ICSE Class 10 Mathematics Board Exam Question Paper 2024 with Solutions

CISCE Class 10 Mathematics Board Exam Question Paper with Solutions

SECTION-A

Question 1. Choose the correct Solutions to the questions from the given options. (Do not copy the questions. Write the correct Solutions only.)

1.1. For an Intra-state sale, the CGST paid by a dealer to the Central government is Rs. 120. If the marked price of the article is Rs. 2000, the rate of GST is ______ [1 Mark]
(A) 6%
(B) 10%
(C) 12%
(D) 16.67%

Answer: (C) 12%

Given CGST = Rs. 120. For an intra-state sale, SGST = CGST = Rs. 120. Total GST = CGST + SGST = Rs. 240. Rate of GST = \((240 \div 2000) \times 100 = 12\%\).

Teacher's Note:
a) In intra-state transactions, the total GST is shared equally between the Central and State governments as CGST and SGST.
b) Students must remember that the GST rate is always calculated on the taxable value (marked price or selling price after discount).

1.2. What must be subtracted from the polynomial \( x^{3} + x^{2} - 2x + 1 \), so that the result is exactly divisible by \( (x - 3) \)? [1 Mark]
(A) 31
(B) - 30
(C) 30
(D) 31 (Note: duplicate option value in paper, evaluated as 31)

Answer: (A) 31

By the remainder theorem, the remainder when \( P(x) \) is divided by \( (x - 3) \) is \( P(3) = (3)^{3} + (3)^{2} - 2(3) + 1 = 27 + 9 - 6 + 1 = 31 \).

Teacher's Note:
a) To make a polynomial divisible by a linear factor, the polynomial must be reduced by the exact value of its remainder.
b) Ensure proper substitution of signs when evaluating polynomial values for negative integers.

1.3. The roots of the quadratic equation \( px^{2} - qx + r = 0 \) are real and equal if ______ [1 Mark]
(A) \( p^{2} = 4qr \)
(B) \( q^{2} = 4pr \)
(C) \( - q^{2} = 4pr \)
(D) \( p^{2} \gt 4pr \)

Answer: (B) \( q^{2} = 4pr \)

For real and equal roots, the discriminant \( D = b^{2} - 4ac = 0 \). Here \( a = p, b = -q, c = r \). Thus \( (-q)^{2} - 4pr = 0 \implies q^{2} = 4pr \).

Teacher's Note:
a) The condition for real and equal roots is always Discriminant \( D = 0 \).
b) Pay close attention to the coefficients from the standard form \( ax^{2} + bx + c = 0 \).

1.4. If matrix \( A = \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix} \) and \( A^{2} = \begin{bmatrix} 4 & x \\ 0 & 4 \end{bmatrix} \) then the value of x is ______ [1 Mark]
(A) 2
(B) 4
(C) 8
(D) 10

Answer: (C) 8

\( A^{2} = \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (2\times 2 + 2\times 0) & (2\times 2 + 2\times 2) \\ (0\times 2 + 2\times 0) & (0\times 2 + 2\times 2) \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 0 & 4 \end{bmatrix} \). Comparing with \( \begin{bmatrix} 4 & x \\ 0 & 4 \end{bmatrix} \), we get \( x = 8 \).

Teacher's Note:
a) Matrix multiplication is performed row-wise by column-wise.
b) Do not multiply individual corresponding elements; always follow the standard row-column product rule.

1.5. The median of the following observations arranged in ascending order is 64. Find the value of x:
27, 31, 46, 52, x, x + 4, 71, 79, 85, 90 [1 Mark]
(A) 60
(B) 61
(C) 62
(D) 66

Answer: (C) 62

Total terms \( n = 10 \) (even). Median is the mean of the 5th and 6th terms: \( \frac{x + (x + 4)}{2} = \frac{2x + 4}{2} = x + 2 \). Given median is 64, so \( x + 2 = 64 \implies x = 62 \).

Teacher's Note:
a) For an even number of observations, the median is the average of the \( (\frac{n}{2}) \)th and \( (\frac{n}{2} + 1) \)th terms.
b) Always verify that the given data set is already sorted in ascending order before applying position formulas.

1.6. Points A(x, y), B(3, -2) and C(4, -5) are collinear. The value of y in terms of x is ______ [1 Mark]
(A) 3x - 11
(B) 11 - 3x
(C) 3x - 7
(D) 7 - 3x

Answer: (D) 7 - 3x

For collinear points, the area of the triangle formed by them is zero: \( \frac{1}{2} [x(-2 - (-5)) - y(3 - 4) + 1(3(-5) - (-2)(4))] = 0 \implies 3x + y - 7 = 0 \implies y = 7 - 3x \).

Teacher's Note:
a) Collinearity of three points can be established by proving the area of the triangle is zero or by showing that the slope of AB equals the slope of BC.
b) Be careful with negative signs when expanding the determinant or area formula coordinates.

1.7. The given table shows the distance covered and the time taken by a train moving at a uniform speed along a straight track:
Distance (in m) 60 90 y
Time (in sec) 2 x 5
The values of x and y are: [1 Mark]
(A) x = 4, y = 150
(B) x = 3, y = 100
(C) x = 4, y = 100
(D) x = 3, y = 150

Answer: (D) x = 3, y = 150

Speed is uniform, so \( \frac{60}{2} = \frac{90}{x} = \frac{y}{5} \). Thus \( 30 = \frac{90}{x} \implies x = 3 \), and \( 30 = \frac{y}{5} \implies y = 150 \).

Teacher's Note:
a) Uniform speed implies that the ratio of distance to time remains constant throughout the motion.
b) Set up proportions directly using corresponding table values to solve for unknowns swiftly.

1.8. The 7th term of the given Arithmetic Progression (A.P.): \( \frac{1}{a}, (\frac{1}{a} + 1), (\frac{1}{a} + 2) ... \) is: [1 Mark]
(A) \( (\frac{1}{a} + 6) \)
(B) \( (\frac{1}{a} + 7) \)
(C) \( (\frac{1}{a} + 8) \)
(D) \( (\frac{1}{a} + 77) \)

Answer: (A) \( (\frac{1}{a} + 6) \)

First term \( A = \frac{1}{a} \), Common difference \( D = 1 \). Seventh term \( T_{7} = A + (7 - 1)D = \frac{1}{a} + 6(1) = \frac{1}{a} + 6 \).

Teacher's Note:
a) The formula for the nth term of an A.P. is \( T_{n} = a + (n - 1)d \).
b) Fractions in terms do not change the fundamental arithmetic progression common difference rule.

1.9. The sum invested to purchase 15 shares of a company of nominal value Rs. 75 available at a discount of 20% is ______ [1 Mark]
(A) Rs. 60
(B) Rs. 90
(C) Rs. 1350
(D) Rs. 900

Answer: (D) Rs. 900

Market value of each share = \( 75 - 20\% \text{ of } 75 = 75 - 15 = \text{Rs. } 60 \). Total investment for 15 shares = \( 15 \times 60 = \text{Rs. } 900 \).

Teacher's Note:
a) Investment is always calculated as the number of shares multiplied by the market value per share.
b) Discount reduces the nominal value to arrive at the market value.

1.10. The circumcentre of a triangle is the point which is ______ [1 Mark]
(A) at equal distance from the three sides of the triangle.
(B) at equal distance from the three vertices of the triangle.
(C) the point of intersection of the three medians.
(D) the point of intersection of the three altitudes of the triangle.

Answer: (B) at equal distance from the three vertices of the triangle.

The circumcentre is the center of the circumscribed circle passing through all three vertices, making it equidistant from each vertex.

Teacher's Note:
a) Do not confuse circumcentre (equidistant from vertices, intersection of perpendicular bisectors) with incentre (equidistant from sides, intersection of angle bisectors).
b) Centroid is the intersection of medians, and orthocentre is the intersection of altitudes.

1.11. Statement 1: \( \sin^{2}\theta + \cos^{2}\theta = 1 \)
Statement 2: \( \csc^{2}\theta + \cot^{2}\theta = 1 \)
Which of the following is valid? [1 Mark]
(A) Only 1
(B) Only 2
(C) Both 1 and 2
(D) Neither 1 nor 2

Answer: (A) Only 1

Statement 1 is the fundamental trigonometric identity which is universally true. Statement 2 is incorrect because the correct identity is \( \csc^{2}\theta - \cot^{2}\theta = 1 \).

Teacher's Note:
a) Always verify trigonometric identities carefully by checking signs and squared terms.
b) The three core identities involve \( \sin^{2}\theta + \cos^{2}\theta = 1 \), \( 1 + \tan^{2}\theta = \sec^{2}\theta \), and \( 1 + \cot^{2}\theta = \csc^{2}\theta \).

1.12. In the given diagram, PS and PT are the tangents to the circle. SQ || PT and \( \angle SPT = 80^{\circ} \). The value of \( \angle QST \) is ______ [1 Mark]
[Figure: Circle with external point P, tangents PS and PT, chord SQ parallel to PT, and angle SPT = 80 degrees]
(A) \( 140^{\circ} \)
(B) \( 90^{\circ} \)
(C) \( 80^{\circ} \)
(D) \( 50^{\circ} \)

Answer: (D) \( 50^{\circ} \)

Tangents from P are equal, so \( PS = PT \) and \( \angle PST = \angle PTS = (180^{\circ} - 80^{\circ})\div 2 = 50^{\circ} \). Since \( SQ \parallel PT \) and ST is a transversal, alternate interior angles are equal, so \( \angle QST = \angle PTS = 50^{\circ} \).

Teacher's Note:
a) Tangents drawn from an external point to a circle are equal in length, forming an isosceles triangle with the chord of contact.
b) Parallel line properties like alternate interior angles help transfer angle measures across secants and transversals.

1.13. Assertion (A): A die is thrown once and the probability of getting an even number is \( \frac{2}{3} \).
Reason (R): The sample space for even numbers on a die is \( \{2, 4, 6\} \). [1 Mark]
(A) A is true, R is false.
(B) A is false, R is true.
(C) Both A and R are true.
(D) Both A and R are false.

Answer: (B) A is false, R is true.

Even numbers on a die are \( \{2, 4, 6\} \) (total 3 outcomes), so probability is \( \frac{3}{6} = \frac{1}{2} \), making Assertion false. Reason correctly lists the even numbers set, making Reason true.

Teacher's Note:
a) Probability is defined as the ratio of favorable outcomes to total possible outcomes.
b) Always evaluate assertion and reason statements independently before checking their logical linkage.

1.14. A rectangular sheet of paper of size \( 11\text{ cm} \times 7\text{ cm} \) is first rotated about the side 11 cm and then about the side 7 cm to form a cylinder, as shown in the diagram. The ratio of their curved surface areas is ______ [1 Mark]
[Figure: Rectangular sheet 11 cm by 7 cm, and two resulting cylindrical shapes formed by rotating along different sides]
(A) \( 1:1 \)
(B) \( 7:11 \)
(C) \( 11:7 \)
(D) \( \frac{11\pi}{7} : \frac{7\pi}{11} \)

Answer: (A) \( 1:1 \)

The curved surface area of a cylinder formed by rolling a rectangular sheet is equal to the total area of the rectangle itself, regardless of which side it is rolled along. Thus, both curved surface areas equal \( 77\text{ cm}^{2} \), giving a ratio of \( 1:1 \).

Teacher's Note:
a) The curved surface area of a cylinder formed from a sheet is identically equal to the area of the rectangle.
b) Recognizing geometric invariance saves considerable calculation time in objective questions.

1.15. In the given diagram, \( \Delta ABC \sim \Delta PQR \). If AD and PS are bisectors of \( \angle BAC \) and \( \angle QPR \) respectively then ______ [1 Mark]
[Figure: Two similar triangles ABC and PQR with angle bisectors AD and PS respectively]
(A) \( \Delta ABC \sim \Delta PQS \)
(B) \( \Delta ABD \sim \Delta PQS \)
(C) \( \Delta ABD \sim \Delta PSR \)
(D) \( \Delta ABC \sim \Delta PSR \)

Answer: (B) \( \Delta ABD \sim \Delta PQS \)

Since \( \Delta ABC \sim \Delta PQR \), \( \angle B = \angle Q \) and \( \angle A = \angle P \). Since AD and PS are angle bisectors, half of equal angles are equal, so \( \angle BAD = \angle QPS \). By AA similarity criterion, \( \Delta ABD \sim \Delta PQS \).

Teacher's Note:
a) Angle bisectors of corresponding equal angles in similar triangles maintain proportional angular divisions.
b) Triangle similarity statements must always maintain correct vertex correspondence.

Question 2.

2.1. If \( A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 4 & 0 \\ y & 1 \end{bmatrix} \) and \( C = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \). Find the values of x and y, if \( AB = C \). [4 Marks]

Answer:
Given \( AB = C \):
\( \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 4 & 0 \\ y & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \)
\( \begin{bmatrix} 4x + 0 & 0 + 0 \\ 4 + y & 0 + 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \)
\( \begin{bmatrix} 4x & 0 \\ 4 + y & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \)
By equality of matrices:
\( 4x = 4 \implies x = 1 \)
\( 4 + y = x \implies 4 + y = 1 \implies y = -3 \)
Hence, \( x = 1 \) and \( y = -3 \).

Teacher's Note:
a) Perform matrix multiplication carefully by taking the dot product of rows and columns.
b) Equate corresponding elements of equal matrices to form linear equations for the unknowns.

2.2. A solid metallic cylinder is cut into two identical halves along its height (as shown in the diagram). The diameter of the cylinder is 7 cm and the height is 10 cm. Find:
a. The total surface area (both the halves).
b. The total cost of painting the two halves at the rate of Rs. 30 per \( \text{cm}^{2} \) (Use \( \pi = \frac{22}{7} \)) [4 Marks]
[Figure: Solid cylinder and its two identical half-cylinder sections cut along the height]

Answer:
Radius \( r = \frac{7}{2}\text{ cm} \), height \( h = 10\text{ cm} \).
a. Total surface area of a half cylinder = curved surface area of half cylinder + two flat rectangular cross-sections.
\( \text{T.S.A.} = \frac{2\pi rh}{2} + 2(\pi r^{2}/2) + d \times h = \pi rh + \pi r^{2} + d \times h \)
\( = (\frac{22}{7} \times \frac{7}{2} \times 10) + (\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}) + (7 \times 10) \)
\( = 110 + 38.5 + 70 = 218.5\text{ cm}^{2} \)
Total surface area of both halves = \( 218.5 \times 2 = 437\text{ cm}^{2} \). (Alternatively, total surface area of both halves equals the original cylinder plus two rectangular cut faces: \( 2\pi rh + 2\pi r^{2} + 2(d \times h) = 2(\frac{22}{7}\frac{7}{2}10) + 2(\frac{22}{7}\frac{49}{4}) + 2(70) = 220 + 77 + 140 = 437\text{ cm}^{2} \)).
b. Total cost of painting = \( 437 \times 30 = \text{Rs. } 13,110 \).

Teacher's Note:
a) Cutting a solid cylinder along its axis exposes two rectangular faces of dimensions diameter by height.
b) Double-check surface area components including curved areas, circular bases, and newly exposed rectangular faces.

2.3. 15, 30, 60, 120 .... are in G.P. (Geometric Progression):
a) Find the nth term of this G.P. in terms of n.
b) How many terms of the above G.P. will give the sum 945? [4 Marks]

Answer:
a. First term \( a = 15 \), common ratio \( r = \frac{30}{15} = 2 \).
\( T_{n} = a r^{n-1} = 15(2)^{n-1} \).
b. Sum of n terms \( S_{n} = \frac{a(r^{n} - 1)}{r - 1} = 945 \)
\( \frac{15(2^{n} - 1)}{2 - 1} = 945 \)
\( 15(2^{n} - 1) = 945 \)
\( 2^{n} - 1 = \frac{945}{15} = 63 \)
\( 2^{n} = 64 = 2^{6} \)
\( n = 6 \).
Hence, 6 terms are needed.

Teacher's Note:
a) The nth term formula for G.P. is \( T_{n} = ar^{n-1} \) and the sum formula is \( S_{n} = \frac{a(r^{n}-1)}{r-1} \) when \( r \gt 1 \).
b) Express numbers as powers of the common ratio to solve exponential equations smoothly.

Question 3.

3.1. Factorize: \( \sin^{3}\theta + \cos^{3}\theta \)
Hence, prove the following identity:
\( \frac{\sin^{3}\theta + \cos^{3}\theta}{\sin\theta + \cos\theta} + \sin\theta \cos\theta = 1 \) [4 Marks]

Answer:
\( \sin^{3}\theta + \cos^{3}\theta = (\sin\theta + \cos\theta)(\sin^{2}\theta + \cos^{2}\theta - \sin\theta\cos\theta) = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta) \).
For the identity, taking L.H.S.:
\( \frac{(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)}{\sin\theta + \cos\theta} + \sin\theta\cos\theta \)
\( = (1 - \sin\theta\cos\theta) + \sin\theta\cos\theta = 1 = \text{R.H.S.} \)

Teacher's Note:
a) Use the algebraic expansion formula \( a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}) \).
b) Substitute fundamental trigonometric identities like \( \sin^{2}\theta + \cos^{2}\theta = 1 \) to simplify expressions.

3.2. In the given diagram, O is the centre of the circle. PR and PT are two tangents drawn from the external point P and touching the circle at Q and S respectively. MN is a diameter of the circle. Given \( \angle PQM = 42^{\circ} \) and \( \angle PSM = 25^{\circ} \).
Find:
a) \( \angle OQM \)
b) \( \angle QNS \)
c) \( \angle QOS \)
d) \( \angle QMS \) [4 Marks]
[Figure: Circle with center O, diameter MN, external point P with tangents PR and PT touching at Q and S, chords QM, SM, NS, etc.]

Answer:
a. Radius is perpendicular to tangent at point of contact, so \( \angle OQP = 90^{\circ} \).
\( \angle OQM = \angle OQP - \angle PQM = 90^{\circ} - 42^{\circ} = 48^{\circ} \).
b. By alternate segment theorem, \( \angle QNM = \angle PQM = 42^{\circ} \) and \( \angle SNM = \angle PSM = 25^{\circ} \).
\( \angle QNS = \angle QNM + \angle SNM = 42^{\circ} + 25^{\circ} = 67^{\circ} \).
c. Angle subtended by an arc at the centre is twice the angle subtended at the remaining circumference: \( \angle QOS = 2 \angle QNS = 2 \times 67^{\circ} = 134^{\circ} \).
d. Since QNSM is a cyclic quadrilateral, opposite angles sum to \( 180^{\circ} \):
\( \angle QMS = 180^{\circ} - \angle QNS = 180^{\circ} - 67^{\circ} = 113^{\circ} \).

Teacher's Note:
a) Combine angle properties of tangents, alternate segment theorem, and cyclic quadrilaterals systematically.
b) Remember that the angle between a tangent and radius at the point of contact is always \( 90^{\circ} \).

3.3. Use graph sheet for this question. Take 2 cm = 1 unit along the axes.
a. Plot A(0, 3), B(2, 1) and C(4, -1).
b. Reflect point B and C in y-axis and name their images as B' and C' respectively. Plot and write coordinates of the points B' and C'.
c. Reflect point A in the line BB' and name its images as A'.
d. Plot and write coordinates of point A'.
e. Join the points ABA'B' and give the geometrical name of the closed figure so formed. [4 Marks]
[Figure: Cartesian plane graph showing points A, B, C, B', C', A' and plotted shape ABA'B']

Answer:
a. Plotted points \( A(0, 3), B(2, 1), C(4, -1) \).
b. Reflection in y-axis negates the x-coordinate: \( B'(-2, 1), C'(-4, -1) \).
c. & d. Line BB' is the line \( y = 1 \). Reflecting point \( A(0, 3) \) across \( y = 1 \): the vertical distance from A to line \( y = 1 \) is \( 3 - 1 = 2 \) units, so point A' is 2 units below the line at \( (0, -1) \).
e. Joining points \( A(0, 3), B(2, 1), A'(0, -1), B'(-2, 1) \) forms a rhombus.

Teacher's Note:
a) Reflection across the y-axis transforms \( (x, y) \) to \( (-x, y) \).
b) Reflection across a horizontal line \( y = k \) keeps the x-coordinate unchanged while reflecting the y-coordinate as \( 2k - y \).

SECTION-B

Question 4.

4.1. Suresh has a recurring deposit account in a bank. He deposits Rs. 2000 per month and the bank pays interest at the rate of 8% per annum. If he gets Rs. 1040 as interest at the time of maturity, find in years total time for which the account was held. [3 Marks]

Answer:
Deposit per month \( P = \text{Rs. } 2000 \), Rate \( R = 8\% \), Interest \( I = \text{Rs. } 1040 \).
Let time in months be \( n \).
\( I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{R}{100} \)
\( 1040 = 2000 \times \frac{n(n+1)}{24} \times \frac{8}{100} \)
\( 1040 = \frac{20 \times n(n+1)}{3} \)
\( 52 \times 3 = n^{2} + n \implies n^{2} + n - 156 = 0 \)
\( (n - 12)(n + 13) = 0 \implies n = 12 \) (since \( n \) cannot be negative).
Time in years = \( \frac{12}{12} = 1 \) year.

Teacher's Note:
a) The recurring deposit interest formula is \( I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \).
b) Always solve the resulting quadratic equation and reject negative or impractical time values.

4.2. The following table gives the duration of movies in minutes:
Duration 100 - 110 | 110 - 120 | 120 - 130 | 130 - 140 | 140 - 150 | 150 - 160
No. of movies 5 | 10 | 17 | 8 | 6 | 4
Using step-deviation method, find the mean duration of the movies. [3 Marks]

Answer:

Class width \( h = 10 \), Assumed mean \( A = 135 \).
Mean \( \bar{x} = A + (\frac{\sum f_{i}u_{i}}{\sum f_{i}}) \times h = 135 + (\frac{-38}{50}) \times 10 = 135 - 7.6 = 127.4 \).

Teacher's Note:
a) The step-deviation formula is \( \bar{x} = A + \frac{\sum f_{i}u_{i}}{\sum f_{i}} \times h \), where \( u_{i} = \frac{x_{i} - A}{h} \).
b) Choose the assumed mean \( A \) corresponding to the middle class with the highest frequency to simplify calculations.

4.3.
If \( \frac{(a+b)^{3}}{(a-b)^{3}} = \frac{64}{27} \).
a. Find \( \frac{a+b}{a-b} \)
b. Hence using properties of proportion, find a : b. [4 Marks]

Answer:
a. Taking the cube root on both sides:
\( \frac{a+b}{a-b} = \sqrt[3]{\frac{64}{27}} = \frac{4}{3} \).
b. Applying componendo and dividendo on \( \frac{a+b}{a-b} = \frac{4}{3} \):
\( \frac{(a+b) + (a-b)}{(a+b) - (a-b)} = \frac{4 + 3}{4 - 3} \)
\( \frac{2a}{2b} = \frac{7}{1} \)
\( \frac{a}{b} = \frac{7}{1} \)
Hence, \( a : b = 7 : 1 \).

Teacher's Note:
a) Componendo and dividendo states that if \( \frac{x}{y} = \frac{p}{q} \), then \( \frac{x+y}{x-y} = \frac{p+q}{p-q} \).
b) Cube rooting powers first makes ratio reduction straightforward.

Question 5.

5.1. The given graph with a histogram represents the number of plants of different heights grown in a school campus. Study the graph carefully and answer the following questions:
a. Make a frequency table with respect to the class boundaries and their corresponding frequencies.
b. State the modal class.
c. Identify and note down the mode of the distribution.
d. Find the number of plants whose height range is between 80 cm to 90 cm. [4 Marks]
[Figure: Histogram of plant heights with frequency bars and mode determination lines]

Answer:
a. Frequency table:

b. Modal class is 60 - 70 (having the highest frequency of 12).
c. Mode of the distribution from the histogram = 64.
d. Number of plants with height between 80 cm to 90 cm = 3.

Teacher's Note:
a) The modal class is the class interval with the maximum frequency.
b) Mode can be graphically estimated from the histogram using diagonal intersection lines from the top corners of the modal bar.

5.2. The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are \( 52^{\circ} \) and \( 45^{\circ} \) respectively. Find the distance AB, to the nearest metre. [3 Marks]
[Figure: Vertical tree CD of height 100 m with points A and B on opposite sides at angles of elevation 52 degrees and 45 degrees respectively]

Answer:
Let CD = 100 m be the height of the tree. Let C be the base of the tree between A and B.
In right triangle ACD: \( \tan 52^{\circ} = \frac{CD}{AC} \implies AC = \frac{100}{\tan 52^{\circ}} = \frac{100}{1.2799} \approx 78.13\text{ m} \).
In right triangle BCD: \( \tan 45^{\circ} = \frac{CD}{BC} \implies BC = \frac{100}{1} = 100\text{ m} \).
Total distance \( AB = AC + BC = 78.13 + 100 = 178.13\text{ m} \approx 178\text{ m} \).

Teacher's Note:
a) When points are on opposite sides of a tower, the total distance between them is the sum of their individual distances from the base.
b) Use standard trigonometric table values for tangents accurately.

Question 6.

6.1. Solve the following equation for x and give, in the following case, your answer correct to 2 decimal places:
\( 2x^{2} - 10x + 5 = 0 \) [3 Marks]

Answer:
Comparing with \( ax^{2} + bx + c = 0 \), we get \( a = 2, b = -10, c = 5 \).
\( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \)
\( x = \frac{-(-10) \pm \sqrt{(-10)^{2} - 4(2)(5)}}{2(2)} \)
\( x = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm \sqrt{60}}{4} = \frac{10 \pm 7.746}{4} \)
\( x_{1} = \frac{17.746}{4} = 4.4365 \approx 4.44 \)
\( x_{2} = \frac{2.254}{4} = 0.5635 \approx 0.56 \)
(Note: Using textbook approximation values where \( \sqrt{15} \approx 3.873 \), gives \( 4.44 \) and \( 0.56 \)).

Teacher's Note:
a) Quadratic formula is indispensable when an equation cannot be easily factorized.
b) Round off answers strictly to two decimal places as requested in the question.

6.2. The nth term of an Arithmetic Progression (A.P.) is given by the relation \( T_{n} = 6(7 - n) \).
Find:
a. its first term and common difference
b. sum of its first 25 terms [3 Marks]

Answer:
a. First term (\( n = 1 \)): \( T_{1} = a = 6(7 - 1) = 6(6) = 36 \).
Second term (\( n = 2 \)): \( T_{2} = 6(7 - 2) = 6(5) = 30 \).
Common difference \( d = T_{2} - T_{1} = 30 - 36 = -6 \).
b. Sum of first 25 terms: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( S_{25} = \frac{25}{2}[2(36) + (25 - 1)(-6)] \)
\( S_{25} = \frac{25}{2}[72 + 24(-6)] = \frac{25}{2}[72 - 144] = \frac{25}{2}(-72) = 25 \times (-36) = -900 \).

Teacher's Note:
a) Substitute \( n = 1 \) and \( n = 2 \) into the general term expression to determine the first term and common difference.
b) The sum formula for an A.P. is \( S_{n} = \frac{n}{2}[2a + (n-1)d] \).

6.3. In the given diagram \( \Delta ADB \) and \( \Delta ACB \) are two right angled triangles with \( \angle ADB = \angle BCA = 90^{\circ} \). If AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm.
a. Prove that \( \Delta APD \sim \Delta BPC \)
b. Find the length of BD and PB
c. Hence, find the length of PA
d. Find area \( \Delta APD : \text{area } \Delta BPC \) [4 Marks]
[Figure: Intersecting right triangles ADB and ACB sharing hypotenuse AB, with point P on DB]

Answer:
a. In \( \Delta APD \) and \( \Delta BPC \):
\( \angle APD = \angle BPC \) (Vertically opposite angles)
\( \angle ADP = \angle BCP = 90^{\circ} \)
Therefore, \( \Delta APD \sim \Delta BPC \) (by AA similarity criterion).
b. In right triangle ADB: \( AB^{2} = AD^{2} + BD^{2} \implies 10^{2} = 6^{2} + BD^{2} \implies 100 = 36 + BD^{2} \implies BD = 8\text{ cm} \).
\( PB = BD - DP = 8 - 4.5 = 3.5\text{ cm} \).
c. In right triangle APD: \( AP^{2} = AD^{2} + DP^{2} = 6^{2} + (4.5)^{2} = 36 + 20.25 = 56.25 \implies AP = \sqrt{56.25} = 7.5\text{ cm} \).
d. The ratio of areas of similar triangles is the square of the ratio of corresponding sides: \( \frac{\text{ar}(\Delta APD)}{\text{ar}(\Delta BPC)} = (\frac{AD}{BC})^{2} = (\frac{6}{2.4})^{2} = (\frac{2.5}{1})^{2} = \frac{6.25}{5.76} = \frac{25}{4} \) (or \( 25:4 \)).

Teacher's Note:
a) Establish triangle similarity using AA (Angle-Angle) criteria.
b) The ratio of areas of two similar triangles equals the square of the ratio of any pair of corresponding sides.

Question 7.

7.1. In the given diagram an isosceles \( \Delta ABC \) is inscribed in a circle with centre O. PQ is a tangent to the circle at C. OM is perpendicular to chord AC and \( \angle COM = 65^{\circ} \).
Find:
a. \( \angle ABC \)
b. \( \angle BAC \)
c. \( \angle BCQ \) [3 Marks]
[Figure: Isosceles triangle ABC inscribed in circle with center O, OM perpendicular to AC, tangent PQ at C]

Answer:
a. \( \angle AOC = 2 \angle COM = 65^{\circ} + 65^{\circ} = 130^{\circ} \) (since perpendicular from centre bisects the chord and central angle).
\( \angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 130^{\circ} = 65^{\circ} \).
b. In isosceles \( \Delta ABC \ (\text{with } AB = AC) \), \( \angle ACB = \angle ABC = 65^{\circ} \).
\( \angle BAC = 180^{\circ} - (65^{\circ} + 65^{\circ}) = 180^{\circ} - 130^{\circ} = 50^{\circ} \).
c. Radius is perpendicular to tangent at C, so \( \angle OCQ = 90^{\circ} \).
In right triangle OMC, \( \angle OCM = 90^{\circ} - 65^{\circ} = 25^{\circ} \).
\( \angle OCB = \angle ACB - \angle OCM = 65^{\circ} - 25^{\circ} = 40^{\circ} \).
\( \angle BCQ = \angle OCQ - \angle OCB = 90^{\circ} - 40^{\circ} = 50^{\circ} \).

Teacher's Note:
a) The perpendicular from the centre of a circle to a chord bisects the chord and the vertical angle at the centre.
b) The angle between a tangent and radius at the point of contact is always \( 90^{\circ} \).

7.2. Solve the following inequation, write down the solution set and represent it on the real number line.
\( -3 + x \leq \frac{7x}{2} + 2 \lt 8 + 2x, x \in I \) [3 Marks]

Answer:
Split into two inequalities:
1) \( -3 + x \leq \frac{7x}{2} + 2 \)
\( -3 - 2 \leq \frac{7x}{2} - x \)
\( -5 \leq \frac{5x}{2} \implies -10 \leq 5x \implies x \geq -2 \).
2) \( \frac{7x}{2} + 2 \lt 8 + 2x \)
\( \frac{7x}{2} - 2x \lt 8 - 2 \)
\( \frac{3x}{2} \lt 6 \implies 3x \lt 12 \implies x \lt 4 \).
Combining both: \( -2 \leq x \lt 4 \).
Solution set in integers \( x \in I \): \( \{-2, -1, 0, 1, 2, 3\} \).
[Figure: Number line marked from -3 to 5 showing closed circle at -2 and open circle at 4 with points -2, -1, 0, 1, 2, 3 highlighted]

Teacher's Note:
a) Solve compound inequalities by splitting them into two separate inequalities and finding the intersection of their solution sets.
b) Pay close attention to strict (\(\lt\)) versus non-strict (\(\leq\)) inequalities when plotting on the number line.

7.3. In the given diagram, ABC is a triangle, where B(4, -4) and C(-4, -2). D is a point on AC.
a. Write down the coordinates of A and D.
b. Find the coordinates of the centroid of \( \Delta ABC \).
c. If D divides AC in the ratio k : 1, find the value of k.
d. Find the equation of the line BD. [4 Marks]
[Figure: Cartesian coordinate plane showing triangle ABC with vertices A, B(4, -4), C(-4, -2) and point D on AC]

Answer:
a. From the graph, coordinates of \( A = (0, 6) \) and \( D = (-3, 0) \).
b. Centroid of \( \Delta ABC \) with \( A(0, 6), B(4, -4), C(-4, -2) \):
\( (\frac{0 + 4 + (-4)}{3}, \frac{6 + (-4) + (-2)}{3}) = (\frac{0}{3}, \frac{0}{3}) = (0, 0) \).
c. Using section formula for point \( D(-3, 0) \) dividing segment AC where \( A(0, 6) \) and \( C(-4, -2) \) in ratio \( k:1 \):
\( x = \frac{k(x_{2}) + 1(x_{1})}{k + 1} \implies -3 = \frac{k(-4) + 1(0)}{k + 1} \)
\( -3(k + 1) = -4k \implies -3k - 3 = -4k \implies k = 3 \).
d. Equation of line BD passing through \( B(4, -4) \) and \( D(-3, 0) \):
Slope \( m = \frac{0 - (-4)}{-3 - 4} = \frac{4}{-7} = -\frac{4}{7} \).
Using point-slope form: \( y - 0 = -\frac{4}{7}(x - (-3)) \)
\( 7y = -4(x + 3) \implies 7y = -4x - 12 \implies 4x + 7y + 12 = 0 \) (Note: depending on exact D coordinates from graph, standard form yields \( 4x + y + 12 = 0 \) or similar linear form).

Teacher's Note:
a) The centroid formula is \( (\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}) \).
b) The section formula allows finding the division ratio when internal point coordinates are known.

Question 8.

8.1. The polynomial \( 3x^{3} + 8x^{2} - 15x + k \) has \( (x - 1) \) as a factor. Find the value of k. Hence factorize the resulting polynomial completely. [4 Marks]

Answer:
Let \( P(x) = 3x^{3} + 8x^{2} - 15x + k \). Since \( (x - 1) \) is a factor, by factor theorem, \( P(1) = 0 \):
\( 3(1)^{3} + 8(1)^{2} - 15(1) + k = 0 \)
\( 3 + 8 - 15 + k = 0 \implies -4 + k = 0 \implies k = 4 \).
Polynomial is \( 3x^{3} + 8x^{2} - 15x + 4 \).
Dividing by \( (x - 1) \), quotient is \( 3x^{2} + 11x - 4 \).
Factorizing the quadratic: \( 3x^{2} + 12x - x - 4 = 3x(x + 4) - 1(x + 4) = (3x - 1)(x + 4) \).
Completely factorized form: \( (x - 1)(3x - 1)(x + 4) \).

Teacher's Note:
a) Use the factor theorem to find unknown coefficients by equating \( P(a) = 0 \).
b) Complete factorization requires dividing the cubic polynomial by the linear factor and splitting the middle term of the resulting quadratic.

8.2. The following letters A, D, M, N, O, S, U, Y of the English alphabet are written on separate cards and put in a box. The cards are well shuffled and one card is drawn at random. What is the probability that the card drawn is a letter of the word,
a. MONDAY?
b. Which does not appear in MONDAY?
c. Which appears both in SUNDAY and MONDAY? [3 Marks]

Answer:
Total sample space \( n(S) = 8 \) (letters: A, D, M, N, O, S, U, Y).
a. Letters of MONDAY present in the box: M, O, n(E) = 6 (M, O, N, D, A, Y). Probability = \( \frac{6}{8} = \frac{3}{4} \).
b. Letters in the box that do not appear in MONDAY: S, U. Probability = \( \frac{2}{8} = \frac{1}{4} \).
c. Letters appearing in both SUNDAY and MONDAY: N, D, A, Y (4 letters). Probability = \( \frac{4}{8} = \frac{1}{2} \).

Teacher's Note:
a) Identify set intersections carefully when filtering words against the master letter set.
b) Probability is always the count of favorable outcomes divided by total outcomes.

8.3. Oil is stored in a spherical vessel occupying \( \frac{3}{4} \) of its full capacity. Radius of this spherical vessel is 28 cm. This oil is then poured into a cylindrical vessel with a radius of 21 cm. Find the height of the oil in the cylindrical vessel (correct to the nearest cm). Take \( \pi = \frac{22}{7} \) [3 Marks]
[Figure: Spherical vessel of radius 28 cm and cylindrical vessel of radius 21 cm containing oil]

Answer:
Radius of sphere \( R = 28\text{ cm} \). Volume of oil in sphere = \( \frac{3}{4} \times \frac{4}{3}\pi R^{3} = \pi R^{3} \).
Radius of cylinder \( r = 21\text{ cm} \). Let height of oil be \( h \).
Volume of cylinder = Volume of oil \( \implies \pi r^{2}h = \pi R^{3} \)
\( \pi (21)^{2}h = \pi (28)^{3} \)
\( h = \frac{28 \times 28 \times 28}{21 \times 21} = \frac{4 \times 4 \times 28}{3 \times 3} = \frac{448}{9} = 49.78\text{ cm} \approx 50\text{ cm} \).

Teacher's Note:
a) When liquid is transferred between containers, its volume remains conserved.
b) Equate the volume of oil stored (\( \frac{3}{4} \) of sphere) to the volume of the cylinder \( \pi r^{2}h \).

Question 9.

9.1. The figure shows a circle of radius 9 cm with O as the centre. The diameter AB produced meets the tangent PQ at P. If PA = 24 cm, find the length of tangent PQ: [3 Marks]
[Figure: Circle with center O, diameter AB, produced to point P, tangent PQ]

Answer:
Radius \( r = 9\text{ cm} \), so diameter \( AB = 18\text{ cm} \).
Given \( PA = 24\text{ cm} \). Since O is center and AB is diameter, \( OA = OB = 9\text{ cm} \).
Distance from P to center \( OP = PA + OA = 24 + 9 = 33\text{ cm} \), or using secant-tangent theorem: \( PB \times PA = PQ^{2} \).
Here \( PB = PA - AB = 24 - 18 = 6\text{ cm} \).
\( PQ^{2} = PB \times PA = 6 \times 24 = 144 \implies PQ = \sqrt{144} = 12\text{ cm} \).

Teacher's Note:
a) The tangent-secant theorem states that when a secant and tangent are drawn from an external point, the product of the external segment and the entire secant equals the square of the tangent segment (\( PB \times PA = PQ^{2} \)).
b) Ensure correct segment lengths are measured from the external point P.

9.2. Mr. Gupta invested Rs. 33000 in buying Rs. 100 shares of a company at 10% premium. The dividend declared by the company is 12%.
Find:
a. the number of shares purchased by him
b. his annual dividend. [4 Marks]

Answer:
Investment = Rs. 33,000, Nominal Value (N.V.) = Rs. 100, Premium = 10%.
Market Value (M.V.) = \( 100 + 10\% \text{ of } 100 = \text{Rs. } 110 \).
a. Number of shares purchased = \( \frac{\text{Total Investment}}{\text{Market Value per share}} = \frac{33000}{110} = 300 \text{ shares} \).
b. Annual dividend = \( \text{Number of shares} \times \text{Rate of dividend} \times \text{N.V.} \)
\( = 300 \times \frac{12}{100} \times 100 = 300 \times 12 = \text{Rs. } 3,600 \).

Teacher's Note:
a) Premium increases the market value above the nominal value.
b) Dividend is always calculated on the total nominal (face) value of the shares held, never on the market investment amount.

9.3. A life insurance agent found the following data for distribution of ages of 100 policy holders.
Age in years | Policy Holders (frequency) | Cumulative frequency
20 - 25 | 2 | 2
25 - 30 | 4 | 6
30 - 35 | 12 | 18
35 - 40 | 20 | 38
40 - 45 | 28 | 66
45 - 50 | 22 | 88
50 - 55 | 8 | 96
55 - 60 | 4 | 100
On a graph sheet draw an ogive using the given data. Take 2 cm = 5 years along one axis and 2 cm = 10 policy holders along the other axis.
Use your graph to find:
a. The median age.
b. Number of policy holders whose age is above 52 years. [4 Marks]
[Figure: Ogive (cumulative frequency curve) for age distribution of policy holders]

Answer:
Total frequency \( N = 100 \). Median position \( \frac{N}{2} = \frac{100}{2} = 50 \).
a. From the ogive graph at cumulative frequency 50, the corresponding median age = 43 years.
b. At age 52 years, find the corresponding cumulative frequency from the graph (approx. 85 policy holders). Number of policy holders above 52 years = \( 100 - 85 = 15 \).

Teacher's Note:
a) Median is located on the ogive at cumulative frequency \( N/2 \).
b) "Age above 52 years" requires subtracting the cumulative frequency at age 52 from the total number of observations \( N \).

Question 10.

10.1. Rohan bought the following eatables for his friends:
Soham Sweet Mart: Bill
S.N. Item | Price | Quantity | Rate of GST
1 | Laddu | Rs. 500 per kg | 2 kg | 5%
2 | Pastries | Rs. 100 per kg | 12 pieces | 18%
Calculate:
a. Total GST paid.
b. Total bill amount including GST. [4 Marks]

Answer:

a. Total GST paid = \( \text{Rs. } 50 + \text{Rs. } 216 = \text{Rs. } 266 \).
b. Total bill including GST = \( \text{Rs. } 1000 + \text{Rs. } 50 + \text{Rs. } 1200 + \text{Rs. } 216 = \text{Rs. } 2,466 \).

Teacher's Note:
a) Calculate the taxable value for each item first by multiplying unit price by quantity.
b) Compute individual GST amounts by applying the respective GST percentage to each item's total price.

10.2.a If the lines \( kx - y + 4 = 0 \) and \( 2y = 6x + 7 \) are perpendicular to each other, find the value of k. [3 Marks]

Answer:
First line: \( kx - y + 4 = 0 \implies y = kx + 4 \). Slope \( m_{1} = k \).
Second line: \( 2y = 6x + 7 \implies y = 3x + \frac{7}{2} \). Slope \( m_{2} = 3 \).
Since lines are perpendicular, \( m_{1} \times m_{2} = -1 \):
\( k \times 3 = -1 \implies k = -\frac{1}{3} \).

Teacher's Note:
a) Two lines are perpendicular if and only if the product of their slopes is \( -1 \).
b) Convert equations to slope-intercept form \( y = mx + c \) to easily identify slopes.

10.2.b Find the equation of a line parallel to \( 2y = 6x + 7 \) and passing through (-1, 1). [3 Marks]

Answer:
Given line: \( 2y = 6x + 7 \implies y = 3x + \frac{7}{2} \). Slope \( m = 3 \).
Parallel lines have equal slopes, so the required line has slope \( m = 3 \) and passes through \( (-1, 1) \).
Using point-slope form: \( y - y_{1} = m(x - x_{1}) \)
\( y - 1 = 3(x - (-1)) \)
\( y - 1 = 3(x + 1) \)
\( y - 1 = 3x + 3 \implies y = 3x + 4 \) (or \( 3x - y + 4 = 0 \)).

Teacher's Note:
a) Parallel lines always share the exact same slope.
b) Use the point-slope form with the given point coordinates to establish the final linear equation.

10.3. Use ruler and compass to answer this question. Construct \( \angle ABC = 90^{\circ} \), where AB = 6 cm, BC = 8 cm.
a. Construct the locus of points equidistant from B and C.
b. Construct the locus of points equidistant from A and B.
c. Mark the point which satisfies both the conditions (a) and (b) as O. Construct the locus of points keeping a fixed distance OA from the fixed point O.
d. Construct the locus of points which are equidistant from BA and BC. [4 Marks]
[Figure: Geometric construction showing triangle ABC, perpendicular bisectors, circumcentre O, circumcircle, and angle bisector]

Answer:
a. The locus of points equidistant from two points B and C is the perpendicular bisector of segment BC.
b. The locus of points equidistant from A and B is the perpendicular bisector of segment AB.
c. Point O, which is the intersection of the perpendicular bisectors of AB and BC, is the circumcentre of triangle ABC. The locus of points at a fixed distance OA from O is the circumcircle of triangle ABC.
d. The locus of points equidistant from two intersecting lines BA and BC is the angle bisector of \( \angle ABC \).

Teacher's Note:
a) Standard locus theorems form the basis of geometric constructions.
b) Ensure all construction arcs and lines are clearly drawn using a sharp pencil, ruler, and compass.