; then the other is $15 - x$ .
  ½ mark
2. Given: $\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10}$ .
  1 mark
3. $\frac{15}{x ( 15 - x )} = \frac{3}{10} \Rightarrow 150 = 45 x - 3 x^{2} \Rightarrow 3 x^{2} - 45 x + 150 = 0 \Rightarrow x^{2} - 15 x + 50 = 0$ .
  ½ mark
4. $(x - 10) (x - 5) = 0 \Rightarrow x = 10$ or $x = 5$ .
  ½ mark
5. If $x = 10$ the other number is 5, and if $x = 5$ it is 10. So the numbers are 10 and 5.
  ½ mark
Answer: The numbers are 10 and 5
Where the marks are
- Takes the numbers as x and 15 − x (½ mark)
- Forms the reciprocal equation (1 mark)
- Reduces it to $x^{2} - 15 x + 50 = 0$ (½ mark)
- Solves to get x = 10, 5 (½ mark)
- States the two numbers (½ mark)
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerQuadratic Equations
If $α$ and $β$ are roots of the quadratic equation $x^{2} - 7 x + 10 = 0$ , find the quadratic equation whose roots are $α^{2}$ and $β^{2}$ .
Try it first, then show our solutionSolution
1. For $x^{2} - 7 x + 10 = 0$ : $α + β = 7$ and $α β = 10$ .
  ½ mark
2. Sum of new roots: $α^{2} + β^{2} = (α + β)^{2} - 2 α β = 49 - 20 = 29$ .
  1 mark
3. Product of new roots: $α^{2} β^{2} = (α β)^{2} = 100$ .
  1 mark
4. The required equation is $x^{2} - (sum) x + product = 0$ , i.e. $x^{2} - 29 x + 100 = 0$ .
  ½ mark
Answer: $x^{2} - 29 x + 100 = 0$
Where the marks are
- Writes $α + β = 7$ and $α β = 10$ (½ mark)
- Finds $α^{2} + β^{2} = 29$ (1 mark)
- Finds $α^{2} β^{2} = 100$ (1 mark)
- Forms $x^{2} - 29 x + 100 = 0$ (½ mark)
Open Q29(B) on its own page
Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{1 + s e c A}{s e c A} = \frac{s i n ^{2} A}{1 - c o s A}$ .
Try it first, then show our solutionSolution
1. LHS $= \frac{1 + s e c A}{s e c A} = \frac{1 + \frac{1}{c o s A}}{\frac{1}{c o s A}} = \frac{c o s A + 1}{c o s A} \times cos A = 1 + cos A$ .
  1 mark
2. Multiply and divide by $(1 - cos A)$ : $= \frac{( 1 + c o s A ) ( 1 - c o s A )}{1 - c o s A} = \frac{1 - c o s ^{2} A}{1 - c o s A}$ .
  1 mark
3. Using $sin^{2} A + cos^{2} A = 1$ : $= \frac{s i n ^{2} A}{1 - c o s A} =$ RHS. Hence proved.
  1 mark
Answer: LHS simplifies to $1 + cos A = \frac{s i n ^{2} A}{1 - c o s A}$ = RHS.
Where the marks are
- Simplifies LHS to $1 + cos A$ (1 mark)
- Multiplies and divides by $1 - cos A$ (1 mark)
- Uses $1 - cos^{2} A = sin^{2} A$ to reach RHS (1 mark)
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Q31 · 3 marks · Short answerAreas Related to Circles
In a circle of radius 21 cm, an arc subtends an angle of $6 0^{\circ}$ at the centre. Find the area of the sector formed by the arc. Also, find the length of the arc.
Try it first, then show our solutionSolution
1. Area of sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21 = 231$ cm².
  1½ marks
2. Length of arc $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22$ cm.
  1½ marks
Answer: Area of sector = 231 cm²; length of arc = 22 cm
Where the marks are
- Uses the sector area formula and gets 231 cm² (1½ marks)
- Uses the arc length formula and gets 22 cm (1½ marks)
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Section D

Long Answer (LA)

Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerCircles
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $∠ P T Q = 2 ∠ O P Q$ .
Figure in the paper: Circle with centre O. External point T on the left with tangents TP (touching at upper point P) and TQ (touching at lower point Q). OP and chord PQ are drawn; angle PTQ at T and angle OPQ at P are the relevant angles. See it in the official paper.
Try it first, then show our solutionSolution
1. TP = TQ [tangents from an external point are equal], so $∠ T P Q = ∠ T QP$ [angles opposite equal sides].
  1 mark
2. Let $∠ P T Q = θ$ . In $△ T P Q$ , $∠ T P Q + ∠ T QP = 18 0^{\circ} - θ$ , so $∠ T P Q = ∠ T QP = \frac{18 0 ^{\circ} - θ}{2} = 9 0^{\circ} - \frac{θ}{2}$ .
  1½ marks
3. $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ O P Q = ∠ O P T - ∠ T P Q = 9 0^{\circ} - (9 0^{\circ} - \frac{θ}{2}) = \frac{θ}{2}$ .
  1½ marks
4. Therefore $∠ P T Q = θ = 2∠ O P Q$ . Hence proved.
  1 mark
Answer: $∠ O P Q = \frac{θ}{2}$ where $θ = ∠ P T Q$ , so $∠ P T Q = 2∠ O P Q$ .
Where the marks are
- TP = TQ, hence $∠ T P Q = ∠ T QP$ (1 mark)
- Expresses $∠ T P Q = 9 0^{\circ} - \frac{θ}{2}$ (1½ marks)
- Uses $∠ O P T = 9 0^{\circ}$ to get $∠ O P Q = \frac{θ}{2}$ (1½ marks)
- Concludes the result (1 mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerCircles
A circle touches the side BC of a $△ A B C$ at a point P and touches AB and AC when produced at Q and R respectively. Show that $A Q = \frac{1}{2}$ (Perimeter of $△ A B C$ ).
Figure in the paper: Triangle ABC with apex A at top; side BC is horizontal at the bottom with P on BC. A circle lies below BC touching BC at P, and touches the extended sides AB (at Q, beyond B) and AC (at R, beyond C). See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents from an external point to a circle are equal in length: AQ = AR (from A), BQ = BP (from B), CP = CR (from C).
  1 mark
2. So $2 A Q = A Q + A R$ .
  1 mark
3. Since Q lies on AB produced and R on AC produced, $A Q = A B + B Q$ and $A R = A C + C R$ . So $2 A Q = A B + B Q + A C + C R$ .
  ½ mark
4. Replacing BQ by BP and CR by CP: $2 A Q = A B + A C + (B P + C P)$ .
  ½ mark
5. $B P + P C = B C$ , so $2 A Q = A B + A C + B C$ .
  1 mark
6. Therefore $A Q = \frac{1}{2} (A B + B C + C A) = \frac{1}{2}$ (Perimeter of $△ A B C$ ).
  1 mark
Answer: $A Q = \frac{1}{2} (A B + B C + C A)$ , i.e. half the perimeter of $△ A B C$ .
Where the marks are
- Uses equal tangents AQ = AR, BQ = BP, CP = CR (1 mark)
- Writes 2AQ = AQ + AR (1 mark)
- Splits into AB + BQ + AC + CR (½ mark)
- Replaces with BP + CP (½ mark)
- Uses BP + CP = BC (1 mark)
- Concludes AQ = ½ perimeter (1 mark)
Open Q32(B) on its own page
Q33 · 5 marks · Long answerSurface Areas and Volumes
A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid.
Try it first, then show our solutionSolution
1. Radius of cone = radius of hemisphere $r = 7$ cm. Height of cone = diameter $= 14$ cm.
  1 mark
2. Volume of solid = volume of hemisphere + volume of cone $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{2}{3} π (7)^{3} + \frac{1}{3} π (7)^{2} (14)$ .
  3 marks
3. $= \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 (14 + 14) = \frac{154}{3} \times 28 = \frac{4312}{3} \approx 1437.33$ cm³.
  1 mark
Answer: Volume = $\frac{4312}{3}$ cm³ ≈ 1437.33 cm³
Where the marks are
- Takes r = 7 cm and cone height 14 cm (1 mark)
- Correct volume of hemisphere and of cone substituted (1½ + 1½ marks)
- Final volume $\frac{4312}{3}$ cm³ ≈ 1437.33 cm³ (1 mark)
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Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerArithmetic Progressions
The ratio of the 11th term to the 18th term of an A.P. is 2 : 3. Find the ratio of the 5th term to the 21st term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms.
Try it first, then show our solutionSolution
1. $\frac{a _{11}}{a _{18}} = \frac{a + 10 d}{a + 17 d} = \frac{2}{3}$ .
  1 mark
2. $3 a + 30 d = 2 a + 34 d \Rightarrow a = 4 d$ .
  1 mark
3. $\frac{a _{5}}{a _{21}} = \frac{a + 4 d}{a + 20 d} = \frac{4 d + 4 d}{4 d + 20 d} = \frac{8 d}{24 d} = \frac{1}{3}$ , so $a_{5} : a_{21} = 1 : 3$ .
  1 mark
4. $\frac{S _{5}}{S _{21}} = \frac{\frac{5}{2} [ 2 a + 4 d ]}{\frac{21}{2} [ 2 a + 20 d ]} = \frac{5 [ 8 d + 4 d ]}{21 [ 8 d + 20 d ]}$ .
  1 mark
5. $= \frac{5 \times 12 d}{21 \times 28 d} = \frac{60}{588} = \frac{5}{49}$ , so $S_{5} : S_{21} = 5 : 49$ .
  1 mark
Answer: $a_{5} : a_{21} = 1 : 3$ ; $S_{5} : S_{21} = 5 : 49$
Where the marks are
- Forms $\frac{a + 10 d}{a + 17 d} = \frac{2}{3}$ (1 mark)
- Gets a = 4d (1 mark)
- Ratio of 5th to 21st term = 1 : 3 (1 mark)
- Substitutes into $S_{n}$ for $S_{5}$ and $S_{21}$ (1 mark)
- Gets 5 : 49 (1 mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerArithmetic Progressions
If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Try it first, then show our solutionSolution
1. $S_{6} = 36 \Rightarrow \frac{6}{2} [2 a + 5 d] = 36$ .
  1 mark
2. $\Rightarrow 2 a + 5 d = 12$ …(1)
  1 mark
3. $S_{16} = 256 \Rightarrow \frac{16}{2} [2 a + 15 d] = 256 \Rightarrow 2 a + 15 d = 32$ …(2)
  1 mark
4. Subtracting (1) from (2): $10 d = 20 \Rightarrow d = 2$ . Then from (1), $2 a = 12 - 10 = 2 \Rightarrow a = 1$ .
  1 mark
5. $S_{10} = \frac{10}{2} [2 (1) + 9 (2)] = 5 \times 20 = 100$ .
  1 mark
Answer: $S_{10} = 100$
Where the marks are
- Uses the $S_{n}$ formula for $S_{6}$ (1 mark)
- Gets 2a + 5d = 12 (1 mark)
- Gets 2a + 15d = 32 (1 mark)
- Solves to get d = 2, a = 1 (1 mark)
- Finds $S_{10} = 100$ (1 mark)
Open Q34(B) on its own page
Q35 · 5 marks · Long answerStatistics
250 apples of a box were weighed and the distribution of masses of the apples is given in the following table:
Mass (in grams) 80–100 100–120 120–140 140–160 160–180
Number of apples 20 60 70 $x$ 60
1. (i)
  Find the value of x and the mean mass of the apples.
  3 marks
2. (ii)
  Find the modal mass of the apples.
  2 marks
Try it first, then show our solutionSolution
1. (i) The total number of apples is 250, so $20 + 60 + 70 + x + 60 = 250$ , which gives $210 + x = 250$ and $x = 40$ .
  1 mark
2. (i) Table of class marks $x_{i}$ and products $f_{i} x_{i}$ :
  Mass (g) $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  80–100 20 90 1800
  100–120 60 110 6600
  120–140 70 130 9100
  140–160 40 150 6000
  160–180 60 170 10200
  Total 250 33700
  1 mark
3. (i) Mean $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{33700}{250} = 134.8$ .
  So the mean mass is $134.8$ g.
  1 mark
4. (ii) The highest frequency is 70, so the modal class is 120–140.
  Here $l = 120$ , $f_{1} = 70$ , $f_{0} = 60$ , $f_{2} = 40$ , $h = 20$ .
  Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 120 + \frac{70 - 60}{140 - 60 - 40} \times 20$ .
  1 mark
5. (ii) Mode $= 120 + \frac{10}{40} \times 20 = 120 + 5 = 125$ .
  So the modal mass is $125$ g.
  1 mark
Answer: (i) x = 40, mean mass = 134.8 g; (ii) modal mass = 125 g
Where the marks are
- Uses the total of 250 to get $x = 40$ (1 mark)
- Correct table of $x_{i}$ and $f_{i} x_{i}$ with $\sum f_{i} x_{i} = 33700$ (1 mark)
- Mean $= 33700/250 = 134.8$ g (1 mark)
- Identifies modal class 120–140 and substitutes correctly in the mode formula (1 mark)
- Modal mass $= 125$ g (1 mark)
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Mass (in grams)	80–100	100–120	120–140	140–160	160–180
Number of apples	20	60	70	$x$	60

Mass (g)	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
80–100	20	90	1800
100–120	60	110	6600
120–140	70	130	9100
140–160	40	150	6000
160–180	60	170	10200
Total	250		33700

Section E

Case study / passage based, internal choice in 2-mark subparts

Q36 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
A coaching institute of Mathematics conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, there are 20 poor and 5 rich children, whereas in batch II, there are 5 poor and 25 rich children. The total monthly collection of fees from batch I is ₹ 9000 and from batch II is ₹ 26,000. Assume that each poor child pays ₹ x per month and each rich child pays ₹ y per month.
[Photograph of a classroom with students seated in rows facing a lecture/presentation screen.]
Based on the above information, answer the following questions:
1. (i)
  Represent the information given above in terms of x and y.
  1 mark
2. (ii)
  Find the monthly fee paid by a poor child.
  2 marks
3. OR
  (ii)
  Find the difference in the monthly fee paid by a poor child and a rich child.
  2 marks
4. (iii)
  If there are 10 poor and 20 rich children in batch II, what is the total monthly collection of fees from batch II?
  1 mark
Figure in the paper: Photograph of a classroom: an audience of students seated in rows, facing a front presentation screen, with framed pictures on the side walls. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Batch I has 20 poor and 5 rich children, so $20 x + 5 y = 9000$ , i.e. $4 x + y = 1800$ .
  Batch II has 5 poor and 25 rich children, so $5 x + 25 y = 26000$ , i.e. $x + 5 y = 5200$ .
  1 mark
2. (ii) From $4 x + y = 1800$ we get $y = 1800 - 4 x$ .
  Substituting in $x + 5 y = 5200$ : $x + 5 (1800 - 4 x) = 5200$ , so $9000 - 19 x = 5200$ .
  This gives $19 x = 3800$ , so $x = 200$ .
  1½ marks
3. (ii) The monthly fee paid by a poor child is ₹ 200.
  ½ mark
4. (ii) OR: Solve as above to get $x = 200$ . Then $y = 1800 - 4 (200) = 1000$ .
  1½ marks
5. (ii) OR: Difference in monthly fee $= y - x = 1000 - 200 =$ ₹ 800.
  ½ mark
6. (iii) Total collection $= 10 x + 20 y = 10 (200) + 20 (1000) = 2000 + 20000 =$ ₹ 22000.
  1 mark
Answer: (i) 20x + 5y = 9000, 5x + 25y = 26000; (ii) ₹ 200 OR difference ₹ 800; (iii) ₹ 22000
Where the marks are
- Writes both equations $20 x + 5 y = 9000$ and $5 x + 25 y = 26000$ (1 mark)
- Solves the pair correctly to get $x = 200$ (1½ marks)
- States the poor child's fee as ₹ 200 (½ mark)
- OR part: obtains $x = 200$ and $y = 1000$ (1½ marks)
- OR part: difference ₹ 800 (½ mark)
- Computes $10 x + 20 y =$ ₹ 22000 (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two Sections A and B. Tower is supported by wires from a point O.
Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of the Section B is $3 0^{\circ}$ and the angle of elevation of the top of Section A is $4 5^{\circ}$ .
Based on the above information, answer the following questions:
1. (i)
  Find the length of the wire from the point O to the top of Section B.
  1 mark
2. (ii)
  Find the distance AB.
  2 marks
3. OR
  (ii)
  Find the area of $△ O P B$ .
  2 marks
4. (iii)
  Find the height of the Section A from the base of the tower.
  1 mark
Figure in the paper: Vertical tower PA on the left with base P on the ground. Section B is the lower part with top at B on the tower, Section A the upper part with top at A. Point O is on the ground 36 cm from P. Wires run from O to A and from O to B. Angle of 45° is marked at O between OP and OA; angle of 30° is marked at O between OP and OB. Segment PO = 36 cm is labelled on the ground. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In right $△ O P B$ (right-angled at P), $cos 3 0^{\circ} = \frac{O P}{O B}$ .
  ½ mark
2. (i) $\frac{3}{2} = \frac{36}{O B}$ , so $O B = \frac{72}{3} = 243$ cm.
  The wire from O to the top of Section B is $243$ cm long ( $\approx 41.57$ cm).
  ½ mark
3. (ii) In $△ O P B$ , $tan 3 0^{\circ} = \frac{P B}{O P}$ .
  $\frac{1}{3} = \frac{P B}{36}$ , so $P B = \frac{36}{3} = 123$ cm.
  1 mark
4. (ii) In right $△ O P A$ , $tan 4 5^{\circ} = \frac{A P}{O P}$ .
  $1 = \frac{A P}{36}$ , so $A P = 36$ cm.
  ½ mark
5. (ii) $A B = A P - P B = 36 - 123 = 12 (3 - 3)$ cm ( $\approx 15.22$ cm).
  ½ mark
6. (ii) OR: Area of $△ O P B = \frac{1}{2} \times O P \times P B$ , where $P B = 36 tan 3 0^{\circ} = 123$ cm.
  1 mark
7. (ii) OR: Area $= \frac{1}{2} \times 36 \times 123 = 2163$ cm $^{2}$ ( $\approx 374.1$ cm $^{2}$ ).
  1 mark
8. (iii) The height of Section A from the base is $A P$ . In $△ O P A$ , $A P = 36 tan 4 5^{\circ} = 36$ cm.
  1 mark
Answer: (i) OB = 24√3 cm; (ii) AB = 12(3 − √3) cm OR area = 216√3 cm²; (iii) 36 cm
Where the marks are
- Uses $cos 3 0^{\circ} = O P / O B$ (½ mark)
- Gets $O B = 243$ cm (½ mark)
- Finds $P B = 123$ cm using $tan 3 0^{\circ}$ (1 mark)
- Finds $A P = 36$ cm using $tan 4 5^{\circ}$ (½ mark)
- Gets $A B = 36 - 123$ cm (½ mark)
- OR part: writes the area formula with $P B = 123$ cm (1 mark)
- OR part: area $= 2163$ cm $^{2}$ (1 mark)
- Height of Section A, $A P = 36$ cm (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedProbability
"Eight Ball" is a game played on a pool table with 15 balls numbered 1 to 15 and a "cue ball" that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8 and seven are striped balls numbered 9 to 15.
[Photograph of pool balls lined up on a pool table with a cue stick.]
The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random.
Based on the above information, answer the following questions:
1. (i)
  What is the probability that the drawn ball bears number 8?
  1 mark
2. (ii)
  What is the probability that the drawn ball bears an even number?
  2 marks
3. OR
  (ii)
  What is the probability that the drawn ball bears a number, which is a multiple of 3?
  2 marks
4. (iii)
  What is the probability that the drawn ball is a solid coloured and bears an even number?
  1 mark
Figure in the paper: Photograph of pool balls lined up on a green pool table, with a cue stick at the left and a large ball in the foreground; the balls are numbered and show solid and striped designs. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Total outcomes = 15 balls, and only one ball bears number 8.
  P(number 8) $= \frac{1}{15}$ .
  1 mark
2. (ii) The even numbers from 1 to 15 are 2, 4, 6, 8, 10, 12, 14, so there are 7 favourable outcomes.
  ½ mark
3. (ii) P(even number) $= \frac{7}{15}$ .
  1½ marks
4. (ii) OR: The multiples of 3 from 1 to 15 are 3, 6, 9, 12, 15, so there are 5 favourable outcomes.
  ½ mark
5. (ii) OR: P(multiple of 3) $= \frac{5}{15} = \frac{1}{3}$ .
  1½ marks
6. (iii) The solid coloured balls are numbered 1 to 8. Of these, the even-numbered ones are 2, 4, 6, 8, so there are 4 favourable outcomes.
  P(solid and even) $= \frac{4}{15}$ .
  1 mark
Answer: (i) 1/15; (ii) 7/15 OR 1/3; (iii) 4/15
Where the marks are
- P(number 8) $= 1/15$ (1 mark)
- Lists the 7 even-numbered balls (½ mark)
- P(even) $= 7/15$ (1½ marks)
- OR part: lists the multiples of 3: 3, 6, 9, 12, 15 (½ mark)
- OR part: P(multiple of 3) $= 1/3$ (1½ marks)
- P(solid and even) $= 4/15$ (1 mark)
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