MathsCBSE Class 10 · Standard · Sample paper, session 2024-25

CBSE Class 10 Mathematics Standard: sample paper 2024-25

The sample question paper CBSE published for the 2024-25 session, 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 sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassX_2024_25/MathsStandard-SQP.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 based; subparts of 1, 1 and 2 marks

Instructions in brief: All questions compulsory; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D, and in all 2-mark subparts of Section E. Calculators not allowed; take π = 22/7 where required if not stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPolynomials
The graph of a quadratic polynomial $p (x)$ passes through the points $(- 6, 0)$ , $(0, - 30)$ , $(4, - 20)$ and $(6, 0)$ . The zeroes of the polynomial are
1. (A) $- 6, 0$
2. (B) $4, 6$
3. (C) $- 30, - 20$
4. (D) $- 6, 6$
Try it first, then show our solutionSolution
1. The zeroes of $p (x)$ are the $x$ -coordinates of the points where its graph meets the $x$ -axis, that is, where $y = 0$ .
2. Among the given points, only $(- 6, 0)$ and $(6, 0)$ have $y = 0$ . So the zeroes are $- 6$ and $6$ , which is option (D).
  1 mark
Answer: $- 6, 6$ , option (D)
Where the marks are
- Picks the $x$ -coordinates of the points with $y = 0$ , giving zeroes $- 6, 6$ (1 mark)
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Q2 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of $k$ for which the system of equations $3 x - k y = 7$ and $6 x + 10 y = 3$ is inconsistent, is
1. (A) $- 10$
2. (B) $- 5$
3. (C) $5$
4. (D) $7$
Try it first, then show our solutionSolution
1. A pair of linear equations is inconsistent when $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
2. Here $\frac{3}{6} = \frac{- k}{10}$ , so $- k = 5$ and $k = - 5$ .
  1 mark
3. Check: $\frac{c _{1}}{c _{2}} = \frac{- 7}{- 3} = \frac{7}{3} \neq = \frac{1}{2}$ , so the condition holds. The answer is option (B).
Answer: $k = - 5$ , option (B)
Where the marks are
- Uses $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ and obtains $k = - 5$ (1 mark)
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Q3 · 1 mark · MCQCircles
Which of the following statements is not true?
1. (A)A number of secants can be drawn at any point on the circle.
2. (B)Only one tangent can be drawn at any point on a circle.
3. (C)A chord is a line segment joining two points on the circle
4. (D)From a point inside a circle only two tangents can be drawn.
Try it first, then show our solutionSolution
1. Statements (A), (B) and (C) are standard facts about circles, so they are true.
2. From a point inside a circle, no tangent can be drawn, because every line through an interior point cuts the circle at two points. So statement (D) is not true. The answer is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Identifies that no tangent can be drawn from a point inside a circle, so (D) is false (1 mark)
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Q4 · 1 mark · MCQArithmetic Progressions
If $n$ th term of an A.P. is $7 n - 4$ then the common difference of the A.P. is
1. (A) $7$
2. (B) $7 n$
3. (C) $- 4$
4. (D) $4$
Try it first, then show our solutionSolution
1. $a_{1} = 7 (1) - 4 = 3$ and $a_{2} = 7 (2) - 4 = 10$ .
2. $d = a_{2} - a_{1} = 10 - 3 = 7$ . In general $a_{n + 1} - a_{n} = 7$ . The answer is option (A).
  1 mark
Answer: $d = 7$ , option (A)
Where the marks are
- Finds $d = a_{2} - a_{1} = 7$ (1 mark)
Q5 · 1 mark · MCQSurface Areas and Volumes
The radius of the base of a right circular cone and the radius of a sphere are each $5$ cm in length. If the volume of the cone is equal to the volume of the sphere then the height of the cone is
1. (A) $5$ cm
2. (B) $20$ cm
3. (C) $10$ cm
4. (D) $4$ cm
Try it first, then show our solutionSolution
1. Volume of cone = volume of sphere gives $\frac{1}{3} π r^{2} h = \frac{4}{3} π r^{3}$ .
2. So $h = 4 r = 4 \times 5 = 20$ cm. The answer is option (B).
  1 mark
Answer: $20$ cm, option (B)
Where the marks are
- Equates the two volumes and gets $h = 4 r = 20$ cm (1 mark)
Q6 · 1 mark · MCQIntroduction to Trigonometry
If $tan θ = \frac{5}{2}$ then $\frac{4 s i n θ + c o s θ}{4 s i n θ - c o s θ}$ is equal to
1. (A) $\frac{11}{9}$
2. (B) $\frac{3}{2}$
3. (C) $\frac{9}{11}$
4. (D) $4$
Try it first, then show our solutionSolution
1. Divide the numerator and the denominator by $cos θ$ : $\frac{4 t a n θ + 1}{4 t a n θ - 1}$ .
2. $= \frac{4 \cdot \frac{5}{2} + 1}{4 \cdot \frac{5}{2} - 1} = \frac{11}{9}$ . The answer is option (A).
  1 mark
Answer: $\frac{11}{9}$ , option (A)
Where the marks are
- Converts the expression to $tan θ$ and evaluates it as $\frac{11}{9}$ (1 mark)
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Q7 · 1 mark · MCQCircles
In the given figure, a tangent has been drawn at a point P on the circle centred at O. If $∠ T P Q = 11 0^{\circ}$ then $∠ P O Q$ is equal to
1. (A) $11 0^{\circ}$
2. (B) $7 0^{\circ}$
3. (C) $14 0^{\circ}$
4. (D) $5 5^{\circ}$
Figure in the paper: A circle with centre O. A point P lies on the circle; a tangent line passes through P, with T on the tangent on the left of P and an arrow on the right. Q is another point on the circle, joined to P by a chord. O is joined to P and to Q by radii. The angle TPQ is marked between the tangent (toward T) and chord PQ. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius]. So $∠ O P Q = 11 0^{\circ} - 9 0^{\circ} = 2 0^{\circ}$ .
2. $O P = O Q$ [radii], so $∠ O QP = 2 0^{\circ}$ . Then $∠ P O Q = 18 0^{\circ} - 2 0^{\circ} - 2 0^{\circ} = 14 0^{\circ}$ . The answer is option (C).
  1 mark
Answer: $14 0^{\circ}$ , option (C)
Where the marks are
- Uses tangent ⟂ radius and the isosceles triangle $O P Q$ to get $∠ P O Q = 14 0^{\circ}$ (1 mark)
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Q8 · 1 mark · MCQPolynomials
A quadratic polynomial having zeroes $- \frac{5}{2}$ and $\frac{5}{2}$ is
1. (A) $x^{2} - 52 x + 1$
2. (B) $8 x^{2} - 20$
3. (C) $15 x^{2} - 6$
4. (D) $x^{2} - 25 x - 1$
Try it first, then show our solutionSolution
1. Sum of zeroes $= 0$ and product of zeroes $= - \frac{5}{2}$ .
2. So the polynomial is $k (x^{2} - \frac{5}{2})$ . Taking $k = 8$ gives $8 x^{2} - 20$ . The answer is option (B).
  1 mark
Answer: $8 x^{2} - 20$ , option (B)
Where the marks are
- Forms $x^{2} - \frac{5}{2}$ from the sum and product of zeroes and matches it with $8 x^{2} - 20$ (1 mark)
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Q9 · 1 mark · MCQStatistics
Consider the frequency distribution of 45 observations.
Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 9 15 10 6
The upper limit of median class is
1. (A) $20$
2. (B) $10$
3. (C) $30$
4. (D) $40$
Try it first, then show our solutionSolution
1. $\frac{n}{2} = \frac{45}{2} = 22.5$ . The cumulative frequencies are $5, 14, 29, 39, 45$ .
2. The first cumulative frequency greater than $22.5$ is $29$ , so the median class is $20$ – $30$ . Its upper limit is $30$ . The answer is option (C).
  1 mark
Answer: $30$ , option (C)
Where the marks are
- Uses cumulative frequency with $\frac{n}{2} = 22.5$ to get median class $20$ – $30$ , upper limit $30$ (1 mark)
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Q10 · 1 mark · MCQCircles
$O$ is the point of intersection of two chords $A B$ and $C D$ of a circle. If $∠ B O C = 8 0^{\circ}$ and $O A = O D$ then $△ O D A$ and $△ O B C$ are
1. (A)equilateral and similar
2. (B)isosceles and similar
3. (C)isosceles but not similar
4. (D)not similar
Figure in the paper: A circle with two chords AB and CD intersecting at the point O inside the circle. Points on the circle: A (bottom left), B (top right), C (bottom right), D (top left). Segment OA and OD are marked equal (tick marks). The angle BOC at O is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. $O A = O D$ , so $△ O D A$ is isosceles and $∠ O A D = ∠ O D A$ .
2. $∠ A O D = ∠ B O C = 8 0^{\circ}$ [vertically opposite]. Also $∠ O A D = ∠ O C B$ and $∠ O D A = ∠ O B C$ [angles in the same segment]. So $△ O D A \sim △ O C B$ (AA).
3. Hence $∠ O B C = ∠ O C B$ , so $△ O B C$ is also isosceles. The base angles are $5 0^{\circ}$ each, so neither triangle is equilateral. Both are isosceles and similar: option (B).
  1 mark
Answer: Isosceles and similar, option (B)
Where the marks are
- Shows both triangles are isosceles and similar (AA) using vertically opposite angles and angles in the same segment (1 mark)
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Q11 · 1 mark · MCQQuadratic Equations
The roots of the quadratic equation $x^{2} + x - 1 = 0$ are
1. (A)Irrational and distinct
2. (B)not real
3. (C)rational and distinct
4. (D)real and equal
Try it first, then show our solutionSolution
1. $D = b^{2} - 4 a c = 1^{2} - 4 (1) (- 1) = 5$ .
2. $D > 0$ and $5$ is not a perfect square, so the roots are real, distinct and irrational. The answer is option (A).
  1 mark
Answer: Irrational and distinct, option (A)
Where the marks are
- Finds $D = 5 > 0$ , not a perfect square, so the roots are irrational and distinct (1 mark)
Q12 · 1 mark · MCQIntroduction to Trigonometry
If $θ = 3 0^{\circ}$ then the value of $3 tan θ$ is
1. (A) $1$
2. (B) $\frac{1}{3}$
3. (C) $\frac{3}{3}$
4. (D)not defined
Try it first, then show our solutionSolution
1. $tan 3 0^{\circ} = \frac{1}{3}$ .
2. $3 tan 3 0^{\circ} = \frac{3}{3}$ (which equals $3$ ). The answer is option (C).
  1 mark
Answer: $\frac{3}{3}$ , option (C)
Where the marks are
- Uses $tan 3 0^{\circ} = \frac{1}{3}$ to get $\frac{3}{3}$ (1 mark)
Q13 · 1 mark · MCQSurface Areas and Volumes
The volume of a solid hemisphere is $\frac{396}{7} c m^{3}$ . The total surface area of the solid hemisphere (in sq.cm) is
1. (A) $\frac{396}{7}$
2. (B) $\frac{594}{7}$
3. (C) $\frac{549}{7}$
4. (D) $\frac{604}{7}$
Try it first, then show our solutionSolution
1. $\frac{2}{3} \cdot \frac{22}{7} r^{3} = \frac{396}{7}$ gives $r^{3} = \frac{396 \times 3}{44} = 27$ , so $r = 3$ cm.
2. TSA $= 3 π r^{2} = 3 \times \frac{22}{7} \times 9 = \frac{594}{7}$ cm². The answer is option (B).
  1 mark
Answer: $\frac{594}{7}$ sq cm, option (B)
Where the marks are
- Finds $r = 3$ cm and TSA $= 3 π r^{2} = \frac{594}{7}$ cm² (1 mark)
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Q14 · 1 mark · MCQProbability
In a bag containing 24 balls, 4 are blue, 11 are green and the rest are white. One ball is drawn at random. The probability that drawn ball is white in colour is
1. (A) $\frac{1}{6}$
2. (B) $\frac{3}{8}$
3. (C) $\frac{11}{24}$
4. (D) $\frac{5}{8}$
Try it first, then show our solutionSolution
1. Number of white balls $= 24 - 4 - 11 = 9$ .
2. $P (white) = \frac{9}{24} = \frac{3}{8}$ . The answer is option (B).
  1 mark
Answer: $\frac{3}{8}$ , option (B)
Where the marks are
- Finds 9 white balls and the probability $\frac{3}{8}$ (1 mark)
Q15 · 1 mark · MCQCoordinate Geometry
The point on the $x$ -axis nearest to the point $(- 4, - 5)$ is
1. (A) $(0, 0)$
2. (B) $(- 4, 0)$
3. (C) $(- 5, 0)$
4. (D) $(41, 0)$
Try it first, then show our solutionSolution
1. The point on the $x$ -axis nearest to a given point is the foot of the perpendicular from that point to the $x$ -axis.
2. For $(- 4, - 5)$ the foot is $(- 4, 0)$ , at distance $5$ . Any other point $(x, 0)$ is at distance $(x + 4)^{2} + 25 > 5$ . The answer is option (B).
  1 mark
Answer: $(- 4, 0)$ , option (B)
Where the marks are
- Uses the perpendicular to the $x$ -axis to get $(- 4, 0)$ (1 mark)
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Q16 · 1 mark · MCQStatistics
Which of the following gives the middle most observation of the data?
1. (A)Median
2. (B)Mean
3. (C)Range
4. (D)Mode
Try it first, then show our solutionSolution
1. The median is the value that lies exactly in the middle when the observations are arranged in ascending or descending order. Half of the data lies below it and half lies above it.
  1 mark
2. The mean is the average, the mode is the most frequent value and the range is the spread (highest minus lowest). None of these is the middle-most observation, so the answer is option (A) Median.
Answer: Median, option (A)
Where the marks are
- Identifies the median as the middle-most observation of ordered data (1 mark)
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Q17 · 1 mark · MCQCoordinate Geometry
A point on the $x$ -axis divides the line segment joining the points $A (2, - 3)$ and $B (5, 6)$ in the ratio $1 : 2$ . The point is
1. (A) $(4, 0)$
2. (B) $(\frac{7}{2}, \frac{3}{2})$
3. (C) $(3, 0)$
4. (D) $(0, 3)$
Try it first, then show our solutionSolution
1. By the section formula with ratio $1 : 2$ : $x = \frac{1 ( 5 ) + 2 ( 2 )}{1 + 2} = 3$ and $y = \frac{1 ( 6 ) + 2 ( - 3 )}{3} = 0$ .
2. The point is $(3, 0)$ , which lies on the $x$ -axis. The answer is option (C).
  1 mark
Answer: $(3, 0)$ , option (C)
Where the marks are
- Applies the section formula to get $(3, 0)$ (1 mark)
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Q18 · 1 mark · MCQProbability
A card is drawn from a well shuffled deck of playing cards. The probability of getting red face card is
1. (A) $\frac{3}{13}$
2. (B) $\frac{1}{2}$
3. (C) $\frac{3}{52}$
4. (D) $\frac{3}{26}$
Try it first, then show our solutionSolution
1. Red face cards are J, Q, K of hearts and of diamonds, so there are $6$ of them out of $52$ cards.
2. $P = \frac{6}{52} = \frac{3}{26}$ . The answer is option (D).
  1 mark
Answer: $\frac{3}{26}$ , option (D)
Where the marks are
- Counts 6 red face cards and gets $\frac{3}{26}$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Direction: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion (A): HCF of any two consecutive even natural numbers is always 2.
Reason (R): Even natural numbers are divisible by 2.
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 and 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. Consecutive even numbers are $2 n$ and $2 n + 2$ . Their HCF is $2 \times HCF (n, n + 1) = 2 \times 1 = 2$ , so (A) is true.
2. (R) is true. However, divisibility by 2 only shows that 2 is a common factor; it does not show that there is no larger common factor. So (R) does not explain (A). The answer is option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Judges both A and R true, with R not the correct explanation of A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedAreas Related to Circles
Assertion (A): If the radius of sector of a circle is reduced to its half and angle is doubled then the perimeter of the sector remains the same.
Reason (R): The length of the arc subtending angle $θ$ at the centre of a circle of radius $r = \frac{π r θ}{180}$ .
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 and 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. Original perimeter of the sector $= 2 r + \frac{π r θ}{180}$ .
2. With radius $\frac{r}{2}$ and angle $2 θ$ : perimeter $= r + \frac{π ( r /2 ) ( 2 θ )}{180} = r + \frac{π r θ}{180}$ . This is less than before, so (A) is false.
3. (R) is the correct formula for arc length, so it is true. The answer is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Shows the new perimeter is $r + \frac{π r θ}{180}$ , so A is false while R is true (1 mark)
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Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	9	15	10	6

Section B

very short answer (VSA)

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerReal Numbers
Find the H.C.F and L.C.M of $480$ and $720$ using the Prime factorisation method.
Try it first, then show our solutionSolution
1. $480 = 2^{5} \times 3 \times 5$
  ½ mark
2. $720 = 2^{4} \times 3^{2} \times 5$
  ½ mark
3. LCM = product of the greatest powers of the prime factors $= 2^{5} \times 3^{2} \times 5 = 1440$
  ½ mark
4. HCF = product of the smallest powers of the common prime factors $= 2^{4} \times 3 \times 5 = 240$
  ½ mark
Answer: HCF = 240, LCM = 1440
Where the marks are
- Prime factorisation of 480 (½ mark)
- Prime factorisation of 720 (½ mark)
- LCM = 1440 (½ mark)
- HCF = 240 (½ mark)
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OR
Q21(B) · 2 marks · Very short answerReal Numbers
The H.C.F of $85$ and $238$ is expressible in the form $85 m - 238$ . Find the value of $m$ .
Try it first, then show our solutionSolution
1. $85 = 5 \times 17$ and $238 = 2 \times 7 \times 17$ , so HCF $(85, 238) = 17$ .
  1 mark
2. $85 m - 238 = 17 \Rightarrow 85 m = 255 \Rightarrow m = 3$ .
  1 mark
Answer: $m = 3$
Where the marks are
- Finds HCF = 17 (1 mark)
- Solves $85 m - 238 = 17$ to get $m = 3$ (1 mark)
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Answer one of these 2 alternatives.
Q22(A) · 2 marks · Very short answerProbability
Two dice are rolled together bearing numbers $4, 6, 7, 9, 11, 12$ . Find the probability that the product of numbers obtained is an odd number.
Try it first, then show our solutionSolution
1. Total number of outcomes $= 6 \times 6 = 36$ .
  ½ mark
2. The product is odd only when both numbers are odd. The odd faces are $7, 9, 11$ . The favourable outcomes are $(7, 7), (7, 9), (7, 11), (9, 7), (9, 9), (9, 11), (11, 7), (11, 9), (11, 11)$ , which is $9$ outcomes.
  1 mark
3. $P (odd product) = \frac{9}{36} = \frac{1}{4}$
  ½ mark
Answer: $\frac{1}{4}$
Where the marks are
- Total outcomes = 36 (½ mark)
- Notes both numbers must be odd and lists the 9 favourable outcomes (1 mark)
- Probability $= \frac{1}{4}$ (½ mark)
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OR
Q22(B) · 2 marks · Very short answerProbability
How many positive three digit integers have the hundredths digit $8$ and unit's digit $5$ ? Find the probability of selecting one such number out of all three digit numbers.
Try it first, then show our solutionSolution
1. Three-digit numbers run from 100 to 999, so there are $900$ of them.
  ½ mark
2. Numbers with hundreds digit 8 and units digit 5 are $805, 815, 825, \dots, 895$ . Only the tens digit can vary ( $0$ to $9$ ), so there are $10$ such numbers.
  1 mark
3. $P = \frac{10}{900} = \frac{1}{90}$
  ½ mark
Answer: 10 such numbers; probability $= \frac{1}{90}$
Where the marks are
- Total three-digit numbers = 900 (½ mark)
- Identifies the 10 numbers 805 to 895 (1 mark)
- Probability $= \frac{1}{90}$ (½ mark)
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Q23 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $\frac{2 s i n ^{2} 6 0 ^{\circ} - t a n ^{2} 3 0 ^{\circ}}{s e c ^{2} 4 5 ^{\circ}}$
Try it first, then show our solutionSolution
1. Standard values: $sin 6 0^{\circ} = \frac{3}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ , $sec 4 5^{\circ} = 2$ .
2. Substituting: $\frac{2 ( \frac{3}{2} ) ^{2} - ( \frac{1}{3} ) ^{2}}{( 2 ) ^{2}} = \frac{2 \cdot \frac{3}{4} - \frac{1}{3}}{2} = \frac{\frac{3}{2} - \frac{1}{3}}{2}$
  1½ marks
3. $= \frac{\frac{9 - 2}{6}}{2} = \frac{7}{6} \times \frac{1}{2} = \frac{7}{12}$
  ½ mark
Answer: $\frac{7}{12}$
Where the marks are
- Substitutes the correct values of $sin 6 0^{\circ}$ , $tan 3 0^{\circ}$ and $sec 4 5^{\circ}$ into the expression (1½ marks)
- Simplifies to $\frac{7}{12}$ (½ mark)
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Q24 · 2 marks · Very short answerCoordinate Geometry
Find the point(s) on the $x$ -axis which is at a distance of $41$ units from the point $(8, - 5)$ .
Try it first, then show our solutionSolution
1. Any point on the $x$ -axis has $y$ -coordinate 0, so let the required point be $P (x, 0)$ .
  ½ mark
2. By the distance formula, the distance from $(8, - 5)$ is $(8 - x)^{2} + (- 5 - 0)^{2} = 41$ , that is, $(8 - x)^{2} + 25 = 41$ .
  ½ mark
3. Squaring both sides: $(8 - x)^{2} + 25 = 41 \Rightarrow (8 - x)^{2} = 16 \Rightarrow 8 - x = \pm 4$ .
4. So $x = 8 - 4 = 4$ or $x = 8 + 4 = 12$ . The required points are $(4, 0)$ and $(12, 0)$ .
  1 mark
Answer: $(4, 0)$ and $(12, 0)$
Where the marks are
- Takes the point on the $x$ -axis as $(x, 0)$ (½ mark)
- Sets up the distance equation correctly (½ mark)
- Solves to get both points $(4, 0)$ and $(12, 0)$ (1 mark)
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Q25 · 2 marks · Very short answerCoordinate Geometry
Show that the points $A (- 5, 6)$ , $B (3, 0)$ and $C (9, 8)$ are the vertices of an isosceles triangle.
Try it first, then show our solutionSolution
1. $A B = (3 + 5)^{2} + (0 - 6)^{2} = 64 + 36 = 100 = 10$ units
  ½ mark
2. $B C = (9 - 3)^{2} + (8 - 0)^{2} = 36 + 64 = 100 = 10$ units
  ½ mark
3. $A C = (9 + 5)^{2} + (8 - 6)^{2} = 196 + 4 = 200 = 102$ units
  ½ mark
4. $A B + B C = 20 > 102 = A C$ , so the points are not collinear and form a triangle. Since $A B = B C = 10$ units, $△ A B C$ is isosceles.
  ½ mark
Answer: $A B = B C = 10$ , $A C = 102$ , so $△ A B C$ is isosceles.
Where the marks are
- Finds $A B = 10$ (½ mark)
- Finds $B C = 10$ (½ mark)
- Finds $A C = 102$ (½ mark)
- Concludes the triangle is isosceles because two sides are equal (½ mark)
Open Q25 on its own page

Section C

short answer (SA)

Answer one of these 2 alternatives.
Q26(A) · 3 marks · Short answerTriangles
In $△ A B C$ , $D$ , $E$ and $F$ are midpoints of $B C$ , $C A$ and $A B$ respectively. Prove that $△ F B D \sim △ D E F$ and $△ D E F \sim △ A B C$ .
Figure in the paper: Triangle ABC with A at the top, B bottom-left, C bottom-right. D is the midpoint of BC (bottom), E is the midpoint of CA (right side), F is the midpoint of AB (left side). Segments FD, DE and FE are drawn, forming the midpoint triangle DEF inside ABC. See it in the official paper.
Try it first, then show our solutionSolution
1. Given: $D, E, F$ are the midpoints of $B C, C A, A B$ of $△ A B C$ . To prove: $△ F B D \sim △ D E F$ and $△ D E F \sim △ A B C$ .
2. By the midpoint theorem (the segment joining the midpoints of two sides is parallel to the third side and half of it):
  $F E ∥ B C$ , $F E = \frac{1}{2} B C$ ; $D E ∥ A B$ , $D E = \frac{1}{2} A B$ ; $F D ∥ A C$ , $F D = \frac{1}{2} A C$ .
  Also $F B = \frac{1}{2} A B$ and $B D = \frac{1}{2} B C$ [ $F$ , $D$ are midpoints].
  1 mark
3. In $△ F B D$ and $△ D E F$ : $F B = \frac{1}{2} A B = D E$ , $B D = \frac{1}{2} B C = E F$ , $D F = F D$ (common).
  So $\frac{F B}{D E} = \frac{B D}{E F} = \frac{D F}{F D} = 1$ . Hence $△ F B D \sim △ D E F$ [SSS similarity]. (In fact the two triangles are congruent.)
  1 mark
4. In $△ D E F$ and $△ A B C$ : $\frac{D E}{A B} = \frac{1}{2}$ , $\frac{E F}{B C} = \frac{1}{2}$ , $\frac{F D}{C A} = \frac{1}{2}$ .
  Since all three ratios are equal, $△ D E F \sim △ A B C$ [SSS similarity]. Hence proved.
  1 mark
Answer: $△ F B D \sim △ D E F$ and $△ D E F \sim △ A B C$ (SSS similarity, using the midpoint theorem).
Where the marks are
- Uses the midpoint theorem to get the parallel sides and half-lengths (1 mark)
- Proves $△ F B D \sim △ D E F$ with a valid criterion (1 mark)
- Proves $△ D E F \sim △ A B C$ with a valid criterion (1 mark)
Open Q26(A) on its own page
OR
Q26(B) · 3 marks · Short answerTriangles
In $△ A B C$ , $P$ and $Q$ are points on $A B$ and $A C$ respectively such that $P Q$ is parallel to $B C$ . Prove that the median $A D$ drawn from $A$ on $B C$ bisects $P Q$ .
Figure in the paper: Triangle with apex A at top, B bottom-left, C bottom-right, D on BC (bottom). P lies on AB and Q lies on AC, with segment PQ drawn horizontally. R is the point where median AD meets PQ. The median AD is drawn from A to D. See it in the official paper.
Try it first, then show our solutionSolution
1. Given: $P Q ∥ B C$ , $A D$ is a median ( $B D = D C$ ), and $A D$ meets $P Q$ at $R$ . To prove: $P R = R Q$ .
2. In $△ A P R$ and $△ A B D$ : $∠ P A R = ∠ B A D$ (common), and $∠ A P R = ∠ A B D$ [corresponding angles, $P Q ∥ B C$ ].
  So $△ A P R \sim △ A B D$ [AA], which gives $\frac{A P}{A B} = \frac{P R}{B D} = \frac{A R}{A D}$ …(i)
  1 mark
3. Similarly, in $△ A QR$ and $△ A C D$ : $∠ Q A R = ∠ C A D$ (common), and $∠ A QR = ∠ A C D$ [corresponding angles].
  So $△ A QR \sim △ A C D$ [AA], which gives $\frac{A Q}{A C} = \frac{R Q}{D C} = \frac{A R}{A D}$ …(ii)
  1 mark
4. From (i) and (ii), both ratios equal $\frac{A R}{A D}$ , so $\frac{P R}{B D} = \frac{R Q}{D C}$ .
  Since $B D = D C$ [ $A D$ is a median], $P R = R Q$ . Hence the median $A D$ bisects $P Q$ .
  1 mark
Answer: $P R = R Q$ , so the median $A D$ bisects $P Q$ .
Where the marks are
- Proves $△ A P R \sim △ A B D$ and writes the ratio of sides (1 mark)
- Proves $△ A QR \sim △ A C D$ and writes the ratio of sides (1 mark)
- Equates the ratios and uses $B D = D C$ to conclude $P R = R Q$ (1 mark)
Open Q26(B) on its own page
Q27 · 3 marks · Short answerQuadratic Equations
The sum of two numbers is $18$ and the sum of their reciprocals is $\frac{9}{40}$ . Find the numbers.
Try it first, then show our solutionSolution
1. Let the numbers be $x$ and $18 - x$ .
  ½ mark
2. Given $\frac{1}{x} + \frac{1}{18 - x} = \frac{9}{40}$ , so $\frac{18 - x + x}{x ( 18 - x )} = \frac{9}{40}$ , that is, $\frac{18}{x ( 18 - x )} = \frac{9}{40}$ .
  1 mark
3. Cross-multiplying: $720 = 9 x (18 - x) \Rightarrow 80 = 18 x - x^{2} \Rightarrow x^{2} - 18 x + 80 = 0$ .
  Factorising: $(x - 10) (x - 8) = 0 \Rightarrow x = 10$ or $x = 8$ .
  1 mark
4. If $x = 10$ , the other number is $8$ ; if $x = 8$ , the other is $10$ . The numbers are $8$ and $10$ .
  Check: $\frac{1}{8} + \frac{1}{10} = \frac{5 + 4}{40} = \frac{9}{40}$ ✓
  ½ mark
Answer: 8 and 10
Where the marks are
- Takes the numbers as $x$ and $18 - x$ (½ mark)
- Forms the equation from the sum of reciprocals (1 mark)
- Reduces it to $x^{2} - 18 x + 80 = 0$ and solves (1 mark)
- States the numbers 8 and 10 (½ mark)
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Q28 · 3 marks · Short answerPolynomials
If $α$ and $β$ are zeroes of a polynomial $6 x^{2} - 5 x + 1$ then form a quadratic polynomial whose zeroes are $α^{2}$ and $β^{2}$ .
Try it first, then show our solutionSolution
1. For $6 x^{2} - 5 x + 1$ : $α + β = - \frac{b}{a} = \frac{5}{6}$ and $α β = \frac{c}{a} = \frac{1}{6}$ .
  1 mark
2. Sum of the new zeroes: $α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{25}{36} - \frac{2}{6} = \frac{25 - 12}{36} = \frac{13}{36}$ .
  1 mark
3. Product of the new zeroes: $α^{2} β^{2} = (α β)^{2} = (\frac{1}{6})^{2} = \frac{1}{36}$ .
  ½ mark
4. Required polynomial: $k [x^{2} - (sum) x + product] = k (x^{2} - \frac{13}{36} x + \frac{1}{36})$ . Taking $k = 36$ , the polynomial is $36 x^{2} - 13 x + 1$ .
  ½ mark
Answer: $36 x^{2} - 13 x + 1$ (or any non-zero multiple of it)
Where the marks are
- Writes $α + β = \frac{5}{6}$ and $α β = \frac{1}{6}$ (1 mark)
- Finds $α^{2} + β^{2} = \frac{13}{36}$ (1 mark)
- Finds $α^{2} β^{2} = \frac{1}{36}$ (½ mark)
- Forms the polynomial $36 x^{2} - 13 x + 1$ (½ mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerIntroduction to Trigonometry
If $cos θ + sin θ = 1$ , then prove that $cos θ - sin θ = \pm 1$ .
Try it first, then show our solutionSolution
1. Expanding and adding: $(cos θ + sin θ)^{2} + (cos θ - sin θ)^{2} = 2 (cos^{2} θ + sin^{2} θ) = 2$ [since $sin^{2} θ + cos^{2} θ = 1$ ].
  1½ marks
2. Substituting $cos θ + sin θ = 1$ : $1^{2} + (cos θ - sin θ)^{2} = 2 \Rightarrow (cos θ - sin θ)^{2} = 1$ .
  1 mark
3. Taking square roots: $cos θ - sin θ = \pm 1$ . Hence proved.
  ½ mark
Answer: $cos θ - sin θ = \pm 1$ (proved)
Where the marks are
- Uses $(cos θ + sin θ)^{2} + (cos θ - sin θ)^{2} = 2$ through the identity (1½ marks)
- Substitutes the given value to get $(cos θ - sin θ)^{2} = 1$ (1 mark)
- Concludes $cos θ - sin θ = \pm 1$ (½ mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(A) · 3 marks · Short answerCompetency-basedAreas Related to Circles
The minute hand of a wall clock is $18$ cm long. Find the area of the face of the clock described by the minute hand in $35$ minutes.
Try it first, then show our solutionSolution
1. The minute hand turns $36 0^{\circ}$ in 60 minutes, so it turns $6^{\circ}$ per minute. In 35 minutes, $θ = 35 \times 6^{\circ} = 21 0^{\circ}$ . The radius is $r = 18$ cm.
  1 mark
2. Area swept = area of the sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{210}{360} \times \frac{22}{7} \times 18 \times 18$
  1 mark
3. $= \frac{7}{12} \times \frac{22}{7} \times 324 = 22 \times 27 = 594 cm^{2}$
  1 mark
Answer: $594 cm^{2}$
Where the marks are
- Finds the angle swept in 35 minutes, $21 0^{\circ}$ (1 mark)
- Uses the sector area formula with $r = 18$ cm (1 mark)
- Gets the area $594 cm^{2}$ (1 mark)
Open Q30(A) on its own page
OR
Q30(B) · 3 marks · Short answerAreas Related to Circles
$A B$ is a chord of a circle centred at $O$ such that $∠ A O B = 6 0^{\circ}$ . If $O A = 14$ cm then find the area of the minor segment. (take $3 = 1.73$ )
Figure in the paper: A circle with centre O at the top. A and B are two points on the circle at the bottom, with chord AB joined. OA and OB are radii forming the angle AOB at the centre. See it in the official paper.
Try it first, then show our solutionSolution
1. Area of the minor segment = area of sector $O A B$ − area of $△ O A B$ .
2. Area of sector $O A B = \frac{60}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{6} \times 616 \approx 102.67 cm^{2}$
  1 mark
3. $O A = O B$ (radii) and $∠ A O B = 6 0^{\circ}$ , so $△ O A B$ is equilateral with side 14 cm.
  Area $= \frac{3}{4} \times 1 4^{2} = 493 = 49 \times 1.73 = 84.77 cm^{2}$
  1 mark
4. Area of the minor segment $= 102.67 - 84.77 \approx 17.90 cm^{2}$
  1 mark
Answer: $\approx 17.9 cm^{2}$ (17.90 cm²)
Where the marks are
- Finds the area of the sector, $\approx 102.67 cm^{2}$ (1 mark)
- Finds the area of equilateral $△ O A B$ , $84.77 cm^{2}$ (1 mark)
- Subtracts to get the segment area $\approx 17.9 cm^{2}$ (1 mark)
Open Q30(B) on its own page
Q31 · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $3$ is rational. Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are co-prime.
  ½ mark
2. Squaring: $3 q^{2} = p^{2}$ . So 3 divides $p^{2}$ , and since 3 is prime, 3 divides $p$ …(i) [if a prime divides $a^{2}$ , it divides $a$ ].
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $3 q^{2} = 9 a^{2} \Rightarrow q^{2} = 3 a^{2}$ . So 3 divides $q^{2}$ , and hence 3 divides $q$ …(ii)
  1 mark
4. From (i) and (ii), 3 is a common factor of $p$ and $q$ . This contradicts the fact that $p$ and $q$ are co-prime. So our assumption is wrong, and $3$ is irrational.
  ½ mark
Answer: $3$ is irrational (proved by contradiction).
Where the marks are
- Assumes $3 = \frac{p}{q}$ with $p, q$ co-prime and $q \neq = 0$ (½ mark)
- Shows that 3 divides $p$ (1 mark)
- Shows that 3 divides $q$ (1 mark)
- Reaches the contradiction and concludes (½ mark)
Open Q31 on its own page

Section D

long answer (LA)

Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerPair of Linear Equations in Two Variables
Solve the following system of linear equations graphically: $x + 2 y = 3$ , $2 x - 3 y + 8 = 0$
Try it first, then show our solutionSolution
1. Line 1: $x + 2 y = 3 \Rightarrow y = \frac{3 - x}{2}$ .
  Table: $x = 3, y = 0$ ; $x = 1, y = 1$ ; $x = - 1, y = 2$ .
  Plot $(3, 0)$ , $(1, 1)$ , $(- 1, 2)$ and join them to get the line for $x + 2 y = 3$ .
  2 marks
2. Line 2: $2 x - 3 y + 8 = 0 \Rightarrow y = \frac{2 x + 8}{3}$ .
  Table: $x = - 4, y = 0$ ; $x = - 1, y = 2$ ; $x = 2, y = 4$ .
  Plot $(- 4, 0)$ , $(- 1, 2)$ , $(2, 4)$ on the same axes and join them.
  2 marks
3. The two lines intersect at $(- 1, 2)$ .
  Check: $- 1 + 2 (2) = 3$ ✓ and $2 (- 1) - 3 (2) + 8 = 0$ ✓.
  So the solution is $x = - 1$ , $y = 2$ .
  1 mark
Answer: $x = - 1, y = 2$
Where the marks are
- Makes a correct table and draws the line for $x + 2 y = 3$ (2 marks)
- Makes a correct table and draws the line for $2 x - 3 y + 8 = 0$ (2 marks)
- Reads the intersection point, $x = - 1$ , $y = 2$ (1 mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables
Places A and B are $180$ km apart on a highway. One car starts from A and another from B at the same time. If the car travels in the same direction at different speeds, they meet in $9$ hours. If they travel towards each other with the same speeds as before, they meet in an hour. What are the speeds of the two cars?
Try it first, then show our solutionSolution
1. Let the speed of the car from A be $x$ km/h and the speed of the car from B be $y$ km/h, with $x > y$ .
2. Same direction: the car from A catches up with the car from B in 9 hours, so the difference in the distances covered equals AB:
  $9 x - 9 y = 180 \Rightarrow x - y = 20$ …(i)
  2 marks
3. Towards each other: they meet in 1 hour, so the sum of the distances covered equals AB:
  $x + y = 180$ …(ii)
  2 marks
4. Adding (i) and (ii): $2 x = 200 \Rightarrow x = 100$ . Then $y = 180 - 100 = 80$ .
  The speeds of the cars are 100 km/h (from A) and 80 km/h (from B).
  1 mark
Answer: 100 km/h and 80 km/h
Where the marks are
- Forms the equation $x - y = 20$ for the same-direction case (2 marks)
- Forms the equation $x + y = 180$ for the opposite-direction case (2 marks)
- Solves to get the speeds 100 km/h and 80 km/h (1 mark)
Open Q32(B) on its own page
Q33 · 5 marks · Long answerCircles
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Using above result, find the length $B C$ of $△ A B C$ . Given that, a circle is inscribed in $△ A B C$ touching the sides $A B$ , $B C$ and $C A$ at $R$ , $P$ and $Q$ respectively and $A B = 10$ cm, $A Q = 7$ cm, $C Q = 5$ cm.
Figure in the paper: Triangle ABC with A at the top, B bottom-left, C bottom-right. A circle is inscribed in the triangle, touching AB at R (left side), CA at Q (right side) and BC at P (bottom). See it in the official paper.
Try it first, then show our solutionSolution
1. Given: A circle with centre $O$ . $P Q$ and $P R$ are two tangents drawn from an external point $P$ , touching the circle at $Q$ and $R$ .
  To prove: $P Q = P R$ .
  Construction: Join $O P$ , $O Q$ and $O R$ .
  Figure: A circle with centre $O$ and a point $P$ outside it. Tangent segments $P Q$ and $P R$ are drawn, and the radii $O Q$ , $O R$ and the segment $O P$ are joined.
  1 mark
2. Proof: In $△ O QP$ and $△ O R P$ :
  $∠ O QP = ∠ O R P = 9 0^{\circ}$ [tangent $⊥$ radius at the point of contact]
  $O Q = O R$ [radii of the same circle]
  $O P = O P$ [common hypotenuse]
  1 mark
3. So $△ O QP ≅ △ O R P$ [RHS congruence rule].
  Hence $P Q = P R$ [CPCT].
  Therefore the lengths of the tangents drawn from an external point to a circle are equal.
  1 mark
4. Application: Tangents from $A$ are $A R$ and $A Q$ . So $A R = A Q = 7$ cm.
  ½ mark
5. Tangents from $B$ are $B R$ and $B P$ . So $B P = B R = A B - A R = 10 - 7 = 3$ cm.
  ½ mark
6. Tangents from $C$ are $C P$ and $C Q$ . So $C P = C Q = 5$ cm.
  ½ mark
7. $B C = B P + P C = 3 + 5 = 8$ cm.
  ½ mark
Answer: Theorem proved (tangents from an external point are equal); BC = 8 cm
Where the marks are
- Writes the given, to prove and construction, with a correct figure (1 mark)
- Proves $△ O QP ≅ △ O R P$ using the right angles, equal radii and common side (RHS), and concludes $P Q = P R$ (2 marks)
- States $A R = A Q = 7$ cm (½ mark)
- Finds $B R = B P = 3$ cm (½ mark)
- States $C P = C Q = 5$ cm (½ mark)
- Gets $B C = 8$ cm (½ mark)
Open Q33 on its own page
Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A boy whose eye level is $1.35$ m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is $6 0^{\circ}$ . After $12$ seconds, the angle of elevation reduces to $3 0^{\circ}$ . If the speed of the wind is $3$ m/s then find the height of the balloon from the ground. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: Let $A$ be the boy's eye, with $A D = 1.35$ m the eye height above the ground. $B$ and $C$ are the two positions of the balloon, both on a horizontal line at height $h$ above eye level. $G$ and $F$ are the feet of the perpendiculars from $B$ and $C$ to the horizontal line through $A$ . Then $∠ B A G = 6 0^{\circ}$ , $∠ C A F = 3 0^{\circ}$ , $A G = x$ and $B G = C F = h$ .
  1 mark
2. Distance moved by the balloon in 12 s $= 3 \times 12 = 36$ m.
  So $B C = GF = 36$ m and $A F = x + 36$ .
  1 mark
3. In right $△ A GB$ : $tan 6 0^{\circ} = \frac{h}{x}$ , so $3 = \frac{h}{x}$ and $h = 3 x$ ...(i)
  1 mark
4. In right $△ A F C$ : $tan 3 0^{\circ} = \frac{h}{x + 36}$ , so $\frac{1}{3} = \frac{h}{x + 36}$ and $h = \frac{x + 36}{3}$ ...(ii)
  1 mark
5. From (i) and (ii): $3 x = \frac{x + 36}{3}$ .
  This gives $3 x = x + 36$ , so $x = 18$ m.
  Then $h = 183 = 18 \times 1.73 = 31.14$ m.
  Height of the balloon from the ground $= 31.14 + 1.35 = 32.49$ m.
  1 mark
Answer: Height of balloon from ground = 18√3 + 1.35 ≈ 32.49 m
Where the marks are
- Draws a correct figure with the eye level, both balloon positions and both angles (1 mark)
- Finds the distance moved $= 36$ m and sets $B C = GF = 36$ m (1 mark)
- Forms $h = 3 x$ from $tan 6 0^{\circ}$ (1 mark)
- Forms $h = (x + 36) / 3$ from $tan 3 0^{\circ}$ (1 mark)
- Solves to get $h = 183 = 31.14$ m and adds 1.35 m to get 32.49 m (1 mark)
Open Q34 on its own page

Answer one of these 2 alternatives.

Q35(A) · 5 marks · Long answerStatistics

Find the mean and median of the following data:

Class	85-90	90-95	95-100	100-105	105-110	110-115
Frequency	15	22	20	18	20	25

Try it first, then show our solutionSolution

Take the assumed mean $a = 102.5$ and class size $h = 5$ , with $u_{i} = \frac{x _{i} - 102.5}{5}$ .
Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$ c.f.
85-90 87.5 15 -3 -45 15
90-95 92.5 22 -2 -44 37
95-100 97.5 20 -1 -20 57
100-105 102.5 18 0 0 75
105-110 107.5 20 1 20 95
110-115 112.5 25 2 50 120
Total 120 -39
2 marks
Mean $\overset{x}{ˉ} = a + h \cdot \frac{\sum f _{i} u _{i}}{\sum f _{i}}$
$= 102.5 + 5 \times \frac{- 39}{120}$
$= 102.5 - 1.625 = 100.875$
1 mark
$N = 120$ , so $\frac{N}{2} = 60$ .
½ mark
The first cumulative frequency greater than 60 is 75, so the median class is $100$ – $105$ .
Here $l = 100$ , $c f = 57$ , $f = 18$ , $h = 5$ .
½ mark
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h$
$= 100 + \frac{60 - 57}{18} \times 5$
$= 100 + 0.83 = 100.83$ (approx.)
1 mark

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$	c.f.
85-90	87.5	15	-3	-45	15
90-95	92.5	22	-2	-44	37
95-100	97.5	20	-1	-20	57
100-105	102.5	18	0	0	75
105-110	107.5	20	1	20	95
110-115	112.5	25	2	50	120
Total		120		-39

Answer: Mean = 100.875; Median ≈ 100.83

Where the marks are

Draws a correct table with class marks, $u_{i}$ , $f_{i} u_{i}$ and cumulative frequencies (2 marks)
Applies the step-deviation formula to get mean $= 100.875$ (1 mark)
Finds $N /2 = 60$ (½ mark)
Identifies the median class 100–105 (½ mark)
Applies the median formula to get $\approx 100.83$ (1 mark)

Open Q35(A) on its own page

Q35(B) · 5 marks · Long answerCompetency-basedStatistics

The monthly expenditure on milk in $200$ families of a Housing Society is given below.

Monthly Expenditure (in Rs.)	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of families	24	40	33	$x$	30	22	16	7

Find the value of $x$ and also find the mean expenditure.

Try it first, then show our solutionSolution

Expenditure (Rs.)	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
1000-1500	24	1250	30000
1500-2000	40	1750	70000
2000-2500	33	2250	74250
2500-3000	$x = 28$	2750	77000
3000-3500	30	3250	97500
3500-4000	22	3750	82500
4000-4500	16	4250	68000
4500-5000	7	4750	33250
Total	200		532500

2 marks

Total number of families $= 200$ :
$24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 200$
$172 + x = 200$ , so $x = 28$ .
1 mark
$\sum f_{i} x_{i} = 532500$ and $\sum f_{i} = 200$ .
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{532500}{200}$
1 mark
Mean $= 2662.5$
The mean monthly expenditure on milk is Rs. 2662.50.
1 mark

Answer: x = 28; mean expenditure = Rs. 2662.50

Where the marks are

Draws a correct table with class marks and $f_{i} x_{i}$ (2 marks)
Forms $172 + x = 200$ and gets $x = 28$ (1 mark)
Writes mean $= 532500/200$ (1 mark)
Gets mean $=$ Rs. 2662.5 (1 mark)

Open Q35(B) on its own page

Section E

case study based; subparts of 1, 1 and 2 marks

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Ms. Sheela visited a store near her house and found that the glass jars are arranged one above the other in a specific pattern. On the top layer there are 3 jars. In the next layer there are 6 jars. In the 3rd layer from the top there are 9 jars and so on till the 8th layer.
On the basis of the above situation answer the following questions.
1. (i)
  Write an A.P whose terms represent the number of jars in different layers starting from top. Also, find the common difference.
  1 mark
2. (ii)
  Is it possible to arrange 34 jars in a layer if this pattern is continued? Justify your answer.
  1 mark
3. (iii)(A)
  If there are ' $n$ ' number of rows in a layer then find the expression for finding the total number of jars in terms of $n$ . Hence find $S_{8}$ .
  2 marks
4. OR
  (iii)(B)
  The shopkeeper added 3 jars in each layer. How many jars are there in the 5th layer from the top?
  2 marks
Try it first, then show our solutionSolution
1. (i) The numbers of jars in the layers from the top form the A.P. $3, 6, 9, 12, \dots, 24$ (8 layers).
  ½ mark
2. (i) Common difference $d = 6 - 3 = 3$ .
  ½ mark
3. (ii) Suppose a layer has 34 jars. Using $a_{n} = a + (n - 1) d$ :
  $34 = 3 + (n - 1) \times 3$
  $3 n = 34$ , so $n = \frac{34}{3} = 11 \frac{1}{3}$ .
  ½ mark
4. (ii) $n$ is not a positive integer. So 34 is not a term of the A.P., and a layer of 34 jars is not possible in this pattern.
  ½ mark
5. (iii)(a) Take $n$ as the number of layers. Here $a = 3$ and $d = 3$ .
  $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [2 \times 3 + (n - 1) 3]$
  ½ mark
6. (iii)(a) Simplifying:
  $S_{n} = \frac{n}{2} [6 + 3 n - 3] = \frac{n}{2} (3 n + 3)$
  $S_{n} = \frac{3 n ( n + 1 )}{2}$
  1 mark
7. (iii)(a) $S_{8} = \frac{3 \times 8 \times 9}{2} = 108$ jars.
  ½ mark
8. (iii)(b) After adding 3 jars to each layer, the A.P. becomes $6, 9, 12, \dots$ with $a = 6$ and $d = 3$ .
  ½ mark
9. (iii)(b) $a_{5} = a + 4 d = 6 + 4 \times 3$
  1 mark
10. (iii)(b) $a_{5} = 18$ jars.
  ½ mark
Answer: (i) 3, 6, 9, …, 24; d = 3; (ii) Not possible, since n = 34/3 is not a natural number; (iii)(a) Sₙ = 3n(n+1)/2, S₈ = 108; (iii)(b) 18 jars
Where the marks are
- (i) Writes the A.P. 3, 6, 9, … (½ mark)
- (i) Gets $d = 3$ (½ mark)
- (ii) Forms $34 = 3 + (n - 1) 3$ and gets $n = 34/3$ (½ mark)
- (ii) Concludes it is not possible because $n$ is not a natural number (½ mark)
- (iii)(a) Writes the sum formula with $a = 3$ , $d = 3$ (½ mark)
- (iii)(a) Simplifies to $S_{n} = 3 n (n + 1) /2$ (1 mark)
- (iii)(a) Gets $S_{8} = 108$ (½ mark)
- (iii)(b) Identifies the new A.P. with $a = 6$ , $d = 3$ (½ mark)
- (iii)(b) Uses $a_{5} = 6 + 4 \times 3$ (1 mark)
- (iii)(b) Gets 18 jars (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedTriangles
Triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer. Here the largest triangle is represented by $△ A B C$ and smallest one with shelf is represented by $△ D E F$ . $P Q$ is parallel to $E F$ .
1. (i)
  Show that $△ D P Q \sim △ D E F$ .
  1 mark
2. (ii)
  If $D P = 50$ cm and $P E = 70$ cm then find $\frac{P Q}{E F}$ .
  1 mark
3. (iii)(A)
  If $2 A B = 5 D E$ and $△ A B C \sim △ D E F$ then show that $\frac{perimeter of △ A B C}{perimeter of △ D E F}$ is constant.
  2 marks
4. OR
  (iii)(B)
  If $A M$ and $D N$ are medians of triangles $A B C$ and $D E F$ respectively then prove that $△ A B M \sim △ D E N$ .
  2 marks
Figure in the paper: Top: a photograph of a triangular wooden shelf cabinet with shelves and decorative objects. Below: two triangles. Left triangle ABC is large with A at top, B bottom-left, C bottom-right. Right triangle DEF is small with D at top, E bottom-left, F bottom-right; P lies on DE and Q lies on DF, with segment PQ parallel to EF. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In $△ D P Q$ and $△ D E F$ :
  $∠ D P Q = ∠ D E F$ [corresponding angles, $P Q ∥ E F$ ]
  $∠ P D Q = ∠ E D F$ [common angle]
  So $△ D P Q \sim △ D E F$ [AA similarity].
  1 mark
2. (ii) $D E = D P + P E = 50 + 70 = 120$ cm.
  Since $△ D P Q \sim △ D E F$ , we have $\frac{D P}{D E} = \frac{P Q}{E F}$ .
  ½ mark
3. (ii) $\frac{P Q}{E F} = \frac{50}{120} = \frac{5}{12}$
  ½ mark
4. (iii)(a) $2 A B = 5 D E$ gives $\frac{A B}{D E} = \frac{5}{2}$ .
  As $△ A B C \sim △ D E F$ , corresponding sides are proportional:
  $\frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F} = \frac{5}{2}$
  So $A B = \frac{5}{2} D E$ , $B C = \frac{5}{2} E F$ and $A C = \frac{5}{2} D F$ .
  1 mark
5. (iii)(a) $\frac{perimeter of △ A B C}{perimeter of △ D E F} = \frac{\frac{5}{2} ( D E + E F + F D )}{D E + E F + F D} = \frac{5}{2}$
  This is a constant.
  1 mark
6. (iii)(b) Figure: $△ A B C$ with median $A M$ , where $M$ is the midpoint of $B C$ . $△ D E F$ with median $D N$ , where $N$ is the midpoint of $E F$ .
  ½ mark
7. (iii)(b) Since $△ A B C \sim △ D E F$ :
  $\frac{A B}{D E} = \frac{B C}{E F} = \frac{B C /2}{E F /2} = \frac{B M}{E N}$
  Also $∠ B = ∠ E$ [corresponding angles of similar triangles].
  1 mark
8. (iii)(b) So $△ A B M \sim △ D E N$ [SAS similarity].
  ½ mark
Answer: (i) △DPQ ~ △DEF by AA; (ii) PQ/EF = 5/12; (iii)(a) ratio of perimeters = 5/2 (constant); (iii)(b) △ABM ~ △DEN by SAS
Where the marks are
- (i) Gives two pairs of equal angles and concludes AA similarity (1 mark)
- (ii) Finds $D E = 120$ cm and writes $D P / D E = P Q / E F$ (½ mark)
- (ii) Gets $P Q / E F = 5/12$ (½ mark)
- (iii)(a) Writes the proportional sides with ratio $5/2$ (1 mark)
- (iii)(a) Shows the perimeter ratio $= 5/2$ , a constant (1 mark)
- (iii)(b) Draws a correct figure with the medians (½ mark)
- (iii)(b) Shows $A B / D E = B M / E N$ and $∠ B = ∠ E$ (1 mark)
- (iii)(b) Concludes SAS similarity (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Metallic silos are used by farmers for storing grains. Farmer Girdhar has decided to build a new metallic silo to store his harvested grains. It is in the shape of a cylinder mounted by a cone.
Dimensions of the conical part of a silo is as follows: Radius of base = $1.5$ m, Height = $2$ m.
Dimensions of the cylindrical part of a silo is as follows: Radius = $1.5$ m, Height = $7$ m.
On the basis of the above information answer the following questions.
1. (i)
  Calculate the slant height of the conical part of one silo.
  1 mark
2. (ii)
  Find the curved surface area of the conical part of one silo.
  1 mark
3. (iii)(A)
  Find the cost of metal sheet used to make the curved cylindrical part of 1 silo at the rate of ₹ $2000$ per $m^{2}$ .
  2 marks
4. OR
  (iii)(B)
  Find the total capacity of one silo to store grains.
  2 marks
Try it first, then show our solutionSolution
1. (i) Slant height $l = r^{2} + h^{2} = (1.5)^{2} + 2^{2} = 2.25 + 4$
  ½ mark
2. (i) $l = 6.25 = 2.5$ m
  ½ mark
3. (ii) CSA of cone $= π r l = \frac{22}{7} \times 1.5 \times 2.5$
  ½ mark
4. (ii) CSA of cone $= \frac{82.5}{7} \approx 11.79$ m $^{2}$
  ½ mark
5. (iii)(a) CSA of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66$ m $^{2}$
  1 mark
6. (iii)(a) Cost of metal sheet $= 66 \times 2000 =$ ₹ $1, 32, 000$
  1 mark
7. (iii)(b) Volume of cylinder $= π r^{2} h = \frac{22}{7} \times (1.5)^{2} \times 7 = 49.5$ m $^{3}$
  ½ mark
8. (iii)(b) Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (1.5)^{2} \times 2 \approx 4.71$ m $^{3}$
  1 mark
9. (iii)(b) Total capacity $= 49.5 + 4.71 = 54.21$ m $^{3}$ (approx.)
  ½ mark
Answer: (i) 2.5 m; (ii) ≈ 11.79 m²; (iii)(a) ₹1,32,000; (iii)(b) ≈ 54.21 m³
Where the marks are
- (i) Uses $l = r^{2} + h^{2}$ with the values substituted (½ mark)
- (i) Gets $l = 2.5$ m (½ mark)
- (ii) Writes $π r l$ with substitution (½ mark)
- (ii) Gets $\approx 11.79$ m $^{2}$ (½ mark)
- (iii)(a) Gets CSA of cylinder $= 66$ m $^{2}$ (1 mark)
- (iii)(a) Gets cost ₹1,32,000 (1 mark)
- (iii)(b) Gets volume of cylinder $= 49.5$ m $^{3}$ (½ mark)
- (iii)(b) Gets volume of cone $\approx 4.71$ m $^{3}$ (1 mark)
- (iii)(b) Adds to get total $\approx 54.21$ m $^{3}$ (½ mark)
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Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.