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

CBSE Class 10 Mathematics Basic: 2025 board paper, set 430/6/1

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.

37 of 38 questions (counting each OR alternative) have our step-by-step solution so far. We show a solution only after an independent re-solve agrees with it; the rest say "Solution being checked".

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 430-6-1_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
C	26-31	Short answer
D	32-35	Long answer
E	36-38	Case-based integrated units

Instructions in brief: All 38 questions are compulsory. Internal choice (OR) is given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 two-mark subparts in Section E. There is no overall choice. Use of calculators is not allowed. Take π = 22/7 unless stated otherwise.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPolynomials
In the given figure, graph of polynomial $p (x)$ is shown. Number of zeroes of $p (x)$ is
1. (A)3
2. (B)2
3. (C)1
4. (D)4
Figure in the paper: Graph of a cubic-shaped polynomial p(x) on x-x' and y-y' axes with origin O. The curve starts at the origin, rises to a maximum above the x-axis, crosses the x-axis, dips to a minimum below the x-axis, then rises again crossing the x-axis to the right of the minimum. See it in the official paper.
Try it first, then show our solutionSolution
1. The number of zeroes equals the number of points where the graph meets the $x$ -axis.
2. The curve meets the $x$ -axis at O, again after the maximum, and again after the minimum. That is 3 points, so $p (x)$ has 3 zeroes.
  1 mark
Answer: 3, option (A)
Where the marks are
- Counts the 3 points where the graph meets the x-axis (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQArithmetic Progressions
22nd term of the A.P.: $\frac{3}{2}, \frac{1}{2}, \frac{- 1}{2}, \frac{- 3}{2}, \dots$ is
1. (A) $\frac{45}{2}$
2. (B) $- 9$
3. (C) $\frac{- 39}{2}$
4. (D) $- 21$
Try it first, then show our solutionSolution
1. Here $a = \frac{3}{2}$ and $d = \frac{1}{2} - \frac{3}{2} = - 1$ .
2. $a_{22} = a + 21 d = \frac{3}{2} - 21 = - \frac{39}{2}$
  1 mark
Answer: $- \frac{39}{2}$ , option (C)
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ to get $- 39/2$ (1 mark)
Q3 · 1 mark · MCQCoordinate Geometry
The line $2 x - 3 y = 6$ intersects $x$ -axis at
1. (A) $(0, - 2)$
2. (B) $(0, 3)$
3. (C) $(- 2, 0)$
4. (D) $(3, 0)$
Try it first, then show our solutionSolution
1. On the $x$ -axis $y = 0$ , so $2 x = 6$ , giving $x = 3$ . The point is $(3, 0)$ .
  1 mark
Answer: (3, 0), option (D)
Where the marks are
- Puts $y = 0$ and gets $(3, 0)$ (1 mark)
Q4 · 1 mark · MCQSurface Areas and Volumes
Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm, then height of the solid is
1. (A)8 cm
2. (B) $45$ cm
3. (C) $25$ cm
4. (D)12 cm
Figure in the paper: Two identical cones placed base to base-free, apex to apex: one cone with circular base at top and apex at centre, another cone with circular base at bottom and apex at the same central point, forming an hourglass-like solid. See it in the official paper.
Try it first, then show our solutionSolution
1. Height of one cone: $h = l^{2} - r^{2} = 36 - 16 = 20 = 25$ cm.
2. The solid is made of two such cones, so its height is $2 \times 25 = 45$ cm.
  1 mark
Answer: $45$ cm, option (B)
Where the marks are
- Finds the height of one cone and doubles it to get $45$ cm (1 mark)
Q5 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of k for which the system of equations $3 x - 7 y = 1$ and $k x + 14 y = 6$ is inconsistent, is
1. (A) $- 6$
2. (B) $\frac{2}{3}$
3. (C) $6$
4. (D) $\frac{- 3}{2}$
Try it first, then show our solutionSolution
1. For an inconsistent system, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
2. $\frac{3}{k} = \frac{- 7}{14} = - \frac{1}{2}$ gives $k = - 6$ . Also $\frac{c _{1}}{c _{2}} = \frac{1}{6} \neq = - \frac{1}{2}$ , so the system is indeed inconsistent.
  1 mark
Answer: k = −6, option (A)
Where the marks are
- Uses the inconsistency condition to get $k = - 6$ (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQProbability
Two dice are rolled together. The probability of getting a sum more than 9 is
1. (A) $\frac{5}{6}$
2. (B) $\frac{5}{18}$
3. (C) $\frac{1}{6}$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. There are 36 outcomes. A sum of 10 occurs in 3 ways, 11 in 2 ways and 12 in 1 way, giving 6 favourable outcomes.
2. $P = \frac{6}{36} = \frac{1}{6}$
  1 mark
Answer: $\frac{1}{6}$ , option (C)
Where the marks are
- Counts 6 favourable outcomes out of 36 to get 1/6 (1 mark)
Q7 · 1 mark · MCQCoordinate Geometry
ABCD is a rectangle with its vertices at $(2, - 2)$ , $(8, 4)$ , $(4, 8)$ and $(- 2, 2)$ taken in order. Length of its diagonal is
1. (A) $42$
2. (B) $62$
3. (C) $426$
4. (D) $226$
Try it first, then show our solutionSolution
1. Diagonal AC joins $(2, - 2)$ and $(4, 8)$ .
2. $A C = (4 - 2)^{2} + (8 + 2)^{2} = 4 + 100 = 104 = 226$
  1 mark
Answer: $226$ , option (D)
Where the marks are
- Applies the distance formula to AC to get $226$ (1 mark)
Q8 · 1 mark · MCQCircles
In the given figure, PA is tangent to a circle with centre O. If $∠ A P O = 3 0^{\circ}$ and $O A = 2.5$ cm, then OP is equal to
1. (A)2.5 cm
2. (B)5 cm
3. (C) $\frac{5}{3}$ cm
4. (D)2 cm
Figure in the paper: Circle with centre O. Point A on the circle, tangent line through A extends to external point P. Segment AO is a radius, segment OP joins centre to P. Angle APO is marked as 30 degrees at P. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius].
2. $sin 3 0^{\circ} = \frac{O A}{O P}$ , so $\frac{1}{2} = \frac{2.5}{O P}$ , giving $O P = 5$ cm.
  1 mark
Answer: 5 cm, option (B)
Where the marks are
- Uses the right angle at A and $sin 3 0^{\circ}$ to get OP = 5 cm (1 mark)
Q9 · 1 mark · MCQProbability
If probability of happening of an event is 57%, then probability of non-happening of the event is
1. (A)0.43
2. (B)0.57
3. (C)53%
4. (D) $\frac{1}{57}$
Try it first, then show our solutionSolution
1. $P (\overset{ˉ}{E}) = 1 - P (E) = 1 - 0.57 = 0.43$
  1 mark
Answer: 0.43, option (A)
Where the marks are
- Uses $P (\overset{ˉ}{E}) = 1 - P (E)$ (1 mark)
Q10 · 1 mark · MCQAreas Related to Circles
OAB is sector of a circle with centre O and radius 7 cm. If length of arc $A B = \frac{22}{3}$ cm, then $∠$ AOB is equal to
1. (A) $(\frac{120}{7})^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $3 0^{\circ}$
Try it first, then show our solutionSolution
1. Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r$ , so $\frac{θ}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{22}{3}$ .
2. This simplifies to $\frac{44 θ}{360} = \frac{22}{3}$ , giving $θ = 6 0^{\circ}$ .
  1 mark
Answer: 60°, option (C)
Where the marks are
- Applies the arc length formula to get 60° (1 mark)
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Q11 · 1 mark · MCQTriangles
In $△$ ABC, DE $∥$ BC. If AE = $(2 x + 1)$ cm, EC = 4 cm, AD = $(x + 1)$ cm and DB = 3 cm, then value of $x$ is
1. (A)1
2. (B) $\frac{1}{2}$
3. (C) $- 1$
4. (D) $\frac{1}{3}$
Figure in the paper: Triangle ABC with vertex A at top. D lies on AB and E lies on AC, with DE parallel to BC (arrows marked). Labels: AD = x+1, DB = 3 cm, AE = 2x+1, EC = 4 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ , so $\frac{x + 1}{3} = \frac{2 x + 1}{4}$ .
2. Cross-multiplying, $4 x + 4 = 6 x + 3$ , giving $x = \frac{1}{2}$ .
  1 mark
Answer: $x = \frac{1}{2}$ , option (B)
Where the marks are
- Uses the Basic Proportionality Theorem to get $x = 1/2$ (1 mark)
Q12 · 1 mark · MCQProbability
Three coins are tossed together. The probability that exactly one coin shows head, is
1. (A) $\frac{1}{8}$
2. (B) $\frac{1}{4}$
3. (C)1
4. (D) $\frac{3}{8}$
Try it first, then show our solutionSolution
1. There are 8 outcomes. Exactly one head occurs as HTT, THT or TTH, which is 3 outcomes.
2. $P = \frac{3}{8}$
  1 mark
Answer: $\frac{3}{8}$ , option (D)
Where the marks are
- Counts 3 favourable outcomes out of 8 (1 mark)
Q13 · 1 mark · MCQCircles
In two concentric circles centred at O, a chord AB of the larger circle touches the smaller circle at C. If OA = 3.5 cm, OC = 2.1 cm, then AB is equal to
1. (A)5.6 cm
2. (B)2.8 cm
3. (C)3.5 cm
4. (D)4.2 cm
Figure in the paper: Two concentric circles centred at O. Chord AB of the larger circle is horizontal and touches the smaller circle at C; OC is perpendicular to AB. Dashed line joins A to O. See it in the official paper.
Try it first, then show our solutionSolution
1. $O C ⊥ A B$ [tangent ⟂ radius], and the perpendicular from the centre bisects the chord AB.
2. $A C = 3. 5^{2} - 2. 1^{2} = 12.25 - 4.41 = 7.84 = 2.8$ cm, so $A B = 2 \times 2.8 = 5.6$ cm.
  1 mark
Answer: 5.6 cm, option (A)
Where the marks are
- Uses Pythagoras to get AC = 2.8 cm and doubles it to get AB = 5.6 cm (1 mark)
Open Q13 on its own page
Q14 · 1 mark · MCQIntroduction to Trigonometry
If $3 sin θ = cos θ$ , then value of $θ$ is
1. (A) $3$
2. (B) $6 0^{\circ}$
3. (C) $\frac{1}{3}$
4. (D) $3 0^{\circ}$
Try it first, then show our solutionSolution
1. Dividing both sides by $cos θ$ gives $tan θ = \frac{1}{3}$ , so $θ = 3 0^{\circ}$ .
  1 mark
Answer: 30°, option (D)
Where the marks are
- Obtains $tan θ = 1/ 3$ and hence $θ = 3 0^{\circ}$ (1 mark)
Q15 · 1 mark · MCQStatistics
To calculate mean of a grouped data, Rahul used assumed mean method. He used $d = (x - A)$ , where A is assumed mean. Then $\overset{x}{ˉ}$ is equal to
1. (A) $A + \overset{ˉ}{d}$
2. (B) $A + h \overset{ˉ}{d}$
3. (C) $h (A + \overset{ˉ}{d})$
4. (D) $A - h \overset{ˉ}{d}$
Try it first, then show our solutionSolution
1. With $d_{i} = x_{i} - A$ , the mean is $\overset{x}{ˉ} = A + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = A + \overset{ˉ}{d}$ . No division by $h$ was made, so $h$ does not appear.
  1 mark
Answer: $A + \overset{ˉ}{d}$ , option (A)
Where the marks are
- States $\overset{x}{ˉ} = A + \overset{ˉ}{d}$ (1 mark)
Q16 · 1 mark · MCQArithmetic Progressions
If the sum of first $n$ terms of an A.P. is given by $S_{n} = \frac{n}{2} (3 n + 1)$ , then the first term of the A.P. is
1. (A)2
2. (B) $\frac{3}{2}$
3. (C)4
4. (D) $\frac{5}{2}$
Try it first, then show our solutionSolution
1. $a_{1} = S_{1} = \frac{1}{2} (3 + 1) = 2$
  1 mark
Answer: 2, option (A)
Where the marks are
- Uses $a = S_{1}$ to get 2 (1 mark)
Q17 · 1 mark · MCQIntroduction to Trigonometry
In $△$ ABC, $∠$ B = $9 0^{\circ}$ . If $\frac{A B}{A C} = \frac{1}{2}$ , then $cos C$ is equal to
1. (A) $\frac{3}{2}$
2. (B) $\frac{1}{2}$
3. (C) $\frac{3}{2}$
4. (D) $\frac{1}{3}$
Try it first, then show our solutionSolution
1. Since $∠ B = 9 0^{\circ}$ , $sin C = \frac{A B}{A C} = \frac{1}{2}$ , so $C = 3 0^{\circ}$ .
2. $cos C = cos 3 0^{\circ} = \frac{3}{2}$
  1 mark
Answer: $\frac{3}{2}$ , option (C)
Where the marks are
- Gets $sin C = 1/2$ and hence $cos C = 3 /2$ (1 mark)
Q18 · 1 mark · MCQSurface Areas and Volumes
The volume of air in a hollow cylinder is $450 c m^{3}$ . A cone of same height and radius as that of cylinder is kept inside it. The volume of empty space in the cylinder is
1. (A) $225 c m^{3}$
2. (B) $150 c m^{3}$
3. (C) $250 c m^{3}$
4. (D) $300 c m^{3}$
Figure in the paper: A hollow cylinder with a cone inside it. The cone has its base on the cylinder's base and its apex at the centre of the top circle, with the same radius and height as the cylinder. See it in the official paper.
Try it first, then show our solutionSolution
1. Volume of cone $= \frac{1}{3} \times$ volume of cylinder $= \frac{1}{3} \times 450 = 150$ cm³.
2. Empty space $= 450 - 150 = 300$ cm³.
  1 mark
Answer: 300 cm³, option (D)
Where the marks are
- Uses cone = ⅓ of cylinder to get 300 cm³ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Directions: In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion (A): $(a + b) \cdot (a - b)$ is a rational number, where $a$ and $b$ are positive integers.
Reason (R): Product of two irrationals is always rational.
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. $(a + b) (a - b) = a^{2} - b$ , which is an integer and hence rational. So A is true.
2. R is false: for example, $2 \times 3 = 6$ is irrational. Hence option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows A is true and gives a counterexample to R (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedTriangles
Directions: In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion (A): $△$ ABC $\sim$ $△$ PQR such that $∠$ A = $6 5^{\circ}$ , $∠$ C = $6 0^{\circ}$ . Hence $∠$ Q = $5 5^{\circ}$ .
Reason (R): Sum of all angles of a triangle is $18 0^{\circ}$ .
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.
Solution being checked. Our step-by-step solution is written and is waiting for an independent check before we show it.

Section B

Very short answer

Q21(a) · 2 marks · Very short answerQuadratic Equations
Answer the following.
1. (a)
  Solve the equation $4 x^{2} - 9 x + 3 = 0$ , using quadratic formula.
  0 marks
2. (b)
  Find the nature of roots of the equation $3 x^{2} - 43 x + 4 = 0$ .
  0 marks
Try it first, then show our solutionSolution
1. Compare $4 x^{2} - 9 x + 3 = 0$ with $a x^{2} + b x + c = 0$ : $a = 4, b = - 9, c = 3$ .
  Discriminant $D = b^{2} - 4 a c = (- 9)^{2} - 4 (4) (3) = 81 - 48 = 33$ .
  1 mark
2. By the quadratic formula, $x = \frac{- b \pm D}{2 a} = \frac{9 \pm 33}{8}$ .
  So the roots are $\frac{9 + 33}{8}$ and $\frac{9 - 33}{8}$ .
  1 mark
Answer: $x = \frac{9 + 33}{8}$ or $x = \frac{9 - 33}{8}$
Where the marks are
- Finds the discriminant $D = 33$ (1 mark)
- Applies the quadratic formula to get $x = \frac{9 \pm 33}{8}$ (1 mark)
Open Q21(a) on its own page
Q22 · 2 marks · Very short answerTriangles
In a trapezium ABCD, AB $∥$ DC and its diagonals intersect at O. Prove that $\frac{O A}{O C} = \frac{O B}{O D}$ .
Try it first, then show our solutionSolution
1. Draw trapezium ABCD with AB ∥ DC and diagonals AC and BD meeting at O.
  In $△ O A B$ and $△ O C D$ :
  $∠ O A B = ∠ O C D$ [alternate angles, AB ∥ DC, AC is a transversal]
  $∠ O B A = ∠ O D C$ [alternate angles, AB ∥ DC, BD is a transversal]
  Therefore $△ O A B \sim △ O C D$ [AA similarity criterion].
  1½ marks
2. Corresponding sides of similar triangles are proportional, so $\frac{O A}{O C} = \frac{O B}{O D}$ . Hence proved.
  ½ mark
Answer: $△ O A B \sim △ O C D$ (AA), hence $\frac{O A}{O C} = \frac{O B}{O D}$
Where the marks are
- Proves $△ O A B \sim △ O C D$ using alternate angles and the AA criterion (1½ marks)
- Uses proportionality of corresponding sides to conclude $\frac{O A}{O C} = \frac{O B}{O D}$ (½ mark)
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Q23 · 2 marks · Very short answerProbability
A box contains 120 discs, which are numbered from 1 to 120. If one disc is drawn at random from the box, find the probability that
1. (i)
  it bears a 2-digit number.
  1 mark
2. (ii)
  the number is a perfect square.
  1 mark
Try it first, then show our solutionSolution
1. (i) Total outcomes = 120. Two-digit numbers are 10, 11, …, 99, so their count is $99 - 10 + 1 = 90$ .
  $P (2-digit number) = \frac{90}{120} = \frac{3}{4}$ .
  1 mark
2. (ii) Perfect squares from 1 to 120 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, i.e. 10 numbers ( $1 1^{2} = 121 > 120$ ).
  $P (perfect square) = \frac{10}{120} = \frac{1}{12}$ .
  1 mark
Answer: (i) $\frac{3}{4}$ ; (ii) $\frac{1}{12}$
Where the marks are
- Counts 90 two-digit numbers and gets probability $\frac{90}{120} = \frac{3}{4}$ (1 mark)
- Counts 10 perfect squares and gets probability $\frac{10}{120} = \frac{1}{12}$ (1 mark)
Open Q23 on its own page
Q24(a) · 2 marks · Very short answerIntroduction to Trigonometry
Answer the following.
1. (a)
  Evaluate: $\frac{cos 4 5 ^{\circ}}{tan 3 0 ^{\circ} + sin 6 0 ^{\circ}}$
  0 marks
2. (b)
  Verify that $sin 2 A = \frac{2 tan A}{1 + tan ^{2} A}$ , for $A = 3 0^{\circ}$ .
  0 marks
Try it first, then show our solutionSolution
1. Substitute $cos 4 5^{\circ} = \frac{1}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ , $sin 6 0^{\circ} = \frac{3}{2}$ :
  $\frac{\frac{1}{2}}{\frac{1}{3} + \frac{3}{2}} = \frac{\frac{1}{2}}{\frac{2 + 3}{2 3}} = \frac{1}{2} \times \frac{2 3}{5}$
  1½ marks
2. $= \frac{2 3}{5 2} = \frac{6}{5}$
  ½ mark
Answer: $\frac{2 3}{5 2} = \frac{6}{5}$
Where the marks are
- Substitutes correct values of $cos 4 5^{\circ}, tan 3 0^{\circ}, sin 6 0^{\circ}$ and simplifies the denominator (1½ marks)
- Final value $\frac{6}{5}$ (½ mark)
Open Q24(a) on its own page
Q25 · 2 marks · Very short answerReal Numbers
Using prime factorisation, find the HCF of 180, 140 and 210.
Try it first, then show our solutionSolution
1. Prime factorisations:
  $180 = 2^{2} \times 3^{2} \times 5$
  $140 = 2^{2} \times 5 \times 7$
  $210 = 2 \times 3 \times 5 \times 7$
  1½ marks
2. HCF = product of the smallest powers of the common prime factors $= 2^{1} \times 5^{1} = 10$ .
  ½ mark
Answer: HCF = 10
Where the marks are
- Correct prime factorisation of 180, 140 and 210 (1½ marks)
- HCF $= 2 \times 5 = 10$ (½ mark)
Open Q25 on its own page

Section C

Short answer

Q26(a) · 3 marks · Short answerPolynomials
Answer the following.
1. (a)
  If $α, β$ are zeroes of the polynomial $8 x^{2} - 5 x - 1$ , then form a quadratic polynomial in $x$ whose zeroes are $\frac{2}{α}$ and $\frac{2}{β}$ .
  0 marks
2. (b)
  Find the zeroes of the polynomial $p (x) = 3 x^{2} + x - 10$ and verify the relationship between zeroes and its coefficients.
  0 marks
Try it first, then show our solutionSolution
1. For $8 x^{2} - 5 x - 1$ : $α + β = - \frac{- 5}{8} = \frac{5}{8}$ and $α β = \frac{- 1}{8}$ .
  1 mark
2. Sum of new zeroes: $\frac{2}{α} + \frac{2}{β} = \frac{2 ( α + β )}{α β} = \frac{2 \times \frac{5}{8}}{- \frac{1}{8}} = - 10$ .
  ½ mark
3. Product of new zeroes: $\frac{2}{α} \times \frac{2}{β} = \frac{4}{α β} = \frac{4}{- \frac{1}{8}} = - 32$ .
  ½ mark
4. Required polynomial $= x^{2} - (sum) x + product = x^{2} + 10 x - 32$ , or $k (x^{2} + 10 x - 32)$ for any non-zero real $k$ .
  1 mark
Answer: $x^{2} + 10 x - 32$ (or $k (x^{2} + 10 x - 32)$ , $k \neq = 0$ )
Where the marks are
- Writes $α + β = \frac{5}{8}$ and $α β = - \frac{1}{8}$ (½ + ½ mark)
- Sum of new zeroes $= - 10$ (½ mark)
- Product of new zeroes $= - 32$ (½ mark)
- Forms polynomial $x^{2} + 10 x - 32$ (1 mark)
Open Q26(a) on its own page
Q27 · 3 marks · Short answerCompetency-basedQuadratic Equations
Find length and breadth of a rectangular park whose perimeter is 100 m and area is $600 m^{2}$ .
Try it first, then show our solutionSolution
1. Let length = $a$ m and breadth = $b$ m.
  Perimeter: $2 (a + b) = 100 \Rightarrow a + b = 50$ …(i)
  ½ mark
2. Area: $ab = 600$ …(ii)
  ½ mark
3. From (i), $b = 50 - a$ . Substituting in (ii): $a (50 - a) = 600 \Rightarrow a^{2} - 50 a + 600 = 0$ .
  1 mark
4. $(a - 30) (a - 20) = 0 \Rightarrow a = 30$ or $a = 20$ .
  ½ mark
5. Then $b = 20$ or $b = 30$ . So the length is 30 m and the breadth is 20 m.
  ½ mark
Answer: Length = 30 m, breadth = 20 m
Where the marks are
- Forms $2 (a + b) = 100$ (½ mark)
- Forms $ab = 600$ (½ mark)
- Obtains quadratic $a^{2} - 50 a + 600 = 0$ (1 mark)
- Solves to get $a = 30$ or $20$ (½ mark)
- Gets corresponding breadth and states dimensions 30 m × 20 m (½ mark)
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Q28 · 3 marks · Short answerCompetency-basedReal Numbers
Three measuring rods are of lengths 120 cm, 100 cm and 150 cm. Find the least length of a fence that can be measured an exact number of times, using any of the rods. How many times each rod will be used to measure the length of the fence?
Try it first, then show our solutionSolution
1. The required length must be a multiple of all three rod lengths, so we need their LCM.
  $120 = 2^{3} \times 3 \times 5$ , $100 = 2^{2} \times 5^{2}$ , $150 = 2 \times 3 \times 5^{2}$
  1½ marks
2. LCM $= 2^{3} \times 3 \times 5^{2} = 600$ . So the least length of the fence is 600 cm (= 6 m).
  1 mark
3. Number of times each rod is used: $600 \div 120 = 5$ , $600 \div 100 = 6$ , $600 \div 150 = 4$ .
  ½ mark
Answer: 600 cm; the rods are used 5, 6 and 4 times respectively
Where the marks are
- Correct prime factorisation of 120, 100, 150 (1½ marks)
- LCM = 600 cm as least length (1 mark)
- Rods used 5, 6 and 4 times (½ mark)
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Q29 · 3 marks · Short answerAreas Related to Circles
AB and CD are diameters of a circle with centre O and radius 7 cm. If $∠$ BOD = $3 0^{\circ}$ , then find the area and perimeter of the shaded region.
Figure in the paper: Circle with centre O and two diameters AB and CD. Points D and B are on the upper right, A and C on the lower left. The shaded regions are the two segments-like triangular regions between chords: one between D, B and the arc near the upper right, and one between A, C and the arc near the lower left. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ A O C = ∠ B O D = 3 0^{\circ}$ [vertically opposite angles], so the shaded region consists of two equal sectors of radius 7 cm and angle $3 0^{\circ}$ .
  Area $= 2 (\frac{θ}{36 0 ^{\circ}} \times π r^{2}) = 2 (\frac{30}{360} \times \frac{22}{7} \times 7 \times 7)$
  1 mark
2. $= 2 \times \frac{77}{6} = \frac{77}{3} cm^{2} \approx 25.67 cm^{2}$
  ½ mark
3. Perimeter of each sector $= 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r$ .
  Perimeter of shaded region $= 2 (14 + \frac{30}{360} \times 2 \times \frac{22}{7} \times 7) = 2 (14 + \frac{11}{3})$
  1 mark
4. $= 2 \times \frac{53}{3} = \frac{106}{3} cm \approx 35.33 cm$
  ½ mark
Answer: Area $= \frac{77}{3}$ cm² ≈ 25.67 cm²; Perimeter $= \frac{106}{3}$ cm ≈ 35.33 cm
Where the marks are
- Uses area of two sectors of angle 30° correctly (1 mark)
- Area $= \frac{77}{3}$ cm² (½ mark)
- Perimeter expression $2 (14 + arc length)$ (1 mark)
- Perimeter $= \frac{106}{3}$ cm (½ mark)
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Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = sec θ cosec θ + 1$ .
Try it first, then show our solutionSolution
1. LHS $= \frac{\frac{s i n θ}{c o s θ}}{1 - \frac{c o s θ}{s i n θ}} + \frac{\frac{c o s θ}{s i n θ}}{1 - \frac{s i n θ}{c o s θ}}$
  ½ mark
2. $= \frac{sin ^{2} θ}{cos θ ( sin θ - cos θ )} + \frac{cos ^{2} θ}{sin θ ( cos θ - sin θ )} = \frac{sin ^{2} θ}{cos θ ( sin θ - cos θ )} - \frac{cos ^{2} θ}{sin θ ( sin θ - cos θ )}$
  ½ mark
3. $= \frac{sin ^{3} θ - cos ^{3} θ}{( sin θ - cos θ ) sin θ cos θ} = \frac{( sin θ - cos θ ) ( sin ^{2} θ + sin θ cos θ + cos ^{2} θ )}{( sin θ - cos θ ) sin θ cos θ}$
  1 mark
4. $= \frac{1 + sin θ cos θ}{sin θ cos θ}$ [using $sin^{2} θ + cos^{2} θ = 1$ ]
  ½ mark
5. $= \frac{1}{sin θ cos θ} + 1 = sec θ cosec θ + 1 =$ RHS. Hence proved.
  ½ mark
Answer: LHS simplifies to $sec θ cosec θ + 1$ = RHS
Where the marks are
- Writes tan and cot in terms of sin and cos (½ mark)
- Simplifies each fraction (½ mark)
- Combines to $\frac{s i n ^{3} θ - c o s ^{3} θ}{( s i n θ - c o s θ ) s i n θ c o s θ}$ (1 mark)
- Uses $a^{3} - b^{3}$ and $sin^{2} θ + cos^{2} θ = 1$ to get $\frac{1 + s i n θ c o s θ}{s i n θ c o s θ}$ (½ mark)
- Concludes $= sec θ cosec θ + 1$ (½ mark)
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Q31(a) · 3 marks · Short answerArithmetic Progressions
Answer the following.
1. (a)
  Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.
  0 marks
2. (b)
  Find the sum of first 20 terms of an A.P. whose $n^{t h}$ term is given by $a_{n} = 5 + 2 n$ . Can 52 be a term of this A.P.?
  0 marks
Try it first, then show our solutionSolution
1. Let first term = $a$ , common difference = $d$ . Third term: $a + 2 d = 16$ …(i)
  ½ mark
2. $a_{7} = a_{5} + 12 \Rightarrow a + 6 d = a + 4 d + 12$ …(ii)
  ½ mark
3. From (ii), $2 d = 12 \Rightarrow d = 6$ ; from (i), $a = 16 - 12 = 4$ .
  ½ mark
4. The A.P. is 4, 10, 16, 22, …
  ½ mark
5. $S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{29} = \frac{29}{2} [8 + 28 \times 6]$
  ½ mark
6. $= \frac{29}{2} \times 176 = 29 \times 88 = 2552$
  ½ mark
Answer: A.P.: 4, 10, 16, …; $S_{29} = 2552$
Where the marks are
- Forms $a + 2 d = 16$ (½ mark)
- Forms $a + 6 d = a + 4 d + 12$ (½ mark)
- Solves to get $d = 6$ , $a = 4$ (½ mark)
- Writes the A.P. 4, 10, 16, … (½ mark)
- Correct substitution in $S_{29}$ formula (½ mark)
- $S_{29} = 2552$ (½ mark)
Open Q31(a) on its own page

Section D

Long answer

Q32(a) · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables
Answer the following.
1. (a)
  Solve the following pair of linear equations by graphical method: $2 x + y = 9$ and $x - 2 y = 2$ .
  0 marks
2. (b)
  Nidhi received simple interest of ₹ 1,200 when invested ₹ $x$ at 6% p.a. and ₹ $y$ at 5% p.a. for 1 year. Had she invested ₹ $x$ at 3% p.a. and ₹ $y$ at 8% p.a. for that year, she would have received simple interest of ₹ 1,260. Find the values of $x$ and $y$ .
  0 marks
Try it first, then show our solutionSolution
1. For $2 x + y = 9$ , i.e. $y = 9 - 2 x$ :
  $x$ 0 2 4
  $y$ 9 5 1
  Plot (0, 9), (2, 5), (4, 1) on graph paper and join them to draw the line $2 x + y = 9$ .
  2 marks
2. For $x - 2 y = 2$ , i.e. $y = \frac{x - 2}{2}$ :
  $x$ 0 2 4
  $y$ −1 0 1
  Plot (0, −1), (2, 0), (4, 1) and join them to draw the line $x - 2 y = 2$ on the same axes.
  2 marks
3. The two lines intersect at the point (4, 1). Hence the solution is $x = 4, y = 1$ .
  (Check: $2 (4) + 1 = 9$ and $4 - 2 (1) = 2$ .)
  1 mark
Answer: $x = 4, y = 1$
Where the marks are
- Correct table and graph of $2 x + y = 9$ (2 marks)
- Correct table and graph of $x - 2 y = 2$ (2 marks)
- Reads intersection point (4, 1) as solution (1 mark)
Open Q32(a) on its own page
Q33(a) · 5 marks · Long answerCircles
Answer the following.
1. (a)
  The given figure shows a circle with centre O and radius 4 cm circumscribed by $△$ ABC. BC touches the circle at D such that BD = 6 cm and DC = 10 cm. Find the length of AE.
  0 marks
2. (b)
  PA and PB are tangents drawn to a circle with centre O. If $∠$ AOB = $12 0^{\circ}$ and OA = 10 cm, then:
  0 marks
Try it first, then show our solutionSolution
1. Let $A E = x$ cm. Tangents drawn from an external point to a circle are equal, so $A F = A E = x$ , $C E = C D = 10$ cm, $B F = B D = 6$ cm.
  Hence $B C = 16$ , $C A = x + 10$ , $A B = x + 6$ .
  1 mark
2. Semi-perimeter $s = \frac{16 + ( x + 10 ) + ( x + 6 )}{2} = 16 + x$ .
  ½ mark
3. By Heron's formula, $ar (△ A B C) = s (s - a) (s - b) (s - c) = (16 + x) (x) (6) (10)$ …(i)
  1 mark
4. Also, joining O to A, B, C (OD ⟂ BC, OE ⟂ CA, OF ⟂ AB, radius 4 cm) [tangent ⟂ radius]:
  $ar (△ A B C) = ar (O B C) + ar (O C A) + ar (O A B) = \frac{1}{2} [16 \times 4 + (10 + x) \times 4 + (6 + x) \times 4] = 4 (16 + x)$ …(ii)
  1 mark
5. Equating (i) and (ii) and squaring: $60 x (16 + x) = 16 (16 + x)^{2}$ .
  Since $16 + x \neq = 0$ ( $x = - 16$ is rejected), $60 x = 16 (16 + x) \Rightarrow 44 x = 256 \Rightarrow x = \frac{64}{11}$ .
  So $A E = \frac{64}{11}$ cm $\approx 5.8$ cm.
  1½ marks
Answer: $A E = \frac{64}{11}$ cm ≈ 5.8 cm
Where the marks are
- Uses equal tangents to write AF = AE = x, CE = 10, BF = 6 (1 mark)
- Semi-perimeter $s = 16 + x$ (½ mark)
- Area by Heron's formula $60 x (16 + x)$ (1 mark)
- Area as sum of three triangles $= \frac{1}{2} \cdot 4 \cdot (32 + 2 x)$ (1 mark)
- Equates and solves to get $x = \frac{64}{11}$ cm (1½ marks)
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Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
The angles of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: Let CD be the multi-storeyed building of height $h$ m (D at top, C at foot) and AB the 9 m building (B top, A foot), with $A C = x$ m. Draw BE ⟂ CD, so $E C = 9$ m, $D E = (h - 9)$ m, $B E = A C = x$ . Angles of depression from D: to B is $3 0^{\circ}$ , to A is $6 0^{\circ}$ ; so $∠ D B E = 3 0^{\circ}$ and $∠ D A C = 6 0^{\circ}$ [alternate angles].
  1 mark
2. In right $△ D E B$ : $tan 3 0^{\circ} = \frac{D E}{B E} = \frac{h - 9}{x} \Rightarrow \frac{1}{3} = \frac{h - 9}{x} \Rightarrow x = (h - 9) 3$ …(i)
  1½ marks
3. In right $△ D C A$ : $tan 6 0^{\circ} = \frac{D C}{A C} = \frac{h}{x} \Rightarrow h = x 3$ …(ii)
  1½ marks
4. Substituting (ii) in (i): $x = (x 3 - 9) 3 = 3 x - 93 \Rightarrow 2 x = 93 \Rightarrow x = 4.5 3 = 4.5 \times 1.73 \approx 7.79$ m.
  Then $h = x 3 = 4.5 \times 3 = 13.5$ m.
  Height of the multi-storeyed building = 13.5 m; distance between the buildings ≈ 7.79 m.
  1 mark
Answer: Height = 13.5 m; distance between buildings ≈ 7.79 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 3 0^{\circ} = \frac{h - 9}{x}$ to get $x = (h - 9) 3$ (1½ marks)
- Uses $tan 6 0^{\circ} = \frac{h}{x}$ to get $h = x 3$ (1½ marks)
- Solves: $x \approx 7.79$ m and $h = 13.5$ m (½ + ½ mark)
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Q35 · 5 marks · Long answerStatistics
Find 'mean' and 'mode' of the following data:
Class 15–20 20–25 25–30 30–35 35–40 40–45
Frequency 12 10 15 11 7 5
Try it first, then show our solutionSolution
1. Take assumed mean $a = 27.5$ , class size $h = 5$ , $u_{i} = \frac{x _{i} - 27.5}{5}$ .
  Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  15–20 17.5 12 −2 −24
  20–25 22.5 10 −1 −10
  25–30 27.5 15 0 0
  30–35 32.5 11 1 11
  35–40 37.5 7 2 14
  40–45 42.5 5 3 15
  Total 60 6
  1½ marks
2. Mean $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 27.5 + 5 \times \frac{6}{60} = 27.5 + 0.5 = 28$ .
  1½ marks
3. Highest frequency is 15, so the modal class is 25–30: $l = 25$ , $f_{1} = 15$ , $f_{0} = 10$ , $f_{2} = 11$ , $h = 5$ .
  Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 25 + \frac{15 - 10}{30 - 10 - 11} \times 5 = 25 + \frac{25}{9}$
  1½ marks
4. $= 25 + 2.78 = 27.78$ (approx.)
  ½ mark
Answer: Mean = 28; Mode ≈ 27.78
Where the marks are
- Correct table with $x_{i}$ , $u_{i}$ , $f_{i} u_{i}$ , $\sum f_{i} = 60$ , $\sum f_{i} u_{i} = 6$ (1½ marks)
- Mean $= 27.5 + 5 \times \frac{6}{60} = 28$ (1½ marks)
- Identifies modal class 25–30 and substitutes in mode formula (1½ marks)
- Mode ≈ 27.78 (½ mark)
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$x$	0	2	4
$y$	9	5	1

$x$	0	2	4
$y$	−1	0	1

Class	15–20	20–25	25–30	30–35	35–40	40–45
Frequency	12	10	15	11	7	5

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
15–20	17.5	12	−2	−24
20–25	22.5	10	−1	−10
25–30	27.5	15	0	0
30–35	32.5	11	1	11
35–40	37.5	7	2	14
40–45	42.5	5	3	15
Total		60		6

Section E

Case-based integrated units

Q36 · 4 marks · Case-basedCompetency-basedTriangles
A triangular window of a building is shown above. Its diagram represents a $△$ ABC with $∠$ A = $9 0^{\circ}$ and AB = AC. Points P and R trisect AB and PQ $∥$ RS $∥$ AC.
Based on the above, answer the following questions:
1. (i)
  Show that $△$ BPQ $\sim$ $△$ BAC.
  1 mark
2. (ii)
  Prove that $P Q = \frac{1}{3} A C$ .
  1 mark
3. (iii)(a)
  If AB = 3 m, find length BQ and BS. Verify that $B Q = \frac{1}{2} B S$ .
  2 marks
4. OR
  (iii)(b)
  Prove that $B R^{2} + R S^{2} = \frac{4}{9} B C^{2}$ .
  2 marks
Figure in the paper: Right-angled isosceles triangle ABC with right angle at A (bottom left), AC vertical, AB horizontal. Points P and R lie on AB with A, R, P, B in order, and S and Q lie on hypotenuse CB. Segments RS and PQ are drawn vertically, parallel to AC, with small arrows marked. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In $△ B P Q$ and $△ B A C$ : $∠ B P Q = ∠ B A C = 9 0^{\circ}$ [corresponding angles, PQ ∥ AC], and $∠ B$ is common. Hence $△ B P Q \sim △ B A C$ [AA similarity criterion].
  1 mark
2. (ii) Since P and R trisect AB (order A, R, P, B), $B P = \frac{1}{3} A B$ . From $△ B P Q \sim △ B A C$ : $\frac{P Q}{A C} = \frac{B P}{B A} = \frac{1}{3}$
  ½ mark
3. $\Rightarrow P Q = \frac{1}{3} A C$ . Hence proved.
  ½ mark
4. (iii)(a) $A B = A C = 3$ m, so $B P = 1$ m. From $\frac{B P}{B A} = \frac{P Q}{A C}$ and $B A = A C$ , $P Q = B P = 1$ m. In right $△ B P Q$ ( $∠ P = 9 0^{\circ}$ ): $B Q = 1^{2} + 1^{2} = 2$ m.
  1 mark
5. (iii)(a) Similarly $△ B R S \sim △ B A C$ with $B R = 2$ m, so $R S = B R = 2$ m and $B S = 2^{2} + 2^{2} = 22$ m.
  ½ mark
6. (iii)(a) $\frac{1}{2} B S = \frac{1}{2} \times 22 = 2 = B Q$ . Hence verified.
  ½ mark
7. (iii)(b) $△ B R S \sim △ B A C$ [RS ∥ AC, AA], and $B R = \frac{2}{3} A B$ , so $R S = \frac{2}{3} A C$ . Hence $B R^{2} + R S^{2} = (\frac{2}{3} A B)^{2} + (\frac{2}{3} A C)^{2}$
  1 mark
8. (iii)(b) $= \frac{4}{9} (A B^{2} + A C^{2})$
  ½ mark
9. (iii)(b) $= \frac{4}{9} B C^{2}$ [Pythagoras theorem in $△ A B C$ , $∠ A = 9 0^{\circ}$ ]. Hence proved.
  ½ mark
Answer: (i) AA similarity; (ii) $P Q = \frac{1}{3} A C$ ; (iii)(a) $B Q = 2$ m, $B S = 22$ m, so $B Q = \frac{1}{2} B S$ ; (iii)(b) $B R^{2} + R S^{2} = \frac{4}{9} B C^{2}$
Where the marks are
- (i) Shows $△ B P Q \sim △ B A C$ by AA (1 mark)
- (ii) Writes $\frac{P Q}{A C} = \frac{B P}{B A} = \frac{1}{3}$ (½ mark) and concludes $P Q = \frac{1}{3} A C$ (½ mark)
- (iii)(a) Finds PQ = 1 m and $B Q = 2$ m (1 mark)
- (iii)(a) Finds $B S = 22$ m (½ mark) and verifies $B Q = \frac{1}{2} B S$ (½ mark)
- (iii)(b) Writes $B R = \frac{2}{3} A B$ , $R S = \frac{2}{3} A C$ (½ + ½ mark)
- (iii)(b) Gets $\frac{4}{9} (A B^{2} + A C^{2})$ (½ mark) and uses Pythagoras to get $\frac{4}{9} B C^{2}$ (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is 10 cm. Outer radius of the bowl is 10.5 cm.
Based on the above, answer the following questions:
1. (i)
  Find the dimensions of the cuboidal box.
  1 mark
2. (ii)
  Find the total outer surface area of the box.
  1 mark
3. (iii)(a)
  Find the difference between the capacity of the bowl and the volume of the box. (use $π = 3.14$ )
  2 marks
4. OR
  (iii)(b)
  The inner surface of the bowl and the thickness is to be painted. Find the area to be painted.
  2 marks
Figure in the paper: Left: an open cuboidal box with a hemispherical bowl inside. Right: a hemispherical bowl sitting inside a cuboid, with the cuboid's top face at the bowl's rim level, showing a dashed rim ellipse and the bowl's curved bottom touching the base of the cuboid. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The bowl just fits, so the box's length and breadth equal the outer diameter $= 2 \times 10.5 = 21$ cm and its height equals the outer radius $= 10.5$ cm. Dimensions: $21 cm \times 21 cm \times 10.5 cm$ .
  1 mark
2. (ii) TSA $= 2 (l b + bh + h l) = 2 (21 \times 21 + 21 \times 10.5 + 10.5 \times 21) = 2 (441 + 220.5 + 220.5)$
  ½ mark
3. (ii) $= 2 \times 882 = 1764 cm^{2}$
  ½ mark
4. (iii)(a) Capacity of bowl (inner radius 10 cm) $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 1 0^{3}$
  ½ mark
5. (iii)(a) $= \frac{6280}{3} \approx 2093.33 cm^{3}$
  ½ mark
6. (iii)(a) Volume of box $= 21 \times 21 \times 10.5 = 4630.5 cm^{3}$
  ½ mark
7. (iii)(a) Difference $= 4630.5 - 2093.33 = \frac{15223}{6} \approx 2537.17 cm^{3}$
  ½ mark
8. (iii)(b) Area to be painted = inner curved surface + ring at the rim $= 2 π r^{2} + π (R^{2} - r^{2}) = 2 \times \frac{22}{7} \times 1 0^{2} + \frac{22}{7} \times (10. 5^{2} - 1 0^{2})$
  1 mark
9. (iii)(b) $= \frac{4400}{7} + \frac{22}{7} \times 10.25 = \frac{4400}{7} + \frac{451}{14} = \frac{9251}{14} \approx 660.79 cm^{2}$
  1 mark
Answer: (i) 21 cm × 21 cm × 10.5 cm; (ii) 1764 cm²; (iii)(a) ≈ 2537.17 cm³; (iii)(b) ≈ 660.79 cm²
Where the marks are
- (i) Box dimensions 21 cm × 21 cm × 10.5 cm (1 mark)
- (ii) Correct TSA formula/substitution (½ mark) and 1764 cm² (½ mark)
- (iii)(a) Capacity of bowl $\frac{2}{3} \times 3.14 \times 1 0^{3}$ (½ mark) $= 2093.33$ cm³ (½ mark)
- (iii)(a) Volume of box 4630.5 cm³ (½ mark); difference ≈ 2537.17 cm³ (½ mark)
- (iii)(b) Expression $2 π (10)^{2} + π (10. 5^{2} - 1 0^{2})$ (1 mark); area ≈ 660.79 cm² (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Gurveer and Arushi built a robot that can paint a path as it moves on a graph paper. Some co-ordinates of points are marked on it. It starts from (0, 0), moves to the points listed in order (in straight lines) and ends at (0, 0).
Arushi entered the points P(8, 6), Q(12, 2) and S( $- 6$ , 6) in order. The path drawn by robot is shown in the figure.
Based on the above, answer the following questions:
1. (i)
  Determine the distance OP.
  1 mark
2. (ii)
  QS is represented by equation $2 x + 9 y = 42$ . Find the co-ordinates of the point where it intersects y-axis.
  1 mark
3. (iii)(a)
  Point R(4.8, y) divides the line segment OP in a certain ratio, find the ratio. Hence, find the value of y.
  2 marks
4. OR
  (iii)(b)
  Using distance formula, show that $\frac{P Q}{O S} = \frac{2}{3}$ .
  2 marks
Figure in the paper: Coordinate graph with x-axis and y-axis through origin O. Points S(-6, 6) in the second quadrant, P(8, 6) in the first quadrant, and Q(12, 2) in the first quadrant. The robot path goes O to S (arrow), S to Q (straight line, arrow pointing left toward S), Q to P (arrow), and P to O (through point R on the line, arrow), with R marked on segment OP. A small robot illustration is shown at left. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $O P = (8 - 0)^{2} + (6 - 0)^{2} = 64 + 36 = 100 = 10$ units.
  1 mark
2. (ii) On the y-axis $x = 0$ : $9 y = 42 \Rightarrow y = \frac{14}{3}$ . Point of intersection is $(0, \frac{14}{3})$ .
  1 mark
3. (iii)(a) Let R divide OP in the ratio $k : 1$ . By the section formula, $x$ -coordinate: $4.8 = \frac{8 k + 0}{k + 1} \Rightarrow 4.8 k + 4.8 = 8 k \Rightarrow 3.2 k = 4.8 \Rightarrow k = \frac{3}{2}$ . So $O R : R P = 3 : 2$ .
  1½ marks
4. (iii)(a) $y = \frac{3 \times 6 + 2 \times 0}{3 + 2} = \frac{18}{5} = 3.6$ .
  ½ mark
5. (iii)(b) $P Q = (12 - 8)^{2} + (2 - 6)^{2} = 16 + 16 = 32 = 42$ .
  ½ mark
6. (iii)(b) $O S = (- 6 - 0)^{2} + (6 - 0)^{2} = 36 + 36 = 72 = 62$ .
  ½ mark
7. (iii)(b) $\frac{P Q}{O S} = \frac{4 2}{6 2} = \frac{2}{3}$ . Hence shown.
  1 mark
Answer: (i) 10 units; (ii) $(0, \frac{14}{3})$ ; (iii)(a) 3 : 2, $y = \frac{18}{5}$ ; (iii)(b) $P Q = 42$ , $O S = 62$ , ratio $\frac{2}{3}$
Where the marks are
- (i) OP = 10 using distance formula (1 mark)
- (ii) Intersection with y-axis $(0, \frac{14}{3})$ (1 mark)
- (iii)(a) Section formula on x-coordinate giving ratio 3 : 2 (1½ marks)
- (iii)(a) $y = \frac{18}{5}$ (½ mark)
- (iii)(b) $P Q = 32$ (½ mark), $O S = 72$ (½ mark), ratio $\frac{2}{3}$ (1 mark)
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