MathsCBSE Class 10 · Basic · Sample paper, session 2025-26

CBSE Class 10 Mathematics Basic: sample paper 2025-26

The sample question paper CBSE published for the 2025-26 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_2025_26/MathsBasic-SQP.pdf. Questions are transcribed from that PDF; figures are described in words, so check the official paper for them. Questions printed only for visually impaired candidates are left out (7).

Practise a paper like this, timedMy board paper sets a full paper in this exact format from real board questions, with a timer.Get your written answers markedThe Answer Coach marks a photo of your handwritten answer step by step, like the examiner.

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer
C	26-31	Short answer
D	32-35	Long answer
E	36-38	Case study based, sub-parts of 1, 1 and 2 marks

Instructions in brief: No overall choice; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D, and in all 2-mark sub-parts of Section E. Some questions have visually impaired (VI) alternatives printed in lieu of the original question.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The exponent of 3 in the prime factorization of 2025 is
1. (A)1
2. (B)2
3. (C)3
4. (D)4
Try it first, then show our solutionSolution
1. $2025 = 81 \times 25 = 3^{4} \times 5^{2}$ .
  1 mark
2. So the exponent of 3 is 4, option (D).
Answer: 4, option (D)
Where the marks are
- Prime factorises 2025 and reads exponent of 3 as 4 (1 mark)
Q2 · 1 mark · MCQPair of Linear Equations in Two Variables
If $2024 x + 2025 y = 1$ ; $2025 x + 2024 y = - 1$ , then $x - y =$
1. (A)0
2. (B) $- 2$
3. (C)2
4. (D) $- 1$
Try it first, then show our solutionSolution
1. Subtract the first equation from the second: $(2025 - 2024) x + (2024 - 2025) y = - 1 - 1$ .
2. So $x - y = - 2$ , option (B).
  1 mark
Answer: $x - y = - 2$ , option (B)
Where the marks are
- Subtracts the equations to get $x - y = - 2$ (1 mark)
Q3 · 1 mark · MCQPolynomials
The number of polynomials having $- 2$ and $5$ as its zeroes is
1. (A)one
2. (B)two
3. (C)three
4. (D)Infinitely many
Try it first, then show our solutionSolution
1. Any polynomial $k (x + 2) (x - 5) = k (x^{2} - 3 x - 10)$ with $k \neq = 0$ has zeroes $- 2$ and $5$ .
2. Since $k$ can take infinitely many real values, there are infinitely many such polynomials, option (D).
  1 mark
Answer: Infinitely many, option (D)
Where the marks are
- Recognises $k (x + 2) (x - 5)$ works for every non-zero $k$ (1 mark)
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Q4 · 1 mark · MCQQuadratic Equations
Which of the following is not a quadratic equation?
1. (A) $(x + 2)^{2} = 2 (x + 3)$
2. (B) $x^{2} + 3 x = (- 1) (1 - 3 x^{2})$
3. (C) $(x + 2) (x - 1) = x^{2} - 2 x - 3$
4. (D) $x^{3} - x^{2} + 2 x + 1 = (x + 1)^{3}$
Try it first, then show our solutionSolution
1. (A) simplifies to $x^{2} + 2 x - 2 = 0$ , (B) to $2 x^{2} - 3 x - 1 = 0$ and (D) to $4 x^{2} + x = 0$ . All three are quadratic.
2. (C): $x^{2} + x - 2 = x^{2} - 2 x - 3$ gives $3 x + 1 = 0$ . This is linear, so it is not quadratic. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Simplifies (C) to the linear equation $3 x + 1 = 0$ (1 mark)
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Q5 · 1 mark · MCQArithmetic Progressions
The value of $x$ for which $2 x$ , $(x + 10)$ and $(3 x + 2)$ are the three consecutive terms of an AP is
1. (A)6
2. (B) $- 6$
3. (C) $- 2$
4. (D)2
Try it first, then show our solutionSolution
1. For three terms in AP, twice the middle term equals the sum of the other two: $2 (x + 10) = 2 x + 3 x + 2$ .
2. $2 x + 20 = 5 x + 2 \Rightarrow 3 x = 18 \Rightarrow x = 6$ , option (A).
  1 mark
Answer: $x = 6$ , option (A)
Where the marks are
- Uses $2 b = a + c$ and gets $x = 6$ (1 mark)
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Q6 · 1 mark · MCQArithmetic Progressions
If $1 + 2 + 3 + 4 + \dots + 50 = 25 k$ , then $k =$
1. (A)50
2. (B)51
3. (C)49
4. (D)26
Try it first, then show our solutionSolution
1. Using $\frac{n ( n + 1 )}{2}$ : $1 + 2 + \dots + 50 = \frac{50 \times 51}{2} = 1275$ .
2. $25 k = 1275 \Rightarrow k = 51$ , option (B).
  1 mark
Answer: $k = 51$ , option (B)
Where the marks are
- Uses the sum formula and finds $k = 51$ (1 mark)
Q7 · 1 mark · MCQCoordinate Geometry
The distance between the points $(cos 3 0^{\circ}, sin 3 0^{\circ})$ and $(cos 6 0^{\circ}, - sin 6 0^{\circ})$ is
1. (A)0 unit
2. (B) $3$ units
3. (C)1 unit
4. (D) $2$ units
Try it first, then show our solutionSolution
1. The points are $(\frac{3}{2}, \frac{1}{2})$ and $(\frac{1}{2}, - \frac{3}{2})$ .
2. $d^{2} = (\frac{3 - 1}{2})^{2} + (\frac{1 + 3}{2})^{2} = \frac{( 4 - 2 3 ) + ( 4 + 2 3 )}{4} = 2$ .
3. So $d = 2$ units, option (D).
  1 mark
Answer: $2$ units, option (D)
Where the marks are
- Substitutes the trigonometric values in the distance formula and gets $2$ (1 mark)
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Q8 · 1 mark · MCQCoordinate Geometry
The co-ordinates of the point which is mirror image of the point $(- 3, 5)$ about $x$ -axis are
1. (A) $(3, 5)$
2. (B) $(3, - 5)$
3. (C) $(- 3, - 5)$
4. (D) $(- 3, 5)$
Try it first, then show our solutionSolution
1. Reflecting in the x-axis keeps the x-coordinate and changes the sign of the y-coordinate.
2. So the image of $(- 3, 5)$ is $(- 3, - 5)$ , option (C).
  1 mark
Answer: $(- 3, - 5)$ , option (C)
Where the marks are
- Changes the sign of the ordinate only (1 mark)
Q9 · 1 mark · MCQTriangles
If in $△$ ABC and $△$ DEF, $\frac{A B}{E F} = \frac{A C}{D E}$ then they will be similar when
1. (A) $∠ A = ∠ D$
2. (B) $∠ A = ∠ E$
3. (C) $∠ C = ∠ F$
4. (D) $∠ B = ∠ E$
Try it first, then show our solutionSolution
1. By the SAS similarity criterion, the included angles must be equal.
2. The angle included between AB and AC is $∠ A$ . The angle included between EF and DE is $∠ E$ .
3. So we need $∠ A = ∠ E$ , option (B).
  1 mark
Answer: $∠ A = ∠ E$ , option (B)
Where the marks are
- Identifies the included angles for SAS similarity (1 mark)
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Q10 · 1 mark · MCQTriangles
If $△$ ABC $\sim$ $△$ PQR, then perimeter of the triangle PQR (in cm) is
1. (A)12
2. (B)24
3. (C)18
4. (D)20
Figure in the paper: Two right-angled triangles. Triangle ABC: right angle at B, AB = 3 cm (vertical), BC = 4 cm (horizontal), hypotenuse AC = 5 cm. Triangle PQR: right angle at Q, PQ vertical, QR horizontal, hypotenuse PR = 10 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $△ A B C \sim △ P QR$ : $\frac{A C}{P R} = \frac{5}{10} = \frac{1}{2}$ . So $P Q = 6$ cm and $QR = 8$ cm.
2. Perimeter $= 6 + 8 + 10 = 24$ cm, option (B).
  1 mark
Answer: 24 cm, option (B)
Where the marks are
- Uses the ratio of sides to get the sides and perimeter 24 cm (1 mark)
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Q11 · 1 mark · MCQCircles
In the figure given below, radius $r$ of the circle which touches the sides of the triangle is
1. (A)3 cm
2. (B)6 cm
3. (C)7 cm
4. (D)4 cm
Figure in the paper: Right-angled triangle ABC with right angle at C. Side AC (vertical) = 24 cm, base BC (horizontal) = 7 cm, hypotenuse AB = 25 cm. A circle with centre O and radius r touches all three sides at D (on BC), E (on AC) and F (on AB); radii OD, OE and OF are marked r. See it in the official paper.
Try it first, then show our solutionSolution
1. OECD is a square of side $r$ [tangent ⟂ radius, $∠ C = 9 0^{\circ}$ ]. So $C E = C D = r$ .
2. Tangents from an external point are equal: $A F = A E = 24 - r$ and $B F = B D = 7 - r$ .
3. $A B = A F + B F$ : $(24 - r) + (7 - r) = 25 \Rightarrow r = 3$ cm, option (A).
  1 mark
Answer: 3 cm, option (A)
Where the marks are
- Uses equal tangents to get $31 - 2 r = 25$ , so $r = 3$ cm (1 mark)
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Q12 · 1 mark · MCQIntroduction to Trigonometry
Which one of the following is not equal to Unity?
1. (A) $sin^{2} x + cos^{2} x$
2. (B) $cot^{2} x - cosec^{2} x$
3. (C) $sec^{2} x - tan^{2} x$
4. (D) $tan x \cdot cot x$
Try it first, then show our solutionSolution
1. $1 + cot^{2} x = cosec^{2} x$ , so $cot^{2} x - cosec^{2} x = - 1$ .
2. Options (A), (C) and (D) all equal 1. So the answer is (B).
  1 mark
Answer: Option (B)
Where the marks are
- Uses the identity to get $- 1$ (1 mark)
Q13 · 1 mark · MCQStatistics
Consider the following frequency distribution
Class 0 – 5 5 – 10 10 – 15 15 – 20 20 – 25
Frequency 11 12 13 9 11
The upper limit of median class is
1. (A)10
2. (B)13
3. (C)15
4. (D)20
Try it first, then show our solutionSolution
1. $N = 56$ , so $\frac{N}{2} = 28$ . Cumulative frequencies are 11, 23, 36, 45, 56.
2. 28 first falls in the cumulative frequency 36, so the median class is 10–15. Its upper limit is 15, option (C).
  1 mark
Answer: 15, option (C)
Where the marks are
- Finds the median class 10–15 from cumulative frequencies (1 mark)
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Q14 · 1 mark · MCQStatistics
Let empirical relationship between the three measures of central tendency be $a (Median) = Mode + b (Mean)$ , then $(2 b + 3 a) =$
1. (A)11
2. (B)12
3. (C)13
4. (D)14
Try it first, then show our solutionSolution
1. The empirical relation is $3 Median = Mode + 2 Mean$ . So $a = 3$ and $b = 2$ .
2. $2 b + 3 a = 4 + 9 = 13$ , option (C).
  1 mark
Answer: 13, option (C)
Where the marks are
- Uses the empirical relation to get $a = 3$ , $b = 2$ and the value 13 (1 mark)
Q15 · 1 mark · MCQCircles
From an external point Q, the length of tangent to a circle is 12 cm and the distance of Q from the centre of circle is 13 cm. The radius of circle (in cm) is
1. (A)10
2. (B)5
3. (C)12
4. (D)7
Try it first, then show our solutionSolution
1. The radius is perpendicular to the tangent at the point of contact [tangent ⟂ radius].
2. $r = 1 3^{2} - 1 2^{2} = 25 = 5$ cm, option (B).
  1 mark
Answer: 5 cm, option (B)
Where the marks are
- Applies Pythagoras theorem to get $r = 5$ (1 mark)
Q16 · 1 mark · MCQCircles
In the given figure, PA is a tangent from an external point P to a circle with centre O and diameter AB. If $∠ P O B = 11 5^{\circ}$ , then measure of $∠ A P O$ is
1. (A) $2 5^{\circ}$
2. (B) $3 0^{\circ}$
3. (C) $2 0^{\circ}$
4. (D) $6 5^{\circ}$
Figure in the paper: A circle with centre O and diameter AB (A at top-left, B at bottom-right). External point P lies to the lower-left; PA is a tangent touching the circle at A. Segment PO is drawn to the centre; the angle POB at O is marked as 115°. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ P A O = 9 0^{\circ}$ [tangent ⟂ radius].
2. $∠ P O B$ is an exterior angle of $△ P A O$ , so $∠ A P O = 11 5^{\circ} - 9 0^{\circ} = 2 5^{\circ}$ , option (A).
  1 mark
Answer: 25°, option (A)
Where the marks are
- Uses the right angle and the exterior angle property to get 25° (1 mark)
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Q17 · 1 mark · MCQAreas Related to Circles
The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is
1. (A)3 : 4
2. (B)4 : 3
3. (C)9 : 16
4. (D)16 : 9
Try it first, then show our solutionSolution
1. Circumference is proportional to the radius, so $r_{1} : r_{2} = 3 : 4$ .
2. Areas are in the ratio $r_{1}^{2} : r_{2}^{2} = 9 : 16$ , option (C).
  1 mark
Answer: 9 : 16, option (C)
Where the marks are
- Squares the ratio of radii (1 mark)
Q18 · 1 mark · MCQProbability
An event is most unlikely to happen. Its probability is
1. (A)0.0001
2. (B)0.001
3. (C)0.01
4. (D)0.1
Try it first, then show our solutionSolution
1. A most unlikely event has a probability very close to 0.
2. The smallest value among the options is 0.0001, option (A).
  1 mark
Answer: 0.0001, option (A)
Where the marks are
- Chooses the probability closest to 0 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedTriangles
Each of the following questions contains two statements i.e., ASSERTION and REASON, and has following four choices. Only one of which is the correct answer.
ASSERTION (A): Line joining the midpoints of two sides of triangle is parallel to the third side.
REASON (R): If a line divides two sides of a triangle in the same ratio then it is parallel to the third side.
1. (A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
2. (B)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
3. (C)Assertion (A) is true but reason (R) is false.
4. (D)Assertion (A) is false but reason (R) is true.
Try it first, then show our solutionSolution
1. R is true: it is the converse of the Basic Proportionality Theorem.
2. The midpoints divide both sides in the ratio 1 : 1, so by R the line joining them is parallel to the third side. Hence A is true and R explains it, option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises that both statements are true and that R explains A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedProbability
Each of the following questions contains two statements i.e., ASSERTION and REASON, and has following four choices. Only one of which is the correct answer.
ASSERTION (A): Two coins are tossed simultaneously. Possible outcomes are two heads, one head and one tail, two tails. Hence, the probability of getting two heads is $\frac{1}{3}$ .
REASON (R): Probabilities of 'equally likely' outcomes of an experiment are always equal.
1. (A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
2. (B)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
3. (C)Assertion (A) is true but reason (R) is false.
4. (D)Assertion (A) is false but reason (R) is true.
Try it first, then show our solutionSolution
1. The outcomes HH, HT, TH and TT are equally likely. So $P (two heads) = \frac{1}{4}$ , not $\frac{1}{3}$ , and A is false.
2. R is a true statement. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Shows $P (two heads) = \frac{1}{4}$ , so A is false while R is true (1 mark)

Class	0 – 5	5 – 10	10 – 15	15 – 20	20 – 25
Frequency	11	12	13	9	11

Section B

Very short answer

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerReal Numbers
Show that the number $2 \times 5 \times 7 \times 11 + 11 \times 13$ is a composite number.
Try it first, then show our solutionSolution
1. Take 11 common: $2 \times 5 \times 7 \times 11 + 11 \times 13 = 11 \times (2 \times 5 \times 7 + 13) = 11 \times (70 + 13) = 11 \times 83$
  1 mark
2. The number equals $11 \times 83 = 913$ . It has factors 11 and 83 besides 1 and itself, so it is a composite number.
  1 mark
Answer: $2 \times 5 \times 7 \times 11 + 11 \times 13 = 11 \times 83$ , so it is composite.
Where the marks are
- Takes 11 common and writes the number as $11 \times 83$ (1 mark)
- Concludes it is composite because it has factors other than 1 and itself (1 mark)
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OR
Q21(B) · 2 marks · Very short answerReal Numbers
Find the smallest number which is divisible by both 306 and 657.
Try it first, then show our solutionSolution
1. The smallest number divisible by both 306 and 657 is their LCM.
  ½ mark
2. Prime factorisation: $306 = 2 \times 3^{2} \times 17$ and $657 = 3^{2} \times 73$
  1 mark
3. LCM $= 2 \times 3^{2} \times 17 \times 73 = 22338$ . So the required number is 22338.
  ½ mark
Answer: 22338
Where the marks are
- Recognises that the required number is the LCM (½ mark)
- Correct prime factorisations of 306 and 657 (1 mark)
- LCM = 22338 (½ mark)
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Q22 · 2 marks · Very short answerCircles
Find the radius of the circle with centre at origin, if line $l$ given by $x + y = 5$ is tangent to the circle at point P.
Figure in the paper: A circle with centre C(0, 0). A horizontal line $l$ touches the circle at the point P(3, a) at the bottom of the circle; a dashed radius joins C to P. See it in the official paper.
Try it first, then show our solutionSolution
1. P(3, a) lies on the line $x + y = 5$ , so $3 + a = 5 \Rightarrow a = 2$ . Hence P is (3, 2).
  1 mark
2. Radius = CP $= (3 - 0)^{2} + (2 - 0)^{2} = 9 + 4 = 13$ units.
  1 mark
Answer: Radius = $13$ units
Where the marks are
- Substitutes P in the line to get $a = 2$ (1 mark)
- Uses distance formula to get radius $13$ units (1 mark)
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Q23 · 2 marks · Very short answerPolynomials
If the zeroes of the quadratic polynomial $x^{2} + (a + 1) x + b$ are 2 and $- 3$ , then find the values of $a$ and $b$ .
Try it first, then show our solutionSolution
1. For $x^{2} + (a + 1) x + b$ : sum of zeroes $= - \frac{a + 1}{1}$ . So $2 + (- 3) = - (a + 1) \Rightarrow - 1 = - (a + 1) \Rightarrow a = 0$
  1 mark
2. Product of zeroes $= \frac{b}{1}$ . So $2 \times (- 3) = b \Rightarrow b = - 6$ . Hence $a = 0, b = - 6$ .
  1 mark
Answer: $a = 0, b = - 6$
Where the marks are
- Uses sum of zeroes to get $a = 0$ (1 mark)
- Uses product of zeroes to get $b = - 6$ (1 mark)
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Q24 · 2 marks · Very short answerQuadratic Equations
Find the nature of roots of the quadratic equation $x^{2} + 4 x - 32 = 0$ .
Try it first, then show our solutionSolution
1. Here $a = 1, b = 4, c = - 32$ . Discriminant $D = b^{2} - 4 a c = 16 - 4 (1) (- 32) = 16 + 122$
  1 mark
2. Since $D = 16 + 122 > 0$ , the equation has two distinct real roots.
  1 mark
Answer: $D = 16 + 122 > 0$ ; roots are real and distinct.
Where the marks are
- Correct discriminant $16 + 122$ (1 mark)
- Concludes roots are real and distinct since $D > 0$ (1 mark)
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Answer one of these 2 alternatives.
Q25(A) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $2 sin 3 0^{\circ} tan 6 0^{\circ} - 3 cos^{2} 6 0^{\circ} sec^{2} 3 0^{\circ}$
Try it first, then show our solutionSolution
1. Substitute $sin 3 0^{\circ} = \frac{1}{2}, tan 6 0^{\circ} = 3, cos 6 0^{\circ} = \frac{1}{2}, sec 3 0^{\circ} = \frac{2}{3}$ : $2 (\frac{1}{2}) (3) - 3 (\frac{1}{2})^{2} (\frac{2}{3})^{2} = 3 - 3 \cdot \frac{1}{4} \cdot \frac{4}{3}$
  1½ marks
2. $= 3 - 1$
  ½ mark
Answer: $3 - 1$
Where the marks are
- Substitutes correct values of all four trigonometric ratios (1½ marks)
- Simplifies to $3 - 1$ (½ mark)
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OR
Q25(B) · 2 marks · Very short answerIntroduction to Trigonometry
If $sin x = \frac{7}{25}$ , where $x$ is an acute angle, then find the value of $sin x \cdot cos x (tan x + cot x)$ .
Try it first, then show our solutionSolution
1. $sin x cos x (tan x + cot x) = sin x cos x (\frac{s i n x}{c o s x} + \frac{c o s x}{s i n x})$
  ½ mark
2. $= sin x cos x \cdot \frac{s i n ^{2} x + c o s ^{2} x}{s i n x c o s x} = sin^{2} x + cos^{2} x = 1$ . The value is 1 for every acute $x$ , so with $sin x = \frac{7}{25}$ it is 1. (Check: $cos x = \frac{24}{25}$ , $\frac{7}{25} \cdot \frac{24}{25} (\frac{7}{24} + \frac{24}{7}) = \frac{168}{625} \cdot \frac{625}{168} = 1$ .)
  1½ marks
Answer: 1
Where the marks are
- Writes $tan x$ and $cot x$ in terms of $sin x$ and $cos x$ (½ mark)
- Simplifies using $sin^{2} x + cos^{2} x = 1$ to get 1 (1½ marks)
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Section C

Short answer

Q26 · 3 marks · Short answerReal Numbers
Show that $2 - 5$ is an irrational number.
Try it first, then show our solutionSolution
1. Suppose, to the contrary, that $2 - 5 = x$ , where $x$ is rational ( $x \neq = 0$ ). Then $2 = x + 5$ ; squaring, $2 = x^{2} + 5 + 25 x$ .
  1 mark
2. So $25 x = - x^{2} - 3 \Rightarrow 5 = \frac{- x ^{2} - 3}{2 x}$ ...(1)
  1 mark
3. Since $x$ is rational, $\frac{- x ^{2} - 3}{2 x}$ is rational, but $5$ is irrational (square root of a prime). A rational number cannot equal an irrational number, so our assumption is wrong. Hence $2 - 5$ is irrational.
  1 mark
Answer: $2 - 5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $2 - 5$ rational and squares correctly (1 mark)
- Expresses $5$ as a rational expression in $x$ (1 mark)
- Gives the contradiction and concludes (1 mark)
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Answer one of these 2 alternatives.
Q27(A) · 3 marks · Short answerStatistics
The frequency distribution table of agriculture holdings in a village is given below:
Area of land (in hectares) 1 – 3 3 – 5 5 – 7 7 – 9 9 – 11 11 – 13
No. of families 20 45 80 55 40 12
Find the modal agriculture holdings of the village.
Try it first, then show our solutionSolution
1. Maximum frequency is 80, so modal class = 5 – 7. Here $l = 5$ , $h = 2$ , $f_{1} = 80$ , $f_{0} = 45$ , $f_{2} = 55$ .
  1 mark
2. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) h = 5 + \frac{80 - 45}{160 - 45 - 55} \times 2 = 5 + \frac{35}{60} \times 2 = 5 + 1.17 \approx 6.17$
  2 marks
3. Hence, the modal agriculture holding of the village is about 6.17 hectares.
Answer: Mode ≈ 6.17 hectares
Where the marks are
- Identifies modal class 5–7 with correct $l, h, f_{0}, f_{1}, f_{2}$ (1 mark)
- Substitutes in mode formula and gets ≈ 6.17 hectares (2 marks)
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OR
Q27(B) · 3 marks · Short answerStatistics
If the mean of the following distribution is 54, find the value of $p$ .
Class Interval 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
Frequency 7 $p$ 10 9 13
Try it first, then show our solutionSolution
1. Take assumed mean $A = 30$ , $h = 20$ , $u_{i} = \frac{x _{i} - 30}{20}$ .
  Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
  0–20 7 10 −1 −7
  20–40 $p$ 30 0 0
  40–60 10 50 1 10
  60–80 9 70 2 18
  80–100 13 90 3 39
  Total $39 + p$ 60
  2 marks
2. Mean $= A + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h \Rightarrow 54 = 30 + \frac{60}{39 + p} \times 20 \Rightarrow \frac{1200}{39 + p} = 24 \Rightarrow 39 + p = 50 \Rightarrow p = 11$
  1 mark
Answer: $p = 11$
Where the marks are
- Correct table with $\sum f_{i} = 39 + p$ and $\sum f_{i} u_{i} = 60$ (2 marks)
- Forms the mean equation and gets $p = 11$ (1 mark)
Open Q27(B) on its own page
Q28 · 3 marks · Short answerCircles
A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the given figure. Show that $\frac{A B + C D}{A D + B C} = 1$
Figure in the paper: Quadrilateral ABCD circumscribing a circle. Vertices: D top-left, C top-right, B bottom-right, A bottom-left. The circle touches the sides at R (on DC), Q (on CB), P (on AB) and S (on AD). See it in the official paper.
Try it first, then show our solutionSolution
1. Lengths of tangents drawn from an external point to a circle are equal. So $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ .
  1½ marks
2. Adding: $(A P + P B) + (C R + R D) = (A S + B Q) + (C Q + D S) = (A S + D S) + (B Q + C Q)$ , i.e. $A B + C D = A D + B C$ .
  1 mark
3. Hence $\frac{A B + C D}{A D + B C} = 1$ .
  ½ mark
Answer: $A B + C D = A D + B C$ , so $\frac{A B + C D}{A D + B C} = 1$ .
Where the marks are
- Writes the four pairs of equal tangents with reason (1½ marks)
- Adds to obtain $A B + C D = A D + B C$ (1 mark)
- Concludes the ratio equals 1 (½ mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
On a particular day, 50000 people attended a Cricket Test Match between India and Australia in Sydney Cricket Ground. Let $x$ be the number of adults attended the cricket match and $y$ be the number of children attended the cricket match. Cost of an adult ticket was ₹1000 while cost of a child ticket was ₹200. On that day Revenue earned by selling all 50,000 tickets, was ₹4,20,00,000. Find how many adults and how many children attended the cricket match?
Try it first, then show our solutionSolution
1. Revenue: $1000 x + 200 y = 42000000 \Rightarrow 5 x + y = 210000$ ...(1)
  1 mark
2. Total people: $x + y = 50000$ ...(2)
  ½ mark
3. (1) − (2): $4 x = 160000 \Rightarrow x = 40000$
  1 mark
4. From (2): $y = 50000 - 40000 = 10000$ . So 40000 adults and 10000 children attended.
  ½ mark
Answer: Adults = 40000, Children = 10000
Where the marks are
- Forms revenue equation $5 x + y = 210000$ (1 mark)
- Forms $x + y = 50000$ (½ mark)
- Solves to get $x = 40000$ (1 mark)
- Gets $y = 10000$ (½ mark)
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve for $x$ and $y$ , graphically: $2 x + y = 6$ ; $x + y = 5$
Try it first, then show our solutionSolution
1. For $2 x + y = 6$ , i.e. $y = 6 - 2 x$ : points (0, 6), (2, 2), (3, 0). For $x + y = 5$ , i.e. $y = 5 - x$ : points (0, 5), (2, 3), (5, 0).
  1 mark
2. Plot these points on a graph paper and draw the two lines through them; the lines intersect at the point (1, 4).
  1 mark
3. Hence the solution is $x = 1, y = 4$ . (Check: $2 (1) + 4 = 6$ , $1 + 4 = 5$ .)
  1 mark
Answer: $x = 1, y = 4$
Where the marks are
- Correct tables of values for both equations (1 mark)
- Correct graph of both lines (1 mark)
- Reads intersection point (1, 4) as the solution (1 mark)
Open Q29(B) on its own page
Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $(sin x - cos x + 1) \cdot (sec x - tan x) = (sin x + cos x - 1)$
Try it first, then show our solutionSolution
1. LHS $= (sin x - cos x + 1) (\frac{1}{c o s x} - \frac{s i n x}{c o s x}) = (sin x - cos x + 1) \cdot \frac{1 - s i n x}{c o s x}$
  1 mark
2. $= (1 + sin x) \frac{1 - s i n x}{c o s x} - cos x \cdot \frac{1 - s i n x}{c o s x} = \frac{1 - s i n ^{2} x}{c o s x} - (1 - sin x)$
  1 mark
3. $= \frac{c o s ^{2} x}{c o s x} - 1 + sin x = sin x + cos x - 1 =$ RHS. Hence proved.
  1 mark
Answer: LHS = $sin x + cos x - 1$ = RHS (proved)
Where the marks are
- Writes $sec x - tan x = \frac{1 - s i n x}{c o s x}$ (1 mark)
- Splits and simplifies to $\frac{1 - s i n ^{2} x}{c o s x} - (1 - sin x)$ (1 mark)
- Uses $1 - sin^{2} x = cos^{2} x$ to reach RHS (1 mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerArithmetic Progressions
The sum of first $n$ terms of an AP is $5 n^{2} - n$ . Find the $n^{t h}$ term of the AP.
Try it first, then show our solutionSolution
1. $n$ th term: $a_{n} = S_{n} - S_{n - 1}$
  ½ mark
2. $a_{n} = [5 n^{2} - n] - [5 (n - 1)^{2} - (n - 1)] = 5 [n^{2} - (n - 1)^{2}] - [n - (n - 1)]$
  1 mark
3. $= 5 (2 n - 1) - 1 = 10 n - 6$ . So $a_{n} = 10 n - 6$ . (Check: $a_{1} = S_{1} = 4$ .)
  1½ marks
Answer: $a_{n} = 10 n - 6$
Where the marks are
- Uses $a_{n} = S_{n} - S_{n - 1}$ (½ mark)
- Substitutes $S_{n - 1}$ correctly (1 mark)
- Simplifies to $a_{n} = 10 n - 6$ (1½ marks)
Open Q31 on its own page

Area of land (in hectares)	1 – 3	3 – 5	5 – 7	7 – 9	9 – 11	11 – 13
No. of families	20	45	80	55	40	12

Class Interval	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Frequency	7	$p$	10	9	13

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
0–20	7	10	−1	−7
20–40	$p$	30	0	0
40–60	10	50	1	10
60–80	9	70	2	18
80–100	13	90	3	39
Total	$39 + p$			60

Section D

Long answer

Q32 · 5 marks · Long answerTriangles
Prove that a line drawn parallel to one side of a triangle intersecting other two sides in distinct points, divides the other two sides in the same ratio.
Try it first, then show our solutionSolution
1. Given: In $△ A B C$ , a line $l ∥ B C$ meets AB at D and AC at E.
  ½ mark
2. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  ½ mark
3. Construction: Join BE and CD. Draw $D M ⊥ A C$ and $E N ⊥ A B$ .
  ½ mark
4. Figure: Triangle ABC with D on AB, E on AC, DE ∥ BC; dashed segments BE and CD; N the foot of perpendicular from E on AB, M the foot of perpendicular from D on AC.
  ½ mark
5. Proof: $\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{\frac{1}{2} \cdot A D \cdot E N}{\frac{1}{2} \cdot D B \cdot E N} = \frac{A D}{D B}$ ...(1)
  1 mark
6. $\frac{a r ( △ A D E )}{a r ( △ C E D )} = \frac{\frac{1}{2} \cdot A E \cdot D M}{\frac{1}{2} \cdot E C \cdot D M} = \frac{A E}{E C}$ ...(2)
  ½ mark
7. $a r (△ B D E) = a r (△ C E D)$ ...(3) [triangles on the same base DE and between the same parallels DE and BC]
  ½ mark
8. From (1), (2) and (3): $\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{a r ( △ A D E )}{a r ( △ C E D )} \Rightarrow \frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: $\frac{A D}{D B} = \frac{A E}{E C}$ (Basic Proportionality Theorem proved)
Where the marks are
- Given (½ mark)
- To prove (½ mark)
- Construction (½ mark)
- Correct figure (½ mark)
- Area ratio giving $\frac{A D}{D B}$ (1 mark)
- Area ratio giving $\frac{A E}{E C}$ (½ mark)
- Equal areas of $△ B D E$ and $△ C E D$ with reason (½ mark)
- Combining to conclude (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerQuadratic Equations
The numerator of a fraction is 3 less than its denominator. If 2 is added to both of its numerator and denominator then the sum of the new fraction and original fraction is $\frac{29}{20}$ . Find the original fraction.
Try it first, then show our solutionSolution
1. Let the denominator be $x$ . Then numerator $= x - 3$ and the fraction is $\frac{x - 3}{x}$ .
  1 mark
2. New fraction $= \frac{x - 1}{x + 2}$ . Given: $\frac{x - 1}{x + 2} + \frac{x - 3}{x} = \frac{29}{20}$
  1 mark
3. $\frac{x ( x - 1 ) + ( x - 3 ) ( x + 2 )}{x ( x + 2 )} = \frac{29}{20} \Rightarrow 20 (2 x^{2} - 2 x - 6) = 29 (x^{2} + 2 x) \Rightarrow 11 x^{2} - 98 x - 120 = 0$
  1½ marks
4. $11 x^{2} - 110 x + 12 x - 120 = 0 \Rightarrow (x - 10) (11 x + 12) = 0 \Rightarrow x = 10$ or $x = - \frac{12}{11}$
  1 mark
5. $x = - \frac{12}{11}$ is rejected (denominator must be a natural number), so $x = 10$ and the fraction is $\frac{7}{10}$ .
  ½ mark
Answer: $\frac{7}{10}$
Where the marks are
- Assumes fraction as $\frac{x - 3}{x}$ (1 mark)
- Forms the equation $\frac{x - 1}{x + 2} + \frac{x - 3}{x} = \frac{29}{20}$ (1 mark)
- Simplifies to $11 x^{2} - 98 x - 120 = 0$ (1½ marks)
- Solves to get $x = 10$ or $- \frac{12}{11}$ (1 mark)
- Rejects negative value and states fraction $\frac{7}{10}$ (½ mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerQuadratic Equations
A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hr, it takes 2 hours less in the journey. Find the original speed of the train.
Try it first, then show our solutionSolution
1. Let the original speed be $x$ km/hr. Time taken $= \frac{300}{x}$ hr.
  ½ mark
2. New speed $= (x + 5)$ km/hr; new time $= \frac{300}{x + 5}$ hr.
  ½ mark
3. Given: $\frac{300}{x} - \frac{300}{x + 5} = 2$
  1 mark
4. $\frac{300 ( x + 5 ) - 300 x}{x ( x + 5 )} = 2 \Rightarrow \frac{1500}{x ^{2} + 5 x} = 2 \Rightarrow x^{2} + 5 x - 750 = 0$
  1½ marks
5. $x^{2} + 30 x - 25 x - 750 = 0 \Rightarrow (x + 30) (x - 25) = 0 \Rightarrow x = 25$ or $x = - 30$
  1 mark
6. Speed cannot be negative, so $x = 25$ . The original speed of the train is 25 km/hr.
  ½ mark
Answer: 25 km/hr
Where the marks are
- Original time $\frac{300}{x}$ (½ mark)
- New time $\frac{300}{x + 5}$ (½ mark)
- Forms equation $\frac{300}{x} - \frac{300}{x + 5} = 2$ (1 mark)
- Reduces to $x^{2} + 5 x - 750 = 0$ (1½ marks)
- Factorises to get $x = 25$ or $- 30$ (1 mark)
- Rejects $- 30$ , speed 25 km/hr (½ mark)
Open Q33(B) on its own page
Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerSome Applications of Trigonometry
The angle of elevation of the top of a chimney from the foot of a tower is $6 0^{\circ}$ and the angle of depression of the foot of the chimney from the top of the tower is $3 0^{\circ}$ . If the height of the tower is 40 meters, find the height of the chimney. Also, find the length of the wire tied from the top of the chimney to the top of tower.
Try it first, then show our solutionSolution
1. Figure: Let AB be the chimney (B its foot) and CD the tower (C its foot), CD = 40 m, BC horizontal. $∠ A C B = 6 0^{\circ}$ (elevation of A from C) and $∠ D B C = 3 0^{\circ}$ (equal to the angle of depression of B from D, alternate angles). Draw DE ⟂ AB, so EBCD is a rectangle: EB = 40 m, ED = BC.
  1 mark
2. In right $△ D C B$ : $tan 3 0^{\circ} = \frac{C D}{B C} \Rightarrow \frac{1}{3} = \frac{40}{B C} \Rightarrow B C = 403$ m. In right $△ A B C$ : $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow 3 = \frac{A B}{40 3} \Rightarrow A B = 120$ m. Height of chimney = 120 m.
  1½ marks
3. AE $= A B - E B = 120 - 40 = 80$ m and ED $= B C = 403$ m. In right $△ A E D$ : $A D = A E^{2} + E D^{2} = 6400 + 4800 = 11200 = 407$ m.
  1½ marks
4. Hence the chimney is 120 m high and the wire from the top of the chimney to the top of the tower is $407$ m (≈ 105.83 m) long.
  1 mark
Answer: Height of chimney = 120 m; length of wire = $407$ m (≈ 105.83 m)
Where the marks are
- Correct labelled figure (1 mark)
- Finds BC = $403$ m and AB = 120 m (1½ marks)
- Finds AE = 80 m, ED = $403$ m and AD = $407$ m by Pythagoras (1½ marks)
- Correct final statements (1 mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerSome Applications of Trigonometry
The angles of depression of the top and bottom of a 50m high building from the top of a tower are $4 5^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: Let $E C$ be the tower, with $E$ its top and $C$ its foot. Let $A B = 50$ m be the building, with $A$ its top and $B$ its foot. Draw $A D ⊥ E C$ , so $D C = A B = 50$ m. Let $E D = h$ m and $B C = A D = x$ m. From $E$ , the angle of depression of $A$ is $4 5^{\circ}$ , so $∠ E A D = 4 5^{\circ}$ . The angle of depression of $B$ is $6 0^{\circ}$ , so $∠ E B C = 6 0^{\circ}$ [alternate angles].
  1 mark
2. In right $△ E D A$ : $tan 4 5^{\circ} = \frac{E D}{A D} = \frac{h}{x}$ , so $1 = \frac{h}{x}$ and hence $h = x$ .
  1½ marks
3. In right $△ E C B$ : $tan 6 0^{\circ} = \frac{E C}{B C} = \frac{h + 50}{x}$ . Substituting $x = h$ gives $3 h = h + 50$ .
  So $h (3 - 1) = 50$ , giving $h = \frac{50}{3 - 1} = \frac{50 ( 3 + 1 )}{2} = 25 (3 + 1)$ m.
  1½ marks
4. Using $3 = 1.73$ : $h = 25 \times 2.73 = 68.25$ m. So the horizontal distance between the tower and the building is $x = h = 68.25$ m.
  ½ mark
5. Height of the tower $= E C = h + 50 = 68.25 + 50 = 118.25$ m.
  ½ mark
Answer: Height of tower = 118.25 m; horizontal distance = 68.25 m
Where the marks are
- Correct labelled figure showing the tower, the building and both angles of depression (1 mark)
- Uses $tan 4 5^{\circ}$ in $△ E D A$ to get $h = x$ (1½ marks)
- Uses $tan 6 0^{\circ}$ in $△ E C B$ to get $h = 25 (3 + 1)$ m (1½ marks)
- Horizontal distance = 68.25 m (½ mark)
- Height of tower = 118.25 m (½ mark)
Open Q34(B) on its own page
Q35 · 5 marks · Long answerSurface Areas and Volumes
A solid toy is in the form of a hemisphere surmounted by a right circular cone of height 2cm and diameter of base 4cm. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. [Use $π = 3.14$ ]
Try it first, then show our solutionSolution
1. Figure: A hemisphere of radius $r = 2$ cm, surmounted by a cone of base radius 2 cm and height $h = 2$ cm, all enclosed in a cylinder. The cylinder has radius $r = 2$ cm and height $H = r + h = 2 + 2 = 4$ cm.
  1 mark
2. Volume of toy $=$ volume of hemisphere $+$ volume of cone
  $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (2 r + h) = \frac{1}{3} \times 3.14 \times 4 \times (4 + 2) = 8 \times 3.14 = 25.12 cm^{3}$
  2 marks
3. Volume of circumscribing cylinder $= π r^{2} H = 3.14 \times 4 \times 4 = 50.24 cm^{3}$
  1 mark
4. Required difference $= 50.24 - 25.12 = 25.12 cm^{3}$
  1 mark
Answer: 25.12 cm³
Where the marks are
- Correct figure with $r = 2$ cm, cone height 2 cm and cylinder height 4 cm (1 mark)
- Volume of toy (hemisphere + cone) = 25.12 cm³ (2 marks)
- Volume of cylinder = 50.24 cm³ (1 mark)
- Difference = 25.12 cm³ (1 mark)
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Section E

Case study based, sub-parts of 1, 1 and 2 marks

Q36 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Carpooling is the sharing of car journeys so that more than one person travels in a car, and prevents the need for others to have to drive to a location themselves. By having more people using one vehicle, carpooling reduces each person's travel costs such as: fuel costs, tolls, and the stress of driving. Carpooling is also a more environmentally friendly and sustainable way to travel as sharing journeys reduces air pollution, carbon emissions, traffic congestion on the roads, and the need for parking spaces.
Three friends Amar, Bhavin and Chetanya live in societies represented by the points A(4,5), B(6,2) and C(2,6) respectively. They all work in offices located in a same building represented by the point O(0,0). Since they all go to same building every day, they decided to do carpooling to save money on petrol. Based on the above information, answer the following questions.
1. (i)
  What is the distance between B and C?
  1 mark
2. (ii)
  If Bhavin and Chetanya planned to meet at a club situated at the mid-point of the line joining the points B and C, find the coordinates of this point.
  1 mark
3. (iii)
  (A) Which society is farthest from the office? Also find its distance from the office.
  2 marks
4. OR
  (iii)
  (B) Out of B and C which society is nearer to A? Also find their distances.
  2 marks
Figure in the paper: Coordinate graph with axes marked 5, 10, 15 on both axes, origin O. Points plotted: C(2,6) in blue, A(4,5) in red, B(6,2) in green, and O(0,0). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $B C = (2 - 6)^{2} + (6 - 2)^{2} = 16 + 16 = 32 = 42$ units
  1 mark
2. (ii) Mid-point of $B C = (\frac{6 + 2}{2}, \frac{2 + 6}{2}) = (4, 4)$
  1 mark
3. (iii)(A) Using the distance formula from $O (0, 0)$ :
  $O A = 4^{2} + 5^{2} = 41$ units
  $O B = 6^{2} + 2^{2} = 40$ units
  $O C = 2^{2} + 6^{2} = 40$ units
  1½ marks
4. (iii)(A) Since $41 > 40$ , society A is the farthest from the office, at a distance of $41$ units.
  ½ mark
5. (iii)(B) $A B = (6 - 4)^{2} + (2 - 5)^{2} = 4 + 9 = 13$ units
  $A C = (2 - 4)^{2} + (6 - 5)^{2} = 4 + 1 = 5$ units
  1½ marks
6. (iii)(B) Since $5 < 13$ , society C is nearer to A.
  ½ mark
Answer: (i) $42$ units; (ii) (4, 4); (iii)(A) Society A, $41$ units; (iii)(B) C is nearer, with AB = $13$ units and AC = $5$ units
Where the marks are
- (i) BC = $42$ units (1 mark)
- (ii) Mid-point (4, 4) (1 mark)
- (iii)(A) OA = $41$ , OB = OC = $40$ units (1½ marks); concludes A is farthest (½ mark)
- (iii)(B) AB = $13$ , AC = $5$ units (1½ marks); concludes C is nearer (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
A water sprinkler is a device used to irrigate agricultural crops, lawns, landscapes, golf courses, and other areas. Water sprinklers can be used for residential, industrial, and agricultural usage.
A water sprinkler is set to shoot a stream of water a distance of 21 m and rotate through an angle which is equal to complementary angle of $1 0^{\circ}$ .
1. (i)
  What is the area of sector in terms of arc length?
  1 mark
2. (ii)
  What is the area of the watered region (in terms of $π$ )?
  1 mark
3. (iii)
  (A) If the radius( $r$ ) changes to 28m, find the angle $θ$ so that the area of the watered region remains the same.
  2 marks
4. OR
  (iii)
  (B) If the radius( $r$ ) is increased from 21m to 28m and the angle remains the same, what is the increase in the area of the watered region?
  2 marks
Figure in the paper: A sector of a circle with the sprinkler at the vertex on the right. The arc on the left is labelled $l$ , the bounding radii are straight lines, the central angle is marked $θ$ , and the shaded region is labelled 'Watered region'. The vertex is labelled 'Water sprinkler'. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) For a sector with arc length $l$ and radius $r$ : area of sector $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{1}{2} \times l \times r$ , i.e. $\frac{l \times r}{2}$ .
  1 mark
2. (ii) The angle of rotation is the complement of $1 0^{\circ}$ , so $θ = 9 0^{\circ} - 1 0^{\circ} = 8 0^{\circ}$ , with $r = 21$ m.
  Area watered $= \frac{80}{360} \times π \times 2 1^{2} = \frac{2}{9} \times 441 π = 98 π m^{2}$
  1 mark
3. (iii)(A) For the area to stay the same with $r = 28$ m: $\frac{θ}{360} \times π \times 2 8^{2} = 98 π$
  1 mark
4. (iii)(A) So $θ = \frac{98 \times 360}{784} = 4 5^{\circ}$ .
  1 mark
5. (iii)(B) Increase in area $= \frac{80}{360} π (2 8^{2} - 2 1^{2}) = \frac{2}{9} π (784 - 441) = \frac{2}{9} \times 343 π$
  1 mark
6. (iii)(B) $= \frac{686}{9} \times \frac{22}{7} = \frac{2156}{9} \approx 239.56 m^{2}$ $(i.e. \frac{686 π}{9} m^{2})$
  1 mark
Answer: (i) Area = $\frac{l \times r}{2}$ ; (ii) $98 π$ m²; (iii)(A) $θ = 4 5^{\circ}$ ; (iii)(B) ≈ 239.56 m² ( $\frac{686 π}{9}$ m²)
Where the marks are
- (i) Area of sector = ½ × arc length × radius (1 mark)
- (ii) Uses $θ = 8 0^{\circ}$ to get area $98 π$ m² (1 mark)
- (iii)(A) Equates $\frac{θ}{360} π (28)^{2}$ to $98 π$ (1 mark); gets $θ = 4 5^{\circ}$ (1 mark)
- (iii)(B) Writes increase as $\frac{80}{360} π (784 - 441)$ (1 mark); gets ≈ 239.56 m² (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedProbability
One of four main blood types can be found in a human body. They are known as A, B, AB and O. Each blood type can be further classified as either a Rhesus positive (+) or Rhesus negative (−). For example, a possible combination is blood type O and Rhesus negative which is written as O $^{-}$ .
The data below shows the distribution of the blood types and Rhesus types of given blood type for a Blood Donation Center recorded (in percentages) for the year 2023.
Blood Group Rhesus Factor Number of Persons (in %)
O O $^{-}$ $x$
O O $^{+}$ 30
A A $^{-}$ 8
A A $^{+}$ 24
B B $^{-}$ 6
B B $^{+}$ 18
AB AB $^{-}$ 1
AB AB $^{+}$ 3
1. (i)
  Find the value of $x$ .
  1 mark
2. (ii)
  Find the probability that a randomly selected person has a Rhesus negative blood type.
  1 mark
3. (iii)
  (A) What is the probability that the person selected from the record is Rhesus positive but neither blood type A nor B?
  2 marks
4. OR
  (iii)
  (B) People with blood type AB positive (AB $^{+}$ ) are known as the universal recipient and with blood type O negative (O $^{-}$ ) are known as universal donor. Find the probability of a selected person to be neither universal recipient nor universal donor.
  2 marks
Try it first, then show our solutionSolution
1. (i) The percentages total 100, so $x = 100 - (30 + 8 + 24 + 6 + 18 + 1 + 3) = 100 - 90 = 10$ .
  1 mark
2. (ii) Rhesus negative persons: $O^{-} + A^{-} + B^{-} + A B^{-} = 10 + 8 + 6 + 1 = 25$ (per cent).
  So $P (Rh^{-}) = \frac{25}{100} = \frac{1}{4}$ .
  1 mark
3. (iii)(A) Rhesus positive but neither A nor B means $O^{+}$ or $A B^{+}$ , which is $30 + 3 = 33$ (per cent).
  1 mark
4. (iii)(A) Required probability $= \frac{33}{100}$
  1 mark
5. (iii)(B) Universal recipients ( $A B^{+}$ ) $= 3$ and universal donors ( $O^{-}$ ) $= 10$ , giving 13 in total.
  $P (neither) = 1 - \frac{3 + 10}{100} = 1 - \frac{13}{100}$
  1½ marks
6. (iii)(B) $= \frac{87}{100}$
  ½ mark
Answer: (i) x = 10; (ii) 1/4; (iii)(A) 33/100; (iii)(B) 87/100
Where the marks are
- (i) x = 10 (1 mark)
- (ii) P(Rh negative) = 25/100 = 1/4 (1 mark)
- (iii)(A) Identifies $O^{+}$ and $A B^{+}$ , total 33 (1 mark); probability 33/100 (1 mark)
- (iii)(B) Uses $1 - \frac{3 + 10}{100}$ (1½ marks); gets 87/100 (½ mark)
Open Q38 on its own page

Blood Group	Rhesus Factor	Number of Persons (in %)
O	O $^{-}$	$x$
O	O $^{+}$	30
A	A $^{-}$	8
A	A $^{+}$	24
B	B $^{-}$	6
B	B $^{+}$	18
AB	AB $^{-}$	1
AB	AB $^{+}$	3

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