MathsCBSE Class 10 · Standard · Sample paper, session 2023-24

CBSE Class 10 Mathematics Standard: sample paper 2023-24

The sample question paper CBSE published for the 2023-24 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_2023_24/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	Internal choice in 24 and 25
C	26-31	Internal choice in 28, 29
D	32-35	Internal choice in 32, 34
E	36-38	Case-based units with sub-parts of 1, 1 and 2 marks; internal choice in 2-mark sub-parts

Instructions in brief: All questions compulsory; internal choice in 2 questions of 5 marks, 2 of 3 marks, 2 of 2 marks, and in the 2-mark sub-parts of Section E.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If two positive integers $a$ and $b$ are written as $a = x^{3} y^{2}$ and $b = x y^{3}$ , where $x, y$ are prime numbers, then the result obtained by dividing the product of the positive integers by the LCM $(a, b)$ is
1. (A) $x y$
2. (B) $x y^{2}$
3. (C) $x^{3} y^{3}$
4. (D) $x^{2} y^{2}$
Try it first, then show our solutionSolution
1. Product: $ab = x^{3} y^{2} \cdot x y^{3} = x^{4} y^{5}$ .
2. LCM takes the highest power of each prime: $LCM (a, b) = x^{3} y^{3}$ .
3. $\frac{ab}{LCM} = \frac{x ^{4} y ^{5}}{x ^{3} y ^{3}} = x y^{2}$ . This is the HCF, as expected. Option (B).
  1 mark
Answer: $x y^{2}$ , option (B)
Where the marks are
- Uses product ÷ LCM (= HCF) to get $x y^{2}$ , option (B) (1 mark)
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Q2 · 1 mark · MCQPolynomials
The given linear polynomial $y = f (x)$ has
1. (A)2 zeros
2. (B)1 zero and the zero is '3'
3. (C)1 zero and the zero is '4'
4. (D)No zero
Figure in the paper: Graph of a straight line $y = f (x)$ on a coordinate grid (x from -4 to 5, y from -2 to 5). The line passes through the labelled points (0, 4) on the y-axis and (3, 0) on the x-axis, and is labelled $y = f (x)$ . See it in the official paper.
Try it first, then show our solutionSolution
1. A zero of $f (x)$ is the x-coordinate of the point where the graph meets the x-axis.
2. The line meets the x-axis only at $(3, 0)$ . So there is exactly 1 zero, and it is $3$ . The point $(0, 4)$ is the y-intercept, not a zero. Option (B).
  1 mark
Answer: 1 zero, which is 3, option (B)
Where the marks are
- Identifies the single x-intercept $(3, 0)$ as the zero, option (B) (1 mark)
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Q3 · 1 mark · Assertion-reasonCompetency-basedPair of Linear Equations in Two Variables
The lines representing the given pair of linear equations are non-intersecting. Which of the following statements is true?
1. (A) $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
2. (B) $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$
3. (C) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
4. (D) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$
Figure in the paper: Two parallel straight lines on a coordinate grid, labelled $a_{1} x + b_{1} y + c_{1} = 0$ (upper line) and $a_{2} x + b_{2} y + c_{2} = 0$ (lower line). See it in the official paper.
Try it first, then show our solutionSolution
1. Non-intersecting lines are parallel, so the pair has no solution (it is inconsistent).
2. The condition for parallel lines is $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ . Option (B).
  1 mark
Answer: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , option (B)
Where the marks are
- States the parallel-lines (no solution) condition, option (B) (1 mark)
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Q4 · 1 mark · MCQQuadratic Equations
The nature of roots of the quadratic equation $9 x^{2} - 6 x - 2 = 0$ is:
1. (A)No real roots
2. (B)2 equal real roots
3. (C)2 distinct real roots
4. (D)More than 2 real roots
Try it first, then show our solutionSolution
1. Here $a = 9$ , $b = - 6$ , $c = - 2$ .
2. Discriminant: $D = b^{2} - 4 a c = 36 - 4 (9) (- 2) = 36 + 72 = 108 > 0$ .
3. Since $D > 0$ , the equation has 2 distinct real roots. Option (C).
  1 mark
Answer: 2 distinct real roots, option (C)
Where the marks are
- Finds $D = 108 > 0$ and concludes 2 distinct real roots, option (C) (1 mark)
Q5 · 1 mark · MCQArithmetic Progressions
Two APs have the same common difference. The first term of one of these is $- 1$ and that of the other is $- 8$ . The difference between their 4th terms is
1. (A)1
2. (B)-7
3. (C)7
4. (D)9
Try it first, then show our solutionSolution
1. The 4th terms are $- 1 + 3 d$ and $- 8 + 3 d$ .
2. Difference: $(- 1 + 3 d) - (- 8 + 3 d) = 7$ . Option (C).
  1 mark
Answer: 7, option (C)
Where the marks are
- Writes both 4th terms and subtracts to get 7, option (C) (1 mark)
Q6 · 1 mark · MCQCoordinate Geometry
What is the ratio in which the line segment joining $(2, - 3)$ and $(5, 6)$ is divided by x-axis?
1. (A)1:2
2. (B)2:1
3. (C)2:5
4. (D)5:2
Try it first, then show our solutionSolution
1. Let the x-axis divide the segment in the ratio $k : 1$ . A point on the x-axis has y-coordinate $0$ .
2. By the section formula, $\frac{6 k + ( - 3 )}{k + 1} = 0$ . So $6 k = 3$ , giving $k = \frac{1}{2}$ .
3. The ratio is $\frac{1}{2} : 1 = 1 : 2$ . Option (A).
  1 mark
Answer: 1:2, option (A)
Where the marks are
- Sets the y-coordinate from the section formula to 0 and gets 1:2, option (A) (1 mark)
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Q7 · 1 mark · MCQCoordinate Geometry
A point $(x, y)$ is at a distance of 5 units from the origin. How many such points lie in the third quadrant?
1. (A)0
2. (B)1
3. (C)2
4. (D)infinitely many
Try it first, then show our solutionSolution
1. Distance 5 from the origin means $x^{2} + y^{2} = 25$ . This is a circle of radius 5 centred at the origin.
2. Part of this circle (an arc) lies in the third quadrant, where $x < 0, y < 0$ . An arc contains infinitely many points. Option (D).
  1 mark
Answer: infinitely many, option (D)
Where the marks are
- Recognises the locus as a circle whose third-quadrant arc has infinitely many points, option (D) (1 mark)
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Q8 · 1 mark · MCQTriangles
In $Δ$ ABC, DE $∥$ AB. If AB = a, DE = x, BE = b and EC = c. Then x expressed in terms of a, b and c is:
1. (A) $\frac{a c}{b}$
2. (B) $\frac{a c}{b + c}$
3. (C) $\frac{ab}{c}$
4. (D) $\frac{ab}{b + c}$
Figure in the paper: Triangle ABC with A at top left, B at bottom left, C at bottom right. D lies on side AC and E lies on side BC, with DE parallel to AB. Labels: AB = a, DE = x, BE = b, EC = c. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $D E ∥ A B$ , $△ C D E \sim △ C A B$ (AA similarity: $∠ C$ is common and corresponding angles are equal).
2. So $\frac{D E}{A B} = \frac{C E}{C B}$ , which gives $\frac{x}{a} = \frac{c}{b + c}$ .
3. Therefore $x = \frac{a c}{b + c}$ . Option (B).
  1 mark
Answer: $x = \frac{a c}{b + c}$ , option (B)
Where the marks are
- Uses $△ C D E \sim △ C A B$ to get $x = \frac{a c}{b + c}$ , option (B) (1 mark)
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Q9 · 1 mark · MCQCircles
If O is centre of a circle and Chord PQ makes an angle $5 0^{\circ}$ with the tangent PR at the point of contact P, then the angle subtended by the chord at the centre is
1. (A) $13 0^{\circ}$
2. (B) $10 0^{\circ}$
3. (C) $5 0^{\circ}$
4. (D) $3 0^{\circ}$
Figure in the paper: Circle with centre O. Chord PQ, with P on the circle. Tangent PR touches the circle at P, extending to the right. O is joined to P and to Q, forming triangle OPQ. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O P R = 9 0^{\circ}$ [tangent ⟂ radius]. So $∠ O P Q = 9 0^{\circ} - 5 0^{\circ} = 4 0^{\circ}$ .
2. $O P = O Q$ (radii), so $∠ O QP = 4 0^{\circ}$ .
3. $∠ P O Q = 18 0^{\circ} - 4 0^{\circ} - 4 0^{\circ} = 10 0^{\circ}$ . Option (B).
  1 mark
Answer: $10 0^{\circ}$ , option (B)
Where the marks are
- Uses tangent ⟂ radius and the isosceles triangle OPQ to get $10 0^{\circ}$ , option (B) (1 mark)
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Q10 · 1 mark · MCQCircles
A quadrilateral PQRS is drawn to circumscribe a circle. If PQ = 12 cm, QR = 15 cm and RS = 14 cm, then find the length of SP is
1. (A)15 cm
2. (B)14 cm
3. (C)12 cm
4. (D)11 cm
Figure in the paper: Quadrilateral PQRS drawn around a circle (circumscribed). Sides: PQ = 12 (top), QR = 15 (right), RS = 14 (bottom), SP is the left side (unlabelled). See it in the official paper.
Try it first, then show our solutionSolution
1. For a quadrilateral circumscribing a circle, $P Q + R S = QR + S P$ (tangents from an external point are equal).
2. $12 + 14 = 15 + S P$ , so $S P = 11$ cm. Option (D).
  1 mark
Answer: 11 cm, option (D)
Where the marks are
- Uses $P Q + R S = QR + S P$ to get SP = 11 cm, option (D) (1 mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
Given that $sin θ = \frac{a}{b}$ , then $cos θ$ is.
1. (A) $\frac{b}{b ^{2} - a ^{2}}$
2. (B) $\frac{b}{a}$
3. (C) $\frac{b ^{2} - a ^{2}}{b}$
4. (D) $\frac{a}{b ^{2} - a ^{2}}$
Try it first, then show our solutionSolution
1. Take a right triangle with opposite side $a$ and hypotenuse $b$ . The adjacent side is $b^{2} - a^{2}$ (Pythagoras).
2. $cos θ = \frac{b ^{2} - a ^{2}}{b}$ . Option (C).
  1 mark
Answer: $\frac{b ^{2} - a ^{2}}{b}$ , option (C)
Where the marks are
- Finds the adjacent side by Pythagoras and gets $cos θ$ , option (C) (1 mark)
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Q12 · 1 mark · MCQIntroduction to Trigonometry
$(sec A + tan A) (1 - sin A)$ equals:
1. (A) $sec A$
2. (B) $sin A$
3. (C) $cosec A$
4. (D) $cos A$
Try it first, then show our solutionSolution
1. $sec A + tan A = \frac{1 + sin A}{cos A}$ .
2. $\frac{( 1 + sin A ) ( 1 - sin A )}{cos A} = \frac{1 - sin ^{2} A}{cos A} = \frac{cos ^{2} A}{cos A} = cos A$ . Option (D).
  1 mark
Answer: $cos A$ , option (D)
Where the marks are
- Converts to sin/cos and simplifies to $cos A$ , option (D) (1 mark)
Q13 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
If a pole 6 m high casts a shadow $23$ m long on the ground, then the Sun's elevation is
1. (A) $6 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $3 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan θ = \frac{height}{shadow} = \frac{6}{2 3} = 3$ .
2. $θ = 6 0^{\circ}$ . Option (A).
  1 mark
Answer: $6 0^{\circ}$ , option (A)
Where the marks are
- Uses $tan θ = 6/ (23) = 3$ to get $6 0^{\circ}$ , option (A) (1 mark)
Q14 · 1 mark · MCQAreas Related to Circles
If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
1. (A)2 units
2. (B) $π$ units
3. (C)4 units
4. (D)7 units
Try it first, then show our solutionSolution
1. Set perimeter equal to area: $2 π r = π r^{2}$ .
2. Divide by $π r$ (since $r \neq = 0$ ): $r = 2$ units. Option (A).
  1 mark
Answer: 2 units, option (A)
Where the marks are
- Equates $2 π r = π r^{2}$ to get r = 2, option (A) (1 mark)
Q15 · 1 mark · MCQCompetency-basedAreas Related to Circles
It is proposed to build a new circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park is
1. (A)10 m
2. (B)15 m
3. (C)20 m
4. (D)24 m
Try it first, then show our solutionSolution
1. The radii are 8 m and 6 m. So $π R^{2} = π (8)^{2} + π (6)^{2}$ .
2. $R^{2} = 64 + 36 = 100$ , so $R = 10$ m. Option (A).
  1 mark
Answer: 10 m, option (A)
Where the marks are
- Equates the areas and gets R = 10 m, option (A) (1 mark)
Q16 · 1 mark · MCQCompetency-basedProbability
There is a square board of side '2a' units circumscribing a red circle. Jayadev is asked to keep a dot on the above said board. The probability that he keeps the dot on the shaded region is.
1. (A) $\frac{π}{4}$
2. (B) $\frac{4 - π}{4}$
3. (C) $\frac{π - 4}{4}$
4. (D) $\frac{4}{π}$
Figure in the paper: A square board with a circle inscribed in it (touching all four sides). The region outside the circle but inside the square is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. The inscribed circle has radius $a$ . Area of square $= 4 a^{2}$ and area of circle $= π a^{2}$ .
2. Shaded area $= 4 a^{2} - π a^{2}$ .
3. $P = \frac{( 4 - π ) a ^{2}}{4 a ^{2}} = \frac{4 - π}{4}$ . Option (B).
  1 mark
Answer: $\frac{4 - π}{4}$ , option (B)
Where the marks are
- Finds shaded area ÷ square area $= \frac{4 - π}{4}$ , option (B) (1 mark)
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Q17 · 1 mark · MCQProbability
2 cards of hearts and 4 cards of spades are missing from a pack of 52 cards. A card is drawn at random from the remaining pack. What is the probability of getting a black card?
1. (A) $\frac{22}{52}$
2. (B) $\frac{22}{46}$
3. (C) $\frac{24}{52}$
4. (D) $\frac{24}{46}$
Try it first, then show our solutionSolution
1. Cards remaining $= 52 - 6 = 46$ .
2. Black cards remaining $= 26 - 4 = 22$ (the hearts removed are red).
3. $P (black) = \frac{22}{46}$ . Option (B).
  1 mark
Answer: $\frac{22}{46}$ , option (B)
Where the marks are
- Counts 46 cards with 22 black, so $\frac{22}{46}$ , option (B) (1 mark)
Q18 · 1 mark · MCQCompetency-basedStatistics
The upper limit of the modal class of the given distribution is:
Height [in cm] Below 140 Below 145 Below 150 Below 155 Below 160 Below 165
Number of girls 4 11 29 40 46 51
1. (A)165
2. (B)160
3. (C)155
4. (D)150
Try it first, then show our solutionSolution
1. Convert the cumulative frequencies to class frequencies: below 140: 4; 140–145: 7; 145–150: 18; 150–155: 11; 155–160: 6; 160–165: 5.
2. The highest frequency is 18, so the modal class is 145–150. Its upper limit is 150. Option (D).
  1 mark
Answer: 150, option (D)
Where the marks are
- Finds class frequencies, identifies modal class 145–150 with upper limit 150, option (D) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedSurface Areas and Volumes
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.
Statement A (Assertion): Total Surface area of the top is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.
Statement R (Reason): Top is obtained by joining the plane surfaces of the hemisphere and cone together.
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.
Figure in the paper: A top (toy-like solid) made of a hemisphere on top of a cone, with the flat circular face of the hemisphere joined to the circular base of the cone; the cone point faces downward. See it in the official paper.
Try it first, then show our solutionSolution
1. The top is formed by joining the flat face of the hemisphere to the base of the cone, so these plane faces are hidden. R is true.
2. Only the curved surfaces stay exposed, so TSA = CSA of hemisphere + CSA of cone. A is true, and R explains why. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises both statements are true and R explains A, option (A) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedArithmetic Progressions
Statement A (Assertion): $- 5, \frac{- 5}{2}, 0, \frac{5}{2}, \dots$ is in Arithmetic Progression.
Statement R (Reason): The terms of an Arithmetic Progression cannot have both positive and negative rational numbers.
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. The differences are $- \frac{5}{2} - (- 5) = \frac{5}{2}$ , $0 - (- \frac{5}{2}) = \frac{5}{2}$ and $\frac{5}{2} - 0 = \frac{5}{2}$ . They are constant, so A is true.
2. This very AP contains both negative and positive terms, so R is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows the common difference is $\frac{5}{2}$ (A true) and R is false, option (C) (1 mark)
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Height [in cm]	Below 140	Below 145	Below 150	Below 155	Below 160	Below 165
Number of girls	4	11	29	40	46	51

Section B

Internal choice in 24 and 25

Q21 · 2 marks · Very short answerReal Numbers
Prove that $2$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $2$ is rational. Then $2 = \frac{a}{b}$ , where $a, b$ are integers, $b \neq = 0$ , and $a, b$ are coprime.
  ½ mark
2. Then $b 2 = a$ . Squaring gives $2 b^{2} = a^{2}$ . So 2 divides $a^{2}$ , and hence 2 divides $a$ (2 is prime).
  ½ mark
3. Write $a = 2 c$ . Then $2 b^{2} = 4 c^{2}$ , so $b^{2} = 2 c^{2}$ . So 2 divides $b^{2}$ , and hence 2 divides $b$ .
  ½ mark
4. Now 2 is a common factor of $a$ and $b$ . This contradicts the fact that $a, b$ are coprime. The contradiction comes from our assumption, so $2$ is irrational.
  ½ mark
Answer: $2$ is irrational (proved by contradiction).
Where the marks are
- Assumes $2 = a / b$ with a, b coprime (½ mark)
- Gets $2 b^{2} = a^{2}$ and deduces 2 | a (½ mark)
- Substitutes $a = 2 c$ and deduces 2 | b (½ mark)
- States the contradiction and concludes (½ mark)
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Q22 · 2 marks · Very short answerTriangles
ABCD is a parallelogram. Point P divides AB in the ratio 2:3 and point Q divides DC in the ratio 4:1. Prove that OC is half of OA.
Figure in the paper: Parallelogram ABCD with A bottom-left, B bottom-right, C top-right, D top-left. P lies on AB and Q lies on DC. Lines AQ and CP are drawn, meeting at O inside the parallelogram; O lies on AQ and on PC. See it in the official paper.
Try it first, then show our solutionSolution
1. Let $A B = D C = a$ (opposite sides of a parallelogram). P divides AB in the ratio 2:3, so $A P = \frac{2}{5} a$ .
  ½ mark
2. Q divides DC in the ratio 4:1, so $D Q = \frac{4}{5} a$ and $C Q = \frac{1}{5} a$ .
  ½ mark
3. O is the point where the diagonal AC meets PQ. In $△ A P O$ and $△ C QO$ : $∠ P A O = ∠ QC O$ [alternate angles, $A B ∥ D C$ ] and $∠ A O P = ∠ C O Q$ [vertically opposite]. So $△ A P O \sim △ C QO$ [AA similarity].
  ½ mark
4. Hence $\frac{A O}{C O} = \frac{A P}{C Q} = \frac{\frac{2}{5} a}{\frac{1}{5} a} = 2$ . So $O C = \frac{1}{2} O A$ . Proved.
  ½ mark
Answer: $△ A P O \sim △ C QO$ gives $A O : O C = 2 : 1$ , so $O C = \frac{1}{2} O A$ .
Where the marks are
- Expresses AP in terms of side a (½ mark)
- Expresses CQ in terms of side a (½ mark)
- Proves $△ A P O \sim △ C QO$ by AA (½ mark)
- Uses the ratio of sides to conclude $O C = \frac{1}{2} O A$ (½ mark)
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Q23 · 2 marks · Very short answerCircles
From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At a point E on the circle, a tangent is drawn to intersect PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of $Δ$ PCD.
Figure in the paper: Circle with centre O. External point P with tangents PA (touching at A, upper) and PB (touching at B, lower). A third tangent touches the circle at E, meeting PA at C and PB at D; the triangle PCD is formed. See it in the official paper.
Try it first, then show our solutionSolution
1. $P A = P B$ , $C A = C E$ and $D B = D E$ [tangents from an external point are equal].
  ½ mark
2. Perimeter of $△ P C D = P C + C D + P D = P C + C E + E D + P D$ . Replacing CE by CA and ED by DB gives $P C + C A + D B + P D = P A + P B$ .
  1 mark
3. Perimeter $= 2 P A = 2 \times 10 = 20$ cm.
  ½ mark
Answer: 20 cm
Where the marks are
- States the equal-tangent pairs (½ mark)
- Rewrites the perimeter as PA + PB (1 mark)
- Gets 20 cm (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerIntroduction to Trigonometry
If $tan (A + B) = 3$ and $tan (A - B) = \frac{1}{3}$ ; $0^{\circ} < A + B < 9 0^{\circ}$ ; $A > B$ , find A and B.
Try it first, then show our solutionSolution
1. $tan (A + B) = 3 = tan 6 0^{\circ}$ , so $A + B = 6 0^{\circ}$ …(1)
  ½ mark
2. $tan (A - B) = \frac{1}{3} = tan 3 0^{\circ}$ , so $A - B = 3 0^{\circ}$ …(2)
  ½ mark
3. Adding (1) and (2): $2 A = 9 0^{\circ}$ , so $A = 4 5^{\circ}$ .
  ½ mark
4. Subtracting (2) from (1): $2 B = 3 0^{\circ}$ , so $B = 1 5^{\circ}$ .
  ½ mark
Answer: $A = 4 5^{\circ}$ , $B = 1 5^{\circ}$
Where the marks are
- Gets $A + B = 6 0^{\circ}$ (½ mark)
- Gets $A - B = 3 0^{\circ}$ (½ mark)
- Finds $A = 4 5^{\circ}$ (½ mark)
- Finds $B = 1 5^{\circ}$ (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerIntroduction to Trigonometry
Find the value of x if $2 cosec^{2} 3 0^{\circ} + x sin^{2} 6 0^{\circ} - \frac{3}{4} tan^{2} 3 0^{\circ} = 10$
Try it first, then show our solutionSolution
1. Use the standard values: $cosec 3 0^{\circ} = 2$ , $sin 6 0^{\circ} = \frac{3}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ . Substituting them:
  $2 (2)^{2} + x (\frac{3}{2})^{2} - \frac{3}{4} (\frac{1}{3})^{2} = 10$
  1 mark
2. Simplify each term: $2 (4) + x \cdot \frac{3}{4} - \frac{3}{4} \cdot \frac{1}{3} = 10$ , that is,
  $8 + \frac{3 x}{4} - \frac{1}{4} = 10$
  ½ mark
3. Multiply both sides by 4: $32 + 3 x - 1 = 40$ . So $3 x = 9$ , which gives $x = 3$ .
  ½ mark
Answer: $x = 3$
Where the marks are
- Substitutes the correct values of cosec 30°, sin 60° and tan 30° (1 mark)
- Simplifies to $8 + \frac{3 x}{4} - \frac{1}{4} = 10$ (1/2 mark)
- Obtains $x = 3$ (1/2 mark)
Open Q24(b) on its own page
Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerAreas Related to Circles
With vertices A, B and C of $Δ$ ABC as centres, arcs are drawn with radii 14 cm and the three portions of the triangle so obtained are removed. Find the total area removed from the triangle.
Try it first, then show our solutionSolution
1. Each removed portion is a sector of radius $r = 14$ cm. Its angle is the angle of the triangle at that vertex.
  $Total area removed = \frac{∠ A}{36 0 ^{\circ}} π r^{2} + \frac{∠ B}{36 0 ^{\circ}} π r^{2} + \frac{∠ C}{36 0 ^{\circ}} π r^{2}$
  ½ mark
2. Take $π r^{2}$ common: $= \frac{∠ A + ∠ B + ∠ C}{36 0 ^{\circ}} π r^{2} = \frac{18 0 ^{\circ}}{36 0 ^{\circ}} π r^{2}$ [angle sum property of a triangle].
  ½ mark
3. Substitute the values: $= \frac{180}{360} \times \frac{22}{7} \times (14)^{2}$ .
  ½ mark
4. $= \frac{1}{2} \times \frac{22}{7} \times 196 = 308 cm^{2}$
  So the total area removed is $308 cm^{2}$ .
  ½ mark
Answer: 308 cm²
Where the marks are
- Writes the total area as the sum of three sector areas (1/2 mark)
- Uses ∠A + ∠B + ∠C = 180° to get $\frac{180}{360} π r^{2}$ (1/2 mark)
- Substitutes r = 14 cm and π = 22/7 (1/2 mark)
- Gets 308 cm² (1/2 mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerAreas Related to Circles
Find the area of the unshaded region shown in the given figure.
Figure in the paper: A square of side 14 cm. Inside it is a cloverleaf-shaped unshaded figure made of four semicircular-type bulges, each marked 3 cm from the nearest side of the square (3 cm from the top, bottom, left and right sides). The rest of the square is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. The unshaded region is a square of side $a$ in the centre. On each of its four sides stands a semicircle of diameter $a$ .
  $Area = a^{2} + 4 \times (area of a semicircle of diameter a)$
  ½ mark
2. Across the square, the unshaded region leaves a 3 cm gap on each side. So its full width is $14 - 3 - 3 = 8$ cm.
  ½ mark
3. This width is made of a radius, the side of the square and another radius:
  $\frac{a}{2} + a + \frac{a}{2} = 8 \Rightarrow 2 a = 8 \Rightarrow a = 4 cm$
  So the radius of each semicircle is $2$ cm.
  ½ mark
4. $Area = 4^{2} + 4 \times \frac{1}{2} π (2)^{2} = 16 + 8 π cm^{2}$
  With $π = \frac{22}{7}$ , this is $16 + \frac{176}{7} \approx 41.14 cm^{2}$ .
  ½ mark
Answer: $(16 + 8 π)$ cm² ≈ 41.14 cm²
Where the marks are
- Identifies the region as a square plus four semicircles with diameter equal to the side of the square (1/2 mark)
- Finds the width of the unshaded region as 14 − 3 − 3 = 8 cm (1/2 mark)
- Gets the side of the square a = 4 cm (1/2 mark)
- Gets the area (16 + 8π) cm² (1/2 mark)
Open Q25(b) on its own page

Section C

Internal choice in 28, 29

Q26 · 3 marks · Short answerCompetency-basedReal Numbers
National Art convention got registrations from students from all parts of the country, of which 60 are interested in music, 84 are interested in dance and 108 students are interested in handicrafts. For optimum cultural exchange, organisers wish to keep them in minimum number of groups such that each group consists of students interested in the same artform and the number of students in each group is the same. Find the number of students in each group. Find the number of groups in each art form. How many rooms are required if each group will be allotted a room?
Try it first, then show our solutionSolution
1. Every group must have the same number of students, and the number of groups must be as small as possible. So the group size must be the largest number that divides 60, 84 and 108, that is, HCF(60, 84, 108).
  ½ mark
2. Prime factorisation: $60 = 2^{2} \times 3 \times 5$ , $84 = 2^{2} \times 3 \times 7$ , $108 = 2^{2} \times 3^{3}$ .
  So HCF $= 2^{2} \times 3 = 12$ , and each group has 12 students.
  ½ mark
3. Number of groups in music $= \frac{60}{12} = 5$
  ½ mark
4. Number of groups in dance $= \frac{84}{12} = 7$
  ½ mark
5. Number of groups in handicrafts $= \frac{108}{12} = 9$
  ½ mark
6. Each group gets one room. Total rooms required $= 5 + 7 + 9 = 21$ .
  ½ mark
Answer: 12 students per group; Music 5, Dance 7, Handicrafts 9 groups; 21 rooms
Where the marks are
- Recognises that the group size is HCF(60, 84, 108) (1/2 mark)
- Finds HCF = 12 (1/2 mark)
- Music groups = 5 (1/2 mark)
- Dance groups = 7 (1/2 mark)
- Handicraft groups = 9 (1/2 mark)
- Total rooms = 21 (1/2 mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerPolynomials
If $α, β$ are zeroes of quadratic polynomial $5 x^{2} + 5 x + 1$ , find the value of
1. $α^{2} + β^{2}$
2. $α^{- 1} + β^{- 1}$
Try it first, then show our solutionSolution
1. For $p (x) = 5 x^{2} + 5 x + 1$ : $a = 5$ , $b = 5$ , $c = 1$ .
  ½ mark
2. Sum of zeroes: $α + β = - \frac{b}{a} = - \frac{5}{5} = - 1$
  ½ mark
3. Product of zeroes: $α β = \frac{c}{a} = \frac{1}{5}$
  ½ mark
4. 1. Use $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ :
  $= (- 1)^{2} - 2 (\frac{1}{5})$
  ½ mark
5. $= 1 - \frac{2}{5} = \frac{3}{5}$
  ½ mark
6. 2. $α^{- 1} + β^{- 1} = \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- 1}{1/5} = - 5$
  ½ mark
Answer: 1. $α^{2} + β^{2} = \frac{3}{5}$ ; 2. $α^{- 1} + β^{- 1} = - 5$
Where the marks are
- Identifies the coefficients a = 5, b = 5, c = 1 (1/2 mark)
- Finds α + β = −1 (1/2 mark)
- Finds αβ = 1/5 (1/2 mark)
- Uses the identity for α² + β² (1/2 mark)
- Gets α² + β² = 3/5 (1/2 mark)
- Gets α⁻¹ + β⁻¹ = −5 (1/2 mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerPair of Linear Equations in Two Variables
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Try it first, then show our solutionSolution
1. Let the tens digit be $x$ and the units digit be $y$ .
  The number is $10 x + y$ . The number with its digits reversed is $10 y + x$ .
  ½ mark
2. Given condition: $(10 x + y) + (10 y + x) = 66$
  $\Rightarrow 11 (x + y) = 66 \Rightarrow x + y = 6$ ...(1)
  ½ mark
3. The digits differ by 2, so either $x - y = 2$ ...(2)
  ½ mark
4. or $y - x = 2$ ...(3)
  ½ mark
5. Case 1: adding (1) and (2) gives $2 x = 8$ , so $x = 4$ and $y = 2$ . The number is 42.
  ½ mark
6. Case 2: adding (1) and (3) gives $2 y = 8$ , so $y = 4$ and $x = 2$ . The number is 24.
  So there are two such numbers: 42 and 24.
  ½ mark
Answer: The number is 42 or 24; there are 2 such numbers
Where the marks are
- Writes the number and its reverse as 10x + y and 10y + x (1/2 mark)
- Gets x + y = 6 (1/2 mark)
- Writes x − y = 2 (1/2 mark)
- Writes y − x = 2 as the second possibility (1/2 mark)
- Solves to get 42 (1/2 mark)
- Solves to get 24 and states there are two numbers (1/2 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve: $\frac{2}{x} + \frac{3}{y} = 2$ ; $\frac{4}{x} - \frac{9}{y} = - 1$ , $x, y > 0$
Try it first, then show our solutionSolution
1. Let $\frac{1}{x} = m$ and $\frac{1}{y} = n$ .
  ½ mark
2. The equations become:
  $2 m + 3 n = 2 ...(1)$
  $4 m - 9 n = - 1 ...(2)$
  ½ mark
3. Multiply (1) by 2: $4 m + 6 n = 4$ . Subtract (2) from this: $15 n = 5$ , so $n = \frac{1}{3}$ .
  ½ mark
4. Substitute $n = \frac{1}{3}$ in (1): $2 m + 1 = 2$ , so $2 m = 1$ and $m = \frac{1}{2}$ .
  ½ mark
5. $\frac{1}{x} = \frac{1}{2} \Rightarrow x = 2 \Rightarrow x = 4$
  $\frac{1}{y} = \frac{1}{3} \Rightarrow y = 3 \Rightarrow y = 9$
  1 mark
Answer: $x = 4$ , $y = 9$
Where the marks are
- Substitutes 1/√x = m and 1/√y = n (1/2 mark)
- Forms the linear equations 2m + 3n = 2 and 4m − 9n = −1 (1/2 mark)
- Gets n = 1/3 (1/2 mark)
- Gets m = 1/2 (1/2 mark)
- Gets x = 4 and y = 9 (1 mark)
Open Q28(b) on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCircles
PA and PB are tangents drawn to a circle of centre O from an external point P. Chord AB makes an angle of $3 0^{\circ}$ with the radius at the point of contact. If length of the chord is 6 cm, find the length of the tangent PA and the length of the radius OA.
Figure in the paper: Circle with centre O. External point P with tangents PA (touching at A, top) and PB (touching at B, bottom). Chord AB is drawn, and radii OA and OB are joined. See it in the official paper.
Try it first, then show our solutionSolution
1. Given $∠ O A B = 3 0^{\circ}$ . Also $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  So $∠ P A B = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .
  ½ mark
2. $P A = P B$ [tangents from an external point are equal]. So $∠ P B A = ∠ P A B = 6 0^{\circ}$ [angles opposite equal sides].
  ½ mark
3. In $△ A B P$ : $∠ A P B = 18 0^{\circ} - 6 0^{\circ} - 6 0^{\circ} = 6 0^{\circ}$ [angle sum property]. So $△ A B P$ is equilateral.
  ½ mark
4. Therefore $P A = A B = 6$ cm.
  ½ mark
5. OP bisects $∠ A P B$ [centre lies on the bisector of the angle between the tangents], so $∠ O P A = 3 0^{\circ}$ .
  In right $△ O A P$ :
  $tan 3 0^{\circ} = \frac{O A}{P A} \Rightarrow \frac{1}{3} = \frac{O A}{6}$
  ½ mark
6. $O A = \frac{6}{3} = 23$ cm
  ½ mark
Answer: PA = 6 cm, OA = 2√3 cm
Where the marks are
- Uses tangent ⟂ radius to get ∠PAB = 60° (1/2 mark)
- Uses PA = PB to get ∠PBA = 60° (1/2 mark)
- Shows ∠APB = 60°, so triangle ABP is equilateral (1/2 mark)
- PA = 6 cm (1/2 mark)
- Uses tan 30° = OA/PA in triangle OAP (1/2 mark)
- OA = 2√3 cm (1/2 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCircles
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $∠ P T Q = 2∠ O P Q$ .
Try it first, then show our solutionSolution
1. Given: TP and TQ are tangents from T to a circle with centre O. To prove: $∠ P T Q = 2∠ O P Q$ .
  Let $∠ T P Q = θ$ . Since $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact], we get $∠ O P Q = 9 0^{\circ} - θ$ ...(1)
  ½ mark
2. $T P = T Q$ [tangents from an external point are equal]
  ½ mark
3. So $∠ T QP = ∠ T P Q = θ$ [angles opposite equal sides of a triangle].
  ½ mark
4. In $△ P T Q$ : $∠ T P Q + ∠ T QP + ∠ P T Q = 18 0^{\circ}$ [angle sum property]
  $\Rightarrow θ + θ + ∠ P T Q = 18 0^{\circ}$
  ½ mark
5. $\Rightarrow ∠ P T Q = 18 0^{\circ} - 2 θ = 2 (9 0^{\circ} - θ)$
  ½ mark
6. Using (1), $∠ P T Q = 2∠ O P Q$ . Hence proved.
  ½ mark
Answer: Proved: ∠PTQ = 2∠OPQ
Where the marks are
- Uses tangent ⟂ radius to write ∠OPQ = 90° − θ (1/2 mark)
- States TP = TQ, giving the reason (1/2 mark)
- Gets ∠TQP = ∠TPQ = θ (1/2 mark)
- Applies the angle sum property in triangle PTQ (1/2 mark)
- Gets ∠PTQ = 180° − 2θ = 2(90° − θ) (1/2 mark)
- Concludes ∠PTQ = 2∠OPQ (1/2 mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerIntroduction to Trigonometry
If $1 + sin^{2} θ = 3 sin θ cos θ$ , then prove that $tan θ = 1$ or $\frac{1}{2}$
Try it first, then show our solutionSolution
1. Given $1 + sin^{2} θ = 3 sin θ cos θ$ .
  Here $cos θ \neq = 0$ : if $cos θ = 0$ , the left side would be 2 and the right side 0. So we may divide both sides by $cos^{2} θ$ :
  $\frac{1}{cos ^{2} θ} + tan^{2} θ = 3 tan θ$
  ½ mark
2. $\Rightarrow sec^{2} θ + tan^{2} θ = 3 tan θ$
  ½ mark
3. Use $sec^{2} θ = 1 + tan^{2} θ$ : $1 + tan^{2} θ + tan^{2} θ = 3 tan θ$
  ½ mark
4. $\Rightarrow 2 tan^{2} θ - 3 tan θ + 1 = 0$
  ½ mark
5. Put $tan θ = x$ : $2 x^{2} - 3 x + 1 = 0$ , so $(x - 1) (2 x - 1) = 0$ and $x = 1$ or $\frac{1}{2}$ .
  Hence $tan θ = 1$ or $\frac{1}{2}$ . Proved.
  1 mark
Answer: $tan θ = 1$ or $\frac{1}{2}$ (proved)
Where the marks are
- Divides both sides by cos²θ (1/2 mark)
- Writes sec²θ + tan²θ = 3 tan θ (1/2 mark)
- Uses sec²θ = 1 + tan²θ (1/2 mark)
- Forms 2tan²θ − 3tanθ + 1 = 0 (1/2 mark)
- Factorises to get tan θ = 1 or 1/2 (1 mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerStatistics
The length of 40 leaves of a plant are measured correct to nearest millimetre, and the data obtained is represented in the following table. Find the mean length of the leaves.
Length [in mm] Number of leaves
118 – 126 3
127 – 135 5
136 – 144 9
145 – 153 12
154 – 162 5
163 – 171 4
172 – 180 2
Try it first, then show our solutionSolution
1. Take the assumed mean $a = 149$ and $d_{i} = x_{i} - 149$ . The class marks are the midpoints, for example $(118 + 126) /2 = 122$ .
  Length (mm) $f_{i}$ $x_{i}$ $d_{i}$ $f_{i} d_{i}$
  118–126 3 122 −27 −81
  127–135 5 131 −18 −90
  136–144 9 140 −9 −81
  145–153 12 149 0 0
  154–162 5 158 9 45
  163–171 4 167 18 72
  172–180 2 176 27 54
  Total 40 −81
  So $\sum f_{i} = 40$ and $\sum f_{i} d_{i} = - 252 + 171 = - 81$ .
  2 marks
2. $Mean = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 149 + \frac{- 81}{40}$
  ½ mark
3. $= 149 - 2.025 = 146.975$
  The mean length of the leaves is 146.975 mm.
  ½ mark
Answer: Mean length = 146.975 mm
Where the marks are
- Correct table of class marks, deviations and f·d, with Σf = 40 and Σfd = −81 (2 marks)
- Applies the assumed-mean formula correctly (1/2 mark)
- Gets the mean as 146.975 mm (1/2 mark)
Open Q31 on its own page

Length [in mm]	Number of leaves
118 – 126	3
127 – 135	5
136 – 144	9
145 – 153	12
154 – 162	5
163 – 171	4
172 – 180	2

Length (mm)	$f_{i}$	$x_{i}$	$d_{i}$	$f_{i} d_{i}$
118–126	3	122	−27	−81
127–135	5	131	−18	−90
136–144	9	140	−9	−81
145–153	12	149	0	0
154–162	5	158	9	45
163–171	4	167	18	72
172–180	2	176	27	54
Total	40			−81

Section D

Internal choice in 32, 34

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of stream.
Try it first, then show our solutionSolution
1. Let the speed of the stream be $x$ km/h.
  Upstream speed $= (18 - x)$ km/h. Downstream speed $= (18 + x)$ km/h.
  1 mark
2. Using time = distance ÷ speed:
  Time upstream $= \frac{24}{18 - x}$ h. Time downstream $= \frac{24}{18 + x}$ h.
  1 mark
3. The upstream journey takes 1 hour more:
  $\frac{24}{18 - x} - \frac{24}{18 + x} = 1$
  1 mark
4. $24 (18 + x) - 24 (18 - x) = (18 - x) (18 + x)$
  $\Rightarrow 48 x = 324 - x^{2}$
  $\Rightarrow x^{2} + 48 x - 324 = 0$
  $\Rightarrow (x + 54) (x - 6) = 0$ , so $x = 6$ or $x = - 54$ .
  1 mark
5. A speed cannot be negative, so we reject $x = - 54$ .
  The speed of the stream is 6 km/h.
  1 mark
Answer: Speed of the stream = 6 km/h
Where the marks are
- Writes the upstream and downstream speeds as 18 − x and 18 + x (1 mark)
- Writes both times as distance ÷ speed (1 mark)
- Forms the equation 24/(18−x) − 24/(18+x) = 1 (1 mark)
- Reduces it to x² + 48x − 324 = 0 and solves to get x = 6 or −54 (1 mark)
- Rejects the negative value and states 6 km/h (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerCompetency-basedQuadratic Equations
Two water taps together can fill a tank in $9 \frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Try it first, then show our solutionSolution
1. Let the smaller tap take $x$ hours to fill the tank. Then the larger tap takes $(x - 10)$ hours.
  ½ mark
2. Part of the tank filled in 1 hour by the smaller tap $= \frac{1}{x}$ , and by the larger tap $= \frac{1}{x - 10}$ .
  1 mark
3. Together the taps fill the tank in $9 \frac{3}{8} = \frac{75}{8}$ hours.
  ½ mark
4. So the part filled by both taps in 1 hour $= \frac{8}{75}$ .
  ½ mark
5. $\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$
  ½ mark
6. $75 (2 x - 10) = 8 x (x - 10)$
  $\Rightarrow 8 x^{2} - 230 x + 750 = 0$
  $\Rightarrow 4 x^{2} - 115 x + 375 = 0$
  $x = \frac{115 \pm 13225 - 6000}{8} = \frac{115 \pm 85}{8}$
  So $x = 25$ or $x = \frac{30}{8} = 3.75$ .
  1 mark
7. If $x = 3.75$ , the larger tap's time $x - 10$ would be negative, which is impossible. So $x = 25$ .
  ½ mark
8. The smaller tap fills the tank in 25 hours, and the larger tap in $25 - 10 = 15$ hours.
  ½ mark
Answer: Smaller tap: 25 hours; larger tap: 15 hours
Where the marks are
- Lets the smaller tap's time be x and the larger tap's time be x − 10 (1/2 mark)
- Writes the one-hour work of each tap as 1/x and 1/(x−10) (1 mark)
- Converts 9 3/8 h to 75/8 h (1/2 mark)
- Writes the combined one-hour work as 8/75 (1/2 mark)
- Forms the equation 1/x + 1/(x−10) = 8/75 (1/2 mark)
- Gets the quadratic 8x² − 230x + 750 = 0 with roots 25 and 30/8 (1 mark)
- Rejects 30/8 with a reason (1/2 mark)
- States 25 h and 15 h (1/2 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerTriangles
(a) State and prove Basic Proportionality theorem.
(b) In the given figure $∠ C E F = ∠ C F E$ . F is the midpoint of DC. Prove that $\frac{A B}{B D} = \frac{A E}{F D}$
Figure in the paper: Vertical line AB with A at top and B at bottom. D lies on AB. E lies on AC and C is to the right of D at the same level as the horizontal line DC. F is the midpoint of DC; segment EF is drawn, forming triangle CEF. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) Statement (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle and it meets the other two sides in distinct points, then it divides those two sides in the same ratio.
  ½ mark
2. Given: In $△ P QR$ , a line $X Y ∥ QR$ meets $P Q$ at $X$ and $P R$ at $Y$ .
  To prove: $\frac{P X}{X Q} = \frac{P Y}{Y R}$
  ½ mark
3. Figure and construction: Draw $△ P QR$ with $X$ on $P Q$ and $Y$ on $P R$ , and $X Y ∥ QR$ . Join $Q Y$ and $R X$ . Draw $Y N ⊥ P Q$ (N on PQ) and $X M ⊥ P R$ (M on PR).
  ½ mark
4. Proof: Using area $= \frac{1}{2} \times$ base $\times$ height:
  $ar (P X Y) = \frac{1}{2} P X \cdot Y N$ and $ar (QX Y) = \frac{1}{2} X Q \cdot Y N$ .
  So $\frac{ar ( P X Y )}{ar ( QX Y )} = \frac{P X}{X Q}$ ...(1)
  ½ mark
5. Similarly, $ar (P X Y) = \frac{1}{2} P Y \cdot X M$ and $ar (R X Y) = \frac{1}{2} Y R \cdot X M$ .
  So $\frac{ar ( P X Y )}{ar ( R X Y )} = \frac{P Y}{Y R}$ ...(2)
  ½ mark
6. $△ QX Y$ and $△ R X Y$ stand on the same base $X Y$ and lie between the same parallels $X Y$ and $QR$ . Hence $ar (QX Y) = ar (R X Y)$ ...(3)
  From (1), (2) and (3): $\frac{P X}{X Q} = \frac{P Y}{Y R}$ . Hence proved.
  ½ mark
7. (b) In the figure, D lies on AB, E lies on AC, F is the midpoint of DC, and B, F, E are collinear (the line BE passes through F).
  Construction: Through D, draw $D G ∥ B E$ , meeting AE at G.
  In $△ A B E$ , $D G ∥ B E$ , so by BPT $\frac{A D}{D B} = \frac{A G}{GE}$ .
  Adding 1 to both sides gives $\frac{A D + D B}{D B} = \frac{A G + GE}{GE}$ , that is, $\frac{A B}{B D} = \frac{A E}{GE}$ ...(i)
  ½ mark
8. F is the midpoint of DC, so $C F = F D$ ...(ii)
  ½ mark
9. In $△ C D G$ , F is the midpoint of CD and $F E ∥ D G$ (FE is part of BE).
  By the converse of the mid-point theorem, E is the midpoint of CG. So $GE = C E$ ...(iii)
  ½ mark
10. Given $∠ C E F = ∠ C F E$ , so $C F = C E$ [sides opposite equal angles are equal] ...(iv)
  From (ii), (iii) and (iv): $GE = C E = C F = F D$ .
  Putting $GE = F D$ in (i): $\frac{A B}{B D} = \frac{A E}{F D}$ . Hence proved.
  ½ mark
Answer: (a) BPT stated and proved; (b) $\frac{A B}{B D} = \frac{A E}{F D}$ proved (construction DG ∥ BE, then BPT and the converse of the mid-point theorem)
Where the marks are
- Correct statement of BPT (½ mark)
- Given and To Prove written correctly (½ mark)
- Figure with construction (join QY, RX and draw the perpendiculars) (½ mark)
- Area ratios with common heights, equal areas of triangles on the same base between the same parallels, and the conclusion (1½ marks)
- Constructs DG ∥ BE and applies BPT to get AB/BD = AE/GE (½ mark)
- Uses F as the midpoint: CF = FD (½ mark)
- Converse of the mid-point theorem in △CDG gives GE = CE (½ mark)
- CF = CE from the equal angles, hence GE = FD and the result (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm? What should be the speed of water if the rise in water level is to be attained in 1 hour?
Try it first, then show our solutionSolution
1. Volume of water needed in the pond $= l \times b \times h$ , with $h = 21$ cm $= \frac{21}{100}$ m:
  $V = 50 \times 44 \times \frac{21}{100} = 462 m^{3}$
  1 mark
2. Radius of the pipe $r = \frac{14}{2} = 7$ cm $= \frac{7}{100}$ m.
  Area of cross-section $= π r^{2} = \frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} = \frac{154}{10000} m^{2}$
  1 mark
3. Speed of flow $= 15$ km/h $= 15000$ m/h. In 1 hour a column of water 15000 m long leaves the pipe.
  Volume flowing in 1 hour $= \frac{154}{10000} \times 15000 = 231 m^{3}$
  1 mark
4. Time required $= \frac{volume of pond water}{volume per hour} = \frac{462}{231} = 2$ hours
  1 mark
5. To fill $462 m^{3}$ in 1 hour, the length of the water column per hour must be $\frac{462}{154/10000} = 30000$ m.
  So the speed must be 30000 m/h $= 30$ km/h. (Doubling the speed halves the time.)
  1 mark
Answer: 2 hours; required speed = 30 km/h
Where the marks are
- Volume of the pond's water $= 462 m^{3}$ (1 mark)
- Cross-section area of the pipe $= \frac{154}{10000} m^{2}$ (1 mark)
- Converts 15 km/h to 15000 m/h and finds the volume per hour (1 mark)
- Time = 2 hours (1 mark)
- Speed for 1 hour = 30 km/h (1 mark)
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OR
Q34(b) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, find the area of the canvas required for making the tent, keeping a provision of $26 m^{2}$ of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of ₹ 500 per $m^{2}$ .
Try it first, then show our solutionSolution
1. Radius $r = 14$ m, height of the cylinder $= 3$ m.
  Height of the cone $h = 13.5 - 3 = 10.5$ m
  ½ mark
2. Slant height $l = h^{2} + r^{2} = (10.5)^{2} + 1 4^{2} = 110.25 + 196 = 306.25 = 17.5$ m
  1 mark
3. CSA of the cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 14 \times 3 = 264 m^{2}$
  1 mark
4. CSA of the cone $= π r l = \frac{22}{7} \times 14 \times 17.5 = 770 m^{2}$
  1 mark
5. Total curved surface (canvas for the tent) $= 264 + 770 = 1034 m^{2}$
  ½ mark
6. Canvas to be purchased $= 1034 + 26 = 1060 m^{2}$
  ½ mark
7. Cost $= 1060 \times ₹500 = ₹5, 30, 000$
  ½ mark
Answer: Canvas required = 1060 m²; cost = ₹5,30,000
Where the marks are
- Height of the cone = 10.5 m (½ mark)
- Slant height = 17.5 m (1 mark)
- CSA of the cylinder = 264 m² (1 mark)
- CSA of the cone = 770 m² (1 mark)
- Total CSA = 1034 m² (½ mark)
- Adds the wastage: 1060 m² (½ mark)
- Cost = ₹5,30,000 (½ mark)
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Q35 · 5 marks · Long answerCompetency-basedStatistics
The median of the following data is 50. Find the values of 'p' and 'q', if the sum of all frequencies is 90. Also find the mode of the data.
Marks obtained Number of students
20 – 30 p
30 – 40 15
40 – 50 25
50 – 60 20
60 – 70 q
70 – 80 8
80 – 90 10
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Marks f cf
  20–30 p p
  30–40 15 p+15
  40–50 25 p+40
  50–60 20 p+60
  60–70 q p+q+60
  70–80 8 p+q+68
  80–90 10 p+q+78
  1 mark
2. Sum of frequencies $= 90$ :
  $p + q + 78 = 90 \Rightarrow p + q = 12$ ...(1)
  ½ mark
3. The median is 50, so the median class is 50–60, with $\frac{n}{2} = 45$ .
  Here $l = 50$ , $c f = p + 40$ , $f = 20$ , $h = 10$ .
  ½ mark
4. Median $= l + \frac{\frac{n}{2} - c f}{f} \times h$
  $\Rightarrow 50 = 50 + \frac{45 - ( p + 40 )}{20} \times 10$
  ½ mark
5. $\Rightarrow 45 - (p + 40) = 0 \Rightarrow p = 5$
  ½ mark
6. From (1): $q = 12 - 5 = 7$
  ½ mark
7. The highest frequency is 25, so the modal class is 40–50.
  Here $l = 40$ , $f_{1} = 25$ , $f_{0} = 15$ , $f_{2} = 20$ , $h = 10$ .
  Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
  ½ mark
8. Mode $= 40 + \frac{25 - 15}{50 - 15 - 20} \times 10 = 40 + \frac{100}{15} = 40 + 6.67 = 46.67$ marks
  1 mark
Answer: p = 5, q = 7; mode ≈ 46.67
Where the marks are
- Cumulative frequency column (1 mark)
- p + q = 12 (½ mark)
- Identifies the median class 50–60 with n/2 = 45 (½ mark)
- Substitutes correctly into the median formula (½ mark)
- p = 5 (½ mark)
- q = 7 (½ mark)
- Modal class 40–50 and the mode formula (½ mark)
- Mode = 46.67 (1 mark)
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Marks obtained	Number of students
20 – 30	p
30 – 40	15
40 – 50	25
50 – 60	20
60 – 70	q
70 – 80	8
80 – 90	10

Marks	f	cf
20–30	p	p
30–40	15	p+15
40–50	25	p+40
50–60	20	p+60
60–70	q	p+q+60
70–80	8	p+q+68
80–90	10	p+q+78

Section E

Case-based units with sub-parts of 1, 1 and 2 marks; internal choice in 2-mark sub-parts

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Manpreet Kaur is the national record holder for women in the shot-put discipline. Her throw of 18.86 m at the Asian Grand Prix in 2017 is the maximum distance for an Indian female athlete. Keeping her as a role model, Sanjitha is determined to earn gold in Olympics one day. Initially her throw reached 7.56 m only. Being an athlete in school, she regularly practiced both in the mornings and in the evenings and was able to improve the distance by 9 cm every week. During the special camp for 15 days, she started with 40 throws and every day kept increasing the number of throws by 12 to achieve this remarkable progress.
1. (i)
  How many throws Sanjitha practiced on 11th day of the camp?
  1 mark
2. (ii)
  What would be Sanjitha's throw distance at the end of 6 weeks?
  2 marks
3. OR
  (ii)
  When will she be able to achieve a throw of 11.16 m?
  2 marks
4. (iii)
  How many throws did she do during the entire camp of 15 days?
  1 mark
Figure in the paper: Photograph of a female athlete in a red top and gloves, holding a shot-put at the side of her neck, in throwing position. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Throws per day form an AP with $a = 40$ , $d = 12$ .
  $a_{11} = a + 10 d = 40 + 10 \times 12 = 160$ throws
  1 mark
2. (ii)(a) Distances form an AP with $a = 7.56$ m and $d = 9$ cm $= 0.09$ m. The initial throw is at the start, so the end of 6 weeks is the 7th term ( $n - 1 = 6$ ).
  ½ mark
3. (ii)(a) $a_{n} = a + (n - 1) d$
  ½ mark
4. (ii)(a) $a_{7} = 7.56 + 6 (0.09)$
  ½ mark
5. (ii)(a) $a_{7} = 7.56 + 0.54 = 8.1$ m
  ½ mark
6. (ii)(b) $a = 7.56$ m, $d = 0.09$ m, $a_{n} = 11.16$ m
  ½ mark
7. (ii)(b) $a_{n} = a + (n - 1) d \Rightarrow 11.16 = 7.56 + (n - 1) (0.09)$
  ½ mark
8. (ii)(b) $3.6 = (n - 1) (0.09)$
  ½ mark
9. (ii)(b) $n - 1 = \frac{3.6}{0.09} = 40 \Rightarrow n = 41$ .
  So 11.16 m is the 41st term, which she reaches after 40 weeks of improvement (in the 41st term of the AP).
  ½ mark
10. (iii) $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
  $S_{15} = \frac{15}{2} [2 (40) + 14 (12)]$
  ½ mark
11. (iii) $S_{15} = \frac{15}{2} [80 + 168] = \frac{15}{2} \times 248 = 1860$ throws
  ½ mark
Answer: (i) 160 throws; (ii)(a) 8.1 m; (ii)(b) 41st term, i.e. after 40 weeks (scheme: 41 weeks); (iii) 1860 throws
Where the marks are
- (i) Uses $a_{11} = a + 10 d$ to get 160 (1 mark)
- (ii)(a) Correct a and d in metres (½ mark)
- (ii)(a) Writes the nth-term formula (½ mark)
- (ii)(a) Substitutes $n - 1 = 6$ (½ mark)
- (ii)(a) Gets 8.1 m (½ mark)
- (ii)(b) Correct a, d and $a_{n}$ (½ mark)
- (ii)(b) Forms the equation $11.16 = 7.56 + (n - 1) (0.09)$ (½ mark)
- (ii)(b) Simplifies to $3.6 = (n - 1) (0.09)$ (½ mark)
- (ii)(b) Gets n = 41 (½ mark)
- (iii) Sum formula with substitution (½ mark)
- (iii) Gets 1860 (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Tharunya was thrilled to know that the football tournament is fixed with a monthly timeframe from 20th July to 20th August 2023 and for the first time in the FIFA Women's World Cup's history, two nations host in 10 venues. Her father felt that the game can be better understood if the position of players is represented as points on a coordinate plane.
1. (i)
  At an instance, the midfielders and forward formed a parallelogram. Find the position of the central midfielder (D) if the position of other players who formed the parallelogram are: A(1,2), B(4,3) and C(6,6).
  1 mark
2. (ii)
  Check if the Goal keeper G(-3,5), Sweeper H(3,1) and Wing-back K(0,3) fall on a same straight line.
  2 marks
3. OR
  (ii)
  Check if the Full-back J(5,-3) and centre-back I(-4,6) are equidistant from forward C(0,1) and if C is the mid-point of IJ.
  2 marks
4. (iii)
  If Defensive midfielder A(1,4), Attacking midfielder B(2,-3) and Striker E(a,b) lie on the same straight line and B is equidistant from A and E, find the position of E.
  1 mark
Figure in the paper: Football pitch diagram with player positions labelled: J Full-back, I Centre-back, G Goalkeeper, H Sweeper, K Wing-back, A Defensive midfielder, D Central midfielder, B Attacking midfielder, C Forward, E Striker, F Winger. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let $D = (a, b)$ . The diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD:
  $(\frac{1 + 6}{2}, \frac{2 + 6}{2}) = (\frac{4 + a}{2}, \frac{3 + b}{2})$
  ½ mark
2. (i) $4 + a = 7 \Rightarrow a = 3$ and $3 + b = 8 \Rightarrow b = 5$ .
  The central midfielder is at D(3, 5).
  ½ mark
3. (ii)(a) $G H = (3 + 3)^{2} + (1 - 5)^{2} = 36 + 16 = 52 = 213$
  ½ mark
4. (ii)(a) $G K = (0 + 3)^{2} + (3 - 5)^{2} = 9 + 4 = 13$
  ½ mark
5. (ii)(a) $H K = (3 - 0)^{2} + (1 - 3)^{2} = 13$
  ½ mark
6. (ii)(a) $G K + H K = 13 + 13 = 213 = G H$ .
  So G, H and K lie on the same straight line, with K between G and H.
  ½ mark
7. (ii)(b) $C J = (5 - 0)^{2} + (- 3 - 1)^{2} = 25 + 16 = 41$
  ½ mark
8. (ii)(b) $C I = (- 4 - 0)^{2} + (6 - 1)^{2} = 16 + 25 = 41$ .
  So $C J = C I$ : J and I are equidistant from C.
  ½ mark
9. (ii)(b) Midpoint of IJ $= (\frac{5 - 4}{2}, \frac{- 3 + 6}{2}) = (\frac{1}{2}, \frac{3}{2})$
  ½ mark
10. (ii)(b) $(\frac{1}{2}, \frac{3}{2}) \neq = (0, 1)$ , so C is not the midpoint of IJ.
  ½ mark
11. (iii) A, B, E are collinear and $A B = B E$ , so B is the midpoint of AE:
  $(\frac{1 + a}{2}, \frac{4 + b}{2}) = (2, - 3)$
  ½ mark
12. (iii) $1 + a = 4 \Rightarrow a = 3$ and $4 + b = - 6 \Rightarrow b = - 10$ .
  So E is (3, −10).
  ½ mark
Answer: (i) D(3, 5); (ii)(a) Yes, collinear since GK + HK = GH; (ii)(b) Equidistant (both √41), but C is not the midpoint of IJ (midpoint is (1/2, 3/2)); (iii) E(3, −10)
Where the marks are
- (i) Equates the midpoints of the diagonals (½ mark)
- (i) Gets D(3,5) (½ mark)
- (ii)(a) GH = 2√13 (½ mark)
- (ii)(a) GK = √13 (½ mark)
- (ii)(a) HK = √13 (½ mark)
- (ii)(a) GK + HK = GH, so collinear (½ mark)
- (ii)(b) CJ = √41 (½ mark)
- (ii)(b) CI = √41, so equidistant (½ mark)
- (ii)(b) Midpoint of IJ = (1/2, 3/2) (½ mark)
- (ii)(b) Concludes C is not the midpoint (½ mark)
- (iii) B is the midpoint of AE; sets up the equations (½ mark)
- (iii) Gets E(3, −10) (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of $4 5^{\circ}$ . When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of $3 0^{\circ}$ and the ball flying towards him at the same height at an angle of elevation of $6 0^{\circ}$ .
1. (i)
  At what distance from the foot of the tree was he observing the bird sitting on the tree?
  1 mark
2. (ii)
  How far did the bird fly in the mentioned time?
  2 marks
3. OR
  (ii)
  After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball?
  2 marks
4. (iii)
  What is the speed of the bird in m/min if it had flown $20 (3 + 1)$ m?
  1 mark
Figure in the paper: Diagram of Kaushik at ground point C. Tree AB is 80 m tall, with B on the ground at the foot of the tree. A bird sits at the top A. Another bird is at D, at the same height 80 m, above point E on the ground. Ball is at F, at the same height, above ground point G, with FG vertical and G directly below F; the line from C to F makes $6 0^{\circ}$ with the ground, the line CA makes $4 5^{\circ}$ and the line CD makes $3 0^{\circ}$ with the ground. Labels: Tree 80 m, Height 80 m, Ground level, points C, G, B, E on the ground line. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In right $△ A B C$ (tree AB = 80 m, angle at C = 45°):
  $tan 4 5^{\circ} = \frac{A B}{C B} \Rightarrow 1 = \frac{80}{C B} \Rightarrow C B = 80$ m
  1 mark
2. (ii)(a) The bird flies from A to D at the same height of 80 m, with DE = 80 m. In right $△ D E C$ :
  $tan 3 0^{\circ} = \frac{D E}{C E}$
  ½ mark
3. (ii)(a) $\frac{1}{3} = \frac{80}{C E}$
  ½ mark
4. (ii)(a) $C E = 803$ m
  ½ mark
5. (ii)(a) Since $A D ∥ B E$ at the same height, the distance flown is $A D = B E = C E - C B = 803 - 80 = 80 (3 - 1)$ m $\approx 58.56$ m
  ½ mark
6. (ii)(b) The ball is at F, 80 m high above G, with the angle at C = 60°. In right $△ F GC$ :
  $tan 6 0^{\circ} = \frac{F G}{C G}$
  ½ mark
7. (ii)(b) $3 = \frac{80}{C G}$
  ½ mark
8. (ii)(b) $C G = \frac{80}{3}$ m
  ½ mark
9. (ii)(b) Distance travelled by the ball after hitting the tree $= F A = GB = C B - C G = 80 - \frac{80}{3} = 80 (1 - \frac{1}{3})$ m $\approx 33.81$ m
  ½ mark
10. (iii) Speed $= \frac{distance}{time} = \frac{20 ( 3 + 1 )}{2} = 10 (3 + 1)$ m/s
  ½ mark
11. (iii) In m/min: $10 (3 + 1) \times 60 = 600 (3 + 1)$ m/min $\approx 1639.2$ m/min
  ½ mark
Answer: (i) 80 m; (ii)(a) 80(√3 − 1) m; (ii)(b) 80(1 − 1/√3) m; (iii) 600(√3 + 1) m/min
Where the marks are
- (i) tan 45° gives CB = 80 m (1 mark)
- (ii)(a) Uses tan 30° = DE/CE (½ mark)
- (ii)(a) Substitutes values (½ mark)
- (ii)(a) CE = 80√3 m (½ mark)
- (ii)(a) Distance = CE − CB = 80(√3 − 1) m (½ mark)
- (ii)(b) Uses tan 60° = FG/CG (½ mark)
- (ii)(b) Substitutes values (½ mark)
- (ii)(b) CG = 80/√3 m (½ mark)
- (ii)(b) Distance = CB − CG = 80(1 − 1/√3) m (½ mark)
- (iii) Speed in m/s = 10(√3+1) (½ mark)
- (iii) Converts to 600(√3+1) m/min (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.