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

CBSE Class 10 Mathematics Standard: 2023 board paper, set 30/4/1

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_STANDARD.zip (a zip; this paper is the file 30_4_1 Maths Standard.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very Short Answer; internal choice in 23 and 25
C	26-31	Short Answer; internal choice in 26 and 29
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: All 38 questions are compulsory with no overall choice; internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (one in each case study). Use of calculators is not allowed; take π = 22/7 unless stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The ratio of HCF to LCM of the least composite number and the least prime number is:
1. (A)1:2
2. (B)2:1
3. (C)1:1
4. (D)1:3
Try it first, then show our solutionSolution
1. The least composite number is $4$ and the least prime number is $2$ .
2. $HCF (4, 2) = 2$ and $LCM (4, 2) = 4$ .
3. $HCF : LCM = 2 : 4 = 1 : 2$ . So the answer is option (A).
  1 mark
Answer: 1 : 2, option (A)
Where the marks are
- Chooses option (A) 1 : 2 (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQQuadratic Equations
The roots of the equation $x^{2} + 3 x - 10 = 0$ are:
1. (A)2, -5
2. (B)-2, 5
3. (C)2, 5
4. (D)-2, -5
Try it first, then show our solutionSolution
1. Split the middle term: $x^{2} + 5 x - 2 x - 10 = 0$ .
2. $x (x + 5) - 2 (x + 5) = 0 \Rightarrow (x - 2) (x + 5) = 0$ .
3. $x = 2$ or $x = - 5$ . So the answer is option (A).
  1 mark
Answer: 2, −5, option (A)
Where the marks are
- Chooses option (A) 2, −5 (1 mark)
Q3 · 1 mark · MCQArithmetic Progressions
The next term of the A.P.: $6, 24, 54$ is:
1. (A) $60$
2. (B) $96$
3. (C) $72$
4. (D) $216$
Try it first, then show our solutionSolution
1. Simplify each term: $6$ , $24 = 26$ , $54 = 36$ .
2. The common difference is $d = 6$ , so the next term is $36 + 6 = 46$ .
3. $46 = 16 \times 6 = 96$ . So the answer is option (B).
  1 mark
Answer: $96$ , option (B)
Where the marks are
- Chooses option (B) $96$ (1 mark)
Open Q3 on its own page
Q4 · 1 mark · MCQCoordinate Geometry
The distance of the point $(- 1, 7)$ from x-axis is:
1. (A)-1
2. (B)7
3. (C)6
4. (D) $50$
Try it first, then show our solutionSolution
1. The distance of a point $(x, y)$ from the x-axis is $∣ y ∣$ .
2. For $(- 1, 7)$ the distance is $∣7∣ = 7$ units. So the answer is option (B).
  1 mark
Answer: 7, option (B)
Where the marks are
- Chooses option (B) 7 (1 mark)
Q5 · 1 mark · MCQAreas Related to Circles
What is the area of a semi-circle of diameter 'd'?
1. (A) $\frac{1}{16} π d^{2}$
2. (B) $\frac{1}{4} π d^{2}$
3. (C) $\frac{1}{8} π d^{2}$
4. (D) $\frac{1}{2} π d^{2}$
Try it first, then show our solutionSolution
1. The radius is $r = \frac{d}{2}$ .
2. Area of the semicircle $= \frac{1}{2} π r^{2} = \frac{1}{2} π \cdot \frac{d ^{2}}{4} = \frac{1}{8} π d^{2}$ . So the answer is option (C).
  1 mark
Answer: $\frac{1}{8} π d^{2}$ , option (C)
Where the marks are
- Chooses option (C) $\frac{1}{8} π d^{2}$ (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQStatistics
The empirical relation between the mode, median and mean of a distribution is:
1. (A)Mode = 3 Median - 2 Mean
2. (B)Mode = 3 Mean - 2 Median
3. (C)Mode = 2 Median - 3 Mean
4. (D)Mode = 2 Mean - 3 Median
Try it first, then show our solutionSolution
1. The empirical relation is $3 Median = Mode + 2 Mean$ .
2. Rearranging gives $Mode = 3 Median - 2 Mean$ . So the answer is option (A).
  1 mark
Answer: Mode = 3 Median − 2 Mean, option (A)
Where the marks are
- Chooses option (A) (1 mark)
Open Q6 on its own page
Q7 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of linear equations $2 x = 5 y + 6$ and $15 y = 6 x - 18$ represents two lines which are:
1. (A)intersecting
2. (B)parallel
3. (C)coincident
4. (D)either intersecting or parallel
Try it first, then show our solutionSolution
1. Write both equations in standard form: $2 x - 5 y - 6 = 0$ and $6 x - 15 y - 18 = 0$ .
2. Compare the coefficients: $\frac{a _{1}}{a _{2}} = \frac{2}{6} = \frac{1}{3}$ , $\frac{b _{1}}{b _{2}} = \frac{- 5}{- 15} = \frac{1}{3}$ , $\frac{c _{1}}{c _{2}} = \frac{- 6}{- 18} = \frac{1}{3}$ .
3. All three ratios are equal, so the lines are coincident. The answer is option (C).
  1 mark
Answer: Coincident, option (C)
Where the marks are
- Chooses option (C) coincident (1 mark)
Open Q7 on its own page
Q8 · 1 mark · MCQPolynomials
If $α, β$ are zeroes of the polynomial $x^{2} - 1$ , then value of $(α + β)$ is:
1. (A)2
2. (B)1
3. (C)-1
4. (D)0
Try it first, then show our solutionSolution
1. For $x^{2} + 0 \cdot x - 1$ : $a = 1$ , $b = 0$ , $c = - 1$ .
2. Sum of zeroes $α + β = - \frac{b}{a} = 0$ . (Check: the zeroes are $1$ and $- 1$ .) So the answer is option (D).
  1 mark
Answer: 0, option (D)
Where the marks are
- Chooses option (D) 0 (1 mark)
Open Q8 on its own page
Q9 · 1 mark · MCQSome Applications of Trigonometry
If a pole 6 m high casts a shadow $23$ m long on the ground, then sun's elevation is:
1. (A) $6 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $3 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. Let $θ$ be the sun's elevation. Then $tan θ = \frac{height of pole}{length of shadow} = \frac{6}{2 3} = 3$ .
2. $tan θ = 3 \Rightarrow θ = 6 0^{\circ}$ . So the answer is option (A).
  1 mark
Answer: $6 0^{\circ}$ , option (A)
Where the marks are
- Chooses option (A) $6 0^{\circ}$ (1 mark)
Open Q9 on its own page
Q10 · 1 mark · MCQIntroduction to Trigonometry
$sec θ$ when expressed in terms of $cot θ$ , is equal to:
1. (A) $\frac{1 + c o t ^{2} θ}{c o t θ}$
2. (B) $1 + cot^{2} θ$
3. (C) $\frac{1 + c o t ^{2} θ}{c o t θ}$
4. (D) $\frac{1 - c o t ^{2} θ}{c o t θ}$
Try it first, then show our solutionSolution
1. Use $sec^{2} θ = 1 + tan^{2} θ = 1 + \frac{1}{cot ^{2} θ} = \frac{cot ^{2} θ + 1}{cot ^{2} θ}$ .
2. Taking the square root (for acute $θ$ ): $sec θ = \frac{1 + cot ^{2} θ}{cot θ}$ . So the answer is option (C).
  1 mark
Answer: $\frac{1 + c o t ^{2} θ}{c o t θ}$ , option (C)
Where the marks are
- Chooses option (C) (1 mark)
Open Q10 on its own page
Q11 · 1 mark · MCQProbability
Two dice are thrown together. The probability of getting the difference of numbers on their upper faces equals to 3 is:
1. (A) $\frac{1}{9}$
2. (B) $\frac{2}{9}$
3. (C) $\frac{1}{6}$
4. (D) $\frac{1}{12}$
Try it first, then show our solutionSolution
1. Throwing two dice gives $6 \times 6 = 36$ equally likely outcomes.
2. The outcomes with difference 3 are $(1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3)$ , which is 6 outcomes.
3. $P = \frac{6}{36} = \frac{1}{6}$ . So the answer is option (C).
  1 mark
Answer: $\frac{1}{6}$ , option (C)
Where the marks are
- Chooses option (C) $\frac{1}{6}$ (1 mark)
Open Q11 on its own page
Q12 · 1 mark · MCQTriangles
In the given figure, $△ A B C \sim △ QP R$ . If AC = 6 cm, BC = 5 cm, QR = 3 cm and PR = x; then the value of x is:
1. (A)3.6 cm
2. (B)2.5 cm
3. (C)10 cm
4. (D)3.2 cm
Figure in the paper: Two triangles side by side. Triangle ABC: vertices A (top left), C (top right), B (bottom). AC = 6 cm (top side), BC = 5 cm (slanted side). Triangle QPR: vertices R (top), Q (bottom left), P (bottom right). QR = 3 cm (left slanted side), PR = x (right slanted side). Base QP is horizontal. See it in the official paper.
Try it first, then show our solutionSolution
1. From $△ A B C \sim △ QP R$ the vertices correspond as $A \leftrightarrow Q$ , $B \leftrightarrow P$ , $C \leftrightarrow R$ . So $A C \leftrightarrow QR$ and $B C \leftrightarrow P R$ .
2. $\frac{A C}{QR} = \frac{B C}{P R} \Rightarrow \frac{6}{3} = \frac{5}{x}$ .
3. $x = \frac{5 \times 3}{6} = 2.5$ cm. So the answer is option (B).
  1 mark
Answer: 2.5 cm, option (B)
Where the marks are
- Chooses option (B) 2.5 cm (1 mark)
Open Q12 on its own page
Q13 · 1 mark · MCQCoordinate Geometry
The distance of the point $(- 6, 8)$ from origin is:
1. (A)6
2. (B)-6
3. (C)8
4. (D)10
Try it first, then show our solutionSolution
1. Distance from the origin $= x^{2} + y^{2} = (- 6)^{2} + 8^{2}$ .
2. $= 36 + 64 = 100 = 10$ units. So the answer is option (D).
  1 mark
Answer: 10, option (D)
Where the marks are
- Chooses option (D) 10 (1 mark)
Q14 · 1 mark · MCQCircles
In the given figure, PQ is a tangent to the circle with centre O. If $∠ O P Q = x$ , $∠ P O Q = y$ , then $x + y$ is:
1. (A) $4 5^{\circ}$
2. (B) $9 0^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $18 0^{\circ}$
Figure in the paper: A circle with centre O. A tangent line passes through Q on the circle and the point P outside the circle, so PQ is the tangent. P, O are on a horizontal line; the segment PO joins P to the centre O. Angle at P between PQ and PO is marked x, and angle at O between OP and OQ is marked y. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
2. In $△ O P Q$ : $x + y + 9 0^{\circ} = 18 0^{\circ}$ [angle sum property].
3. $x + y = 9 0^{\circ}$ . So the answer is option (B).
  1 mark
Answer: $9 0^{\circ}$ , option (B)
Where the marks are
- Chooses option (B) $9 0^{\circ}$ (1 mark)
Open Q14 on its own page
Q15 · 1 mark · MCQCircles
In the given figure, TA is a tangent to the circle with centre O such that OT = 4 cm, $∠ O T A = 3 0^{\circ}$ , then length of TA is:
1. (A) $23$ cm
2. (B)2 cm
3. (C) $22$ cm
4. (D) $3$ cm
Figure in the paper: A circle with centre O. T is a point outside the circle on a horizontal line; TA is a tangent touching the circle at A. The segment OT is marked 4 cm, and the angle at T between TO and TA is $3 0^{\circ}$ . See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A T = 9 0^{\circ}$ [tangent ⟂ radius], so $△ O A T$ is right-angled at A.
2. $cos 3 0^{\circ} = \frac{T A}{O T} \Rightarrow T A = 4 \times \frac{3}{2}$ .
3. $T A = 23$ cm. So the answer is option (A).
  1 mark
Answer: $23$ cm, option (A)
Where the marks are
- Chooses option (A) $23$ cm (1 mark)
Open Q15 on its own page
Q16 · 1 mark · MCQTriangles
In $△ A B C$ , $P Q ∥ B C$ . If PB = 6 cm, AP = 4 cm, AQ = 8 cm, find the length of AC.
1. (A)12 cm
2. (B)20 cm
3. (C)6 cm
4. (D)14 cm
Figure in the paper: Triangle ABC with A at the top. P lies on side AB and Q lies on side AC, with PQ drawn horizontally parallel to base BC. Marked lengths: PB = 6 cm, AP = 4 cm, AQ = 8 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $P Q ∥ B C$ , by the Basic Proportionality Theorem $\frac{A P}{P B} = \frac{A Q}{QC}$ .
2. $\frac{4}{6} = \frac{8}{QC} \Rightarrow QC = 12$ cm.
3. $A C = A Q + QC = 8 + 12 = 20$ cm. So the answer is option (B).
  1 mark
Answer: 20 cm, option (B)
Where the marks are
- Chooses option (B) 20 cm (1 mark)
Open Q16 on its own page
Q17 · 1 mark · MCQPolynomials
If $α, β$ are the zeroes of the polynomial $p (x) = 4 x^{2} - 3 x - 7$ , then $(\frac{1}{α} + \frac{1}{β})$ is equal to:
1. (A) $\frac{7}{3}$
2. (B) $\frac{- 7}{3}$
3. (C) $\frac{3}{7}$
4. (D) $\frac{- 3}{7}$
Try it first, then show our solutionSolution
1. For $4 x^{2} - 3 x - 7$ : $α + β = - \frac{b}{a} = \frac{3}{4}$ and $α β = \frac{c}{a} = - \frac{7}{4}$ .
2. $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{3/4}{- 7/4} = - \frac{3}{7}$ . So the answer is option (D).
  1 mark
Answer: $- \frac{3}{7}$ , option (D)
Where the marks are
- Chooses option (D) $- \frac{3}{7}$ (1 mark)
Open Q17 on its own page
Q18 · 1 mark · MCQProbability
A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the card drawn is not an ace is:
1. (A) $\frac{1}{13}$
2. (B) $\frac{9}{13}$
3. (C) $\frac{4}{13}$
4. (D) $\frac{12}{13}$
Try it first, then show our solutionSolution
1. A pack has 4 aces, so the number of cards that are not aces is $52 - 4 = 48$ .
2. $P (not an ace) = \frac{48}{52} = \frac{12}{13}$ . So the answer is option (D).
  1 mark
Answer: $\frac{12}{13}$ , option (D)
Where the marks are
- Chooses option (D) $\frac{12}{13}$ (1 mark)
Open Q18 on its own page
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
DIRECTIONS: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option out of the following.
Assertion (A): The probability that a leap year has 53 Sundays is $\frac{2}{7}$ .
Reason (R): The probability that a non-leap year has 53 Sundays is $\frac{5}{7}$ .
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 and Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true but Reason (R) is false.
4. (D)Assertion (A) is false but Reason (R) is true.
Try it first, then show our solutionSolution
1. A leap year has 366 days, which is 52 weeks and 2 extra days. The extra pair of days can be one of 7 equally likely pairs: (Sun, Mon), (Mon, Tue), …, (Sat, Sun). Two of these pairs contain a Sunday, so $P = \frac{2}{7}$ . Hence A is true.
2. A non-leap year has 365 days, which is 52 weeks and 1 extra day. That extra day is a Sunday in 1 case out of 7, so $P = \frac{1}{7}$ , not $\frac{5}{7}$ . Hence R is false.
3. A is true but R is false. The answer is option (C).
  1 mark
Answer: Option (C): A true, R false
Where the marks are
- Chooses option (C) (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedArithmetic Progressions
DIRECTIONS: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option out of the following.
Assertion (A): a, b, c are in A.P. if and only if $2 b = a + c$ .
Reason (R): The sum of first n odd natural numbers is $n^{2}$ .
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 and Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true but Reason (R) is false.
4. (D)Assertion (A) is false but Reason (R) is true.
Try it first, then show our solutionSolution
1. a, b, c are in A.P. $⟺ b - a = c - b ⟺ 2 b = a + c$ . Hence A is true.
2. $1 + 3 + \dots + (2 n - 1) = \frac{n}{2} [1 + (2 n - 1)] = n^{2}$ . Hence R is true.
3. R is about the sum of odd numbers and does not explain the condition for three numbers to be in A.P. The answer is option (B).
  1 mark
Answer: Option (B): both true, R not the correct explanation of A
Where the marks are
- Chooses option (B) (1 mark)
Open Q20 on its own page

Section B

Very Short Answer; internal choice in 23 and 25

Q21 · 2 marks · Very short answerReal Numbers
Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers?
Try it first, then show our solutionSolution
1. Let the numbers be $2 x$ and $3 x$ . Since 2 and 3 are co-prime, $LCM = 6 x$ .
2. $6 x = 180 \Rightarrow x = 30$ .
  1 mark
3. The numbers are $60$ and $90$ , so $HCF (60, 90) = 30$ .
  1 mark
4. Therefore the HCF of the numbers is $30$ .
Answer: HCF = 30 (numbers 60 and 90)
Where the marks are
- Takes the numbers as $2 x, 3 x$ , uses LCM $= 6 x = 180$ and gets $x = 30$ (1 mark)
- Finds the numbers 60, 90 and their HCF $= 30$ (1 mark)
Open Q21 on its own page
Q22 · 2 marks · Very short answerPolynomials
If one zero of the polynomial $p (x) = 6 x^{2} + 37 x - (k - 2)$ is reciprocal of the other, then find the value of k.
Try it first, then show our solutionSolution
1. Let the zeroes be $α$ and $\frac{1}{α}$ .
  ½ mark
2. Product of zeroes $= \frac{c}{a}$ : $α \cdot \frac{1}{α} = \frac{- ( k - 2 )}{6} \Rightarrow 1 = \frac{- ( k - 2 )}{6}$ .
  1 mark
3. $6 = - k + 2 \Rightarrow k = - 4$ .
  ½ mark
Answer: k = −4
Where the marks are
- Takes the zeroes as $α$ and $1/ α$ (½ mark)
- Uses product of zeroes $= c / a$ to write $1 = - (k - 2) /6$ (1 mark)
- Solves to get $k = - 4$ (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerQuadratic Equations
Find the sum and product of the roots of the quadratic equation $2 x^{2} - 9 x + 4 = 0$ .
Try it first, then show our solutionSolution
1. Comparing with $a x^{2} + b x + c = 0$ : $a = 2$ , $b = - 9$ , $c = 4$ .
2. Sum of roots $= - \frac{b}{a} = - \frac{- 9}{2} = \frac{9}{2}$ .
  1 mark
3. Product of roots $= \frac{c}{a} = \frac{4}{2} = 2$ .
  1 mark
Answer: Sum = 9/2, Product = 2
Where the marks are
- Sum of roots $= - b / a = 9/2$ (1 mark)
- Product of roots $= c / a = 2$ (1 mark)
Open Q23(A) on its own page
OR
Q23(B) · 2 marks · Very short answerQuadratic Equations
Find the discriminant of the quadratic equation $4 x^{2} - 5 = 0$ and hence comment on the nature of roots of the equation.
Try it first, then show our solutionSolution
1. Comparing $4 x^{2} + 0 \cdot x - 5 = 0$ with $a x^{2} + b x + c = 0$ : $a = 4$ , $b = 0$ , $c = - 5$ .
2. Discriminant $D = b^{2} - 4 a c = 0^{2} - 4 (4) (- 5) = 80$ .
  1½ marks
3. Since $D = 80 > 0$ , the equation has two distinct real roots (they are $x = \pm \frac{5}{2}$ ).
  ½ mark
Answer: D = 80 > 0; roots are real and distinct
Where the marks are
- Identifies $a, b, c$ and computes $D = b^{2} - 4 a c = 80$ (1½ marks)
- Concludes that the roots are real and distinct since $D > 0$ (½ mark)
Open Q23(B) on its own page
Q24 · 2 marks · Very short answerProbability
If a fair coin is tossed twice, find the probability of getting 'atmost one head'.
Try it first, then show our solutionSolution
1. When a fair coin is tossed twice, the sample space is ${H H, H T, T H, T T}$ . So the total number of outcomes is $4$ .
  ½ mark
2. 'At most one head' means 0 heads or 1 head. The favourable outcomes are $H T, T H, T T$ , so there are $3$ of them.
  ½ mark
3. $P (at most one head) = \frac{favourable outcomes}{total outcomes} = \frac{3}{4}$
  1 mark
Answer: $\frac{3}{4}$
Where the marks are
- Lists all 4 outcomes HH, HT, TH, TT (½ mark)
- Identifies the 3 favourable outcomes HT, TH, TT (½ mark)
- Gets the probability $\frac{3}{4}$ (1 mark)
Open Q24 on its own page
Answer one of these 2 alternatives.
Q25(A) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate $\frac{5 cos ^{2} 6 0 ^{\circ} + 4 sec ^{2} 3 0 ^{\circ} - tan ^{2} 4 5 ^{\circ}}{sin ^{2} 3 0 ^{\circ} + cos ^{2} 3 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. Use the standard values: $cos 6 0^{\circ} = \frac{1}{2}$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $tan 4 5^{\circ} = 1$ .
  In the denominator, $sin^{2} 3 0^{\circ} + cos^{2} 3 0^{\circ} = 1$ by the identity $sin^{2} θ + cos^{2} θ = 1$ .
2. Substituting these values:
  $\frac{5 ( \frac{1}{2} ) ^{2} + 4 ( \frac{2}{3} ) ^{2} - ( 1 ) ^{2}}{1} = \frac{5}{4} + \frac{16}{3} - 1$
  1½ marks
3. Take the LCM 12:
  $\frac{15}{12} + \frac{64}{12} - \frac{12}{12} = \frac{67}{12}$
  ½ mark
Answer: $\frac{67}{12}$
Where the marks are
- Substitutes the correct values of $cos 6 0^{\circ}$ , $sec 3 0^{\circ}$ , $tan 4 5^{\circ}$ and gets denominator $= 1$ (1½ marks)
- Simplifies to $\frac{67}{12}$ (½ mark)
Open Q25(A) on its own page
OR
Q25(B) · 2 marks · Very short answerIntroduction to Trigonometry
If A and B are acute angles such that $sin (A - B) = 0$ and $2 cos (A + B) - 1 = 0$ , then find angles A and B.
Try it first, then show our solutionSolution
1. $sin (A - B) = 0 = sin 0^{\circ}$ , so $A - B = 0^{\circ}$ ...(1)
  ½ mark
2. $2 cos (A + B) - 1 = 0$ gives $cos (A + B) = \frac{1}{2} = cos 6 0^{\circ}$ , so $A + B = 6 0^{\circ}$ ...(2)
  ½ mark
3. Adding (1) and (2): $2 A = 6 0^{\circ}$ , so $A = 3 0^{\circ}$ .
  Then from (1), $B = 3 0^{\circ}$ .
  1 mark
Answer: $A = 3 0^{\circ}$ , $B = 3 0^{\circ}$
Where the marks are
- Gets $A - B = 0^{\circ}$ (½ mark)
- Gets $A + B = 6 0^{\circ}$ (½ mark)
- Solves to $A = 3 0^{\circ}$ , $B = 3 0^{\circ}$ (1 mark)
Open Q25(B) on its own page

Section C

Short Answer; internal choice in 26 and 29

Answer one of these 2 alternatives.
Q26(A) · 3 marks · Short answerArithmetic Progressions
How many terms are there in an A.P. whose first and fifth terms are -14 and 2, respectively and the last term is 62.
Try it first, then show our solutionSolution
1. Here $a = - 14$ and $a_{5} = 2$ .
  Using $a_{n} = a + (n - 1) d$ : $a_{5} = a + 4 d = 2$ .
  1 mark
2. $- 14 + 4 d = 2$ , so $4 d = 16$ and $d = 4$ .
  1 mark
3. The last term is $a_{n} = 62$ :
  $- 14 + (n - 1) (4) = 62$
  $(n - 1) \cdot 4 = 76$ , so $n - 1 = 19$ and $n = 20$ .
  The A.P. has 20 terms.
  1 mark
Answer: 20 terms
Where the marks are
- Writes $a + 4 d = 2$ using the fifth term (1 mark)
- Finds $d = 4$ (1 mark)
- Uses $a_{n} = 62$ to get $n = 20$ (1 mark)
Open Q26(A) on its own page
OR
Q26(B) · 3 marks · Short answerArithmetic Progressions
Which term of the A.P.: 65, 61, 57, 53, .................. is the first negative term?
Try it first, then show our solutionSolution
1. For the A.P. 65, 61, 57, 53, …: $a = 65$ and $d = 61 - 65 = - 4$ .
  ½ mark
2. Let $a_{n}$ be the first negative term. Then $a_{n} < 0$ :
  $a + (n - 1) d < 0$
  $65 + (n - 1) (- 4) < 0$
  $69 - 4 n < 0$
  1 mark
3. $4 n > 69$ , so $n > \frac{69}{4} = 17 \frac{1}{4}$
  1 mark
4. The least whole number greater than $17 \frac{1}{4}$ is $n = 18$ .
  Check: $a_{17} = 65 - 64 = 1 > 0$ and $a_{18} = 65 - 68 = - 3 < 0$ .
  So the 18th term is the first negative term.
  ½ mark
Answer: 18th term (which equals $- 3$ )
Where the marks are
- Identifies $a = 65$ , $d = - 4$ (½ mark)
- Sets up $65 + (n - 1) (- 4) < 0$ (1 mark)
- Obtains $n > \frac{69}{4}$ (1 mark)
- Concludes the 18th term is the first negative term (½ mark)
Open Q26(B) on its own page
Q27 · 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. Substituting: $25 a^{2} = 5 q^{2}$ , so $q^{2} = 5 a^{2}$ .
  Thus $q^{2}$ is divisible by 5, and hence $q$ is divisible by 5.
  ½ mark
4. So 5 is a common factor of $p$ and $q$ . This contradicts the assumption that $p$ and $q$ are co-prime.
  The assumption is wrong. Hence $5$ is irrational.
  1 mark
Answer: $5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 = \frac{p}{q}$ with $p, q$ co-prime and $q \neq = 0$ (½ mark)
- Shows $5 ∣ p^{2}$ , hence $5 ∣ p$ , and writes $p = 5 a$ (1 mark)
- Shows $q^{2} = 5 a^{2}$ , hence $5 ∣ q$ (½ mark)
- States the contradiction with co-primeness and concludes (1 mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerCircles
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Try it first, then show our solutionSolution
1. Given: A circle with centre O. PA and PB are tangents from an external point P, touching the circle at A and B.
  To prove: $∠ A P B + ∠ A O B = 18 0^{\circ}$ .
  Figure: Draw the circle with centre O and a point P outside it. Draw tangents PA and PB, and join OA and OB to form the quadrilateral OAPB.
  1 mark
2. In quadrilateral OAPB, the angles add up to $36 0^{\circ}$ :
  $∠ O A P + ∠ A P B + ∠ O B P + ∠ A O B = 36 0^{\circ}$
  1 mark
3. $∠ O A P = ∠ O B P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  So $9 0^{\circ} + ∠ A P B + 9 0^{\circ} + ∠ A O B = 36 0^{\circ}$
  ½ mark
4. $∠ A P B + ∠ A O B = 18 0^{\circ}$
  Hence the angle between the tangents is supplementary to the angle subtended at the centre by AB.
  ½ mark
Answer: $∠ A P B + ∠ A O B = 18 0^{\circ}$ , i.e., the two angles are supplementary.
Where the marks are
- Correct labelled figure with tangents PA, PB and radii OA, OB (1 mark)
- Uses the angle sum of quadrilateral OAPB $= 36 0^{\circ}$ (1 mark)
- Uses tangent ⟂ radius to get $∠ O A P = ∠ O B P = 9 0^{\circ}$ (½ mark)
- Concludes $∠ A P B + ∠ A O B = 18 0^{\circ}$ (½ mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{sin A - 2 sin ^{3} A}{2 cos ^{3} A - cos A} = tan A$
Try it first, then show our solutionSolution
1. Take out the common factors:
  $LHS = \frac{sin A - 2 sin ^{3} A}{2 cos ^{3} A - cos A} = \frac{sin A ( 1 - 2 sin ^{2} A )}{cos A ( 2 cos ^{2} A - 1 )}$
  1 mark
2. Use $sin^{2} A = 1 - cos^{2} A$ :
  $1 - 2 sin^{2} A = 1 - 2 (1 - cos^{2} A) = 1 - 2 + 2 cos^{2} A = 2 cos^{2} A - 1$
  1 mark
3. $LHS = \frac{sin A ( 2 cos ^{2} A - 1 )}{cos A ( 2 cos ^{2} A - 1 )} = \frac{sin A}{cos A} = tan A = RHS$
  Hence proved.
  1 mark
Answer: LHS simplifies to $tan A$ = RHS (proved).
Where the marks are
- Factorises the numerator and denominator as $sin A (1 - 2 sin^{2} A)$ and $cos A (2 cos^{2} A - 1)$ (1 mark)
- Converts $1 - 2 sin^{2} A$ to $2 cos^{2} A - 1$ (1 mark)
- Cancels the common factor and gets $tan A$ (1 mark)
Open Q29(A) on its own page
OR
Q29(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 Q29(B) on its own page
Q30 · 3 marks · Short answerCircles
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Try it first, then show our solutionSolution
1. Figure: Draw two concentric circles with centre O, of radii 5 cm and 3 cm. Draw chord AB of the larger circle touching the smaller circle at P. Join OP and OA.
  Here $O A = 5$ cm and $O P = 3$ cm.
  ½ mark
2. AB is a tangent to the smaller circle at P, so $O P ⊥ A B$ [tangent ⟂ radius].
  Also, the perpendicular from the centre to a chord bisects the chord. So $A P = P B$ .
  1 mark
3. In right $△ O P A$ , by Pythagoras theorem:
  $A P^{2} = O A^{2} - O P^{2} = 5^{2} - 3^{2} = 16$
  So $A P = 4$ cm.
  1 mark
4. $A B = 2 A P = 2 \times 4 = 8$ cm
  ½ mark
Answer: 8 cm
Where the marks are
- Correct figure (½ mark)
- States $O P ⊥ A B$ (tangent ⟂ radius) and hence $A P = P B$ (1 mark)
- Uses Pythagoras to get $A P = 4$ cm (1 mark)
- Gets $A B = 8$ cm (½ mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerQuadratic Equations
Find the value of 'p' for which the quadratic equation $p x (x - 2) + 6 = 0$ has two equal real roots.
Try it first, then show our solutionSolution
1. Expand: $p x (x - 2) + 6 = 0$ becomes $p x^{2} - 2 p x + 6 = 0$ .
  So $a = p$ , $b = - 2 p$ , $c = 6$ .
  ½ mark
2. For two equal real roots, the discriminant must be zero: $D = b^{2} - 4 a c = 0$ .
  ½ mark
3. $(- 2 p)^{2} - 4 (p) (6) = 0$
  $4 p^{2} - 24 p = 0$
  1 mark
4. $4 p (p - 6) = 0$ , so $p = 0$ or $p = 6$ .
  ½ mark
5. If $p = 0$ , the equation becomes $6 = 0$ , which is not a quadratic. So $p = 0$ is rejected.
  Hence $p = 6$ .
  ½ mark
Answer: $p = 6$
Where the marks are
- Writes the equation in standard form $p x^{2} - 2 p x + 6 = 0$ (½ mark)
- Uses the condition $D = 0$ for equal roots (½ mark)
- Forms $4 p^{2} - 24 p = 0$ (1 mark)
- Factorises to get $p = 0$ or $p = 6$ (½ mark)
- Rejects $p = 0$ and gives $p = 6$ (½ mark)
Open Q31 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
A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high tower observes two cars at angles of depression of $3 0^{\circ}$ and $6 0^{\circ}$ , which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: Let AB be the tower, with $A B = 75$ m, A its foot and B its top. P and Q are the two cars on the highway, on the same side of the tower, with P nearer to A.
  Draw the horizontal line BX through B. The angles of depression are $∠ X B Q = 3 0^{\circ}$ and $∠ X B P = 6 0^{\circ}$ .
  As BX ∥ AQ (alternate angles), $∠ B Q A = 3 0^{\circ}$ and $∠ B P A = 6 0^{\circ}$ .
  1 mark
2. In right $△ B A P$ :
  $tan 6 0^{\circ} = \frac{A B}{A P}$ , so $3 = \frac{75}{A P}$
  $A P = \frac{75}{3} = 253$ m
  1½ marks
3. In right $△ B A Q$ :
  $tan 3 0^{\circ} = \frac{A B}{A Q}$ , so $\frac{1}{3} = \frac{75}{A Q}$
  $A Q = 753$ m
  1½ marks
4. Distance between the cars:
  $P Q = A Q - A P = 753 - 253 = 503$ m
  ½ mark
5. $P Q = 50 \times 1.73 = 86.5$ m
  ½ mark
Answer: 86.5 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ}$ to get $A P = 253$ m (1½ marks)
- Uses $tan 3 0^{\circ}$ to get $A Q = 753$ m (1½ marks)
- Writes $P Q = A Q - A P = 503$ (½ mark)
- Gets $P Q = 86.5$ 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 $3 0^{\circ}$ . Determine the height of the tower.
Try it first, then show our solutionSolution
1. Figure: Let DE be the building, with D its foot and E its top, so $D E = 7$ m. Let AC be the cable tower, with C its foot and A its top, and $A C = h$ m.
  Draw EB ⟂ AC, so EB is horizontal and BCDE is a rectangle. Then $B C = D E = 7$ m and $A B = (h - 7)$ m.
  The angle of elevation is $∠ A E B = 6 0^{\circ}$ and the angle of depression is $∠ B E C = 3 0^{\circ}$ . Since EB ∥ CD (alternate angles), $∠ E C D = 3 0^{\circ}$ .
  1 mark
2. Let $C D = x$ m. In right $△ E D C$ :
  $tan 3 0^{\circ} = \frac{D E}{C D}$ , so $\frac{1}{3} = \frac{7}{x}$
  $x = 73$ m
  So $B E = C D = 73$ m.
  1½ marks
3. In right $△ A B E$ :
  $tan 6 0^{\circ} = \frac{A B}{B E}$
  1 mark
4. $\frac{h - 7}{7 3} = 3$ , so $h - 7 = 7 \times 3 = 21$
  1 mark
5. $h = 28$
  The height of the tower is 28 m.
  ½ mark
Answer: 28 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 3 0^{\circ}$ to find the horizontal distance $73$ m (1½ marks)
- Sets up $tan 6 0^{\circ} = \frac{A B}{B E}$ (1 mark)
- Substitutes to get $h - 7 = 21$ (1 mark)
- Height $= 28$ m (½ mark)
Open Q32(B) on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerTriangles
D is a point on the side BC of a triangle ABC such that $∠ A D C = ∠ B A C$ , prove that $C A^{2} = C B . C D$
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ with D a point on BC. Join AD. Mark $∠ B A C$ at vertex A and $∠ A D C$ at vertex D as equal angles.
  Given: $∠ A D C = ∠ B A C$ . To prove: $C A^{2} = C B \cdot C D$ .
  1 mark
2. Consider $△ C B A$ and $△ C A D$ .
  $∠ B C A = ∠ A C D$ , since $∠ C$ is common to both triangles.
  1 mark
3. $∠ B A C = ∠ A D C$ [given]
  1 mark
4. So $△ C B A \sim △ C A D$ [AA similarity criterion].
  The correspondence is $C \leftrightarrow C$ , $B \leftrightarrow A$ , $A \leftrightarrow D$ .
  1 mark
5. Corresponding sides of similar triangles are proportional:
  $\frac{C B}{C A} = \frac{C A}{C D}$
  $\Rightarrow C A^{2} = C B \cdot C D$ . Hence proved.
  1 mark
Answer: $△ C B A \sim △ C A D$ (AA), so $\frac{C B}{C A} = \frac{C A}{C D}$ , giving $C A^{2} = C B \cdot C D$ .
Where the marks are
- Correct labelled figure with the equal angles marked (1 mark)
- Notes that $∠ C$ is common (1 mark)
- Uses the given $∠ B A C = ∠ A D C$ (1 mark)
- Concludes $△ C B A \sim △ C A D$ by AA (1 mark)
- Writes the proportion of sides and gets $C A^{2} = C B \cdot C D$ (1 mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerTriangles
If AD and PM are medians of triangles ABC and PQR, respectively where $△ A B C \sim △ P QR$ , prove that $\frac{A B}{P Q} = \frac{A D}{P M}$ .
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ with median AD (D is the midpoint of BC). Draw $△ P QR$ with median PM (M is the midpoint of QR).
  Given: $△ A B C \sim △ P QR$ . To prove: $\frac{A B}{P Q} = \frac{A D}{P M}$ .
  1 mark
2. Since $△ A B C \sim △ P QR$ , their corresponding sides are proportional:
  $\frac{A B}{P Q} = \frac{B C}{QR}$
  D and M are midpoints, so $B C = 2 B D$ and $QR = 2 QM$ .
  $\Rightarrow \frac{A B}{P Q} = \frac{2 B D}{2 QM} = \frac{B D}{QM}$ ...(1)
  1½ marks
3. Also $∠ B = ∠ Q$ , because corresponding angles of similar triangles are equal.
  In $△ A B D$ and $△ P QM$ :
  $\frac{A B}{P Q} = \frac{B D}{QM}$ [from (1)]
  $∠ B = ∠ Q$
  $\Rightarrow △ A B D \sim △ P QM$ [SAS similarity criterion]
  1½ marks
4. Corresponding sides of similar triangles are proportional, so
  $\frac{A B}{P Q} = \frac{A D}{P M}$ . Hence proved.
  1 mark
Answer: $△ A B D \sim △ P QM$ (SAS), hence $\frac{A B}{P Q} = \frac{A D}{P M}$ .
Where the marks are
- Correct figure showing both medians (1 mark)
- Uses similarity to get $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{B D}{QM}$ with the halves of the sides (1½ marks)
- Uses $∠ B = ∠ Q$ and proves $△ A B D \sim △ P QM$ by SAS (1½ marks)
- Concludes $\frac{A B}{P Q} = \frac{A D}{P M}$ (1 mark)
Open Q33(B) on its own page
Q34 · 5 marks · Long answerSurface Areas and Volumes
A student was asked to make a model shaped like a cylinder with two cones attached to its ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its total length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model.
Try it first, then show our solutionSolution
1. Radius of the cylinder = radius of each cone: $r = \frac{3}{2}$ cm.
  Height of each cone: $H = 2$ cm.
  Height of the cylinder: $h = 12 - 2 \times 2 = 8$ cm.
  1 mark
2. Volume of air = volume of cylinder + volume of 2 cones
  $= π r^{2} h + 2 \times \frac{1}{3} π r^{2} H = π r^{2} (h + \frac{2}{3} H)$
  1½ marks
3. Substitute the values:
  $= \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} (8 + \frac{2}{3} \times 2) = \frac{22}{7} \times \frac{9}{4} \times \frac{28}{3}$
  1½ marks
4. $= \frac{22 \times 9 \times 28}{7 \times 4 \times 3} = 66 cm^{3}$
  So the volume of air in the model is $66 cm^{3}$ .
  1 mark
Answer: $66 cm^{3}$
Where the marks are
- Finds $r = 1.5$ cm and cylinder height $8$ cm (1 mark)
- Writes volume = cylinder + 2 cones (1½ marks)
- Substitutes the values correctly (1½ marks)
- Gets $66 cm^{3}$ (1 mark)
Open Q34 on its own page

Q35 · 5 marks · Long answerStatistics

The monthly expenditure on milk in 200 families of a Housing Society is given below:

Monthly Expenditure (in ₹)	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of families	24	40	33	x	30	22	16	7

Find the value of x and also, find the median and mean expenditure on milk.

Try it first, then show our solutionSolution

Take assumed mean $a = 2750$ , class width $h = 500$ and $u_{i} = \frac{x _{i} - 2750}{500}$ .

Class	$x_{i}$	$f_{i}$	c.f.	$u_{i}$	$f_{i} u_{i}$
1000–1500	1250	24	24	−3	−72
1500–2000	1750	40	64	−2	−80
2000–2500	2250	33	97	−1	−33
2500–3000	2750	28	125	0	0
3000–3500	3250	30	155	1	30
3500–4000	3750	22	177	2	44
4000–4500	4250	16	193	3	48
4500–5000	4750	7	200	4	28
Total		200			−35

2 marks

Sum of frequencies: $24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 172 + x$ .
$172 + x = 200$
$\Rightarrow x = 28$
1 mark
Median: $\frac{N}{2} = 100$ . The c.f. just above 100 is 125, so the median class is 2500–3000.
Here $l = 2500$ , $c f = 97$ , $f = 28$ , $h = 500$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 2500 + \frac{100 - 97}{28} \times 500 = 2500 + 53.57 \approx ₹ 2553.6$
1 mark
Mean: Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 2750 + \frac{- 35}{200} \times 500 = 2750 - 87.5 = ₹ 2662.5$
1 mark

Answer: x = 28; median ≈ ₹2553.6; mean = ₹2662.5

Where the marks are

Correct table with class marks, cumulative frequencies and $f_{i} u_{i}$ (2 marks)
Finds $x = 28$ from total 200 (1 mark)
Identifies median class 2500–3000 and gets median ≈ ₹2553.6 (1 mark)
Gets mean = ₹2662.5 (1 mark)

Open Q35 on its own page

Section E

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

Q36 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
Two schools 'P' and 'Q' decided to award prizes to their students for two games of Hockey ₹ x per student and Cricket ₹ y per student. School 'P' decided to award a total of ₹ 9,500 for the two games to 5 and 4 students respectively; while school 'Q' decided to award ₹ 7,370 for the two games to 4 and 3 students respectively.
Based on the above information, answer the following questions:
1. (i)
  Represent the following information algebraically (in terms of x and y).
  1 mark
2. (ii)(a)
  What is the prize amount for hockey?
  2 marks
3. OR
  (ii)(b)
  Prize amount on which game is more and by how much?
  2 marks
4. (iii)
  What will be the total prize amount if there are 2 students each from two games?
  1 mark
Figure in the paper: Photograph of two silver-coloured trophies placed side by side, shown beside the case study text. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) School P gives 5 hockey prizes and 4 cricket prizes worth ₹9500 in all:
  $5 x + 4 y = 9500$ ...(1)
  ½ mark
2. (i) School Q gives 4 hockey prizes and 3 cricket prizes worth ₹7370 in all:
  $4 x + 3 y = 7370$ ...(2)
  ½ mark
3. (ii)(a) Multiply (1) by 3 and (2) by 4:
  $15 x + 12 y = 28500$
  $16 x + 12 y = 29480$
  Subtracting the first from the second gives $x = 980$ .
  So the prize amount for hockey is ₹980 per student.
  2 marks
4. (ii)(b) Solving as in (a) gives $x = 980$ .
  Putting this in (1): $4 y = 9500 - 4900 = 4600$ , so $y = 1150$ .
  1 mark
5. (ii)(b) Since $1150 > 980$ , the cricket prize is more, by $₹ (1150 - 980) = ₹170$ .
  1 mark
6. (iii) For 2 students in each game:
  Total $= 2 x + 2 y = 2 (980 + 1150) = 2 \times 2130 = ₹4260$
  1 mark
Answer: (i) 5x + 4y = 9500, 4x + 3y = 7370; (ii)(a) ₹980; (ii)(b) Cricket, by ₹170; (iii) ₹4260
Where the marks are
- Forms $5 x + 4 y = 9500$ (½ mark)
- Forms $4 x + 3 y = 7370$ (½ mark)
- (ii)(a) Solves to get hockey prize ₹980 (2 marks)
- (ii)(b) Finds $x = 980$ , $y = 1150$ (1 mark)
- (ii)(b) States the cricket prize is more, by ₹170 (1 mark)
- (iii) Total $2 (x + y) = ₹4260$ (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Jagdish has a field which is in the shape of a right angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O.
Based on the above information, answer the following questions:
1. (i)
  Taking O as origin, coordinates of P are (-200, 0) and of Q are (200, 0). PQRS being a square, what are the coordinates of R and S?
  1 mark
2. (ii)(a)
  What is the area of square PQRS?
  2 marks
3. OR
  (ii)(b)
  What is the length of diagonal PR in square PQRS?
  2 marks
4. (iii)
  If S divides CA in the ratio K:1, what is the value of K, where point A is (200, 800)?
  1 mark
Figure in the paper: Coordinate grid with X'OX horizontal and YOY' vertical, O at origin. Right-angled triangle AQC: C on the negative x-axis, Q on the positive x-axis at (200, 0), A directly above Q at (200, 800). Hypotenuse CA is a straight line passing through B on the y-axis. Square PQRS sits on the x-axis with P (-200, 0), Q (200, 0), R and S at the top corners (R on the right, S on the left), so the square has side 400. Points labelled on the figure: P, Q, R, S, O, A, B, C. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Side $P Q = 200 - (- 200) = 400$ . R lies directly above Q and S directly above P, each at height 400.
  So $R (200, 400)$ and $S (- 200, 400)$ .
  1 mark
2. (ii)(a) Side of the square: $P Q = 200 + 200 = 400$ units.
  1 mark
3. (ii)(a) Area of square PQRS $= 400 \times 400 = 160000$ sq. units.
  1 mark
4. (ii)(b) Using the distance formula with $P (- 200, 0)$ and $R (200, 400)$ :
  $P R = (200 + 200)^{2} + (400 - 0)^{2} = 40 0^{2} + 40 0^{2}$
  1 mark
5. (ii)(b) $P R = 320000 = 4002$ units ( $\approx 565.6$ units)
  1 mark
6. (iii) Line CA passes through $A (200, 800)$ and $S (- 200, 400)$ . Its slope is $\frac{800 - 400}{200 + 200} = 1$ , so the line is $y = x + 600$ .
  C is on the x-axis ( $y = 0$ ), so $C (- 600, 0)$ .
  Let S divide CA in the ratio $k : 1$ . By the section formula:
  $- 200 = \frac{200 k - 600}{k + 1}$
  $\Rightarrow - 200 k - 200 = 200 k - 600$
  $\Rightarrow 400 k = 400$
  $\Rightarrow k = 1$
  Check with the y-coordinate: $\frac{800 \cdot 1 + 0}{2} = 400$ ✓
  1 mark
Answer: (i) R(200, 400), S(−200, 400); (ii)(a) 160000 sq. units; (ii)(b) 400√2 units; (iii) K = 1
Where the marks are
- (i) R(200, 400) and S(−200, 400) (½ + ½ mark)
- (ii)(a) Side = 400 (1 mark)
- (ii)(a) Area = 160000 sq. units (1 mark)
- (ii)(b) Applies the distance formula / Pythagoras (1 mark)
- (ii)(b) PR = $4002$ (1 mark)
- (iii) Uses the section formula with C(−600, 0) to get K = 1 (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking.
After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats.
Based on the above information, answer the following questions:
1. (i)
  What is the total perimeter of the parking area?
  1 mark
2. (ii)(a)
  What is the total area of parking and the two quadrants?
  2 marks
3. OR
  (ii)(b)
  What is the ratio of area of playground to the area of parking area?
  2 marks
4. (iii)
  Find the cost of fencing the playground and parking area at the rate of ₹ 2 per unit.
  1 mark
Figure in the paper: Rectangular playground with length 14 units (bottom side) and breadth 7 units (right side marked). Two quadrants of radius 2 units, labelled I (top left corner) and II (bottom left corner), drawn at the left corners of the rectangle. A semicircular region labelled PARKING AREA is attached to the right side of the rectangle, with its diameter equal to the 7-unit side. The rectangle is labelled PLAYGROUND. The boundary is drawn with a zigzag fence line. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The parking area is a semicircle on the 7-unit side, so its diameter is 7 and $r = \frac{7}{2}$ units.
  Perimeter $= π r + 2 r = \frac{22}{7} \times \frac{7}{2} + 7 = 11 + 7 = 18$ units
  1 mark
2. (ii)(a) Area of parking $= \frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{4}$ sq. units
  1 mark
3. (ii)(a) Area of two quadrants of radius 2 $= 2 \times \frac{1}{4} \times \frac{22}{7} \times 2 \times 2 = \frac{44}{7}$ sq. units
  ½ mark
4. (ii)(a) Total area $= \frac{77}{4} + \frac{44}{7} = \frac{539 + 176}{28} = \frac{715}{28} \approx 25.54$ sq. units
  ½ mark
5. (ii)(b) Area of playground $= 14 \times 7 = 98$ sq. units.
  Area of parking $= \frac{77}{4}$ sq. units.
  1 mark
6. (ii)(b) Ratio $= \frac{98}{77/4} = \frac{392}{77} = \frac{56}{11}$
  So the ratio of playground area to parking area is $56 : 11$ .
  1 mark
7. (iii) Fence the whole rectangle and the curved edge of the parking area:
  Length $= 2 (l + b) + π r = 2 (14 + 7) + \frac{22}{7} \times \frac{7}{2} = 42 + 11 = 53$ units
  ½ mark
8. (iii) Cost of fencing $= 53 \times ₹2 = ₹106$
  ½ mark
Answer: (i) 18 units; (ii)(a) 715/28 ≈ 25.54 sq. units; (ii)(b) 56 : 11; (iii) ₹106
Where the marks are
- (i) Perimeter $π r + 2 r = 18$ units (1 mark)
- (ii)(a) Area of semicircle $= 77/4$ (1 mark)
- (ii)(a) Area of two quadrants $= 44/7$ (½ mark)
- (ii)(a) Total $715/28 \approx 25.54$ sq. units (½ mark)
- (ii)(b) Areas 98 and 77/4 (1 mark)
- (ii)(b) Ratio 56 : 11 (1 mark)
- (iii) Fencing length 53 units (½ mark)
- (iii) Cost ₹106 (½ mark)
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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.