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

CBSE Class 10 Mathematics Basic: 2023 board paper, set 430/4/1

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_4_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	Very short answer; internal choice in 22 and 23
C	26-31	Short answer; internal choice in 28 and 31
D	32-35	Long answer; internal choice in 32 and 33
E	36-38	Case based; internal choice in 2-mark subparts of each case

Instructions in brief: 38 compulsory questions in five sections (A-E) with no overall choice; internal choice is given in 2 questions in Section B, 2 in Section C, 2 in Section D and 3 in Section E (the 2-mark subparts of each case study).

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPolynomials
A quadratic polynomial the sum and product of whose zeroes are $- 3$ and $2$ respectively, is:
1. (A) $x^{2} + 3 x + 2$
2. (B) $x^{2} - 3 x + 2$
3. (C) $x^{2} - 3 x - 2$
4. (D) $x^{2} + 3 x - 2$
Try it first, then show our solutionSolution
1. A quadratic polynomial with sum of zeroes $S$ and product $P$ is $x^{2} - S x + P$ .
2. $x^{2} - (- 3) x + 2 = x^{2} + 3 x + 2$ , which is option (A).
  1 mark
Answer: $x^{2} + 3 x + 2$ , option (A)
Where the marks are
- Uses $x^{2} - (sum) x + product$ to get $x^{2} + 3 x + 2$ (1 mark)
Q2 · 1 mark · MCQReal Numbers
$(HCF \times LCM)$ for the numbers $70$ and $40$ is:
1. (A) $10$
2. (B) $280$
3. (C) $2800$
4. (D) $70$
Try it first, then show our solutionSolution
1. For any two positive integers, HCF × LCM = product of the numbers.
2. $70 \times 40 = 2800$ , which is option (C).
  1 mark
Answer: 2800, option (C)
Where the marks are
- Uses HCF × LCM = product to get 2800 (1 mark)
Q3 · 1 mark · MCQAreas Related to Circles
If the radius of a semi-circular protractor is $7$ cm, then its perimeter is:
1. (A) $11$ cm
2. (B) $14$ cm
3. (C) $22$ cm
4. (D) $36$ cm
Try it first, then show our solutionSolution
1. Perimeter of a semicircle $= π r + 2 r$ , which is the curved arc plus the diameter.
2. $\frac{22}{7} \times 7 + 14 = 22 + 14 = 36$ cm, which is option (D).
  1 mark
Answer: 36 cm, option (D)
Where the marks are
- Uses $π r + 2 r$ to get 36 cm (1 mark)
Q4 · 1 mark · MCQReal Numbers
The number $(5 - 35 + 5)$ is:
1. (A)an integer
2. (B)a rational number
3. (C)an irrational number
4. (D)a whole number
Try it first, then show our solutionSolution
1. Simplify: $5 - 35 + 5 = 5 - 25$ .
2. $25$ is irrational, and a rational number minus an irrational number is irrational. So the number is irrational, option (C).
  1 mark
Answer: An irrational number, option (C)
Where the marks are
- Simplifies to $5 - 25$ and concludes it is irrational (1 mark)
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Q5 · 1 mark · MCQPolynomials
If $p (x) = x^{2} + 5 x + 6$ , then $p (- 2)$ is:
1. (A) $20$
2. (B) $0$
3. (C) $- 8$
4. (D) $8$
Try it first, then show our solutionSolution
1. $p (- 2) = (- 2)^{2} + 5 (- 2) + 6 = 4 - 10 + 6 = 0$ , which is option (B).
  1 mark
Answer: 0, option (B)
Where the marks are
- Substitutes $x = - 2$ correctly to get 0 (1 mark)
Q6 · 1 mark · MCQProbability
Which of the following cannot be the probability of an event?
1. (A) $0.1$
2. (B) $\frac{5}{3}$
3. (C) $3%$
4. (D) $\frac{1}{3}$
Try it first, then show our solutionSolution
1. The probability of any event lies between 0 and 1, i.e. $0 \leq P (E) \leq 1$ .
2. $\frac{5}{3} > 1$ , so it cannot be a probability. The other options, $0.1$ , $3% = 0.03$ and $\frac{1}{3}$ , all lie in $[0, 1]$ . Answer: option (B).
  1 mark
Answer: $\frac{5}{3}$ , option (B)
Where the marks are
- Uses $0 \leq P (E) \leq 1$ to reject $\frac{5}{3}$ (1 mark)
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Q7 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of linear equations $x + 2 y + 5 = 0$ and $- 3 x - 6 y + 1 = 0$ has:
1. (A)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. Compare the ratios: $\frac{a _{1}}{a _{2}} = \frac{1}{- 3}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{- 6} = - \frac{1}{3}$ , $\frac{c _{1}}{c _{2}} = \frac{5}{1} = 5$ .
2. Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the lines are parallel, so there is no solution. Answer: option (D).
  1 mark
Answer: No solution, option (D)
Where the marks are
- Compares ratios of coefficients and concludes no solution (1 mark)
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Q8 · 1 mark · MCQTriangles
If $△ A B C \sim △ D E F$ and $∠ A = 4 7^{\circ}$ , $∠ E = 8 3^{\circ}$ , then $∠ C$ is equal:
1. (A) $4 7^{\circ}$
2. (B) $5 0^{\circ}$
3. (C) $8 3^{\circ}$
4. (D) $13 0^{\circ}$
Try it first, then show our solutionSolution
1. In similar triangles, corresponding angles are equal, so $∠ B = ∠ E = 8 3^{\circ}$ .
2. By the angle sum property, $∠ C = 18 0^{\circ} - 4 7^{\circ} - 8 3^{\circ} = 5 0^{\circ}$ . Answer: option (B).
  1 mark
Answer: $5 0^{\circ}$ , option (B)
Where the marks are
- Uses $∠ B = ∠ E$ and the angle sum property to get $5 0^{\circ}$ (1 mark)
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Q9 · 1 mark · MCQPair of Linear Equations in Two Variables
If the pair of linear equations $x - y = 1$ , $x + k y = 5$ has a unique solution $x = 2$ , $y = 1$ , then the value of $k$ is:
1. (A) $- 2$
2. (B) $- 3$
3. (C) $3$
4. (D) $4$
Try it first, then show our solutionSolution
1. The solution must satisfy $x + k y = 5$ , so substitute $x = 2$ , $y = 1$ : $2 + k = 5$ , giving $k = 3$ . Answer: option (C).
  1 mark
Answer: $k = 3$ , option (C)
Where the marks are
- Substitutes the solution into the second equation to get $k = 3$ (1 mark)
Q10 · 1 mark · MCQIntroduction to Trigonometry
The value of $5 sin^{2} 9 0^{\circ} - 2 cos^{2} 0^{\circ}$ is:
1. (A) $- 2$
2. (B) $5$
3. (C) $3$
4. (D) $- 3$
Try it first, then show our solutionSolution
1. $sin 9 0^{\circ} = 1$ and $cos 0^{\circ} = 1$ , so the value is $5 (1) - 2 (1) = 3$ . Answer: option (C).
  1 mark
Answer: 3, option (C)
Where the marks are
- Uses $sin 9 0^{\circ} = cos 0^{\circ} = 1$ to get 3 (1 mark)
Q11 · 1 mark · MCQAreas Related to Circles
The length of the arc of a circle of radius $14$ cm which subtends an angle of $6 0^{\circ}$ at the centre of the circle is:
1. (A) $\frac{44}{3}$ cm
2. (B) $\frac{88}{3}$ cm
3. (C) $\frac{308}{3}$ cm
4. (D) $\frac{616}{3}$ cm
Try it first, then show our solutionSolution
1. Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r$ .
2. $\frac{60}{360} \times 2 \times \frac{22}{7} \times 14 = \frac{88}{6} = \frac{44}{3}$ cm. Answer: option (A).
  1 mark
Answer: $\frac{44}{3}$ cm, option (A)
Where the marks are
- Applies the arc length formula to get $\frac{44}{3}$ cm (1 mark)
Q12 · 1 mark · MCQSome Applications of Trigonometry
The angle of elevation of the top of a $30$ m high tower at a point $30$ m away from the base of the tower is:
1. (A) $3 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan θ = \frac{height}{distance} = \frac{30}{30} = 1$ , so $θ = 4 5^{\circ}$ . Answer: option (B).
  1 mark
Answer: $4 5^{\circ}$ , option (B)
Where the marks are
- Uses $tan θ = 1$ to get $4 5^{\circ}$ (1 mark)
Q13 · 1 mark · MCQStatistics
The mode of the numbers $2, 3, 3, 4, 5, 4, 4, 5, 3, 4, 2, 6, 7$ is:
1. (A) $2$
2. (B) $3$
3. (C) $4$
4. (D) $5$
Try it first, then show our solutionSolution
1. Frequencies: 2 occurs twice, 3 occurs three times, 4 occurs four times, 5 occurs twice, and 6 and 7 occur once each.
2. 4 occurs most often, so the mode is 4. Answer: option (C).
  1 mark
Answer: 4, option (C)
Where the marks are
- Counts frequencies and identifies 4 as the mode (1 mark)
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Q14 · 1 mark · MCQProbability
From a well-shuffled deck of $52$ playing cards, a card is drawn at random. What is the probability of getting a red queen?
1. (A) $\frac{1}{52}$
2. (B) $\frac{1}{26}$
3. (C) $\frac{1}{13}$
4. (D) $\frac{12}{13}$
Try it first, then show our solutionSolution
1. There are 2 red queens (hearts and diamonds) in a deck of 52 cards.
2. $P = \frac{2}{52} = \frac{1}{26}$ . Answer: option (B).
  1 mark
Answer: $\frac{1}{26}$ , option (B)
Where the marks are
- Counts 2 red queens and gets $\frac{1}{26}$ (1 mark)
Q15 · 1 mark · MCQQuadratic Equations
A quadratic equation whose one root is $2$ and the sum of whose roots is zero, is:
1. (A) $x^{2} + 4 = 0$
2. (B) $x^{2} - 2 = 0$
3. (C) $4 x^{2} - 1 = 0$
4. (D) $x^{2} - 4 = 0$
Try it first, then show our solutionSolution
1. The sum of the roots is 0, so the other root is $- 2$ . Product of roots $= - 4$ .
2. The equation is $x^{2} - 0 x + (- 4) = 0$ , i.e. $x^{2} - 4 = 0$ . Answer: option (D).
  1 mark
Answer: $x^{2} - 4 = 0$ , option (D)
Where the marks are
- Finds the other root $- 2$ and forms $x^{2} - 4 = 0$ (1 mark)
Q16 · 1 mark · MCQQuadratic Equations
Which of the following is not a quadratic equation?
1. (A) $2 (x - 1)^{2} = 4 x^{2} - 2 x + 1$
2. (B) $2 x - x^{2} = x^{2} + 5$
3. (C) $(2 x + 3)^{2} + x^{2} = 3 x^{2} - 5 x$
4. (D) $(x^{2} + 2 x)^{2} = x^{4} + 3 + 4 x^{3}$
Try it first, then show our solutionSolution
1. For (C): $2 x^{2} + 26 x + 3 + x^{2} = 3 x^{2} - 5 x$ , which simplifies to $(26 + 5) x + 3 = 0$ . The $x^{2}$ terms cancel, so it is linear, not quadratic.
2. Each of the other options keeps an $x^{2}$ term with highest degree 2 after simplifying. Answer: option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Expands (C) and shows the $x^{2}$ terms cancel (1 mark)
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Q17 · 1 mark · MCQCircles
How many tangents can be drawn to a circle from a point on it?
1. (A)One
2. (B)Two
3. (C)Infinite
4. (D)Zero
Try it first, then show our solutionSolution
1. At a point on the circle, there is exactly one tangent: the line perpendicular to the radius at that point. Answer: option (A).
  1 mark
Answer: One, option (A)
Where the marks are
- States exactly one tangent (1 mark)
Q18 · 1 mark · MCQCircles
The length of the tangent from an external point A to a circle, of radius $3$ cm, is $4$ cm. The distance of A from the centre of the circle is:
1. (A) $7$ cm
2. (B) $5$ cm
3. (C) $7$ cm
4. (D) $25$ cm
Try it first, then show our solutionSolution
1. The radius is perpendicular to the tangent, so $O A^{2} = 3^{2} + 4^{2} = 25$ .
2. $O A = 5$ cm. Answer: option (B).
  1 mark
Answer: 5 cm, option (B)
Where the marks are
- Applies Pythagoras with radius ⟂ tangent to get 5 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 an external point to a circle are equal.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true but Reason (R) does not give 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. Both statements are standard theorems, so both A and R are true.
2. R (equal tangent lengths) does not explain why the tangent is perpendicular to the radius. Answer: option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Identifies both as true and R not explaining A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedQuadratic Equations
Assertion (A): If one root of the quadratic equation $4 x^{2} - 10 x + (k - 4) = 0$ is reciprocal of the other, then value of $k$ is $8$ .
Reason (R): Roots of the quadratic equation $x^{2} - x + 1 = 0$ are real.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true but Reason (R) does not give 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. If the roots are reciprocals, their product is 1, so $\frac{k - 4}{4} = 1$ , giving $k = 8$ . A is true.
2. For $x^{2} - x + 1 = 0$ : $D = 1 - 4 = - 3 < 0$ , so the roots are not real and R is false. Answer: option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows $k = 8$ using product of roots = 1, and $D < 0$ for R (1 mark)
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Section B

Very short answer; internal choice in 22 and 23

Q21 · 2 marks · Very short answerIntroduction to Trigonometry
If $sin α = \frac{1}{2}$ , then find the value of $(3 cos α - 4 cos^{3} α)$ .
Try it first, then show our solutionSolution
1. $sin α = \frac{1}{2} = sin 3 0^{\circ} \Rightarrow α = 3 0^{\circ}$
  ½ mark
2. $3 cos 3 0^{\circ} - 4 cos^{3} 3 0^{\circ} = 3 (\frac{3}{2}) - 4 (\frac{3}{2})^{3} = \frac{3 3}{2} - \frac{4 \cdot 3 3}{8}$
  1 mark
3. $= \frac{3 3}{2} - \frac{3 3}{2} = 0$
  ½ mark
Answer: 0
Where the marks are
- Identifies $α = 3 0^{\circ}$ (½ mark)
- Substitutes $cos 3 0^{\circ} = \frac{3}{2}$ and simplifies the cube correctly (1 mark)
- Obtains value 0 (½ mark)
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Answer one of these 2 alternatives.
Q22(A) · 2 marks · Very short answerCoordinate Geometry
Find the coordinates of the point which divides the join of $A (- 1, 7)$ and $B (4, - 3)$ in the ratio $2 : 3$ .
Try it first, then show our solutionSolution
1. By the section formula with $m_{1} : m_{2} = 2 : 3$ , $A (- 1, 7)$ , $B (4, - 3)$ : $x = \frac{2 ( 4 ) + 3 ( - 1 )}{2 + 3} = \frac{8 - 3}{5} = 1$
  1 mark
2. $y = \frac{2 ( - 3 ) + 3 ( 7 )}{2 + 3} = \frac{- 6 + 21}{5} = 3$ . Required point is $(1, 3)$ .
  1 mark
Answer: (1, 3)
Where the marks are
- Correct x-coordinate 1 using section formula (1 mark)
- Correct y-coordinate 3 (1 mark)
Open Q22(A) on its own page
OR
Q22(B) · 2 marks · Very short answerCoordinate Geometry
If the points $A (2, 3)$ , $B (- 5, 6)$ , $C (6, 7)$ and $D (p, 4)$ are the vertices of a parallelogram $A B C D$ , find the value of $p$ .
Try it first, then show our solutionSolution
1. Diagonals of a parallelogram bisect each other, so mid-point of AC = mid-point of BD: $(\frac{2 + 6}{2}, \frac{3 + 7}{2}) = (\frac{- 5 + p}{2}, \frac{6 + 4}{2})$ , i.e. $(4, 5) = (\frac{p - 5}{2}, 5)$
  1 mark
2. $\frac{p - 5}{2} = 4 \Rightarrow p = 13$
  1 mark
Answer: p = 13
Where the marks are
- Equates mid-points of diagonals AC and BD (1 mark)
- Solves to get $p = 13$ (1 mark)
Open Q22(B) on its own page
Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerQuadratic Equations
Find the discriminant of the quadratic equation $3 x^{2} - 2 x + \frac{1}{3} = 0$ and hence find the nature of its roots.
Try it first, then show our solutionSolution
1. Comparing with $a x^{2} + b x + c = 0$ : $a = 3$ , $b = - 2$ , $c = \frac{1}{3}$
  ½ mark
2. $D = b^{2} - 4 a c = (- 2)^{2} - 4 (3) (\frac{1}{3}) = 4 - 4 = 0$
  1 mark
3. Since $D = 0$ , the roots are real and equal.
  ½ mark
Answer: D = 0; roots are real and equal
Where the marks are
- Identifies a, b, c (½ mark)
- Computes $D = 0$ (1 mark)
- States roots are real and equal (½ mark)
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OR
Q23(B) · 2 marks · Very short answerQuadratic Equations
Find the roots of the quadratic equation $x^{2} - x - 2 = 0$ .
Try it first, then show our solutionSolution
1. Splitting the middle term: $x^{2} - 2 x + x - 2 = 0 \Rightarrow x (x - 2) + 1 (x - 2) = 0 \Rightarrow (x - 2) (x + 1) = 0$
  1 mark
2. $x - 2 = 0 \Rightarrow x = 2$ ; $x + 1 = 0 \Rightarrow x = - 1$
  1 mark
Answer: x = 2, x = −1
Where the marks are
- Factorises as $(x - 2) (x + 1) = 0$ (1 mark)
- Each root 2 and −1 (½ + ½ mark)
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Q24 · 2 marks · Very short answerCircles
In the adjoining figure, PT is a tangent at T to the circle with centre O. If $∠ T P O = 3 0^{\circ}$ , find the value of $x$ .
Figure in the paper: A circle with centre O. P is a point outside the circle, and PT is a tangent touching the circle at T. OT is a radius and OP is joined. The angle TPO = 30° is marked at P. The angle x is marked at the centre O, below the line OP. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O T P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]
  1 mark
2. In $△ O T P$ : $∠ T O P = 18 0^{\circ} - 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ . Since TO produced is a straight line, $x + ∠ T O P = 18 0^{\circ}$ [linear pair] $\Rightarrow x = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$
  1 mark
Answer: x = 120°
Where the marks are
- Uses $∠ O T P = 9 0^{\circ}$ (tangent ⟂ radius) (1 mark)
- Finds $∠ T O P = 6 0^{\circ}$ and $x = 12 0^{\circ}$ by linear pair (1 mark)
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Q25 · 2 marks · Very short answerTriangles
In the adjoining figure, A, B and C are points on OP, OQ and OR respectively such that $A B ∥ P Q$ and $A C ∥ P R$ . Show that $B C ∥ QR$ .
Figure in the paper: A triangle PQR with P at the top, Q at bottom-left and R at bottom-right. O is a point inside the triangle, and P, O, A lie on a line from P, with A between P and O. Lines OP, OQ and OR are drawn from O. A lies on OP, B lies on OQ and C lies on OR. AB is drawn parallel to PQ and AC is drawn parallel to PR. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ P O Q$ , $A B ∥ P Q$ , so $\frac{O A}{A P} = \frac{O B}{B Q}$ ...(i) [Basic Proportionality (Thales) Theorem]
  ½ mark
2. In $△ P O R$ , $A C ∥ P R$ , so $\frac{O A}{A P} = \frac{O C}{C R}$ ...(ii) [BPT]
  ½ mark
3. From (i) and (ii): $\frac{O B}{B Q} = \frac{O C}{C R}$
  ½ mark
4. Hence in $△ QO R$ , B and C divide OQ and OR in the same ratio, so $B C ∥ QR$ [converse of BPT].
  ½ mark
Answer: BC ∥ QR (proved)
Where the marks are
- Applies BPT in △POQ (½ mark)
- Applies BPT in △POR (½ mark)
- Equates the ratios $\frac{O B}{B Q} = \frac{O C}{C R}$ (½ mark)
- Concludes BC ∥ QR by converse of BPT (½ mark)
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Section C

Short answer; internal choice in 28 and 31

Q26 · 3 marks · Short answerPolynomials
Find the zeroes of the quadratic polynomial $x^{2} + 6 x + 8$ and verify the relationship between the zeroes and the coefficients.
Try it first, then show our solutionSolution
1. $x^{2} + 6 x + 8 = x^{2} + 4 x + 2 x + 8 = (x + 4) (x + 2)$
  1 mark
2. Zeroes: $x = - 4$ and $x = - 2$
  1 mark
3. Sum of zeroes $= - 4 + (- 2) = - 6 = \frac{- 6}{1} = \frac{- coeff. of x}{coeff. of x ^{2}}$
  ½ mark
4. Product of zeroes $= (- 4) (- 2) = 8 = \frac{8}{1} = \frac{constant term}{coeff. of x ^{2}}$ . Relationship verified.
  ½ mark
Answer: Zeroes −4 and −2; sum −6 = −b/a, product 8 = c/a (verified)
Where the marks are
- Factorises $(x + 4) (x + 2)$ (1 mark)
- Writes zeroes −4, −2 (1 mark)
- Verifies sum = −b/a (½ mark)
- Verifies product = c/a (½ mark)
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Q27 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{1 + t a n ^{2} A}{1 + c o t ^{2} A} = sec^{2} A - 1$
Try it first, then show our solutionSolution
1. LHS $= \frac{1 + \frac{s i n ^{2} A}{c o s ^{2} A}}{1 + \frac{c o s ^{2} A}{s i n ^{2} A}}$
  1 mark
2. $= \frac{\frac{c o s ^{2} A + s i n ^{2} A}{c o s ^{2} A}}{\frac{s i n ^{2} A + c o s ^{2} A}{s i n ^{2} A}}$
  ½ mark
3. $= \frac{\frac{1}{c o s ^{2} A}}{\frac{1}{s i n ^{2} A}} = \frac{s i n ^{2} A}{c o s ^{2} A} = \frac{1 - c o s ^{2} A}{c o s ^{2} A}$ [using $sin^{2} A + cos^{2} A = 1$ ]
  1 mark
4. $= \frac{1}{c o s ^{2} A} - 1 = sec^{2} A - 1 =$ RHS
  ½ mark
Answer: LHS = sec²A − 1 = RHS (proved)
Where the marks are
- Writes tan and cot in terms of sin and cos (1 mark)
- Combines into single fractions (½ mark)
- Uses $sin^{2} A + cos^{2} A = 1$ to reach $\frac{1 - c o s ^{2} A}{c o s ^{2} A}$ (1 mark)
- Concludes $sec^{2} A - 1$ (½ mark)
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Answer one of these 2 alternatives.
Q28(A) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
A lending library has a fixed charge for first three days and an additional charge for each day thereafter. Rittik paid ₹ $27$ for a book kept for $7$ days and Manmohan paid ₹ $21$ for a book kept for $5$ days. Find the fixed charges and the charge for each extra day.
Try it first, then show our solutionSolution
1. Let fixed charge (first 3 days) be ₹ $x$ and charge per extra day be ₹ $y$ . Rittik: 7 days = 3 + 4 extra days, so $x + 4 y = 27$ ...(1)
  1 mark
2. Manmohan: 5 days = 3 + 2 extra days, so $x + 2 y = 21$ ...(2)
  1 mark
3. Subtracting (2) from (1): $2 y = 6 \Rightarrow y = 3$ ; then $x = 21 - 6 = 15$
  1 mark
4. Fixed charge = ₹15, charge for each extra day = ₹3.
Answer: Fixed charge ₹15; extra charge ₹3 per day
Where the marks are
- Forms $x + 4 y = 27$ (1 mark)
- Forms $x + 2 y = 21$ (1 mark)
- Solves $x = 15$ , $y = 3$ (½ + ½ mark)
Open Q28(A) on its own page
OR
Q28(B) · 3 marks · Short answerPair of Linear Equations in Two Variables
Find the values of ' $a$ ' and ' $b$ ' for which the system of linear equations $3 x + 4 y = 12$ , $(a + b) x + 2 (a - b) y = 24$ has infinite number of solutions.
Try it first, then show our solutionSolution
1. For infinitely many solutions, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ : $\frac{3}{a + b} = \frac{4}{2 ( a - b )} = \frac{12}{24}$
  1 mark
2. $\frac{3}{a + b} = \frac{1}{2} \Rightarrow a + b = 6$
  ½ mark
3. $\frac{4}{2 ( a - b )} = \frac{1}{2} \Rightarrow \frac{2}{a - b} = \frac{1}{2} \Rightarrow a - b = 4$
  ½ mark
4. Adding: $2 a = 10 \Rightarrow a = 5$ ; then $b = 6 - 5 = 1$
  1 mark
Answer: a = 5, b = 1
Where the marks are
- Writes condition $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ with values (1 mark)
- Gets $a + b = 6$ (½ mark)
- Gets $a - b = 4$ (½ mark)
- Solves $a = 5$ , $b = 1$ (½ + ½ mark)
Open Q28(B) on its own page
Q29 · 3 marks · Short answerProbability
A die is rolled once. Find the probability of getting:
1. (i)
  an even prime number.
  0 marks
2. (ii)
  a number greater than $4$ .
  0 marks
3. (iii)
  an odd number.
  0 marks
Try it first, then show our solutionSolution
1. Sample space $S = {1, 2, 3, 4, 5, 6}$ , total outcomes = 6. (i) Even prime number: only 2, so 1 favourable outcome. $P = \frac{1}{6}$
  1 mark
2. (ii) Numbers greater than 4: 5, 6, so 2 favourable outcomes. $P = \frac{2}{6} = \frac{1}{3}$
  1 mark
3. (iii) Odd numbers: 1, 3, 5, so 3 favourable outcomes. $P = \frac{3}{6} = \frac{1}{2}$
  1 mark
Answer: (i) 1/6; (ii) 1/3; (iii) 1/2
Where the marks are
- (i) $\frac{1}{6}$ (1 mark)
- (ii) $\frac{1}{3}$ (1 mark)
- (iii) $\frac{1}{2}$ (1 mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerAreas Related to Circles
Find the area of the sector of a circle of radius $7$ cm and of central angle $9 0^{\circ}$ . Also, find the area of corresponding major sector.
Try it first, then show our solutionSolution
1. Area of sector $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{90}{360} \times \frac{22}{7} \times 7 \times 7$
  1 mark
2. $= \frac{77}{2} = 38.5 cm^{2}$
  ½ mark
3. Area of circle $= π r^{2} = \frac{22}{7} \times 7 \times 7 = 154 cm^{2}$
  1 mark
4. Area of major sector $= 154 - 38.5 = 115.5 cm^{2}$
  ½ mark
Answer: Minor sector 38.5 cm²; major sector 115.5 cm²
Where the marks are
- Uses sector formula with correct substitution (1 mark)
- Gets 38.5 cm² (½ mark)
- Area of circle 154 cm² (1 mark)
- Major sector 115.5 cm² (½ mark)
Open Q30 on its own page
Answer one of these 2 alternatives.
Q31(A) · 3 marks · Short answerCircles
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Try it first, then show our solutionSolution
1. Figure: circle with centre O; from external point P draw tangents PQ and PR touching the circle at Q and R; join OP, OQ, OR.
  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 congruence].
  1 mark
4. Hence $P Q = P R$ [CPCT].
  ½ mark
Answer: PQ = PR (proved)
Where the marks are
- Correct labelled figure (1 mark)
- Given, To prove, Construction (½ mark)
- Proves △OQP ≅ △ORP by RHS with reasons (1 mark)
- Concludes PQ = PR by CPCT (½ mark)
Open Q31(A) on its own page
OR
Q31(B) · 3 marks · Short answerCircles
Two concentric circles with centre $O$ are of radii $3$ cm and $5$ cm. Find the length of chord $A B$ of the larger circle which touches the smaller circle at $P$ .
Figure in the paper: Two concentric circles with centre O. The smaller circle has radius 3 cm and the larger circle has radius 5 cm. A chord AB of the larger circle is drawn horizontally, touching the smaller circle at P, which lies at the midpoint of AB directly below O. See it in the official paper.
Try it first, then show our solutionSolution
1. Join OA and OP. AB touches the smaller circle at P, so $O P ⊥ A B$ [radius ⟂ tangent at point of contact].
  ½ mark
2. AB is a chord of the larger circle and $O P ⊥ A B$ , so $A P = P B$ [perpendicular from centre bisects the chord].
  1 mark
3. In right $△ O P A$ : $A P^{2} = O A^{2} - O P^{2} = 5^{2} - 3^{2} = 16$
  ½ mark
4. $A P = 4$ cm $= P B$
  ½ mark
5. $A B = 2 \times 4 = 8$ cm
  ½ mark
Answer: AB = 8 cm
Where the marks are
- States OP ⟂ AB (½ mark)
- AP = PB as perpendicular from centre bisects chord (1 mark)
- Pythagoras: $A P^{2} = 25 - 9 = 16$ (½ mark)
- AP = 4 cm (½ mark)
- AB = 8 cm (½ mark)
Open Q31(B) on its own page

Section D

Long answer; internal choice in 32 and 33

Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerSome Applications of Trigonometry
The shadow of a tower standing on a level ground is found to be $40$ m longer when the Sun's altitude is $3 0^{\circ}$ than when it was $6 0^{\circ}$ . Find the height of the tower.
Try it first, then show our solutionSolution
1. Figure: tower AB = $h$ m with foot A; C and D on the ground on the same side, $A C = x$ m (shadow at $6 0^{\circ}$ ), $C D = 40$ m, $∠ A C B = 6 0^{\circ}$ , $∠ A D B = 3 0^{\circ}$ .
  1 mark
2. In right $△ B A C$ : $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow 3 = \frac{h}{x} \Rightarrow h = 3 x$ ...(1)
  1½ marks
3. In right $△ B A D$ : $tan 3 0^{\circ} = \frac{h}{x + 40} \Rightarrow \frac{1}{3} = \frac{h}{x + 40} \Rightarrow x + 40 = 3 h$ ...(2)
  1½ marks
4. Substituting (1) in (2): $x + 40 = 3 (3 x) = 3 x \Rightarrow 2 x = 40 \Rightarrow x = 20$ m
  ½ mark
5. $h = 203$ m. Height of the tower is $203$ m (≈ 34.64 m).
  ½ mark
Answer: Height of tower = 20√3 m (≈ 34.64 m)
Where the marks are
- Correct figure (1 mark)
- $tan 6 0^{\circ} = \frac{h}{x}$ giving $h = 3 x$ (1½ marks)
- $tan 3 0^{\circ} = \frac{h}{x + 40}$ giving $x + 40 = 3 h$ (1½ marks)
- Finds $x = 20$ m (½ mark)
- Height $= 203$ m (½ mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerSome Applications of Trigonometry
From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . Determine the height of the tower.
Try it first, then show our solutionSolution
1. Figure: building AB = 7 m (A on ground, B top), tower PQ with foot P and top Q, $A P = x$ . Draw $B C ∥ A P$ meeting PQ at C, so $B C = x$ , $C P = 7$ m, $C Q = h$ . $∠ QB C = 6 0^{\circ}$ (elevation), $∠ C B P = 4 5^{\circ} = ∠ B P A$ (depression = alternate angle).
  1 mark
2. In right $△ B A P$ : $tan 4 5^{\circ} = \frac{A B}{A P} = \frac{7}{x} \Rightarrow 1 = \frac{7}{x} \Rightarrow x = 7$ m
  1½ marks
3. In right $△ B C Q$ : $tan 6 0^{\circ} = \frac{C Q}{B C} = \frac{h}{x} \Rightarrow h = 3 x$
  1½ marks
4. $h = 73$ m
  ½ mark
5. Height of tower $P Q = P C + C Q = 7 + 73 = 7 (1 + 3)$ m (≈ 19.12 m)
  ½ mark
Answer: Height of tower = 7(1 + √3) m (≈ 19.12 m)
Where the marks are
- Correct figure (1 mark)
- $tan 4 5^{\circ} = \frac{7}{x}$ giving $x = 7$ m (1½ marks)
- $tan 6 0^{\circ} = \frac{h}{x}$ giving $h = 3 x$ (1½ marks)
- $h = 73$ m (½ mark)
- Tower height $7 (1 + 3)$ m (½ mark)
Open Q32(B) on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerArithmetic Progressions
Find the sum of first $25$ terms of the A.P. whose $n^{t h}$ term is given by $a_{n} = 5 + 6 n$ . Also, find the ratio of $2 0^{t h}$ term to $4 5^{t h}$ term.
Try it first, then show our solutionSolution
1. $a_{1} = 5 + 6 (1) = 11$
  ½ mark
2. $a_{2} = 5 + 6 (2) = 17$
  ½ mark
3. $d = a_{2} - a_{1} = 17 - 11 = 6$
  ½ mark
4. $S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{25} = \frac{25}{2} [2 (11) + 24 (6)] = \frac{25}{2} [22 + 144]$
  1 mark
5. $S_{25} = \frac{25}{2} \times 166 = 2075$
  ½ mark
6. $a_{20} = 5 + 6 (20) = 125$ and $a_{45} = 5 + 6 (45) = 275$
  1 mark
7. $\frac{a _{20}}{a _{45}} = \frac{125}{275} = \frac{5}{11}$ , so the ratio is $5 : 11$ .
  1 mark
Answer: S₂₅ = 2075; a₂₀ : a₄₅ = 5 : 11
Where the marks are
- $a_{1} = 11$ (½ mark)
- $a_{2} = 17$ (½ mark)
- $d = 6$ (½ mark)
- Correct substitution in $S_{n}$ formula (1 mark)
- $S_{25} = 2075$ (½ mark)
- Finds $a_{20} = 125$ and $a_{45} = 275$ (1 mark)
- Ratio $5 : 11$ (1 mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerArithmetic Progressions
In an A.P., if $S_{n} = 3 n^{2} + 5 n$ and $a_{k} = 164$ , find the value of $k$ .
Try it first, then show our solutionSolution
1. $S_{1} = 3 (1)^{2} + 5 (1) = 8$ , so $a_{1} = 8$
  1 mark
2. $S_{2} = 3 (2)^{2} + 5 (2) = 22$ (sum of first two terms)
  1 mark
3. $a_{1} + a_{2} = 22 \Rightarrow a_{2} = 22 - 8 = 14$
  ½ mark
4. $d = a_{2} - a_{1} = 14 - 8 = 6$
  1 mark
5. $a_{k} = a + (k - 1) d = 164 \Rightarrow 8 + (k - 1) 6 = 164$
  1 mark
6. $6 (k - 1) = 156 \Rightarrow k - 1 = 26 \Rightarrow k = 27$
  ½ mark
Answer: k = 27
Where the marks are
- $S_{1} = a_{1} = 8$ (1 mark)
- $S_{2} = 22$ (1 mark)
- $a_{2} = 14$ (½ mark)
- $d = 6$ (1 mark)
- Forms $8 + (k - 1) 6 = 164$ (1 mark)
- $k = 27$ (½ mark)
Open Q33(B) on its own page
Q34 · 5 marks · Long answerStatistics
The following table gives the monthly consumption of electricity of $100$ families:
Monthly Consumption (in units) 130-140 140-150 150-160 160-170 170-180 180-190 190-200
Number of families 5 9 17 28 24 10 7
Find the median of the above data.
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Monthly consumption Families ( $f$ ) $c f$
  130–140 5 5
  140–150 9 14
  150–160 17 31
  160–170 28 59
  170–180 24 83
  180–190 10 93
  190–200 7 100
  $N = 100$
  1 mark
2. $\frac{N}{2} = \frac{100}{2} = 50$ . The first $c f$ greater than 50 is 59, so the median class is $160$ – $170$ .
  1 mark
3. For the median class: $l = 160$ , $h = 10$ , $f = 28$ . The $c f$ of the class before it is $31$ .
  1 mark
4. Median $= l + (\frac{\frac{N}{2} - c f}{f}) \times h = 160 + \frac{50 - 31}{28} \times 10 = 160 + \frac{190}{28}$
  1½ marks
5. $= 160 + 6.79 \approx 166.8$ units
  ½ mark
Answer: Median ≈ 166.8 units
Where the marks are
- Correct cumulative frequency table (1 mark)
- Finds $N /2 = 50$ and identifies median class 160–170 (1 mark)
- Writes $l = 160$ , $h = 10$ , $c f = 31$ , $f = 28$ (1 mark)
- Substitutes correctly in the median formula (1½ marks)
- Final answer 166.8 (½ mark)
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Q35 · 5 marks · Long answerSurface Areas and Volumes
The boilers are used in thermal power plants to store water and then used to produce steam. One such boiler consists of a cylindrical part in middle and two hemispherical parts at its both ends. Length of the cylindrical part is $7$ m and radius of cylindrical part is $\frac{7}{2}$ m. Find the total surface area and the volume of the boiler. Also, find the ratio of the volume of cylindrical part to the volume of one hemispherical part.
Figure in the paper: A boiler shaped as a cylinder standing upright with a hemisphere attached to its top and a hemisphere attached to its bottom. The cylindrical middle part has height 7 m, and its radius is marked as $\frac{7}{2}$ m from the centre of the top circular face to its rim. The height of the cylindrical part is labelled 7 m. See it in the official paper.
Try it first, then show our solutionSolution
1. Here $h = 7$ m and $r = \frac{7}{2}$ m.
  TSA of boiler = CSA of cylinder + CSA of 2 hemispheres $= 2 π r h + 2 (2 π r^{2}) = 2 π r (h + 2 r)$
  $= 2 \times \frac{22}{7} \times \frac{7}{2} (7 + 2 \times \frac{7}{2}) = 22 \times 14$
  1½ marks
2. Total surface area $= 308 m^{2}$
  ½ mark
3. Volume of boiler $= π r^{2} h + 2 (\frac{2}{3} π r^{3})$
  $= \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7 + \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$
  $= 269.5 + \frac{1078}{6}$
  1½ marks
4. $= \frac{1617}{6} + \frac{1078}{6} = \frac{2695}{6} m^{3} \approx 449.17 m^{3}$
  ½ mark
5. $\frac{Volume of cylindrical part}{Volume of one hemisphere} = \frac{π r ^{2} h}{\frac{2}{3} π r ^{3}}$
  ½ mark
6. $= \frac{3 h}{2 r} = \frac{3 \times 7}{2 \times \frac{7}{2}} = \frac{21}{7} = \frac{3}{1}$
  So the required ratio is $3 : 1$ .
  ½ mark
Answer: TSA = 308 m²; Volume = 2695/6 m³ ≈ 449.17 m³; ratio 3 : 1
Where the marks are
- Correct TSA formula with substitution (1½ marks)
- TSA = 308 m² (½ mark)
- Correct volume formula with substitution (1½ marks)
- Volume = 2695/6 m³ ≈ 449.17 m³ (½ mark)
- Sets up the ratio $π r^{2} h : \frac{2}{3} π r^{3}$ (½ mark)
- Ratio = 3 : 1 (½ mark)
Open Q35 on its own page

Monthly Consumption (in units)	130-140	140-150	150-160	160-170	170-180	180-190	190-200
Number of families	5	9	17	28	24	10	7

Monthly consumption	Families ( $f$ )	$c f$
130–140	5	5
140–150	9	14
150–160	17	31
160–170	28	59
170–180	24	83
180–190	10	93
190–200	7	100

Section E

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

Q36 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Use of mobile screen for long hours makes your eye sight weak and give you headaches. Children who are addicted to play "PUBG" can get easily stressed out. To raise social awareness about ill effects of playing PUBG, a school decided to start 'BAN PUBG' campaign, in which students are asked to prepare campaign board in the shape of a rectangle. One such campaign board made by class X student of the school is shown in the figure.
Based on the above information, answer the following questions:
1. (i)
  Find the coordinates of the point of intersection of diagonals AC and BD.
  1 mark
2. (ii)
  Find the length of the diagonal AC.
  1 mark
3. (iii)(a)
  Find the area of the campaign Board ABCD.
  2 marks
4. OR
  (iii)(b)
  Find the ratio of the length of side AB to the length of the diagonal AC.
  2 marks
Figure in the paper: Coordinate axes with X' on the left and Y upwards. A rectangle ABCD with A(1, 1) at bottom-left, B(7, 1) at bottom-right, C(7, 5) at top-right and D(1, 5) at top-left. The text 'BAN PUBG & SAVE KIDS' is written inside the rectangle. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The diagonals of a rectangle bisect each other, so they meet at the midpoint of AC.
  Midpoint $= (\frac{1 + 7}{2}, \frac{1 + 5}{2}) = (4, 3)$
  1 mark
2. (ii) $A C = (7 - 1)^{2} + (5 - 1)^{2} = 36 + 16 = 52 = 213$ units
  1 mark
3. (iii)(a) $A B = (7 - 1)^{2} + (1 - 1)^{2} = 6$ units
  ½ mark
4. (iii)(a) $B C = (7 - 7)^{2} + (5 - 1)^{2} = 4$ units
  ½ mark
5. (iii)(a) Area of ABCD $= A B \times B C = 6 \times 4 = 24$ sq. units
  1 mark
6. (iii)(b) $A B = (7 - 1)^{2} + (1 - 1)^{2} = 6$ units
  1 mark
7. (iii)(b) $\frac{A B}{A C} = \frac{6}{2 13} = \frac{3}{13}$
  So the required ratio is $3 : 13$ .
  1 mark
Answer: (i) (4, 3); (ii) 2√13 units; (iii)(a) 24 sq. units; (iii)(b) 3 : √13
Where the marks are
- (i) Uses the midpoint of AC to get (4, 3) (1 mark)
- (ii) Distance formula gives AC = 2√13 (1 mark)
- (iii)(a) AB = 6 and BC = 4 (½ + ½ marks)
- (iii)(a) Area = 24 sq. units (1 mark)
- (iii)(b) AB = 6 (1 mark)
- (iii)(b) Ratio AB : AC = 3 : √13 (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedReal Numbers
Khushi wants to organize her birthday party. Being health conscious, she decided to serve only fruits in her birthday party. She bought $36$ apples and $60$ bananas and decided to distribute fruits equally among all.
Based on the above information, answer the following questions:
1. (i)
  How many guests Khushi can invite at the most?
  1 mark
2. (ii)
  How many apples and bananas will each guest get?
  1 mark
3. (iii)(a)
  If Khushi decides to add $42$ mangoes, how many guests Khushi can invite at the most?
  2 marks
4. OR
  (iii)(b)
  If the cost of $1$ dozen of bananas is ₹ $60$ , the cost of $1$ apple is ₹ $15$ and cost of $1$ mango is ₹ $20$ , find the total amount spent on $60$ bananas, $36$ apples and $42$ mangoes.
  2 marks
Figure in the paper: A photograph of a basket containing apples, bananas and other fruits, with bananas at the top and apples in the lower part of the basket. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The largest number of guests is the HCF of 36 and 60.
  $36 = 2^{2} \times 3^{2}$ and $60 = 2^{2} \times 3 \times 5$ , so HCF $= 2^{2} \times 3 = 12$ .
  Khushi can invite at most 12 guests.
  1 mark
2. (ii) Apples per guest $= 36 \div 12 = 3$ and bananas per guest $= 60 \div 12 = 5$ .
  Each guest gets 3 apples and 5 bananas.
  1 mark
3. (iii)(a) $42 = 2 \times 3 \times 7$ . Using the factorisations above, HCF $(36, 60, 42) = 2 \times 3 = 6$ .
  Khushi can invite at most 6 guests.
  2 marks
4. (iii)(b) 60 bananas $= 5$ dozen. Total cost $= 5 \times 60 + 36 \times 15 + 42 \times 20 = 300 + 540 + 840$
  1 mark
5. (iii)(b) Total amount spent $=$ ₹ $1680$
  1 mark
Answer: (i) 12 guests; (ii) 3 apples and 5 bananas each; (iii)(a) 6 guests; (iii)(b) ₹1680
Where the marks are
- (i) HCF(36, 60) = 12, so 12 guests (1 mark)
- (ii) 3 apples and 5 bananas per guest (1 mark)
- (iii)(a) HCF(36, 60, 42) = 6, so 6 guests (2 marks)
- (iii)(b) Correct cost expression 5 × 60 + 36 × 15 + 42 × 20 (1 mark)
- (iii)(b) Total = ₹1680 (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedTriangles
Observe the figures given below carefully and answer the questions:
1. (i)
  Name the figure(s) wherein two figures are similar.
  1 mark
2. (ii)
  Name the figure(s) wherein the figures are congruent.
  1 mark
3. (iii)(a)
  Prove that congruent triangles are also similar but not the converse.
  2 marks
4. OR
  (iii)(b)
  What more is least needed for two similar triangles to be congruent?
  2 marks
Figure in the paper: Figure A: A(i) is a quadrilateral ABCD with AD = 2 cm, DC = 2.4 cm, CB = 2.5 cm, AB = 1.2 cm; angles D = 85°, C = 70°, B = 100°, A = 105°. A(ii) is a quadrilateral PQRS with SR = 4.8 cm, RQ = 5 cm, QP = 2.4 cm, PS = 4 cm; angles S = 85°, R = 70°, Q = 100°, P = 105°. Figure B: B(iii) is a square ABCD with sides 3.5 cm on the top and bottom and 3 cm on the left and right, all angles right angles. B(iv) is a square PQRS with sides 3 cm on all four sides, all angles right angles. Figure C: C(v) is triangle ABC with AB = 2 cm, BC = 3 cm, AC = 4.5 cm. C(vi) is triangle PQR with PQ = 2 cm, QR = 3 cm, PR = 4.5 cm, with matching tick marks on corresponding sides. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Figure A: the corresponding angles are equal (85°, 70°, 100°, 105°). The corresponding sides are in the same ratio: $\frac{1.2}{2.4} = \frac{2.4}{4.8} = \frac{2}{4} = \frac{2.5}{5} = \frac{1}{2}$ . So A(i) and A(ii) are similar.
  ½ mark
2. (i) Figure C: the triangles have equal corresponding sides (2 cm, 3 cm, 4.5 cm), so they are congruent and hence similar. Figure B is not similar, because $\frac{3.5}{3} \neq = \frac{3}{3}$ .
  Answer: Figure A and Figure C.
  ½ mark
3. (ii) Only Figure C has congruent figures: $△ A B C ≅ △ P QR$ by SSS, since $A B = P Q = 2$ cm, $B C = QR = 3$ cm and $A C = P R = 4.5$ cm.
  1 mark
4. (iii)(a) If $△ A B C ≅ △ P QR$ , then by CPCT $∠ A = ∠ P$ , $∠ B = ∠ Q$ , $∠ C = ∠ R$ , and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P} = 1$ .
  1 mark
5. (iii)(a) Hence the corresponding angles are equal and the corresponding sides are proportional, so the triangles are similar.
  ½ mark
6. (iii)(a) The converse is not true. Take an equilateral triangle of side 2 cm and one of side 4 cm. All their angles are 60°, so they are similar, but their sides are not equal ( $\frac{2}{4} = \frac{1}{2} \neq = 1$ ). So similar triangles need not be congruent.
  ½ mark
7. (iii)(b) Similar triangles already have equal angles and proportional sides. If at least one pair of corresponding sides is equal, the ratio of proportionality becomes 1, so all corresponding sides are equal and the triangles are congruent.
  So what is least needed is one pair of equal corresponding sides.
  2 marks
Answer: (i) Figure A and Figure C; (ii) Figure C; (iii)(a) congruent ⇒ equal angles and sides in ratio 1 ⇒ similar; the converse fails (e.g. equilateral triangles of sides 2 cm and 4 cm); (iii)(b) one pair of corresponding sides must be equal
Where the marks are
- (i) Names Figure A (½ mark) and Figure C (½ mark)
- (ii) Figure C (1 mark)
- (iii)(a) Congruent ⇒ corresponding angles equal (1 mark)
- (iii)(a) Hence similar (½ mark)
- (iii)(a) Shows the converse fails, with a reason or counter-example (½ mark)
- (iii)(b) At least one pair of corresponding sides must be equal (2 marks)
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