MathsCBSE Class 10 · Basic · Board exam 2026 (session 2025-26)

CBSE Class 10 Mathematics Basic: 2026 board paper, set 430/2/1

The question paper set in the CBSE Class 10 board exam of 2026, 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/2026/X/Mathematics_Standard.zip (a zip; this paper is the file 430-2-1_Mathematics 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	Very Short Answer; internal choice in 21 and 25
C	26-31	Short Answer; internal choice in 26 and 28
D	32-35	Long Answer; internal choice in 32 and 35
E	36-38	Case study based; internal choice in each case study for 2-mark subparts

Instructions in brief: All questions compulsory, no overall choice; internal choices in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 case-study questions in Section E; use π = 22/7 where not stated; calculator not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The value of $(HCF - LCM)$ for the two numbers 3 and 5 is :
1. (A)2
2. (B)4
3. (C)14
4. (D)-14
Try it first, then show our solutionSolution
1. 3 and 5 are co-prime, so HCF $= 1$ and LCM $= 3 \times 5 = 15$ .
2. HCF $-$ LCM $= 1 - 15 = - 14$ , so the answer is option (D).
  1 mark
Answer: −14, option (D)
Where the marks are
- Finds HCF = 1 and LCM = 15, giving −14 (1 mark)
Q2 · 1 mark · MCQReal Numbers
The number $2^{n}$ , where n is a natural number, cannot end with the digit :
1. (A)4
2. (B)6
3. (C)2
4. (D)0
Try it first, then show our solutionSolution
1. For a number to end in 0 it must have both 2 and 5 as prime factors.
2. The only prime factor of $2^{n}$ is 2, so $2^{n}$ can never end with 0. The answer is option (D).
  1 mark
Answer: 0, option (D)
Where the marks are
- Uses prime factorisation (no factor 5) to conclude the digit 0 (1 mark)
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Q3 · 1 mark · MCQPair of Linear Equations in Two Variables
If $(0, 0)$ is the solution of the equation $x + y = c - 1$ , then the value of c is :
1. (A)0
2. (B)1
3. (C)-1
4. (D)any real number
Try it first, then show our solutionSolution
1. Substitute $x = 0, y = 0$ : $0 = c - 1$ .
2. So $c = 1$ , which is option (B).
  1 mark
Answer: c = 1, option (B)
Where the marks are
- Substitutes (0,0) to get c = 1 (1 mark)
Q4 · 1 mark · MCQQuadratic Equations
The value of 'a' for which $a x^{2} + 3 x + 1 = 0$ has real and equal roots is :
1. (A) $\frac{4}{9}$
2. (B) $\frac{9}{4}$
3. (C) $\frac{3}{2}$
4. (D) $\frac{2}{3}$
Try it first, then show our solutionSolution
1. For real and equal roots, the discriminant must be zero: $D = b^{2} - 4 a c = 0$ .
2. $3^{2} - 4 (a) (1) = 0 \Rightarrow 9 = 4 a \Rightarrow a = \frac{9}{4}$ , which is option (B).
  1 mark
Answer: a = 9/4, option (B)
Where the marks are
- Sets $9 - 4 a = 0$ to get $a = \frac{9}{4}$ (1 mark)
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Q5 · 1 mark · MCQQuadratic Equations
Which of the following equations is a quadratic equation ?
1. (A) $x^{2} = (x + 1)^{2}$
2. (B) $(x - 1) (x + 2) = 2 x + 1$
3. (C) $(x + 2)^{3} = 2 x (x^{2} - 1)$
4. (D) $x = x^{2}$
Try it first, then show our solutionSolution
1. (A) $x^{2} = x^{2} + 2 x + 1 \Rightarrow 2 x + 1 = 0$ , which is linear. (C) $x^{3} + 6 x^{2} + 12 x + 8 = 2 x^{3} - 2 x \Rightarrow - x^{3} + 6 x^{2} + 14 x + 8 = 0$ , which is cubic. (D) contains $x$ , so it is not a polynomial equation.
2. (B) $x^{2} + x - 2 = 2 x + 1 \Rightarrow x^{2} - x - 3 = 0$ . This has degree 2, so it is a quadratic equation. Option (B).
  1 mark
Answer: Option (B): $(x - 1) (x + 2) = 2 x + 1$
Where the marks are
- Reduces (B) to $x^{2} - x - 3 = 0$ and identifies it as the quadratic (1 mark)
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Q6 · 1 mark · MCQCoordinate Geometry
The distance of the point $(2, 3)$ from the origin is :
1. (A)2
2. (B)3
3. (C)5
4. (D) $13$
Try it first, then show our solutionSolution
1. Distance from the origin $= x^{2} + y^{2} = 4 + 9 = 13$ , which is option (D).
  1 mark
Answer: √13, option (D)
Where the marks are
- Applies the distance formula to get √13 (1 mark)
Q7 · 1 mark · MCQCoordinate Geometry
The mid-point of the line segment joining points $(1, 3)$ and $(1, - 3)$ lies :
1. (A)at the origin
2. (B)in the second quadrant
3. (C)on x-axis
4. (D)on y-axis
Try it first, then show our solutionSolution
1. Mid-point $= (\frac{1 + 1}{2}, \frac{3 - 3}{2}) = (1, 0)$ .
2. Its y-coordinate is 0, so the mid-point lies on the x-axis: option (C).
  1 mark
Answer: On x-axis, option (C)
Where the marks are
- Finds the mid-point (1,0) and places it on the x-axis (1 mark)
Q8 · 1 mark · MCQTriangles
In the given figure, if $D E ∥ B C$ , $A D = 1.5$ cm, $D B = 3$ cm and $E C = 2$ cm, the length of AC is :
1. (A)1.5 cm
2. (B)3 cm
3. (C)3.5 cm
4. (D)4.5 cm
Figure in the paper: Triangle ABC with D on side AB and E on side AC, DE parallel to BC. Marked: AD = 1.5 cm, DB = 3 cm (along side AB), EC = 2 cm (along side AC). See it in the official paper.
Try it first, then show our solutionSolution
1. Since $D E ∥ B C$ , by the Basic Proportionality Theorem: $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{1.5}{3} = \frac{A E}{2}$ , so $A E = 1$ cm.
  ½ mark
2. $A C = A E + E C = 1 + 2 = 3$ cm. Hence option (B).
  ½ mark
Answer: AC = 3 cm, option (B)
Where the marks are
- Applies BPT to get AE = 1 cm and AC = AE + EC = 3 cm, choosing option (B) (1 mark)
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Q9 · 1 mark · MCQCircles
In two concentric circles, a tangent to the smaller circle will intersect the larger circle at :
1. (A)zero point
2. (B)one point
3. (C)two points
4. (D)three points
Try it first, then show our solutionSolution
1. A tangent to the smaller circle passes through points inside the larger circle, so it is a secant of the larger circle.
2. A secant cuts a circle at two points, so the answer is option (C).
  1 mark
Answer: Two points, option (C)
Where the marks are
- Recognises the tangent is a chord/secant of the larger circle (1 mark)
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Q10 · 1 mark · MCQIntroduction to Trigonometry
The value of $\frac{2 t a n 6 0 ^{\circ}}{1 - t a n ^{2} 6 0 ^{\circ}}$ is :
1. (A) $- 3$
2. (B) $3$
3. (C) $- \frac{1}{3}$
4. (D) $- 3$
Try it first, then show our solutionSolution
1. $tan 6 0^{\circ} = 3$ , so the expression is $\frac{2 3}{1 - 3} = \frac{2 3}{- 2} = - 3$ . Option (D).
  1 mark
Answer: $- 3$ , option (D)
Where the marks are
- Substitutes $tan 6 0^{\circ} = 3$ and simplifies to $- 3$ (1 mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
The value of $(sin^{2} A + cos^{2} A) + (sec^{2} A - tan^{2} A) - (cot^{2} A - cosec^{2} A)$ is :
1. (A)1
2. (B)-1
3. (C)3
4. (D)-2
Try it first, then show our solutionSolution
1. Use the identities $sin^{2} A + cos^{2} A = 1$ , $sec^{2} A - tan^{2} A = 1$ and $cot^{2} A - cosec^{2} A = - 1$ .
2. Value $= 1 + 1 - (- 1) = 3$ , which is option (C).
  1 mark
Answer: 3, option (C)
Where the marks are
- Applies the three identities to get 3 (1 mark)
Q12 · 1 mark · MCQSome Applications of Trigonometry
The length of the shadow of a tower when the sun's altitude changes from $3 0^{\circ}$ to $6 0^{\circ}$ will :
1. (A)become shorter
2. (B)become longer
3. (C)remain same
4. (D)be doubled
Try it first, then show our solutionSolution
1. Shadow $= \frac{h}{t a n θ}$ . At 30° it is $h 3$ and at 60° it is $\frac{h}{3}$ .
2. The shadow decreases, so it becomes shorter: option (A).
  1 mark
Answer: Becomes shorter, option (A)
Where the marks are
- Compares shadow lengths $h 3$ and $h / 3$ to conclude 'shorter' (1 mark)
Q13 · 1 mark · MCQAreas Related to Circles
In the given figure, the shaded region represents :
1. (A)minor sector
2. (B)major sector
3. (C)minor segment
4. (D)major segment
Figure in the paper: A circle with centre O. The shaded region is the part of the disc lying below a horizontal chord near the bottom of the circle, bounded by the chord and the arc below it (the smaller side, not containing O). See it in the official paper.
Try it first, then show our solutionSolution
1. The region is bounded by a chord and its arc, so it is a segment, not a sector (a sector is bounded by two radii).
  ½ mark
2. The region lies on the smaller side of the chord and does not contain the centre O, so it is the minor segment. Option (C).
  ½ mark
Answer: Minor segment, option (C)
Where the marks are
- Identifies the shaded region as the minor segment (1 mark)
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Q14 · 1 mark · MCQAreas Related to Circles
The perimeter of a quadrant of a circle of radius r is :
1. (A) $\frac{1}{4} π r^{2}$
2. (B) $\frac{1}{4} π r^{2} + 2 r$
3. (C) $\frac{π r}{2}$
4. (D) $\frac{π r}{2} + 2 r$
Try it first, then show our solutionSolution
1. The perimeter of a quadrant is its arc plus two radii: $\frac{1}{4} (2 π r) + 2 r = \frac{π r}{2} + 2 r$ , which is option (D).
  1 mark
Answer: πr/2 + 2r, option (D)
Where the marks are
- Adds the quarter arc and two radii (1 mark)
Q15 · 1 mark · MCQSurface Areas and Volumes
A cone and a cylinder have the same base radius and volume. The (height of cone : height of cylinder) is :
1. (A)1 : 1
2. (B)3 : 1
3. (C)1 : 3
4. (D)3 : 2
Try it first, then show our solutionSolution
1. Equal volumes give $\frac{1}{3} π r^{2} h_{1} = π r^{2} h_{2}$ , so $h_{1} = 3 h_{2}$ .
2. $h_{1} : h_{2} = 3 : 1$ , which is option (B).
  1 mark
Answer: 3 : 1, option (B)
Where the marks are
- Equates the volumes to get 3 : 1 (1 mark)
Q16 · 1 mark · MCQStatistics
In the formula of mode given by $mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ , $f_{1}$ denotes the :
1. (A)frequency of the modal class
2. (B)frequency of class preceding modal class
3. (C)frequency of class succeeding modal class
4. (D)cumulative frequency of modal class
Try it first, then show our solutionSolution
1. In the mode formula, $f_{1}$ is the frequency of the modal class, $f_{0}$ that of the preceding class and $f_{2}$ that of the succeeding class. So the answer is option (A).
  1 mark
Answer: Frequency of the modal class, option (A)
Where the marks are
- Identifies $f_{1}$ as the modal class frequency (1 mark)
Q17 · 1 mark · MCQStatistics
For a distribution, if mean = median = a, then its mode is :
1. (A)3a
2. (B)2a
3. (C)a
4. (D)0
Try it first, then show our solutionSolution
1. Use the empirical relation Mode = 3 Median − 2 Mean $= 3 a - 2 a = a$ , which is option (C).
  1 mark
Answer: a, option (C)
Where the marks are
- Uses Mode = 3 Median − 2 Mean to get a (1 mark)
Q18 · 1 mark · MCQProbability
The total number of outcomes in the experiment of simultaneous throw of three dice is :
1. (A)6
2. (B)18
3. (C)36
4. (D)216
Try it first, then show our solutionSolution
1. Each die has 6 outcomes, so three dice give $6 \times 6 \times 6 = 216$ outcomes, which is option (D).
  1 mark
Answer: 216, option (D)
Where the marks are
- Computes $6^{3} = 216$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A) : The prime numbers which divide 36 also divide 6. Reason (R) : Any number which divides $p^{2}$ also divides p.
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. Since $36 = 2^{2} \times 3^{2}$ , the primes dividing 36 are 2 and 3. Both divide 6, so (A) is true.
  ½ mark
2. (R) is false: 4 divides $2^{2} = 4$ but 4 does not divide 2. The statement holds only for prime divisors. Option (C).
  ½ mark
Answer: Option (C)
Where the marks are
- Shows A is true and gives a counterexample proving R false (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedTriangles
Assertion (A) : All congruent triangles are similar. Reason (R) : In congruent triangles, the ratio of corresponding sides is $1 : 1$ .
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. Congruent triangles have equal corresponding angles and corresponding sides in the ratio $1 : 1$ , so (R) is true.
  ½ mark
2. Because corresponding sides are proportional (ratio 1) and the angles are equal, congruent triangles are similar. Thus (R) correctly explains (A). Option (A).
  ½ mark
Answer: Option (A)
Where the marks are
- States both are true and R explains A (1 mark)
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Section B

Very Short Answer; internal choice in 21 and 25

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Find the value of c for which the following pair of linear equations has infinitely many solutions : $c x + 3 y = c - 3$ , $12 x + cy = c$
Try it first, then show our solutionSolution
1. Here $a_{1} = c, b_{1} = 3, c_{1} = c - 3$ and $a_{2} = 12, b_{2} = c, c_{2} = c$ . For infinitely many solutions, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , so $\frac{c}{12} = \frac{3}{c} = \frac{c - 3}{c}$ .
  1 mark
2. From $\frac{c}{12} = \frac{3}{c}$ : $c^{2} = 36$ , so $c = \pm 6$ . From $\frac{3}{c} = \frac{c - 3}{c}$ : $3 c = c^{2} - 3 c$ , i.e. $c^{2} - 6 c = 0$ , so $c = 0$ or $c = 6$ .
  ½ mark
3. The only value satisfying both conditions (and $c \neq = 0$ ) is $c = 6$ .
  ½ mark
Answer: c = 6
Where the marks are
- Writes the condition $\frac{c}{12} = \frac{3}{c} = \frac{c - 3}{c}$ (1 mark)
- Obtains $c = \pm 6$ and $c = 0, 6$ (½ mark)
- Concludes $c = 6$ (½ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve for x and y : $3 x + 2 y = 65$ , $2 x + 3 y = 60$
Try it first, then show our solutionSolution
1. Adding the equations: $5 x + 5 y = 125 \Rightarrow x + y = 25$ ...(i). Subtracting the second from the first: $x - y = 5$ ...(ii).
2. Adding (i) and (ii): $2 x = 30 \Rightarrow x = 15$ .
  1 mark
3. From (i): $y = 25 - 15 = 10$ .
  1 mark
Answer: x = 15, y = 10
Where the marks are
- Obtains $x = 15$ (1 mark)
- Obtains $y = 10$ (1 mark)
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Q22 · 2 marks · Very short answerTriangles
D is a point on side BC of $△$ ABC such that $∠ A D C = ∠ B A C$ . Prove that $(C A)^{2} = C B \cdot C D$ .
Try it first, then show our solutionSolution
1. In $△ A D C$ and $△ B A C$ : $∠ A D C = ∠ B A C$ [given] and $∠ A C D = ∠ B C A$ [common angle $C$ ].
  1 mark
2. $∴ △ A D C \sim △ B A C$ [AA similarity]. Corresponding sides are proportional, so $\frac{C A}{C B} = \frac{C D}{C A}$ .
  ½ mark
3. Cross-multiplying gives $(C A)^{2} = C B \cdot C D$ . Hence proved.
  ½ mark
Answer: $△ A D C \sim △ B A C$ (AA), so $C A / C B = C D / C A$ , giving $C A^{2} = C B \cdot C D$ .
Where the marks are
- Shows that the two pairs of angles are equal, giving reasons (1 mark)
- States $△ A D C \sim △ B A C$ and writes the ratio of corresponding sides (½ mark)
- Concludes $C A^{2} = C B \cdot C D$ (½ mark)
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Q23 · 2 marks · Very short answerCircles
Prove that the tangent drawn at any point of the circle is perpendicular to the radius through the point of contact.
Try it first, then show our solutionSolution
1. Figure: circle with centre O, line $l$ touching it at A, radius OA drawn, and another point B on $l$ joined to O. Given: $l$ is the tangent at A. To prove: $O A ⊥ l$ .
  ½ mark
2. Proof: Take any point B on $l$ other than A and join OB.
  ½ mark
3. Since a tangent meets the circle at only one point, B lies outside the circle, so $O B >$ radius, i.e. $O B > O A$ .
  ½ mark
4. Thus OA is the shortest of all segments from O to points of $l$ . The shortest distance from a point to a line is the perpendicular, so $O A ⊥ l$ . Hence proved.
  ½ mark
Answer: OA is the shortest distance from O to the tangent, hence OA ⟂ tangent.
Where the marks are
- Correct figure with given/to prove (½ mark)
- Takes another point B on the tangent (½ mark)
- Shows B outside the circle so OB > OA (½ mark)
- Concludes OA ⟂ l since perpendicular is the shortest distance (½ mark)
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Q24 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $\frac{c o s 4 5 ^{\circ}}{s e c 3 0 ^{\circ} + cosec 3 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. Using $cos 4 5^{\circ} = \frac{1}{2}$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $cosec 3 0^{\circ} = 2$ : $\frac{cos 4 5 ^{\circ}}{sec 3 0 ^{\circ} + cosec 3 0 ^{\circ}} = \frac{\frac{1}{2}}{\frac{2}{3} + 2}$ .
  1½ marks
2. $= \frac{1}{2} \times \frac{3}{2 + 2 3} = \frac{3}{2 2 + 2 6}$ ; rationalising, $= \frac{3 ( 3 - 1 )}{4 2} = \frac{3 2 - 6}{8}$ .
  ½ mark
Answer: $\frac{3}{2 2 + 2 6}$ or $\frac{3 2 - 6}{8}$
Where the marks are
- Substitutes correct values of cos 45°, sec 30°, cosec 30° (1½ marks)
- Simplifies to the final value (½ mark)
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Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerAreas Related to Circles
The area of a smaller circle is equal to the area of a sector of a larger circle with central angle $12 0^{\circ}$ . The radii of the smaller and larger circles are 'r' and 'R' respectively. Find $r : R$ .
Try it first, then show our solutionSolution
1. Area of smaller circle = area of sector of larger circle: $π r^{2} = \frac{120}{360} π R^{2}$ .
  1 mark
2. $\frac{r ^{2}}{R ^{2}} = \frac{1}{3}$ .
  ½ mark
3. $\frac{r}{R} = \frac{1}{3}$ , so $r : R = 1 : 3$ .
  ½ mark
Answer: r : R = 1 : √3
Where the marks are
- Forms the equation $π r^{2} = \frac{120}{360} π R^{2}$ (1 mark)
- Gets $r^{2} / R^{2} = 1/3$ (½ mark)
- Gets $r : R = 1 : 3$ (½ mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerAreas Related to Circles
In the given figure, O is the centre of a circle of radius 7 cm. AB is a chord of the circle. Find the perimeter of the shaded region.
Figure in the paper: Circle with centre O. Chord AB subtends a central angle of $6 0^{\circ}$ at O; OA and OB are each marked 7 cm. The shaded region is the minor segment between chord AB and the arc AB. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ O A B$ , $O A = O B = 7$ cm and $∠ A O B = 6 0^{\circ}$ . So $∠ O A B = ∠ O B A = 6 0^{\circ}$ and the triangle is equilateral. Hence $A B = 7$ cm.
  1 mark
2. Length of arc $A B = \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{22}{3}$ cm.
  ½ mark
3. Perimeter of the shaded segment $= A B + arc A B = 7 + \frac{22}{3} = \frac{43}{3}$ cm $\approx 14.33$ cm.
  ½ mark
Answer: $\frac{43}{3}$ cm ≈ 14.33 cm
Where the marks are
- Shows that $△ O A B$ is equilateral, so $A B = 7$ cm (1 mark)
- Finds the arc length $22/3$ cm (½ mark)
- Adds them to get the perimeter $43/3$ cm (½ mark)
Open Q25(b) on its own page

Section C

Short Answer; internal choice in 26 and 28

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerReal Numbers
Prove that $5$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $5$ is rational.
  Then $5 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are co-prime.
  ½ mark
2. Squaring gives $5 q^{2} = p^{2}$ . So $p^{2}$ is divisible by 5.
  Since 5 is prime, $p$ is divisible by 5 [if a prime divides $p^{2}$ , it divides $p$ ].
  So $p = 5 a$ for some integer $a$ .
  1 mark
3. Substitute: $25 a^{2} = 5 q^{2} \Rightarrow q^{2} = 5 a^{2}$ . So $q^{2}$ is divisible by 5, and hence $q$ is divisible by 5.
  ½ mark
4. Now 5 is a common factor of $p$ and $q$ . This contradicts the assumption that $p$ and $q$ are co-prime.
  Hence our assumption is wrong, and $5$ is irrational.
  1 mark
Answer: $5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 = p / q$ with $p, q$ co-prime (½ mark)
- Shows 5 divides $p$ , writes $p = 5 a$ (1 mark)
- Shows 5 divides $q$ (½ mark)
- States the contradiction and concludes (1 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerReal Numbers
State the 'Fundamental Theorem of Arithmetic' and use it to find LCM of 36 and 54.
Try it first, then show our solutionSolution
1. Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique apart from the order in which the prime factors occur.
  1 mark
2. $36 = 2 \times 2 \times 3 \times 3 = 2^{2} \times 3^{2}$ .
  ½ mark
3. $54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^{3}$ .
  ½ mark
4. LCM = product of the greatest power of each prime factor $= 2^{2} \times 3^{3} = 4 \times 27 = 108$ .
  1 mark
Answer: LCM(36, 54) = 2²×3³ = 108
Where the marks are
- Correct statement of FTA (1 mark)
- Prime factorisation of 36 (½ mark)
- Prime factorisation of 54 (½ mark)
- LCM = 108 (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerPolynomials
Find the zeroes of the polynomial $p (x) = 3 x^{2} - 2 x - 1$ and verify the relationship between the zeroes of $p (x)$ and the coefficients of $p (x)$ .
Try it first, then show our solutionSolution
1. $p (x) = 3 x^{2} - 3 x + x - 1 = 3 x (x - 1) + 1 (x - 1) = (3 x + 1) (x - 1)$ . Zeroes: $x = - \frac{1}{3}$ and $x = 1$ .
  1 mark
2. Sum of zeroes $= - \frac{1}{3} + 1 = \frac{2}{3}$ ; and $\frac{- coefficient of x}{coefficient of x ^{2}} = \frac{- ( - 2 )}{3} = \frac{2}{3}$ . Verified.
  1 mark
3. Product of zeroes $= - \frac{1}{3} \times 1 = - \frac{1}{3}$ ; and $\frac{constant term}{coefficient of x ^{2}} = \frac{- 1}{3}$ . Verified.
  1 mark
Answer: Zeroes are −1/3 and 1; sum = 2/3 = −b/a, product = −1/3 = c/a
Where the marks are
- Factorises and finds zeroes −1/3, 1 (1 mark)
- Verifies sum of zeroes = −b/a (1 mark)
- Verifies product of zeroes = c/a (1 mark)
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Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve graphically the following pair of linear equations : $2 x - y = 2$ and $4 x - y = 4$ . Also, write the coordinates of the points where the lines represented by these equations cut the y-axis.
Try it first, then show our solutionSolution
1. For $2 x - y = 2$ , i.e. $y = 2 x - 2$ : points $(0, - 2), (1, 0), (2, 2)$ . Plot and join them to draw the line.
  1 mark
2. For $4 x - y = 4$ , i.e. $y = 4 x - 4$ : points $(0, - 4), (1, 0), (2, 4)$ . Plot and join them to draw the second line.
  1 mark
3. The two lines intersect at $(1, 0)$ , so the solution is $x = 1, y = 0$ .
  ½ mark
4. The lines cut the y-axis at $(0, - 2)$ and $(0, - 4)$ respectively.
  ½ mark
Answer: x = 1, y = 0; lines cut y-axis at (0, −2) and (0, −4)
Where the marks are
- Correct graph of $2 x - y = 2$ (1 mark)
- Correct graph of $4 x - y = 4$ (1 mark)
- Solution $(1, 0)$ (½ mark)
- y-intercept points $(0, - 2)$ and $(0, - 4)$ (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
An academy offering cricket coaching bought 10 bats and 5 balls for ₹ 32,500. Later, the academy bought 2 bats and 8 balls for ₹ 10,000. If there is no change in the cost of the bat and of the ball, find the cost of 1 bat and 1 ball.
Try it first, then show our solutionSolution
1. Let the cost of 1 bat be ₹ $x$ and the cost of 1 ball be ₹ $y$ . Then $10 x + 5 y = 32500$ , i.e. $2 x + y = 6500$ ...(1), and $2 x + 8 y = 10000$ ...(2).
  1 mark
2. Subtracting (1) from (2): $7 y = 3500 \Rightarrow y = 500$ .
  1 mark
3. Substituting in (1): $2 x = 6500 - 500 = 6000 \Rightarrow x = 3000$ . So one bat costs ₹3000 and one ball costs ₹500.
  1 mark
Answer: Cost of 1 bat = ₹3000; cost of 1 ball = ₹500
Where the marks are
- Forms the two linear equations correctly (1 mark)
- Solves to get $y = 500$ (1 mark)
- Gets $x = 3000$ and states both costs (1 mark)
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Q29 · 3 marks · Short answerCircles
In the given figure, TP and TQ are tangents at points P and Q of the circle respectively. If reflex $∠ P O Q = 25 0^{\circ}$ , find the measure of each angle of quadrilateral POQT.
Figure in the paper: Circle with centre O. Points P and Q on the circle; T is outside the circle with tangents TP and TQ. The quadrilateral POQT is formed by OP, PT, TQ and QO. The reflex angle POQ is marked as $25 0^{\circ}$ at O. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ P O Q$ (inside the quadrilateral) $= 36 0^{\circ} - 25 0^{\circ} = 11 0^{\circ}$ .
  1 mark
2. $∠ O P T = ∠ O QT = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  1 mark
3. The angle sum of quadrilateral POQT is $36 0^{\circ}$ , so $∠ P T Q = 36 0^{\circ} - (9 0^{\circ} + 9 0^{\circ} + 11 0^{\circ}) = 7 0^{\circ}$ .
  1 mark
Answer: ∠POQ = 110°, ∠OPT = 90°, ∠OQT = 90°, ∠PTQ = 70°
Where the marks are
- Finds $∠ P O Q = 11 0^{\circ}$ (1 mark)
- States $∠ O P T = ∠ O QT = 9 0^{\circ}$ with the reason (1 mark)
- Finds $∠ P T Q = 7 0^{\circ}$ (1 mark)
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Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove the following trigonometric identity : $\frac{c o s θ}{1 + s i n θ} + \frac{1 + s i n θ}{c o s θ} = 2 sec θ$
Try it first, then show our solutionSolution
1. LHS $= \frac{cos ^{2} θ + ( 1 + sin θ ) ^{2}}{( 1 + sin θ ) cos θ}$ (taking the LCM).
  1 mark
2. $= \frac{cos ^{2} θ + 1 + 2 sin θ + sin ^{2} θ}{( 1 + sin θ ) cos θ} = \frac{2 + 2 sin θ}{( 1 + sin θ ) cos θ}$ [since $sin^{2} θ + cos^{2} θ = 1$ ].
  1 mark
3. $= \frac{2 ( 1 + sin θ )}{( 1 + sin θ ) cos θ} = \frac{2}{cos θ} = 2 sec θ =$ RHS. Hence proved.
  1 mark
Answer: LHS simplifies to $2 sec θ$ = RHS
Where the marks are
- Combines the two fractions using the LCM (1 mark)
- Expands the numerator and uses $sin^{2} θ + cos^{2} θ = 1$ (1 mark)
- Cancels $(1 + sin θ)$ to reach $2 sec θ$ (1 mark)
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Q31 · 3 marks · Short answerProbability
A game of chance consists of spinning a wheel which comes to rest at one of the numbers from 1 to 10 (as shown in the given figure) with equal probabilities. What is the probability that the wheel stops at :
1. (i)
  a prime number greater than 2 ?
  1 mark
2. (ii)
  an odd number less than 9 ?
  1 mark
3. (iii)
  a multiple of 4 ?
  1 mark
Figure in the paper: A circular wheel divided into 10 equal sectors numbered 1 to 10 in order (3, 2, 1, 10, 9, 8, 7, 6, 5, 4 arranged around the circle). A fixed pointer on the right side points at the wheel. See it in the official paper.
Try it first, then show our solutionSolution
1. Total equally likely outcomes = 10. (i) Prime numbers greater than 2 are 3, 5, 7, so P = $\frac{3}{10}$ .
  1 mark
2. (ii) Odd numbers less than 9 are 1, 3, 5, 7, so P = $\frac{4}{10} = \frac{2}{5}$ .
  1 mark
3. (iii) Multiples of 4 are 4, 8, so P = $\frac{2}{10} = \frac{1}{5}$ .
  1 mark
Answer: (i) 3/10; (ii) 2/5; (iii) 1/5
Where the marks are
- (i) Lists 3, 5, 7 and gets the probability $3/10$ (1 mark)
- (ii) Lists 1, 3, 5, 7 and gets the probability $2/5$ (1 mark)
- (iii) Lists 4, 8 and gets the probability $1/5$ (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 answerQuadratic Equations
The sum of areas of two squares is $2650$ cm $^{2}$ . If the sum of their perimeters is $280$ cm, find the sides of the two given squares.
Try it first, then show our solutionSolution
1. Let the sides of the two squares be $x$ cm and $y$ cm. Sum of perimeters: $4 x + 4 y = 280 \Rightarrow x + y = 70 \Rightarrow y = 70 - x$ .
  1 mark
2. Sum of areas: $x^{2} + y^{2} = 2650 \Rightarrow x^{2} + (70 - x)^{2} = 2650$ .
  1 mark
3. $2 x^{2} - 140 x + 4900 - 2650 = 0 \Rightarrow x^{2} - 70 x + 1125 = 0$ .
  1 mark
4. Factorising: $(x - 45) (x - 25) = 0 \Rightarrow x = 45$ or $x = 25$ .
  1 mark
5. If $x = 45$ then $y = 25$ ; if $x = 25$ then $y = 45$ . So the sides are 45 cm and 25 cm. Check: $4 5^{2} + 2 5^{2} = 2025 + 625 = 2650$ ✓.
  1 mark
Answer: Sides are 45 cm and 25 cm
Where the marks are
- Forms $x + y = 70$ from the perimeters (1 mark)
- Forms $x^{2} + (70 - x)^{2} = 2650$ from the areas (1 mark)
- Reduces it to $x^{2} - 70 x + 1125 = 0$ (1 mark)
- Solves to get $x = 45$ or $25$ (1 mark)
- States both sides, 45 cm and 25 cm (1 mark)
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OR
Q32(b) · 5 marks · Long answerQuadratic Equations
Express the equation $\frac{1}{x} - \frac{1}{x - 2} = 3$ , $(x \neq = 0, 2)$ as a quadratic equation in standard form. Hence, find the roots of the quadratic equation so obtained.
Try it first, then show our solutionSolution
1. $\frac{1}{x} - \frac{1}{x - 2} = \frac{x - 2 - x}{x ( x - 2 )} = 3 \Rightarrow \frac{- 2}{x ^{2} - 2 x} = 3$ .
  1 mark
2. $3 x^{2} - 6 x = - 2 \Rightarrow 3 x^{2} - 6 x + 2 = 0$ (standard form).
  1 mark
3. Here $a = 3, b = - 6, c = 2$ ; Discriminant $D = b^{2} - 4 a c = 36 - 24 = 12$ .
  1 mark
4. $x = \frac{- b \pm D}{2 a} = \frac{6 + 12}{6}$ or $\frac{6 - 12}{6}$ .
  1 mark
5. Since $12 = 23$ , roots are $1 + \frac{3}{3}$ and $1 - \frac{3}{3}$ .
  1 mark
Answer: $3 x^{2} - 6 x + 2 = 0$ ; roots $1 + \frac{3}{3}$ and $1 - \frac{3}{3}$
Where the marks are
- Combines the fractions correctly (1 mark)
- Obtains $3 x^{2} - 6 x + 2 = 0$ (1 mark)
- Finds discriminant = 12 (1 mark)
- Each root correctly found (1 + 1 marks)
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Q33 · 5 marks · Long answerTriangles
State the SAS criteria of similarity of two triangles. In the given figure, it is given that $O A \cdot O C = O B \cdot O D$ . Use the SAS criteria to prove that $A D ∥ C B$ .
Figure in the paper: Two line segments AB and CD intersect at point O. A and D are on one side of O, and C and B on the other; A-O-B and C-O-D are straight segments. Points: A top-left, D bottom-left, C top-right, B bottom-right, O the intersection point. See it in the official paper.
Try it first, then show our solutionSolution
1. SAS similarity criterion: if one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are proportional, then the two triangles are similar.
  1 mark
2. Given $O A \cdot O C = O B \cdot O D$ . Dividing both sides by $O B \cdot O C$ gives $\frac{O A}{O B} = \frac{O D}{O C}$ .
  1 mark
3. In $△ A O D$ and $△ B O C$ : $\frac{O A}{O B} = \frac{O D}{O C}$ (shown above) and $∠ A O D = ∠ B O C$ [vertically opposite angles].
  1 mark
4. $∴ △ A O D \sim △ B O C$ [SAS similarity], so the corresponding angles are equal: $∠ O A D = ∠ O B C$ , i.e. $∠ D A B = ∠ C B A$ .
  1 mark
5. These are alternate interior angles made by the transversal AB with the lines AD and CB. Since they are equal, $A D ∥ C B$ . Hence proved.
  1 mark
Answer: $△ A O D \sim △ B O C$ (SAS), so $∠ D A B = ∠ C B A$ (alternate angles), hence $A D ∥ C B$ .
Where the marks are
- States the SAS similarity criterion correctly (1 mark)
- Converts the given product into the ratio $O A / O B = O D / O C$ (1 mark)
- Uses the vertically opposite angles at O (1 mark)
- Concludes $△ A O D \sim △ B O C$ and the equal corresponding angles (1 mark)
- Uses equal alternate angles to conclude $A D ∥ C B$ (1 mark)
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Q34 · 5 marks · Long answerSurface Areas and Volumes
A wooden article was made by scooping out a hemisphere (of same diameter) from one end of a solid cylinder as shown in the given figure. If the height of the cylinder is 10 cm and the diameter of the cylinder is 14 cm, find the total surface area of the remaining wooden article. (Use $π = \frac{22}{7}$ )
Figure in the paper: A solid cylinder of height 10 cm and diameter 14 cm, with a hemispherical cavity of the same diameter scooped out from the top end (the flat face of the hemisphere is at the top circular face). See it in the official paper.
Try it first, then show our solutionSolution
1. Radius $r = \frac{14}{2} = 7$ cm and height $h = 10$ cm. The TSA of the article is the curved surface of the cylinder + the bottom circular base + the inner curved surface of the hemisphere.
  1 mark
2. CSA of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440$ cm².
  1 mark
3. Area of base $= π r^{2} = \frac{22}{7} \times 7 \times 7 = 154$ cm².
  1 mark
4. CSA of hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 49 = 308$ cm².
  1 mark
5. TSA $= 440 + 154 + 308 = 902$ cm².
  1 mark
Answer: 902 cm²
Where the marks are
- Identifies the three surfaces of the article (1 mark)
- Finds the cylinder CSA, 440 cm² (1 mark)
- Finds the base area, 154 cm² (1 mark)
- Finds the hemisphere CSA, 308 cm² (1 mark)
- Adds them to get 902 cm² (1 mark)
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Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerStatistics
The lengths of 40 leaves of a plant are measured, correct to the nearest millimetre and data obtained is represented in the following table. Find the median length (in mm) of the leaves.
Length in (mm) Number of leaves
100 – 120 8
120 – 140 9
140 – 160 12
160 – 180 5
180 – 200 6
Try it first, then show our solutionSolution
1. Cumulative frequencies: 100–120: 8; 120–140: 17; 140–160: 29; 160–180: 34; 180–200: 40.
  1 mark
2. $N = 40$ , so $\frac{N}{2} = 20$ . The cf first exceeds 20 in the class 140–160, so the median class is 140–160.
  1 mark
3. Here $l = 140$ , $c f = 17$ , $f = 12$ , $h = 20$ .
  1 mark
4. Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 140 + \frac{20 - 17}{12} \times 20$
  1 mark
5. $= 140 + 5 = 145$ mm.
  1 mark
Answer: Median length = 145 mm
Where the marks are
- Writes the correct cumulative frequency column (1 mark)
- Identifies the median class 140–160 (1 mark)
- Writes the values $l, c f, f, h$ correctly (1 mark)
- Substitutes into the median formula (1 mark)
- Gets the median 145 mm (1 mark)
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OR
Q35(b) · 5 marks · Long answerCompetency-basedStatistics
A class teacher has the following absentees record of 30 students of a class.
Number of days 0–4 4–8 8–12 12–16 16–20 20–24
Number of Absent students 1 8 x 6 5 y
If the mean number of days a student was absent is 12, find the values of x and y.
Try it first, then show our solutionSolution
1. Total number of students: $1 + 8 + x + 6 + 5 + y = 30 \Rightarrow x + y = 10$ ...(1).
  1 mark
2. Class marks $x_{i}$ : 2, 6, 10, 14, 18, 22. Products $f_{i} x_{i}$ : 2, 48, $10 x$ , 84, 90, $22 y$ . So $Σ f_{i} x_{i} = 224 + 10 x + 22 y$ .
  2 marks
3. Mean $= \frac{Σ f _{i} x _{i}}{Σ f _{i}}$ : $\frac{224 + 10 x + 22 y}{30} = 12 \Rightarrow 10 x + 22 y = 136 \Rightarrow 5 x + 11 y = 68$ ...(2).
  1 mark
4. From (1), $x = 10 - y$ . Substituting in (2): $50 - 5 y + 11 y = 68 \Rightarrow 6 y = 18 \Rightarrow y = 3$ , and then $x = 7$ .
  1 mark
Answer: x = 7, y = 3
Where the marks are
- Forms $x + y = 10$ from the total (1 mark)
- Finds the class marks and $Σ f_{i} x_{i} = 224 + 10 x + 22 y$ (2 marks)
- Uses mean = 12 to form $5 x + 11 y = 68$ (1 mark)
- Solves to get $x = 7$ , $y = 3$ (1 mark)
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Length in (mm)	Number of leaves
100 – 120	8
120 – 140	9
140 – 160	12
160 – 180	5
180 – 200	6

Number of days	0–4	4–8	8–12	12–16	16–20	20–24
Number of Absent students	1	8	x	6	5	y

Section E

Case study based; internal choice in each case study for 2-mark subparts

Q36 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Kite festival is a popular festival in India which takes place during Makar Sankranti. The festival is celebrated by people flying kites from their rooftops. Reena and Ravi are also flying kites to enjoy the festival. The height of Reena's kite is 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground, and the inclination of the string with the ground is $3 0^{\circ}$ . Ravi is flying a kite from a 10 m high building. His kite is also flying 60 m above the ground and the length of the string used by Ravi is same as that of Reena's. $θ$ is the angle of elevation of Ravi's kite from a point on the rooftop.
Based on the above information, answer the following questions :
1. (i)
  Find the length of string used by Reena.
  1 mark
2. (ii)
  Find the value of $sin θ$ .
  1 mark
3. (iii)
  (a) If $θ$ changes to $6 0^{\circ}$ , without changing the length of the string, what will be the height of Ravi's kite above the ground ? (Use $3 = 1.7$ )
  2 marks
4. OR
  (iii)
  (b) What would have been the height of Ravi's kite above the ground, if the string had an inclination of $3 0^{\circ}$ with the ground, assuming that the length of the string does not change ?
  2 marks
Figure in the paper: Left: a girl (Reena) holding a string from the ground to a pink kite high in the air. Right: Ravi standing on a rooftop balcony of a 10 m high building, flying a kite high above the city, with the string going from his hand to the kite. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let Reena's string length be $L$ . Then $sin 3 0^{\circ} = \frac{60}{L} \Rightarrow \frac{1}{2} = \frac{60}{L} \Rightarrow L = 120$ m.
  1 mark
2. (ii) Ravi's kite is $60 - 10 = 50$ m above the rooftop, and his string is also 120 m long. So $sin θ = \frac{50}{120} = \frac{5}{12}$ .
  1 mark
3. (iii)(a) With $θ = 6 0^{\circ}$ , height of the kite above the roof $= 120 sin 6 0^{\circ} = 120 \times \frac{3}{2} = 603 = 60 \times 1.7 = 102$ m.
  1 mark
4. (iii)(a) Height above the ground $= 102 + 10 = 112$ m.
  1 mark
5. (iii)(b) With an inclination of $3 0^{\circ}$ (measured at the rooftop, from a line parallel to the ground), height above the roof $= 120 sin 3 0^{\circ} = 60$ m. Height above the ground $= 60 + 10 = 70$ m.
  2 marks
Answer: (i) 120 m; (ii) sin θ = 5/12; (iii)(a) 112 m; (iii)(b) 70 m
Where the marks are
- (i) Uses $sin 3 0^{\circ} = 60/ L$ to get $L = 120$ m (1 mark)
- (ii) Uses the height above the roof, 50 m, to get $sin θ = 5/12$ (1 mark)
- (iii)(a) Finds $120 sin 6 0^{\circ} = 102$ m above the roof (1 mark)
- (iii)(a) Adds the 10 m building to get 112 m (1 mark)
- (iii)(b) Finds $120 sin 3 0^{\circ} = 60$ m and adds 10 m to get 70 m (2 marks)
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Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
A woman borrowed ₹ 10,00,000 from her friend and promised to return the borrowed money in monthly instalments beginning from the next month. After one month, she returned ₹ 10,000, the next month she returned ₹ 15,000, the third month she returned ₹ 20,000 and so on, thereby increasing the monthly instalment uniformly.
Based on the above information, answer the following questions :
1. (i)
  Find the amount of instalment paid in the tenth month.
  1 mark
2. (ii)
  In which instalment did she pay ₹ 40,000 ?
  1 mark
3. (iii)
  (a) If she returned ₹ 11,50,000 in all, how many instalments did she pay ?
  2 marks
4. OR
  (iii)
  (b) By which instalment has she returned a total amount of ₹ 3,25,000 ?
  2 marks
Figure in the paper: Illustration of two women talking (one with a thought bubble showing money being handed over), a hand giving a banknote, and a hand holding a calendar with a clock. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The instalments form an AP with $a = 10000$ and $d = 5000$ . So $a_{10} = a + 9 d = 10000 + 9 \times 5000 = ₹55, 000$ .
  1 mark
2. (ii) $a_{n} = 40000 \Rightarrow 10000 + (n - 1) 5000 = 40000 \Rightarrow n - 1 = 6 \Rightarrow n = 7$ . So she paid ₹40,000 in the 7th instalment.
  1 mark
3. (iii)(a) $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [20000 + 5000 (n - 1)] = 2500 n (n + 3)$ . Setting $2500 n (n + 3) = 1150000$ gives $n^{2} + 3 n - 460 = 0$ .
  1 mark
4. (iii)(a) $(n + 23) (n - 20) = 0 \Rightarrow n = 20$ (rejecting $n = - 23$ ). She paid 20 instalments.
  1 mark
5. (iii)(b) $2500 n (n + 3) = 325000 \Rightarrow n^{2} + 3 n - 130 = 0 \Rightarrow (n + 13) (n - 10) = 0 \Rightarrow n = 10$ (rejecting $n = - 13$ ). She had returned ₹3,25,000 by the 10th instalment.
  2 marks
Answer: (i) ₹55,000; (ii) 7th instalment; (iii)(a) 20 instalments; (iii)(b) 10th instalment
Where the marks are
- (i) Uses $a_{10} = a + 9 d$ to get ₹55,000 (1 mark)
- (ii) Solves $a_{n} = 40000$ to get $n = 7$ (1 mark)
- (iii)(a) Forms $n^{2} + 3 n - 460 = 0$ from $S_{n}$ (1 mark)
- (iii)(a) Solves to get $n = 20$ (1 mark)
- (iii)(b) Forms $n^{2} + 3 n - 130 = 0$ and gets $n = 10$ (2 marks)
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Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
A field is in the form of a rectangle. The coordinates of the rectangular field ABCD are A(10, 10), B(40, 10), C(40, 50) and D(x, y). Anil and Anita, two friends decided to have a race. Anita started from point A and moved to point E along the diagonal AC, where E is the point of intersection of both the diagonals of ABCD. From point E, she moved to point B along the other diagonal DB and then moved back to point A along BA. While Anil started from point C and ran to point A via D along the boundary of the field.
Based on the above information, answer the following questions :
1. (i)
  Find the coordinates of point E.
  1 mark
2. (ii)
  Find the distance between the points B and C.
  1 mark
3. (iii)
  (a) Find the coordinates of point D and the distance BD.
  2 marks
4. OR
  (iii)
  (b) Find the total distance travelled by Anita.
  2 marks
Figure in the paper: Rectangle ABCD with A(10, 10) bottom-left, B(40, 10) bottom-right, C(40, 50) top-right, D(x, y) top-left. Diagonals AC and DB intersect at E. Arrows mark Anita's path A to E along AC, E to B along DB, and B to A along BA; Anil's path runs C to D to A along the boundary. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The diagonals of a rectangle bisect each other, so E is the mid-point of AC: $E = (\frac{10 + 40}{2}, \frac{10 + 50}{2}) = (25, 30)$ .
  1 mark
2. (ii) $B C = (40 - 40)^{2} + (50 - 10)^{2} = 1600 = 40$ units.
  1 mark
3. (iii)(a) E is also the mid-point of BD: $\frac{40 + x}{2} = 25 \Rightarrow x = 10$ and $\frac{10 + y}{2} = 30 \Rightarrow y = 50$ . So $D = (10, 50)$ .
  1 mark
4. (iii)(a) $B D = (10 - 40)^{2} + (50 - 10)^{2} = 900 + 1600 = 2500 = 50$ units.
  1 mark
5. (iii)(b) $A C = 3 0^{2} + 4 0^{2} = 50$ , so $A E = 25$ . Also $E B = \frac{B D}{2} = 25$ and $B A = 40 - 10 = 30$ . Total distance travelled by Anita $= 25 + 25 + 30 = 80$ units.
  2 marks
Answer: (i) E(25, 30); (ii) 40 units; (iii)(a) D(10, 50), BD = 50 units; (iii)(b) 80 units
Where the marks are
- (i) Uses the mid-point of AC to get E(25, 30) (1 mark)
- (ii) Uses the distance formula to get BC = 40 (1 mark)
- (iii)(a) Uses E as the mid-point of BD to get D(10, 50) (1 mark)
- (iii)(a) Finds BD = 50 (1 mark)
- (iii)(b) Finds AE = 25, EB = 25, BA = 30 and adds them to get 80 (2 marks)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.