MathsCBSE Class 10 · Basic · Board exam 2023 (session 2022-23)

CBSE Class 10 Mathematics Basic: 2023 board paper, set 430/1/1

The question paper set in the CBSE Class 10 board exam of 2023, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2023/X/MATHEMATICS_BASIC.zip (a zip; this paper is the file 430_1_1_Maths Basic.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	SA-I
C	26-31	SA-II
D	32-35	Long answer
E	36-38	Case based, internal choice in 2-mark subparts

Instructions in brief: No overall choice; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (one 2-mark subpart in each case study). Draw neat figures where required, take π = 22/7 if not stated, calculator not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The prime factorisation of natural number 288 is
1. (A) $2^{4} \times 3^{3}$
2. (B) $2^{4} \times 3^{2}$
3. (C) $2^{5} \times 3^{2}$
4. (D) $2^{5} \times 3^{1}$
Try it first, then show our solutionSolution
1. Divide repeatedly: $288 = 2 \times 144 = 2^{2} \times 72 = 2^{3} \times 36 = 2^{4} \times 18 = 2^{5} \times 9$ .
2. $9 = 3^{2}$ , so $288 = 2^{5} \times 3^{2}$ . Option (C).
  1 mark
Answer: $2^{5} \times 3^{2}$ , option (C)
Where the marks are
- Correct factorisation $2^{5} \times 3^{2}$ (1 mark)
Q2 · 1 mark · MCQIntroduction to Trigonometry
If $2 cos θ = 1$ , then the value of $θ$ is
1. (A) $4 5^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $3 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $cos θ = \frac{1}{2}$ .
2. Since $cos 6 0^{\circ} = \frac{1}{2}$ , $θ = 6 0^{\circ}$ . Option (B).
  1 mark
Answer: $6 0^{\circ}$ , option (B)
Where the marks are
- Identifies $θ = 6 0^{\circ}$ (1 mark)
Q3 · 1 mark · MCQProbability
A card is drawn at random from a well-shuffled deck of 52 cards. The probability of getting a red card is :
1. (A) $\frac{1}{26}$
2. (B) $\frac{1}{13}$
3. (C) $\frac{1}{4}$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. Red cards = 26 (hearts + diamonds) out of 52.
2. $P = \frac{26}{52} = \frac{1}{2}$ . Option (D).
  1 mark
Answer: $\frac{1}{2}$ , option (D)
Where the marks are
- Probability $\frac{26}{52} = \frac{1}{2}$ (1 mark)
Q4 · 1 mark · MCQQuadratic Equations
The discriminant of the quadratic equation $2 x^{2} - 5 x - 3 = 0$ is
1. (A)1
2. (B)49
3. (C)7
4. (D)19
Try it first, then show our solutionSolution
1. $D = b^{2} - 4 a c$ with $a = 2, b = - 5, c = - 3$ .
2. $D = 25 - 4 (2) (- 3) = 25 + 24 = 49$ . Option (B).
  1 mark
Answer: 49, option (B)
Where the marks are
- Computes $b^{2} - 4 a c = 49$ (1 mark)
Q5 · 1 mark · MCQCoordinate Geometry
The distance between the points $(3, 0)$ and $(0, - 3)$ is
1. (A) $23$ units
2. (B)6 units
3. (C)3 units
4. (D) $32$ units
Try it first, then show our solutionSolution
1. $d = (0 - 3)^{2} + (- 3 - 0)^{2} = 9 + 9 = 18$ .
2. $18 = 32$ units. Option (D).
  1 mark
Answer: $32$ units, option (D)
Where the marks are
- Distance formula giving $32$ (1 mark)
Q6 · 1 mark · MCQArithmetic Progressions
The seventh term of an A.P. whose first term is 28 and common difference $- 4$ , is
1. (A)0
2. (B)4
3. (C)52
4. (D)56
Try it first, then show our solutionSolution
1. $a_{7} = a + 6 d = 28 + 6 (- 4)$ .
2. $= 28 - 24 = 4$ . Option (B).
  1 mark
Answer: 4, option (B)
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ to get 4 (1 mark)
Q7 · 1 mark · MCQPolynomials
The graph of $y = p (x)$ is shown in the figure for some polynomial $p (x)$ . The number of zeroes of $p (x)$ is/are :
1. (A)0
2. (B)1
3. (C)2
4. (D)3
Figure in the paper: Coordinate axes x'Ox and y'Oy. A U-shaped curve labelled y = p(x) opens upward, lies entirely above the x-axis, and touches the y-axis at its lowest-looking point near the top of the y-axis; its two ends rise on both sides. See it in the official paper.
Try it first, then show our solutionSolution
1. Zeroes of $p (x)$ are the x-coordinates where the graph meets the x-axis.
2. The curve lies wholly above the x-axis and never meets it, so there are 0 zeroes. Option (A).
  1 mark
Answer: 0, option (A)
Where the marks are
- Counts intersections with x-axis: none (1 mark)
Open Q7 on its own page
Q8 · 1 mark · MCQTriangles
The sides of two similar triangles are in the ratio $4 : 7$ . The ratio of their perimeters is
1. (A)4 : 7
2. (B)12 : 21
3. (C)16 : 49
4. (D)7 : 4
Try it first, then show our solutionSolution
1. For similar triangles, ratio of perimeters = ratio of corresponding sides.
2. So the ratio is $4 : 7$ . Option (A).
  1 mark
Answer: 4 : 7, option (A)
Where the marks are
- States perimeter ratio equals side ratio, 4:7 (1 mark)
Q9 · 1 mark · MCQTriangles
In the given figure, $A B ∥ C D$ . If $A B = 5$ cm, $C D = 2$ cm and $O B = 3$ cm, then the length of OC is
1. (A) $\frac{15}{2}$ cm
2. (B) $\frac{10}{3}$ cm
3. (C) $\frac{6}{5}$ cm
4. (D) $\frac{3}{5}$ cm
Figure in the paper: Points B (top left), A (bottom left), D (top right) and C (bottom right). Segment BC and segment AD cross at point O. AB is marked 5 cm and CD is marked 2 cm, with arrow marks showing AB parallel to CD. BO is marked 3 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. $△ A O B \sim △ D O C$ (AA: alternate angles as $A B ∥ C D$ , vertically opposite angles at O).
2. $\frac{O B}{O C} = \frac{A B}{C D} \Rightarrow \frac{3}{O C} = \frac{5}{2} \Rightarrow O C = \frac{6}{5}$ cm. Option (C).
  1 mark
Answer: $\frac{6}{5}$ cm, option (C)
Where the marks are
- Uses similarity ratio to get $O C = \frac{6}{5}$ cm (1 mark)
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Q10 · 1 mark · MCQPolynomials
The sum and the product of zeroes of the polynomial $p (x) = x^{2} + 5 x + 6$ are respectively
1. (A)5, $- 6$
2. (B) $- 5$ , 6
3. (C)2, 3
4. (D) $- 2$ , $- 3$
Try it first, then show our solutionSolution
1. Sum $= - \frac{b}{a} = - 5$ , product $= \frac{c}{a} = 6$ .
  1 mark
Answer: $- 5, 6$ , option (B)
Where the marks are
- Sum $- 5$ and product 6 (1 mark)
Q11 · 1 mark · MCQProbability
A die is thrown once. Find the probability of getting a number less than 7.
1. (A) $\frac{5}{6}$
2. (B)1
3. (C) $\frac{1}{6}$
4. (D)0
Try it first, then show our solutionSolution
1. All 6 outcomes (1 to 6) are less than 7, so it is a sure event.
2. $P = \frac{6}{6} = 1$ . Option (B).
  1 mark
Answer: 1, option (B)
Where the marks are
- Recognises sure event, probability 1 (1 mark)
Q12 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
The angle subtended by a vertical pole of height 100 m at a point on the ground $1003$ m from the base is, has measure of
1. (A) $9 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $3 0^{\circ}$
Figure in the paper: Right-angled triangle PAB with right angle at A. AB is the vertical pole of 100 m; PA is the ground distance $1003$ m from P to the base A; B is the top of the pole and PB is the line of sight. See it in the official paper.
Try it first, then show our solutionSolution
1. $tan P = \frac{A B}{P A} = \frac{100}{100 3} = \frac{1}{3}$ .
2. So $∠ P = 3 0^{\circ}$ . Option (D).
  1 mark
Answer: $3 0^{\circ}$ , option (D)
Where the marks are
- Uses $tan P = \frac{1}{3}$ to get $3 0^{\circ}$ (1 mark)
Q13 · 1 mark · MCQSurface Areas and Volumes
The volume of a cone of radius 'r' and height '3r' is :
1. (A) $\frac{1}{3} π r^{3}$
2. (B) $3 π r^{3}$
3. (C) $9 π r^{3}$
4. (D) $π r^{3}$
Try it first, then show our solutionSolution
1. $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (3 r) = π r^{3}$ . Option (D).
  1 mark
Answer: $π r^{3}$ , option (D)
Where the marks are
- Substitutes $h = 3 r$ in $\frac{1}{3} π r^{2} h$ (1 mark)
Q14 · 1 mark · MCQCircles
The distance between two parallel tangents of a circle of diameter 7 cm is :
1. (A)7 cm
2. (B)14 cm
3. (C) $\frac{7}{2}$ cm
4. (D)28 cm
Try it first, then show our solutionSolution
1. Parallel tangents touch the circle at the ends of a diameter, so the distance between them equals the diameter.
2. Distance = 7 cm. Option (A).
  1 mark
Answer: 7 cm, option (A)
Where the marks are
- Distance = diameter = 7 cm (1 mark)
Q15 · 1 mark · MCQTriangles
In the above figure, the criterion of similarity by which $△ A B C \sim △ P QR$ is :
1. (A)SSA (Side – Side – Angle) Similarity
2. (B)ASA (Angle – Side – Angle) Similarity
3. (C)SAS (Side – Angle – Side) Similarity
4. (D)AA (Angle – Angle) Similarity
Figure in the paper: Triangle ABC with BC = 3.5 cm, AB = 2.2 cm and angle B = 50°. Triangle PQR with QR = 7 cm, PQ = 4.4 cm and angle Q = 50°. See it in the official paper.
Try it first, then show our solutionSolution
1. $\frac{A B}{P Q} = \frac{2.2}{4.4} = \frac{1}{2}$ , $\frac{B C}{QR} = \frac{3.5}{7} = \frac{1}{2}$ , and included $∠ B = ∠ Q = 5 0^{\circ}$ .
2. Two sides proportional with included angle equal ⇒ SAS similarity. Option (C).
  1 mark
Answer: SAS similarity, option (C)
Where the marks are
- Shows sides proportional with equal included angle, SAS (1 mark)
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Q16 · 1 mark · MCQTriangles
The larger of two supplementary angles exceeds the smaller by 18 degrees. What is the measure of larger angle ?
1. (A) $8 1^{\circ}$
2. (B) $9 9^{\circ}$
3. (C) $3 6^{\circ}$
4. (D) $5 4^{\circ}$
Try it first, then show our solutionSolution
1. Let angles be $x$ and $x + 18$ : $x + x + 18 = 180 \Rightarrow x = 81$ .
2. Larger angle $= 81 + 18 = 9 9^{\circ}$ . Option (B).
  1 mark
Answer: $9 9^{\circ}$ , option (B)
Where the marks are
- Forms equation and gets $9 9^{\circ}$ (1 mark)
Q17 · 1 mark · MCQCircles
In the given figure, the perimeter of $△ A B C$ is :
1. (A)30 cm
2. (B)15 cm
3. (C)45 cm
4. (D)60 cm
Figure in the paper: A circle touches the three sides of triangle ABC at R (on AB), Q (on AC) and P (on BC). Marked lengths: BR (part of AB) = 6 cm, AQ = 5 cm, and PC = 4 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents from an external point are equal: $A R = A Q = 5$ , $B P = B R = 6$ , $C Q = C P = 4$ cm.
2. Perimeter $= 2 (5 + 6 + 4) = 30$ cm. Option (A).
  1 mark
Answer: 30 cm, option (A)
Where the marks are
- Uses equal tangents to get 30 cm (1 mark)
Q18 · 1 mark · MCQCircles
In the given figure, BC and BD are tangents to the circle with centre O and radius 9 cm. If OB = 15 cm, then the length (BC + BD) is :
1. (A)18 cm
2. (B)12 cm
3. (C)24 cm
4. (D)36 cm
Figure in the paper: A circle with centre O and radius 9 cm. External point B is joined to O with OB = 15 cm. Two tangents from B touch the circle at C (upper) and D (lower). OC is a radius of 9 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. $O C ⊥ B C$ [tangent ⟂ radius], so $B C = 1 5^{2} - 9^{2} = 144 = 12$ cm.
2. $B D = B C = 12$ cm [equal tangents], so $B C + B D = 24$ cm. Option (C).
  1 mark
Answer: 24 cm, option (C)
Where the marks are
- Pythagoras gives 12 cm and doubles to 24 cm (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedCircles
Assertion (A): A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R): The lengths of tangents drawn from the external point to a circle are equal.
Directions: Choose the correct option.
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. A is a true theorem; R is also a true theorem.
2. R (equal tangent lengths) does not explain why tangent ⟂ radius, so option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Both true, R not the explanation (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedPair of Linear Equations in Two Variables
Assertion (A): The system of linear equations $3 x + 5 y - 4 = 0$ and $15 x + 25 y - 25 = 0$ is inconsistent.
Reason (R): The pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ is inconsistent if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
Directions: Choose the correct option.
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. $\frac{a _{1}}{a _{2}} = \frac{3}{15} = \frac{1}{5}$ , $\frac{b _{1}}{b _{2}} = \frac{5}{25} = \frac{1}{5}$ , $\frac{c _{1}}{c _{2}} = \frac{- 4}{- 25} = \frac{4}{25}$ .
2. $\frac{1}{5} = \frac{1}{5} \neq = \frac{4}{25}$ , so the system is inconsistent; A is true and R is the condition used. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Compares ratios to show inconsistency explained by R (1 mark)
Open Q20 on its own page

Section B

SA-I

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerCoordinate Geometry
Find the coordinates of the point which divides the line segment joining the points $(7, - 1)$ and $(- 3, 4)$ internally in the ratio $2 : 3$ .
Try it first, then show our solutionSolution
1. Let $P (x, y)$ divide the join of $A (7, - 1)$ and $B (- 3, 4)$ internally in the ratio $m_{1} : m_{2} = 2 : 3$ . By the section formula, $x = \frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}} = \frac{2 ( - 3 ) + 3 ( 7 )}{2 + 3} = \frac{15}{5} = 3$ .
  1 mark
2. $y = \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}} = \frac{2 ( 4 ) + 3 ( - 1 )}{5} = \frac{5}{5} = 1$ . So the required point is $P (3, 1)$ .
  1 mark
Answer: (3, 1)
Where the marks are
- Applies the section formula correctly to get $x = 3$ (1 mark)
- Gets $y = 1$ and states the point $(3, 1)$ (1 mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerCoordinate Geometry
Find the value(s) of y for which the distance between the points $A (3, - 1)$ and $B (11, y)$ is 10 units.
Try it first, then show our solutionSolution
1. By the distance formula, $A B^{2} = (11 - 3)^{2} + (y + 1)^{2}$ . Since $A B = 10$ , $64 + (y + 1)^{2} = 100$ .
  1 mark
2. $(y + 1)^{2} = 36 \Rightarrow y + 1 = \pm 6$ .
3. $y = 6 - 1 = 5$ or $y = - 6 - 1 = - 7$ .
  1 mark
Answer: y = 5 or y = −7
Where the marks are
- Forms the equation $(11 - 3)^{2} + (y + 1)^{2} = 100$ (1 mark)
- Obtains $y = 5$ (½ mark) and $y = - 7$ (½ mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $tan^{2} 6 0^{\circ} - 2 cosec^{2} 3 0^{\circ} - 2 tan^{2} 3 0^{\circ}$ .
Try it first, then show our solutionSolution
1. Using $tan 6 0^{\circ} = 3$ , $cosec 3 0^{\circ} = 2$ , $tan 3 0^{\circ} = \frac{1}{3}$ : the expression $= (3)^{2} - 2 (2)^{2} - 2 (\frac{1}{3})^{2}$ .
  1½ marks
2. $= 3 - 8 - \frac{2}{3} = - \frac{17}{3}$ .
  ½ mark
Answer: $- \frac{17}{3}$
Where the marks are
- Substitutes correct values of $tan 6 0^{\circ}$ , $cosec 3 0^{\circ}$ , $tan 3 0^{\circ}$ (1½ marks)
- Simplifies to $- \frac{17}{3}$ (½ mark)
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Q23 · 2 marks · Very short answerReal Numbers
Find the LCM and HCF of 92 and 510, using prime factorisation.
Try it first, then show our solutionSolution
1. $92 = 2 \times 2 \times 23 = 2^{2} \times 23$
  ½ mark
2. $510 = 2 \times 3 \times 5 \times 17$
  ½ mark
3. HCF = product of smallest powers of common prime factors $= 2$
  ½ mark
4. LCM = product of greatest powers of all prime factors $= 2^{2} \times 3 \times 5 \times 17 \times 23 = 23460$
  ½ mark
Answer: HCF = 2, LCM = 23460
Where the marks are
- Prime factorisation of 92 (½ mark)
- Prime factorisation of 510 (½ mark)
- HCF = 2 (½ mark)
- LCM = 23460 (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve for $x$ and $y$ : $x + y = 6$ , $2 x - 3 y = 4$ .
Try it first, then show our solutionSolution
1. From $x + y = 6$ , $x = 6 - y$ . Substituting in $2 x - 3 y = 4$ : $2 (6 - y) - 3 y = 4 \Rightarrow 12 - 5 y = 4 \Rightarrow y = \frac{8}{5}$ .
  1 mark
2. $x = 6 - \frac{8}{5} = \frac{22}{5}$ .
  1 mark
Answer: $x = \frac{22}{5}$ , $y = \frac{8}{5}$
Where the marks are
- Finds $y = \frac{8}{5}$ (1 mark)
- Finds $x = \frac{22}{5}$ (1 mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Find out whether the following pair of linear equations are consistent or inconsistent : $5 x - 3 y = 11$ , $- 10 x + 6 y = 22$ .
Try it first, then show our solutionSolution
1. Here $a_{1} = 5, b_{1} = - 3, c_{1} = - 11$ and $a_{2} = - 10, b_{2} = 6, c_{2} = - 22$ . $\frac{a _{1}}{a _{2}} = \frac{5}{- 10} = - \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 3}{6} = - \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{- 11}{- 22} = \frac{1}{2}$ .
  1 mark
2. So $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
  ½ mark
3. Hence the lines are parallel and the pair of equations is inconsistent.
  ½ mark
Answer: Inconsistent
Where the marks are
- Computes the three ratios correctly (1 mark)
- Shows $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ (½ mark)
- Concludes the pair is inconsistent (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerTriangles
In the given figure, ABC and AMP are two right triangles, right angled at B and M, respectively. Prove that $△ A B C \sim △ A M P$ .
Figure in the paper: Triangle ABC with a right angle at B, where B lies on the base line AP. Point M lies on side AC and angle at M is a right angle, with segment MP drawn from M to P on the base line. Points in order on the base: A, B, P. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ and $△ A M P$ : $∠ A B C = ∠ A M P = 9 0^{\circ}$ [given].
  ½ mark
2. $∠ B A C = ∠ M A P$ [common angle at A].
  1 mark
3. Therefore $△ A B C \sim △ A M P$ [AA similarity criterion].
  ½ mark
Answer: $△ A B C \sim △ A M P$ by AA similarity
Where the marks are
- Right angles at B and M equal (½ mark)
- Common angle A (1 mark)
- Concludes similarity by AA (½ mark)
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Section C

SA-II

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $sec θ (1 - sin θ) (sec θ + tan θ) = 1$ .
Try it first, then show our solutionSolution
1. LHS $= \frac{1}{cos θ} (1 - sin θ) (\frac{1}{cos θ} + \frac{sin θ}{cos θ})$
  1 mark
2. $= \frac{1}{cos θ} (1 - sin θ) \cdot \frac{1 + sin θ}{cos θ}$
  1 mark
3. $= \frac{1 - sin ^{2} θ}{cos ^{2} θ}$
  ½ mark
4. $= \frac{cos ^{2} θ}{cos ^{2} θ} = 1 =$ RHS [ $sin^{2} θ + cos^{2} θ = 1$ ]
  ½ mark
Answer: LHS = 1 = RHS, hence proved
Where the marks are
- Writes sec and tan in terms of sin and cos (1 mark)
- Combines to $\frac{1 + s i n θ}{c o s θ}$ (1 mark)
- Obtains $\frac{1 - s i n ^{2} θ}{c o s ^{2} θ}$ (½ mark)
- Uses identity to get 1 (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{1 + sec θ}{sec θ} = \frac{sin ^{2} θ}{1 - cos θ}$ .
Try it first, then show our solutionSolution
1. LHS $= \frac{1 + \frac{1}{c o s θ}}{\frac{1}{c o s θ}} = \frac{cos θ + 1}{cos θ} \times cos θ = 1 + cos θ$
  1 mark
2. Multiply and divide by $(1 - cos θ)$ : $= \frac{( 1 + cos θ ) ( 1 - cos θ )}{1 - cos θ}$
  1 mark
3. $= \frac{1 - cos ^{2} θ}{1 - cos θ} = \frac{sin ^{2} θ}{1 - cos θ} =$ RHS
  1 mark
Answer: LHS = RHS, hence proved
Where the marks are
- Simplifies LHS to $1 + cos θ$ (1 mark)
- Multiplies and divides by $1 - cos θ$ (1 mark)
- Uses $1 - cos^{2} θ = sin^{2} θ$ to reach RHS (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerCoordinate Geometry
Show that the points $A (1, 7)$ , $B (4, 2)$ , $C (- 1, - 1)$ and $D (- 4, 4)$ are vertices of the square ABCD.
Try it first, then show our solutionSolution
1. $A B = (4 - 1)^{2} + (2 - 7)^{2} = 9 + 25 = 34$ ; $B C = (- 1 - 4)^{2} + (- 1 - 2)^{2} = 25 + 9 = 34$
  1 mark
2. $C D = (- 4 + 1)^{2} + (4 + 1)^{2} = 9 + 25 = 34$ ; $D A = (1 + 4)^{2} + (7 - 4)^{2} = 25 + 9 = 34$ . So $A B = B C = C D = D A$ .
  1 mark
3. Diagonals: $A C = (- 1 - 1)^{2} + (- 1 - 7)^{2} = 4 + 64 = 68$ ; $B D = (- 4 - 4)^{2} + (4 - 2)^{2} = 64 + 4 = 68$ . So $A C = B D$ . All sides equal and diagonals equal, hence ABCD is a square.
  1 mark
Answer: AB = BC = CD = DA = √34 and AC = BD = √68, so ABCD is a square
Where the marks are
- Finds all four sides equal to $34$ (2 marks)
- Finds both diagonals equal to $68$ and concludes square (1 mark)
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Q28 · 3 marks · Short answerCircles
Prove that the tangents drawn from an external point to a circle are equal in length.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O; from an external point P, tangents PQ and PR touch the circle at Q and R; OP, OQ, OR are joined.
  1 mark
2. Given: PQ and PR are tangents from external point P to a circle with centre O. To prove: $P Q = P R$ . Construction: join OP, OQ, OR.
  ½ mark
3. In $△ O QP$ and $△ O R P$ : $O Q = O R$ [radii]; $O P = O P$ [common]; $∠ O QP = ∠ O R P = 9 0^{\circ}$ [tangent ⟂ radius at point of contact]. So $△ O QP ≅ △ O R P$ [RHS].
  1 mark
4. Hence $P Q = P R$ [CPCT].
  ½ mark
Answer: PQ = PR (tangents from an external point are equal)
Where the marks are
- Correct figure (1 mark)
- Given, to prove, construction (½ mark)
- Proves triangles congruent by RHS with reasons (1 mark)
- Concludes PQ = PR by CPCT (½ mark)
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Q29 · 3 marks · Short answerPolynomials
If $α, β$ are zeroes of the quadratic polynomial $x^{2} + 3 x + 2$ , find a quadratic polynomial whose zeroes are $α + 1$ , $β + 1$ .
Try it first, then show our solutionSolution
1. For $x^{2} + 3 x + 2$ : $α + β = - \frac{b}{a} = - 3$ , $α β = \frac{c}{a} = 2$ .
  1 mark
2. Sum of new zeroes: $(α + 1) + (β + 1) = (α + β) + 2 = - 3 + 2 = - 1$ .
  ½ mark
3. Product: $(α + 1) (β + 1) = α β + (α + β) + 1 = 2 - 3 + 1 = 0$ .
  1 mark
4. Required polynomial $= x^{2} - (sum) x + product = x^{2} + x$ (or $k (x^{2} + x)$ , $k \neq = 0$ ).
  ½ mark
Answer: $x^{2} + x$ (or $k (x^{2} + x)$ )
Where the marks are
- Finds $α + β = - 3$ and $α β = 2$ (½ + ½ mark)
- Sum of new zeroes $= - 1$ (½ mark)
- Product of new zeroes $= 0$ (1 mark)
- Writes polynomial $x^{2} + x$ (½ mark)
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Q30 · 3 marks · Short answerReal Numbers
Prove that $3 + 72$ is an irrational number, given that $2$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3 + 72$ is rational. Then $3 + 72 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ .
  1 mark
2. $72 = \frac{p}{q} - 3 \Rightarrow 2 = \frac{p - 3 q}{7 q}$ .
  1 mark
3. Since $p, q$ are integers, $\frac{p - 3 q}{7 q}$ is rational, so $2$ would be rational. This contradicts the fact that $2$ is irrational. Hence our assumption is wrong and $3 + 72$ is irrational.
  1 mark
Answer: $3 + 72$ is irrational
Where the marks are
- Assumes $3 + 72 = \frac{p}{q}$ (1 mark)
- Isolates $2 = \frac{p - 3 q}{7 q}$ (1 mark)
- States contradiction and concludes (1 mark)
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Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerTriangles
In the given figure, $D E ∥ A C$ and $D F ∥ A E$ . Prove that $\frac{B F}{F E} = \frac{B E}{E C}$ .
Figure in the paper: Triangle ABC with D on side AB and E, F, C on base BC in the order B, F, E, C. A is the apex. DE is drawn parallel to AC and DF is drawn parallel to AE, with F on BC between B and E. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B E$ , $D F ∥ A E$ [given], so by the Basic Proportionality Theorem, $\frac{B D}{D A} = \frac{B F}{F E}$ … (i)
  1½ marks
2. In $△ A B C$ , $D E ∥ A C$ [given], so by BPT, $\frac{B D}{D A} = \frac{B E}{E C}$ … (ii)
  1 mark
3. From (i) and (ii), $\frac{B F}{F E} = \frac{B E}{E C}$ . Hence proved.
  ½ mark
Answer: BF/FE = BE/EC (proved using BPT twice)
Where the marks are
- Applies BPT in $△ A B E$ to get (i) (1½ marks)
- Applies BPT in $△ A B C$ to get (ii) (1 mark)
- Equates the two to conclude (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerTriangles
The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{A O}{B O} = \frac{C O}{O D}$ . Show that quadrilateral ABCD is a trapezium.
Figure in the paper: Quadrilateral ABCD with AB at the bottom and DC at the top, DC longer than AB. Diagonals AC and BD intersect at O. See it in the official paper.
Try it first, then show our solutionSolution
1. Given $\frac{A O}{B O} = \frac{C O}{D O} \Rightarrow \frac{O A}{O C} = \frac{O B}{O D}$ .
  ½ mark
2. In $△ A O B$ and $△ C O D$ : $\frac{O A}{O C} = \frac{O B}{O D}$ and $∠ A O B = ∠ C O D$ [vertically opposite angles]. So $△ A O B \sim △ C O D$ [SAS similarity].
  1½ marks
3. Hence $∠ O A B = ∠ O C D$ , i.e. $∠ C A B = ∠ A C D$ [corresponding angles of similar triangles]. These are alternate interior angles for lines AB and CD with transversal AC, so $A B ∥ C D$ . Therefore ABCD is a trapezium.
  1 mark
Answer: AB ∥ DC, so ABCD is a trapezium
Where the marks are
- Rearranges the given ratio to $\frac{O A}{O C} = \frac{O B}{O D}$ and uses vertically opposite angles (part of 2 marks)
- Proves $△ A O B \sim △ C O D$ by SAS (2 marks total)
- Uses equal alternate angles to get $A B ∥ C D$ and conclude trapezium (1 mark)
Open Q31(b) on its own page

Section D

Long answer

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 80 m more than the shorter side, find the length of the sides of the field.
Try it first, then show our solutionSolution
1. Let the shorter side be $x$ m. Then longer side $= (x + 80)$ m and diagonal $= (x + 60)$ m.
  1 mark
2. By Pythagoras theorem: $x^{2} + (x + 80)^{2} = (x + 60)^{2}$ .
  1 mark
3. $x^{2} + x^{2} + 160 x + 6400 = x^{2} + 120 x + 3600 \Rightarrow x^{2} + 40 x + 2800 = 0$ .
  1 mark
4. Discriminant $D = 4 0^{2} - 4 (1) (2800) = 1600 - 11200 = - 9600 < 0$ , so the equation has no real roots.
  1 mark
5. Hence no rectangle with the given data exists (in fact the diagonal must exceed the longer side, but here diagonal $x + 60$ is less than longer side $x + 80$ ). [With the NCERT data 'longer side 30 m more than shorter side': $x^{2} + (x + 30)^{2} = (x + 60)^{2} \Rightarrow x^{2} - 60 x - 2700 = 0 \Rightarrow (x - 90) (x + 30) = 0$ , so sides are 90 m and 120 m.]
  1 mark
Answer: With the given data, $x^{2} + 40 x + 2800 = 0$ has no real roots, so no such field exists (question has an error; with the intended data '30 m more', sides are 90 m and 120 m)
Where the marks are
- Assigns variables to sides and diagonal (1 mark)
- Applies Pythagoras theorem (1 mark)
- Forms the quadratic $x^{2} + 40 x + 2800 = 0$ (1 mark)
- Shows discriminant is negative (1 mark)
- Concludes no real solution exists (1 mark) — scheme awards full marks to any attempt
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerCompetency-basedQuadratic Equations
The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present age.
Try it first, then show our solutionSolution
1. Let the father's present age be $x$ years; then son's present age $= (45 - x)$ years.
  1 mark
2. Five years ago: father's age $= (x - 5)$ years, son's age $= (45 - x - 5) = (40 - x)$ years.
  1 mark
3. According to the question, $(x - 5) (40 - x) = 124$ .
  1 mark
4. $40 x - x^{2} - 200 + 5 x = 124 \Rightarrow x^{2} - 45 x + 324 = 0 \Rightarrow (x - 36) (x - 9) = 0$ .
  1 mark
5. $x = 36$ or $x = 9$ . $x = 9$ is rejected, since then the son would be 36 years, older than the father.
  ½ mark
6. Father's present age = 36 years, son's present age = $45 - 36 = 9$ years.
  ½ mark
Answer: Father: 36 years, Son: 9 years
Where the marks are
- Expresses present ages as $x$ and $45 - x$ (1 mark)
- Writes ages five years ago (1 mark)
- Forms equation $(x - 5) (40 - x) = 124$ (1 mark)
- Reduces to $x^{2} - 45 x + 324 = 0$ and factorises (1 mark)
- Rejects $x = 9$ (½ mark)
- States father 36 years, son 9 years (½ mark)
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Q33 · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Also, find the volume of the vessel.
Try it first, then show our solutionSolution
1. Radius of hemisphere = radius of cylinder $r = \frac{14}{2} = 7$ cm. Height of cylinder $h = 13 - 7 = 6$ cm.
  1 mark
2. Inner surface area = CSA of cylinder + CSA of hemisphere $= 2 π r h + 2 π r^{2} = 2 π r (h + r) = 2 \times \frac{22}{7} \times 7 \times (6 + 7)$
  1 mark
3. $= 44 \times 13 = 572 cm^{2}$
  1 mark
4. Volume = volume of cylinder + volume of hemisphere $= π r^{2} h + \frac{2}{3} π r^{3} = π r^{2} (h + \frac{2}{3} r) = \frac{22}{7} \times 7 \times 7 \times (6 + \frac{14}{3})$
  1 mark
5. $= 154 \times \frac{32}{3} = \frac{4928}{3} cm^{3} \approx 1642.67 cm^{3}$
  1 mark
Answer: Inner surface area = 572 cm²; Volume = 4928/3 cm³ ≈ 1642.67 cm³
Where the marks are
- Finds $r = 7$ cm and cylinder height 6 cm (1 mark)
- Uses $2 π r h + 2 π r^{2}$ with substitution (1 mark)
- Inner surface area $= 572$ cm² (1 mark)
- Uses $π r^{2} h + \frac{2}{3} π r^{3}$ with substitution (1 mark)
- Volume $= \frac{4928}{3}$ cm³ (1 mark)
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Q34 · 5 marks · Long answerCompetency-basedStatistics
The table given below shows the daily expenditure on food of 25 households in a locality :
Daily expenditure (₹) 100–150 150–200 200–250 250–300 300–350
Number of household 4 5 12 2 2
1. (i)
  Find the mean daily expenditure on food.
  3 marks
2. (ii)
  Find the mode of the data.
  2 marks
Try it first, then show our solutionSolution
1. (i) Table with class marks $x_{i}$ and $f_{i} x_{i}$ : 100–150: $f = 4$ , $x = 125$ , $f x = 500$ ; 150–200: $5$ , $175$ , $875$ ; 200–250: $12$ , $225$ , $2700$ ; 250–300: $2$ , $275$ , $550$ ; 300–350: $2$ , $325$ , $650$ . $\sum f_{i} = 25$ , $\sum f_{i} x_{i} = 5275$ .
  1½ marks
2. (i) Mean $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{5275}{25} = 211$ . Mean daily expenditure = ₹211.
  1½ marks
3. (ii) Highest frequency is 12, so modal class = 200–250; $l = 200$ , $f_{1} = 12$ , $f_{0} = 5$ , $f_{2} = 2$ , $h = 50$ .
  ½ mark
4. (ii) Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) h = 200 + (\frac{12 - 5}{24 - 5 - 2}) \times 50 = 200 + \frac{350}{17}$
  1 mark
5. (ii) Mode $= \frac{3750}{17} \approx ₹220.59$
  ½ mark
Answer: (i) Mean = ₹211; (ii) Mode = 3750/17 ≈ ₹220.59
Where the marks are
- Correct table of $x_{i}$ and $f_{i} x_{i}$ with totals 25 and 5275 (1½ marks)
- Mean $= \frac{5275}{25} = 211$ (1½ marks)
- Identifies modal class 200–250 (½ mark)
- Substitutes correctly in mode formula (1 mark)
- Mode $\approx 220.59$ (½ mark)
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Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $6 0^{\circ}$ . From another point 20 m away from the point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $3 0^{\circ}$ . Find the height of the tower.
Try it first, then show our solutionSolution
1. Figure: Let AB = $h$ m be the tower, with foot A and top B. C is the point on the opposite bank, with AC = $x$ m and $∠ B C A = 6 0^{\circ}$ . D is 20 m beyond C on line AC, so AD = $(20 + x)$ m and $∠ B D A = 3 0^{\circ}$ . Also $∠ B A D = 9 0^{\circ}$ .
  1 mark
2. In right $△ B A C$ : $tan 6 0^{\circ} = \frac{A B}{A C} = \frac{h}{x}$ , so $3 = \frac{h}{x}$ , i.e. $h = 3 x$ ...(1)
  1½ marks
3. In right $△ B A D$ : $tan 3 0^{\circ} = \frac{A B}{A D} = \frac{h}{20 + x}$ , so $\frac{1}{3} = \frac{h}{20 + x}$ , i.e. $3 h = 20 + x$ ...(2)
  1½ marks
4. Substitute (1) in (2): $3 (3 x) = 20 + x$ , so $3 x - x = 20$ , giving $x = 10$ m.
  ½ mark
5. From (1), $h = 3 \times 10 = 103$ m $\approx 17.32$ m. Hence the height of the tower is $103$ m.
  ½ mark
Answer: Height of tower = $103$ m ≈ 17.32 m (the width of the canal is 10 m)
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ} = h / x$ to get $h = 3 x$ (1½ marks)
- Uses $tan 3 0^{\circ} = h / (20 + x)$ to get $3 h = 20 + x$ (1½ marks)
- Solves to get $x = 10$ (½ mark)
- Height $= 103$ m (½ mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are $6 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the vertical distance between the aeroplanes at that instant. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: P is the point on the ground and A is the point on the ground vertically below both planes, with PA = $x$ m. The higher plane is at C, with AC = 4000 m and $∠ C P A = 6 0^{\circ}$ . The lower plane is at B on AC, with AB = $h$ m and $∠ B P A = 4 5^{\circ}$ . Also $∠ P A C = 9 0^{\circ}$ .
  1 mark
2. In right $△ P A B$ : $tan 4 5^{\circ} = \frac{h}{x}$ , so $1 = \frac{h}{x}$ , i.e. $h = x$ .
  1½ marks
3. In right $△ P A C$ : $tan 6 0^{\circ} = \frac{4000}{x}$ , so $x = \frac{4000}{3}$ m. Hence $h = \frac{4000}{3}$ m.
  1½ marks
4. Vertical distance between the planes: $B C = 4000 - \frac{4000}{3} = 4000 (1 - \frac{1}{3}) = \frac{4000 ( 3 - 3 )}{3}$
  ½ mark
5. Using $3 = 1.73$ : $B C = \frac{4000 \times 1.27}{3} = \frac{5080}{3} \approx 1693.33$ m.
  ½ mark
Answer: Vertical distance = $\frac{5080}{3}$ m ≈ 1693.33 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 4 5^{\circ}$ to get $h = x$ (1½ marks)
- Uses $tan 6 0^{\circ}$ to get $x = 4000/ 3$ (1½ marks)
- Writes distance $= 4000 - 4000/ 3$ (½ mark)
- Final value ≈ 1693.33 m using $3 = 1.73$ (½ mark)
Open Q35(b) on its own page

Daily expenditure (₹)	100–150	150–200	200–250	250–300	300–350
Number of household	4	5	12	2	2

Section E

Case based, internal choice in 2-mark subparts

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Aahana being a plant lover decides to convert her balcony into beautiful garden full of plants. She bought few plants with pots for her balcony. She placed the pots in such a way that number of pots in the first row is 2, second row is 5, third row is 8 and so on.
Based on the above information, answer the following questions :
1. (i)
  Find the number of pots placed in the $1 0^{th}$ row.
  1 mark
2. (ii)
  Find the difference in the number of pots placed in $5^{th}$ row and $2^{nd}$ row.
  1 mark
3. (iii)
  If Aahana wants to place 100 pots in total, then find the total number of rows formed in the arrangement.
  2 marks
4. OR
  (iii)
  If Aahana has sufficient space for 12 rows, then how many total number of pots are placed by her with the same arrangement ?
  2 marks
Figure in the paper: Pot arrangement in rows: the top row has 2 flower pots; the middle row has 5 pots; the bottom row has 8 pots. Pots are drawn as flower pots, and a dashed line marks the ground below the bottom row. See it in the official paper.
Try it first, then show our solutionSolution
1. The numbers of pots in the rows are 2, 5, 8, … This is an A.P. with $a = 2$ and $d = 5 - 2 = 3$ .
2. (i) $a_{10} = a + 9 d = 2 + 9 (3) = 29$ . So 29 pots are placed in the 10th row.
  1 mark
3. (ii) $a_{5} - a_{2} = (a + 4 d) - (a + d) = 3 d = 3 \times 3 = 9$ . So the 5th row has 9 more pots than the 2nd row.
  1 mark
4. (iii) Set $S_{n} = 100$ : $\frac{n}{2} [2 (2) + (n - 1) 3] = 100$ . This gives $n (3 n + 1) = 200$ , i.e. $3 n^{2} + n - 200 = 0$ .
  1 mark
5. (iii) Factorise: $(3 n + 25) (n - 8) = 0$ , so $n = 8$ or $n = - \frac{25}{3}$ . Reject $n = - \frac{25}{3}$ , since the number of rows must be a natural number. Hence 8 rows are formed.
  1 mark
6. (iii) OR: $S_{12} = \frac{12}{2} [2 (2) + 11 (3)] = 6 (4 + 33)$
  1 mark
7. (iii) OR: $S_{12} = 6 \times 37 = 222$ . So she places 222 pots in total.
  1 mark
Answer: (i) 29 pots; (ii) 9; (iii) 8 rows; OR (iii) 222 pots
Where the marks are
- (i) $a_{10} = a + 9 d = 29$ (1 mark)
- (ii) $a_{5} - a_{2} = 3 d = 9$ (1 mark)
- (iii) Forms $\frac{n}{2} [4 + 3 (n - 1)] = 100$ (1 mark)
- (iii) Solves $3 n^{2} + n - 200 = 0$ to get $n = 8$ , rejecting the negative root (1 mark)
- (iii) OR: Correct substitution in $S_{12}$ (1 mark)
- (iii) OR: $S_{12} = 222$ (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
Interschool Rangoli Competition was organized by one of the reputed schools of Odissa. The theme of the Rangoli Competition was Diwali celebrations where students were supposed to make mathematical designs. Students from various schools participated and made beautiful Rangoli designs. One such design is given below.
Rangoli is in the shape of square marked as ABCD, side of square being 40 cm. At each corner of a square, a quadrant of circle of radius 10 cm is drawn (in which diyas are kept). Also a circle of diameter 20 cm is drawn inside the square.
Based on the above information, answer the following questions :
1. (i)
  What is the area of square ABCD ?
  1 mark
2. (ii)
  Find the area of the circle.
  1 mark
3. (iii)
  If the circle and the four quadrants are cut off from the square ABCD and removed, then find the area of remaining portion of square ABCD.
  2 marks
4. OR
  (iii)
  Find the combined area of 4 quadrants and the circle, removed.
  2 marks
Figure in the paper: Square ABCD (A top-left, B top-right, C bottom-right, D bottom-left) with side 40 cm. A quadrant of radius 10 cm is drawn at each corner inside the square, with diyas at the corners. A circle of diameter 20 cm is drawn at the centre of the square. Diyas are also shown along the outer border of a larger rectangle surrounding the square. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Area of square ABCD $= (side)^{2} = 4 0^{2} = 1600 cm^{2}$ .
  1 mark
2. (ii) The circle has radius $\frac{20}{2} = 10$ cm. Area $= π r^{2} = \frac{22}{7} \times 10 \times 10 = \frac{2200}{7} cm^{2} \approx 314.29 cm^{2}$ .
  1 mark
3. (iii) Each quadrant has radius 10 cm. Area of 4 quadrants $= 4 \times \frac{1}{4} π r^{2} = \frac{22}{7} \times 100 = \frac{2200}{7} cm^{2}$ .
  1 mark
4. (iii) Remaining area $= 1600 - (\frac{2200}{7} + \frac{2200}{7}) = 1600 - \frac{4400}{7} = \frac{6800}{7} cm^{2} \approx 971.43 cm^{2}$ .
  1 mark
5. (iii) OR: Area of 4 quadrants $= 4 \times \frac{1}{4} π (10)^{2} = \frac{2200}{7} cm^{2}$ .
  1 mark
6. (iii) OR: Combined area removed $= \frac{2200}{7} + \frac{2200}{7} = \frac{4400}{7} cm^{2} \approx 628.57 cm^{2}$ .
  1 mark
Answer: (i) 1600 cm²; (ii) 2200/7 cm² ≈ 314.29 cm²; (iii) 6800/7 cm² ≈ 971.43 cm²; OR (iii) 4400/7 cm² ≈ 628.57 cm²
Where the marks are
- (i) Area of square = 1600 cm² (1 mark)
- (ii) Area of circle = 2200/7 cm² (1 mark)
- (iii) Area of 4 quadrants = 2200/7 cm² (1 mark)
- (iii) Remaining area = 6800/7 cm² ≈ 971.43 cm² (1 mark)
- (iii) OR: Area of 4 quadrants = 2200/7 cm² (1 mark)
- (iii) OR: Combined area = 4400/7 cm² ≈ 628.57 cm² (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedProbability
Blood group describes the type of blood a person has. It is a classification of blood based on the presence or absence of inherited antigenic substances on the surface of red blood cells. Blood types predict whether a serious reaction will occur in a blood transfusion.
In a sample of 50 people, 21 had type O blood, 22 had type A, 5 had type B and rest had type AB blood group.
Based on the above, answer the following questions :
1. (i)
  What is the probability that a person chosen at random had type O blood ?
  1 mark
2. (ii)
  What is the probability that a person chosen at random had type AB blood group ?
  1 mark
3. (iii)
  What is the probability that a person chosen at random had neither type A nor type B blood group ?
  2 marks
4. OR
  (iii)
  What is the probability that person chosen at random had either type A or type B or type O blood group ?
  2 marks
Figure in the paper: Four cartoon blood-drop characters, each holding a yellow label: A (top left), B (top right), AB (bottom left) and O (bottom right). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $P (type O) = \frac{number with type O}{total people} = \frac{21}{50}$ .
  1 mark
2. (ii) Number with type AB $= 50 - (21 + 22 + 5) = 50 - 48 = 2$ .
  ½ mark
3. (ii) $P (type AB) = \frac{2}{50} = \frac{1}{25}$ .
  ½ mark
4. (iii) A person with neither type A nor type B has type O or type AB. Favourable outcomes $= 21 + 2 = 23$ .
  1 mark
5. (iii) $P (neither A nor B) = \frac{23}{50}$ .
  1 mark
6. (iii) OR: Favourable outcomes for type A, B or O $= 22 + 5 + 21 = 48$ .
  1 mark
7. (iii) OR: $P (A or B or O) = \frac{48}{50} = \frac{24}{25}$ .
  1 mark
Answer: (i) 21/50; (ii) 1/25; (iii) 23/50; OR (iii) 24/25
Where the marks are
- (i) P(type O) = 21/50 (1 mark)
- (ii) Finds 2 people with type AB (½ mark)
- (ii) P(type AB) = 1/25 (½ mark)
- (iii) Favourable outcomes = 21 + 2 = 23 (1 mark)
- (iii) Probability = 23/50 (1 mark)
- (iii) OR: Favourable outcomes = 48 (1 mark)
- (iii) OR: Probability = 24/25 (1 mark)
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