MathsCBSE Class 10 · Standard · Board exam 2024 (session 2023-24)

CBSE Class 10 Mathematics Standard: 2024 board paper, set 30/3/1

The question paper set in the CBSE Class 10 board exam of 2024, 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/2024/X/MATHEMATICS_STANDARD.zip (a zip; this paper is the file 30-3-1(Mathematics Standard).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 22 and 23
C	26-31	Short answer; internal choice in 26 and 29
D	32-35	Long answer; internal choice in 32 and 35
E	36-38	Case study based; internal choice in 36, 37 and 38

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (case studies).

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of linear equations $x + 2 y + 5 = 0$ and $- 3 x = 6 y - 1$ has
1. (A)unique solution
2. (B)exactly two solutions
3. (C)infinitely many solutions
4. (D)no solution
Try it first, then show our solutionSolution
1. Write both equations in standard form: $x + 2 y + 5 = 0$ and $3 x + 6 y - 1 = 0$ .
2. Compare the ratios: $\frac{a _{1}}{a _{2}} = \frac{1}{3}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{6} = \frac{1}{3}$ , $\frac{c _{1}}{c _{2}} = \frac{5}{- 1} = - 5$ .
3. Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the lines are parallel. The pair has no solution, so the answer is option (D).
  1 mark
Answer: No solution, option (D)
Where the marks are
- Identifies that the pair has no solution, option (D) (1 mark)
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Q2 · 1 mark · MCQArithmetic Progressions
The common difference of the A.P. $\frac{1}{2 x}, \frac{1 - 4 x}{2 x}, \frac{1 - 8 x}{2 x}, \dots$ is:
1. (A) $- 2 x$
2. (B) $- 2$
3. (C) $2$
4. (D) $2 x$
Try it first, then show our solutionSolution
1. Use $d = a_{2} - a_{1}$ : $d = \frac{1 - 4 x}{2 x} - \frac{1}{2 x} = \frac{- 4 x}{2 x}$ .
2. So $d = - 2$ , which is option (B).
  1 mark
Answer: $- 2$ , option (B)
Where the marks are
- Finds $d = - 2$ , option (B) (1 mark)
Q3 · 1 mark · MCQProbability
Two dice are thrown together. The probability that they show different numbers is:
1. (A) $\frac{1}{6}$
2. (B) $\frac{5}{6}$
3. (C) $\frac{1}{3}$
4. (D) $\frac{2}{3}$
Try it first, then show our solutionSolution
1. When two dice are thrown there are $6 \times 6 = 36$ outcomes. Exactly 6 of them show the same number: (1,1), (2,2), …, (6,6).
2. P(different numbers) $= 1 - \frac{6}{36} = \frac{30}{36} = \frac{5}{6}$ , which is option (B).
  1 mark
Answer: $\frac{5}{6}$ , option (B)
Where the marks are
- Obtains $\frac{5}{6}$ , option (B) (1 mark)
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Q4 · 1 mark · MCQProbability
The probability of guessing the correct answer to a certain test question is $\frac{x}{6}$ . If the probability of not guessing the correct answer to this question is $\frac{2}{3}$ , then the value of $x$ is:
1. (A)2
2. (B)3
3. (C)4
4. (D)6
Try it first, then show our solutionSolution
1. Use $P (E) + P (not E) = 1$ : $\frac{x}{6} = 1 - \frac{2}{3} = \frac{1}{3}$ .
2. So $x = 2$ , which is option (A).
  1 mark
Answer: $x = 2$ , option (A)
Where the marks are
- Uses the complement rule to get $x = 2$ , option (A) (1 mark)
Q5 · 1 mark · MCQReal Numbers
If $a = 2^{2} \times 3^{x}$ , $b = 2^{2} \times 3 \times 5$ , $c = 2^{2} \times 3 \times 7$ and $LCM (a, b, c) = 3780$ , then $x$ is equal to
1. (A)1
2. (B)2
3. (C)3
4. (D)0
Try it first, then show our solutionSolution
1. Factorise: $3780 = 2^{2} \times 3^{3} \times 5 \times 7$ .
2. The LCM takes the highest power of each prime, so LCM $= 2^{2} \times 3^{x} \times 5 \times 7$ .
3. Comparing, $3^{x} = 3^{3}$ , so $x = 3$ , which is option (C).
  1 mark
Answer: $x = 3$ , option (C)
Where the marks are
- Compares prime factorisations to get $x = 3$ , option (C) (1 mark)
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Q6 · 1 mark · MCQPolynomials
The zeroes of the quadratic polynomial $2 x^{2} - 3 x - 9$ are:
1. (A) $3, \frac{- 3}{2}$
2. (B) $- 3, \frac{- 3}{2}$
3. (C) $- 3, \frac{3}{2}$
4. (D) $3, \frac{3}{2}$
Try it first, then show our solutionSolution
1. Split the middle term: $2 x^{2} - 3 x - 9 = 2 x^{2} - 6 x + 3 x - 9 = (x - 3) (2 x + 3)$ .
2. The zeroes are $x = 3$ and $x = - \frac{3}{2}$ , which is option (A).
  1 mark
Answer: $3, - \frac{3}{2}$ , option (A)
Where the marks are
- Factorises the polynomial and gets zeroes $3$ and $- \frac{3}{2}$ , option (A) (1 mark)
Q7 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
From a point on the ground, which is 30 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be $6 0^{\circ}$ . The height (in metres) of the tower is:
1. (A) $103$
2. (B) $303$
3. (C)60
4. (D)30
Try it first, then show our solutionSolution
1. In the right triangle, $tan 6 0^{\circ} = \frac{h}{30}$ , so $h = 303$ m.
  1 mark
2. This is option (B).
Answer: $303$ m, option (B)
Where the marks are
- Uses $tan 6 0^{\circ} = h /30$ to get $303$ , option (B) (1 mark)
Q8 · 1 mark · MCQIntroduction to Trigonometry
If $cos θ = \frac{3}{2}$ and $sin ϕ = \frac{1}{2}$ , then $tan (θ + ϕ)$ is:
1. (A) $3$
2. (B) $\frac{1}{3}$
3. (C)1
4. (D)not defined
Try it first, then show our solutionSolution
1. $cos θ = \frac{3}{2}$ gives $θ = 3 0^{\circ}$ , and $sin ϕ = \frac{1}{2}$ gives $ϕ = 3 0^{\circ}$ .
2. $tan (θ + ϕ) = tan 6 0^{\circ} = 3$ , which is option (A).
  1 mark
Answer: $3$ , option (A)
Where the marks are
- Identifies both angles as $3 0^{\circ}$ and evaluates $tan 6 0^{\circ} = 3$ , option (A) (1 mark)
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Q9 · 1 mark · MCQCircles
Maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is:
1. (A)4
2. (B)3
3. (C)2
4. (D)1
Try it first, then show our solutionSolution
1. When two circles cut at two points, only the two direct (external) common tangents are possible.
2. No transverse tangent can pass between them, because the circles overlap. So the maximum is 2, which is option (C).
  1 mark
Answer: 2, option (C)
Where the marks are
- States that 2 common tangents exist, option (C) (1 mark)
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Q10 · 1 mark · MCQCircles
In the given figure, if PT is a tangent to a circle with centre O and $∠ TPO = 3 5^{\circ}$ , then the measure of $∠ x$ is:
1. (A) $11 0^{\circ}$
2. (B) $11 5^{\circ}$
3. (C) $12 0^{\circ}$
4. (D) $12 5^{\circ}$
Figure in the paper: A circle with centre O. A tangent line touches the circle at T. P lies on the tangent outside the circle; a straight line from P passes through O and continues to the far side of the circle. TO is joined. The angle at P between PT and PO is marked $3 5^{\circ}$ . The angle at O between OT and the segment OX (going to the lower-left part of the circle) is marked $x$ . See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O T P = 9 0^{\circ}$ [tangent ⟂ radius].
2. In $△ O T P$ : $∠ T O P = 18 0^{\circ} - 9 0^{\circ} - 3 5^{\circ} = 5 5^{\circ}$ .
3. Angle $x$ and $∠ T O P$ form a linear pair on line PO, so $x = 18 0^{\circ} - 5 5^{\circ} = 12 5^{\circ}$ . This is option (D).
  1 mark
Answer: $12 5^{\circ}$ , option (D)
Where the marks are
- Uses tangent ⟂ radius and the linear pair to get $12 5^{\circ}$ , option (D) (1 mark)
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Q11 · 1 mark · MCQTriangles
If the diagonals of a quadrilateral divide each other proportionally, then it is a:
1. (A)parallelogram
2. (B)rectangle
3. (C)square
4. (D)trapezium
Try it first, then show our solutionSolution
1. If the diagonals AC and BD meet at O with $\frac{A O}{O C} = \frac{B O}{O D}$ , then $△ A O B \sim △ C O D$ (SAS similarity).
2. So $∠ O A B = ∠ O C D$ (alternate angles), which gives AB ∥ CD. The quadrilateral is therefore a trapezium, option (D).
  1 mark
Answer: Trapezium, option (D)
Where the marks are
- Identifies the quadrilateral as a trapezium, option (D) (1 mark)
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Q12 · 1 mark · MCQTriangles
In $△$ ABC, DE $∥$ BC (as shown in the figure). If AD = 2 cm, BD = 3 cm, BC = 7.5 cm, then the length of DE (in cm) is:
1. (A)2.5
2. (B)3
3. (C)5
4. (D)6
Figure in the paper: Triangle ABC with vertex A at the top, B at bottom left and C at bottom right. D lies on AB and E lies on AC, with DE drawn parallel to BC. See it in the official paper.
Try it first, then show our solutionSolution
1. $A B = 2 + 3 = 5$ cm. Since DE ∥ BC, $△ A D E \sim △ A B C$ (AA).
2. $\frac{D E}{B C} = \frac{A D}{A B}$ , so $D E = \frac{2}{5} \times 7.5 = 3$ cm. This is option (B).
  1 mark
Answer: 3 cm, option (B)
Where the marks are
- Uses similarity to get DE = 3 cm, option (B) (1 mark)
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Q13 · 1 mark · MCQReal Numbers
Given $HCF (2520, 6600) = 40$ , $LCM (2520, 6600) = 252 \times k$ , then the value of $k$ is:
1. (A)1650
2. (B)1600
3. (C)165
4. (D)1625
Try it first, then show our solutionSolution
1. Use HCF × LCM = product of the numbers: LCM $= \frac{2520 \times 6600}{40} = 2520 \times 165$ .
2. $252 k = 2520 \times 165$ , so $k = 10 \times 165 = 1650$ . This is option (A).
  1 mark
Answer: $k = 1650$ , option (A)
Where the marks are
- Uses HCF × LCM = product to get $k = 1650$ , option (A) (1 mark)
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Q14 · 1 mark · MCQReal Numbers
A pair of irrational numbers whose product is a rational number is:
1. (A) $(16, 4)$
2. (B) $(5, 2)$
3. (C) $(3, 27)$
4. (D) $(36, 2)$
Try it first, then show our solutionSolution
1. In (A) and (D), $16 = 4$ and $36 = 6$ are rational, so those pairs are not both irrational. In (B), $5 \times 2 = 10$ is irrational.
2. In (C), both numbers are irrational and $3 \times 27 = 81 = 9$ is rational. So the answer is option (C).
  1 mark
Answer: $(3, 27)$ , option (C)
Where the marks are
- Chooses option (C) with product 9 (1 mark)
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Q15 · 1 mark · MCQProbability
If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9; then the probability that this digit is an odd prime number is:
1. (A) $\frac{1}{3}$
2. (B) $\frac{2}{3}$
3. (C) $\frac{4}{9}$
4. (D) $\frac{5}{9}$
Try it first, then show our solutionSolution
1. The odd primes among 1 to 9 are 3, 5 and 7, so there are 3 favourable outcomes out of 9.
2. P $= \frac{3}{9} = \frac{1}{3}$ , which is option (A).
  1 mark
Answer: $\frac{1}{3}$ , option (A)
Where the marks are
- Counts 3 odd primes and gets $\frac{1}{3}$ , option (A) (1 mark)
Q16 · 1 mark · MCQStatistics
The mean of five observations is 15. If the mean of first three observations is 14 and that of the last three observations is 17, then the third observation is
1. (A)20
2. (B)19
3. (C)18
4. (D)17
Try it first, then show our solutionSolution
1. Sum of all five $= 5 \times 15 = 75$ . Sum of the first three $= 42$ and sum of the last three $= 51$ .
2. The third observation is counted twice: $42 + 51 = 75 + x_{3}$ , so $x_{3} = 18$ . This is option (C).
  1 mark
Answer: 18, option (C)
Where the marks are
- Uses the overlap of sums to get 18, option (C) (1 mark)
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Q17 · 1 mark · MCQAreas Related to Circles
Perimeter of a sector of a circle whose central angle is $9 0^{\circ}$ and radius 7 cm is:
1. (A)35 cm
2. (B)11 cm
3. (C)22 cm
4. (D)25 cm
Try it first, then show our solutionSolution
1. Arc length $= \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 = 11$ cm.
2. Perimeter $= 2 r + arc = 14 + 11 = 25$ cm, which is option (D).
  1 mark
Answer: 25 cm, option (D)
Where the marks are
- Adds the two radii to the arc length to get 25 cm, option (D) (1 mark)
Q18 · 1 mark · MCQCircles
In the given figure, O is the centre of the circle. MN is the chord and the tangent ML at point M makes an angle of $7 0^{\circ}$ with MN. The measure of $∠ MON$ is:
1. (A) $12 0^{\circ}$
2. (B) $14 0^{\circ}$
3. (C) $7 0^{\circ}$
4. (D) $9 0^{\circ}$
Figure in the paper: A circle with centre O. Chord MN is drawn. Radii OM and ON are joined forming triangle OMN. A tangent line through M extends to point L on one side. The angle between the tangent ML and chord MN at M is marked $7 0^{\circ}$ . See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O M L = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ O M N = 9 0^{\circ} - 7 0^{\circ} = 2 0^{\circ}$ .
2. OM = ON (radii), so $∠ O N M = 2 0^{\circ}$ . Then $∠ M O N = 18 0^{\circ} - 4 0^{\circ} = 14 0^{\circ}$ , which is option (B).
  1 mark
Answer: $14 0^{\circ}$ , option (B)
Where the marks are
- Uses tangent ⟂ radius and the isosceles triangle to get $14 0^{\circ}$ , option (B) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedCoordinate Geometry
Assertion (A): The point which divides the line segment joining the points A(1, 2) and B(–1, 1) internally in the ratio 1 : 2 is $(\frac{- 1}{3}, \frac{5}{3})$ .
Reason (R): The coordinates of the point which divides the line segment joining the points A $(x_{1}, y_{1})$ and B $(x_{2}, y_{2})$ in the ratio $m_{1} : m_{2}$ are $(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$ .
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the 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. Using the section formula with $m_{1} : m_{2} = 1 : 2$ : $x = \frac{1 ( - 1 ) + 2 ( 1 )}{3} = \frac{1}{3}$ and $y = \frac{1 ( 1 ) + 2 ( 2 )}{3} = \frac{5}{3}$ .
2. The point is $(\frac{1}{3}, \frac{5}{3})$ , not $(- \frac{1}{3}, \frac{5}{3})$ , so A is false. R is the correct section formula, so R is true. The answer is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Shows A is false and R is true, option (D) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A): In a cricket match, a batsman hits a boundary 9 times out of 45 balls he plays. The probability that in a given ball, he does not hit the boundary is $\frac{4}{5}$ .
Reason (R): $P (E) + P (not E) = 1$
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the 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. P(boundary) $= \frac{9}{45} = \frac{1}{5}$ , so P(no boundary) $= 1 - \frac{1}{5} = \frac{4}{5}$ . A is true.
2. R is true, and R is exactly the rule used to find the answer in A. So the answer is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Verifies A using R, option (A) (1 mark)
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Section B

Very short answer; internal choice in 22 and 23

Q21 · 2 marks · Very short answerProbability
One card is drawn at random from a well shuffled deck of 52 cards. Find the probability that the card drawn
1. (i)
  is queen of hearts;
  1 mark
2. (ii)
  is not a jack.
  1 mark
Try it first, then show our solutionSolution
1. Total number of outcomes = 52.
2. (i) There is only one queen of hearts, so P(queen of hearts) $= \frac{1}{52}$ .
  1 mark
3. (ii) There are 4 jacks, so 48 cards are not jacks. P(not a jack) $= \frac{48}{52} = \frac{12}{13}$ .
  1 mark
Answer: (i) $\frac{1}{52}$ ; (ii) $\frac{12}{13}$
Where the marks are
- (i) Gets $\frac{1}{52}$ (1 mark)
- (ii) Gets $\frac{48}{52} = \frac{12}{13}$ (1 mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerPair of Linear Equations in Two Variables
If $2 x + y = 13$ and $4 x - y = 17$ , find the value of $(x - y)$ .
Try it first, then show our solutionSolution
1. Let $2 x + y = 13$ be (i) and $4 x - y = 17$ be (ii).
2. Adding (i) and (ii): $6 x = 30$ , so $x = 5$ .
  1 mark
3. Substituting in (i): $10 + y = 13$ , so $y = 3$ .
  ½ mark
4. $x - y = 5 - 3 = 2$ .
  ½ mark
Answer: $x - y = 2$
Where the marks are
- Solves the equations to get $x = 5$ , $y = 3$ (1½ marks)
- States $x - y = 2$ (½ mark)
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OR
Q22(b) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Sum of two numbers is 105 and their difference is 45. Find the numbers.
Try it first, then show our solutionSolution
1. Let the numbers be $x$ and $y$ with $x > y$ . Then $x + y = 105$ …(i).
  ½ mark
2. Also $x - y = 45$ …(ii).
  ½ mark
3. Adding: $2 x = 150$ , so $x = 75$ , and then $y = 105 - 75 = 30$ . The numbers are 75 and 30.
  1 mark
Answer: 75 and 30
Where the marks are
- Forms $x + y = 105$ (½ mark)
- Forms $x - y = 45$ (½ mark)
- Solves to get 75 and 30 (1 mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerCoordinate Geometry
Find a relation between $x$ and $y$ such that the point P $(x, y)$ is equidistant from the points A(7, 1) and B(3, 5).
Try it first, then show our solutionSolution
1. PA = PB, so PA² = PB²: $(x - 7)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2}$ .
  1 mark
2. Expanding: $- 14 x - 2 y + 50 = - 6 x - 10 y + 34$ , so $- 8 x + 8 y + 16 = 0$ . The relation is $x - y - 2 = 0$ , i.e. $x - y = 2$ .
  1 mark
Answer: $x - y - 2 = 0$ (i.e. $x - y = 2$ )
Where the marks are
- Writes PA² = PB² using the distance formula (1 mark)
- Simplifies to $x - y - 2 = 0$ (1 mark)
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OR
Q23(b) · 2 marks · Very short answerCoordinate Geometry
Points A(–1, y) and B(5, 7) lie on a circle with centre O(2, –3y) such that AB is a diameter of the circle. Find the value of $y$ . Also, find the radius of the circle.
Try it first, then show our solutionSolution
1. AB is a diameter, so the centre O is the mid-point of AB. Equate the y-coordinate of the mid-point of A(–1, y) and B(5, 7) with that of O(2, –3y): $\frac{y + 7}{2} = - 3 y$ . (The x-coordinate checks: $\frac{- 1 + 5}{2} = 2$ .)
  ½ mark
2. $y + 7 = - 6 y \Rightarrow 7 y = - 7 \Rightarrow y = - 1$ . So A is $(- 1, - 1)$ and O is $(2, 3)$ .
  ½ mark
3. Radius $= O B = (5 - 2)^{2} + (7 - 3)^{2} = 9 + 16 = 25 = 5$ units.
  1 mark
Answer: y = –1; radius = 5 units
Where the marks are
- Uses that the centre is the mid-point of diameter AB and writes $\frac{y + 7}{2} = - 3 y$ (½ mark)
- Gets $y = - 1$ (½ mark)
- Uses the distance formula to get radius = 5 units (1 mark)
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Q24 · 2 marks · Very short answerTriangles
In the given figure, $\frac{EA}{EC} = \frac{EB}{ED}$ , prove that $△ EAB \sim △ ECD$ .
Figure in the paper: Two line segments AC and BD intersect at point E inside, forming triangles EAB and ECD. A is at upper left, B at lower left, C at lower right, D at upper right. E is the intersection point of AC and BD. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ E A B$ and $△ E C D$ : $\frac{E A}{E C} = \frac{E B}{E D}$ [given].
2. $∠ A E B = ∠ C E D$ [vertically opposite angles, since AC and BD cross at E].
  1 mark
3. Two sides are in the same ratio and the included angles are equal, so $△ E A B \sim △ E C D$ [SAS similarity criterion].
  1 mark
Answer: △EAB ~ △ECD by the SAS similarity criterion
Where the marks are
- States the given ratio and $∠ A E B = ∠ C E D$ as vertically opposite angles (1 mark)
- Concludes similarity by SAS criterion (1 mark)
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Q25 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $\frac{c o s 4 5 ^{\circ} + s i n 6 0 ^{\circ}}{s e c 3 0 ^{\circ} + cosec 3 0 ^{\circ}}$ .
Try it first, then show our solutionSolution
1. Use the standard values: $cos 4 5^{\circ} = \frac{1}{2}$ , $sin 6 0^{\circ} = \frac{3}{2}$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $cosec 3 0^{\circ} = 2$ . So the expression is $\frac{\frac{1}{2} + \frac{3}{2}}{\frac{2}{3} + 2}$ .
  1½ marks
2. Numerator $= \frac{2 + 6}{2 2}$ and denominator $= \frac{2 + 2 3}{3}$ . Their quotient is $\frac{3 ( 2 + 6 )}{2 2 \cdot 2 ( 1 + 3 )} = \frac{2 3 + 3 2}{4 2 ( 1 + 3 )}$ . Rationalising gives $\frac{3 3 + 3 2 - 6 - 3}{8}$ .
  ½ mark
Answer: $\frac{2 3 + 3 2}{4 2 ( 1 + 3 )}$ (= $\frac{3 3 + 3 2 - 6 - 3}{8}$ )
Where the marks are
- Substitutes all four correct trigonometric values (1½ marks)
- Simplifies to $\frac{2 3 + 3 2}{4 2 ( 1 + 3 )}$ or an equivalent form (½ mark)
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Section C

Short answer; internal choice in 26 and 29

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerArithmetic Progressions
If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first 20 terms.
Try it first, then show our solutionSolution
1. Let the first term be $a$ and the common difference be $d$ . Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . Then $S_{7} = \frac{7}{2} (2 a + 6 d) = 49 \Rightarrow a + 3 d = 7$ ...(i)
  ½ mark
2. $S_{17} = \frac{17}{2} (2 a + 16 d) = 289 \Rightarrow a + 8 d = 17$ ...(ii)
  ½ mark
3. Subtract (i) from (ii): $5 d = 10 \Rightarrow d = 2$ . Then $a = 7 - 6 = 1$ .
  1 mark
4. $S_{20} = \frac{20}{2} [2 (1) + 19 (2)] = 10 \times 40$
  ½ mark
5. $= 400$
  ½ mark
Answer: S₂₀ = 400
Where the marks are
- Forms $a + 3 d = 7$ from $S_{7}$ (½ mark)
- Forms $a + 8 d = 17$ from $S_{17}$ (½ mark)
- Solves to get $a = 1$ , $d = 2$ (1 mark)
- Substitutes correctly in the formula for $S_{20}$ (½ mark)
- Gets $S_{20} = 400$ (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerArithmetic Progressions
The ratio of the 10th term to its 30th term of an A.P. is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference of A.P.
Try it first, then show our solutionSolution
1. Let the first term be $a$ and the common difference be $d$ . Then $a_{10} = a + 9 d$ and $a_{30} = a + 29 d$ , so $\frac{a + 9 d}{a + 29 d} = \frac{1}{3}$ .
  ½ mark
2. $3 a + 27 d = a + 29 d \Rightarrow 2 a = 2 d \Rightarrow a = d$ ...(i)
  ½ mark
3. $S_{6} = \frac{6}{2} (2 a + 5 d) = 42$
  ½ mark
4. $\Rightarrow 2 a + 5 d = 14$ ...(ii)
  ½ mark
5. Put $d = a$ in (ii): $7 a = 14 \Rightarrow a = 2$ , so $d = 2$ .
  1 mark
Answer: First term a = 2, common difference d = 2
Where the marks are
- Writes $\frac{a + 9 d}{a + 29 d} = \frac{1}{3}$ (½ mark)
- Simplifies to $a = d$ (½ mark)
- Writes $S_{6} = \frac{6}{2} (2 a + 5 d) = 42$ (½ mark)
- Simplifies to $2 a + 5 d = 14$ (½ mark)
- Gets $a = 2$ and $d = 2$ (½ + ½ marks)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerPolynomials
Find the zeroes of the quadratic polynomial $x^{2} - 15$ and verify the relationship between the zeroes and the coefficients of the polynomial.
Try it first, then show our solutionSolution
1. Let $p (x) = x^{2} - 15 = x^{2} - (15)^{2} = (x - 15) (x + 15)$ .
  1 mark
2. $p (x) = 0$ when $x = 15$ or $x = - 15$ . So the zeroes are $15$ and $- 15$ .
  1 mark
3. Here $a = 1$ , $b = 0$ , $c = - 15$ . Sum of zeroes $= 15 + (- 15) = 0$ , and $- \frac{b}{a} = - \frac{0}{1} = 0$ . These are equal.
  ½ mark
4. Product of zeroes $= 15 \times (- 15) = - 15$ , and $\frac{c}{a} = \frac{- 15}{1} = - 15$ . These are equal. Hence the relationship is verified.
  ½ mark
Answer: Zeroes are √15 and –√15; sum = 0 = –b/a and product = –15 = c/a (verified)
Where the marks are
- Factorises $x^{2} - 15$ as $(x - 15) (x + 15)$ (1 mark)
- States the zeroes $\pm 15$ (1 mark)
- Verifies sum of zeroes $= - b / a$ (½ mark)
- Verifies product of zeroes $= c / a$ (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve the following system of linear equations graphically:
$x - y + 1 = 0$
$x + y = 5$
Try it first, then show our solutionSolution
1. For $x - y + 1 = 0$ , i.e. $y = x + 1$ , make a table of points: $(0, 1)$ , $(- 1, 0)$ , $(2, 3)$ . Plot them on graph paper and join them to get line $l_{1}$ .
  1 mark
2. For $x + y = 5$ , i.e. $y = 5 - x$ , make a table of points: $(0, 5)$ , $(5, 0)$ , $(2, 3)$ . Plot them on the same axes and join them to get line $l_{2}$ .
  1 mark
3. The two lines intersect at the point $(2, 3)$ . Check: $2 - 3 + 1 = 0$ and $2 + 3 = 5$ . So the solution is $x = 2$ , $y = 3$ .
  1 mark
Answer: x = 2, y = 3
Where the marks are
- Correct table and line for $x - y + 1 = 0$ (1 mark)
- Correct table and line for $x + y = 5$ (1 mark)
- Reads the intersection point and states $x = 2$ , $y = 3$ (1 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCoordinate Geometry
Find the ratio in which the line segment joining the points (5, 3) and (–1, 6) is divided by Y-axis.
Try it first, then show our solutionSolution
1. Any point on the Y-axis has x-coordinate 0. Let the Y-axis meet the segment at $P (0, y)$ , and let P divide the segment joining $(5, 3)$ and $(- 1, 6)$ in the ratio $k : 1$ .
  ½ mark
2. By the section formula, x-coordinate of P $= \frac{k ( - 1 ) + 1 ( 5 )}{k + 1} = 0$ .
  1 mark
3. $- k + 5 = 0 \Rightarrow k = 5$ .
  1 mark
4. So the required ratio is $5 : 1$ . (The point of division is $(0, \frac{5 \cdot 6 + 3}{6}) = (0, \frac{11}{2})$ .)
  ½ mark
Answer: 5 : 1
Where the marks are
- Takes the point on the Y-axis as $(0, y)$ (½ mark)
- Applies the section formula to the x-coordinate with ratio $k : 1$ (1 mark)
- Gets $k = 5$ (1 mark)
- States the ratio $5 : 1$ (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCoordinate Geometry
P(–2, 5) and Q(3, 2) are two points. Find the coordinates of the point R on line segment PQ such that PR = 2QR.
Try it first, then show our solutionSolution
1. Let $R (x, y)$ lie on PQ. Since $P R = 2 QR$ , $P R : R Q = 2 : 1$ .
  ½ mark
2. By the section formula with $m_{1} = 2$ , $m_{2} = 1$ , P(–2, 5) and Q(3, 2): $x = \frac{2 ( 3 ) + 1 ( - 2 )}{2 + 1} = \frac{4}{3}$ .
  1 mark
3. $y = \frac{2 ( 2 ) + 1 ( 5 )}{2 + 1} = \frac{9}{3} = 3$ .
  1 mark
4. Hence the coordinates of R are $(\frac{4}{3}, 3)$ .
  ½ mark
Answer: R(4/3, 3)
Where the marks are
- Writes $P R : R Q = 2 : 1$ (½ mark)
- Correct x-coordinate $\frac{4}{3}$ by the section formula (1 mark)
- Correct y-coordinate 3 (1 mark)
- States R as $(\frac{4}{3}, 3)$ (½ mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{s i n θ - 2 s i n ^{3} θ}{2 c o s ^{3} θ - c o s θ} = tan θ$ .
Try it first, then show our solutionSolution
1. LHS $= \frac{sin θ - 2 sin ^{3} θ}{2 cos ^{3} θ - cos θ} = \frac{sin θ ( 1 - 2 sin ^{2} θ )}{cos θ ( 2 cos ^{2} θ - 1 )}$ [taking common factors]
  ½ mark
2. $= \frac{tan θ ( 1 - 2 sin ^{2} θ )}{2 ( 1 - sin ^{2} θ ) - 1}$ [using $\frac{s i n θ}{c o s θ} = tan θ$ and $cos^{2} θ = 1 - sin^{2} θ$ ]
  1 mark
3. $= \frac{tan θ ( 1 - 2 sin ^{2} θ )}{1 - 2 sin ^{2} θ}$
  1 mark
4. $= tan θ =$ RHS. Hence proved.
  ½ mark
Answer: LHS simplifies to tan θ = RHS (proved)
Where the marks are
- Takes out $sin θ$ and $cos θ$ as common factors (½ mark)
- Uses $cos^{2} θ = 1 - sin^{2} θ$ and $tan θ = sin θ / cos θ$ (1 mark)
- Simplifies the denominator to $1 - 2 sin^{2} θ$ (1 mark)
- Cancels and concludes $= tan θ$ (½ mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCircles
Prove that the tangents drawn at the end points of a chord of a circle makes equal angles with the chord.
Try it first, then show our solutionSolution
1. Figure: draw a circle with centre O and a chord AB. Draw tangents at A and B that meet at an external point P. Join PA and PB.
  Given: AB is a chord, and PA and PB are tangents at A and B.
  To prove: $∠ P A B = ∠ P B A$ .
  1 mark
2. In $△ P A B$ : $P A = P B$ [lengths of tangents drawn from an external point to a circle are equal].
  1 mark
3. So $△ P A B$ is isosceles. Hence $∠ P A B = ∠ P B A$ [angles opposite equal sides are equal]. Therefore the tangents make equal angles with the chord.
  (Alternative: $O A ⊥ P A$ and $O B ⊥ P B$ [tangent ⟂ radius], and $∠ O A B = ∠ O B A$ since $O A = O B$ . So $∠ P A B = 9 0^{\circ} - ∠ O A B = 9 0^{\circ} - ∠ O B A = ∠ P B A$ . This version also works when AB is a diameter and the tangents are parallel.)
  1 mark
Answer: Since PA = PB, ∠PAB = ∠PBA; the tangents make equal angles with the chord (proved)
Where the marks are
- Correct labelled figure with given and to prove (1 mark)
- States $P A = P B$ with the reason (equal tangents from an external point) (1 mark)
- Concludes $∠ P A B = ∠ P B A$ (angles opposite equal sides) (1 mark)
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Section D

Long answer; internal choice in 32 and 35

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and by doing so, the time of flight is increased by 30 minutes. Find the original duration of the flight.
Try it first, then show our solutionSolution
1. Let the original speed be $x$ km/h. The reduced speed is $(x - 100)$ km/h. Time = distance ÷ speed, and 30 min $= \frac{1}{2}$ h. So $\frac{2800}{x - 100} - \frac{2800}{x} = \frac{1}{2}$ .
  2 marks
2. $2800 \cdot \frac{x - ( x - 100 )}{x ( x - 100 )} = \frac{1}{2} \Rightarrow 560000 = x^{2} - 100 x \Rightarrow x^{2} - 100 x - 560000 = 0$ .
  ½ mark
3. $x^{2} - 800 x + 700 x - 560000 = 0 \Rightarrow (x - 800) (x + 700) = 0$ .
  1 mark
4. $x = 800$ or $x = - 700$ . Speed cannot be negative, so $x = 800$ km/h.
  ½ mark
5. Original duration $= \frac{2800}{800} = \frac{7}{2}$ h $= 3$ h 30 min.
  1 mark
Answer: Original duration = 3.5 hours (3 h 30 min)
Where the marks are
- Forms the equation $\frac{2800}{x - 100} - \frac{2800}{x} = \frac{1}{2}$ (2 marks)
- Reduces it to $x^{2} - 100 x - 560000 = 0$ (½ mark)
- Factorises as $(x - 800) (x + 700) = 0$ (1 mark)
- Rejects the negative value and takes $x = 800$ (½ mark)
- Finds the original duration $\frac{7}{2}$ h (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerQuadratic Equations
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $2 \frac{16}{21}$ , find the fraction.
Try it first, then show our solutionSolution
1. Let the numerator be $x$ . Then the denominator is $2 x + 1$ .
  ½ mark
2. Fraction $= \frac{x}{2 x + 1}$ , and its reciprocal is $\frac{2 x + 1}{x}$ .
  ½ mark
3. Given: $\frac{x}{2 x + 1} + \frac{2 x + 1}{x} = 2 \frac{16}{21} = \frac{58}{21}$ .
  1½ marks
4. $21 [x^{2} + (2 x + 1)^{2}] = 58 x (2 x + 1) \Rightarrow 105 x^{2} + 84 x + 21 = 116 x^{2} + 58 x \Rightarrow 11 x^{2} - 26 x - 21 = 0$ .
  ½ mark
5. $11 x^{2} - 33 x + 7 x - 21 = 0 \Rightarrow 11 x (x - 3) + 7 (x - 3) = 0 \Rightarrow (x - 3) (11 x + 7) = 0$ .
  1 mark
6. $x = 3$ or $x = - \frac{7}{11}$ . The numerator must be a whole number here, so $x \neq = - \frac{7}{11}$ . Hence $x = 3$ .
  ½ mark
7. Fraction $= \frac{3}{2 ( 3 ) + 1} = \frac{3}{7}$ . Check: $\frac{3}{7} + \frac{7}{3} = \frac{9 + 49}{21} = \frac{58}{21}$ ✓
  ½ mark
Answer: The fraction is 3/7
Where the marks are
- Takes the numerator as x and the denominator as 2x + 1 (½ mark)
- Writes the fraction $\frac{x}{2 x + 1}$ (½ mark)
- Forms the equation $\frac{x}{2 x + 1} + \frac{2 x + 1}{x} = \frac{58}{21}$ (1½ marks)
- Simplifies to $11 x^{2} - 26 x - 21 = 0$ (½ mark)
- Factorises as $(x - 3) (11 x + 7) = 0$ (1 mark)
- Rejects $- \frac{7}{11}$ and takes $x = 3$ (½ mark)
- States the fraction $\frac{3}{7}$ (½ mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerTriangles
State and prove Basic Proportionality theorem.
Try it first, then show our solutionSolution
1. Statement (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle so that it meets the other two sides in distinct points, then it divides those two sides in the same ratio.
  1 mark
2. Given: In $△ A B C$ , a line parallel to $B C$ meets $A B$ at $D$ and $A C$ at $E$ , so $D E ∥ B C$ .
  ½ mark
3. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  ½ mark
4. Construction: Join $B E$ and $C D$ . Draw $E N ⊥ A B$ (with $N$ on $A B$ ) and $D M ⊥ A C$ (with $M$ on $A C$ ).
  ½ mark
5. Figure: Draw $△ A B C$ with $A$ at the top. Mark $D$ on $A B$ and $E$ on $A C$ with $D E ∥ B C$ . Show the segments $B E$ and $C D$ , and the perpendiculars $E N$ to $A B$ and $D M$ to $A C$ .
  ½ mark
6. Proof: Take $E N$ as the height. Then $ar (△ A D E) = \frac{1}{2} \cdot A D \cdot E N$ and $ar (△ B D E) = \frac{1}{2} \cdot D B \cdot E N$ .
  So $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ ...(1)
  Now take $D M$ as the height. Then $ar (△ A D E) = \frac{1}{2} \cdot A E \cdot D M$ and $ar (△ D E C) = \frac{1}{2} \cdot E C \cdot D M$ .
  So $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ...(2)
  1 mark
7. $△ B D E$ and $△ D E C$ stand on the same base $D E$ and lie between the same parallels $D E$ and $B C$ . Hence $ar (B D E) = ar (D E C)$ ...(3)
  From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: Statement of BPT and proof that $\frac{A D}{D B} = \frac{A E}{E C}$ when $D E ∥ B C$
Where the marks are
- States the theorem correctly (1 mark)
- Writes the Given, To prove and Construction, and draws a correct labelled figure, ½ mark each (2 marks)
- Expresses the two area ratios as $A D / D B$ and $A E / E C$ using the perpendicular heights (1 mark)
- Uses ar(BDE) = ar(DEC) [same base, between same parallels] and concludes (1 mark)
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Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are $4 5^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower.
Try it first, then show our solutionSolution
1. Figure: Let $P$ be the point on the ground and $A$ the foot of the building. Let $A B = 20$ m be the building and $B C = h$ m the tower on top of it. Let $P A = x$ m. Then $∠ B P A = 4 5^{\circ}$ , $∠ C P A = 6 0^{\circ}$ and $∠ P A C = 9 0^{\circ}$ .
  1 mark
2. In right $△ B P A$ : $tan 4 5^{\circ} = \frac{A B}{P A}$ , so $1 = \frac{20}{x}$ .
  1 mark
3. Therefore $x = 20$ m ...(i)
  ½ mark
4. In right $△ C P A$ : $tan 6 0^{\circ} = \frac{A C}{P A}$ , so $3 = \frac{h + 20}{x}$ .
  1 mark
5. Therefore $h + 20 = x 3$ ...(ii)
  ½ mark
6. Substitute (i) in (ii): $h + 20 = 203$ , so $h = 20 (3 - 1)$ m $\approx 14.64$ m.
  Hence the height of the tower is $20 (3 - 1)$ m.
  1 mark
Answer: Height of tower $= 20 (3 - 1)$ m $\approx 14.64$ m
Where the marks are
- Draws a correct labelled figure (1 mark)
- Uses $tan 4 5^{\circ} = 20/ x$ (1 mark)
- Obtains $x = 20$ m (½ mark)
- Uses $tan 6 0^{\circ} = (h + 20) / x$ (1 mark)
- Gets $h + 20 = x 3$ (½ mark)
- Finds $h = 20 (3 - 1)$ m (1 mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A solid iron pole consists of a solid cylinder of height 200 cm and base diameter 28 cm, which is surmounted by another cylinder of height 50 cm and radius 7 cm. Find the mass of the pole, given that $1 cm^{3}$ of iron has approximately 8 g mass.
Try it first, then show our solutionSolution
1. Radius of the lower cylinder $= \frac{28}{2} = 14$ cm, and its height is $200$ cm.
  The upper cylinder has radius $7$ cm and height $50$ cm.
  ½ mark
2. Volume of the lower cylinder $= π r^{2} h = \frac{22}{7} \times 14 \times 14 \times 200 = 123200 cm^{3}$
  1 mark
3. Volume of the upper cylinder $= \frac{22}{7} \times 7 \times 7 \times 50 = 7700 cm^{3}$
  1 mark
4. Total volume of the pole $= 123200 + 7700 = 130900 cm^{3}$
  1 mark
5. Mass $= 8 \times 130900$
  1 mark
6. $= 1047200$ g $= 1047.2$ kg
  ½ mark
Answer: Mass of pole $= 1047200$ g $= 1047.2$ kg
Where the marks are
- Takes the radius of the lower cylinder as 14 cm (½ mark)
- Finds the volume of the lower cylinder (1 mark)
- Finds the volume of the upper cylinder (1 mark)
- Gets the total volume 130900 cm³ (1 mark)
- Multiplies by 8 g per cm³ (1 mark)
- Final mass 1047200 g, i.e. 1047.2 kg (½ mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 4 mm, find its surface area. Also, find its volume.
Figure in the paper: A capsule shape: a cylinder with a hemisphere at each end. Overall length marked 14 mm (horizontal) and diameter marked 4 mm (vertical) at the left end. See it in the official paper.
Try it first, then show our solutionSolution
1. Radius of each hemisphere $=$ radius of the cylinder $= \frac{4}{2} = 2$ mm
  ½ mark
2. Length of the cylindrical part $= 14 - 2 - 2 = 10$ mm
  ½ mark
3. Surface area $=$ CSA of cylinder $+ 2 \times$ CSA of hemisphere $= 2 π r h + 2 (2 π r^{2})$
  $= 2 \times \frac{22}{7} \times 2 \times 10 + 2 \times 2 \times \frac{22}{7} \times 2 \times 2 = \frac{880}{7} + \frac{352}{7}$
  1 mark
4. $= \frac{1232}{7} = 176 mm^{2}$
  1 mark
5. Volume $=$ volume of cylinder $+ 2 \times$ volume of hemisphere $= π r^{2} h + 2 \times \frac{2}{3} π r^{3}$
  $= \frac{22}{7} \times 2 \times 2 \times 10 + 2 \times \frac{2}{3} \times \frac{22}{7} \times 2 \times 2 \times 2 = \frac{880}{7} + \frac{704}{21}$
  1 mark
6. $= \frac{2640 + 704}{21} = \frac{3344}{21} \approx 159.24 mm^{3}$
  1 mark
Answer: Surface area $= 176$ mm²; Volume $= \frac{3344}{21}$ mm³ $\approx 159.24$ mm³
Where the marks are
- Radius 2 mm (½ mark)
- Cylinder length 10 mm (½ mark)
- Correct surface-area expression (1 mark)
- Surface area 176 mm² (1 mark)
- Correct volume expression (1 mark)
- Volume 3344/21 ≈ 159.24 mm³ (1 mark)
Open Q35(b) on its own page

Section E

Case study based; internal choice in 36, 37 and 38

Q36 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
A stable owner has four horses. He usually tie these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze.
The figure shows a square field ABCD with side 20 m, with a horse tied at each corner by a rope of length 7 m, grazing in a quarter circle.
Based on the above, answer the following questions:
1. (i)
  Find the area of the square shaped grass field.
  1 mark
2. (ii)
  (a) Find the area of the total field in which these horses can graze.
  2 marks
3. OR
  (ii)
  (b) If the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse. (Use $π = 3.14$ )
  2 marks
4. (iii)
  What is area of the field that is left ungrazed, if the length of the rope of each horse is 7 cm?
  1 mark
Figure in the paper: Square field ABCD with A at top left, D at top right, B at bottom left and C at bottom right. Side AB is marked 20 m and the top side AD has a 7 m arrow marked from A. Each corner has a horse tied with a rope drawn as a quarter-circle arc of radius 7 m inside the square. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Area of the square field $= (side)^{2} = 20 \times 20 = 400 m^{2}$
  1 mark
2. (ii)(a) Each horse grazes a quadrant of radius 7 m (the corner angle is $9 0^{\circ}$ ). So the total grazed area $= 4 \times \frac{1}{4} π r^{2} = 4 \times \frac{1}{4} \times \frac{22}{7} \times 7 \times 7$
  1 mark
3. (ii)(a) $= 154 m^{2}$
  1 mark
4. (ii)(b) With a 10 m rope, area grazed by one horse $= \frac{1}{4} π r^{2} = \frac{1}{4} \times 3.14 \times 10 \times 10$
  1 mark
5. (ii)(b) $= 78.5 m^{2}$
  1 mark
6. (iii) The rope is 7 cm $= 0.07$ m. Area grazed by the four horses $= 4 \times \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 154 cm^{2} = 0.0154 m^{2}$
  ½ mark
7. (iii) Ungrazed area $= 400 - 0.0154 = 399.9846 m^{2}$
  ½ mark
Answer: (i) 400 m²; (ii)(a) 154 m², (ii)(b) 78.5 m²; (iii) 399.9846 m²
Where the marks are
- (i) Area 400 m² (1 mark)
- (ii)(a) Writes 4 quadrants of radius 7 m (1 mark) and gets 154 m² (1 mark)
- (ii)(b) Writes a quadrant of radius 10 m with π = 3.14 (1 mark) and gets 78.5 m² (1 mark)
- (iii) Grazed area 154 cm² = 0.0154 m² (½ mark); ungrazed area 399.9846 m² (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedStatistics
Vocational training complements traditional education by providing practical skills and hands-on experience. While education equips individuals with a broad knowledge base, vocational training focuses on job-specific skills, enhancing employability thus making the student self-reliant. Keeping this in view, a teacher made the following table giving the frequency distribution of students/adults undergoing vocational training from the training institute.
Age (in years) 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54
Number of participants 62 132 96 37 13 11 10 4
From the above answer the following questions:
1. (i)
  What is the lower limit of the modal class of the above data?
  1 mark
2. (ii)
  (a) Find the median class of the above data.
  2 marks
3. OR
  (ii)
  (b) Find the number of participants of age less than 50 years who undergo vocational training.
  2 marks
4. (iii)
  Give the empirical relationship between mean, median and mode.
  1 mark
Try it first, then show our solutionSolution
1. (i) The highest frequency is 132, which belongs to the class 20–24. Make the classes continuous by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit. The modal class becomes 19.5–24.5, so its lower limit is 19.5.
  1 mark
2. (ii)(a) Cumulative frequency table, using continuous classes:
  Age 14.5–19.5 19.5–24.5 24.5–29.5 29.5–34.5 34.5–39.5 39.5–44.5 44.5–49.5 49.5–54.5
  f 62 132 96 37 13 11 10 4
  cf 62 194 290 327 340 351 361 365
  1 mark
3. (ii)(a) Here $n = 365$ , so $\frac{n}{2} = 182.5$ .
  ½ mark
4. (ii)(a) The first cumulative frequency greater than 182.5 is 194. So the median class is 19.5–24.5 (that is, 20–24).
  ½ mark
5. (ii)(b) Participants younger than 50 years are those in all classes up to 45–49. Their number $= 62 + 132 + 96 + 37 + 13 + 11 + 10 = 361$ . (Equivalently, $365 - 4 = 361$ .)
  2 marks
6. (iii) Empirical relationship: $3 Median = Mode + 2 Mean$
  1 mark
Answer: (i) 19.5; (ii)(a) 19.5–24.5, (ii)(b) 361; (iii) 3 Median = Mode + 2 Mean
Where the marks are
- (i) Identifies the modal class 19.5–24.5 with lower limit 19.5 (1 mark)
- (ii)(a) Correct cumulative frequency table (1 mark), n/2 = 182.5 (½ mark), median class 19.5–24.5 (½ mark)
- (ii)(b) Adds the frequencies of the classes below 50 to get 361 (2 marks)
- (iii) States 3 Median = Mode + 2 Mean (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedReal Numbers
Teaching Mathematics through activities is a powerful approach that enhances students' understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announces the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to second student. Second student also multiplied it by a prime number and passed it to third student. In this way by multiplying to a prime number, the last student got 173250.
Now, Mukta asked some questions as given below to the students:
1. (i)
  What is the least prime number used by students?
  1 mark
2. (ii)
  (a) How many students are in the class?
  2 marks
3. OR
  (ii)
  (b) What is the highest prime number used by students?
  2 marks
4. (iii)
  Which prime number has been used maximum times?
  1 mark
Try it first, then show our solutionSolution
1. Prime factorisation: $173250 = 2 \times 5^{3} \times 3^{2} \times 7 \times 11$ .
  Check: $2 \times 125 \times 9 \times 77 = 173250$ .
  The factor 2 is the number announced by the teacher. The remaining primes, $5, 5, 5, 3, 3, 7, 11$ , were multiplied in by the students.
2. (i) The primes used by the students are 3, 5, 7 and 11. The least of these is 3.
  1 mark
3. (ii)(a) $173250 = 2 \times 5^{3} \times 3^{2} \times 7 \times 11$ . Leaving out the starting 2, the students used $5^{3} \times 3^{2} \times 7 \times 11$ .
  1 mark
4. (ii)(a) Each student multiplied by one prime, so the number of students $= 3 + 2 + 1 + 1 = 7$ .
  1 mark
5. (ii)(b) From $173250 = 2 \times 5^{3} \times 3^{2} \times 7 \times 11$ , the primes used by the students are 3, 5, 7 and 11.
  1 mark
6. (ii)(b) The highest prime used is 11.
  1 mark
7. (iii) The prime 5 appears 3 times (power $5^{3}$ ), more than any other prime. So 5 was used the maximum number of times.
  1 mark
Answer: (i) 3; (ii)(a) 7 students, (ii)(b) 11; (iii) 5
Where the marks are
- (i) Least prime used by the students is 3 (1 mark)
- (ii)(a) Correct prime factorisation of 173250 (1 mark); counts 7 students (1 mark)
- (ii)(b) Correct prime factorisation (1 mark); highest prime is 11 (1 mark)
- (iii) The prime 5 is used the most times (1 mark)
Open Q38 on its own page

Age (in years)	15-19	20-24	25-29	30-34	35-39	40-44	45-49	50-54
Number of participants	62	132	96	37	13	11	10	4

Age	14.5–19.5	19.5–24.5	24.5–29.5	29.5–34.5	34.5–39.5	39.5–44.5	44.5–49.5	49.5–54.5
f	62	132	96	37	13	11	10	4
cf	62	194	290	327	340	351	361	365

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.