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

CBSE Class 10 Mathematics Basic: 2025 board paper, set 430/4/1

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/241_Mathematics_Basic.zip (a zip; this paper is the file 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 23 and 24
C	26-31	Short answer; internal choice in 27 and 29
D	32-35	Long answer; internal choice in 32 and 34
E	36-38	Case-based; internal choice in (iii) of each case

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 two-mark subparts of Section E. Use of calculators is not allowed. Take π = 22/7 where required.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If $HCF (x, 20) = 2$ and $LCM (x, 20) = 60$ , then value of $x$ is:
1. (A)3
2. (B)6
3. (C)20
4. (D)10
Try it first, then show our solutionSolution
1. Use $HCF \times LCM = x \times 20$ : $2 \times 60 = 20 x$ , so $x = 6$ . Hence option (B).
  1 mark
Answer: x = 6, option (B)
Where the marks are
- Uses HCF × LCM = product of the numbers to get x = 6 (1 mark)
Q2 · 1 mark · MCQCoordinate Geometry
The distance between the points $(- 6, 9)$ and $(2, 7)$ is:
1. (A) $217$
2. (B) $417$
3. (C) $25$
4. (D) $215$
Try it first, then show our solutionSolution
1. By the distance formula, $d = (2 + 6)^{2} + (7 - 9)^{2} = 64 + 4 = 68 = 217$ . Hence option (A).
  1 mark
Answer: $217$ , option (A)
Where the marks are
- Applies the distance formula correctly to get $217$ (1 mark)
Q3 · 1 mark · MCQArithmetic Progressions
If $n^{th}$ term of an A.P. is $5 n - 6$ , then its common difference is:
1. (A) $- 6$
2. (B) $5 n$
3. (C)5
4. (D)6
Try it first, then show our solutionSolution
1. $a_{1} = 5 (1) - 6 = - 1$ and $a_{2} = 5 (2) - 6 = 4$ , so $d = a_{2} - a_{1} = 4 - (- 1) = 5$ . Hence option (C).
  1 mark
Answer: d = 5, option (C)
Where the marks are
- Finds $d = a_{2} - a_{1} = 5$ (1 mark)
Q4 · 1 mark · MCQPolynomials
One of the zeroes of the polynomial $p (x) = k x^{2} - 9 x + 3$ is $(- \frac{3}{2})$ . The value of $k$ is:
1. (A) $\frac{22}{3}$
2. (B) $- \frac{14}{3}$
3. (C) $\frac{14}{3}$
4. (D) $- \frac{22}{3}$
Try it first, then show our solutionSolution
1. Put $x = - \frac{3}{2}$ in $p (x)$ and set it to zero: $p (- \frac{3}{2}) = \frac{9}{4} k + \frac{27}{2} + 3 = 0$ .
  ½ mark
2. So $\frac{9}{4} k = - \frac{33}{2}$ , which gives $k = - \frac{66}{9} = - \frac{22}{3}$ . Hence option (D).
  ½ mark
Answer: $k = - \frac{22}{3}$ , option (D)
Where the marks are
- Substitutes the zero and equates $p (- \frac{3}{2})$ to 0 (½ mark)
- Solves to get $k = - \frac{22}{3}$ (½ mark)
Open Q4 on its own page
Q5 · 1 mark · MCQProbability
Three coins are tossed together. The probability that only one coin shows tail, is:
1. (A) $\frac{1}{2}$
2. (B) $\frac{3}{8}$
3. (C) $\frac{7}{8}$
4. (D)1
Try it first, then show our solutionSolution
1. Tossing three coins gives 8 equally likely outcomes.
2. Exactly one tail occurs in TFF-type cases: THH, HTH, HHT, which is 3 outcomes. So $P = \frac{3}{8}$ . Hence option (B).
  1 mark
Answer: $\frac{3}{8}$ , option (B)
Where the marks are
- Counts 3 favourable outcomes out of 8 to get $\frac{3}{8}$ (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQSurface Areas and Volumes
Two right circular cylinders of equal volumes have their heights in the ratio $1 : 2$ . The ratio of their radii is:
1. (A) $2 : 1$
2. (B) $1 : 2$
3. (C) $1 : 4$
4. (D) $1 : 2$
Try it first, then show our solutionSolution
1. Equal volumes give $π r_{1}^{2} h_{1} = π r_{2}^{2} h_{2}$ , so $\frac{r _{1}^{2}}{r _{2}^{2}} = \frac{h _{2}}{h _{1}} = \frac{2}{1}$ .
  ½ mark
2. Taking square roots, $r_{1} : r_{2} = 2 : 1$ . Hence option (A).
  ½ mark
Answer: $2 : 1$ , option (A)
Where the marks are
- Equates the two volumes to get $r_{1}^{2} / r_{2}^{2} = 2$ (½ mark)
- Concludes $r_{1} : r_{2} = 2 : 1$ (½ mark)
Open Q6 on its own page
Q7 · 1 mark · MCQIntroduction to Trigonometry
If $2 sin θ = 1$ , then $cot θ \times cosec θ$ is equal to:
1. (A) $\frac{1}{2}$
2. (B) $\frac{1}{2 2}$
3. (C) $2$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. $sin θ = \frac{1}{2}$ , so $θ = 4 5^{\circ}$ .
  ½ mark
2. Then $cot 4 5^{\circ} \times cosec 4 5^{\circ} = 1 \times 2 = 2$ . Hence option (C).
  ½ mark
Answer: $2$ , option (C)
Where the marks are
- Finds $θ = 4 5^{\circ}$ (½ mark)
- Evaluates the product as $2$ (½ mark)
Q8 · 1 mark · MCQCircles
PQ and PR are tangents to the circle of radius 3 cm and centre O. If length of each tangent is 4 cm, then perimeter of $Δ$ OQP is:
1. (A)5 cm
2. (B)12 cm
3. (C)9 cm
4. (D)8 cm
Figure in the paper: A circle with centre O and two tangents from external point P touching the circle at Q (top) and R (bottom). OP passes through the centre; OQ and OR are radii, with PQ and PR tangents. Q is marked with a right-angle symbol at the point of contact. See it in the official paper.
Try it first, then show our solutionSolution
1. $O Q ⊥ P Q$ [tangent ⟂ radius], so $O P = 3^{2} + 4^{2} = 5$ cm.
  ½ mark
2. The perimeter of $△ O QP$ is $3 + 4 + 5 = 12$ cm. Hence option (B).
  ½ mark
Answer: 12 cm, option (B)
Where the marks are
- Uses Pythagoras to get OP = 5 cm (½ mark)
- Adds the three sides to get a perimeter of 12 cm (½ mark)
Q9 · 1 mark · MCQPolynomials
$α$ , $β$ are zeroes of the polynomial $2 x^{2} + 5 x + 1$ . The value of $(\frac{1}{α} + \frac{1}{β})$ is:
1. (A) $- \frac{5}{4}$
2. (B)5
3. (C) $\frac{5}{4}$
4. (D) $- 5$
Try it first, then show our solutionSolution
1. For $2 x^{2} + 5 x + 1$ : $α + β = - \frac{5}{2}$ and $α β = \frac{1}{2}$ .
  ½ mark
2. $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- 5/2}{1/2} = - 5$ . Hence option (D).
  ½ mark
Answer: −5, option (D)
Where the marks are
- Writes the sum and product of the zeroes (½ mark)
- Gets $\frac{1}{α} + \frac{1}{β} = - 5$ (½ mark)
Open Q9 on its own page
Q10 · 1 mark · MCQArithmetic Progressions
The 20 $^{th}$ term of the A.P.: $102, 62, 22, \dots$ is:
1. (A) $- 76 + 102$
2. (B) $- 622$
3. (C) $- 662$
4. (D) $862$
Try it first, then show our solutionSolution
1. $a = 102$ and $d = 62 - 102 = - 42$ .
  ½ mark
2. $a_{20} = a + 19 d = 102 - 762 = - 662$ . Hence option (C).
  ½ mark
Answer: $- 662$ , option (C)
Where the marks are
- Finds $d = - 42$ (½ mark)
- Uses $a_{20} = a + 19 d$ to get $- 662$ (½ mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
If $sec θ - tan θ = 2$ , then $sec θ + tan θ$ is equal to:
1. (A) $\frac{1}{2}$
2. (B) $2$
3. (C) $\frac{1}{2}$
4. (D)2
Try it first, then show our solutionSolution
1. $(sec θ - tan θ) (sec θ + tan θ) = sec^{2} θ - tan^{2} θ = 1$ .
  ½ mark
2. So $sec θ + tan θ = \frac{1}{2}$ . Hence option (A).
  ½ mark
Answer: $\frac{1}{2}$ , option (A)
Where the marks are
- Uses the identity $sec^{2} θ - tan^{2} θ = 1$ (½ mark)
- Gets $\frac{1}{2}$ (½ mark)
Q12 · 1 mark · MCQProbability
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that drawn card shows number '9' is:
1. (A) $\frac{1}{26}$
2. (B) $\frac{4}{13}$
3. (C) $\frac{1}{52}$
4. (D) $\frac{1}{13}$
Try it first, then show our solutionSolution
1. A deck has four 9s, one in each suit, so $P = \frac{4}{52} = \frac{1}{13}$ . Hence option (D).
  1 mark
Answer: $\frac{1}{13}$ , option (D)
Where the marks are
- Counts 4 favourable cards out of 52 to get $\frac{1}{13}$ (1 mark)
Q13 · 1 mark · MCQAreas Related to Circles
The length of arc subtending an angle of $21 0^{\circ}$ at the centre of the circle, is $\frac{44}{3}$ cm. The radius of the circle is:
1. (A) $22$ cm
2. (B)4 cm
3. (C)8 cm
4. (D) $\frac{1}{4}$ cm
Try it first, then show our solutionSolution
1. Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r$ , so $\frac{210}{360} \times 2 \times \frac{22}{7} \times r = \frac{44}{3}$ .
  ½ mark
2. This simplifies to $\frac{11 r}{3} = \frac{44}{3}$ , so $r = 4$ cm. Hence option (B).
  ½ mark
Answer: 4 cm, option (B)
Where the marks are
- Applies the arc length formula (½ mark)
- Solves to get r = 4 cm (½ mark)
Open Q13 on its own page
Q14 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of $m$ for which lines $14 x + m y = 20$ and $- 3 x + 2 y = 16$ are parallel, is:
1. (A) $- \frac{3}{14}$
2. (B) $- \frac{7}{3}$
3. (C) $- \frac{28}{3}$
4. (D) $- \frac{3}{28}$
Try it first, then show our solutionSolution
1. For parallel lines, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , so $\frac{14}{- 3} = \frac{m}{2}$ .
  ½ mark
2. This gives $m = - \frac{28}{3}$ . Hence option (C).
  ½ mark
Answer: $m = - \frac{28}{3}$ , option (C)
Where the marks are
- Uses the condition for parallel lines (½ mark)
- Gets $m = - \frac{28}{3}$ (½ mark)
Q15 · 1 mark · MCQSurface Areas and Volumes
The curved surface area of a cone with base radius 7 cm, is $550 c m^{2}$ . The slant height of the cone is:
1. (A)25 cm
2. (B)14 cm
3. (C)20 cm
4. (D)24 cm
Try it first, then show our solutionSolution
1. CSA $= π r l$ , so $\frac{22}{7} \times 7 \times l = 550$ , giving $22 l = 550$ and $l = 25$ cm. Hence option (A).
  1 mark
Answer: 25 cm, option (A)
Where the marks are
- Uses CSA $= π r l$ to get l = 25 cm (1 mark)
Q16 · 1 mark · MCQIntroduction to Trigonometry
If $sin A = \frac{2}{3}$ , then $cos A$ is equal to:
1. (A) $\frac{3}{2}$
2. (B) $\frac{5}{3}$
3. (C) $\frac{1}{3}$
4. (D) $\frac{1}{3}$
Try it first, then show our solutionSolution
1. $cos A = 1 - sin^{2} A = 1 - \frac{4}{9} = \frac{5}{3}$ . Hence option (B).
  1 mark
Answer: $\frac{5}{3}$ , option (B)
Where the marks are
- Uses $sin^{2} A + cos^{2} A = 1$ to get $\frac{5}{3}$ (1 mark)
Q17 · 1 mark · MCQCircles
PA and PB are tangents to a circle with centre O. If $∠ A O B = 10 5^{\circ}$ then $∠ O A P + ∠ A P B$ is equal to:
1. (A) $7 5^{\circ}$
2. (B) $17 5^{\circ}$
3. (C) $18 0^{\circ}$
4. (D) $16 5^{\circ}$
Figure in the paper: A circle with centre O, external point P with tangents PA (top) and PB (bottom) touching the circle at A and B. Chord-like segment AB joins the points of contact and angle AOB = 105° is marked at O. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius].
  ½ mark
2. $∠ A P B = 18 0^{\circ} - 10 5^{\circ} = 7 5^{\circ}$ , so the sum is $9 0^{\circ} + 7 5^{\circ} = 16 5^{\circ}$ . Hence option (D).
  ½ mark
Answer: 165°, option (D)
Where the marks are
- States $∠ O A P = 9 0^{\circ}$ (½ mark)
- Finds $∠ A P B = 7 5^{\circ}$ and the sum 165° (½ mark)
Q18 · 1 mark · MCQTriangles
In $Δ$ ABC, PQ $∥$ BC. It is given that AP = 2.4 cm, PB = 3.6 cm and BC = 5.4 cm. PQ is equal to:
1. (A)2.7 cm
2. (B)1.8 cm
3. (C)3.6 cm
4. (D)2.16 cm
Figure in the paper: Triangle ABC with A at the top, P on side AB and Q on side AC; PQ is drawn parallel to the base BC, with arrows marking the parallel lines. See it in the official paper.
Try it first, then show our solutionSolution
1. $△ A P Q \sim △ A B C$ (AA), so $\frac{P Q}{B C} = \frac{A P}{A B} = \frac{2.4}{6}$ .
  ½ mark
2. $P Q = 0.4 \times 5.4 = 2.16$ cm. Hence option (D).
  ½ mark
Answer: 2.16 cm, option (D)
Where the marks are
- Uses similarity with AB = 6 cm (½ mark)
- Gets PQ = 2.16 cm (½ mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedStatistics
Directions: Question numbers 19 and 20 are Assertion and Reason based questions. Select the correct answer from the codes (a), (b), (c) and (d). Assertion (A): Median marks of students in a class test is 16. It means half of the class got marks less than 16. Reason (R): Median divides the distribution in two equal parts.
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. The median is the middle value, so it splits the data into two equal halves; R is true.
  ½ mark
2. Hence half the class scored below 16, so A is true, and R explains A. Option (A).
  ½ mark
Answer: Option (A)
Where the marks are
- Recognises that both statements are true and R explains A (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A): If E is an event such that $P (E) = \frac{1}{999}$ , then $P (\overset{ˉ}{E}) = 0.001$ . Reason (R): $P (E) + P (\overset{ˉ}{E}) = 1$
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. $P (E) + P (\overset{ˉ}{E}) = 1$ is a true result, so R is true.
  ½ mark
2. $P (\overset{ˉ}{E}) = 1 - \frac{1}{999} = \frac{998}{999} \neq = 0.001$ , so A is false. Option (D).
  ½ mark
Answer: Option (D)
Where the marks are
- Computes $P (\overset{ˉ}{E}) = \frac{998}{999}$ and concludes A is false, R is true (1 mark)

Section B

Very short answer; internal choice in 23 and 24

Q21 · 2 marks · Very short answerReal Numbers
Show that $4 5^{n}$ can not end with the digit 0, $n$ being a natural number. Write the prime number ' $a$ ' which on multiplying with $4 5^{n}$ makes the product end with the digit 0.
Try it first, then show our solutionSolution
1. Prime factorisation: $45 = 3^{2} \times 5$ , so $4 5^{n} = (3 \times 3)^{n} \times 5^{n} = 3^{2 n} \times 5^{n}$ .
  1 mark
2. A number ends with the digit 0 only if it is divisible by 10, i.e. its prime factorisation contains both 2 and 5. The prime factorisation of $4 5^{n}$ has no 2. By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), $4 5^{n}$ cannot end with the digit 0 for any natural number $n$ .
  ½ mark
3. Multiplying by the prime 2 gives $2 \times 4 5^{n} = 2 \times 3^{2 n} \times 5^{n}$ . This contains both 2 and 5, so the product ends with 0. Hence $a = 2$ .
  ½ mark
Answer: $4 5^{n} = 3^{2 n} \times 5^{n}$ has no factor 2, so it cannot end with 0; $a = 2$
Where the marks are
- Writes $4 5^{n} = 3^{2 n} \times 5^{n}$ (1 mark)
- Argues that 2 is absent, so the number cannot end with 0 (½ mark)
- States $a = 2$ (½ mark)
Open Q21 on its own page
Q22 · 2 marks · Very short answerCoordinate Geometry
Point P $(x, 0)$ divides the line segment joining the points $(2, 8)$ and $(- 3, - 5)$ in a certain ratio. Find the ratio and hence find the value of $x$ .
Try it first, then show our solutionSolution
1. Let P $(x, 0)$ divide the join of A $(2, 8)$ and B $(- 3, - 5)$ in the ratio $k : 1$ . By the section formula, the $y$ -coordinate of P is $\frac{k ( - 5 ) + 1 ( 8 )}{k + 1} = 0$ .
2. $- 5 k + 8 = 0 \Rightarrow k = \frac{8}{5}$ . So the required ratio is $8 : 5$ .
  1½ marks
3. $x = \frac{8 ( - 3 ) + 5 ( 2 )}{8 + 5} = \frac{- 24 + 10}{13} = - \frac{14}{13}$
  ½ mark
Answer: Ratio $8 : 5$ ; $x = - \frac{14}{13}$
Where the marks are
- Uses the section formula on the $y$ -coordinate and gets the ratio $8 : 5$ (1½ marks)
- Finds $x = - \frac{14}{13}$ (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerProbability
A coin is dropped at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with radius 0.7 m?
Figure in the paper: A rectangle of length 3 m (top) and width 2 m (right side), with a circle drawn inside it. See it in the official paper.
Try it first, then show our solutionSolution
1. Area of circle $= π r^{2} = \frac{22}{7} \times \frac{7}{10} \times \frac{7}{10} = \frac{154}{100} = 1.54$ m²
  ½ mark
2. Area of rectangle $= 3 \times 2 = 6$ m²
  ½ mark
3. P(coin lands inside the circle) $= \frac{area of circle}{area of rectangle} = \frac{1.54}{6} = \frac{154}{600} = \frac{77}{300}$
  1 mark
Answer: $\frac{77}{300}$
Where the marks are
- Area of circle $= 1.54$ m² (½ mark)
- Area of rectangle $= 6$ m² (½ mark)
- Probability $= \frac{77}{300}$ (1 mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerProbability
A die is thrown twice. What is the probability that (i) difference between two numbers obtained is 3? (ii) sum of the numbers obtained is 8?
Try it first, then show our solutionSolution
1. A die thrown twice gives $6 \times 6 = 36$ equally likely outcomes.
2. (i) Outcomes with difference 3: $(1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3)$ , which is 6 outcomes. P(difference is 3) $= \frac{6}{36} = \frac{1}{6}$
  1 mark
3. (ii) Outcomes with sum 8: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ , which is 5 outcomes. P(sum is 8) $= \frac{5}{36}$
  1 mark
Answer: (i) $\frac{1}{6}$ ; (ii) $\frac{5}{36}$
Where the marks are
- Lists the 6 favourable outcomes and gets $\frac{6}{36} = \frac{1}{6}$ (1 mark)
- Lists the 5 favourable outcomes and gets $\frac{5}{36}$ (1 mark)
Open Q23(b) on its own page
Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerCircles
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O. PA and PB are tangents from the external point P, touching the circle at A and B. Join OA and OB.
  ½ mark
2. $∠ O A P = ∠ O B P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]
  ½ mark
3. In quadrilateral OAPB, $∠ A P B + ∠ P B O + ∠ B O A + ∠ O A P = 36 0^{\circ}$ [angle sum of a quadrilateral]. So $∠ A P B + 9 0^{\circ} + ∠ A O B + 9 0^{\circ} = 36 0^{\circ}$ , which gives $∠ A P B + ∠ A O B = 18 0^{\circ}$ .
  ½ mark
4. Hence $∠ A P B$ and $∠ A O B$ are supplementary.
  ½ mark
Answer: $∠ A P B + ∠ A O B = 18 0^{\circ}$ , so the two angles are supplementary.
Where the marks are
- Correct labelled figure (½ mark)
- $∠ O A P = ∠ O B P = 9 0^{\circ}$ with reason (½ mark)
- Angle sum of quadrilateral OAPB $= 36 0^{\circ}$ (½ mark)
- Concludes $∠ A P B + ∠ A O B = 18 0^{\circ}$ (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerCircles
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O and diameter AB. Line PQ is the tangent at A (P and Q on either side of A), and line RS is the tangent at B (R above P, S above Q).
  ½ mark
2. $∠ O A P = 9 0^{\circ}$ and $∠ O B R = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]
  ½ mark
3. $∠ O A P + ∠ O B R = 9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ . These are co-interior angles made by the transversal AB with the lines PQ and RS, and they are supplementary.
  ½ mark
4. Therefore $P Q ∥ R S$ , i.e. the tangents at the ends of a diameter are parallel.
  ½ mark
Answer: The co-interior angles at A and B add to $18 0^{\circ}$ , so the tangents PQ and RS are parallel.
Where the marks are
- Correct figure (½ mark)
- Both angles $= 9 0^{\circ}$ with reason (½ mark)
- Sum $= 18 0^{\circ}$ , co-interior angles supplementary (½ mark)
- Concludes $P Q ∥ R S$ (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerTriangles
In the given figure $∠ A D E = ∠ A C B$ and $\frac{A D}{D B} = \frac{A E}{E C}$ . Prove that $Δ$ ABC is an isosceles triangle.
Figure in the paper: Triangle ABC with A at the top. D is on side AB and E on side AC; DE is drawn parallel to BC. Angles at D and C are marked equal. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $\frac{A D}{D B} = \frac{A E}{E C}$ , we get $D E ∥ B C$ [converse of Basic Proportionality Theorem].
  ½ mark
2. Therefore $∠ A D E = ∠ A B C$ [corresponding angles, DE ∥ BC, AB is the transversal].
  ½ mark
3. It is given that $∠ A D E = ∠ A C B$ , so $∠ A B C = ∠ A C B$ .
  ½ mark
4. Sides opposite equal angles are equal, so $A C = A B$ . Hence $Δ A B C$ is isosceles.
  ½ mark
Answer: $∠ A B C = ∠ A C B \Rightarrow A B = A C$ , so $Δ A B C$ is isosceles.
Where the marks are
- $D E ∥ B C$ by converse of BPT (½ mark)
- $∠ A D E = ∠ A B C$ (corresponding angles) (½ mark)
- $∠ A B C = ∠ A C B$ (½ mark)
- $A B = A C$ , so the triangle is isosceles (½ mark)
Open Q25 on its own page

Section C

Short answer; internal choice in 27 and 29

Q26 · 3 marks · Short answerPolynomials
Find the zeroes of the polynomial $p (x) = 6 x^{2} + 13 x - 5$ and verify the relationship between its zeroes and the coefficients.
Try it first, then show our solutionSolution
1. Split the middle term: $6 x^{2} + 13 x - 5 = 6 x^{2} + 15 x - 2 x - 5 = 3 x (2 x + 5) - 1 (2 x + 5) = (3 x - 1) (2 x + 5)$ . So the zeroes are $x = \frac{1}{3}$ and $x = - \frac{5}{2}$ .
  1 mark
2. Sum of zeroes $= \frac{1}{3} - \frac{5}{2} = \frac{2 - 15}{6} = - \frac{13}{6}$ , and $- \frac{coefficient of x}{coefficient of x ^{2}} = - \frac{13}{6}$ . Verified.
  1 mark
3. Product of zeroes $= \frac{1}{3} \times (- \frac{5}{2}) = - \frac{5}{6}$ , and $\frac{constant term}{coefficient of x ^{2}} = \frac{- 5}{6}$ . Verified.
  1 mark
Answer: Zeroes $\frac{1}{3}$ and $- \frac{5}{2}$ ; sum $= - \frac{13}{6}$ , product $= - \frac{5}{6}$ (relationship verified)
Where the marks are
- Factorises and finds the zeroes $\frac{1}{3}, - \frac{5}{2}$ (1 mark)
- Verifies the sum of zeroes $= - b / a$ (1 mark)
- Verifies the product of zeroes $= c / a$ (1 mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerArithmetic Progressions
Find the sum of the A.P. $7, 10 \frac{1}{2}, 14, \dots, 84$ .
Try it first, then show our solutionSolution
1. $a = 7$ , $d = \frac{21}{2} - 7 = \frac{7}{2}$ , last term $l = 84$
  1 mark
2. $a_{n} = a + (n - 1) d$ : $84 = 7 + (n - 1) \times \frac{7}{2} \Rightarrow (n - 1) \times \frac{7}{2} = 77 \Rightarrow n - 1 = 22 \Rightarrow n = 23$
  1 mark
3. $S_{23} = \frac{n}{2} (a + l) = \frac{23}{2} (7 + 84) = \frac{23 \times 91}{2} = \frac{2093}{2}$ , i.e. $1046.5$
  1 mark
Answer: Sum $= \frac{2093}{2} = 1046.5$
Where the marks are
- Identifies $a = 7$ , $d = \frac{7}{2}$ (1 mark)
- Finds $n = 23$ (1 mark)
- Finds $S_{23} = \frac{2093}{2}$ (1 mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerArithmetic Progressions
If the sum of first $n$ terms of an A.P. is given by $S_{n} = \frac{n}{2} (2 n + 8)$ . Then, find its first term and common difference. Hence, find its 15 $^{th}$ term.
Try it first, then show our solutionSolution
1. $S_{n} = \frac{n}{2} (2 n + 8) = n (n + 4)$ . First term $a = S_{1} = 1 \times 5 = 5$
  ½ mark
2. $S_{2} = 2 \times 6 = 12$
  ½ mark
3. $a_{2} = S_{2} - S_{1} = 12 - 5 = 7$ , so $d = a_{2} - a_{1} = 7 - 5 = 2$
  1 mark
4. $a_{15} = a + 14 d = 5 + 14 \times 2 = 33$
  1 mark
Answer: $a = 5$ , $d = 2$ , $a_{15} = 33$
Where the marks are
- $S_{1} = a = 5$ (½ mark)
- $S_{2} = 12$ (½ mark)
- $d = 2$ (1 mark)
- $a_{15} = 33$ (1 mark)
Open Q27(b) on its own page
Q28 · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3$ is rational.
  Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are coprime.
  Squaring: $3 q^{2} = p^{2}$ .
  ½ mark
2. So $p^{2}$ is divisible by 3. Since 3 is prime, 3 divides $p^{2}$ means 3 divides $p$ . ... (1)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2}$ , so $q^{2} = 3 a^{2}$ .
  So $q^{2}$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)
  1 mark
4. From (1) and (2), 3 divides both $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong. Hence $3$ is irrational.
  ½ mark
Answer: Assuming $3 = p / q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $3$ is irrational
Where the marks are
- Assumes $3 = \frac{p}{q}$ with p, q coprime and gets $3 q^{2} = p^{2}$ (1/2 mark)
- Shows 3 divides p (1 mark)
- Substitutes $p = 3 a$ and shows 3 divides q (1 mark)
- States the contradiction and concludes (1/2 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCoordinate Geometry
If points A $(- 5, y)$ , B $(2, - 2)$ , C $(8, 4)$ and D $(x, 5)$ taken in order, form a parallelogram ABCD, then find the values of $x$ and $y$ . Hence, find lengths of sides of the parallelogram.
Try it first, then show our solutionSolution
1. The diagonals of a parallelogram bisect each other, so the mid-point of BD = the mid-point of AC: $(\frac{2 + x}{2}, \frac{- 2 + 5}{2}) = (\frac{- 5 + 8}{2}, \frac{y + 4}{2})$
  1 mark
2. $\frac{2 + x}{2} = \frac{3}{2} \Rightarrow x = 1$ ; $\frac{3}{2} = \frac{y + 4}{2} \Rightarrow y = - 1$
  1 mark
3. So A $(- 5, - 1)$ , B $(2, - 2)$ , C $(8, 4)$ , D $(1, 5)$ . $A B = (2 + 5)^{2} + (- 2 + 1)^{2} = 49 + 1 = 50 = 52$ units $= C D$
  ½ mark
4. $B C = (8 - 2)^{2} + (4 + 2)^{2} = 36 + 36 = 72 = 62$ units $= A D$
  ½ mark
Answer: $x = 1$ , $y = - 1$ ; sides AB = CD = $52$ units, BC = AD = $62$ units
Where the marks are
- Equates the mid-points of the diagonals AC and BD (1 mark)
- Gets $x = 1$ and $y = - 1$ (½ + ½ marks)
- $A B = 52$ (½ mark)
- $B C = 62$ (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCoordinate Geometry
A(6, $- 3$ ), B(0, 5) and C $(- 2, 1)$ are vertices of $Δ$ ABC. Points P(3, 1) and Q(2, $- 1$ ) lie on sides AB and AC respectively. Check whether $\frac{A P}{P B} = \frac{A Q}{QC}$ .
Try it first, then show our solutionSolution
1. $A P = (3 - 6)^{2} + (1 + 3)^{2} = 9 + 16 = 5$
  ½ mark
2. $P B = (0 - 3)^{2} + (5 - 1)^{2} = 9 + 16 = 5$
  ½ mark
3. $A Q = (2 - 6)^{2} + (- 1 + 3)^{2} = 16 + 4 = 25$
  ½ mark
4. $QC = (- 2 - 2)^{2} + (1 + 1)^{2} = 16 + 4 = 25$
  ½ mark
5. $\frac{A P}{P B} = \frac{5}{5} = 1$ and $\frac{A Q}{QC} = \frac{2 5}{2 5} = 1$ . Therefore $\frac{A P}{P B} = \frac{A Q}{QC}$ . Yes, the ratios are equal.
  1 mark
Answer: Yes, $\frac{A P}{P B} = \frac{A Q}{QC} = 1$
Where the marks are
- $A P = 5$ (½ mark)
- $P B = 5$ (½ mark)
- $A Q = 25$ (½ mark)
- $QC = 25$ (½ mark)
- Both ratios equal 1, hence equal (1 mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerAreas Related to Circles
A chord of a circle of radius 10 cm subtends an angle of $6 0^{\circ}$ at the centre O. Find the area of the shaded region. (Use $3 = 1.73$ , $2 = 1.41$ and $π = 3.14$ )
Figure in the paper: A circle with centre O and a chord AB. Angle AOB = 60° is marked at O. The minor segment between chord AB and the arc is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. Area of sector OAB $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{60}{360} \times 3.14 \times 1 0^{2} = \frac{314}{6} \approx 52.33$ cm²
  1 mark
2. OA = OB = 10 cm and $∠ A O B = 6 0^{\circ}$ , so $Δ O A B$ is equilateral. Area of $Δ O A B = \frac{3}{4} \times 1 0^{2} = \frac{1.73 \times 100}{4} = \frac{173}{4} = 43.25$ cm²
  1 mark
3. Area of shaded segment $= \frac{314}{6} - \frac{173}{4} = \frac{628 - 519}{12} = \frac{109}{12} \approx 9.08$ cm²
  1 mark
Answer: $\frac{109}{12}$ cm² $\approx 9.08$ cm²
Where the marks are
- Area of sector $= \frac{314}{6}$ cm² (1 mark)
- Area of equilateral $Δ O A B = \frac{173}{4}$ cm² (1 mark)
- Shaded area $= 9.08$ cm² (1 mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCircles
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Try it first, then show our solutionSolution
1. Figure: quadrilateral ABCD circumscribes a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S respectively. Join OA, OB, OC, OD and OP, OQ, OR, OS. Name the angles at O: $∠1 = ∠ A O P$ , $∠2 = ∠ P O B$ , $∠3 = ∠ B O Q$ , $∠4 = ∠ QO C$ , $∠5 = ∠ C O R$ , $∠6 = ∠ R O D$ , $∠7 = ∠ D O S$ , $∠8 = ∠ S O A$ . To prove: $∠ A O B + ∠ C O D = 18 0^{\circ}$ .
  ½ mark
2. In $Δ O A P$ and $Δ O A S$ : OP = OS [radii], AP = AS [tangents from an external point are equal], OA = OA [common]. So $Δ O A P ≅ Δ O A S$ [SSS].
  ½ mark
3. Hence $∠1 = ∠8$ [CPCT]. Similarly $∠2 = ∠3$ , $∠4 = ∠5$ , $∠6 = ∠7$ .
  1 mark
4. Angles round O: $∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 36 0^{\circ}$ , so $2 (∠1 + ∠2 + ∠5 + ∠6) = 36 0^{\circ}$ , giving $∠1 + ∠2 + ∠5 + ∠6 = 18 0^{\circ}$ .
  ½ mark
5. Since $∠1 + ∠2 = ∠ A O B$ and $∠5 + ∠6 = ∠ C O D$ , we get $∠ A O B + ∠ C O D = 18 0^{\circ}$ . Similarly $∠ B O C + ∠ A O D = 18 0^{\circ}$ . Hence opposite sides subtend supplementary angles at the centre.
  ½ mark
Answer: $∠ A O B + ∠ C O D = 18 0^{\circ}$ and $∠ B O C + ∠ A O D = 18 0^{\circ}$
Where the marks are
- Correct figure (½ mark)
- Proves $Δ O A P ≅ Δ O A S$ (½ mark)
- Obtains the four pairs of equal angles (1 mark)
- Uses the angle sum $36 0^{\circ}$ at O (½ mark)
- Concludes $∠ A O B + ∠ C O D = 18 0^{\circ}$ (½ mark)
Open Q31 on its own page

Section D

Long answer; internal choice in 32 and 34

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerQuadratic Equations
It is given that $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$ ; $(p \neq = 0)$ . (i) Show that the discriminant (D) of above equation is a perfect square. (ii) Find the roots of the equation.
Try it first, then show our solutionSolution
1. (i) Here $a = p^{2}$ , $b = p^{2} - q^{2}$ , $c = - q^{2}$ . $D = b^{2} - 4 a c = (p^{2} - q^{2})^{2} - 4 (p^{2}) (- q^{2}) = (p^{2} - q^{2})^{2} + 4 p^{2} q^{2}$
  1 mark
2. $= p^{4} - 2 p^{2} q^{2} + q^{4} + 4 p^{2} q^{2} = p^{4} + 2 p^{2} q^{2} + q^{4} = (p^{2} + q^{2})^{2}$ , which is a perfect square.
  1 mark
3. (ii) $x = \frac{- b \pm D}{2 a} = \frac{- ( p ^{2} - q ^{2} ) \pm ( p ^{2} + q ^{2} ) ^{2}}{2 p ^{2}} = \frac{- ( p ^{2} - q ^{2} ) \pm ( p ^{2} + q ^{2} )}{2 p ^{2}}$
  1 mark
4. Taking +: $x = \frac{- p ^{2} + q ^{2} + p ^{2} + q ^{2}}{2 p ^{2}} = \frac{2 q ^{2}}{2 p ^{2}} = \frac{q ^{2}}{p ^{2}}$
  1 mark
5. Taking −: $x = \frac{- p ^{2} + q ^{2} - p ^{2} - q ^{2}}{2 p ^{2}} = \frac{- 2 p ^{2}}{2 p ^{2}} = - 1$
  1 mark
Answer: (i) $D = (p^{2} + q^{2})^{2}$ , a perfect square; (ii) roots $\frac{q ^{2}}{p ^{2}}$ and $- 1$
Where the marks are
- Writes $D = (p^{2} - q^{2})^{2} + 4 p^{2} q^{2}$ (1 mark)
- Simplifies to $(p^{2} + q^{2})^{2}$ (1 mark)
- Applies the quadratic formula correctly (1 mark)
- Root $\frac{q ^{2}}{p ^{2}}$ (1 mark)
- Root $- 1$ (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerQuadratic Equations
Three consecutive positive integers are such that the sum of the square of smallest and product of other two is 67. Find the numbers, using quadratic equation.
Try it first, then show our solutionSolution
1. Let the three consecutive positive integers be $x$ , $x + 1$ and $x + 2$ .
  1 mark
2. According to the question, $x^{2} + (x + 1) (x + 2) = 67$ .
  1 mark
3. $x^{2} + x^{2} + 3 x + 2 = 67 \Rightarrow 2 x^{2} + 3 x - 65 = 0$
  1 mark
4. $2 x^{2} + 13 x - 10 x - 65 = 0 \Rightarrow x (2 x + 13) - 5 (2 x + 13) = 0 \Rightarrow (2 x + 13) (x - 5) = 0$ , so $x = 5$ or $x = - \frac{13}{2}$ . Reject $- \frac{13}{2}$ since $x$ must be a positive integer, so $x = 5$ .
  1 mark
5. The numbers are 5, 6 and 7. Check: $5^{2} + 6 \times 7 = 25 + 42 = 67$ .
  1 mark
Answer: 5, 6 and 7
Where the marks are
- Takes the integers as $x, x + 1, x + 2$ (1 mark)
- Forms the equation $x^{2} + (x + 1) (x + 2) = 67$ (1 mark)
- Reduces to $2 x^{2} + 3 x - 65 = 0$ (1 mark)
- Solves to get $x = 5$ , rejecting $- \frac{13}{2}$ (1 mark)
- States the numbers 5, 6, 7 (1 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerStatistics
Find 'mean' and 'mode' of the following data:
Try it first, then show our solutionSolution
1. Take assumed mean $a = 27.5$ , $h = 5$ , $u_{i} = \frac{x _{i} - 27.5}{5}$ .
  Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  15–20 17.5 6 −2 −12
  20–25 22.5 16 −1 −16
  25–30 27.5 17 0 0
  30–35 32.5 4 1 4
  35–40 37.5 5 2 10
  40–45 42.5 2 3 6
  $\sum f_{i} = 50$ $\sum f_{i} u_{i} = - 8$
  1½ marks
2. Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 27.5 + \frac{- 8}{50} \times 5$
  1 mark
3. $= 27.5 - 0.8 = 26.7$
  ½ mark
4. The highest frequency is 17, so the modal class is 25–30: $l = 25$ , $f_{1} = 17$ , $f_{0} = 16$ , $f_{2} = 4$ , $h = 5$ . Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 25 + \frac{17 - 16}{34 - 16 - 4} \times 5$
  1½ marks
5. $= 25 + \frac{5}{14} \approx 25 + 0.36 = 25.36$
  ½ mark
Answer: Mean $= 26.7$ ; Mode $\approx 25.36$
Where the marks are
- Correct table with $\sum f_{i} = 50$ and $\sum f_{i} u_{i} = - 8$ (1½ marks)
- Correct substitution in the mean formula (1 mark)
- Mean $= 26.7$ (½ mark)
- Identifies the modal class 25–30 and substitutes in the mode formula (1½ marks)
- Mode $\approx 25.36$ (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(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 Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerTriangles
In $Δ$ ABC, P and Q are points on sides AB and AC respectively such that PQ $∥$ BC. Prove that the median AD, drawn from A to BC, bisects PQ.
Try it first, then show our solutionSolution
1. Figure: $Δ A B C$ with P on AB and Q on AC, PQ ∥ BC. Median AD meets BC at D (BD = DC) and cuts PQ at R.
  ½ mark
2. Since PQ ∥ BC, PR ∥ BD. In $Δ A R P$ and $Δ A D B$ : $∠ P A R = ∠ B A D$ [common], $∠ A P R = ∠ A B D$ [corresponding angles]. So $Δ A R P \sim Δ A D B$ [AA similarity].
  1 mark
3. Hence $\frac{A R}{A D} = \frac{P R}{B D}$ ...(i)
  ½ mark
4. Similarly, in $Δ A R Q$ and $Δ A D C$ : $∠ Q A R = ∠ C A D$ [common], $∠ A QR = ∠ A C D$ [corresponding angles], so $Δ A R Q \sim Δ A D C$ [AA]. This gives $\frac{A R}{A D} = \frac{R Q}{D C}$ ...(ii)
  1 mark
5. From (i) and (ii): $\frac{P R}{B D} = \frac{R Q}{D C}$
  ½ mark
6. AD is a median, so $B D = D C$ . Therefore $P R = R Q$ .
  1 mark
7. Hence R is the mid-point of PQ, i.e. the median AD bisects PQ.
  ½ mark
Answer: $P R = R Q$ , so the median AD bisects PQ.
Where the marks are
- Correct figure (½ mark)
- $Δ A R P \sim Δ A D B$ by AA (1 mark)
- Ratio $\frac{A R}{A D} = \frac{P R}{B D}$ (½ mark)
- $Δ A R Q \sim Δ A D C$ and $\frac{A R}{A D} = \frac{R Q}{D C}$ (1 mark)
- Equates the two ratios (½ mark)
- Uses BD = DC to get PR = RQ (1 mark)
- Conclusion (½ mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerSome Applications of Trigonometry
From a point on the ground, the angle of elevation of the top of a pedestal is $3 0^{\circ}$ and that of the top of the flagstaff fixed on the pedestal is $6 0^{\circ}$ . If the length of the flagstaff is 5 m, then find the height of the pedestal and its distance from the point of observation on ground. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: AB is the pedestal of height $h$ m, with foot B on the ground. AD is the flagstaff of 5 m on top, so BD = $(h + 5)$ m. C is the point of observation, with BC = $x$ m and $∠ A B C = 9 0^{\circ}$ . $∠ A C B = 3 0^{\circ}$ and $∠ D C B = 6 0^{\circ}$ .
  1 mark
2. In right $Δ A B C$ : $tan 3 0^{\circ} = \frac{A B}{B C} \Rightarrow \frac{1}{3} = \frac{h}{x} \Rightarrow x = h 3$ ...(i)
  1½ marks
3. In right $Δ D B C$ : $tan 6 0^{\circ} = \frac{D B}{B C} \Rightarrow 3 = \frac{h + 5}{x} \Rightarrow h + 5 = x 3$ ...(ii)
  1½ marks
4. Substituting (i) in (ii): $h + 5 = h 3 \times 3 = 3 h \Rightarrow 2 h = 5 \Rightarrow h = 2.5$ m
  ½ mark
5. $x = 2.5 \times 1.73 = 4.325 \approx 4.33$ m. So the pedestal is 2.5 m high and the point of observation is 4.33 m from its foot.
  ½ mark
Answer: Height of pedestal $= 2.5$ m; distance $\approx 4.33$ m
Where the marks are
- Correct figure (1 mark)
- $tan 3 0^{\circ} = \frac{h}{x} \Rightarrow x = h 3$ (1½ marks)
- $tan 6 0^{\circ} = \frac{h + 5}{x} \Rightarrow h + 5 = x 3$ (1½ marks)
- $h = 2.5$ m (½ mark)
- $x = 4.33$ m (½ mark)
Open Q35 on its own page

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
15–20	17.5	6	−2	−12
20–25	22.5	16	−1	−16
25–30	27.5	17	0	0
30–35	32.5	4	1	4
35–40	37.5	5	2	10
40–45	42.5	2	3	6
		$\sum f_{i} = 50$		$\sum f_{i} u_{i} = - 8$

Section E

Case-based; internal choice in (iii) of each case

Q36 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
A telecommunication company came up with two plans– plan A and plan B for its customers. The plans are represented by linear equations where ' $t$ ' represents the time (in minutes) bought and ' $C$ ' represents the cost. The equations are: Plan A: $3 C = 20 t$ ; Plan B: $3 C = 10 t + 300$ . Based on above information, answer the following questions. (Image: a calculator and a mobile tower.)
1. (i)
  If you purchase plan B, how much initial amount you have to pay?
  1 mark
2. (ii)
  Charu purchased plan A. How many minutes she bought for ₹ 250?
  1 mark
3. (iii)(a)
  At how many minutes, do both the plans charge the same amount? What is that amount?
  2 marks
4. OR
  (iii)(b)
  Which plan is better if you want to buy 60 minutes? Give reason for your answer.
  2 marks
Figure in the paper: Decorative picture of a calculator in front of a mobile phone tower with antennas and dishes. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The initial amount is the cost when no minutes are used. Put $t = 0$ in Plan B: $3 C = 10 (0) + 300 = 300 \Rightarrow C = 100$ . So ₹100 must be paid initially.
  1 mark
2. (ii) For Plan A put $C = 250$ : $3 \times 250 = 20 t \Rightarrow 750 = 20 t \Rightarrow t = 37.5$ . Charu bought 37.5 minutes.
  1 mark
3. (iii)(a) Both plans cost the same when their values of $3 C$ are equal: $20 t = 10 t + 300 \Rightarrow 10 t = 300 \Rightarrow t = 30$ minutes.
  1 mark
4. (iii)(a) Amount: $C = \frac{20 \times 30}{3} = ₹200$ . Both plans charge ₹200 for 30 minutes.
  1 mark
5. (iii)(b) Plan A at $t = 60$ : $3 C = 20 \times 60 = 1200 \Rightarrow C = ₹400$ .
  ½ mark
6. (iii)(b) Plan B at $t = 60$ : $3 C = 10 \times 60 + 300 = 900 \Rightarrow C = ₹300$ .
  ½ mark
7. (iii)(b) Plan B is better for 60 minutes, because it costs ₹100 less than Plan A.
  1 mark
Answer: (i) ₹100; (ii) 37.5 minutes; (iii)(a) 30 minutes, ₹200; (iii)(b) Plan B (₹300 against ₹400, saves ₹100)
Where the marks are
- Puts $t = 0$ in Plan B and gets ₹100 (1 mark)
- Solves $750 = 20 t$ to get $t = 37.5$ minutes (1 mark)
- Equates the plans to get $t = 30$ minutes (1 mark)
- Finds the common amount ₹200 (1 mark)
- Cost under Plan A for 60 minutes is ₹400 (½ mark)
- Cost under Plan B for 60 minutes is ₹300 (½ mark)
- Concludes Plan B is better, with the ₹100 saving as the reason (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Playing in a ball pool is good entertainment for kids. Suhana bought 600 new balls of diameter 7 cm to fill in the pool for her kids. The cuboidal box containing 600 balls has dimensions 42 cm × 91 cm × 50 cm ( $l \times b \times h$ ). Based on above information, answer the following questions.
1. (i)
  Find the volume of one ball.
  1 mark
2. (ii)
  10 balls are painted with neon colours. Determine the area of painted surface.
  1 mark
3. (iii)(a)
  Find the volume of empty space in the box.
  2 marks
4. OR
  (iii)(b)
  The lowermost layer of the balls covers the base of the box edge to edge when balls are placed evenly adjacent to each other. (A) How much area is covered by one ball? (B) How many balls are there in lowermost layer?
  2 marks
Figure in the paper: Photograph of a cuboidal box filled with colourful balls, with a label on its front. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Radius $r = \frac{7}{2}$ cm. Volume of one ball $= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} = \frac{539}{3} \approx 179.67 cm^{3}$ .
  1 mark
2. (ii) Surface area of 10 balls $= 10 \times 4 π r^{2} = 10 \times 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 10 \times 154 = 1540 cm^{2}$ .
  1 mark
3. (iii)(a) Volume of box $= l \times b \times h = 42 \times 91 \times 50 = 191100 cm^{3}$ .
  ½ mark
4. (iii)(a) Volume of 600 balls $= 600 \times \frac{539}{3} = 107800 cm^{3}$ .
  ½ mark
5. (iii)(a) Volume of empty space $= 191100 - 107800 = 83300 cm^{3}$ .
  1 mark
6. (iii)(b)(A) Balls placed edge to edge each take up a square of side equal to the diameter, 7 cm. Area covered by one ball $= 7 \times 7 = 49 cm^{2}$ .
  1 mark
7. (iii)(b)(B) Balls along the length $= \frac{42}{7} = 6$ and along the breadth $= \frac{91}{7} = 13$ . Number of balls in the lowermost layer $= 6 \times 13 = 78$ . (Check: $\frac{42 \times 91}{49} = 78$ .)
  1 mark
Answer: (i) $\frac{539}{3} \approx 179.67$ cm³; (ii) 1540 cm²; (iii)(a) 83300 cm³; (iii)(b) (A) 49 cm², (B) 78 balls
Where the marks are
- Uses $\frac{4}{3} π r^{3}$ with $r = 3.5$ cm to get $\frac{539}{3}$ cm³ (1 mark)
- Uses $10 \times 4 π r^{2}$ to get 1540 cm² (1 mark)
- Volume of box is 191100 cm³ (½ mark)
- Volume of 600 balls is 107800 cm³ (½ mark)
- Empty space is 83300 cm³ (1 mark)
- Area covered by one ball is $7 \times 7 = 49$ cm² (1 mark)
- Number of balls in the bottom layer is 78 (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Rahim and Nadeem are two friends whose plots are adjacent to each other. Rahim's son made a drawing of the plots with necessary details. It is decided that Rahim will fence the triangular plot ABC and Nadeem will fence along the sides AF, FE and BE. Observe the diagram carefully and answer the following questions. (Use $2 = 1.41$ and $3 = 1.73$ )
1. (i)
  Find length BC.
  1 mark
2. (ii)
  Find length AG.
  1 mark
3. (iii)(a)
  Calculate perimeter of $Δ$ ABC.
  2 marks
4. OR
  (iii)(b)
  Calculate length of (AF + FE + EB).
  2 marks
Figure in the paper: Plan of plots. Points A, G, D, H, B lie on one horizontal line (A left, B right). C is above D, with angle C = 60° and angle at A = 45°, angle at B = 30° (and 45° marked near B on the upper side). F is below G, with AF = 53.58 m, FE = 53 m, and E lies between F and H/B. Lengths marked: BH = 40 m, HD = 10 m, DG = 40 m, GF = 38 m, FE = 53 m, AF = 53.58 m. Vertical segments CD and HE are drawn; GF is vertical with a right-angle mark at G; DH is 10 m, HE is vertical. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $D B = D H + H B = 10 + 40 = 50$ m. In right $△ C D B$ , $∠ C B D = 4 5^{\circ}$ , so $cos 4 5^{\circ} = \frac{D B}{B C} \Rightarrow \frac{1}{2} = \frac{50}{B C} \Rightarrow B C = 502 = 50 \times 1.41 = 70.5$ m.
  1 mark
2. (ii) In right $△ A GF$ , $∠ F A G = 4 5^{\circ}$ , so $tan 4 5^{\circ} = \frac{GF}{A G} \Rightarrow 1 = \frac{38}{A G} \Rightarrow A G = 38$ m.
  1 mark
3. (iii)(a) $A D = A G + G D = 38 + 40 = 78$ m. In right $△ A D C$ , $∠ A C D = 6 0^{\circ}$ , so $sin 6 0^{\circ} = \frac{A D}{A C} \Rightarrow \frac{3}{2} = \frac{78}{A C} \Rightarrow A C = \frac{156}{3} = 523 = 52 \times 1.73 = 89.96$ m.
  1 mark
4. (iii)(a) $A B = A G + G D + D H + H B = 38 + 40 + 10 + 40 = 128$ m. Perimeter of $△ A B C = A B + B C + C A = 128 + 70.5 + 89.96 = 288.46$ m.
  1 mark
5. (iii)(b) In right $△ E H B$ , $∠ H B E = 3 0^{\circ}$ and $B H = 40$ m, so $cos 3 0^{\circ} = \frac{B H}{B E} \Rightarrow \frac{3}{2} = \frac{40}{B E} \Rightarrow B E = \frac{80}{3} = \frac{80 3}{3} = \frac{80 \times 1.73}{3} = 46.13$ m.
  1 mark
6. (iii)(b) $A F + F E + E B = 53.58 + 53 + 46.13 = 152.71$ m.
  1 mark
Answer: (i) BC = 70.5 m; (ii) AG = 38 m; (iii)(a) 288.46 m; (iii)(b) 152.71 m
Where the marks are
- Uses $cos 4 5^{\circ} = \frac{50}{B C}$ to get BC = 70.5 m (1 mark)
- Uses $tan 4 5^{\circ} = \frac{38}{A G}$ to get AG = 38 m (1 mark)
- Uses $sin 6 0^{\circ} = \frac{78}{A C}$ to get AC = 89.96 m (1 mark)
- Adds the sides to get perimeter 288.46 m (1 mark)
- Uses $cos 3 0^{\circ} = \frac{40}{B E}$ to get BE = 46.13 m (1 mark)
- Adds to get AF + FE + EB = 152.71 m (1 mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.