ICSE Paper 2023 (Class 10): Mathematics

(Maximum Marks: 80)

(Time allowed: Two hours and a half)

Attempt all questions from Section A and any four questions from Section B.
All work, including rough work, must be clearly shown and must be done on the same sheet as the rest of the Solution.
Omission of essential work will result in a loss of marks.
The intended marks for questions or parts of questions are given in brackets [ ].
Mathematical tables and graph papers are provided.

SECTION-A (40 Marks)

(Attempt all questions from this Section)

Question 1.

Choose the correct answer to the questions from the given options. [15]

(Do not copy the questions, write the correct answer only)

$\displaystyle \text {(i) If } \begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ -8 \end{bmatrix}, \text{ the value of } x \text{ and } y \text{ respectively are: }$

$\displaystyle (a) \ 1, -2 \hspace{1.0cm} (b) \ -2, 1 \hspace{1.0cm}(c) \ 1, 2 \hspace{1.0cm}(d) \ -2, -1$

Answer:

$\fbox{ a }$

$\displaystyle \text{ Given} \begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ -8 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x+0y \\ 0x+4y \end{bmatrix} = \begin{bmatrix} 2 \\ -8 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x \\ 4y \end{bmatrix} = \begin{bmatrix} 2 \\ -8 \end{bmatrix}$

On equating the corresponding terms

$\displaystyle 2x=2 \Rightarrow x = 1$

$\displaystyle 4y = -8 \Rightarrow y = - 2$

$\\$

$\displaystyle \text {(ii) If } x-2 \text { is a factor of } x^3-kx-12, \text { then the value of } k \text { is }$

$\displaystyle (a) \ 3 \hspace{1.0cm} (b) \ 2 \hspace{1.0cm}(c) \ -2 \hspace{1.0cm}(d) \ -3$

Answer:

$\fbox{ c }$

$\displaystyle (x-2) \text{ is a factor of } x^3-kx-12$

$\displaystyle \text{ Then putting } x-2 = 0$

$\displaystyle \text{ i.e. } x-2 \text{ in given expression, }$

$\displaystyle \text{ Remainder } = 0$

$\displaystyle \Rightarrow 2^3 - 2k - 12 = 0$

$\displaystyle \Rightarrow 8-2k-12 = 0$

$\displaystyle \Rightarrow 2k = -4$

$\displaystyle \Rightarrow k = -2$

$\\$

$\text {(iii) If in the given diagram RT is a tangent touching the circle at S. If } \angle{PST}= 30^{\circ} \text { and } \angle{SPQ}= 60^{\circ} \text{ then } \angle{PSQ} \text{ is equal to: }$

$\displaystyle (a) \ 40^{\circ} \hspace{1.0cm} (b) \ 30^{\circ} \hspace{1.0cm}(c) \ 60^{\circ} \hspace{1.0cm}(d) \ 90^{\circ}$

Answer:

$\fbox{ d }$

$\displaystyle \angle PQS = \angle PST = 30^{\circ} (\angle S \text{ in alternate segment })$

$\displaystyle \text{In } \triangle PQS,$

$\displaystyle \angle PQS + \angle PSQ + \angle SPQ = 180^{\circ} \text{ (sum of all angles of a triangle )}$

$\displaystyle 30^{\circ} + \angle PSQ + 60^{\circ} = 180^{\circ}$

$\displaystyle \angle PSQ = 180^{\circ} - 90^{\circ} = 90^{\circ}$

$\\$

$\text {(iv) A letter is chosen at random from all the letters of the English alphabets. } \\ \text{The probability that the letter chosen is a vowel, is}$

$\displaystyle (a) \ \frac{4}{26} \hspace{1.0cm} (b) \ \frac{5}{26} \hspace{1.0cm}(c) \ \frac{21}{26} \hspace{1.0cm}(d) \ \frac{5}{24}$

Answer:

$\fbox{ b }$

$\displaystyle \text{Total number of letters in English alphabets } n(S) = 26$

$\displaystyle \text{Let } E = \text{Letter chosen is a vowel } = \{ a, e, i, o, u \}$

$\displaystyle \therefore n(E) = 5$

$\displaystyle \therefore \text{Required Probability }$

$\displaystyle P(E) = \frac{n(E)}{n(S)} = \frac{5}{26}$

$\\$

$\displaystyle \text {(v) If 3 is a root of the quadratic equation } x^2 - px + 3 = 0 \text{ then } p \text{ is equal to }$

$\displaystyle (a) \ 4 \hspace{1.0cm} (b) \ 3 \hspace{1.0cm}(c) \ 5 \hspace{1.0cm}(d) \ 2$

Answer:

$\fbox{ a }$

$\displaystyle x=3 \text { is a root of the equation }$

$\displaystyle x^2 - px + 3 = 0$

$\displaystyle \text {Then } x = 3 \text { will satisfy the above equation }$

$\displaystyle \therefore 3^2 - 3p + 3 = 0$

$\displaystyle \Rightarrow 9 - 3p + 3 = 0$

$\displaystyle \Rightarrow 3p = 12$

$\displaystyle \Rightarrow p = 4$

$\\$

$\displaystyle \text {(vi) In the given figure } \angle BAP = \angle DCP = 70^{\circ}, PC = 6 \text{ cm and } CA = 4 \text{ cm, then } PD : DB \text{ is }$

$\displaystyle (a) \ 5:3 \hspace{1.0cm} (b) \ 3:5 \hspace{1.0cm}(c) \ 3:2 \hspace{1.0cm}(d) \ 2:3$

Answer:

$\fbox{ c }$

$\displaystyle \text{In } \triangle APB \text{ and } \triangle CPD$

$\displaystyle \angle A = \angle C = 70^{\circ} \hspace{1.0cm} \text{(given) }$

$\displaystyle \text{and } \angle P = \angle P \hspace{1.0cm} \text{(Common angle) }$

$\displaystyle\therefore \triangle APB \sim \triangle CPD \hspace{1.0cm} \text{( By AA axiom) }$

$\displaystyle\therefore \frac{PD}{DB} = \frac{PC}{CA} = \frac{6}{4} = \frac{3}{2}$

$\displaystyle\therefore PD : DB = 3 : 2$

$\\$

$\displaystyle \text {(vii) The printed price of an article is Rs.} 3080. \text{ If the rate of GST is } 10%, \text{ then the GST charged is}$

$\displaystyle (a) \ Rs.154 \hspace{1.0cm} (b) \ Rs.308 \hspace{1.0cm}(c) \ Rs.30.80 \hspace{1.0cm}(d) \ Rs.15.40$

Answer:

$\fbox{ b }$

$\displaystyle \text{Printed Price } = \text{ Rs. } 3080$

$\displaystyle \text{Then GST } @ 10\% = 3080 \times \frac{10}{100} = \text{ Rs. } 308$

$\\$

$\displaystyle \text {(viii) } (1 + \sin A)(1 - \sin A) \text{ is equal to }$

$\displaystyle (a) \ \mathrm{cosec}^2 A \hspace{1.0cm} (b) \ \sin^2 A \hspace{1.0cm}(c) \ \sec^2 A \hspace{1.0cm}(d) \ \cos^2 A$

Answer:

$\fbox{ d }$

$\displaystyle (1+ \sin A)(1 - \sin A) = 1 - \sin^2 A = \cos^2 A$

$\\$

$\displaystyle \text {(ix) The coordinates of the vertices of } \triangle ABC \text { are respectively } (-4, -2), (6, 2) \text { and } (4, 6). \\ \text { The centroid } G \text { of } \triangle ABC \text { is }$

$\displaystyle (a) \ (2, 2) \hspace{1.0cm} (b) \ (2, 3) \hspace{1.0cm}(c) \ (3, 3) \hspace{1.0cm}(d) \ (0, -1)$

Answer:

$\fbox{ a }$

$\displaystyle \text{Coordinates of the centroid } G \text{ of } \triangle ABC$

$\displaystyle = \Big( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \Big)$

$\displaystyle = \Big( \frac{-4+6+4}{3}, \frac{-2+2+6}{3} \Big)$

$\displaystyle = (2, 2)$

$\\$

$\displaystyle \text {(x) The } n^{th} \text{ term of an Arithmetic Progression (A.P.) is } 2n + 5. \text{ The } 10^{th} \text{ term is }$

$\displaystyle (a) \ 7 \hspace{1.0cm} (b) \ 15 \hspace{1.0cm}(c) \ 25 \hspace{1.0cm}(d) \ 45$

Answer:

$\fbox{ c }$

$\displaystyle n^{th} \text{ term of A.P. } = 2n + 5$

$\displaystyle \text{Put } n = 10$

$\displaystyle \therefore 10^{th} \text{ term of A.P. } = 2 \times 10 + 5 = 20 + 5 = 25$

$\\$

$\displaystyle \text {(xi) The mean proportional between 4 and 9 is.}$

$\displaystyle (a) \ 4 \hspace{1.0cm} (b) \ 6 \hspace{1.0cm}(c) \ 9 \hspace{1.0cm}(d) \ 36$

Answer:

$\fbox{ b }$

$\displaystyle \text{Mean proportional } = \sqrt{4 \times 9} = \sqrt{36} = 6$

$\\$

$\displaystyle \text {(xii) Which of the following cannot be determined graphically for a grouped frequency } \\ \text{ distribution? }$

$\displaystyle (a) \ \text {Median } \hspace{1.0cm} (b) \ \text {Mode } \hspace{1.0cm}(c) \ \text {Quartiles } \hspace{1.0cm}(d) \ \text {Mean }$

Answer:

$\fbox{ d }$

A histogram can be used to determine the mode, and an ogive can be used to graphically determine the median and quartiles.

But mean cannot be expressed graphically.

$\\$

$\displaystyle \text {(xiii) Volume of a cylinder of height 3 cm is a } 48\pi . \text{ Radius of the cylinder is}$

$\displaystyle (a) \ 48 \ cm \hspace{1.0cm} (b) \ 16 \ cm \hspace{1.0cm}(c) \ 4 \ cm \hspace{1.0cm}(d) \ 24 \ cm$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given height of a cylinder } h = 3 \text{ cm }$

$\displaystyle \text{Volume of cylinder } V = 48 \pi$

$\displaystyle \text{Let Radius of cylinder } = r$

$\displaystyle V = \pi r^2 h$

$\displaystyle 48 \pi = \pi r^2 \times 3$

$\displaystyle r^2 = \frac{48}{3} = 16$

$\displaystyle r = \sqrt{16} = 4 \text{ cm }$

$\\$

$\displaystyle \text {(xiv) Naveen deposits Rs. 800 every month in a recurring deposit account for 6 months. } \\ \text{ If he receives Rs. 4884 at the time of maturity, then the interest he earns is }$

$\displaystyle (a) \ Rs.\ 84 \hspace{1.0cm} (b) \ Rs.\ 42 \hspace{1.0cm}(c) \ Rs.\ 24 \hspace{1.0cm}(d) \ Rs.\ 284$

Answer:

$\fbox{ a }$

$\displaystyle \text{Maturity value } = \text{ Rs. } 4884$

$\displaystyle \text{Deposited value } = \text{ Rs. } 800 \times 6 = \text{ Rs. } 4800$

$\displaystyle \therefore \text{Interest earned by Naveen } = \text{ Rs. } 4884 - \text{ Rs. } 4800 = \text{ Rs. } 84$

$\\$

$\displaystyle \text {(xv) The solution set for the inequation } 2x + 4 \leq 14, x \in W \text{ is }$

$\displaystyle (a) \ \{ 1,2,3,4,5\} \hspace{1.0cm} (b) \ \{ 0,1,2,3,4,5 \} \hspace{1.0cm}(c) \ \{ 1,2,3,4 \} \hspace{1.0cm}(d) \ \{0,1,2,3,4 \}$

Answer:

$\fbox{ b }$

In equation

$\displaystyle 2x + 4 \leq 14 \text{ where }x \in W$

$\displaystyle \Rightarrow 2x \leq 10$

$\displaystyle \Rightarrow x \leq 5$

$\displaystyle \text{ Hence the solution set } = \{ 0, 1, 2, 3, 4 , 5 \}$

$\\$

Question 2:

$\displaystyle \text{(i) Find the value of } 'a' \text{ if } x - a \text{ is a factor of the polynomial } \\ 3x^3 + x^2 - ax - 81.$

Answer:

$\displaystyle \text{Let polynomial } P(x) = 3x^3+x^2-ax-81$

$\displaystyle (x-a) \text{ is a factor of } P(x)$

$\displaystyle \text{Then putting } x-a = 0 \Rightarrow x=a \text{ in } P(x) \text{ we get } P(a) = 0$

$\displaystyle \therefore 3a^3 + a^2 - a \times a - 81 = 0$

$\displaystyle \Rightarrow 3a^3 = 81$

$\displaystyle \Rightarrow a^3 = 27$

$\displaystyle \Rightarrow a = 3$

$\\$

$\displaystyle \text{(ii) Salman deposits Rs. } 1000 \text{ every month in a recurring deposit account for 2 years.} \\ \text{ If he receives Rs. } 26000 \text{ on maturity, find: }$
$\displaystyle (a) \ \text{ the total interest Salman earns } \hspace{1.0cm} (b) \ \text{ the rate of interest }$

Answer:

$\displaystyle \text{P } = \text{Rs. } 1000$

$\displaystyle n = 2 \text{ years } = 24 \text{ months }$

$\displaystyle \text{Maturity value } = \text{ Rs. } 26000$

$\displaystyle \text{(a) Total interest = Maturity value - Deposited value }$

$\displaystyle = \text{Rs. } 26000 = \text{Rs. } 1000 \times 24 = \text{Rs. } 26000 - \text{Rs. } 24000 = \text{Rs. } 2000$

$\displaystyle \text{(b) Simple Interest (S.I) } = P \times \frac{n(n+1)}{2} \times \frac{1}{12} \times\frac{r}{100}$

$\displaystyle 2000 = 1000 \times \frac{24 \times 25}{2} \times \frac{1}{12} \times\frac{r}{100}$

$\displaystyle 2000 = 250r$

$\displaystyle r = \frac{2000}{250} = 8\% \text{ Per Annum }$

$\\$

$\displaystyle \text{(iii) In the given figure } O, \text{ is the centre of the circle. } CE \text{ is a tangent to the circle at } A. \\ \text{ If } \angle ABD = 26^{\circ} \text{ find: }$

$\displaystyle (a) \ \angle BDA \hspace{1.0cm} (b) \ \angle BAD \hspace{1.0cm}(c) \ \angle CAD \hspace{1.0cm}(d) \ \angle ODB$

Answer:

$\displaystyle \text{Given } \angle ABD = 26^{\circ}$

$\displaystyle \text{a) Angle in a semi circle is a right angle }$

$\displaystyle \therefore \angle BDA = 90^{\circ}$

$\displaystyle \text{b) Sum of all angles in a triangle } = 180^{\circ}$

$\displaystyle \therefore \angle BAD + \angle BDA + \angle ABD = 180^{\circ}$

$\displaystyle \Rightarrow \angle BAD + 90^{\circ}+ 26^{\circ} = 180^{\circ}$

$\displaystyle \Rightarrow \angle BAD = 180^{\circ} 0 116^{\circ} = 64^{\circ}$

$\displaystyle \text{c) Angles in alternate segments are equal }$

$\displaystyle \therefore \angle CAD = \angle ABD = 26^{\circ}$

$\displaystyle \text{d) in } \triangle ODB,$

$\displaystyle OB = OD \hspace{1.0cm} \text{ Radius of the same circle }$

$\displaystyle \therefore \angle ODB = \angle OBD = \angle ABD = 26^{\circ}$

$\\$

Question 3:

$\displaystyle \text{(i) Solve the following quadratic equation: } x^2 + 4x-8 = 0$

$\displaystyle \text{Give your solution correct to one decimal place. } \text{(Use mathematical tables if necessary.) }$

Answer:

$\displaystyle \text{(i) } x^2 + 4x - 8 = 0$

$\displaystyle \text{Here } a = 1, b = 4 , c = -8$

$\displaystyle D = b^2 - 4ac = 4^2 - 4 \times 1 \times (-8) = 16 + 32 = 48 > 0$

$\displaystyle \therefore \text{ Roots are real }$

$\displaystyle \text{Hence, } x = \frac{-b \pm \sqrt{D}}{2a}$

$\displaystyle = \frac{-4 \pm \sqrt{48}}{2 \times 1}$

$\displaystyle = -2 \pm 2\sqrt{3}$

$\displaystyle = -2 \pm (3.46)$

$\displaystyle \Rightarrow x = 1.46 , - 5.46$

$\displaystyle \text{or } x = 1.5 , -5.5 \hspace{1.0cm} \text{ (Correct to one decimal place) }$

$\\$

$\displaystyle \text{(ii) Prove the following identity: }$

$\displaystyle (\sin^2 \theta - 1)(\tan 2\theta + 1) + 1 = 0$

Answer:

$\displaystyle \text{LHS } = (\sin^2 \theta - 1)(\tan 2\theta + 1) + 1$

$\displaystyle = (- \cos^2 \theta) \sec^2 \theta + 1$

$\displaystyle = - \cos^2 \theta \times \frac{1}{\cos^2 \theta}+ 1$

$\displaystyle = -1+1 = 0 = \text{ RHS }$

Hence Proved.

$\\$

$\displaystyle \text{(iii) Use graph sheet to Solution this question. Take 2 cm = 1 unit along both the axes. }$

$\displaystyle \text{a. Plot A, B, C where A(0, 4), B(1, 1) and C(4, 0) }$

$\displaystyle \text{b. Reflect A and B on the x-axis and name them as E and D respectively. }$

$\displaystyle \text{c. Reflect B through the origin and name it F. Write down the coordinates of F. }$

$\displaystyle \text{d. Reflect B and C on the y-axis and name them as H and G respectively. }$

$\displaystyle \text{e. Join points A, B, C, D, E, F, G, H and A in order and name the closed figure formed. }$

Answer:

$\\$

SECTION-B (40 Marks)

(Attempt any four questions from this Section)

Question 4:

$\displaystyle \text{(i) If } A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}, C = \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$

$\displaystyle \text{Find } A(B+C) - 14I$ [3]

Answer:

$\displaystyle A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}, C = \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$

$\displaystyle A(B+C) - 14I$

$\displaystyle = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \Bigg( \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} \Bigg) - 14 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \Bigg( \begin{bmatrix} 5 & 3 \\ 3 & 9 \end{bmatrix} \Bigg) - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 5+9 & 3+27 \\ 10 + 12 & 6 + 36 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 30 \\ 22 & 28 \end{bmatrix}$

$\\$

$\displaystyle \text{(ii)}$ [3]

$\displaystyle \text{a) ABC is a triangle whose vertices are A(1, -1), B(0, 4) and C(- 6, 4). } \\ \text{D is the midpoint of BC. Find the coordinates of D.}$

$\displaystyle \text{b) ABC is a triangle whose vertices are A(1, -1), B(0, 4) and C(- 6, 4). } \\ \text{D is the midpoint of BC. Find the equation of the median AD.}$

Answer:

$\displaystyle \text{a) } D = \Big( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \Big) = \Big( \frac{0-6}{2}, \frac{4+4}{2} \Big) = (-3, 4)$

$\displaystyle \text{Therefore the coordinates for D is (-3, 4) }$

$\displaystyle \text{b) Slope of median } AD = \frac{y_2-y_1}{x_2-x_1} = \frac{4+1}{-3-1} = \frac{-5}{4}$

$\displaystyle \text{Equation of AD:}$

$\displaystyle y-y_1 = m(x-x_1)$

$\displaystyle \Rightarrow y+1 = \frac{-5}{4} (x-1)$

$\displaystyle \Rightarrow 5x+ 4y - 1 = 0$

$\\$

$\displaystyle \text{(iii) In the given figure, O is the center of the circle. PQ is a tangent } \\ \text{to the circle at T. Chord AB produced meets the tangent at P. }$
$\displaystyle \text{AB = 9 cm, BP = 16 cm, } \angle PTB = 50^{\circ} \angle OBA = 45^{\circ} \text{ Find: }$ [4]

$\displaystyle \text{(a) Length of } PT \hspace{1.0cm} (b) \ \angle BAT \hspace{1.0cm}(c) \ \angle BOT \hspace{1.0cm}(d) \ \angle ABT$

Answer:

$\displaystyle \text{a) } PT^2 = PA \times PB = (16+9) \times 16$

$\displaystyle \therefore PT = \sqrt{25\times16} = 5 \times 4 = 20$

$\displaystyle \text{b) } \angle BAT = \angle PTB = 50^{\circ} \hspace{1.0cm} \text{(Alternate segment angle)}$

$\displaystyle \text{c) } \angle BOT = 2 \angle BAT \hspace{1.0cm} \text{(Degree measure theorem)}$

$\displaystyle \angle BOT = 2 \times 50^{\circ} = 100^{\circ}$

$\displaystyle \text{d) In } \triangle OBT$

$\displaystyle \angle OBT + \angle OTB + \angle BOT = 180^{\circ} \text{(sum of all angles in a triangle)}$

$\displaystyle \angle OBT + \angle OBT + 100^{\circ}= 180^{\circ}$

$\displaystyle \angle OBT = \angle OTB \hspace{1.0cm} \text{(angles opposite to equal sides of a triangle are equal)}$

$\displaystyle 2 \angle OBT = 180^{\circ} - 100^{\circ} = 80^{\circ}$

$\displaystyle \angle OBT = 40^{\circ}$

$\displaystyle \angle ABT = \angle ABO + \angle OBT = 45^{\circ} + 40^{\circ} = 85^{\circ}$

$\\$

Question 5:

$\displaystyle \text{(i) Mrs. Arora bought the following articles from a departmental store: }$ [3]

$\begin{array}{ |c|c|c|c|c| } \hline \text{S. No. } & \text{Item } & \text{Price } & \text{Rate of GST } & \text{Discount } \\ \hline 1. & \text{Hair Oil } & \text{Rs. } 1200 & 10\% & \text{Rs. } 100 \\ \hline 2. & \text{Cashew nuts } & \text{Rs. } 600 & 12\% & - \\ \hline \end{array}$

$\displaystyle \text{Find the: }$
$\displaystyle \text{a. Total GST paid, }$
$\displaystyle \text{b. Total bill amount including GST }$

Answer:

$\displaystyle \text{(i) Mrs. Arora paid for }$

$\displaystyle \text{1. Hair oil Rs. (1200-100) = Rs. 1100 }$

$\displaystyle \text{GST } = \frac{18}{100} \times 1100 = \text{ Rs. } 198$

$\displaystyle \text{2. Cashew nuts }$

$\displaystyle \text{GST } = 600 \times \frac{12}{100} = \text{ Rs. } 72$

$\displaystyle \text{(a) Total GST paid } = \text{ Rs. } 198 + \text{ Rs. } 72 = \text{ Rs. } 270$

$\displaystyle \text{(b) Total bill amount including GST } \\ = \text{ Rs. } 1100 + \text{ Rs. } 600 + \text{ Rs. } 270 = \text{ Rs. } 1970$

$\\$

$\displaystyle \text{(ii) Solve the following inequation. } \\ \text{Write down the solution set and represent it on the real number line. }$ [3]

$\displaystyle -5(x - 9) \geq 17 - 9x > x + 2, x \in R$

Answer:

$\displaystyle -5(x-9) \geq 17 - 9x > x+2$

$\displaystyle -5(x-9) \geq 17-9x \text{ and } 17-9x > x+2$

$\displaystyle -5x + 9x \geq 17-45 \Rightarrow 4x \geq -28 \Rightarrow x \geq -7$

$\displaystyle 17-9x > x+2 \Rightarrow -9x-x > 2-17 \Rightarrow -10x > -15 \Rightarrow x < 1.5$

Solution set $\displaystyle -7 \leq x < 1.5$

$\\$

$\displaystyle \text{(iii) In the given figure, } AC \parallel DE \parallel BF. \text{If AC = 24 cm, EG = 8 cm, } \\ \text{GB = 16 cm, BF = 30 cm. }$ [4]

$\displaystyle \text{(a) Prove } \triangle GED \sim \triangle GBF$

$\displaystyle \text{(b) Find } DE$

$\displaystyle \text{(c) } DB:AB$

Answer:

$\displaystyle \text{(a) In } \triangle GED \text{ and } \triangle BGF,$

$\displaystyle \angle EGD = \angle BGF \text{ (Vertically opposite angles) }$

$\displaystyle \angle DEG = \angle FBG \text{ (Alternative angles } DE \parallel BF )$

$\displaystyle \therefore \triangle GED \sim \triangle GBF \text{ (AA Similarity Criterion) Hence proved. }$

$\displaystyle \text{(b) } \triangle GED \sim \triangle GBF$

$\displaystyle \frac{GE}{GB} = \frac{ED}{BF} = \frac{GD}{GF} \text{ (by BPT) }$

$\displaystyle \frac{ED}{BF} = \frac{GE}{GB}$

$\displaystyle \frac{ED}{30} = \frac{8}{16}$

$\displaystyle ED = \frac{8}{16} \times 30 = 15 \text{cm }$

$\displaystyle DE = 15 \text{cm }$

$\displaystyle \text{(c) In } \triangle ABC, DE \parallel AC$

$\displaystyle \angle BDE = \angle BAC$

$\displaystyle \text{and } \angle BED = \angle BCA$

$\displaystyle \text{by AA Similarity criterion, } \triangle BDE \sim \triangle BAC$

$\displaystyle \frac{DB}{AB} = \frac{BE}{BC} = \frac{DE}{AC} \text{ (by BPT) }$

$\displaystyle \frac{DB}{AB} = \frac{DE}{AC}$

$\displaystyle \frac{DB}{AB} = \frac{15}{24} = \frac{5}{8}$

$\displaystyle \therefore DB : AB = 5: 8$

$\\$

Question 6:

(i) The following distribution gives the daily wages of 60 workers of a factory

$\begin{array}{ |c|c|} \hline \text{Daily Income in Rs. } & \text{Number of Workers (f) } \\ \hline 200-300 & 6 \\ \hline 300-400 & 10\\ \hline 400-500 & 14\\ \hline 500-600 & 16\\ \hline 600-700 & 10\\ \hline 700-800 & 4 \\ \hline \end{array}$

Use graph paper to Solution this question. Take 2 cm = Rs. 100 along one axis and 2 cm = 2 workers along the other axis. Draw a histogram and hence find the mode of the given distribution

[3]

Answer:

$\\$

(ii) The 5th term and the 9th term of an Arithmetic Progression are 4 and – 12 respectively. Find:
a) the first term
b) common difference
c) sum of 16 terms of the AP. [3]

Answer:

$\displaystyle \text{Given, } t_5 = 4 \text{ and } t_9 = -12$

$\displaystyle \text{Let the first term be a and common difference be } d \text{ respectively. }$

$\displaystyle a_n = a + (n-1) d$

$\displaystyle t_5 = 4$

$\displaystyle a + (5-1) d = 4$

$\displaystyle a+4d = 4\ldots (i)$

$\displaystyle \text{Similarly, } a+8d = -12 \ldots (ii)$

$\displaystyle \text{From (i) and (ii) we get } d = -4 \text{ and } a = 20$

$\displaystyle \text{(a) first term is } 20$

$\displaystyle \text{(b) Common difference } = -4$

$\displaystyle \text{(c) } S_n = \frac{n}{2} \Big[2a+ (n+1)d \Big]$

$\displaystyle \therefore S_{16} = \frac{16}{2} \Big[2 \times 20 + (16-1)(-4) \Big] = 8[40-60] = -160$

$\displaystyle \text{Therefore Sum of 16 terms is } =-160$

$\\$

(iii) A and B are two points on the x-axis and y-axis respectively.
a) Write down the coordinates of A and B.
b) P is a point on AB such that AP : PB = 1 : 1
c) Find the equation of a line passing through P and perpendicular to AB [4]

Answer:

$\displaystyle \text{a) Coordinates of A(4, 0) and Coordinates of B(0,4) }$

$\displaystyle \text{b) } P = \Big( \frac{m_1x_2+m_2x_1}{m_1+m_2}, \frac{m_1y_2+m_2y_1}{m_1+m_2} \Big)$

$\displaystyle = \Big( \frac{3 \times 0 + 1 \times 4}{3+1}, \frac{3 \times 4+1 \times 0}{3+1} \Big)$

$\displaystyle P= (1, 3)$

$\displaystyle \text{(c) Slope of line } AB = \frac{4-0}{0-4} = -1$

$\displaystyle \text{Therefore Slope of perpendicular line AB } = 1$

$\displaystyle y-y_1 = m (x-x_1)$

$\displaystyle y-3 = 1 (x-1)$

$\displaystyle y-3 = x-1$

$\displaystyle x-y+2 = 0$

$\\$

Question 7:

$\displaystyle \text{(i) A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. } \\ \text{What is the probability that the number on the card drawn is multiple of 5? } \hspace{1.0cm} [3]$

Answer:

$\displaystyle S = \{ 1, 2, 3, 4, \ldots \ldots 25 \}$

$\displaystyle \therefore n(S) = 25$

$\displaystyle \text{(a) multiple of } 5$

$\displaystyle E = \{ 5, 10, 15, 20, 25 \}$

$\displaystyle n(E) = 5$

$\displaystyle \text{Probability of multiple of } 5 = \frac{n(E)}{n(S)} = \frac{5}{25} = \frac{1}{5}$

$\displaystyle \text{(b) Perfect square } = \{ 1, 4, 9, 16, 25 \}$

$\displaystyle n(E) = 5$

$\displaystyle \text{Probability of perfect square } = \frac{n(E)}{n(S)} = \frac{5}{25} = \frac{1}{5}$

$\displaystyle \text{(c) Prime number } = \{ 2, 3, 5, 7, 11, 13, 17, 19, 23 \}$

$\displaystyle n(E) = 9$

$\displaystyle \text{Probability of Prime number } = \frac{n(E)}{n(S)} = \frac{9}{25} = \frac{1}{5}$

$\\$

$\displaystyle \text{(ii) A man covers a distance of 100 km, travelling with a uniform speed of } x \ km/hr. \\ \text{ Had the speed been } 5\ km/hr \text{ more it would have taken 1 hour less. Find } \\ x \text{\ the original speed.} \hspace{10.0cm} [3]$

Answer:

$\displaystyle \text{Let the original speed be } x \ \text{km/hr}$

$\displaystyle Time = \frac{Distance}{Speed}$

$\displaystyle t_1 = \frac{100}{x} \text{ hour}$

$\displaystyle t_2 = \frac{100}{x+5} \text{ hour}$

$\displaystyle t_1 - t_2 = 1 \hspace{1.0cm} \text{ (Given)}$

$\displaystyle \frac{100}{x} - \frac{100}{x+5} = 1$

$\displaystyle \frac{100x + 500 - 100x}{x(x+5)} = 1$

$\displaystyle x^2 + 5x - 500 = 0$

$\displaystyle (x+25)(x-20) = 0$

$\displaystyle x = -25 or x = 20$

Negative speed is not possible.

$\displaystyle \text{Therefore original speed } x = 20 \ \text{km/hr}$

$\\$

$\displaystyle \text{(iii) A solid is in the shape of a hemisphere of radius 7 cm, surmounted by a cone } \\ \text{of height 4 cm. The solid is immersed completely in a cylindrical container filled } \\ \text{with water to a certain height. If the radius of the cylinder is 14 cm, find the } \\ \text{rise in the water level. }$ [4]

Answer:

Volume of Solid = Volume of Hemisphere + Volume of Cone

$\displaystyle = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h$

$\displaystyle = \frac{\pi r^2}{3} (2r+h)$

$\displaystyle = \frac{\pi \times 7 \times 7}{3} (2 \times 7 + 4)$

$\displaystyle = \frac{49\pi}{3} (18)$

$\displaystyle = 294 \pi \ cm^3$

$\displaystyle \text{Let the level of water rise in the cylinder be } x \text{ cm }$

$\displaystyle \text{Therefore Rise in volume of water } = 294\pi$

$\displaystyle \Rightarrow \pi R^2 x = 294 \pi$

$\displaystyle \Rightarrow 14^2 x = 294$

$\displaystyle \Rightarrow x = \frac{294}{14 \times 14} = \frac{21}{14}$

$\displaystyle \Rightarrow x = \frac{3}{2} = 1/5 \text{ cm}$

$\\$

Question 8:

$\displaystyle \text{(i) The following table gives the marks scored by a set of students in an examination. } \\ \text{Calculate the mean of the distribution by using the short cut method. }$

$\begin{array}{ |c|c|} \hline \text{Marks } & \text{Number of Students (f) } \\ \hline 0-10 & 3 \\ \hline 10-20 & 8\\ \hline 20-30 & 14\\ \hline 30-40 & 9\\ \hline 40-50 & 4\\ \hline 50-60 & 2 \\ \hline \end{array}$

Answer:

$\begin{array}{ |c|c|c|c|c|} \hline \text{Marks } & \text{Number of Students (f) } & x & d=x-A& fd \\ \hline 0-10 & 3 & 5 & -20 & -60 \\ \hline 10-20 & 8 &15 & -10 &-80 \\ \hline 20-30 & 14 & 25=A & 0 & 0\\ \hline 30-40 & 9 & 35 & 10 & 90 \\ \hline 40-50 & 4 & 45 & 20 & 80 \\ \hline 50-60 & 2 &55 & 30 & 60\\ \hline & 40 && &90 \\ \hline \end{array}$

$\displaystyle \overline{x} = A + \frac{\Sigma fd}{\Sigma f} = 25 + \frac{90}{40} = 25 + 2.25 = 27.25$

$\displaystyle \text{Hence, Mean }= 27.25$

$\\$

$\displaystyle \text{(ii) What number must be added to each of the numbers 4, 6, 8, 11 in order to } \\ \text{ get the four numbers in proportion? }$

Answer:

$\displaystyle \text{Let } x \text{ be added to get the four numbers in proportion.}$

$\displaystyle \therefore (4+x): (6+x) :: (8+x):(11+x)$

$\displaystyle (4+x)(11+x) = (6+x)(8+x)$

$\displaystyle x^2 + 15x + 44 = x^2 + 14x + 48$

$\displaystyle 15x-14x=48-44$

$\displaystyle x=4$

$\displaystyle \text{Therefore } 4 \text{ should be added to get the four numbers in proportion.}$

$\\$

$\displaystyle \text{(iii) Using ruler and compass construct a } \triangle ABC \text{ in which } AB = 6 cm, \angle BAC = 120^{\circ} \text{ and } AC = 5 \text{ cm. Construct a circle passing through A, B and C. Measure } \\ \text{and write down the radius of the circle }$

Answer:

Steps of construction:

Draw AB = 6 cm
$\displaystyle \text{Make } \angle BAD = 120^{\circ}$
Taking center A and radius of AC = 5 cm, draw an arc which intersects at point C
Join BC. Triangle ABC is the required triangle.
Draw perpendicular bisectors of side AB and AC respectively which intersects at point O
Taking O as the center and OA as radius draw a circle which passes through the points A, B and C respectively.
Now measure the radius OA = 5.5 cm

$\\$

Question 9:

(i) Using Componendo and Dividendo solve for x:

$\displaystyle \frac{\sqrt{2x+2}+ \sqrt{2x-1}}{\sqrt{2x+2} - \sqrt{2x-1}} = 3$

Answer:

$\displaystyle \frac{\sqrt{2x+2}+ \sqrt{2x-1}}{\sqrt{2x+2} - \sqrt{2x-1}} = 3$

Applying Componendo and Dividendo

$\displaystyle \frac{\sqrt{2x+2}+ \sqrt{2x-1}+ \sqrt{2x+2} - \sqrt{2x-1}}{\sqrt{2x+2}+ \sqrt{2x-1}- \sqrt{2x+2} + \sqrt{2x-1}} =\frac{3+1}{3-1}$

$\displaystyle \frac{2\sqrt{2x+2}}{2\sqrt{2x-1}} = \frac{4}{2} = 2$

$\displaystyle \frac{\sqrt{2x+2}}{\sqrt{2x-1}} = 2$

Squaring both sides:

$\displaystyle \frac{2x+2}{2x-1} = 4$

$\displaystyle 2x+2 = 8x-4$

$\displaystyle 2x-8x = -4 - 2$

$\displaystyle -6x = -6$

$\displaystyle x=1$

$\\$

(ii) Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60 . . . . . is 300? Hence find the sum of all the terms of the Arithmetic Progression (A.P.)

Answer:

$\displaystyle 15, 30, 45, 60 . \ldots \ldots 300$

$\displaystyle a = 15, d = 30-15 = 15$

$\displaystyle a_n = 300$

$\displaystyle a+ (n-1) d = 300$

$\displaystyle 15 + (n-1) (15) = 300$

$\displaystyle 1 + (n-1) = 20$

$\displaystyle n = 20$

$\displaystyle 20^{th} \text{ term is } 300$

$\displaystyle S_n = \frac{n}{2} \Big[ a + a_n \Big]$

$\displaystyle S_{20}= \frac{20}{2}\Big[ 15 + 300 \Big] = 3150$

$\\$

(iii) From the top of a tower 100 m high a man observes the angles of depression of two ships A and B, on opposite sides of the tower as 45° and 38° respectively. If the foot of the tower and the ships are in the same horizontal line find the distance between the two ships A and B to the nearest meter. (Use Mathematical Tables for this question)

Answer:

$\displaystyle \text{In } \triangle ACD$

$\displaystyle \tan A = \frac{CD}{AC}$

$\displaystyle \Rightarrow \tan 45^{\circ} = \frac{100}{AC}$

$\displaystyle \Rightarrow 1 = \frac{100}{AC}$

$\displaystyle \therefore AC = 100 \ m$

$\displaystyle \text{In } \triangle BCD$

$\displaystyle \tan B = \frac{CD}{BC}$

$\displaystyle \Rightarrow \tan 38^{\circ} = \frac{100}{BC}$

$\displaystyle \Rightarrow 0.7813 = \frac{100}{BC}$

$\displaystyle \therefore BC = \frac{100}{0.7813} = 127.99 \ m$

$\displaystyle \text{Distance between two ships} = AC +BC = 100 + 127.99 = 227.99 = 228 m$

$\\$

Question 10:

(i) Factorize completely using factor theorem: [4]

$\displaystyle 2x^3 - x^2 - 13x - 6$

Answer:

$\displaystyle 2x^3 - x^2 - 13x - 6$

$\displaystyle \text{Put }x= -2$

$\displaystyle \Rightarrow 2(-2)^3-(-2)^2-13(-2)-6 = -16-4 + 26 - 6 = 0$

$\displaystyle \text{Therefore } (x+2) \text{is a factor of the polynomial. }$

$\displaystyle x+2 ) \overline{2x^3 - x^2 - 13x - 6} ( 2x^2-5x-3 \\ \hspace*{0.5cm}(-) { 2x^3+4x^2} \\ \hspace*{2cm} \overline{-5x^2-13x-6} \\ \hspace*{1.5cm}(-) \underline{ -5x^2-10x} \\ \hspace*{3.5cm} {-3x-6} \\ \hspace*{3.0cm}(-) \underline{ -3x-6} \\ \hspace*{4.5cm} {\times}$

$\displaystyle 2x^3 - x^2 - 13x - 6 = (x+2)(2x^2-5x-3)$

$\displaystyle = (x+2)(2x^2 - 6x + x - 3)$

$\displaystyle = (x+2)\{ 2x(x-3)+ 1 (x-3) \}$

$\displaystyle = (x+2)(x-3)(2x+1)$

$\displaystyle \text{Hence } 2x^3 - x^2 - 13x - 6 = (x+2)(x-3)(2x+1)$

$\\$

(ii) Use graph paper to Solution this question. During a medical checkup of 60 students in a school, weights were recorded as follows: [6]

$\begin{array}{ |c|c|} \hline \text{Weight in Kg. } & \text{Number of Students (f) } \\ \hline 28-30 & 2 \\ \hline 30-32 & 4\\ \hline 32-34 & 10\\ \hline 34-36 & 13\\ \hline 36-38 & 15\\ \hline 38-40 & 9 \\ \hline 40-42 & 5 \\ \hline 42-44 & 2 \\ \hline \end{array}$

Taking 2 cm = 2 kg along one axis and 2 cm = 10 students along the other axis draw an ogive. Use your graph to find the:
a. median
b. upper Quartile
c. number of students whose weights is above 37 kg

Answer:

$\begin{array}{ |c|c|c|} \hline \text{Weight in Kg. } & \text{Number of Students (f) } & \text{Cumulative frequency (cf) }\\ \hline 28-30 & 2 &2\\ \hline 30-32 & 4&6\\ \hline 32-34 & 10&16\\ \hline 34-36 & 13&29\\ \hline 36-38 & 15&44\\ \hline 38-40 & 9& 53\\ \hline 40-42 & 5 & 58\\ \hline 42-44 & 2 &60\\ \hline \end{array}$

$\begin{array}{ |c|c|} \hline \text{Weight in Kg. } & \text{Number of Students (f) } \\ \hline 28 & 0 \\ \hline 30 & 2\\ \hline 32 & 6\\ \hline 34 & 16\\ \hline 36 & 29\\ \hline 38 & 44 \\ \hline 40 & 53 \\ \hline 42 & 58 \\ \hline 44 & 60 \\ \hline \end{array}$

(a) median = 36.2 kg

(b) upper quartile = 38.2 kg

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ICSE Paper 2023 (Class 10): Mathematics

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