MathsCBSE Class 10 · Basic · Sample paper, session 2023-24

CBSE Class 10 Mathematics Basic: sample paper 2023-24

The sample question paper CBSE published for the 2023-24 session, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

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

Source: official sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassX_2023_24/MathsBasic-SQP.pdf. Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	SA-I; internal choice in 22 and 25
C	26-31	SA-II; internal choice in 28 and 30
D	32-35	LA; internal choice in 32 and 34
E	36-38	Case/source based; sub-parts of 1, 1 and 2 marks; internal choice within sub-parts of 36, 37 and 38

Instructions in brief: All questions are compulsory; internal choice is given in 2 questions of 2 marks, 2 questions of 3 marks, 2 questions of 5 marks, and in the 2-mark sub-parts of Section E; take π = 22/7 unless stated otherwise.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If two positive integers a and b are written as $a = x^{3} y^{2}$ and $b = x y^{3}$ , where x, y are prime numbers, then HCF (a, b) is:
1. (A) $x y$
2. (B) $x y^{2}$
3. (C) $x^{3} y^{3}$
4. (D) $x^{2} y^{2}$
Try it first, then show our solutionSolution
1. HCF = product of the smallest power of each common prime factor: $x^{m i n (3, 1)} y^{m i n (2, 3)} = x y^{2}$ , option (B).
  1 mark
Answer: $x y^{2}$ , option (B)
Where the marks are
- Takes the lowest powers of the common primes to get $x y^{2}$ (1 mark)
Q2 · 1 mark · MCQReal Numbers
The LCM of smallest two digit composite number and smallest composite number is:
1. (A)12
2. (B)4
3. (C)20
4. (D)44
Try it first, then show our solutionSolution
1. The smallest two-digit composite number is 10 and the smallest composite number is 4.
2. $10 = 2 \times 5$ and $4 = 2^{2}$ , so LCM $= 2^{2} \times 5 = 20$ , option (C).
  1 mark
Answer: 20, option (C)
Where the marks are
- Identifies 10 and 4 and finds LCM = 20 (1 mark)
Q3 · 1 mark · MCQQuadratic Equations
If $x = 3$ is one of the roots of the quadratic equation $x^{2} - 2 k x - 6 = 0$ , then the value of $k$ is
1. (A) $- \frac{1}{2}$
2. (B) $\frac{1}{2}$
3. (C)3
4. (D)2
Try it first, then show our solutionSolution
1. Substitute $x = 3$ : $9 - 6 k - 6 = 0 \Rightarrow 3 = 6 k \Rightarrow k = \frac{1}{2}$ , option (B).
  1 mark
Answer: $k = \frac{1}{2}$ , option (B)
Where the marks are
- Substitutes the root and solves to get $k = \frac{1}{2}$ (1 mark)
Q4 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of equations $y = 0$ and $y = - 7$ has:
1. (A)one solution
2. (B)two solutions
3. (C)infinitely many solutions
4. (D)no solution
Try it first, then show our solutionSolution
1. $y = 0$ is the x-axis and $y = - 7$ is a line parallel to it. Parallel lines never meet, so there is no solution, option (D).
  1 mark
Answer: No solution, option (D)
Where the marks are
- Recognises that the two lines are parallel, so there is no solution (1 mark)
Q5 · 1 mark · MCQQuadratic Equations
Value(s) of $k$ for which the quadratic equation $2 x^{2} - k x + k = 0$ has equal roots is:
1. (A)0 only
2. (B)4
3. (C)8 only
4. (D)0, 8
Try it first, then show our solutionSolution
1. For equal roots, $D = b^{2} - 4 a c = 0$ : $k^{2} - 4 (2) (k) = 0$ .
2. $k (k - 8) = 0 \Rightarrow k = 0$ or $k = 8$ , option (D).
  1 mark
Answer: 0, 8, option (D)
Where the marks are
- Sets the discriminant to zero and gets $k = 0, 8$ (1 mark)
Q6 · 1 mark · MCQCoordinate Geometry
The distance of the point $(3, 5)$ from x-axis is $k$ units, then $k$ equals:
1. (A)3
2. (B)4
3. (C)5
4. (D)8
Try it first, then show our solutionSolution
1. The distance of a point from the x-axis is $∣ y ∣$ , so $k = 5$ , option (C).
  1 mark
Answer: k = 5, option (C)
Where the marks are
- Uses distance from the x-axis = |y-coordinate| to get 5 (1 mark)
Q7 · 1 mark · MCQTriangles
If in $Δ$ ABC and $Δ$ PQR, $\frac{A B}{QR} = \frac{B C}{P R} = \frac{C A}{P Q}$ , then:
1. (A) $Δ$ PQR $\sim Δ$ CAB
2. (B) $Δ$ PQR $\sim Δ$ ABC
3. (C) $Δ$ CBA $\sim Δ$ PQR
4. (D) $Δ$ BCA $\sim Δ$ PQR
Try it first, then show our solutionSolution
1. Match the sides in proportion: AB↔QR, BC↔RP, CA↔PQ. So the vertex correspondence is A↔Q, B↔R, C↔P.
2. Hence $Δ A B C \sim Δ QR P$ , which is the same as $Δ P QR \sim Δ C A B$ (SSS similarity), option (A).
  1 mark
Answer: $Δ P QR \sim Δ C A B$ , option (A)
Where the marks are
- Gets the correct vertex correspondence, giving option (A) (1 mark)
Open Q7 on its own page
Q8 · 1 mark · MCQTriangles
Which of the following is NOT a similarity criterion of triangles?
1. (A)AA
2. (B)SAS
3. (C)AAA
4. (D)RHS
Try it first, then show our solutionSolution
1. AA, AAA and SAS are similarity criteria. RHS is a congruence criterion, not a similarity criterion, so the answer is option (D).
  1 mark
Answer: RHS, option (D)
Where the marks are
- Identifies RHS as the criterion that is not a similarity criterion (1 mark)
Q9 · 1 mark · MCQCircles
In figure, if TP and TQ are the two tangents to a circle with centre O so that $∠ P O Q = 11 0^{\circ}$ , then $∠ P T Q$ is equal to
1. (A) $6 0^{\circ}$
2. (B) $7 0^{\circ}$
3. (C) $8 0^{\circ}$
4. (D) $9 0^{\circ}$
Figure in the paper: A circle with centre O. Points P and Q lie on the circle. T is an external point from which tangents TP and TQ are drawn touching the circle at P and Q. The angle POQ = 110° is marked at the centre O. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O P T = ∠ O QT = 9 0^{\circ}$ [tangent ⟂ radius].
2. In quadrilateral OPTQ, the angles add up to $36 0^{\circ}$ : $∠ P T Q = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 11 0^{\circ} = 7 0^{\circ}$ , option (B).
  1 mark
Answer: $7 0^{\circ}$ , option (B)
Where the marks are
- Uses tangent ⟂ radius and the angle sum of a quadrilateral to get $7 0^{\circ}$ (1 mark)
Open Q9 on its own page
Q10 · 1 mark · MCQIntroduction to Trigonometry
If $cos A = \frac{4}{5}$ then $tan A$ is:
1. (A) $\frac{3}{5}$
2. (B) $\frac{3}{4}$
3. (C) $\frac{4}{3}$
4. (D) $\frac{1}{8}$
Try it first, then show our solutionSolution
1. Take base = 4 and hypotenuse = 5. Then perpendicular = $25 - 16 = 3$ .
2. $tan A = \frac{3}{4}$ , option (B).
  1 mark
Answer: $\frac{3}{4}$ , option (B)
Where the marks are
- Finds the third side as 3 and gets $tan A = 3/4$ (1 mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
If the height of the tower is equal to the length of its shadow, then the angle of elevation of the sun 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. $tan θ = \frac{height}{shadow} = 1 \Rightarrow θ = 4 5^{\circ}$ , 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)
Q12 · 1 mark · MCQIntroduction to Trigonometry
$(1 - cos^{2} A)$ is equal to
1. (A) $sin^{2} A$
2. (B) $tan^{2} A$
3. (C) $1 - sin^{2} A$
4. (D) $sec^{2} A$
Try it first, then show our solutionSolution
1. Since $sin^{2} A + cos^{2} A = 1$ , we get $1 - cos^{2} A = sin^{2} A$ , option (A).
  1 mark
Answer: $sin^{2} A$ , option (A)
Where the marks are
- Applies the identity $sin^{2} A + cos^{2} A = 1$ (1 mark)
Q13 · 1 mark · MCQAreas Related to Circles
The radius of a circle is same as the side of a square. Their perimeters are in the ratio
1. (A) $1 : 1$
2. (B) $2 : π$
3. (C) $π : 2$
4. (D) $π : 2$
Try it first, then show our solutionSolution
1. Circle perimeter = $2 π r$ and square perimeter = $4 r$ . Ratio $= 2 π r : 4 r = π : 2$ , option (C).
  1 mark
Answer: $π : 2$ , option (C)
Where the marks are
- Writes both perimeters and simplifies the ratio to $π : 2$ (1 mark)
Q14 · 1 mark · MCQAreas Related to Circles
The area of the circle is $154 c m^{2}$ . The radius of the circle is
1. (A)7 cm
2. (B)14 cm
3. (C)3.5 cm
4. (D)17.5 cm
Try it first, then show our solutionSolution
1. $\frac{22}{7} r^{2} = 154 \Rightarrow r^{2} = 49 \Rightarrow r = 7$ cm, option (A).
  1 mark
Answer: 7 cm, option (A)
Where the marks are
- Solves $π r^{2} = 154$ to get r = 7 cm (1 mark)
Q15 · 1 mark · MCQProbability
When a dice is thrown once, the probability of getting an even number less than 4 is
1. (A) $\frac{1}{4}$
2. (B)0
3. (C) $\frac{1}{2}$
4. (D) $\frac{1}{6}$
Try it first, then show our solutionSolution
1. The only even number less than 4 on a die is 2, so there is 1 favourable outcome out of 6. P = $\frac{1}{6}$ , option (D).
  1 mark
Answer: $\frac{1}{6}$ , option (D)
Where the marks are
- Identifies the single favourable outcome (2) and gets $1/6$ (1 mark)
Q16 · 1 mark · MCQStatistics
For the following distribution:
Class 0-5 5-10 10-15 15-20 20-25
Frequency 10 15 12 20 9
The lower limit of modal class is:
1. (A)15
2. (B)20
3. (C)10
4. (D)5
Try it first, then show our solutionSolution
1. The highest frequency is 20, for the class 15-20. So the modal class is 15-20 and its lower limit is 15, option (A).
  1 mark
Answer: 15, option (A)
Where the marks are
- Identifies modal class 15-20 and its lower limit 15 (1 mark)
Q17 · 1 mark · MCQSurface Areas and Volumes
A rectangular sheet of paper $40 cm \times 22 cm$ , is rolled to form a hollow cylinder of height $40 cm$ . The radius of the cylinder (in cm) is:
1. (A)3.5
2. (B)7
3. (C) $\frac{80}{7}$
4. (D)5
Try it first, then show our solutionSolution
1. Height = 40 cm, so the 22 cm side becomes the circumference: $2 π r = 22$ .
2. $2 \times \frac{22}{7} \times r = 22 \Rightarrow r = \frac{7}{2} = 3.5$ cm, option (A).
  1 mark
Answer: 3.5 cm, option (A)
Where the marks are
- Uses circumference = 22 cm to get r = 3.5 cm (1 mark)
Open Q17 on its own page
Q18 · 1 mark · MCQStatistics
Consider the following frequency distribution:
Class 0-6 6-12 12-18 18-24 24-30
Frequency 12 10 15 8 11
The median class is:
1. (A)6-12
2. (B)12-18
3. (C)18-24
4. (D)24-30
Try it first, then show our solutionSolution
1. $N = 56$ , so $\frac{N}{2} = 28$ . Cumulative frequencies: 12, 22, 37, 45, 56.
2. The first cumulative frequency greater than 28 is 37, for the class 12-18. So the median class is 12-18, option (B).
  1 mark
Answer: 12-18, option (B)
Where the marks are
- Finds N/2 = 28 and uses cumulative frequency to pick 12-18 (1 mark)
Open Q18 on its own page
Q19 · 1 mark · Assertion-reasonCompetency-basedCoordinate Geometry
Assertion (A): The point $(0, 4)$ lies on y-axis.
Reason (R): The x-coordinate of a point on y-axis is zero.
1. (A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
2. (B)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
3. (C)Assertion (A) is true but reason (R) is false.
4. (D)Assertion (A) is false but reason (R) is true.
Try it first, then show our solutionSolution
1. R is true: every point on the y-axis has x-coordinate 0.
2. The point $(0, 4)$ has x = 0, so it lies on the y-axis. A is true and R explains it, option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both statements true and R explains A (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A): The HCF of two numbers is 5 and their product is 150. Then their LCM is 40.
Reason (R): For any two positive integers a and b, $HCF (a, b) \times LCM (a, b) = a \times b$ .
1. (A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
2. (B)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
3. (C)Assertion (A) is true but reason (R) is false.
4. (D)Assertion (A) is false but reason (R) is true.
Try it first, then show our solutionSolution
1. R is true: HCF × LCM = product of the two numbers.
2. LCM $= \frac{150}{5} = 30$ , not 40. So A is false while R is true, option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Computes LCM = 30, so A is false and R is true (1 mark)

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

Class	0-6	6-12	12-18	18-24	24-30
Frequency	12	10	15	8	11

Section B

SA-I; internal choice in 22 and 25

Q21 · 2 marks · Very short answerPair of Linear Equations in Two Variables
Find whether the following pair of linear equations is consistent or inconsistent:
$3 x + 2 y = 8$
$6 x - 4 y = 9$
Try it first, then show our solutionSolution
1. Compare each equation with $a x + b y = c$ :
  $a_{1} = 3, b_{1} = 2, c_{1} = 8$ and $a_{2} = 6, b_{2} = - 4, c_{2} = 9$
  1 mark
2. Ratios: $\frac{a _{1}}{a _{2}} = \frac{3}{6} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{- 4} = - \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{8}{9}$
  ½ mark
3. Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect at exactly one point. The pair has a unique solution, so it is consistent.
  ½ mark
Answer: Consistent (unique solution), since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$
Where the marks are
- Identifies $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}$ correctly (1 mark)
- Computes the three ratios (1/2 mark)
- Uses $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ to conclude the pair is consistent (1/2 mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerTriangles
In the given figure, if ABCD is a trapezium in which $A B ∥ C D ∥ E F$ , then prove that $\frac{A E}{E D} = \frac{B F}{F C}$ .
Figure in the paper: Trapezium ABCD with AB at the bottom and DC at the top, AB parallel to DC. Points E lie on AD (left slanted side) and F lie on BC (right slanted side). A line EF is drawn parallel to AB and DC, crossing the two non-parallel sides. See it in the official paper.
Try it first, then show our solutionSolution
1. Construction: Join diagonal $B D$ ; let it cut $E F$ at $G$ .
  ½ mark
2. In $△ A B D$ , $E G ∥ A B$ [since $E F ∥ A B$ ]. By BPT, $\frac{A E}{E D} = \frac{B G}{G D}$ ...(1)
  ½ mark
3. In $△ D B C$ , $GF ∥ D C$ [since $E F ∥ C D$ ]. By BPT, $\frac{B F}{F C} = \frac{B G}{G D}$ ...(2)
  ½ mark
4. From (1) and (2), $\frac{A E}{E D} = \frac{B F}{F C}$ . Hence proved.
  ½ mark
Answer: $\frac{A E}{E D} = \frac{B F}{F C}$ proved using BPT in $△ A B D$ and $△ D B C$
Where the marks are
- Construction: join BD meeting EF at G (1/2 mark)
- BPT in $△ A B D$ gives $\frac{A E}{E D} = \frac{B G}{G D}$ (1/2 mark)
- BPT in $△ D B C$ gives $\frac{B F}{F C} = \frac{B G}{G D}$ (1/2 mark)
- Equates the two ratios to conclude (1/2 mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerTriangles
In figure, if $A D = 6$ cm, $D B = 9$ cm, $A E = 8$ cm and $E C = 12$ cm and $∠ A D E = 4 8^{\circ}$ . Find $∠ A B C$ .
Figure in the paper: Triangle ABC with A at the top. D lies on side AB and E lies on side AC; DE is joined. Marked: AD = 6 cm, DB = 9 cm, AE = 8 cm, EC = 12 cm, and angle ADE = 48° at D. Angle ABC at B is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ : $\frac{A D}{D B} = \frac{6}{9} = \frac{2}{3}$ and $\frac{A E}{E C} = \frac{8}{12} = \frac{2}{3}$ , so $\frac{A D}{D B} = \frac{A E}{E C}$ .
  1 mark
2. Therefore $D E ∥ B C$ [converse of BPT]. So $∠ A B C = ∠ A D E$ [corresponding angles], giving $∠ A B C = 4 8^{\circ}$ .
  1 mark
Answer: $∠ A B C = 4 8^{\circ}$
Where the marks are
- Shows $\frac{A D}{D B} = \frac{A E}{E C} = \frac{2}{3}$ (1 mark)
- Concludes $D E ∥ B C$ by the converse of BPT, and $∠ A B C = 4 8^{\circ}$ as corresponding angles (1 mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerCircles
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Try it first, then show our solutionSolution
1. Let O be the centre and AT the tangent, with T the point of contact. Then $∠ O T A = 9 0^{\circ}$ [tangent ⟂ radius]. By Pythagoras theorem, $O A^{2} = O T^{2} + A T^{2}$ .
  ½ mark
2. $5^{2} = O T^{2} + 4^{2} \Rightarrow O T^{2} = 25 - 16 = 9$
  ½ mark
3. $O T = 3$ cm, so the radius of the circle is 3 cm.
  1 mark
Answer: Radius = 3 cm
Where the marks are
- Uses $∠ O T A = 9 0^{\circ}$ and writes $O A^{2} = O T^{2} + A T^{2}$ (1/2 mark)
- Substitutes to get $O T^{2} = 9$ (1/2 mark)
- Radius = 3 cm (1 mark)
Open Q23 on its own page
Q24 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $sin^{2} 6 0^{\circ} + 2 tan 4 5^{\circ} - cos^{2} 3 0^{\circ}$ .
Try it first, then show our solutionSolution
1. Substitute the standard values $sin 6 0^{\circ} = \frac{3}{2}$ , $tan 4 5^{\circ} = 1$ and $cos 3 0^{\circ} = \frac{3}{2}$ :
  $(\frac{3}{2})^{2} + 2 (1) - (\frac{3}{2})^{2}$
  1 mark
2. $= \frac{3}{4} + 2 - \frac{3}{4} = 2$
  1 mark
Answer: 2
Where the marks are
- Correct substitution of the trigonometric values (1 mark)
- Simplifies to 2 (1 mark)
Open Q24 on its own page
Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerAreas Related to Circles
Find the diameter of a circle whose area is equal to the sum of the areas of two circles of radii 40 cm and 9 cm.
Try it first, then show our solutionSolution
1. Let R be the radius of the required circle. Then $π R^{2} = π (40)^{2} + π (9)^{2}$
  ½ mark
2. $R^{2} = 4 0^{2} + 9^{2} = 1600 + 81 = 1681$
  ½ mark
3. $R = 41$ cm
  ½ mark
4. Diameter $= 2 R = 2 \times 41 = 82$ cm
  ½ mark
Answer: Diameter = 82 cm
Where the marks are
- Forms the equation $π R^{2} = π (40)^{2} + π (9)^{2}$ (1/2 mark)
- Gets $R^{2} = 1681$ (1/2 mark)
- $R = 41$ cm (1/2 mark)
- Diameter = 82 cm (1/2 mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerAreas Related to Circles
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of minor segment. (Use $π = 3.14$ )
Try it first, then show our solutionSolution
1. Here $r = 10$ cm and $θ = 9 0^{\circ}$ .
  Area of minor segment = Area of sector − Area of right triangle $= \frac{θ}{36 0 ^{\circ}} π r^{2} - \frac{1}{2} \times r \times r$
  ½ mark
2. $= \frac{9 0 ^{\circ}}{36 0 ^{\circ}} \times 3.14 \times 10 \times 10 - \frac{1}{2} \times 10 \times 10$
  ½ mark
3. $= \frac{314}{4} - 50 = 78.5 - 50$
  ½ mark
4. Area of minor segment $= 28.5 cm^{2}$
  ½ mark
Answer: 28.5 cm²
Where the marks are
- Writes segment area = sector area − triangle area (1/2 mark)
- Correct substitution (1/2 mark)
- Sector area 78.5 cm² and triangle area 50 cm² (1/2 mark)
- Final answer 28.5 cm² (1/2 mark)
Open Q25(b) on its own page

Section C

SA-II; internal choice in 28 and 30

Q26 · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3$ is rational.
  Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are coprime.
  Squaring: $3 q^{2} = p^{2}$ .
  ½ mark
2. So $p^{2}$ is divisible by 3. Since 3 is prime, 3 divides $p^{2}$ means 3 divides $p$ . ... (1)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2}$ , so $q^{2} = 3 a^{2}$ .
  So $q^{2}$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)
  1 mark
4. From (1) and (2), 3 divides both $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong. Hence $3$ is irrational.
  ½ mark
Answer: Assuming $3 = p / q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $3$ is irrational
Where the marks are
- Assumes $3 = \frac{p}{q}$ with p, q coprime and gets $3 q^{2} = p^{2}$ (1/2 mark)
- Shows 3 divides p (1 mark)
- Substitutes $p = 3 a$ and shows 3 divides q (1 mark)
- States the contradiction and concludes (1/2 mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerPolynomials
Find the zeroes of the quadratic polynomial $4 s^{2} - 4 s + 1$ and verify the relationship between the zeroes and the coefficients.
Try it first, then show our solutionSolution
1. $4 s^{2} - 4 s + 1 = 4 s^{2} - 2 s - 2 s + 1 = 2 s (2 s - 1) - 1 (2 s - 1) = (2 s - 1) (2 s - 1)$
  So the zeroes are $α = \frac{1}{2}$ and $β = \frac{1}{2}$ .
  1 mark
2. Here $a = 4$ , $b = - 4$ , $c = 1$ .
  Sum: $α + β = \frac{1}{2} + \frac{1}{2} = 1$ and $- \frac{b}{a} = - \frac{- 4}{4} = 1$
  1 mark
3. Product: $α β = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ and $\frac{c}{a} = \frac{1}{4}$ .
  Hence the relationship between the zeroes and the coefficients is verified.
  1 mark
Answer: Zeroes: $\frac{1}{2}, \frac{1}{2}$ ; sum $= 1 = - \frac{b}{a}$ , product $= \frac{1}{4} = \frac{c}{a}$ (verified)
Where the marks are
- Factorises and finds the zeroes $\frac{1}{2}, \frac{1}{2}$ (1 mark)
- Verifies sum of zeroes $= - \frac{b}{a}$ (1 mark)
- Verifies product of zeroes $= \frac{c}{a}$ (1 mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
The coach of a cricket team buys 4 bats and 1 ball for Rs. 2050. Later, she buys 3 bats and 2 balls for Rs. 1600. Find the cost of each bat and each ball.
Try it first, then show our solutionSolution
1. Let the cost of one bat be Rs $x$ and the cost of one ball be Rs $y$ .
  ½ mark
2. $4 x + y = 2050$ ...(1)
  $3 x + 2 y = 1600$ ...(2)
  ½ mark
3. From (1), $y = 2050 - 4 x$ .
  ½ mark
4. Substituting in (2): $3 x + 2 (2050 - 4 x) = 1600 \Rightarrow 3 x + 4100 - 8 x = 1600 \Rightarrow - 5 x = - 2500 \Rightarrow x = 500$
  ½ mark
5. Substituting in (1): $y = 2050 - 4 (500) = 2050 - 2000 = 50$
  ½ mark
6. Cost of one bat = Rs 500; cost of one ball = Rs 50.
  ½ mark
Answer: Bat = Rs 500, Ball = Rs 50
Where the marks are
- Defines the variables (1/2 mark)
- Forms both equations (1/2 mark)
- Expresses $y$ in terms of $x$ (1/2 mark)
- Finds $x = 500$ (1/2 mark)
- Finds $y = 50$ (1/2 mark)
- States the costs (1/2 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Try it first, then show our solutionSolution
1. Let the fixed charge for the first 3 days be Rs $x$ and the charge for each extra day be Rs $y$ .
  ½ mark
2. Saritha (7 days = 3 + 4 extra days): $x + 4 y = 27$ ...(1)
  Susy (5 days = 3 + 2 extra days): $x + 2 y = 21$ ...(2)
  ½ mark
3. Subtracting (2) from (1): $2 y = 6 \Rightarrow y = 3$
  1 mark
4. Substituting in (2): $x + 2 (3) = 21 \Rightarrow x = 15$
  Fixed charge = Rs 15; charge for each extra day = Rs 3.
  1 mark
Answer: Fixed charge = Rs 15; extra charge = Rs 3 per day
Where the marks are
- Defines the variables (1/2 mark)
- Forms $x + 4 y = 27$ and $x + 2 y = 21$ (1/2 mark)
- Finds $y = 3$ (1 mark)
- Finds $x = 15$ and states the answer (1 mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerCircles
A circle touches all the four sides of quadrilateral ABCD. Prove that $A B + C D = A D + B C$ .
Try it first, then show our solutionSolution
1. Given: A circle touches sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively. To prove: $A B + C D = A D + B C$ . (Figure: quadrilateral ABCD with the inscribed circle touching AB at P, BC at Q, CD at R and DA at S.)
  1 mark
2. Tangents drawn from an external point to a circle are equal in length, so:
  $A P = A S$ ...(1), $B P = B Q$ ...(2), $D R = D S$ ...(3), $C R = C Q$ ...(4)
  1 mark
3. Adding (1) to (4): $A P + B P + D R + C R = A S + D S + B Q + C Q$
  $\Rightarrow (A P + P B) + (D R + R C) = (A S + S D) + (B Q + QC)$
  $\Rightarrow A B + C D = A D + B C$ . Hence proved.
  1 mark
Answer: $A B + C D = A D + B C$ proved using equal tangents from external points
Where the marks are
- Given, to prove and labelled figure (1 mark)
- Four pairs of equal tangents with reason (1 mark)
- Adds the equalities and regroups to get the result (1 mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $(cosec θ - cot θ)^{2} = \frac{1 - c o s θ}{1 + c o s θ}$ .
Try it first, then show our solutionSolution
1. LHS $= (cosec θ - cot θ)^{2} = (\frac{1}{s i n θ} - \frac{c o s θ}{s i n θ})^{2}$
  ½ mark
2. $= (\frac{1 - c o s θ}{s i n θ})^{2} = \frac{( 1 - c o s θ ) ^{2}}{s i n ^{2} θ}$
  ½ mark
3. $= \frac{( 1 - c o s θ ) ^{2}}{1 - c o s ^{2} θ}$ [since $sin^{2} θ = 1 - cos^{2} θ$ ]
  1 mark
4. $= \frac{( 1 - c o s θ ) ^{2}}{( 1 - c o s θ ) ( 1 + c o s θ )} = \frac{1 - c o s θ}{1 + c o s θ} =$ RHS. Hence proved.
  1 mark
Answer: LHS = RHS $= \frac{1 - c o s θ}{1 + c o s θ}$ (proved)
Where the marks are
- Writes cosec and cot in terms of sin and cos (1/2 mark)
- Combines to $\frac{( 1 - c o s θ ) ^{2}}{s i n ^{2} θ}$ (1/2 mark)
- Replaces $sin^{2} θ$ by $1 - cos^{2} θ$ (1 mark)
- Factorises, cancels and reaches the RHS (1 mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $sec A (1 - sin A) (sec A + tan A) = 1$ .
Try it first, then show our solutionSolution
1. Write everything in terms of sine and cosine:
  $LHS = \frac{1}{cos A} (1 - sin A) (\frac{1}{cos A} + \frac{sin A}{cos A})$
  1 mark
2. Combine the bracket:
  $= \frac{1}{cos A} (1 - sin A) \cdot \frac{1 + sin A}{cos A}$
  1 mark
3. Multiply out:
  $= \frac{1 - sin ^{2} A}{cos ^{2} A} = \frac{cos ^{2} A}{cos ^{2} A} = 1 = RHS$ [since $1 - sin^{2} A = cos^{2} A$ ]
  Hence proved.
  1 mark
Answer: LHS = 1 = RHS (proved).
Where the marks are
- Converts $sec A$ and $tan A$ to $\frac{1}{c o s A}$ and $\frac{s i n A}{c o s A}$ (1 mark)
- Simplifies the bracket to $\frac{1 + s i n A}{c o s A}$ (1 mark)
- Uses $1 - sin^{2} A = cos^{2} A$ to get 1 (1 mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerProbability
A bag contains 6 red, 4 black and some white balls.
1. (i)
  Find the number of white balls in the bag if the probability of drawing a white ball is $\frac{1}{3}$ .
  1 mark
2. (ii)
  How many red balls should be removed from the bag for the probability of drawing a white ball to be $\frac{1}{2}$ ?
  2 marks
Try it first, then show our solutionSolution
1. (i) Let the number of white balls be $x$ . Total balls $= 6 + 4 + x = 10 + x$ .
  $P (white) = \frac{x}{10 + x} = \frac{1}{3} \Rightarrow 3 x = 10 + x \Rightarrow x = 5$
  There are 5 white balls.
  1 mark
2. (ii) Let $y$ red balls be removed. Total balls $= (6 - y) + 4 + 5 = 15 - y$ .
  $P (white) = \frac{5}{15 - y} = \frac{1}{2}$
  1 mark
3. $\Rightarrow 10 = 15 - y \Rightarrow y = 5$
  So 5 red balls should be removed.
  1 mark
Answer: (i) 5 white balls; (ii) 5 red balls should be removed
Where the marks are
- (i) Forms $\frac{x}{10 + x} = \frac{1}{3}$ and gets $x = 5$ (1 mark)
- (ii) Forms $\frac{5}{15 - y} = \frac{1}{2}$ (1 mark)
- (ii) Solves to get $y = 5$ (1 mark)
Open Q31 on its own page

Section D

LA; internal choice in 32 and 34

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Try it first, then show our solutionSolution
1. Let the uniform speed of the train be $x$ km/h.
  ½ mark
2. Time taken for 360 km $= \frac{360}{x}$ h.
  ½ mark
3. With speed $(x + 5)$ km/h the time is 1 h less:
  $\frac{360}{x} - \frac{360}{x + 5} = 1$
  1 mark
4. $360 (x + 5) - 360 x = x (x + 5) \Rightarrow 1800 = x^{2} + 5 x \Rightarrow x^{2} + 5 x - 1800 = 0$
  1 mark
5. $x^{2} + 45 x - 40 x - 1800 = 0 \Rightarrow x (x + 45) - 40 (x + 45) = 0 \Rightarrow (x + 45) (x - 40) = 0$
  1 mark
6. $x = 40$ or $x = - 45$ . Speed cannot be negative, so the speed of the train is 40 km/h.
  1 mark
Answer: Speed of the train = 40 km/h
Where the marks are
- Assumes the speed as $x$ km/h (1/2 mark)
- Writes the time as $\frac{360}{x}$ (1/2 mark)
- Forms the equation $\frac{360}{x} - \frac{360}{x + 5} = 1$ (1 mark)
- Reduces it to $x^{2} + 5 x - 1800 = 0$ (1 mark)
- Factorises correctly (1 mark)
- Rejects the negative root; speed = 40 km/h (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerCompetency-basedQuadratic Equations
A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Try it first, then show our solutionSolution
1. Let the speed of the stream be $x$ km/h.
  Upstream speed $= (18 - x)$ km/h. Downstream speed $= (18 + x)$ km/h.
  1 mark
2. Using time = distance ÷ speed:
  Time upstream $= \frac{24}{18 - x}$ h. Time downstream $= \frac{24}{18 + x}$ h.
  1 mark
3. The upstream journey takes 1 hour more:
  $\frac{24}{18 - x} - \frac{24}{18 + x} = 1$
  1 mark
4. $24 (18 + x) - 24 (18 - x) = (18 - x) (18 + x)$
  $\Rightarrow 48 x = 324 - x^{2}$
  $\Rightarrow x^{2} + 48 x - 324 = 0$
  $\Rightarrow (x + 54) (x - 6) = 0$ , so $x = 6$ or $x = - 54$ .
  1 mark
5. A speed cannot be negative, so we reject $x = - 54$ .
  The speed of the stream is 6 km/h.
  1 mark
Answer: Speed of the stream = 6 km/h
Where the marks are
- Writes the upstream and downstream speeds as 18 − x and 18 + x (1 mark)
- Writes both times as distance ÷ speed (1 mark)
- Forms the equation 24/(18−x) − 24/(18+x) = 1 (1 mark)
- Reduces it to x² + 48x − 324 = 0 and solves to get x = 6 or −54 (1 mark)
- Rejects the negative value and states 6 km/h (1 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerTriangles
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
In $Δ$ PQR, S and T are points on PQ and PR respectively. $\frac{P S}{S Q} = \frac{P T}{T R}$ and $∠ P S T = ∠ P R Q$ . Prove that PQR is an isosceles triangle.
Try it first, then show our solutionSolution
1. Given: In $△ A B C$ , $D E ∥ B C$ , with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
  Construction: Join BE and CD. Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  ½ mark
2. $ar (A D E) = \frac{1}{2} \times A D \times E N$ and $ar (D B E) = \frac{1}{2} \times D B \times E N$
  $\Rightarrow \frac{ar ( A D E )}{ar ( D B E )} = \frac{A D}{D B}$ ...(1)
  1 mark
3. $ar (A D E) = \frac{1}{2} \times A E \times D M$ and $ar (D E C) = \frac{1}{2} \times E C \times D M$
  $\Rightarrow \frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ...(2)
  1 mark
4. $△ D B E$ and $△ D E C$ lie on the same base DE and between the same parallels DE and BC, so $ar (D B E) = ar (D E C)$ .
  Hence, from (1) and (2), $\frac{A D}{D B} = \frac{A E}{E C}$ . Proved.
  ½ mark
5. Second part: In $△ P QR$ , $\frac{P S}{S Q} = \frac{P T}{T R}$ , so $S T ∥ QR$ [converse of BPT].
  Therefore $∠ P S T = ∠ P QR$ [corresponding angles].
  1 mark
6. Also $∠ P S T = ∠ P R Q$ [given], so $∠ P QR = ∠ P R Q$ .
  $\Rightarrow P R = P Q$ [sides opposite equal angles are equal]. Hence $△ P QR$ is isosceles.
  1 mark
Answer: BPT proved; $P Q = P R$ , so $△ P QR$ is isosceles
Where the marks are
- Given, to prove, construction and figure (1/2 mark)
- Ratio of areas gives $\frac{A D}{D B}$ (1 mark)
- Ratio of areas gives $\frac{A E}{E C}$ (1 mark)
- Equal areas of $△ D B E$ and $△ D E C$ used to conclude (1/2 mark)
- $S T ∥ QR$ by the converse of BPT, and $∠ P S T = ∠ P QR$ (1 mark)
- $∠ P QR = ∠ P R Q$ , hence $P Q = P R$ and the triangle is isosceles (1 mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerSurface Areas and Volumes
A medicine capsule is in the shape of a cylinder with two hemispheres stuck at each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Figure in the paper: A capsule shape: a cylinder with a hemispherical cap at each end. The overall length is marked 14 mm and the diameter (width) is marked 5 mm. A vertical line marks the junction between the two hemispherical halves. See it in the official paper.
Try it first, then show our solutionSolution
1. Radius of the cylinder and of each hemisphere: $r = \frac{5}{2}$ mm.
  Height of the cylinder: $h = 14 - 2 r = 14 - 5 = 9$ mm.
  1 mark
2. CSA of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times \frac{5}{2} \times 9 = \frac{990}{7} mm^{2}$
  1 mark
3. CSA of one hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times (\frac{5}{2})^{2} = \frac{275}{7} mm^{2}$
  1 mark
4. CSA of two hemispheres $= 2 \times \frac{275}{7} = \frac{550}{7} mm^{2}$
  1 mark
5. Surface area of capsule $= \frac{990}{7} + \frac{550}{7} = \frac{1540}{7} = 220 mm^{2}$
  1 mark
Answer: Surface area = 220 mm²
Where the marks are
- Radius 2.5 mm and cylinder height 9 mm (1 mark)
- CSA of cylinder $\frac{990}{7}$ mm² (1 mark)
- CSA of one hemisphere $\frac{275}{7}$ mm² (1 mark)
- CSA of two hemispheres $\frac{550}{7}$ mm² (1 mark)
- Total surface area 220 mm² (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
Figure in the paper: Photograph-style figure of a glass (tumbler) filled with syrup, containing several gulab jamuns submerged in it. Each gulab jamun is cylindrical with hemispherical ends, labelled with length 5 cm and diameter 2.8 cm in the question. See it in the official paper.
Try it first, then show our solutionSolution
1. Radius of cylinder = radius of hemisphere $= \frac{2.8}{2} = 1.4$ cm.
  Height of cylinder $= 5 - 2 (1.4) = 5 - 2.8 = 2.2$ cm.
  1 mark
2. Volume of one gulab jamun = volume of cylinder + 2 × volume of hemisphere $= π r^{2} h + 2 \times \frac{2}{3} π r^{3}$
  1 mark
3. $= \frac{22}{7} \times (1.4)^{2} \times 2.2 + 2 \times \frac{2}{3} \times \frac{22}{7} \times (1.4)^{3} \approx 13.55 + 11.50 = 25.05 cm^{3}$
  1 mark
4. Volume of syrup in 45 gulab jamuns $= \frac{30}{100} \times 45 \times 25.05 = 338.175 cm^{3} \approx 338 cm^{3}$
  2 marks
Answer: Approximately 338 cm³ of syrup
Where the marks are
- Radius 1.4 cm and cylinder height 2.2 cm (1 mark)
- Volume formula: cylinder + two hemispheres (1 mark)
- Volume of one gulab jamun ≈ 25.05 cm³ (1 mark)
- 30% of the volume of 45 gulab jamuns ≈ 338 cm³ (2 marks)
Open Q34(b) on its own page

Q35 · 5 marks · Long answerCompetency-basedStatistics

The following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)	Number of lamps
1500-2000	14
2000-2500	56
2500-3000	60
3000-3500	86
3500-4000	74
4000-4500	62
4500-5000	48

Find the average life time of a lamp.

Try it first, then show our solutionSolution

Take assumed mean $a = 3250$ and $d = x - a$ .

Life time (h)	$f$	$x$	$d$	$f d$
1500-2000	14	1750	−1500	−21000
2000-2500	56	2250	−1000	−56000
2500-3000	60	2750	−500	−30000
3000-3500	86	3250	0	0
3500-4000	74	3750	500	37000
4000-4500	62	4250	1000	62000
4500-5000	48	4750	1500	72000
Total	$Σ f = 400$			$Σ f d = 64000$

2 marks

Mean $\overset{x}{ˉ} = a + \frac{Σ f d}{Σ f}$
½ mark
$a = 3250$
½ mark
$\overset{x}{ˉ} = 3250 + \frac{64000}{400} = 3250 + 160$
1 mark
$\overset{x}{ˉ} = 3410$ . The average life time of a lamp is 3410 hours.
1 mark

Answer: Average life time = 3410 hours

Where the marks are

Correct table with class marks, $d$ and $f d$ , and totals $Σ f = 400$ , $Σ f d = 64000$ (2 marks)
Writes the assumed-mean formula (1/2 mark)
Chooses $a = 3250$ (1/2 mark)
Correct substitution (1 mark)
Mean = 3410 hours (1 mark)

Open Q35 on its own page

Section E

Case/source based; sub-parts of 1, 1 and 2 marks; internal choice within sub-parts of 36, 37 and 38

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
CASE STUDY 1
India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.
1. 1)
  In which year, the production is 29,200 sets?
  1 mark
2. 2)
  Find the production in the 8th year.
  1 mark
3. 2)
  Find the production in first 3 years.
  1 mark
4. 3)
  Find the difference of the production in 7th year and 4th year.
  2 marks
Try it first, then show our solutionSolution
1. Since production increases by a fixed number every year, the yearly productions form an AP with first term $a$ and common difference $d$ . Given $a_{6} = 16000$ and $a_{9} = 22600$ :
  $a + 5 d = 16000$ ...(1)
  $a + 8 d = 22600$ ...(2)
2. Subtracting (1) from (2): $3 d = 6600 \Rightarrow d = 2200$ . Then $a = 16000 - 5 (2200) = 16000 - 11000 = 5000$ .
3. 1) Let the production be 29200 in the $n$ th year. Using $a_{n} = a + (n - 1) d$ :
  $29200 = 5000 + (n - 1) (2200)$
  ½ mark
4. $(n - 1) (2200) = 24200 \Rightarrow n - 1 = 11 \Rightarrow n = 12$ .
  The production is 29,200 sets in the 12th year.
  ½ mark
5. 2) Production in the 8th year: $a_{8} = a + 7 d = 5000 + 7 (2200)$
  ½ mark
6. $= 5000 + 15400 = 20400$ . The production in the 8th year is 20,400 sets.
  ½ mark
7. 2) (OR) Production in the first 3 years: $S_{3} = \frac{n}{2} [2 a + (n - 1) d] = \frac{3}{2} [2 (5000) + 2 (2200)]$
  ½ mark
8. $= \frac{3}{2} (10000 + 4400) = \frac{3}{2} (14400) = 21600$ . The production in the first 3 years is 21,600 sets.
  ½ mark
9. 3) $a_{4} = a + 3 d = 5000 + 3 (2200) = 5000 + 6600 = 11600$
  ½ mark
10. $a_{7} = a + 6 d = 5000 + 6 (2200) = 5000 + 13200 = 18200$
  ½ mark
11. Difference: $a_{7} - a_{4} = 18200 - 11600 = 6600$ sets. (Check: $a_{7} - a_{4} = 3 d = 3 \times 2200 = 6600$ .)
  1 mark
Answer: 1) 12th year; 2) 20,400 sets (OR first 3 years: 21,600 sets); 3) 6,600 sets
Where the marks are
- Sets up $a + 5 d = 16000$ , $a + 8 d = 22600$ and gets $d = 2200$ , $a = 5000$ (needed to start)
- Writes $29200 = 5000 + (n - 1) 2200$ (½ mark)
- Gets $n = 12$ , i.e. the 12th year (½ mark)
- Writes $a_{8} = 5000 + 7 (2200)$ (½ mark)
- Gets 20400 sets (½ mark)
- OR: uses $S_{3} = \frac{3}{2} [10000 + 4400]$ (½ mark) and gets 21600 sets (½ mark)
- Finds $a_{4} = 11600$ (½ mark) and $a_{7} = 18200$ (½ mark)
- States the difference as 6600 sets (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
CASE STUDY 2
Alia and Shagun are friends living on the same street in Patel Nagar. Shagun's house is at the intersection of one street with another street on which there is a library. They both study in the same school and that is not far from Shagun's house. Suppose the school is situated at the point O, i.e., the origin, Alia's house is at A. Shagun's house is at B and library is at C. Based on the above information, answer the following questions.
1. (i)
  How far is Alia's house from Shagun's house?
  1 mark
2. (ii)
  How far is the library from Shagun's house?
  1 mark
3. (iii)
  Show that for Shagun, school is farther compared to Alia's house and library.
  2 marks
4. OR
  (iii)
  Show that Alia's house, Shagun's house and library form an isosceles right triangle.
  2 marks
Figure in the paper: Coordinate plane with X and X' axes and Y and Y' axes, origin O. Points plotted: A at (2, 3), B at (2, 1), C at (4, 1). Axes are marked at intervals of 1 from -1 to 4 on X and from -1 to 4 on Y. See it in the official paper.
Try it first, then show our solutionSolution
1. From the figure: school $O (0, 0)$ , Alia's house $A (2, 3)$ , Shagun's house $B (2, 1)$ , library $C (4, 1)$ . Distance formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
2. (i) $A B = (2 - 2)^{2} + (1 - 3)^{2} = 0 + (- 2)^{2}$
  ½ mark
3. $= 4 = 2$ units. Alia's house is 2 units from Shagun's house.
  ½ mark
4. (ii) $C B = (2 - 4)^{2} + (1 - 1)^{2} = (- 2)^{2} + 0$
  ½ mark
5. $= 4 = 2$ units. The library is 2 units from Shagun's house.
  ½ mark
6. (iii)(a) $O B = (2 - 0)^{2} + (1 - 0)^{2} = 4 + 1 = 5$ units
  1 mark
7. From (i) and (ii), $A B = 2$ units and $C B = 2$ units, i.e. $A B = C B = 2 = 4$ .
  ½ mark
8. Since $5 > 4$ , $O B > A B$ and $O B > C B$ . Hence, for Shagun, the school is farther than Alia's house and the library.
  ½ mark
9. (iii)(b) $A C = (2 - 4)^{2} + (3 - 1)^{2} = 4 + 4 = 8 = 22$ units, so $A C^{2} = 8$ .
  1 mark
10. $A B = B C = 2$ units, so the triangle is isosceles.
  ½ mark
11. $A B^{2} + B C^{2} = 4 + 4 = 8 = A C^{2}$ , so by the converse of Pythagoras theorem $∠ B = 9 0^{\circ}$ . Hence A, B and C form an isosceles right triangle.
  ½ mark
Answer: (i) 2 units; (ii) 2 units; (iii)(a) OB = √5 units > AB = CB = 2 units, so the school is farther; (iii)(b) AB = BC = 2, AC = 2√2 and AB² + BC² = AC², so it is an isosceles right triangle
Where the marks are
- Substitutes into the distance formula for AB (½ mark)
- Gets AB = 2 units (½ mark)
- Substitutes into the distance formula for CB (½ mark)
- Gets CB = 2 units (½ mark)
- (iii)(a) Finds OB = √5 units (1 mark)
- Uses AB = CB = 2 units (½ mark)
- Concludes OB is greater, so the school is farther (½ mark)
- (iii)(b) Finds AC = 2√2, i.e. AC² = 8 (1 mark)
- Notes AB = BC = 2 (½ mark)
- Shows AB² + BC² = AC², so it is an isosceles right triangle (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
CASE STUDY 3
A boy is standing on the top of light house. He observed that boat P and boat Q are approaching the light house from opposite directions. He finds that angle of depression of boat P is $4 5^{\circ}$ and angle of depression of boat Q is $3 0^{\circ}$ . He also knows that height of the light house is 100 m.
Based on the above information, answer the following questions.
1. (i)
  What is the measure of $∠ A P D$ ?
  1 mark
2. (ii)
  If $∠ Y A Q = 3 0^{\circ}$ , then $∠ A Q D$ is also $3 0^{\circ}$ . Why?
  1 mark
3. (iii)
  Find length of PD.
  2 marks
4. OR
  (iii)
  Find length of DQ.
  2 marks
Figure in the paper: Left: a lighthouse on a rock with boats P (left) and Q (right) on water; a dashed horizontal line from the top of the lighthouse and the angle of depression marked. Right: schematic with a horizontal line XAY through point A at the top of the lighthouse; angle XAP = 45° and angle YAQ = 30° marked as angles of depression. A vertical line from A to D on the base PQ is marked 100 m. P and Q are on the base line, with D between them. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $X Y ∥ P Q$ and AP is a transversal, so $∠ A P D = ∠ X A P$ [alternate interior angles].
  ½ mark
2. Hence $∠ A P D = 4 5^{\circ}$ .
  ½ mark
3. (ii) $X Y ∥ P Q$ and AQ is a transversal, so $∠ A Q D = ∠ Y A Q = 3 0^{\circ}$ [alternate interior angles are equal].
  1 mark
4. (iii)(a) In right $△ A D P$ ( $∠ D = 9 0^{\circ}$ ): $tan 4 5^{\circ} = \frac{A D}{P D} = \frac{100}{P D}$
  ½ mark
5. $1 = \frac{100}{P D}$
  ½ mark
6. $P D = 100$ m, so boat P is 100 m from the lighthouse.
  1 mark
7. (iii)(b) In right $△ A D Q$ ( $∠ D = 9 0^{\circ}$ ): $tan 3 0^{\circ} = \frac{A D}{D Q}$
  ½ mark
8. $\frac{1}{3} = \frac{100}{D Q}$
  ½ mark
9. $D Q = 1003$ m ( $\approx 173.2$ m).
  1 mark
Answer: (i) 45°; (ii) they are alternate interior angles, since XY ∥ PQ; (iii)(a) PD = 100 m; (iii)(b) DQ = 100√3 m ≈ 173.2 m
Where the marks are
- States ∠APD = ∠XAP as alternate angles (½ mark)
- Gets ∠APD = 45° (½ mark)
- Gives the reason: alternate interior angles, since XY ∥ PQ (1 mark)
- Writes tan 45° = 100/PD (½ mark)
- Simplifies using tan 45° = 1 (½ mark)
- Gets PD = 100 m (1 mark)
- OR: writes tan 30° = 100/DQ (½ mark), uses 1/√3 = 100/DQ (½ mark), gets DQ = 100√3 m (1 mark)
Open Q38 on its own page

Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.